Solutions to Exercises

Appendix D Solutions to Exercises

1 Vectors and Geometry in Two and Three Dimensions
1.1 Points

Exercises

1.1.1.

Solution.

The point $(x,y,z)$ satisfies $x^2 +y^2+z^2= 2x-4y+4$ if and only if it satisfies $x^2-2x +y^2+4y+z^2= 4\text{,}$ or equivalently $(x-1)^2 +(y+2)^2+z^2=9 \text{.}$ Since $\sqrt{(x-1)^2 +(y+2)^2+z^2}$ is the distance from $(1, -2, 0)$ to $(x,y,z)\text{,}$ our point satisfies the given equation if and only if its distance from $(1,-2,0)$ is three. So the set is the sphere of radius 3 centered on $(1,-2,0)\text{.}$
As in part (a), $x^2 +y^2+z^2 \lt 2x-4y+4$ if and only if $(x-1)^2 +(y+2)^2+z^2 \lt 9 \text{.}$ Hence our point satifies the given inequality if and only if its distance from $(1,-2,0)$ is strictly smaller than three. The set is the interior of the sphere of radius 3 centered on $(1,-2,0)\text{.}$

1.1.2.

Solution.

$x=y$ is a straight line and passes through the points $(0,0)$ and $(1,1)\text{.}$ So it is the straight line through the origin that makes an angle $45^\circ$ with the $x$- and $y$-axes. It is sketched in the figure on the left below.
$x+y=1$ is the straight line through the points $(1,0)$ and $(0,1)\text{.}$ It is sketched in the figure on the right above.
$x^2+y^2$ is the square of the distance from $(0,0)$ to $(x,y)\text{.}$ So $x^2+y^2=4$ is the circle with centre $(0,0)$ and radius 2. It is sketched in the figure on the left below.
The equation $x^2+y^2=2y$ is equivalent to $x^2+(y-1)^2=1\text{.}$ As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)\text{,}$ $x^2+(y-1)^2=1$ is the circle with centre $(0,1)$ and radius 1. It is sketched in the figure on the right above.
As in part (d),

\begin{align*} x^2+y^2 \lt 2y & \iff x^2+y^2-2y \lt 0 \iff x^2+y^2-2y+1 \lt 1\\ &\iff x^2+(y-1)^2 \lt 1 \end{align*}

As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)\text{,}$ $x^2+(y-1)^2 \lt 1$ is the set of points whose distance from $(0,1)$ is strictly less than $1\text{.}$ That is, it is the set of points strictly inside the circle with centre $(0,1)$ and radius 1. That set is the shaded region (not including the dashed circle) in the sketch below.

1.1.3.

Solution.

For each fixed $y_0\text{,}$ $z=x,\ y=y_0$ is a straight line that lies in the plane, $y=y_0$ (which is parallel to the plane containing the $x$ and $z$ axes and is a distance $y_0$ from it). This line passes through $x=z=0$ and makes an angle $45^\circ$ with the $xy$-plane. Such a line (with $y_0=0$) is sketched in the figure below. The set $z=x$ is the union of all the lines $z=x,\ y=y_0$ with all values of $y_0\text{.}$ As $y_0$ varies $z=x,\ y=y_0$ sweeps out the plane which contains the $y$-axis and which makes an angle $45^\circ$ with the $xy$-plane. Here is a sketch of the part of the plane that is in the first octant.
$x+y+z=1$ is the plane through the points $(1,0,0)\text{,}$ $(0,1,0)$ and $(0,0,1)\text{.}$ Here is a sketch of the part of the plane that is in the first octant.
$x^2+y^2+z^2$ is the square of the distance from $(0,0,0)$ to $(x,y,z)\text{.}$ So $x^2+y^2+z^2=4$ is the set of points whose distance from $(0,0,0)$ is $2\text{.}$ It is the sphere with centre $(0,0,0)$ and radius 2. Here is a sketch of the part of the sphere that is in the first octant.
$x^2+y^2+z^2=4\text{,}$ $z=1$ or equivalently $x^2+y^2=3\text{,}$ $z=1\text{,}$ is the intersection of the plane $z=1$ with the sphere of centre $(0,0,0)$ and radius 2. It is a circle in the plane $z=1$ that has centre $(0,0,1)$ and radius $\sqrt{3}\text{.}$ The part of the circle in the first octant is the heavy quarter circle in the sketch
For each fixed $z_0\text{,}$ $x^2+y^2=4\text{,}$ $z=z_0$ is a circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $2\text{.}$ So $x^2+y^2=4$ is the union of $x^2+y^2=4,\ z=z_0$ for all possible values of $z_0\text{.}$ It is a vertical stack of horizontal circles. It is the cylinder of radius $2$ centered on the $z$-axis. Here is a sketch of the part of the cylinder that is in the first octant.
For each fixed $z_0\ge 0\text{,}$ the curve $z = x^2 + y^2,\ z=z_0$ is the circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $\sqrt{z_0}\text{.}$ As $z = x^2 + y^2$ is the union of $z = x^2 + y^2,\ z=z_0$ for all possible values of $z_0\ge 0\text{,}$ it is a vertical stack of horizontal circles. The intersection of the surface with the $yz$-plane is the parabola $z=y^2\text{.}$ Here is a sketch of the part of the paraboloid that is in the first octant.

1.1.4.

Solution.

The $z$ coordinate of any point is the signed distance from the point to the $xy$-plane. So the distance from $(2,1,3)$ to the $xy$-plane is $|3|=3\text{.}$
The $y$ coordinate of any point is the signed distance from the point to the $xz$-plane. So the distance from $(2,1,3)$ to the $xz$-plane is $|1|=1\text{.}$
The distance from $(2,1,3)$ to $(x,0,0)$ is
\begin{equation*} \sqrt{(2-x)^2 + (1-0)^2 + (3-0)^2} = \sqrt{(x-2)^2 + 10} \end{equation*}
Since $(x-2)^2\ge 0\text{,}$ the distance $\sqrt{(x-2)^2 + 10}$ is minimized when $x=2\text{.}$ Alternatively,
\begin{align*} \diff{}{x} \sqrt{(x-2)^2 + 10} =\frac{x-2}{\sqrt{(x-2)^2 + 10}} =0 &\iff x=2 \end{align*}
So the point on the $x$-axis that is closest to $A$ is $(2,0,0)\text{.}$
As $(2,0,0)$ is the point on the $x$-axis that is nearest $(2,1,3)\text{,}$ the distance from $A$ to the $x$-axis is
\begin{equation*} \sqrt{(2-2)^2 + (1-0)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{10} \end{equation*}

1.1.5.

Solution.

Call the centre of the circumscribing circle $(\bar x,\bar y)\text{.}$ This centre must be equidistant from the three vertices. So

\begin{equation*} \bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2 \end{equation*}

or, subtracting $\bar x^2+\bar y^2$ from the three equal expressions,

\begin{equation*} 0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y \end{equation*}

which implies

\begin{equation*} \bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c} \end{equation*}

The radius is the distance from the vertex $(0,0)$ to the centre $(\bar x,\bar y)\text{,}$ which is $\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}\text{.}$

1.1.6. (✳).

Solution.

The distance from $P$ to the point $(0,0,1)$ is $\sqrt{x^2+y^2+(z-1)^2}\text{.}$ The distance from $P$ to the specified plane is $|z+1|\text{.}$ Hence the equation of the surface is

\begin{equation*} x^2+y^2+(z-1)^2=(z+1)^2\text{ or } x^2+y^2=4z \end{equation*}

All points on this surface have $z\ge 0\text{.}$ The set of points on the surface that have any fixed value, $z_0\ge 0\text{,}$ of $z$ consists of a circle that is centred on the $z$-axis, is parallel to the $xy$-plane and has radius $2\sqrt{z_0}\text{.}$ The surface consists of a stack of these circles, starting with a point at the origin and with radius increasing vertically. The surface is a paraboloid and is sketched below.

1.1.7.

Solution.

Let $(x,y,z)$ be a point in $P\text{.}$ The distances from $(x,y,z)$ to $(3,-2,3)$ and to $(3/2,1,0)$ are

\begin{equation*} \sqrt{(x-3)^2+(y+2)^2+(z-3)^2}\quad\text{ and }\quad \sqrt{(x-3/2)^2+(y-1)^2+z^2} \end{equation*}

respectively. To be in $P\text{,}$ $(x,y,z)$ must obey

\begin{align*} \sqrt{(x-3)^2+(y+2)^2+(z-3)^2}&=2\sqrt{(x-3/2)^2+(y-1)^2+z^2}\\ (x-3)^2+(y+2)^2+(z-3)^2&=4(x-3/2)^2+4(y-1)^2+4z^2 \end{align*}

Squaring out both sides gives

\begin{align*} &x^2-6x+9+y^2+4y+4+z^2-6z+9\\ &\hskip1in=4x^2-12x+9+4y^2-8y+4+4z^2 \end{align*}

and the simplifying gives

\begin{align*} 3x^2-6 x+3y^2-12y+3z^2+6z-9&=0\\ x^2-2 x+y^2-4y+z^2+2z-3&=0\\ (x-1)^2+(y-2)^2+(z+1)^2&=9 \end{align*}

This is a sphere of radius 3 centered on $(1,2,-1)\text{.}$

1.1.8.

Solution.

For each fixed $c\ge 0\text{,}$ the isobar $p(x,y)=c$ is the curve $x^2-2cx+y^2=3c^2\text{,}$ or equivalently, $(x-c)^2+y^2=4c^2\text{.}$ This is a circle with centre $(c,0)$ and radius $2c\text{.}$ Here is a sketch of the isobars $p(x,y)=c$ with $c=0,1,2,3\text{.}$

1.2 Vectors
1.2.9 Exercises

1.2.9.1.

Solution.

$\va+\vb=\llt 3,1\rgt \text{,}$ $\va+2\vb=\llt 4,2\rgt $ $2\va-\vb=\llt 3,-1\rgt $

1.2.9.2.

Solution.

If three points are collinear, then the vector from the first point to the second point, and the vector from the first point to the third point must both be parallel to the line, and hence must be parallel to each other (i.e. must be multiples of each other).

The vectors $\llt 0,3,7\rgt -\llt 1,2,3\rgt =\llt -1,1,4\rgt $ and $\llt 3,5,11\rgt -\llt 1,2,3\rgt =\llt 2,3,8\rgt $ are not parallel (i.e. are not multiples of each other), so the three points are not on the same line.
The vectors $\llt 1,2,-2\rgt -\llt 0,3,-5\rgt =\llt 1,-1,3\rgt $ and $\llt 3,0,4\rgt -\llt 0,3,-5\rgt =\llt 3,-3,9\rgt $ are parallel (i.e. are multiples of each other), so the three points are on the same line.

1.2.9.3.

Solution.

By property 7 of Theorem 1.2.11,

\begin{alignat*}{1} \llt 1,3,2\rgt \cdot\llt 2,-2,2\rgt &=1\times2-3\times 2+2\times2=0 \tag{a}\\ \llt -3,1,7\rgt \cdot\llt 2,-1,1\rgt &=-3\times2-1\times 1+7\times1=0 \tag{b}\\ \llt 2,1,1\rgt \cdot\llt -1,4,2\rgt &=-2\times1+1\times 4+1\times2=4\ne 0 \tag{c} \end{alignat*}

says that the vectors of parts (a) and (b) are perpendicular, while the vectors of part (c) are not perpendicular.

1.2.9.4.

Solution.

The vector $\va$ has length
\begin{equation*} |\llt 3,4 \rgt| =\sqrt{3^2+4^2} =\sqrt{25} =5 \end{equation*}
So the vector $\frac{1}{5}\llt 3,4 \rgt$ has length $1$ (i.e. is a unit vector) and is in the same direction as $\llt 3,4 \rgt\text{.}$
Recall, from Definition 1.2.5, that a vector is parallel to $\va$ if and only if it is of the form $s\va$ for some nonzero real number $s\text{.}$ Such a vector is a unit vector if and only if
\begin{align*} |s\va|=1 &\iff |s|\,|\llt 3,4 \rgt|=1 \iff |s| = \frac{1}{|\llt 3,4 \rgt|}= \frac{1}{5} \\ &\iff s = \pm \frac{1}{5} \end{align*}
So there are two unit vectors that are parallel to $\va\text{,}$ namely $\pm\frac{1}{5}\llt 3,4 \rgt\text{.}$
We have already found, in part (b), all vectors that are parallel to $\va$ and have length $1\text{,}$ namely $\pm\frac{1}{5}\llt 3,4 \rgt\text{.}$ To increase the lengths of those vectors to $10\text{,}$ we just need to multiply them by $10\text{,}$ giving $\pm\frac{10}{5}\llt 3,4 \rgt=\pm 2\llt 3,4 \rgt=\pm\llt 6,8 \rgt\text{.}$
A vector $\llt x,y \rgt$ is perpendicular to $\va=\llt 3,4\rgt$ if and only if
\begin{equation*} 0=\llt x,y \rgt\cdot\llt 3,4\rgt = 3x+4y \iff y=-\frac{3}{4}x \iff \llt x,y \rgt = \llt x,-\frac{3}{4}x \rgt = \frac{x}{4} \llt 4,-3\rgt \end{equation*}
Such a vector is a unit vector if and only if
\begin{align*} \frac{|x|}{4}\,|\llt 4,-3 \rgt|=1 &\iff \frac{|x|}{4} = \frac{1}{|\llt 4,-3 \rgt|}= \frac{1}{5} \\ &\iff \frac{x}{4} = \pm \frac{1}{5} \end{align*}
So there are two unit vectors that are perpendicular to $\va\text{,}$ namely $\pm\frac{1}{5}\llt 4,-3 \rgt\text{.}$

1.2.9.5.

Solution.

The vector $\vb$ has length
\begin{equation*} |\llt 3,4,0 \rgt| =\sqrt{3^2+4^2+0^2} =\sqrt{25} =5 \end{equation*}
So the vector $\frac{1}{5}\llt 3,4,0 \rgt$ has length $1$ (i.e. is a unit vector) and is in the same direction as $\llt 3,4,0 \rgt\text{.}$
Recall, from Definition 1.2.5, that a vector is parallel to $\vb$ if and only if it is of the form $s\vb$ for some nonzero real number $s\text{.}$ Such a vector is a unit vector if and only if
\begin{align*} |s\vb|=1 &\iff |s|\,|\llt 3,4,0 \rgt|=1 \iff |s| = \frac{1}{|\llt 3,4,0 \rgt|}= \frac{1}{5} \\ &\iff s = \pm \frac{1}{5} \end{align*}
So there are two unit vectors that are parallel to $\vb\text{,}$ namely $\pm\frac{1}{5}\llt 3,4,0 \rgt\text{.}$
A vector $\llt x,y,z \rgt$ is perpendicular to $\va=\llt 3,4,0\rgt$ if and only if
\begin{equation*} 0=\llt x,y,z \rgt\cdot\llt 3,4,0\rgt = 3x+4y \iff y=-\frac{3}{4}x \iff \llt x,y,z \rgt = \llt x,-\frac{3}{4}x,z \rgt \end{equation*}
Such a vector is a unit vector if and only if
\begin{align*} \left|\llt x,-\frac{3}{4}x,z \rgt\right|=1 &\iff \sqrt{x^2+\frac{9}{16}x^2+z^2}=1 \iff \sqrt{\frac{25}{16}x^2+z^2}=1 \end{align*}
There are infinitely many pairs $x\text{,}$ $z$ that obey $\sqrt{\frac{25}{16}x^2+z^2}=1\text{.}$ We can easily get two of them by setting $x=0$ and choosing $z$ to obey $\sqrt{z^2}=1\text{,}$ i.e. choosing $z=\pm 1\text{.}$ We can easily get two more of them by setting $z=0$ and choosing $x$ to obey $\sqrt{\frac{25}{16}x^2}=1\text{,}$ i.e. choosing $x=\pm \frac{4}{5}\text{.}$ This gives us four vectors of length one that are perpendicular to $\vb\text{,}$ namely
\begin{equation*} \pm\llt 0,0,1\rgt\qquad \pm\llt \frac{4}{5}\,,\,-\frac{3}{4}\,\frac{4}{5}\,,\,0\rgt =\pm\frac{1}{5}\llt 4,-3,0\rgt \end{equation*}

1.2.9.6.

Solution.

$\text{proj}_{\hi}\va=(\va\cdot\hi)\hi =a_1\hi$ and $\text{proj}_{\hj}\va=(\va\cdot\hj)\hj =a_2\hj\text{.}$

1.2.9.7.

Solution.

The vector from $(1,2,3)$ to $(4,0,5)$ is $\llt 3,-2,2\rgt\text{.}$ The vector from $(1,2,3)$ to $(3,6,4)$ is $\llt 2,4,1\rgt\text{.}$ The dot product between these two vectors is $\llt 3,-2,2\rgt\cdot\llt 2,4,1\rgt=0\text{,}$ so the vectors are perpendicular and the triangle does contain a right angle.

1.2.9.8.

Solution.

The area of a parallelogram is the length of its base time its height.

We can choose the base to be $\va\text{.}$ Then, if $\theta$ is the angle between its sides $\va$ and $\vb\text{,}$ its height is $|\vb|\sin\theta\text{.}$ So

\begin{equation*} \text{area} = |\va||\vb|\sin\theta=|\va\times\vb| \end{equation*}

1.2.9.9.

Solution.

The volume of a parallelepiped is the area of its base time its height. We can choose the base to be the parallelogram determined by the vectors $\vb$ and $\vc\text{.}$ It has area $|\vb\times\vc|\text{.}$ The vector $\vb\times\vc$ is perpendicular to the base.

Denote by $\theta$ the angle between $\va$ and the perpendicular $\vb\times\vc\text{.}$ The height of the parallelepiped is $|\va| |\cos\theta|\text{.}$ So

\begin{equation*} \text{volume} = |\va|\, |\cos\theta|\, |\vb\times\vc| =|\va\cdot(\vb\times\vc)| \end{equation*}

1.2.9.10.

Solution.

(a)

\begin{alignat*}{4} \hi\times\hj&=\det\left[\begin{matrix}\hi&\hj &\hk \\ 1&0&0 \\ 0&1&0\end{matrix}\right]\\ &=\hi(0\times 0-0\times 1) -\hj(1\times 0-0\times 0) +\hk(1\times 1-0\times 0)\\ &=\hk\\ \hj\times\hk&=\det\left[\begin{matrix}\hi&\hj &\hk \\ 0&1&0 \\ 0&0&1\end{matrix}\right]\\ &=\hi(1\times 1-0\times 0) -\hj(0\times 1-0\times 0) +\hk(0\times 0-1\times 0)\\ &=\hi\\ \hk\times\hi&=\det\left[\begin{matrix}\hi&\hj &\hk \\ 0&0&1 \\ 1&0&0\end{matrix}\right]\\ &=\hi(0\times 0-1\times 0) -\hj(0\times 0-1\times 1) +\hk(0\times 0-0\times 1)\\ &=\hj \end{alignat*}

(b)

\begin{alignat*}{2} \va\cdot(\va\times\vb) &=a_1\big(a_2b_3-a_3b_2\big) -a_2\big(a_1b_3-a_3b_1\big) +a_3\big(a_1b_2-a_2b_1\big) &&=0\\ \vb\cdot(\va\times\vb) &=b_1\big(a_2b_3-a_3b_2\big) -b_2\big(a_1b_3-a_3b_1\big) +b_3\big(a_1b_2-a_2b_1\big) &&=0 \end{alignat*}

1.2.9.11.

Solution.

This statement is false. The two numbers $\va\cdot\vb\text{,}$ $\va\cdot\vc$ are equal if and only if $\va\cdot(\vb-\vc)= 0\text{.}$ This in turn is the case if and only if $\va$ is perpendicular to $\vb-\vc$ (under the convention that $\vZero$ is perpendicular to all vectors). For example, if $\va=\llt 1,0,0\rgt \text{,}$ $\vb=\llt 0,1,0\rgt ,\ \vc=\llt 0,0,1\rgt \text{,}$ then $\vb-\vc=\llt 0,1,-1\rgt$ is perpendicular to $\va$ so that $\va\cdot\vb=\va\cdot\vc\text{.}$

1.2.9.12.

Solution.

This statement is true. In the event that $\vb$ and $\vc$ are parallel, $\vb\times\vc=\vZero$ so that $\va\times(\vb\times\vc)=\vZero=0\vb+0\vc\text{,}$ so we may assume that $\vb$ and $\vc$ are not parallel. Then as $\al$ and $\be$ run over $\bbbr\text{,}$ the vector $\al\vb+\be \vc$ runs over the plane that contains the origin and the vectors $\vb$ and $\vc\text{.}$ Call this plane $P\text{.}$ Because $\vd=\vb\times\vc$ is nonzero and perpendicular to both $\vb$ and $\vc\text{,}$ $P$ is the plane that contains the origin and is perpendicular to $\vd\text{.}$ As $\va\times(\vb\times\vc)=\va\times\vd$ is always perpendicular to $\vd\text{,}$ it lies in $P\text{.}$

1.2.9.13.

Solution.

None. The given equation is nonsense. The left hand side is a number while the right hand side is a vector.

1.2.9.14.

Solution.

If $\vb$ and $\vc$ are parallel, then $\vb\times\vc=\vZero$ and $\va\cdot(\vb\times\vc)=0$ for all $\va\text{.}$ If $\vb$ and $\vc$ are not parallel, $\va\cdot(\vb\times\vc)=0$ if and only if $\va$ is perpendicular to $\vd=\vb\times\vc\text{.}$ As we saw in question 1.2.9.12, the set of all vectors perpendicular to $\vd$ is the plane consisting of all vectors of the form $\al\vb+\be\vc$ with $\al$ and $\be$ real numbers. So $\va$ must be of this form.

1.2.9.15.

Solution.

(a) The sketch for part (a) is on the left below. To sketch the projections, we dropped perpendiculars

from $C$ to the line from $O$ to $A\text{,}$ and
from $C$ to the line from $O$ to $B\text{.}$

By definition,

$\text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC}$ is the vector $\overrightarrow{OP_A}$ from $O$ to the point $P_A\text{,}$ where the perpendicular from $C$ to the line from $O$ to $A$ hits the line, and
$\text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC}$ is the vector $\overrightarrow{OP_B}$ from $O$ to the point $P_B\text{,}$ where the perpendicular from $C$ to the line from $O$ to $B$ hits the line.

To evaluate the projections we observe that the three lines from $C$ to $O\text{,}$ from $C$ to $A$ and from $C$ to $B$ all have exactly the same length (namely the radius of the circumscribing circle). Consequently (see the figure on the right above),

the triangle $OCA$ is an isoceles triangle, so that $P_A$ is exactly the midpoint of the line segement from $O$ to $A\text{.}$ That is, $P_A$ is $(a/2,0)$ and
\begin{equation*} \text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} =\overrightarrow{OP_A}=\llt a/2,0\rgt \end{equation*}
Similarly, the triangle $OCB$ is an isoceles triangle, so that $P_B$ is exactly the midpoint of the line segement from $O$ to $B\text{.}$ That is $P_A$ is $(b/2,c/2)$ and
\begin{equation*} \text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} =\overrightarrow{OP_B}=\llt b/2,c/2\rgt \end{equation*}

(b) Call the centre of the circumscribing circle $(\bar x,\bar y)\text{.}$ This centre must be equidistant from the three vertices. So

\begin{gather*} \bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2 \end{gather*}

or, subtracting $\bar x^2+\bar y^2$ from all three expression,

\begin{gather*} 0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y \end{gather*}

which implies

\begin{equation*} \bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c} \end{equation*}

\begin{align*} \overrightarrow{OA}\cdot\overrightarrow{OC} &=\llt a,0\rgt\cdot\llt\frac{a}{2}\,,\,\frac{b^2+c^2-ab}{2c}\rgt =\frac{a^2}{2}=\frac{1}{2}|\overrightarrow{OA}|^2\\ \overrightarrow{OB}\cdot\overrightarrow{OC} &=\llt b,c\rgt\cdot\llt\frac{a}{2}\,,\,\frac{b^2+c^2-ab}{2c}\rgt =\frac{ab}{2}+\frac{b^2+c^2-ab}{2}=\frac{b^2+c^2}{2}\\ &=\frac{1}{2}|\overrightarrow{OB}|^2 \end{align*}

So, by Equation 1.2.14,

\begin{align*} \text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} &=\frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|^2} \overrightarrow{OA} =\frac{1}{2}\overrightarrow{OA} =\llt a/2,0\rgt\\ \text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} &=\frac{\overrightarrow{OB}\cdot\overrightarrow{OC}}{|\overrightarrow{OB}|^2} \overrightarrow{OB} =\frac{1}{2}\overrightarrow{OB} =\llt b/2,c/2\rgt \end{align*}

1.2.9.16.

Solution.

The center of the sphere is $\half\big\{(2,1,4)+(4,3,10)\big\}=(3,2,7)\text{.}$ The diameter (i.e. twice the radius) is $|(2,1,4)-(4,3,10)|=|(-2,-2,-6)|=2|(1,1,3)|=2\sqrt{11}\text{.}$ So the radius of the sphere is $\sqrt{11}$ and the equation of the sphere is

\begin{equation*} (x-3)^2+(y-2)^2+(z-7)^2=11 \end{equation*}

1.2.9.17.

Solution.

Call the vertices of the triangle $A\text{,}$ $B$ and $C$ with $C$ being the vertex that joins the two sides. We can always choose our coordinate system so that $C$ is at the origin. Let $\va$ be the vector from $C$ to $A$ and $\vb$ be the vector from $C$ to $B\text{.}$

Then the vector from $C$ to the midpoint of the side from $C$ to $A$ is $\half\va$ and
the vector from $C$ to the midpoint of the side from $C$ to $B$ is $\half\vb$ so that
the vector joining the two midpoints is $\half\vb-\half\va\text{.}$

As the vector from $A$ to $B$ is $\vb-\va=2\big[\half\vb-\half\va\big]\text{,}$ the line joining the midpoints is indeed parallel to the third side and half its length.

1.2.9.18.

Solution.

(a) By 1.2.17, the area is

\begin{align*} \left| \det\left[\begin{matrix}-3&1\\ 4&3 \end{matrix}\right] \right| &=\big|-3\times 3-1\times 4\big| = |-13| = 13 \end{align*}

(b) By 1.2.17, the area is

\begin{align*} \left|\det\left[\begin{matrix} 4&2\\ 6&8 \end{matrix}\right]\right| &=\big|4\times 8-2\times 6\big| = 20 \end{align*}

1.2.9.19. (✳).

Solution.

Note that

the point on $W$ with $x=0\text{,}$ $y=0$ obeys $-0+3(0)+3z=6$ and so has $z=2$
the point on $W$ with $x=0\text{,}$ $y=2$ obeys $-0+3(2)+3z=6$ and so has $z=0$
the point on $W$ with $x=3\text{,}$ $y=0$ obeys $-3+3(0)+3z=6$ and so has $z=3$
the point on $W$ with $x=3\text{,}$ $y=2$ obeys $-3+3(2)+3z=6$ and so has $z=1$

So the four corners of the parallelogram are $(0,0,2)\text{,}$ $(0,2,0)\text{,}$ $(3,0,3)$ and $(3,2,1)\text{.}$ The vectors

\begin{align*} \vd_1&=\llt 0-0 \,,\, 2-0 \,,\, 0-2 \rgt = \llt 0 \,,\, 2 \,,\, -2\rgt\\ \vd_2&=\llt 3-0 \,,\, 0-0 \,,\, 3-2 \rgt = \llt 3 \,,\, 0 \,,\, 1\rgt \end{align*}

form two sides of the paralleogram. So the area of the parallelogram is

\begin{align*} \big|\vd_1\times\vd_2\big| =\left|\det\left[\begin{matrix} \hi & \hj & \hk\\ 0 & 2 & -2\\ 3 & 0 & 1 \end{matrix}\right]\right| =\left| 2\,\hi - 6\,\hj -6\hk \right| =\sqrt{76} =2\sqrt{19} \end{align*}

1.2.9.20.

Solution.

(a) By 1.2.18, the volume is

\begin{align*} &\left| \det\left[\begin{matrix} 4&1&-1\\ -1&5&2\\ 1&1&6\end{matrix}\right] \right|\\ &\hskip0.5in=\left| 4\det\left[\begin{matrix} 5&2\\ 1&6 \end{matrix}\right] -1\det\left[\begin{matrix} -1&2\\ 1&6 \end{matrix}\right] +(-1)\det\left[\begin{matrix} -1&5\\ 1&1\end{matrix}\right]\right| \\ &\hskip0.5in= \big|4(30-2)-1(-6-2)-1(-1-5)\big| = 4\times28+8+6\\ &\hskip0.5in=126 \end{align*}

(b) By 1.2.18, the volume is

\begin{align*} &\left|\det\left[\begin{matrix} -2&1&2\\ 3&1&2\\ 0&2&5\end{matrix}\right] \right|\\ &\hskip0.5in=\left|-2\det\left[\begin{matrix} 1&2\\ 2&5\end{matrix}\right] -1\det\left[\begin{matrix} 3&2\\ 0&5\end{matrix}\right] +2\det\left[\begin{matrix} 3&1\\ 0&2\end{matrix}\right] \right|\\ &\hskip0.5in=\big|-2(5-4)-1(15-0)+2(6-0)\big| =\big|-2-15+12\big|=\big|-5\big|\\ &\hskip0.5in=5 \end{align*}

1.2.9.21.

Solution.

\begin{alignat*}{2} \va\cdot\vb&=\llt 1,2\rgt\cdot\llt -2,3\rgt=4\qquad & \cos\theta&=\frac{4}{\sqrt{5}\sqrt{13}}=.4961 \notag\\ & & \implies\theta &= 60.25^\circ \tag{a}\\ \va\cdot\vb&=\llt -1,1\rgt\cdot\llt 1,1\rgt=0\qquad & \cos\theta&=\frac{0}{\sqrt{2}\sqrt{2}}=0 \notag\\ & & \implies\theta &= 90^\circ \tag{b}\\ \va\cdot\vb&=\llt 1,1\rgt\cdot\llt 2,2\rgt=4\qquad & \cos\theta&=\frac{4}{\sqrt{2}\sqrt{8}}=1 \notag\\ & & \implies\theta &= 0^\circ \tag{c}\\ \va\cdot\vb&=\llt 1,2,1\rgt\cdot\llt -1,1,1\rgt=2\qquad & \cos\theta&=\frac{2}{\sqrt{6}\sqrt{3}}=.4714 \notag\\ & & \implies\theta &= 61.87^\circ \tag{d}\\ \va\cdot\vb&=\llt -1,2,3\rgt\cdot\llt 3,0,1\rgt=0\qquad & \cos\theta&=\frac{0}{\sqrt{14}\sqrt{10}}=0 \notag\\ & & \implies\theta &= 90^\circ \tag{e} \end{alignat*}

1.2.9.22.

Solution.

By property 6 of Theorem 1.2.11,

\begin{alignat*}{1} &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{1\times 3+2\times 4}{\sqrt{1+4}\sqrt{9+16}} =\frac{11}{5\sqrt{5}}= .9839 \notag\\ &\hskip1in\implies\ \theta=10.3^\circ \tag{a}\\ &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{2\times 4-1\times 2+4\times 1}{\sqrt{4+1+16}\sqrt{16+4+1}} =\frac{10}{21}= .4762 \notag\\ &\hskip1in\implies\ \theta=61.6^\circ \tag{b}\\ &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{1\times 3-2\times 1+1\times 0}{\sqrt{1+4+1}\sqrt{9+1}} =\frac{1}{\sqrt{60}}= .1291 \notag\\ &\hskip1in\implies\ \theta=82.6^\circ \tag{c} \end{alignat*}

1.2.9.23.

Solution.

\begin{alignat*}{2} &\llt 2,4\rgt \cdot\llt 2,y\rgt =2\times2+4\times y=4+4y=0 &&\ \iff\ y=-1 \tag{a}\\ &\llt 4,-1\rgt \cdot\llt y,y^2\rgt =4\times y-1\times y^2=4y-y^2=0 &&\ \iff\ y=0,4 \tag{b}\\ &\llt 3,1,1\rgt \cdot\llt 2,5y,y^2\rgt =6+5y+y^2=0 &&\ \iff\ y=-2,-3 \tag{c} \end{alignat*}

1.2.9.24.

Solution.

(a) We want $0=\vu\cdot\vv=-2\al-10$ or $\al=-5\text{.}$

(b) We want $-2/\al=5/(-2)$ or $\al=0.8\text{.}$

(c) We want $\vu\cdot\vv=-2\al-10 =|\vu|\,|\vv|\,\cos 60^\circ =\sqrt{29}\,\sqrt{\al^2+4}\,\half\text{.}$ Squaring both sides gives

\begin{alignat*}{2} & & 4\al^2+40\al+100&=\frac{29}{4}(\al^2+4)\\ &\implies\quad & 13\al^2-160\al-284&=0\\ &\implies\quad & \al &=\frac{160\pm\sqrt{160^2+4\times13\times284}}{26}\\ &&&\approx 13.88\text{ or }-1.574 \end{alignat*}

Both of these $\al$’s give $\vu\cdot\vv \lt 0$ so no $\al$ works.

1.2.9.25.

Solution.

(a) The component of $\vb$ in the direction $\va$ is

\begin{equation*} \vb\cdot\frac{\va}{|\va|} =\frac{1\times 4+2\times 10+3\times 6}{\sqrt{1+4+9}} = \frac{42}{\sqrt{14}} \end{equation*}

(b) The projection of $\vb$ on $\va$ is a vector of length $42/\sqrt{14}$ in direction $\va/|\va|\text{,}$ namely $\frac{42}{14}\llt 1,2,3\rgt=\llt 3,6,9\rgt\text{.}$

(c) The projection of $\vb$ perpendicular to $\va$ is $\vb$ minus its projection on $\va\text{,}$ namely $\llt 4,10,6\rgt-\llt 3,6,9\rgt=\llt 1,4,-3\rgt\text{.}$

1.2.9.26.

Solution.

\begin{align*} &\llt 1,2,3\rgt\times\llt 4,5,6\rgt =\det\left[\begin{matrix}\hi&\hj &\hk\\ 1&2&3\\ 4&5&6\end{matrix}\right]\\ &\hskip0.5in=\hi\,(2\times 6-3\times 5) -\hj\,(1\times 6-3\times 4) +\hk\,(1\times 5-2\times 4)\\ &\hskip0.5in=-3\,\hi+6\,\hj-3\,\hk \end{align*}

1.2.9.27.

Solution.

\begin{align*} & \det\left[\begin{matrix}\hi&\hj&\hk\cr1&-5&2\cr-2&1&5\end{matrix}\right] \notag\\ &\hskip0.5in=\hi\det\left[\begin{matrix}-5&2\cr1&5\end{matrix}\right] -\hj\det\left[\begin{matrix}1&2\cr-2&5\end{matrix}\right] +\hk\det\left[\begin{matrix}1&-5\cr-2&1\end{matrix}\right]\notag\\ &\hskip0.5in=\hi(-25-2)-\hj(5+4)+\hk(1-10) = \llt -27,-9,-9\rgt \tag{a}\\ & \det\left[\begin{matrix}\hi&\hj&\hk\cr2&-3&-5\cr4&-2&7\end{matrix}\right] \notag\\ &\hskip0.5in=\hi\det\left[\begin{matrix}-3&-5\cr-2&7\end{matrix}\right] -\hj\det\left[\begin{matrix}2&-5\cr4&7\end{matrix}\right] +\hk\det\left[\begin{matrix}2&-3\cr4&-2\end{matrix}\right]\notag\\ &\hskip0.5in=\hi(-21-10)-\hj(14+20)+\hk(-4+12) = \llt -31,-34,8\rgt \tag{b}\\ & \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&0&1\cr0&4&5\end{matrix}\right] \notag\\ &\hskip0.5in=\hi\det\left[\begin{matrix}0&1\cr4&5\end{matrix}\right] -\hj\det\left[\begin{matrix}-1&1\cr0&5\end{matrix}\right] +\hk\det\left[\begin{matrix}-1&0\cr0&4\end{matrix}\right] \notag\\ &\hskip0.5in=\hi(0-4)-\hj(-5-0)+\hk(-4-0) = \llt -4,5,-4\rgt \tag{c} \end{align*}

1.2.9.28.

Solution.

\begin{align*} \vp\times\vp &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr-1&4&2\end{matrix}\right] \tag{a}\\ &=\hi(4\times2-2\times4)-\hj(2-(-2)) +\hk(-4-(-4)) \notag\\ &= \llt 0,0,0\rgt \notag\\ \vp\times\vq &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr3&1&-1\end{matrix}\right]\tag{b}\\ &=\hi(-4-2)-\hj(1-6) +\hk(-1-12)\notag\\ &= \llt -6,5,-13\rgt\notag\\ \vq\times\vp &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr3&1&-1\cr-1&4&2\end{matrix}\right]\notag\\ &=\hi(2+4)-\hj(6-1) +\hk(12+1)\notag\\ & = \llt 6,-5,13\rgt \notag\\ \vp\!\times\!(3\vr) &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr6&-9&-3\end{matrix}\right]\tag{c}\\ &=\hi(-12+18)-\hj(3-12) +\hk(9-24) \notag\\ & = \llt 6,9,-15\rgt \notag\\ 3(\vp\times\vr) &= 3\det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr2&-3&-1\end{matrix}\right]\notag\\ &=3\Big(\hi(-4+6)-\hj(1-4) +\hk(3-8) \Big) \notag\\ &= \llt 6,9,-15\rgt \notag \end{align*}

(d) As $\vq+\vr=\llt 5,-2,-2\rgt $

\begin{align*} \vp\times(\vq+\vr) &= \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr5&-2&-2\end{matrix}\right]\\ &=\hi(-8+4)-\hj(2-10) +\hk(2-20)\\ &= \llt -4,8,-18\rgt \end{align*}

(e) Using the values of $\vp\times\vq$ and $3(\vp\times\vr)$ computed in parts (b) and (c)

\begin{equation*} \vp\times\vq+\vp\times\vr=\llt -6,5,-13\rgt +\frac{1}{3}\llt 6,9,-15\rgt = \llt -4,8,-18\rgt \end{equation*}

\begin{align*} \vq\times\vr &= \det\left[\begin{matrix}\hi&\hj&\hk\cr3&1&-1\cr2&-3&-1\end{matrix}\right]\\ &=\hi(-1-3)-\hj(-3+2) +\hk(-9-2)\\ &= \llt -4,1,-11\rgt\\ \vp\times(\vq\times\vr) &= \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr-4&1&-11\end{matrix}\right]\\ &=\hi(-44-2)-\hj(11+8) +\hk(-1+16)\\ &= \llt -46,-19,15\rgt\\ (\vp\times\vq)\times\vr &= \det\left[\begin{matrix}\hi&\hj&\hk\cr-6&5&-13\cr2&-3&-1\end{matrix}\right]\\ &=\hi(-5-39)-\hj(6+26) +\hk(18-10)\\ &= \llt -44,-32,8\rgt \end{align*}

1.2.9.29.

Solution.

Denote by $\theta$ the angle between the two vectors $\va=\llt 1,2,3\rgt$ and $\vb=\llt 3,2,1\rgt\text{.}$ The area of the triangle is one half times the length, $|\va|\text{,}$ of its base times its height $h=|\vb|\sin\theta\text{.}$

Thus the area of the triangle is $\half|\va|\,|\vb|\,\sin\theta\text{.}$ By property 2 of the cross product in Theorem 1.2.23, $|\va\times\vb|=|\va|\,|\vb|\,\sin\theta\text{.}$ So

\begin{align*} \text{area} &= \half|\va\times\vb| =\half|\llt 1,2,3\rgt\times\llt 3,2,1\rgt|\\ &=\half | \hi\,(2-6)-\hj\,(1-9) +\hk\,(2-6)|\\ &=\half\sqrt{16+64+16}\\ &=2\sqrt{6} \end{align*}

1.2.9.30. (✳).

Solution.

The derivative of $L$ is

\begin{align*} \diff{L}{t}&=\diff{}{t}\big(\vr(t)\times\vr'(t)\big) =\vr'(t)\times\vr'(t)+\vr(t)\times\vr''(t)\\ &=\vr'(t)\times\vr'(t)+\vr(t)\times\big(\rho(t)\vr(t)\big) \end{align*}

Both terms vanish because the cross product of any two parallel vectors is zero. So $\diff{L}{t}=0$ and $L(t)$ is independent of $t\text{.}$

1.2.9.31.

Solution.

The parallelogram determined by the vectors $\va$ and $\vb$ has vertices $\vZero,\ \va,\ \vb$ and $\va+\vb\text{.}$ As $t$ varies from $0$ to $1\text{,}$ $t(\va+\vb)$ traverses the diagonal from $\vZero$ to $\va+\vb\text{.}$ As $s$ varies from $0$ to $1\text{,}$ $\va+s(\vb-\va)$ traverses the diagonal from $\va$ to $\vb\text{.}$ These two straight lines meet when $s$ and $t$ are such that

\begin{equation*} t(\va+\vb)=\va+s(\vb-\va) \end{equation*}

\begin{equation*} (t+s-1)\va=(s-t)\vb \end{equation*}

Assuming that $\va$ and $\vb$ are not parallel (i.e. the parallelogram has not degenerated to a line segment), this is the case only when $t+s-1=0$ and $s-t=0\text{.}$ That is, $s=t=\half\text{.}$ So the two lines meet at their midpoints.

1.2.9.32.

Solution.

We may choose our coordinate axes so that $A=(0,0,0)\text{,}$ $B=(s,0,0)\text{,}$ $C=(s,s,0)\text{,}$ $D=(0,s,0)$ and $A'=(0,0,s)\text{,}$ $B'=(s,0,s)\text{,}$ $C'=(s,s,s)\text{,}$ $D'=(0,s,s)\text{.}$

(a) Then

\begin{alignat*}{3} |A'C'|&=\big|\llt s,s,s\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt s,s,0\rgt \big|&=\sqrt{2}\,s\\ |A'B|&=\big|\llt s,0,0\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt s,0,-s\rgt \big|&=\sqrt{2}\,s\\ |A'D|&=\big|\llt 0,s,0\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt 0,s,-s\rgt \big|&=\sqrt{2}\,s\\ |C'B|&=\big|\llt s,0,0\rgt -\llt s,s,s\rgt \big| &&=\big|\llt 0,-s,-s\rgt \big|&=\sqrt{2}\,s\\ |C'D|&=\big|\llt 0,s,0\rgt -\llt s,s,s\rgt \big| &&=\big|\llt -s,0,-s\rgt \big|&=\sqrt{2}\,s\\ |BD|&=\big|\llt 0,s,0\rgt -\llt s,0,0\rgt \big| &&=\big|\llt -s,s,0\rgt \big|&=\sqrt{2}\,s \end{alignat*}

(b) $E=\half(s,s,s)$ so that $EA=\llt 0,0,0\rgt -\half\llt s,s,s\rgt =-\half\llt s,s,s\rgt $ and $EC=\llt s,s,0\rgt -\half\llt s,s,s\rgt =\half\llt s,s,-s\rgt \text{.}$

\begin{equation*} \cos\theta =\frac{-\llt s,s,s\rgt \cdot\llt s,s,-s\rgt} {|\llt s,s,s\rgt |\,|\llt s,s,-s\rgt |} =\frac{-s^2}{3s^2} =-\frac{1}{3} \qquad \implies\quad \theta=109.5^\circ \end{equation*}

1.2.9.33.

Solution.

Suppose that the cube has height, length and width $s\text{.}$ We may choose our coordinate axes so that the vertices of the cube are at $(0,0,0)\text{,}$ $(s,0,0)\text{,}$ $(0,s,0)\text{,}$ $(0,0,s)\text{,}$ $(s,s,0)\text{,}$ $(0,s,s)\text{,}$ $(s,0,s)$ and $(s,s,s)\text{.}$

We’ll start with a couple of examples. The diagonal from $(0,0,0)$ to $(s,s,s)$ is $\llt s,s,s\rgt\text{.}$ One face of the cube has vertices $(0,0,0)\text{,}$ $(s,0,0)\text{,}$ $(0,s,0)$ and $(s,s,0)\text{.}$ One diagonal of this face runs from $(0,0,0)$ to $(s,s,0)$ and hence is $\llt s,s,0\rgt\text{.}$ The angle between $\llt s,s,s\rgt$ and $\llt s,s,0\rgt$ is

\begin{align*} \arccos\left(\frac{\llt s,s,s\rgt\cdot\llt s,s,0\rgt} {|\llt s,s,s\rgt|\,|\llt s,s,0\rgt|}\right) &=\arccos\left(\frac{2s^2}{\sqrt{3}s\,\sqrt{2}s}\right) =\arccos\left(\frac{2}{\sqrt{6}}\right)\\ &\approx 35.26^\circ \end{align*}

A second diagonal for the face with vertices $(0,0,0)\text{,}$ $(s,0,0)\text{,}$ $(0,s,0)$ and $(s,s,0)$ is that running from $(s,0,0)$ to $(0,s,0)\text{.}$ This diagonal is $\llt -s,s,0\rgt\text{.}$ The angle between $\llt s,s,s\rgt$ and $\llt -s,s,0\rgt$ is

\begin{align*} \arccos\left(\frac{\llt s,s,s\rgt\cdot\llt -s,s,0\rgt} {|\llt s,s,s\rgt|\,|\llt -s,s,0\rgt|}\right) &=\arccos\left(\frac{0}{\sqrt{3}s\,\sqrt{2}s}\right) =\arccos(0)\\ &=90^\circ \end{align*}

Now we’ll consider the general case. Note that every component of every vertex of the cube is either $0$ or $s\text{.}$ In general, two vertices of the cube are at opposite ends of a diagonal of the cube if all three components of the two vertices are different. For example, if one end of the diagonal is $(s,0,s)\text{,}$ the other end is $(0,s,0)\text{.}$ The diagonals of the cube are all of the form $\llt \pm s,\pm s,\pm s\rgt\text{.}$ All of these diagonals are of length $\sqrt{3}s\text{.}$ Two vertices are on the same face of the cube if one of their components agree. They are on opposite ends of a diagonal for the face if their other two components differ. For example $(0,s,s)$ and $(s,0,s)$ are both on the face with $z=s\text{.}$ Because the $x$ components $0,\ s$ are different and the $y$ components $s,\ 0$ are different, $(0,s,s)$ and $(s,0,s)$ are the ends of a diagonal of the face with $z=s\text{.}$ The diagonals of the faces with $z=0$ or $z=s$ are $\llt \pm s,\pm s,0\rgt\text{.}$ The diagonals of the faces with $y=0$ or $y=s$ are $\llt \pm s,0, \pm s\rgt\text{.}$ The diagonals of the faces with $x=0$ or $x=s$ are $\llt 0,\pm s,\pm s\rgt\text{.}$ All of these diagonals have length $\sqrt{2}s\text{.}$ The dot product of one the cube diagonals $\llt \pm s,\pm s,\pm s\rgt$ with one of the face diagonals $\llt \pm s,\pm s,0\rgt\text{,}$ $\llt \pm s,0, \pm s\rgt\text{,}$ $\llt 0,\pm s,\pm s\rgt$ is of the form $\pm s^2\pm s^2+0$ and hence must be either $2s^2$ or $0$ or $-2s^2\text{.}$ In general, the angle between a cube diagonal and a face diagonal is

\begin{align*} \arccos\left(\frac{2s^2 \text{ or } 0 \text{ or } -2s^2}{\sqrt{3}s\,\sqrt{2}s}\right) &=\arccos\left(\frac{2 \text{ or } 0 \text{ or } or -2}{\sqrt{6}}\right)\\ &\approx 5.26^\circ \text{ or } 90^\circ \text{ or } 144.74^\circ. \end{align*}

1.2.9.34.

Solution.

Denote by $\big(x(t),y(t)\big)$ the position of the skier at time $t\text{.}$ As long as the skier remains on the surface of the hill

\begin{align*} y(t)&=h\big(x(t)\big)\\ \implies y'(t)&=h'\big(x(t)\big)\,x'(t)\\ \implies y''(t)&=h''\big(x(t)\big)\, x'(t)^2+h'\big(x(t)\big)\,x''(t) \end{align*}

So the velocity and acceleration vectors of the skier are

\begin{align*} \vv(t)&=\llt 1,h'\big(x(t)\big)\rgt x'(t)\\ \va(t)&=\llt 1,h'\big(x(t)\big)\rgt x''(t) +\llt 0,h''\big(x(t)\big)\rgt x'(t)^2 \end{align*}

The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill and has a magnitude such that the skier remains on the surface of the hill. From the velocity vector of the skier (which remain tangential to the hill as long as the skier remains of the surface of the hill),we see that one vector normal to the hill at $\big(x(t),y(t)\big)$ is

\begin{equation*} \vn(t)=\llt-h'\big(x(t)\big),1\rgt \end{equation*}

This vector is not a unit vector, but that’s ok. By Newton’s law of motion

\begin{equation*} m\va=-mg\,\hj+p(t)\,\vn(t) \end{equation*}

for some function $p(t)\text{.}$ Dot both sides of this equation with $\vn(t)\text{.}$

\begin{equation*} m\va(t)\cdot\vn(t)=-mg\hj\cdot\vn(t)+p(t)|\vn(t)|^2 \end{equation*}

Substituting in

\begin{align*} mh''\big(x(t)\big)\,x'(t)^2&=-mg+p(t)\left[1+h'\big(x(t)\big)^2\right]\\ \implies p(t)\left[1+h'\big(x(t)\big)^2\right] &=m\Big(g+h''\big(x(t)\big)\,x'(t)^2\Big) \end{align*}

As long as $p(t)\ge 0\text{,}$ the hill is pushing up in order to keep the skier on the surface. When $p(t)$ becomes negative, the hill has to pull on the skier in order to keep her on the surface. But the hill can’t pull, so the skier becomes airborne instead. This happens when

\begin{equation*} g+h''\big(x(t)\big)x'(t)^2=0 \end{equation*}

That is when $x'(t)=\sqrt{-g/h''\big(x(t)\big)}\text{.}$ At this time $x(t)=x_0\text{,}$ $y(t)=y_0$ and the speed of the skier is

\begin{equation*} \sqrt{x'(t)^2+y'(t)^2} =\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)} \end{equation*}

1.2.9.35.

Solution.

The marble is subject to two forces. The first, gravity, is $-mg\,\hk$ with $m$ being the mass of the marble. The second is the normal force imposed by the plane. This forces acts in a direction perpendicular to the plane. One vector normal to the plane is $a\,\hi+b\,\hj+c\,\hk\text{.}$ So the force due to the plane is $T\llt a,b,c\rgt$ with $T$ determined by the property that the net force perpendicular to the plane must be exactly zero, so that the marble remains on the plane, neither digging into nor flying off of it. The projection of the gravitational force onto the normal vector $\llt a,b,c\rgt$ is

\begin{equation*} \frac{-mg\llt 0,0,1\rgt\cdot\llt a,b,c\rgt}{|\llt a,b,c\rgt|^2}\llt a,b,c\rgt =\frac{-mgc}{a^2+b^2+c^2}\llt a,b,c\rgt \end{equation*}

The condition that determines $T$ is thus

\begin{equation*} T\llt a,b,c\rgt+\frac{-mgc}{a^2+b^2+c^2}\llt a,b,c\rgt =0 \implies T=\frac{mgc}{a^2+b^2+c^2} \end{equation*}

The total force on the marble is then (ignoring friction - which will have no effect on the direction of motion)

\begin{align*} T\llt a,b,c\rgt-mg\llt 0,0,1\rgt&=\frac{mgc}{a^2+b^2+c^2}\llt a,b,c\rgt-mg\llt 0,0,1\rgt\cr &=mg\frac{c\llt a,b,c\rgt-\llt 0,0,a^2+b^2+c^2\rgt}{a^2+b^2+c^2}\cr &=mg\frac{\llt ac,bc,-a^2-b^2\rgt}{a^2+b^2+c^2}\cr \end{align*}

The direction of motion $\llt ac,bc,-a^2-b^2\rgt\text{.}$ If you want to turn this into a unit vector, just divide by $\sqrt{(a^2+b^2)(a^2+b^2+c^2)}\text{.}$ Note that the direction vector in perpendicular $\llt a,b,c\rgt$ and hence is parallel to the plane. If $c=0\text{,}$ the plane is vertical. In this case, the marble doesn’t roll - it falls straight down. If $a=b=0\text{,}$ the plane is horizontal. In this case, the marble doesn’t roll — it remains stationary.

1.2.9.36.

Solution.

By definition, the left and right hand sides are

\begin{align*} \va\cdot(\vb\times\vc) &=\llt a_1,a_2,a_3\rgt \cdot\llt b_2c_3-b_3c_2, b_3c_1-b_1c_3, b_1c_2-b_2c_1\rgt\\ &=a_1b_2c_3 - a_1b_3c_2 + a_2b_3c_1 - a_2b_1c_3 + a_3b_1c_2 - a_3b_2c_1 \tag{lhs}\\ (\va\times\vb)\cdot\vc &=\llt a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1\rgt \cdot\llt c_1,c_2,c_3\rgt \notag\\ &=a_2b_3c_1 - a_3b_2c_1 + a_3b_1c_2 - a_1b_3c_2 + a_1b_2c_3 - a_2b_1c_3 \tag{rhs} \end{align*}

(lhs) and (rhs) are the same.

1.2.9.37.

Solution.

By definition,

\begin{align*} \vb\times\vc \ =\ &(b_2c_3-b_3c_2)\hi-(b_1c_3-b_3c_1)\hj +(b_1c_2-b_2c_1)\hk\\ \end{align*}

so that the left and right hand sides are

\begin{align*} &\va\times(\vb\times\vc) \ =\ \det\left[\begin{matrix}\hi&\hj &\hk\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ &\hskip0.5in= \hi\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ &\hskip0.5in\ -\hj\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ &\hskip0.5in\ +\hk\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ &(\va\cdot\vc)\vb-(\va\cdot\vb)\vc\\ &\hskip0.5in=(a_1c_1+a_2c_2+a_3c_3)(b_1\hi+b_2\hj+b_3\hk)\\ &\hskip1.0in -(a_1b_1+a_2b_2+a_3b_3)(c_1\hi+c_2\hj+c_3\hk)\\ &\hskip0.5in= \hi\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1]\\ &\hskip0.5in\ +\hj\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2]\\ &\hskip0.5in\ +\hk\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\cr &\hskip0.5in = \hi\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1]\\ &\hskip0.5in\ +\hj\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2]\\ &\hskip0.5in\ +\hk\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)} \end{align*}

(lhs) and (rhs) are the same.

1.2.9.38.

Solution.

By properties 9 and 10 of Theorem 1.2.23,

\begin{align*} &(\va\times\vb)\cdot(\vc\times\vd) =\va\cdot[\vb\times(\vc\times\vd)] &\text{(by property 9 with } \vc\rightarrow (\vc\times\vd))\\ &\hskip0.5in=\va\cdot[(\vb\cdot\vd)\vc-(\vb\cdot\vc)\vd] &\text{(by property 10)}\cr &\hskip0.5in=(\va\cdot\vc)(\vb\cdot\vd) -(\va\cdot\vd)(\vb\cdot\vc) \end{align*}

\begin{equation*} (\va\times\vb)\cdot(\vc\times\vd)= (\va\cdot\vc)(\vb\cdot\vd) -(\va\cdot\vd)(\vb\cdot\vc) \end{equation*}

1.2.9.39.

Solution.

(a) $AA'=\llt 4,0,1\rgt $ and $BB'=\llt 4,0,1\rgt $ are opposite sides of the quadrilateral $AA'B'B\text{.}$ They have the same length and direction. The same is true for $AB=\llt -1,3,0\rgt $ and $A'B'=\llt -1,3,0\rgt \text{.}$ So $AA'B'B$ is a parallelogram. Because, $AA'\cdot AB=\llt 4,0,1\rgt \cdot\llt -1,3,0\rgt =-4\ne 0\text{,}$ the neighbouring edges of $AA'B'B$ are not perpendicular and so $AA'B'B$ is not a rectangle.

Similarly, the quadilateral $ACC'A'$ has opposing sides $AA'=\llt 4,0,1\rgt =CC'=\llt 4,0,1\rgt $ and $AC=\llt -1,0,4\rgt =A'C'=\llt -1,0,4\rgt $ and so is a parallelogram. Because $AA'\cdot AC=\llt 4,0,1\rgt \cdot\llt -1,0,4\rgt = 0\text{,}$ the neighbouring edges of $ACC'A'$ are perpendicular, so $ACC'A'$ is a rectangle.

Finally, the quadilateral $BCC'B'$ has opposing sides $BB'=\llt 4,0,1\rgt =CC'=\llt 4,0,1\rgt $ and $BC=\llt 0,-3,4\rgt =B'C'=\llt 0,-3,4\rgt $ and so is a parallelogram. Because $BB'\cdot BC=\llt 4,0,1\rgt \cdot\llt 0,-3,4\rgt = 4\ne 0\text{,}$ the neighbouring edges of $BCC'B'$ are not perpendicular, so $BCC'B'$ is not a rectangle.

(b) The length of $AA'$ is $|\llt 4,0,1\rgt |=\sqrt{16+1}=\sqrt{17}\text{.}$

(c) The area of a triangle is one half its base times its height. That is, one half times $|AB|$ times $|AC|\sin\theta\text{,}$ where $\theta$ is the angle between $AB$ and $AC\text{.}$ This is precisely $\half |AB\times AC|=\half|\llt -1,3,0\rgt \times\llt -1,0,4\rgt | =\half |\llt 12,4,3\rgt|=\frac{13}{2}\text{.}$

(d) The volume of the prism is the area of its base $ABC\text{,}$ times its height, which is the length of $AA'$ times the cosine of the angle between $AA'$ and the normal to $ABC\text{.}$ This coincides with $\half \llt 12,4,3\rgt \cdot\llt 4,0,1\rgt =\half(48+3)=\frac{51}{2}\text{,}$ which is one half times the length of $\llt 12,4,3\rgt $ (the area of $ABC$) times the length of $\llt 4,0,1\rgt $ (the length of $AA'$) times the cosine of the angle between $\llt 12,4,3\rgt $ and $\llt 4,0,1\rgt $ (the angle between the normal to $ABC$ and $AA'$).

1.2.9.40.

Solution.

Choose our coordinate axes so that the vertex opposite the face of area $D$ is at the origin. Denote by $\va\text{,}$ $\vb$ and $\vc$ the vertices opposite the sides of area $A\text{,}$ $B$ and $C$ respectively. Then the face of area $A$ has edges $\vb$ and $\vc$ so that $A=\half |\vb\times\vc|\text{.}$ Similarly $B=\half|\vc\times\va|$ and $C=\half|\va\times \vb|\text{.}$ The face of area $D$ is the triangle spanned by $\vb-\va$ and $\vc-\va$ so that

\begin{align*} D&=\half|(\vb-\va)\times(\vc-\va)|\cr &=\half|\vb\times \vc-\va\times\vc-\vb\times\va|\\ &=\half|\vb\times \vc+\vc\times\va+\va\times\vb| \end{align*}

By hypothesis, the vectors $\va\text{,}$ $\vb$ and $\vc$ are all perpendicular to each other. Consequently the vectors $\vb\times \vc$ (which is a scalar times $\va$), $\vc\times\va$ (which is a scalar times $\vb$) and $\va\times\vb$ (which is a scalar times $\vc$) are also mutually perpendicular. So, when we multiply out

\begin{equation*} D^2=\frac{1}{4}\big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot\big[\vb\times \vc+\vc\times\va+\va\times\vb\big] \end{equation*}

all the cross terms vanish, leaving

\begin{align*} D^2&=\frac{1}{4}\big[(\vb\times \vc)\cdot(\vb\times \vc) +(\vc\times\va)\cdot(\vc\times\va) +(\va\times\vb)\cdot(\va\times\vb)\big]\\ &=A^2+B^2+C^2 \end{align*}

1.2.9.41.

Solution.

As in problem 1.2.9.40,

\begin{equation*} D^2=\frac{1}{4} \big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot \big[\vb\times \vc+\vc\times\va+\va\times\vb\big] \end{equation*}

But now $(\vb\times \vc)\cdot(\va\times\vc)\text{,}$ instead of vanishing, is $|\vb\times \vc|=2A$ times $|\va\times\vc|=2B$ times the cosine of the angle between $\vb\times \vc$ (which is perpendicular to the face of area $A$) and $\va\times\vc$ (which is perpendicular to the face of area $B$). That is

\begin{align*} (\vb\times \vc)\cdot(\va\times\vc)&=4 AB\cos \ga\\ (\va\times \vb)\cdot(\vc\times\vb)&=4 AC\cos \be\\ (\vb\times \va)\cdot(\vc\times\va)&=4 BC\cos \al \end{align*}

(If you’re worried about the signs, that is, if you are worried about why $(\vb\times \vc)\cdot(\va\times\vc)=4 AB\cos \ga$ rather than $(\vb\times \vc)\cdot(\vc\times\va)=4 AB\cos \ga\text{,}$ note that when $\va\approx\vb\text{,}$ $(\vb\times \vc)\cdot(\va\times\vc)\approx|\vb\times\vc|^2$ is positive and $(\vb\times \vc)\cdot(\vc\times\va) \approx -|\vb\times\vc|^2$ is negative.) Now, expanding out

\begin{align*} D^2\ =\ &\frac{1}{4} \big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot \big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\\ =\ &\frac{1}{4}\big[(\vb\times \vc)\cdot(\vb\times \vc) +(\vc\times\va)\cdot(\vc\times\va) +(\va\times\vb)\cdot(\va\times\vb)\\ &+2(\vb\times \vc)\cdot(\vc\times \va) +2(\vb\times \vc)\cdot(\va\times \vb) +2(\vc\times \va)\cdot(\va\times \vb)\big]\\ =\ &A^2+B^2+C^2-2 AB\cos \ga-2 AC\cos \be-2 BC\cos \al \end{align*}

1.3 Equations of Lines in 2d

Exercises

1.3.1.

Solution.

Since $t$ can be any real number, these equation describe the same line. They’re both valid. For example, the point given by the first parametric equation with $t=7\text{,}$ namely $\mathbf c + 7\mathbf d \text{,}$ is exactly the same as the point given by the second parametric equation with $t=-7\text{,}$ namely $\mathbf c -(-7)\mathbf d \text{.}$

1.3.2.

Solution.

In contrast to Question 1.3.1, the sign on $\mathbf c$ does generally matter. $\mathbf c$ is required to be a point on the line, but except in particular circumstances, there’s no reason to believe that $-\mathbf c =-\mathbf c +t\mathbf d\big|_{t=0}$ is a point on the line. Indeed $-\mathbf c$ is on the line if and only if there is a $t$ with $\mathbf c + t\mathbf d = -\mathbf c\text{,}$ i.e. $t\mathbf d = -2\mathbf c\text{.}$ That is the case if and only if $\mathbf d$ is parallel to $\mathbf c\text{.}$ So, only the first equation is correct in general.

1.3.3.

Solution.

Here is one answer of many.

Setting $t=0$ in the first equation shows that $(1,9)$ is on the first line. To see that $(1,9)$ is also on the second line, we substitute $x=1\text{,}$ $y=9$ into the second equation to give

\begin{equation*} \llt 1-9,9-13\rgt=t\llt 1,\tfrac12\rgt\qquad \text{or}\qquad \llt -8,-4\rgt=t\llt 1,\tfrac12\rgt \end{equation*}

This equation is satisfied when $t=-8\text{.}$ So $(1,9)$ is on both lines.

Setting $t=0$ in the second equation shows that $(9,13)$ is on the second line. To see that $(9,13)$ is also on the first line, we substitute $x=9\text{,}$ $y=13$ into the first equation to give

\begin{equation*} \llt 9-1,13-9\rgt=t\llt 8,4\rgt\qquad \text{or}\qquad \llt 8,4\rgt=t\llt 8,4\rgt \end{equation*}

This equation is satisfied when $t=1\text{.}$ So $(9,13)$ is on both lines.

Since both lines pass through $(1,9)$ and $(9,13)\text{,}$ the lines are identical.

1.3.4.

Solution.

$\llt d_x,d_y\rgt$ is the direction of the line, so it can be any non-zero scalar multiple of $\llt 9,7\rgt\text{.}$

$\llt x_0,y_0\rgt$ can be any point on the line. Describing these is the same as describing the line itself. We’re trying to find all doubles $\llt x_0,y_0\rgt$ that obey

\begin{align*} \begin{cases} x_0-3&=9t\\ y_0-5&=7t \end{cases} \end{align*}

for some real number $t\text{.}$ That is,

\begin{align*} t=\frac{x_0-3}{9}&=\frac{y_0-5}{7}\\ 7(x_0-3)&=9(y_0-5)\\ 7x_0+24&=9y_0 \end{align*}

Any of these steps could specify the possible values of $\llt x_0,y_0\rgt\text{.}$ Say, they can be any pair satisfying $7x_0+24=9y_0\text{.}$

1.3.5.

Solution.

(a) The vector parametric equation is $\llt x,y\rgt=\llt 1,2\rgt+t\llt 3,2\rgt \text{.}$ The scalar parametric equations are $x=1+3t,\ y=2+2t\text{.}$ The symmetric equation is $\frac{x-1}{3}=\frac{y-2}{2}\text{.}$

(b) The vector parametric equation is $\llt x,y\rgt=\llt 5,4\rgt+t\llt 2,-1\rgt \text{.}$ The scalar parametric equations are $x=5+2t,\ y=4-t\text{.}$ The symmetric equation is $\frac{x-5}{2}=\frac{y-4}{-1}\text{.}$

(c) The vector parametric equation is $\llt x,y\rgt=\llt -1,3\rgt+t\llt -1,2\rgt \text{.}$ The scalar parametric equations are $x=-1-t,\ y=3+2t\text{.}$ The symmetric equation is $\frac{x+1}{-1}=\frac{y-3}{2}\text{.}$

1.3.6.

Solution.

(a) The vector $\llt -2,3\rgt $ is perpendicular to $\llt 3,2\rgt $ (you can verify this by taking the dot product of the two vectors) and hence is a direction vector for the line. The vector parametric equation is $\llt x,y\rgt=\llt 1,2\rgt+t\llt -2,3\rgt \text{.}$ The scalar parametric equations are $x=1-2t,\ y=2+3t\text{.}$ The symmetric equation is $\frac{x-1}{-2}=\frac{y-2}{3}\text{.}$

(b) The vector $\llt 1,2\rgt $ is perpendicular to $\llt 2,-1\rgt $ and hence is a direction vector for the line. The vector parametric equation for the line is $\llt x,y\rgt=\llt 5,4\rgt+t\llt 1,2\rgt \text{.}$ The scalar parametric equations are $x=5+t,\ y=4+2t\text{.}$ The symmetric equation is $x-5=\frac{y-4}{2}\text{.}$

(c) The vector $\llt 2,1\rgt $ is perpendicular to $\llt -1,2\rgt $ and hence is a direction vector for the line. The vector parametric equation is $\llt x,y\rgt=\llt -1,3\rgt+t\llt 2,1\rgt \text{.}$ The scalar parametric equations are the two component equations $x=-1+2t,\ y=3+t\text{.}$ The symmetric equation is $\frac{x+1}{2}=y-3\text{.}$

1.3.7.

Solution.

$(0,1)$ is one point on the line $3x-4y=-4\text{.}$ So $\llt-2-0,3-1\rgt =\llt-2,2\rgt$ is a vector whose tail is on the line and whose head is at $(-2,3)\text{.}$ $\llt 3,-4\rgt$ is a vector perpendicular to the line, so $\frac{1}{5}\llt 3,-4\rgt$ is a unit vector perpendicular to the line. The distance from $(-2,3)$ to the line is the length of the projection of $\llt-2,2\rgt$ on $\frac{1}{5}\llt 3,-4\rgt\text{,}$ which is the absolute value of $\frac{1}{5}\llt 3,-4\rgt\cdot\llt -2,2\rgt\text{.}$ So the distance is $14/5\text{.}$

1.3.8.

Solution.

(a) The midpoint of the side opposite $\va$ is $\half(\vb+\vc)\text{.}$ The vector joining $\va$ to that midpoint is $\half\vb+\half\vc-\va\text{.}$ The vector parametric equation of the line through $\va$ and $\half(\vb+\vc)$ is

\begin{equation*} \vx(t)=\va+t\big(\half\vb+\half\vc-\va\big) \end{equation*}

Similarly, for the other two medians (but using $s$ and $u$ as parameters, rather than $t$)

\begin{align*} \vx(s)&=\vb+s\big(\half\va+\half\vc-\vb\big)\\ \vx(u)&=\vc+u\big(\half\va+\half\vb-\vc\big) \end{align*}

(b) The three medians meet at a common point if there are values of $s,t$ and $u$ such that

\begin{alignat*}{2} \va+t\big(\half\vb+\half\vc-\va\big) &\ =\ \vb+s\big(\half\va+\half\vc-\vb\big) &&\ =\ \vc+u\big(\half\va+\half\vb-\vc\big)\\ (1-t)\va+\frac{t}{2}\vb+\frac{t}{2}\vc &\ =\ \frac{s}{2}\va+(1-s)\vb+\frac{s}{2}\vc &&\ =\ \frac{u}{2}\va+\frac{u}{2}\vb+(1-u)\vc \end{alignat*}

Assuming that the triangle has not degenerated to a line segment, this is the case if and only if the coefficients of $\va,\ \vb$ and $\vc$ match

\begin{alignat*}{2} 1-t&=\frac{s}{2}&&=\frac{u}{2}\\ \frac{t}{2}&=1-s&&=\frac{u}{2}\\ \frac{t}{2}&=\frac{s}{2}&&=1-u \end{alignat*}

\begin{equation*} s=t=u,\ 1-t=\frac{t}{2} \implies s=t=u=\frac{2}{3} \end{equation*}

The medians meet at $\frac{1}{3}(\va+\vb+\vc)\text{.}$

1.3.9.

Solution.

A normal vector to the line is the vector with its tail at the centre of $C\text{,}$ $(2,1)\text{,}$ and its head at $\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)\text{.}$ So, we set $\mathbf{n}=\llt \frac{5}{2},1+\frac{\sqrt3}{2} \rgt-\llt 2,1\rgt = \llt \frac12, \frac{\sqrt3}{2}\rgt\text{.}$

We know one point on the line is $\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)\text{,}$ so following Equation 1.3.3:

\begin{align*} n_xx+n_yy&=n_xx_0+n_yy_0\\ \frac12x+\frac{\sqrt3}{2}y&=\frac12\cdot\frac52+\frac{\sqrt3}{2}\cdot\left(1+\frac{\sqrt3}{2}\right)\\ \frac12x+\frac{\sqrt3}{2}y&=2+\frac{\sqrt3}{2}\\ x+{\sqrt3}y&=4+{\sqrt3} \end{align*}

1.4 Equations of Planes in 3d

Exercises

1.4.1.

Solution.

We are looking for a vector that is perpendicular to $z=0$ and hence is parallel to $\hk\text{.}$ To be parallel of $\hk\text{,}$ the vector has to be of the form $c\,\hk$ for some real number $c\text{.}$ For the vector to be nonzero, we need $c\ne 0$ and for the vector to be different from $\hk\text{,}$ we need $c\ne 1\text{.}$ So three possible choices are $-\hk\text{,}$ $2\,\hk\text{,}$ $7.12345\,\hk\text{.}$

1.4.2.

Solution.

Each point on the $y$-axis is of the form $(0,y,0)\text{.}$ Such a point is on the plane $P$ if
\begin{equation*} 3(0)+\frac{1}{2}y+0=4 \iff y=8 \end{equation*}
So the intersection of $P$ with the $y$-axis is the single point $(0,8,0)\text{.}$
Each point on the $z$-axis is of the form $(0,0,z)\text{.}$ Such a point is on the plane $P$ if
\begin{equation*} 3(0)+\frac{1}{2}(0)+z=4 \iff z=4 \end{equation*}
So the intersection of $P$ with the $z$-axis is the single point $(0,0,4)\text{.}$
The intersection of the plane $P$ with the $yz$-plane is a line. We have shown in parts (a) and (b) that the points $(0,8,0)$ and $(0,0,4)$ are on that line. Here is a sketch of the part of that line that is in the first octant.

1.4.3.

Solution.

If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,0)\text{,}$ namely $\llt x,y,z\rgt\text{,}$ lie in the plane. So the vector $\llt x,y,z\rgt$ must be perpendicular to $\llt 1,2,3\rgt$ and
\begin{equation*} 0=\llt x,y,z\rgt\cdot \llt 1,2,3\rgt =x+2y+3z \end{equation*}
If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,1)\text{,}$ namely $\llt x,y,z-1\rgt\text{,}$ lie in the plane. So the vector $\llt x,y,z-1\rgt$ must be perpendicular to $\llt 1,1,3\rgt$ and
\begin{equation*} 0=\llt x,y,z-1\rgt\cdot \llt 1,1,3\rgt =x+y+3(z-1) \iff x+y+3z=3 \end{equation*}
If both $(1,2,3)$ and $(1,0,0)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(1,0,0)\text{,}$ namely $\llt 0,2,3\rgt\text{,}$ lie in the plane. So the vector $\llt 0,2,3\rgt$ must be perpendicular to $\llt 4,5,6\rgt\text{.}$ As
\begin{equation*} \llt 0,2,3\rgt \cdot \llt 4,5,6\rgt = 28\ne 0 \end{equation*}
the vector $\llt 0,2,3\rgt$ is not perpendicular to $\llt 4,5,6\rgt\text{.}$ So there is no plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\llt 4,5,6\rgt\text{.}$
If both $(1,2,3)$ and $(0,3,4)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(0,3,4)\text{,}$ namely $\llt 1,-1,-1\rgt\text{,}$ lie in the plane. So the vector $\llt 1,-1,-1\rgt$ must be perpendicular to $\llt 2,1,1\rgt\text{.}$ As

\begin{equation*} \llt 1,-1,-1\rgt \cdot \llt 2,1,1\rgt = 0 \end{equation*}

the vector $\llt 1,-1,-1\rgt$ is indeed perpendicular to $\llt 2,1,1\rgt\text{.}$ So there is a plane that passes through both $(1,2,3)$ and $(0,3,4)$ and has normal vector $\llt 2,1,1\rgt\text{.}$ We now just have to build its equation.

If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(1,2,3)\text{,}$ namely $\llt x-1,y-2,z-3\rgt\text{,}$ lie in the plane. So the vector $\llt x-1,y-2,z-3\rgt$ must be perpendicular to $\llt 2,1,1\rgt$ and

\begin{align*} &0=\llt x-1,y-2,z-3\rgt\cdot \llt 2,1,1\rgt =2(x-1)+(y-2)+(z-3)\\ &\iff 2x+y+z=7 \end{align*}

As a check, note that both $(x,y,z)=(1,2,3)$ and $(x,y,z)=(0,3,4)$ obey the equation $2x+y+z=7\text{.}$

1.4.4. (✳).

Solution.

Solution 1: That’s too easy. We just guess. The plane $x+y+z=1$ contains all three given points.

Solution 2: The plane does not pass through the origin. (You can see this by just making a quick sketch.) So the plane has an equation of the form $ax+by+cz=1\text{.}$

For $(1,0,0)$ to be on the plane we need that
\begin{equation*} a(1) +b(0) +c(0) =1 \implies a=1 \end{equation*}
For $(0,1,0)$ to be on the plane we need that
\begin{equation*} a(0) +b(1) +c(0) =1 \implies b=1 \end{equation*}
For $(0,0,1)$ to be on the plane we need that
\begin{equation*} a(0) +b(0) +c(1) =1 \implies c=1 \end{equation*}

So the plane is $x+y+z=1\text{.}$

Solution 3: Both the head and the tail of the vector from $(1,0,0)$ to $(0,1,0)\text{,}$ namely $\llt -1,1,0\rgt\text{,}$ lie in the plane. Similarly, both the head and the tail of the vector from $(1,0,0)$ to $(0,0,1)\text{,}$ namely $\llt -1,0,1\rgt\text{,}$ lie in the plane. So the vector

\begin{align*} \llt -1,1,0\rgt \times \llt -1,0,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{matrix}\right] =\llt 1 , 1 ,1 \rgt \end{align*}

is a normal vector for the plane. As $(1,0,0)$ is a point in the plane,

\begin{gather*} \llt 1 , 1 ,1 \rgt \cdot \llt x-1\,,\,y-0\,,\, z-0 \rgt = 0\qquad \text{or}\quad x+y+z=1 \end{gather*}

is an equation for the plane.

1.4.5.

Solution.

Solution 1: That’s too easy. We just guess. The plane $x+y+z=2$ contains all three given points.

Solution 2: Both the head and the tail of the vector from $(1,0,1)$ to $(0,1,1)\text{,}$ namely $\llt 1,-1,0\rgt\text{,}$ lie in the plane. Similarly, both the head and the tail of the vector from $(1,1,0)$ to $(0,1,1)\text{,}$ namely $\llt 1,0,-1\rgt\text{,}$ lie in the plane. So the vector

\begin{align*} \llt 1,-1,0\rgt \times \llt 1,0,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & -1 & 0\\ 1 & 0 & -1 \end{matrix}\right] =\llt 1 , 1 ,1 \rgt \end{align*}

is a normal vector for the plane. As $(0,1,1)$ is a point in the plane,

\begin{gather*} \llt 1 , 1 ,1 \rgt \cdot \llt x-0\,,\,y-1\,,\, z-1 \rgt = 0\qquad \text{or}\quad x+y+z=2 \end{gather*}

is an equation for the plane.
Since
\begin{equation*} \Big[x+y+z\Big]_{(x,y,z)=(1,1,1)} = 3\ne 2 \end{equation*}
the point $(1,1,1)$ is not on $x+y+z=2\text{.}$
Since
\begin{equation*} \Big[x+y+z\Big]_{(x,y,z)=(0,0,0)} = 0\ne 2 \end{equation*}
the origin is not on $x+y+z=2\text{.}$
Since
\begin{equation*} \Big[x+y+z\Big]_{(x,y,z)=(4,-1,-1)} = 2 \end{equation*}
the point $(4,-1,-1)$ is on $x+y+z=2\text{.}$

1.4.6.

Solution.

The vector from $(1,2,3)$ to $(2,3,4)\text{,}$ namely $\llt 1,1,1\rgt$ is parallel to the vector from $(1,2,3)$ to $(3,4,5)\text{,}$ namely $\llt 2,2,2\rgt\text{.}$ So the three given points are collinear. Precisely, all three points $(1,2,3),\ (2,3,4)$ and $(3,4,5)$ are on the line $\llt x-1,y-2,z-3\rgt=t\llt 1,1,1\rgt\text{.}$ There are many planes through that line.

1.4.7.

Solution.

(a) The plane must be parallel to $\llt 2,4,6\rgt -\llt 1,0,1\rgt =\llt 1,4,5\rgt $ and to $\llt 1,2,-1\rgt -\llt 1,0,1\rgt =\llt 0,2,-2\rgt \text{.}$ So its normal vector must be perpendicular to both $\llt 1,4,5\rgt $ and $\llt 0,2,-2\rgt $ and hence parallel to

\begin{equation*} \llt 1,4,5\rgt \times\llt 0,2,-2\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 4 & 5 \\ 0 & 2 & -2 \end{matrix}\right] =\llt -18,2,2\rgt \end{equation*}

The plane is $9(x-1)-y-(z-1)=0$ or $9x-y-z=8\text{.}$

We can check this by observing that $(1,0,1),\ (2,4,6)$ and $(1,2,-1)$ all satisfy $9x-y-z=8\text{.}$

(b) The plane must be parallel to $\llt 4,-4,4\rgt -\llt 1,-2,-3\rgt =\llt 3,-2,7\rgt $ and to $\llt 3,2,-3\rgt -\llt 1,-2,-3\rgt =\llt 2,4,0\rgt \text{.}$ So its normal vector must be perpendicular to both $\llt 3,-2,7\rgt $ and $\llt 2,4,0\rgt$ and hence parallel to

\begin{equation*} \llt 3,-2,7\rgt \times\llt 2,4,0\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 3 & -2 & 7 \\ 2 & 4 & 0 \end{matrix}\right] =\llt -28,14,16\rgt \end{equation*}

The plane is $14(x-1)-7(y+2)-8(z+3)=0$ or $14x-7y-8z=52\text{.}$

We can check this by observing that $(1,-2,-3),\ (4,-4,4)$ and $(3,2,-3)$ all satisfy $14x-7y-8z=52\text{.}$

(c) The plane must be parallel to $\llt 5,2,1\rgt -\llt 1,-2,-3\rgt =\llt 4,4,4\rgt $ and to $\llt -1,-4,-5\rgt -\llt 1,-2,-3\rgt =\llt -2,-2,-2\rgt \text{.}$ My, my. These two vectors are parallel. So the three points are all on the same straight line. Any plane containing the line contains all three points. If $\llt a,b,c\rgt$ is any vector perpendicular to $\llt 1,1,1\rgt$ (i.e. which obeys $a+b+c=0$) then the plane $a(x-1)+b(y+2)+c(z+3)=0$ or $a(x-1)+b(y+2)-(a+b)(z+3)=0$ or $ax+by-(a+b)z=4a+b$ contains the three given points.

We can check this by observing that $(1,-2,-3),\ (5,2,1)$ and $(-1,-4,-5)$ all satisfy the equation $ax+by-(a+b)z=4a+b$ for all $a$ and $b\text{.}$

1.4.8.

Solution.

(a) One point on the plane is $(0,0,7)\text{.}$ The vector from $(-1,2,3)$ to $(0,0,7)$ is $\llt 0,0,7\rgt -\llt -1,2,3\rgt =\llt 1,-2,4\rgt \text{.}$ A unit vector perpendicular to the plane is $\frac{1}{\sqrt{3}}\llt 1,1,1\rgt \text{.}$ The distance from $(-1,2,3)$ to the plane is the length of the projection of $\llt 1,-2,4\rgt $ on $\frac{1}{\sqrt{3}}\llt 1,1,1\rgt $ which is

\begin{gather*} \frac{1}{\sqrt{3}}\llt 1,1,1\rgt \cdot\llt 1,-2,4\rgt =\frac{3}{\sqrt{3}} =\sqrt{3} \end{gather*}

(b) One point on the plane is $(0,0,5)\text{.}$ The vector from $(1,-4,3)$ to $(0,0,5)$ is $\llt 0,0,5\rgt -\llt 1,-4,3\rgt =\llt -1,4,2\rgt\text{.}$ A unit vector perpendicular to the plane is $\frac{1}{\sqrt{6}}\llt 1,-2,1\rgt \text{.}$ The distance from $(1,-4,3)$ to the plane is the length of the projection of $\llt -1,4,2\rgt $ on $\frac{1}{\sqrt{6}}\llt 1,-2,1\rgt $ which is the absolute value of

\begin{equation*} \frac{1}{\sqrt{6}}\llt 1,-2,1\rgt \cdot \llt -1,4,2\rgt =\frac{-7}{\sqrt{6}} \end{equation*}

or $7/\sqrt{6}\text{.}$

1.4.9. (✳).

Solution.

(a) The vector from $C$ to $A\text{,}$ namely $\llt 1-2\,,\,1-1\,,\,3-0\rgt = \llt -1\,,\,0\,,\,3\rgt$ lies entirely inside $\Pi\text{.}$ The vector from $C$ to $B\text{,}$ namely $\llt 2-2\,,\,0-1\,,\,2-0\rgt = \llt 0\,,\,-1\,,\,2\rgt$ also lies entirely inside $\Pi\text{.}$ Consequently, the vector

\begin{align*} \llt -1\,,\,0\,,\,3\rgt\times \llt 0\,,\,-1\,,\,2\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ -1 & 0 & 3\\ 0 & -1 & 2 \end{matrix}\right] =\llt 3 \,,\, 2 \,,\, 1 \rgt \end{align*}

is perpendicular to $\Pi\text{.}$ The equation of $\Pi$ is then

\begin{gather*} \llt 3 \,,\, 2 \,,\, 1 \rgt \cdot \llt x-2 \,,\, y-1 \,,\, z \rgt = 0 \qquad\text{or}\qquad 3x+2y+z = 8 \end{gather*}

(b) Let $E$ be $(x,y,z)\text{.}$ Then the vector from $D$ to $E\text{,}$ namely $\llt x-6\,,\,y-1\,,\,z-2\rgt$ has to be parallel to the vector $\llt 3 \,,\, 2 \,,\, 1 \rgt\text{,}$ which is perpendicular to $\Pi\text{.}$ That is, there must be a number $t$ such that

\begin{align*} & \llt x-6\,,\,y-1\,,\,z-2\rgt = t \llt 3 \,,\, 2 \,,\, 1 \rgt\\ &\text{or }x=6+3t,\ y=1+2t,\ z=2+t \end{align*}

As $(x,y,z)$ must be in $\Pi\text{,}$

\begin{gather*} 8 = 3x+2y+z = 3(6+3t) + 2(1+2t) +(2+t) = 22 +14 t \end{gather*}

which implies that $t=-1\text{.}$ So $(x,y,z) = \big(6+3(-1)\,,\,1+2(-1)\,,\,2+(-1)\big) = \big(3\,,\,-1\,,\,1\big)\text{.}$

1.4.10. (✳).

Solution.

We are going to need a direction vector for $L$ in both parts (a) and (b). So we find one first.

The vector $\llt 1,1,0\rgt$ is perpendicular to $x + y = 1$ and hence to $L\text{.}$
The vector $\llt 1,2,1\rgt$ is perpendicular to $x + 2y + z = 3$ and hence to $L\text{.}$

So the vector

\begin{align*} \llt 1,1,0\rgt \times \llt 1,2,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 1 & 0\\ 1 & 2 & 1 \end{matrix}\right] =\llt 1 \,,\, -1 \,,\, 1 \rgt \end{align*}

is a direction vector for $L\text{.}$

(a) The plane is to contain the point $(2,3,4)$ and is to have $\llt-1,1,-1 \rgt$ as a normal vector. So

\begin{gather*} \llt-1,1,-1 \rgt\cdot \llt x-2,y-3,z-4 \rgt =0\qquad\text{or}\qquad x-y+z=3 \end{gather*}

does the job.

(b) The plane is to contain the points $A=(2,3,4)$ and $(1,0,2)$ (which is on $L$) so that the vector $\llt 2-1,3-0,4-2 \rgt =\llt 1,3,2 \rgt$ is to be parallel to the plane. The direction vector of $L\text{,}$ namely $\llt -1 , 1 , -1 \rgt\text{,}$ is also to be parallel to the plane. So the vector

\begin{align*} \llt 1,3,2\rgt \times \llt -1,1,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 3 & 2\\ -1 & 1 & -1 \end{matrix}\right] =\llt -5 \,,\, -1 \,,\, 4 \rgt \end{align*}

is to be normal to the plane. So

\begin{gather*} \llt-5,-1,4 \rgt\cdot \llt x-2,y-3,z-4 \rgt =0\qquad\text{or}\qquad 5x+y-4z=-3 \end{gather*}

does the job.

1.4.11. (✳).

Solution.

All planes that are parallel to the plane $4x + 2y - 4z = 3$ must have $\llt 4\,,\,2\,,\,-4\rgt$ as a normal vector and hence must have an equation of the form $4x + 2y - 4z = C$ for some constant $C\text{.}$ We must find the $C$’s for which the distance from $4x + 2y - 4z = 3$ to $4x + 2y - 4z = C$ is $2\text{.}$ One point on $4x + 2y - 4z = 3$ is $\big(0,\frac{3}{2},0\big)\text{.}$ The two points $(x',y',z')$ with

\begin{align*} \overbrace{\llt x'-0\,,\,y'-\frac{3}{2}\,,\,z'-0\rgt}^{\text{vector from} \big(0,\frac{3}{2},0\big)\text{ to } (x',y',z')} &=\pm 2\ \overbrace{\frac{\llt 4\,,\,2\,,\,-4\rgt}{\sqrt{16+4+16}}}^{\text{unit vector}} =\pm \frac{2}{6} \llt 4\,,\,2\,,\,-4\rgt\\ &=\pm\llt \frac{4}{3}\,,\,\frac{2}{3}\,,\,-\frac{4}{3}\rgt \end{align*}

are the two points that are a distance $2$ from $\big(0,\frac{3}{2},0\big)$ in the direction of the normal.The two points $(x',y',z')$ are

\begin{align*} \left(0+\frac{4}{3}\,,\, \frac{3}{2}+\frac{2}{3}\,,\,0-\frac{4}{3}\right) &=\left(\frac{4}{3}\,,\, \frac{13}{6}\,,\,-\frac{4}{3}\right)\\ \text{ and } \left(0-\frac{4}{3}\,,\, \frac{3}{2}-\frac{2}{3}\,,\,0+\frac{4}{3}\right) &=\left(-\frac{4}{3}\,,\, \frac{5}{6}\,,\,\frac{4}{3}\right) \end{align*}

These two points lie on the desired planes, so the two desired planes are

\begin{gather*} 4x + 2y - 4z =\frac{4\times 4}{3} +\frac{2\times 13}{6}-\frac{-4\times 4}{3} =\frac{32+26+32}{6} =15 \end{gather*}

and

\begin{gather*} 4x + 2y - 4z =\frac{4\times(-4)}{3} +\frac{2\times 5}{6}-\frac{4\times 4}{3} =\frac{-32+10-32}{6} =-9 \end{gather*}

1.4.12. (✳).

Solution.

The two vectors

\begin{alignat*}{2} \va&=\llt 1,-1,3\rgt - \llt 0,1,1\rgt &&= \llt 1,-2, 2\rgt\\ \vb&=\llt 2,0,-1\rgt - \llt 0,1,1\rgt &&= \llt 2,-1,-2\rgt \end{alignat*}

both lie entirely inside the plane. So the vector

\begin{equation*} \va\times\vb =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & -2 & 2 \\ 2 & -1 & -2 \end{matrix}\right] =\llt 6,6,3\rgt \end{equation*}

is perpendicular to the plane. The vector $\vc=\frac{1}{3}\llt 6,6,3\rgt =\llt 2,2,1\rgt$ is also perpendicular to the plane. The vector

\begin{equation*} \vd=\llt 1,2,3\rgt-\llt 0,1,1\rgt=\llt 1,1,2\rgt \end{equation*}

joins the point to the plane. So, if $\theta$ is the angle between $\vd$ and $\vc\text{,}$ the distance is

\begin{equation*} |\vd|\cos\theta=\frac{\vc\cdot\vd}{|\vc|}=\frac{6}{\sqrt{9}}=2 \end{equation*}

1.4.13. (✳).

Solution.

(a) Let’s use $z$ as the parameter and rename it to $t\text{.}$ That is, $z=t\text{.}$ Subtracting $2$ times the $W_2$ equation from the $W_1$ equation gives

\begin{equation*} -5y -5z = -5 \implies y = 1-z = 1-t\quad\text{ or }\quad y-1=-t \end{equation*}

Substituting the result into the equation for $W_2$ gives

\begin{gather*} -x +3(1-t) +3t =6 \implies x = -3\quad\text{ or }\quad x+3=0 \end{gather*}

So a parametric equation is

\begin{equation*} \llt x+3,y-1,z\rgt = t \llt 0,-1,1\rgt \end{equation*}

(b) Solution 1

We can also parametrize $M$ using $z=t\text{:}$

\begin{align*} x=2z+10=2t+10,\ &y=2z+12=2t+12\\ &\implies\quad \llt x,y,z\rgt = \llt 10,12,0\rgt +t \llt 2,2,1\rgt \end{align*}

So one point on $M$ is $(10,12,0)$ and one point on $L$ is $(-3,1,0)$ and

\begin{equation*} \vv= \llt (-3)-10\,,\, 1 -12 \,,\, 0-0\rgt =\llt -13 \,,\, -11 \,,\, 0\rgt \end{equation*}

is one vector from a point on $M$ to a point on $L\text{.}$

The direction vectors of $L$ and $M$ are $\llt 0,-1,1\rgt$ and $\llt 2,2,1\rgt\text{,}$ respectively. The vector

\begin{align*} \vn = \llt 0,-1,1\rgt \times \llt 2,2,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 0 & -1 & 1\\ 2 & 2 & 1 \end{matrix}\right] =\llt -3 \,,\, 2 \,,\, 2 \rgt \end{align*}

is then perpendicular to both $L$ and $M\text{.}$

The distance from $L$ to $M$ is then the length of the projection of $\vv$ on $\vn\text{,}$ which is

\begin{gather*} \frac{|\vv\cdot\vn|}{|\vn|} =\frac{|39-22+0|}{\sqrt{9+4+4}} =\sqrt{17} \end{gather*}

(b) Solution 2 We can also parametrize $M$ using $z=s\text{:}$

\begin{align*} x=2z+10=2s+10,\ &y=2z+12=2s+12\\ &\implies\quad \llt x,y,z\rgt = \llt 10,12,0\rgt +s \llt 2,2,1\rgt \end{align*}

The vector from the point $(-3\,,\,1-t\,,\,t)$ on $L$ to the point $(10+2s\,,\,12+2s\,,\,s)$ on $M$ is

\begin{equation*} \llt 13+2s \,,\, 11+2s+t \,,\, s-t \rgt \end{equation*}

So the distance from the point $(-3\,,\,1-t\,,\,t)$ on $L$ to the point $(10+2s\,,\,12+2s\,,\,s)$ on $M$ is the square root of

\begin{equation*} D(s,t) = (13+2s)^2 +(11+2s+t)^2 +(s-t)^2 \end{equation*}

That distance is minimized when

\begin{align*} 0 = \pdiff{D}{s} &=4(13+2s) +4(11+2s+t) +2(s-t)\\ 0 = \pdiff{D}{t} &=2(11+2s+t) -2(s-t) \end{align*}

Cleaning up those equations gives

\begin{align*} 18s +2t &= -96\\ 2s +4t &=-22 \end{align*}

\begin{align*} 9s +t &=-48 \tag{E1}\\ s+2t &=-11 \tag{E2} \end{align*}

Subtracting (E2) from twice (E1) gives

\begin{equation*} 17s = -85 \implies s=-5 \end{equation*}

Substituting that into (E2) gives

\begin{equation*} 2t= -11 +5 \implies t = -3 \end{equation*}

Note that

\begin{align*} 13+2s &= 3\\ 11+2s+t &= -2\\ s-t &= -2 \end{align*}

So the distance is

\begin{align*} \sqrt{D(-5,-3)} &=\sqrt{3^2 + (-2)^2 + (-2)^2} =\sqrt{17} \end{align*}

1.4.14.

Solution.

The two planes $x+y+z=3$ and $x+y+z=9$ are parallel. The centre must be on the plane $x+y+z=6$ half way between them. So, the center is on $x+y+z=6\text{,}$ $2x-y=0$ and $3x-z=0\text{.}$ Solving these three equations, or equvalently,

\begin{equation*} y=2x,\ z=3x,\ x+y+z=6x=6 \end{equation*}

gives $(1,2,3)$ as the centre. $(1,1,1)$ is a point on $x+y+z=3\text{.}$ $(3,3,3)$ is a point on $x+y+z=9\text{.}$ So $\llt 2,2,2\rgt$ is a vector with tail on $x+y+z=3$ and head on $x+y+z=9\text{.}$ Furthermore $\llt 2,2,2\rgt$ is perpendicular to the two planes. So the distance between the planes is $|\llt 2,2,2\rgt|=2\sqrt{3}$ and the radius of the sphere is $\sqrt{3}\text{.}$ The sphere is

\begin{equation*} (x-1)^2+(y-2)^2+(z-3)^2=3 \end{equation*}

1.4.15.

Solution.

Set $y=0$ and then solve $2x+3y-z=0,\ x-4y+2z=-5\text{,}$ i.e. $2x-z=0\text{,}$ $x+2z=-5\text{,}$ or equvalently

\begin{equation*} z=2x,\ x+2z=5x=-5 \end{equation*}

The result, $(-1,0,-2)\text{,}$ is one point on the plane. Set $y=5$ and then solve $2x+3y-z=0,\ x-4y+2z=-5\text{,}$ i.e. $2x+15-z=0\text{,}$ $x-20+2z=-5\text{,}$ or equivalently

\begin{equation*} z=2x+15,\ x-20+4x+30=-5 \end{equation*}

The result, $(-3,5,9)\text{,}$ is another point on the plane. So three points on the plane are $(-2,0,1), \ (-1,0,-2)$ and $(-3,5,9)\text{.}$ $\llt -2+1\,,\,0-0\,,\, 1+2 \rgt=\llt -1,0,3\rgt$ and $\llt -2+3\,,\,0-5\,,\, 1-9 \rgt = \llt 1,-5,-8\rgt$ are two vectors having both head and tail in the plane.

\begin{align*} \llt -1,0,3\rgt \times \llt 1,-5,-8\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ -1 & 0 & 3\\ 1 & -5 & -8 \end{matrix}\right] =\llt 15 , -5 ,5 \rgt \end{align*}

is a vector perpendicular to the plane. $\frac{1}{5}\llt 15 , -5 ,5 \rgt=\llt 3,-1,1\rgt$ is also a vector perpendicular to the plane. The plane is

\begin{equation*} 3(x+1) -(y-0) + (z+2)=0\quad\text{or}\quad 3x-y+z=-5 \end{equation*}

1.4.16.

Solution.

The vector $\vn$ is perpendicular to the plane $\vn\cdot\vx= c\text{.}$ So the line

\begin{equation*} \vx(t)=\vp+t\vn \end{equation*}

passes through $\vp$ and is perpendicular to the plane.

It crosses the plane at the value of $t$ which obeys

\begin{equation*} \vn\cdot\vx(t)= c \hskip .25 in\text{or}\hskip .25 in \vn\cdot[\vp+t\vn]= c \end{equation*}

namely

\begin{equation*} t=[c-\vn\cdot\vp]/|\vn|^2 \end{equation*}

The vector

\begin{equation*} \vx(t)-\vp=t\,\vn=\vn\,[c-\vn\cdot\vp]/|\vn|^2 \end{equation*}

has head on the plane $\vn\cdot\vx= c\text{,}$ tail at $\vp\text{,}$ and is perpendicular to the plane. So the distance is the length of that vector, which is

\begin{equation*} |c-\vn\cdot\vp|/|\vn| \end{equation*}

1.4.17.

Solution.

The distance from the point $(x,y,z)$ to $(1,2,3)$ is $\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}$ and the distance from $(x,y,z)$to $(5,2,7)$ is $\sqrt{(x-5)^2+(y-2)^2+(z-7)^2}\text{.}$ Hence $(x,y,z)$ is equidistant from $(1,2,3)$ and $(5,2,7)$ if and only if

\begin{align*} & & (x-1)^2+(y-2)^2+(z-3)^2&=(x-5)^2+(y-2)^2+(z-7)^2\\ &\iff & x^2-2x+1+z^2-6z+9&=x^2-10x+25+z^2-14z+49\\ &\iff & 8x+8z&=64\\ &\iff & x+z&=8 \end{align*}

This is the plane through $(3,2,5)=\half(1,2,3)+\half(5,2,7)$ with normal vector $\llt 1,0,1\rgt=\frac{1}{4}\big(\llt 5,2,7\rgt-\llt 1,2,3\rgt \big)\text{.}$

1.4.18.

Solution.

The distance from the point $\vx$ to $\va$ is $\sqrt{(\vx-\va)\cdot(\vx-\va)}$ and the distance from $\vx$ to $\vb$ is $\sqrt{(\vx-\vb)\cdot(\vx-\vb)}\text{.}$ Hence $\vx$ is equidistant from $\va$ and $\vb$ if and only if

\begin{alignat*}{2} & & (\vx-\va)\cdot(\vx-\va)&=(\vx-\vb)\cdot(\vx-\vb)\\ &\iff\qquad & |\vx|^2-2\va\cdot\vx+|\va|^2&=|\vx|^2-2\vb\cdot\vx+|\vb|^2\\ &\iff & 2(\vb-\va)\cdot\vx&=|\vb|^2-|\va|^2 \end{alignat*}

This is the plane through $\half\va+\half\vb$ with normal vector $\vb-\va\text{.}$

1.4.19. (✳).

Solution.

(a) One side of the triangle is $\overrightarrow{AB}=\llt 1,0,1\rgt - \llt 0,1,1\rgt = \llt 1,-1,0\rgt\text{.}$ A second side of the triangle is $\overrightarrow{AC}=\llt 1,3,0\rgt - \llt 0,1,1\rgt = \llt 1,2,-1\rgt\text{.}$ If the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $\theta$ and if we take $\overrightarrow{AB}$ as the base of the triangle, then the triangle has base length $b=|\overrightarrow{AB}|$ and height $h=|\overrightarrow{AC}|\sin\theta$ and hence

\begin{align*} \text{area}&=\half bh =\half |\overrightarrow{AB}|\,|\overrightarrow{AC}|\sin\theta =\half|\overrightarrow{AB}\times\overrightarrow{AC}|\\ &=\half|\llt 1,-1,0\rgt \times \llt 1,2,-1\rgt| \end{align*}

\begin{equation*} \llt 1,-1,0\rgt\times\llt 1,2,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & -1 & 0 \\ 1 & 2 & -1 \end{matrix}\right] =\hi+\hj+3\hk \end{equation*}

we have

\begin{equation*} \text{area}=\half|\llt 1,1,3\rgt|=\half\sqrt{11}\approx 1.658 \end{equation*}

(b) A unit vector perpendicular to the plane containing the triangle is

\begin{equation*} \hn=\frac{\overrightarrow{AB}\times\overrightarrow{AC}} {|\overrightarrow{AB}\times\overrightarrow{AC}|} =\frac{\llt 1,1,3\rgt}{\sqrt{11}} \end{equation*}

The distance from $P$ to the plane containing the triangle is the length of the projection of $\overrightarrow{AP}=\llt 5,-10,2\rgt-\llt 0,1,1\rgt =\llt 5,-11,1\rgt$ on $\hn\text{.}$ If $\theta$ the angle between $\overrightarrow{AP}$ and $\hn\text{,}$ then this is

\begin{align*} \text{distance}&= |\overrightarrow{AP}|\,|\cos\theta| =\big|\overrightarrow{AP}\cdot\hn\big| =\left|\llt 5,-11,1\rgt\cdot\frac{\llt 1,1,3\rgt}{\sqrt{11}}\right| =\frac{3}{\sqrt{11}}\\ &\approx 0.9045 \end{align*}

1.4.20. (✳).

Solution.

Switch to a new coordinate system with

\begin{equation*} X=x-1\qquad Y=y-2\qquad Z=z+1 \end{equation*}

In this new coordinate system, the sphere has equation $X^2+Y^2+Z^2=2\text{.}$ So the sphere is centred at $(X,Y,Z)=(0,0,0)$ and has radius $\sqrt{2}\text{.}$ In the new coordinate system, the initial point $(x,y,z)=(2,2,0)$ has $(X,Y,Z)=(1,0,1)$ and our final point $(x,y,z)=(2,1,-1)$ has $(X,Y,Z)=(1,-1,0)\text{.}$ Call the initial point $P$ and the final point $Q\text{.}$ The shortest path will follow the great circle from $P$ to $Q\text{.}$

A great circle on a sphere is the intersection of the sphere with a plane that contains the centre of the sphere. Our strategy for finding the initial direction will be based on two observations.

The shortest path lies on the plane $\Pi$ that contains the origin and the points $P$ and $Q\text{.}$ Since the shortest path lies on $\Pi\text{,}$ our direction vector must also lie on $\Pi$ and hence must be perpendicular to the normal vector to $\Pi\text{.}$
The shortest path also remains on the sphere, so our initial direction must also be perpendicular to the normal vector to the sphere at our initial point $P\text{.}$

As our initial direction is perpendicular to the two normal vectors, it is parallel to their cross product.

So our main job is to find normal vectors to the plane $\Pi$ and to the sphere at $P\text{.}$

One way to find a normal vector to $\Pi$ is to guess an equation for $\Pi\text{.}$ As $(0,0,0)$ is on $\Pi\text{,}$ $(0,0,0)$ must obey $\Pi$’s equation. So $\Pi$’s equation must be of the form $aX+bY+cZ=0\text{.}$ That $(X,Y,Z)=(1,0,1)$ is on $\Pi$ forces $a+c=0\text{.}$ That $(X,Y,Z)=(1,-1,0)$ is on $\Pi$ forces $a-b=0\text{.}$ So we may take $a=1\text{,}$ $b=1$ and $c=-1\text{.}$ That is, $\Pi$ is $X+Y-Z=0\text{.}$ (Check that all three points $(0,0,0)\text{,}$ $(1,0,1)$ and $(1,-1,0)$ do indeed obey $X+Y-Z=0\text{.}$) A normal vector to $\Pi$ is $\llt 1,1,-1\rgt\text{.}$
A second way to find a normal vector to $\Pi$ is to observe that both
- the vector from $(0,0,0)$ to $(1,0,1)\text{,}$ that is $\llt 1,0,1\rgt\text{,}$ lies completely inside $\Pi$ and
- the vector from $(0,0,0)$ to $(1,-1,0)\text{,}$ that is $\llt 1,-1,0\rgt\text{,}$ lies completely inside $\Pi\text{.}$
So the vector

\begin{equation*} \llt 1,0,1\rgt\times\llt 1,-1,0\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix}\right] =\hi + \hj -\hk \end{equation*}

is perpendicular to $\Pi\text{.}$
The vector from the centre of the sphere to the point $P$ on the sphere is perpendicular to the sphere at $P\text{.}$ So a normal vector to the sphere at our initial point $(X,Y,Z)=(1,0,1)$ is $\llt 1,0,1\rgt\text{.}$

Since our initial direction

Note that the change of coordinates $X=x-1\text{,}$ $Y=y-2\text{,}$ $Z=z+1$ has absolutely no effect on any velocity or direction vector. If our position at time $t$ is $(x(t),y(t),z(t))$ in the original coordinate system, then it is $(X(t),Y(t),Z(t)) =(x(t)-1,y(t)-2,z(t)+1)$ in the new coordinate system. The velocity vectors in the two coordinate systems $\llt x'(t),y'(t),z'(t)\rgt =\llt X'(t),Y'(t),Z'(t)\rgt$ are identical.

must be perpendicular to both $\llt 1,1,-1\rgt$ and $\llt 1,0,1\rgt\text{,}$ it must be one of $\pm\llt 1,1,-1\rgt\times\llt 1,0,1\rgt\text{.}$ To get from $(1,0,1)$ to $(1,-1,0)$ by the shortest path, our $Z$ coordinate should decrease from $1$ to $0\text{.}$ So the $Z$ coordinate of our initial direction should be negative. This is the case for

\begin{equation*} \llt 1,1,-1\rgt\times\llt 1,0,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 1 & -1 \\ 1 & 0 & 1 \end{matrix}\right] =\hi - 2\,\hj -\hk \end{equation*}

1.5 Equations of Lines in 3d

Exercises

1.5.1.

Solution.

Note $12\llt \frac13,-\frac12,\frac16 \rgt=\llt 4,-6,2 \rgt\text{.}$ So, we actually only know one normal direction to the line we’re supposed to be describing. That means there are actually infinitely many lines satisfying the given conditions.

1.5.2.

Solution.

There are infinitely many correct answers. One is

\begin{align*} L_1:\ \llt x\,,\,y\,,\,z-1\rgt &= t\llt 1,0,0\rgt & L_2:\ \llt x\,,\,y\,,\,z-2\rgt &= t\llt 0,1,0\rgt\\ L_3:\ \llt x\,,\,y\,,\,z-3\rgt &= t\llt 1,1,0\rgt & L_4:\ \llt x\,,\,y\,,\,z-4\rgt &= t\llt 1,-1,0\rgt \end{align*}

No two of the lines are parallel because

$L_1$ and $L_2$ are not parallel because $\llt 1,0,0\rgt\times \llt 0,1,0\rgt=\llt 0,0,1\rgt\ne\vZero\text{.}$
$L_1$ and $L_3$ are not parallel because $\llt 1,0,0\rgt\times \llt 1,1,0\rgt=\llt 0,0,1\rgt\ne\vZero\text{.}$
$L_1$ and $L_4$ are not parallel because $\llt 1,0,0\rgt\times \llt 1,-1,0\rgt=\llt 0,0,-1\rgt\ne\vZero\text{.}$
$L_2$ and $L_3$ are not parallel because $\llt 0,1,0\rgt\times \llt 1,1,0\rgt=\llt 0,0,-1\rgt\ne\vZero\text{.}$
$L_2$ and $L_4$ are not parallel because $\llt 0,1,0\rgt\times \llt 1,-1,0\rgt=\llt 0,0,-1\rgt\ne\vZero\text{.}$
$L_3$ and $L_4$ are not parallel because $\llt 1,1,0\rgt\times \llt 1,-1,0\rgt=\llt 0,0,-2\rgt\ne\vZero\text{.}$

No two of the lines intersect because

every point on $L_1$ has $z=1$ and
every point on $L_2$ has $z=2$ and
every point on $L_3$ has $z=3$ and
every point on $L_4$ has $z=4\text{.}$

1.5.3.

Solution.

(a) The point $(x,y,z)$ obeys both $x-2z=3$ and $y+\half z=5$ if and only if $\llt x,y,z\rgt = \llt 3+2z, 5-\half z, z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt z\text{.}$ So, introducing a new variable $t$ obeying $t=z\text{,}$ we get the vector parametric equation $\llt x,y,z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt t\text{.}$

(b) The point $(x,y,z)$ obeys

\begin{align*} \left\{\Atop{2x-y-2z=-3}{4x-3y-3 z=-5} \right\} &\iff \left\{\Atop{2x-y=2z-3}{4x-3y=3 z-5} \right\}\\ & \iff\left\{\Atop{4x-2y=4z-6}{4x-3y=3 z-5} \right\}\\ &\iff\left\{\Atop{4x-2y=4z-6}{y=z-1} \right\} \end{align*}

Hence the point $(x,y,z)$ is on the line if and only if

\begin{align*} \llt x,y,z\rgt &= \llt \frac{1}{4}(2y+4z-6), z-1, z\rgt =\llt \frac{3}{2}z-2, z-1, z\rgt\\ &= \llt -2,-1,0\rgt+\llt \frac{3}{2},1,1\rgt z \end{align*}

So, introducing a new variable $t$ obeying $t=z\text{,}$ we get the vector parametric equation $\llt x,y,z\rgt = \llt -2,-1,0\rgt +\llt \frac{3}{2},1,1\rgt t\text{.}$

1.5.4.

Solution.

(a) The normals to the two planes are $\llt 1,1,1\rgt $ and $\llt 1,2,3\rgt $ respectively. The line of intersection must have direction perpendicular to both of these normals. Its direction vector is

\begin{equation*} \llt 1,1,1\rgt \times\llt 1,2,3\rgt =\det\left[\begin{matrix} \hi &\hj &\hk \\ 1&1&1 \\ 1&2&3 \end{matrix}\right] =\llt 1,-2,1\rgt \end{equation*}

Substituting $z=0$ into the equations of the two planes and solving

\begin{align*} \left\{\Atop{x+y=3}{x+2y=7} \right\} &\iff \left\{\Atop{x=3-y}{x+2y=7} \right\} \iff\left\{\Atop{x=3-y}{3-y+2y=7} \right\} \end{align*}

we see that $z=0,\ y=4, x=-1$ lies on both planes. The line of intersection is $\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt \text{.}$ This can be checked by verifying that, for all values of $t\text{,}$ $\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt $ satisfies both $x+y+z=3$ and $x+2y+3z=7\text{.}$

(b) The equation $x+y+z=3$ is equivalent to $2x+2y+2z=6\text{.}$ So the two equations $x+y+z=3$ and $2x+2y+2z=7$ are mutually contradictory. They have no solution. The two planes are parallel and do not intersect.

1.5.5.

Solution.

(a) Note that the value of the parameter $t$ in the equation $\llt x,y,z\rgt = \llt -3,2,4\rgt+t\llt -4,2,1\rgt$ need not have the same value as the parameter $t$ in the equation $\llt x,y,z\rgt = \llt 2,1,2\rgt+t\llt 1,1,-1\rgt\text{.}$ So it is much safer to change the name of the parameter in the first equation from $t$ to $s\text{.}$ In order for a point $(x,y,z)$ to lie on both lines we need

\begin{equation*} \llt -3,2,4\rgt+s\llt -4,2,1\rgt = \llt 2,1,2\rgt+t\llt 1,1,-1\rgt \end{equation*}

or equivalently, writing out the three component equations and moving all $s$’s and $t$’s to the left and constants to the right,

\begin{align*} -4s -t &= 5\\ 2s -t &= -1\\ s +t &= -2 \end{align*}

Adding the last two equations together gives $3s=-3$ or $s=-1\text{.}$ Substituting this into the last equation gives $t=-1\text{.}$ Note that $s=t=-1$ does indeed satisfy all three equations so that $\llt x,y,z\rgt=\llt -3,2,4\rgt-\llt -4,2,1\rgt=\llt 1,0,3\rgt$ lies on both lines. Any plane that contains the two lines must be parallel to both direction vectors $\llt -4,2,1\rgt$ and $\llt 1,1,-1\rgt\text{.}$ So its normal vector must be perpendicular to them, i.e. must be parallel to $\llt -4,2,1\rgt\times\llt 1,1,-1\rgt=\llt -3,-3,-6\rgt=-3\llt 1,1,2\rgt\text{.}$ The plane must contain $(1,0,3)$ and be perpendicular to $\llt 1,1,2\rgt\text{.}$ Its equation is $\llt 1,1,2\rgt\cdot\llt x-1,y,z-3\rgt=0$ or $x+y+2z=7\text{.}$ This can be checked by verifying that $\llt -3,2,4\rgt+s\llt -4,2,1\rgt$ and $\llt 2,1,2\rgt+t\llt 1,1,-1\rgt$ obey $x+y+2z=7$ for all $s$ and $t$ respectively.

(b) In order for a point $(x,y,z)$ to lie on both lines we need

\begin{equation*} \llt -3,2,4\rgt+s\llt -4,2,1\rgt = \llt 2,1,-1\rgt+t\llt 1,1,-1\rgt \end{equation*}

or equivalently, writing out the three component equations and moving all $s$’s and $t$’s to the left and constants to the right,

\begin{align*} -4s -t &= 5\\ 2s -t &= -1\\ s +t &= -5 \end{align*}

Adding the last two equations together gives $3s=-6$ or $s=-2\text{.}$ Substituting this into the last equation gives $t=-3\text{.}$ However, substituting $s=-2,\ t=-3$ into the first equation gives $11=5\text{,}$ which is impossible. The two lines do not intersect. In order for two lines to lie in a common plane and not intersect, they must be parallel. So, in this case no plane contains the two lines.

\begin{equation*} \llt -3,2,4\rgt+s\llt -2,-2,2\rgt = \llt 2,1,-1\rgt+t\llt 1,1,-1\rgt \end{equation*}

or equivalently, writing out the three component equations and moving all $s$’s and $t$’s to the left and constants to the right,

\begin{align*} -2s -t &= 5\\ -2s -t &= -1\\ 2s +t &= -5 \end{align*}

The first two equations are obviously contradictory. The two lines do not intersect. Any plane containing the two lines must be parallel to $\llt 1,1,-1\rgt$ (and hence automatically parallel to $\llt -2,-2,2\rgt=-2\llt 1,1,-1\rgt$) and must also be parallel to the vector from the point $(-3,2,4)\text{,}$ which lies on the first line, to the point $(2,1,-1)\text{,}$ which lies on the second. The vector is $\llt 5,-1,-5\rgt\text{.}$ Hence the normal to the plane is $\llt 5,-1,-5\rgt\times\llt 1,1,-1\rgt=\llt 6,0,6\rgt=6\llt 1,0,1\rgt\text{.}$ The plane perpendicular to $\llt 1,0,1\rgt$ containing $(2,1,-1)$ is $\llt 1,0,1\rgt\cdot\llt x-2,y-1,z+1\rgt=0$ or $x+z=1\text{.}$

(d) Again the two lines are parallel, since $\llt -2,-2,2\rgt=-2\llt 1,1,-1\rgt\text{.}$ Furthermore the point $\llt 3,2,-2\rgt=\llt 3,2,-2\rgt+0\llt -2,-2,2\rgt =\llt 2,1,-1\rgt+1\llt 1,1,-1\rgt$ lies on both lines. So the two lines not only intersect but are identical. Any plane that contains the point $(3,2,-2)$ and is parallel to $\llt 1,1,-1\rgt$ contains both lines. In general, the plane $ax+by+cz=d$ contains $(3,2,-2)$ if and only if $d=3a+2b-2c$ and is parallel to $\llt 1,1,-1\rgt$ if and only if $\llt a,b,c\rgt\cdot\llt 1,1,-1\rgt=a+b-c=0\text{.}$ So, for arbitrary $a$ and $b$ (not both zero) $ax+by+(a+b)z=a$ works.

1.5.6.

Solution.

First observe that

$\llt 1,1,0\rgt$ is perpendicular to $x+y=0$ and hence to the line, and
$\llt 1,-1,2\rgt$ is perpendicular to $x-y+2z=0$ and hence to the line.

Consequently

\begin{align*} \llt 1\,,\,1\,,\,0\rgt \times \llt 1\,,\,-1\,,\,2\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 1 & 0\\ 1 & -1 & 2 \end{matrix}\right] =\llt 2 \,,\, -2 \,,\, -2 \rgt \end{align*}

is perpendicular to both $\llt 1,1,0\rgt$ and $\llt 1,-1,2\rgt\text{.}$ So $\frac{1}{2}\llt 2 \,,\, -2 \,,\, -2 \rgt=\llt 1,-1,-1\rgt\ \gt $ is also perpendicular to both $\llt 1,1,0\rgt$ and $\llt 1,-1,2\rgt$ and hence is parallel to the line. As the point $(2,-1,-1)$ is on the line, the vector equation of the line is

\begin{equation*} \llt x-2,y+1,z+1\rgt= t\llt 1,-1,-1\rgt \end{equation*}

The scalar parametric equations for the line are

\begin{gather*} x-2=t,\ y+1=-t,\ z+1=-t \ \ \hbox{or}\ \ x=2+t,\ y=-1-t,\ z=-1-t \end{gather*}

The symmetric equations for the line are

\begin{gather*} (t=)\frac{x-2}{1}=\frac{y+1}{-1}=\frac{z+1}{-1} \hskip .25 in\hbox{or}\hskip .25 in x-2=-y-1=-z-1 \end{gather*}

1.5.7. (✳).

Solution.

Let’s parametrize $L$ using $y\text{,}$ renamed to $t\text{,}$ as the parameter. Then $y=t\text{,}$ so that

\begin{equation*} x+y=1 \implies x+t=1 \implies x=1-t \end{equation*}

and

\begin{equation*} x+2y+z=3 \implies 1-t + 2t +z =3 \implies z = 2-t \end{equation*}

and

\begin{equation*} \llt x,y,z\rgt = \llt 1,0,2\rgt +t\llt-1,1,-1 \rgt \end{equation*}

is a vector parametric equation for $L\text{.}$

1.5.8.

Solution.

(a) The normal vectors to the two given planes are $\llt 1,2,3\rgt $ and $\llt 1,-2,1\rgt $ respectively. Since the line is to be contained in both planes, its direction vector must be perpendicular to both $\llt 1,2,3\rgt $ and $\llt 1,-2,1\rgt \text{,}$ and hence must be parallel to

\begin{equation*} \llt 1,2,3\rgt \times\llt 1,-2,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 2 & 3 \\ 1 & -2 & 1 \end{matrix}\right] =\llt 8, 2,-4\rgt \end{equation*}

or to $\llt 4,1,-2\rgt \text{.}$ Setting $z=0$ in $x+2y+3z=11,\ x-2y+z=-1$ and solving

\begin{align*} \left\{\Atop{x+2y=11}{x-2y=-1} \right\} &\iff \left\{\Atop{2y=11-x}{x-2y=-1} \right\} \iff \left\{\Atop{2y=11-x}{x-(11-x)=-1} \right\}\\ &\iff \left\{\Atop{2y=11-x}{2x=10} \right\} \end{align*}

we see that $(5,3,0)$ is on the line. So the vector parametric equation of the line is $\llt x,y,z\rgt=\llt 5,3,0\rgt+t\llt 4,1,-2\rgt =\llt 5+4t,3+t,-2t\rgt\text{.}$

(b) The vector from $(1,0,1)$ to the point $(5+4t,3+t,-2t)$ on the line is $\llt 4+4t,3+t,-1-2t\rgt\text{.}$ In order for $(5+4t,3+t,-2t)$ to be the point of the line closest to $(1,0,1)\text{,}$ the vector $\llt 4+4t,3+t,-1-2t\rgt$ joining those two points must be perpendicular to the direction vector $\llt 4,1,-2\rgt $ of the line. (See Example 1.5.4.) This is the case when

\begin{align*} &\llt 4,1,-2\rgt \cdot \llt 4+4t,3+t,-1-2t\rgt =0\\ &\text{or}\quad 16+16t+3+t+2+4t=0\\ &\text{or}\quad t=-1 \end{align*}

The point on the line nearest $(1,0,1)$ is thus $(5+4t,3+t,-2t)\Big|_{t=-1}=(5-4,3-1,2)=(1,2,2)\text{.}$ The distance from the point to the line is the length of the vector from $(1,0,1)$ to the point on the line nearest $(1,0,1)\text{.}$ That vector is $\llt 1,2,2\rgt -\llt 1,0,1\rgt =\llt 0,2,1\rgt\text{.}$ So the distance is $|\llt 0,2,1\rgt|=\sqrt{5}\text{.}$

1.5.9.

Solution.

(a) The plane $P$ must be parallel to both $\llt 2,3,2\rgt $ (since it contains $L_1$) and $\llt 5,2,4\rgt $ (since it is parallel to $L_2$). Hence

\begin{equation*} \llt 2,3,2\rgt \times \llt 5,2,4\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 2 & 3 & 2 \\ 5 & 2 & 4 \end{matrix}\right] =\llt 8,2,-11\rgt \end{equation*}

is normal to $P\text{.}$ As the point $(1,-2,-5)$ is on $P\text{,}$ the equation of $P$ is

\begin{equation*} \llt 8,2,-11\rgt \cdot\llt x-1,y+2,z+5\rgt =0\qquad \text{or}\qquad 8x+2y-11z=59 \end{equation*}

(b) As $L_2$ is parallel to $P\text{,}$ the distance from $L_2$ to $P$ is the same as the distance from any one point of $L_2\text{,}$ for example $(-3,4,-1)\text{,}$ to $P\text{.}$ As $(1,-2,-5)$ is a point on $P\text{,}$ the vector $\llt 1,-2,-5\rgt-\llt -3, 4,-1\rgt =\llt 4,-6,-4\rgt $ has its head on $P$ and tail at $(-3,4,-1)$ on $L_2\text{.}$ The distance from $L_2$ to $P$ is the length of the projection of the vector $\llt 4,-6,-4\rgt $ on the normal to $P\text{.}$ (See Example 1.4.5.) This is

\begin{equation*} \left|\text{proj}_{\llt 8,2,-11\rgt}\llt 4,-6,-4\rgt\right| =\frac{|\llt 4,-6,-4\rgt \cdot\llt 8,2,-11\rgt|} {|\llt 8,2,-11\rgt|} =\frac{64}{\sqrt{189}} \approx4.655 \end{equation*}

1.5.10. (✳).

Solution.

(a) The line $L$ must be perpendicular both to $\llt 2\,,\,1\,,\,-1\rgt\text{,}$ which is a normal vector for the plane $2x + y - z = 5\text{,}$ and to $\llt -1\,,\,-2\,,\,3\rgt\text{,}$ which is a direction vector for the line $x = 3 - t\text{,}$ $y = 1 - 2t$ and $z = 3t\text{.}$ Any such vector must be a nonzero constant times

\begin{align*} \llt 2\,,\,1\,,\,-1\rgt \times \llt -1\,,\,-2\,,\,3\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 2 & 1 & -1\\ -1 & -2 & 3 \end{matrix}\right] =\llt 1 \,,\, -5 \,,\, -3 \rgt \end{align*}

(b) For the point $Q(a, b, c)$

to be a distance $2$ from the $xy$--plane, it is necessary that $|c|=2\text{,}$ and
to be a distance $3$ from the $xz$--plane, it is necessary that $|b|=3\text{,}$ and
to be a distance $4$ from the $yz$--plane, it is necessary that $|a|=4\text{.}$

As $a \lt 0\text{,}$ $b \gt 0\text{,}$ $c \gt 0\text{,}$ the point $Q$ is $(-4, 3, 2)$ and the line $L$ is

\begin{equation*} x = -4 + t \qquad y = 3 - 5t \qquad z = 2 - 3t \end{equation*}

1.5.11. (✳).

Solution.

(a) The line $L$ intersects the $xy$--plane when $x + y + z = 6\text{,}$ $x - y + 2z = 0\text{,}$ and $z=0\text{.}$ When $z=0$ the equations of $L$ reduce to $x + y = 6\text{,}$ $x - y = 0\text{.}$ So the intersection point is $(3,3,0)\text{.}$

The line $L$ intersects the $xz$--plane when $x + y + z = 6\text{,}$ $x - y + 2z = 0\text{,}$ and $y=0\text{.}$ When $y=0$ the equations of $L$ reduce to $x + z = 6\text{,}$ $x +2z = 0\text{.}$ Substituting $x=-2z$ into $x+z=6$ gives $-z=6\text{.}$ So the intersection point is $(12,0,-6)\text{.}$

The line $L$ intersects the $yz$--plane when $x + y + z = 6\text{,}$ $x - y + 2z = 0\text{,}$ and $x=0\text{.}$ When $x=0$ the equations of $L$ reduce to $y + z = 6\text{,}$ $-y +2z = 0\text{.}$ Substituting $y=2z$ into $y+z=6$ gives $3z=6\text{.}$ So the intersection point is $(0,4,2)\text{.}$

(b) Our main job is to find a direction vector $\vd$ for the line.

Since the line is to be parallel to $y=z\text{,}$ $\vd$ must be perpendicular to the normal vector for $y=z\text{,}$ which is $\llt 0,1,-1\rgt\text{.}$
$\vd$ must also be perpendicular to $L\text{.}$ For a point $(x,y,z)$ to be on $L$ it must obey $x+y = 6-z$ and $x-y =-2z\text{.}$ Adding these two equations gives $2x=6-3z$ and subtracting the second equation from the first gives $2y= 6+z\text{.}$ So for a point $(x,y,z)$ to be on $L$ it must obey $x=3-\frac{3z}{2}\text{,}$ $y=3+\frac{z}{2}\text{.}$ The point on $L$ with $z=0$ is $(3,3,0)$ and the point on $L$ with $z=2$ is $(0,4,2)\text{.}$ So $\llt 0-3,4-3,2-0 \rgt=\llt -3,1,2 \rgt$ is a direction vector for $L\text{.}$

So $\vd$ must be perpendicular to both $\llt 0,1,-1\rgt$ and $\llt -3,1,2 \rgt$ and so must be a nonzero constant times

\begin{align*} \llt 0,1,-1\rgt \times \llt -3,1,2 \rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 0 & 1 & -1\\ -3 & 1 & 2 \end{matrix}\right] =\llt 3 \,,\, 3 \,,\, 3 \rgt \end{align*}

We choose $\vd=\frac{1}{3}\llt 3 \,,\, 3 \,,\, 3 \rgt=\llt 1\,,\,1\,,\,1 \rgt\text{.}$ So

\begin{equation*} \llt x,y,z\rgt = \llt 10,11,13\rgt + t \llt 1\,,\,1\,,\,1 \rgt \end{equation*}

is a vector parametric equation for the line. We can also write this as $x=10+t\text{,}$ $y=11+t\text{,}$ $z=13+t\text{.}$

1.5.12. (✳).

Solution.

(a) Since

\begin{alignat*}{2} x&=2+3t &&\implies t=\frac{x-2}{3}\\ y&=4t &&\implies t=\frac{y}{4} \end{alignat*}

we have

\begin{equation*} \frac{x-2}{3}=\frac{y}{4}\qquad z=-1 \end{equation*}

(b) The direction vector for the line $\vr(t) = 2\,\hi-\hk +t(3\,\hi+4\,\hj)$ is $\vd=3\,\hi+4\,\hj\text{.}$ A normal vector for the plane $x-y+2z=0$ is $\vn=\pm\big(\hi-\hj+2\,\hk\big)\text{.}$ The angle $\theta$ between $\vd$ and $\vn$ obeys

\begin{gather*} \cos\theta =\frac{\vd\cdot\vn}{|\vd|\,|\vn|} =\frac{1}{5\sqrt{6}} \implies \theta =\arccos\frac{1}{5\sqrt{6}}\approx 1.49\,\text{radians} \end{gather*}

(We picked $\vn=-\hi+\hj-2\hk$ to make $0\le\theta\le\frac{\pi}{2}\text{.}$) Then the angle between $\vd$ and the plane is

\begin{equation*} \al =\frac{\pi}{2}-\arccos\frac{1}{5\sqrt{6}}\approx 0.08\,\text{radians} \end{equation*}

1.5.13. (✳).

Solution.

Let’s use $z$ as the parameter and call it $t\text{.}$ Then $z=t$ and

\begin{align*} x+y&=11-t\\ x-y&=13+t \end{align*}

Adding the two equations gives $2x=24$ and subtracting the second equation from the first gives $2y=-2-2t\text{.}$ So

\begin{equation*} (x,y,z) = \big(12\,,\,-1-t\,,\,t\big) \end{equation*}

1.5.14. (✳).

Solution.

(a) The point $(0,y,0)\text{,}$ on the $y$--axis, is equidistant from $(2, 5, -3)$ and $(-3, 6, 1)$ if and only if

\begin{align*} &\big|\llt 2, 5, -3\rgt-\llt 0,y,0\rgt\big| =\big|\llt -3, 6, 1\rgt-\llt 0,y,0\rgt\big|\\ &\iff 2^2+(5-y)^2+(-3)^2=(-3)^2+(6-y)^2+1^2\\ &\iff 2y=8\\ &\iff y=4 \end{align*}

(b) The points $(1,3,1)$ and $\vr(0)=(0,0,2)$ are both on the plane. Hence the vector $\llt 1,3,1\rgt-\llt 0,0,2\rgt=\llt 1,3,-1\rgt$ joining them, and the direction vector of the line, namely $\llt 1,1,1\rgt$ are both parallel to the plane. So

\begin{equation*} \llt 1,3,-1\rgt\times \llt 1,1,1\rgt =\det\left[\begin{matrix} \hi &\hj &\hk \\ 1&3&-1 \\ 1&1&1 \end{matrix}\right] = \llt 4,-2,-2\rgt \end{equation*}

is perpendicular to the plane. As the point $(0,0,2)$ is on the plane and the vector $\llt 4,-2,-2\rgt$ is perpendicular to the plane, the equation of the plane is

\begin{equation*} 4(x-0)-2(y-0)-2(z-2)=0\text{ or } 2x-y-z=-2 \end{equation*}

1.5.15. (✳).

Solution.

(a) We are given one point on the line, so we just need a direction vector. That direction vector has to be perpendicular to the triangle $ABC\text{.}$

The fast way to get a direction vector is to observe that all three points $A\text{,}$ $B$ and $C\text{,}$ and consequently the entire triangle $ABC\text{,}$ are contained in the plane $y=2\text{.}$ A normal vector to that plane, and consequently a direction vector for the desired line, is $\hj\text{.}$

Here is another, more mechanical, way to get a direction vector. The vector from $A$ to $B$ is $\llt 2-0\,,\,2-2\,,\,2-2\rgt= \llt 2,0,0\rgt$ and the vector from $A$ to $C$ is $\llt 5-0\,,\,2-2\,,\,1-2\rgt= \llt 5,0,-1\rgt\text{.}$ So a vector perpendicular to the triangle $ABC$ is

\begin{align*} \llt 2,0,0\rgt \times \llt 5,0,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 2 & 0 & 0\\ 5 & 0 & -1 \end{matrix}\right] =\llt 0 \,,\, 2 \,,\, 0 \rgt \end{align*}

The vector $\frac{1}{2}\llt 0 \,,\, 2 \,,\, 0 \rgt=\llt 0 \,,\, 1 \,,\, 0 \rgt$ is also perpendicular to the triangle $ABC\text{.}$

So the specified line has to contain the point $(0,2,2)$ and have direction vector $\llt 0 , 1 , 0 \rgt\text{.}$ The parametric equations

\begin{gather*} \llt x,y,z \rgt = \llt 0 , 2 , 2 \rgt + t\llt 0 , 1 , 0 \rgt \end{gather*}

\begin{gather*} x=0,\ y=2+t,\ z=2 \end{gather*}

do the job.

(b) Let $P$ be the point $(x,y,z)\text{.}$ Then the vector from $P$ to $A$ is $\llt 0-x\,,\,2-y\,,\,2-z\rgt$ and the vector from $P$ to $B$ is $\llt 2-x\,,\,2-y\,,\,2-z\rgt\text{.}$ These two vector are perpendicular if and only if

\begin{align*} 0 &= \llt -x\,,\,2-y\,,\,2-z\rgt \cdot \llt 2-x\,,\,2-y\,,\,2-z\rgt\\ &= x(x-2) +(y-2)^2 +(z-2)^2\\ &= (x-1)^2 -1 +(y-2)^2 +(z-2)^2 \end{align*}

This is a sphere.

(c) The light ray that forms $\tilde A$ starts at the origin, passes through $A$ and then intersects the plane $x+7y+z=32$ at $\tilde A\text{.}$ The line from the origin through $A$ has vector parametric equation

\begin{gather*} \llt x,y,z\rgt = \llt 0,0,0 \rgt +t \llt 0,2,2\rgt =\llt 0,2t,2t\rgt \end{gather*}

This line intersects the plane $x+7y+z=32$ at the point whose value of $t$ obeys

\begin{gather*} (0) +7\overbrace{(2t)}^{y} +\overbrace{(2t)}^{z} =32 \iff t=2 \end{gather*}

So $\tilde A$ is $(0,4,4)\text{.}$

1.5.16.

Solution.

The face opposite $\vp$ is the triangle with vertices $\vq\text{,}$ $\vr$ and $\vs\text{.}$ The centroid of this triangle is $\frac{1}{3}(\vq+\vr+\vs)\text{.}$ The direction vector of the line through $\vp$ and the centroid $\frac{1}{3}(\vq+\vr+\vs)$ is $\frac{1}{3}(\vq+\vr+\vs)-\vp\text{.}$ The points on the line through $\vp$ and the centroid $\frac{1}{3}(\vq+\vr+\vs)$ are those of the form

\begin{equation*} \vx=\vp+ t\left[\frac{1}{3}(\vq+\vr+\vs)-\vp\right] \end{equation*}

for some real number $t\text{.}$ Observe that when $t=\frac{3}{4}$

\begin{equation*} \vp+ t\left[\frac{1}{3}(\vq+\vr+\vs)-\vp\right] =\frac{1}{4}(\vp+\vq+\vr+\vs) \end{equation*}

so that $\frac{1}{4}(\vp+\vq+\vr+\vs)$ is on the line. The other three lines have vector parametric equations

\begin{align*} \vx&=\vq+ t\left[\frac{1}{3}(\vp+\vr+\vs)-\vq\right]\\ \vx&=\vr+ t\left[\frac{1}{3}(\vp+\vq+\vs)-\vr\right]\\ \vx&=\vs+ t\left[\frac{1}{3}(\vp+\vq+\vr)-\vs\right] \end{align*}

When $t=\frac{3}{4}\text{,}$ each of the three right hand sides also reduces to $\frac{1}{4}(\vp+\vq+\vr+\vs)$ so that $\frac{1}{4}(\vp+\vq+\vr+\vs)$ is also on each of these three lines.

1.5.17.

Solution.

We’ll use the procedure of Example 1.5.7. The vector

\begin{equation*} \llt 3,-4,4\rgt \times\llt -3,4,1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 3 & -4 & 4 \\ -3 & 4 & 1 \end{matrix}\right] =\llt -20,-15,0\rgt \end{equation*}

is perpendicular to both lines. Hence so is $\vn=-\frac{1}{5}\llt -20,-15,0\rgt =\llt 4,3,0\rgt \text{.}$ The point $(-2,7,2)$ is on the first line and the point $(1,-2,-1)$ is on the second line. Hence $\vv=\llt -2,7,2 \rgt-\llt1,-2,-1 \rgt=\llt -3,9,3\rgt $ is a vector joining the two lines. The desired distance is the length of the projection of $\vv$ on $\vn\text{.}$ This is

\begin{equation*} \big|{\rm proj}_{\vn}\vv\big| =\frac{|\llt -3,9,3\rgt \cdot\llt 4,3,0\rgt|}{|\llt 4,3,0\rgt|} =\frac{15}{5}=3 \end{equation*}

1.6 Curves and their Tangent Vectors
1.6.2 Exercises

1.6.2.1.

Solution.

We can find the time at which the curve hits a given point by considering the two equations that arise from the two coordinates. For the $y$-coordinate to be 0, we must have $(t-5)^2=0\text{,}$ i.e. $t=5\text{.}$ So, the point $(-1/\sqrt{2},0)$ happens when $t=5\text{.}$

Similarly, for the $y$-coordinate to be $25\text{,}$ we need $(t-5)^2=25\text{,}$ so $(t-5)=\pm 5\text{.}$ When $t=0\text{,}$ the curve hits $(1, 25)\text{;}$ when $t=10\text{,}$ the curve hits $(0,25)\text{.}$

So, in order, the curve passes through the points $(1,25)\text{,}$ $(-1/\sqrt2,0)\text{,}$ and $(0,25)\text{.}$

1.6.2.2.

Solution.

The curve “crosses itself” when the same coordinates occur for different values of $t\text{,}$ say $t_1$ and $t_2\text{.}$ So, we want to know when $\sin t_1=\sin t_2$ and also $t_1^2=t_2^2\text{.}$ Since $t_1$ and $t_2$ should be different, the second equation tells us $t_2=-t_1\text{.}$ Then the first equation tells us $\sin t_1=\sin t_2=\sin(-t_1)=-\sin t_1\text{.}$ That is, $\sin t_1 = -\sin t_1\text{,}$ so $\sin t_1=0\text{.}$ That happens whenever $t_1=\pi n$ for an integer $n\text{.}$

So, the points at which the curve crosses itself are those points $(0,(\pi n)^2)$ where $n$ is an integer. It passes such a point at times $t=\pi n $ and $t=-\pi n\text{.}$ So, the curve hits this point $2\pi n$ time units apart.

1.6.2.3.

Solution.

(a) Since, on the specified part of the circle, $x=\sqrt{a^2-y^2}$ and $y$ runs from $0$ to $a\text{,}$ the parametrization is $\vr(y)=\sqrt{a^2-y^2}\,\hi+ y\,\hj\text{,}$ $0\le y\le a\text{.}$

(b) Let $\theta$ be the angle between

the radius vector from the origin to the point $(a\cos\theta,a\sin\theta)$ on the circle and
the positive $x$-axis.

The tangent line to the circle at $(a\cos\theta,a\sin\theta)$ is perpendicular to the radius vector and so makes angle $\phi=\frac{\pi}{2}+\theta$ with the positive $x$ axis. (See the figure on the left below.) As $\theta =\phi-\frac{\pi}{2}\text{,}$ the desired parametrization is

\begin{equation*} \big(x(\phi),y(\phi)\big) =\big(a\cos(\phi-\tfrac{\pi}{2}),a\sin(\phi-\tfrac{\pi}{2})\big) =\big(a\sin \phi ,-a\cos \phi \big),\ \tfrac{\pi}{2}\le\phi\le\pi \end{equation*}

the radius vector from the origin to the point $(a\cos\theta,a\sin\theta)$ on the circle and
the positive $x$-axis.

The arc from $(0,a)$ to $(a\cos\theta,a\sin\theta)$ subtends an angle $\frac{\pi}{2}-\theta$ and so has length $s=a\big(\frac{\pi}{2}-\theta\big)\text{.}$ (See the figure on the right above.) Thus $\theta=\frac{\pi}{2}-\frac{s}{a}$ and the desired parametrization is

\begin{equation*} \big(x(s),y(s)\big) =\left(a\cos\left(\frac{\pi}{2}-\frac{s}{a}\right)\,,\, a\sin\left(\frac{\pi}{2}-\frac{s}{a}\right)\right) ,\ 0\le s\le\frac{\pi}{2}a \end{equation*}

1.6.2.4.

Solution.

Pretend that the circle is a spool of thread. As the circle rolls it dispenses the thread along the ground. When the circle rolls $\theta$ radians it dispenses the arc length $\theta a$ of thread and the circle advances a distance $\theta a\text{.}$ So centre of the circle has moved $\theta a$ units to the right from its starting point, $x=a\text{.}$ The centre of the circle always has $y$-coordinate $a\text{.}$ So, after rolling $\theta$ radians, the centre of the circle is at position $\vc(\theta)=(a+a\theta,a)\text{.}$

Now, let’s consider the position of $P$ on the circle, after the circle has rolled $\theta$ radians.

From the diagram, we see that $P$ is $a\cos \theta$ units above the centre of the circle, and $a\sin \theta$ units to the right of it. So, the position of $P$ is $(a+a\theta+a\sin\theta,a+a\cos\theta)\text{.}$

Remark: this type of curve is known as a cycloid.

1.6.2.5.

Solution.

We aren’t concerned with $x\text{,}$ so we can eliminate it by solving for it in one equation, and plugging that into the other. Since $C$ lies on the plane, $x=-y-z\text{,}$ so:

\begin{align*} 1&=x^2-\frac{1}{4}y^2+3z^2=(-y-z)^2-\frac14y^2+3z^2\\ &=\frac{3}{4}y^2+4z^2+2yz\\ \end{align*}

Completing the square,

\begin{align*} 1&=\frac{1}{2}y^2+\left(2z+\frac{y}{2}\right)^2\\ 1-\frac{y^2}{2}&=\left(2z+\frac{y}{2}\right)^2\\ \end{align*}

Since $y$ is small, the left hand is close to $1$ and the right hand side is close to $(2z)^2\text{.}$ So $(2z^2)\approx 1\text{.}$ Since $z$ is negative, $z\approx -\frac{1}{2}$ and $2z+\frac{y}{2} \lt 0\text{.}$ Also, $1-\frac{y^2}{2}$ is positive, so it has a real square root.

\begin{align*} -\sqrt{1-\frac{y^2}{2}}&=2z+\frac{y}{2}\\ -\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}&=z \end{align*}

1.6.2.6.

Solution.

To determine whether the particle is rising or falling, we only need to consider its $z$-coordinate: $z(t)=(t-1)^2(t-3)^2\text{.}$ Its derivative with respect to time is $z'(t)=4(t-1)(t-2)(t-3)\text{.}$ This is positive when $1 \lt t \lt 2$ and when $3 \lt t\text{,}$ so the particle is increasing on $(1,2) \cup (3,\infty)$ and decreasing on $(0,1) \cup (2,3)\text{.}$

If $\vr(t)$ is the position of the particle at time $t\text{,}$ then its speed is $|\vr'(t)|\text{.}$ We differentiate:

\begin{equation*} \vr'(t)=-e^{-t}\,\hi-\frac{1}{t^2}\,\hj+4(t-1)(t-2)(t-3)\hk \end{equation*}

So, $\vr(1)=-\frac{1}{e}\,\hi-1\,\hj$ and $\vr(3)=-\frac{1}{e^3}\,\hi-\frac{1}{9}\,\hj\text{.}$ The absolute value of every component of $\vr(1)$ is greater than or equal to that of the corresponding component of $\vr(3)\text{,}$ so $|\vr(1)| \gt |\vr(3)|\text{.}$ That is, the particle is moving more swiftly at $t=1$ than at $t=3\text{.}$

Note: We could also compute the sizes of both vectors directly: $|\vr'(1)|=\sqrt{\left(\frac{1}{e}\right)^2+(-1)^2}\text{,}$ and $|\vr'(3)|=\sqrt{\left(\frac{1}{e^3}\right)^2+\left(-\frac{1}{9}\right)^2}\text{.}$

1.6.2.7.

Solution.

The red vector is $\vr(t+h)-\vr(t)\text{.}$ The arclength of the segment indicated by the blue line is the (scalar) $s(t+h)-s(t)\text{.}$

Remark: as $h$ approaches 0, the curve (if it’s differentiable at $t$) starts to resemble a straight line, with the length of the vector $\vr(t+h)-\vr(t)$ approaching the scalar $s(t+h)-s(t)\text{.}$ This step is crucial to understanding Lemma 1.6.12.

1.6.2.8.

Solution.

Velocity is a vector-valued quantity, so it has both a magnitude and a direction. Speed is a scalar--the magnitude of the velocity. It does not include a direction.

1.6.2.9. (✳).

Solution.

By the product rule

\begin{align*} \diff{}{t}\big[ (\vr \times \vr')\cdot\vr'' \big] &= (\vr' \times \vr')\cdot\vr'' +(\vr \times \vr'')\cdot\vr'' +(\vr \times \vr')\cdot\vr''' \end{align*}

The first term vanishes because $\vr'\times\vr'=\vZero\text{.}$ The second term vanishes because $\vr \times \vr''$ is perpendicular to $\vr''\text{.}$ So

\begin{equation*} \diff{}{t}\big[ (\vr \times \vr')\cdot\vr'' \big] = (\vr \times \vr')\cdot\vr''' \end{equation*}

which is (c).

1.6.2.10. (✳).

Solution.

We have

\begin{equation*} \vv(t) = \vr'(t) = 5 \sqrt{2}\,\hi + 5e^{5t}\,\hj +5 e^{-5t}\,\hk \end{equation*}

and hence

\begin{equation*} |\vv(t)| = |\vr'(t)| = 5 \big|\sqrt{2}\,\hi + e^{5t}\,\hj + e^{-5t}\,\hk\big| = 5\sqrt{2+ e^{10t}+ e^{-10t}} \end{equation*}

Since $2+ e^{10t}+ e^{-10t} = \big(e^{5t}+e^{-5t}\big)^2\text{,}$ that’s (d).

1.6.2.11.

Solution.

(a) By definition,

\begin{align*} \vr(t)&= a \cos t\,\hi+a\sin t\,\hj+ct\,\hk\\ \vv(t)=\vr'(t)&= -a \sin t\,\hi+a\cos t\,\hj+c\,\hk\\ \diff{s}{t}(t)=|\vv(t)|&= \sqrt{a^2+c^2}\\ \va(t)=\vr''(t)&= -a \cos t\,\hi-a\sin t\,\hj \end{align*}

The $(x,y)= a(\cos t,\sin t)$ coordinates go around a circle of radius $a$ and centre $(0,0)$ counterclockwise. One circle is completed for each increase of $t$ by $2\pi\text{.}$ At the same time, the $z$ coordinate increases at a constant rate. Each time the $(x,y)$ coordinates complete one circle, the $z$ coordinate increases by $2 \pi c\text{.}$ The path is a helix with radius $a$ and with each turn having height $2\pi c\text{.}$

(b) By definition,

\begin{align*} \vr(t)&= a \cos t\sin t\,\hi +a\sin^2 t\,\hj+a\cos t\,\hk\\ &= \frac{a}{2} \sin 2t\,\hi+a\frac{1-\cos 2t}{2}\,\hj+a\cos t\,\hk\\ \vv(t)=\vr'(t)&= a \cos 2t\,\hi +a\sin 2t\,\hj-a\sin t\,\hk\\ \diff{s}{t}(t)=|\vv(t)|&= a\sqrt{1+\sin^2t}\\ \va(t)=\vr''(t)&= -2a \sin 2t\,\hi +2a\cos 2t\,\hj-a\cos t\,\hk \end{align*}

Write

\begin{gather*} x(t)=\tfrac{a}{2} \sin 2t \qquad y(t)=a\tfrac{1-\cos 2t}{2} \qquad z(t)=a\cos t \end{gather*}

Then

\begin{gather*} x(t)^2+\big[y(t)-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} \cos^2 2t =\tfrac{a^2}{4} \end{gather*}

Thus the $(x,y)$ coordinates go around a circle of radius $\frac{a}{2}$ and centre $\big(0,\frac{a}{2}\big)$ counterclockwise. At the same time the $z$ coordinate oscillates over the interval between $1$ and $-1$ half as fast. In fact, one can say more about the curve.

\begin{align*} x(t)^2+y(t)^2&=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4}\big(1-\cos 2t\big)^2 \\ &=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} - \tfrac{a^2}{2}\cos 2t + \tfrac{a^2}{4} \cos^2 2t \\ &= \tfrac{a^2}{2} - \tfrac{a^2}{2} (1- 2\sin^2 t) \\ &= a^2\sin^2 t\\ x(t)^2+y(t)^2+z(t)^2 &= a^2\sin^2 t + a^2\cos^2 t =a^2 \end{align*}

So our curve lies on the intersection of the cylinder $x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4}$ and the sphere $x^2+y^2+z^2=a^2\text{.}$ (Don’t worry if you didn’t think of trying to evaluate $x(t)^2+y(t)^2+z(t)^2\text{.}$)

1.6.2.12. (✳).

Solution.

(a) Since $\vr'(t) = (2t,0,t^2)\text{,}$ the specified unit tangent at $t=1$ is

\begin{equation*} \hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}} \end{equation*}

(b) We are to find the arc length between $\vr(0)$ and $\vr(-1)\text{.}$ As $\diff{s}{t}=\sqrt{4t^2+t^4}\text{,}$ the

\begin{align*} \text{arc length} &= \int_{-1}^0 \sqrt{4t^2+t^4}\ \dee{t} \end{align*}

The integrand is even, so

\begin{align*} \text{arc length} &= \int_0^1 \sqrt{4t^2+t^4}\ \dee{t} =\int_0^1 t\sqrt{4+t^2}\ \dee{t} =\Big[\tfrac{1}{3}{(4+t^2)}^{3/2}\Big]_0^1\\ &=\tfrac{1}{3}\big[5^{3/2}-8\big] \end{align*}

1.6.2.13.

Solution.

By Lemma 1.6.12, the arclength of $\vr(t)$ from $t=0$ to $t=1$ is $\int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t\text{.}$ We’ll calculate this in a few pieces to make the steps clearer.

\begin{align*} \vr(t)&=\left(t,\sqrt{\frac{3}{2}}t^2,t^3\right)\\ \diff{\vr}{t}(t)&=\left(1,\sqrt{6}t,3t^2\right)\\ \left|\diff{\vr}{t}(t)\right|&=\sqrt{1^2+(\sqrt{6}t)^2+(3t^2)^2}=\sqrt{1+6t^2+9t^4}\\ &=\sqrt{(3t^2+1)^2}=3t^2+1\\ \int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t&=\int_0^1\left(3t^2+1 \right)\dee t=2 \end{align*}

1.6.2.14.

Solution.

Since $\vr(t)$ is the position of the particle, its acceleration is $\vr''(t)\text{.}$

\begin{align*} \vr(t)&=(t+\sin t, \cos t)\\ \vr'(t)&=(1+\cos t,-\sin t)\\ \vr''(t)&=(-\sin t,-\cos t)\\ |\vr''(t)|&=\sqrt{\sin^2t+\cos^2t}=1 \end{align*}

The magnitude of acceleration is constant, but its direction is changing, since $\vr''(t)$ is a vector with changing direction.

1.6.2.15. (✳).

Solution.

(a) The speed is

\begin{align*} \diff{s}{t}(t) =\big|\vr'(t)\big| & = \left|\left(2\cos t - 2t\sin t\,,\, 2\sin t + 2t \cos t\,,\, t^2\right)\right|\\ &=\sqrt{\big(2\cos t - 2t\sin t\big)^2 +\big(2\sin t + 2t \cos t\big)^2 +t^4}\\ &= \sqrt{4+ 4t^2 +t^4}\\ &= 2+t^2 \end{align*}

so the length of the curve is

\begin{align*} \text{length } &=\int_0^2 \diff{s}{t}\,\dee{t} =\int_0^2 (2+t^2)\,\dee{t} = \left[2t +\frac{t^3}{3}\right]_0^2 =\frac{20}{3} \end{align*}

(b) A tangent vector to the curve at $\vr(\pi)=\big(-2\pi\,,\, 0\,,\, \frac{\pi^3}{3}\big)$ is

\begin{equation*} \vr'(\pi) = \left(2\cos\pi - 2\pi\sin\pi\,,\, 2\sin\pi + 2\pi \cos\pi\,,\, \pi^2\right) = (-2\,,\,-2\pi\,,\,\pi^2) \end{equation*}

So parametric equations for the tangent line at $\vr(\pi)$ are

\begin{align*} x(t) &= -2\pi -2t\\ y(t) &= -2\pi t\\ z(t) &= \frac{\pi^3}{3} + \pi^2 t \end{align*}

1.6.2.16. (✳).

Solution.

(a) As $\vr(t) = \big(3 \cos t, 3 \sin t, 4t\big)\text{,}$ the velocity of the particle is

\begin{equation*} \vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big) \end{equation*}

(b) As $\diff{s}{t}\text{,}$ the rate of change of arc length per unit time, is

\begin{equation*} \diff{s}{t}(t) = |\vr'(t)| = \big|\big(-3 \sin t, 3 \cos t, 4\big)\big| =5 \end{equation*}

the arclength of its path between $t = 1$ and $t = 2$ is

\begin{equation*} \int_1^2\dee{t}\ \diff{s}{t}(t) =\int_1^2\dee{t}\ 5 =5 \end{equation*}

1.6.2.17. (✳).

Solution.

(a) As

\begin{align*} \vr'(t) & = -\sin t\cos^2 t\,\hi + \sin^2 t\cos t\,\hi + 3\sin^2 t\cos t\,\hk\\ &= \sin t\cos t\big(-\cos t\,\hi +\sin t\,\hj +3\sin t\,\hk\big)\\ \diff{s}{t}(t) & = |\sin t\cos t|\sqrt{\cos^2 t + \sin ^2 t + 9\sin^2 t} = |\sin t\cos t|\sqrt{1+ 9\sin^2 t} \end{align*}

the arclength from $t = 0$ to $t = \frac{\pi}{2}$ is

\begin{align*} \int_0^{\pi/2} \diff{s}{t}(t)\,\dee{t} &=\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &=\frac{1}{18}\int_1^{10} \sqrt{u}\ \dee{u}\\ &\hskip0.5in\text{with } u = 1+ 9\sin^2 t,\ \dee{u} = 18\sin t\cos t\,\dee{t}\\ &=\frac{1}{18}\Big[\frac{2}{3}u^{3/2}\Big]_1^{10}\\ &=\frac{1}{27}\big(10\sqrt{10}-1\big) \end{align*}

(b) The arclength from $t = 0$ to $t = \pi$ is

\begin{align*} \int_0^{\pi} \diff{s}{t}(t)\,\dee{t} &=\int_0^{\pi} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &\hskip0.5in\text{Don't forget the absolute value signs!}\\ &=2\int_0^{\pi/2} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &= 2\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t} \end{align*}

since the integrand is invariant under $t\rightarrow\pi-t\text{.}$ So the arc length from $t = 0$ to $t = \pi$ is just twice the arc length from part (a), namely $\frac{2}{27}\big(10\sqrt{10}-1\big)\text{.}$

1.6.2.18. (✳).

Solution.

Since

\begin{align*} \vr(t)&= \frac{t^3}{3}\,\hi + \frac{t^2}{2}\,\hj + \frac{t}{2}\,\hk\\ \vr'(t)&= t^2\,\hi + t\,\hj + \frac{1}{2}\,\hk\\ \diff{s}{t}(t)=|\vr'(t)|&=\sqrt{t^4+t^2+\frac{1}{4}} =\sqrt{\Big(t^2+\frac{1}{2}\Big)^2}=t^2+\frac{1}{2} \end{align*}

the length of the curve is

\begin{gather*} s(t)=\int_0^t \diff{s}{t}(u)\,\dee{u} =\int_0^t \Big(u^2+\frac{1}{2}\Big)\,\dee{u} =\frac{t^3}{3} +\frac{t}{2} \end{gather*}

1.6.2.19. (✳).

Solution.

Since

\begin{align*} \vr(t) & = t^m\,\hi + t^m\,\hj + t^{3m/2}\,\hk\\ \vr'(t) &= mt^{m-1}\,\hi + mt^{m-1}\,\hj +\frac{3m}{2}t^{3m/2-1}\,\hk\\ \diff{s}{t} = |\vr'(t)| & = \sqrt{ 2m^2 t^{2m-2} +\frac{9m^2}{4} t^{3m-2} } = mt^{m-1}\sqrt{2 + \frac{9}{4}t^m } \end{align*}

the arc length is

\begin{align*} \int_a^b \diff{s}{t}(t)\,\dee{t} &=\int_a^b mt^{m-1}\sqrt{2 + \frac{9}{4}t^m }\ \dee{t}\\ &=\frac{4}{9}\int_{2 + \frac{9}{4}a^m}^{2 + \frac{9}{4}b^m}\sqrt{u}\,\dee{u} \qquad\text{with } u = 2 + \frac{9}{4}t^m,\ \dee{u} = \frac{9m}{4}t^{m-1}\\ &=\frac{4}{9}\Big[\frac{2}{3}u^{3/2}\Big]_{2 + \frac{9}{4}a^m} ^{2 + \frac{9}{4}b^m}\\ &=\frac{8}{27}\Big[\Big(2 + \frac{9}{4}b^m\Big)^{3/2} -\Big(2 + \frac{9}{4}a^m\Big)^{3/2}\Big] \end{align*}

1.6.2.20.

Solution.

Given the position of the particle, we can find its velocity:

\begin{equation*} \vv(t)=\vr'(t)=(\cos t, -\sin t , 1) \end{equation*}

Applying the given formula,

\begin{equation*} \vL(t)=\vr \times \vv=(\sin t , \cos t , t) \times (\cos t, -\sin t , 1). \end{equation*}

[Solution 1:] We can first compute the cross product, then differentiate:
\begin{align*} \vL(t)& = (\cos t + t\sin t)\hi + (t\cos t - \sin t)\hj-\hk\\ \vL'(t)&=t\cos t\,\hi -t\sin t\, \hj\\ |\vL'(t)|&=\sqrt{t^2(\sin^2 t + \cos^2 t)}=\sqrt{t^2}=|t| \end{align*}
[Solution 2:] Using the product rule:
\begin{align*} \vL'(t)&=\vr'(t)\times \vv(t) + \vr(t) \times \vv'(t)\\ &=\underbrace{\vr'(t)\times \vr'(t)}_{0} + \vr(t) \times \vv'(t)\\ &= (\sin t ,\cos t,t)\times(-\sin t , -\cos t, 0)\\ &=t\cos t\,\hi-t\sin t\,\hj\\ |\vL'(t)|&=\sqrt{t^2\cos^2 t + t^2\sin t^2 }=|t| \end{align*}

1.6.2.21. (✳).

Solution.

(a) The curve intersects $E$ when

\begin{align*} 2\big(t\sin(\pi t)\big)^2+2\big(t\cos(\pi t)\big)^2 +\big(t^2\big)^2=24 &\iff 2t^2+t^4=24\\ &\iff (t^2-4)(t^2+6)=0 \end{align*}

Since we need $t \gt 0\text{,}$ the desired time is $t=2$ and the corresponding point is $\vr(2)=2\,\hj+4\,\hk\text{.}$

(b) Since

\begin{equation*} \vr'(t)=\big[\sin(\pi t)+\pi t\cos(\pi t)\big]\hi +\big[\cos(\pi t)-\pi t\sin(\pi t)\big]\hj + 2t\hk \end{equation*}

a tangent vector to $\Gamma$ at $P$ is any nonzero multiple of

\begin{equation*} \vr'(2)=2\pi\,\hi+\hj+4\,\hk \end{equation*}

\begin{equation*} \vnabla(2x^2+2y^2+z^2)\big|_{(0,2,4)}= \llt 4x,4y,2z\rgt\big|_{(0,2,4)} =\llt 0,8,8\rgt \end{equation*}

Since $\vr'(2)$ and $\llt 0,8,8\rgt$ are not parallel, $\Gamma$ and $E$ do not intersect at right angles.

1.6.2.22. (✳).

Solution.

\begin{align*} \diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big] &=\diff{}{t}\big[\vr(t)\cdot\vr(t)+\vr'(t)\cdot\vr'(t)\big]\\ &=2\vr(t)\cdot\vr'(t)+2\vr'(t)\cdot\vr''(t)\\ &=2\vr'(t)\cdot\big[\vr(t)+\vr''(t)\big]\\ &=0\qquad \text{ since } \vr''(t)=-\vr(t) \end{align*}

Since $\diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big]=0$ for all $t\text{,}$ $|\vr(t)|^2+|\vr'(t)|^2$ is independent of $t\text{.}$

1.6.2.23. (✳).

Solution.

(a) Since $z=6u\text{,}$ $y=\frac{z^2}{12}=3u^2$ and $x=\frac{yz}{18}=u^3\text{,}$

\begin{align*} \vr(u)&=u^3\,\hi+3u^2\,\hj+6u\,\hk\cr \end{align*}

(b)

\begin{align*} \vr'(u)&=3u^2\,\hi+6u\,\hj+6\,\hk\cr \vr''(u)&=6u\,\hi+6\,\hj\cr \diff{s}{u}(u)=|\vr'(u)| &=\sqrt{9u^4+36u^2+36}=3\big(u^2+2\big)\cr \end{align*}

\begin{equation*} \int_\cC \ ds =\int_0^1 \diff{s}{u}\ \dee{u} =\int_0^1 3\big(u^2+2\big)\ \dee{u} =\big[u^3+6u\big]_0^1 =7 \end{equation*}

\begin{equation*} \vR(t)=\vr\big(u(t)\big)\implies \vR'(t)=\vr'\big(u(t)\big)\diff{u}{t} \end{equation*}

In particular, if the particle is at $(1,3,6)$ at time $t_1\text{,}$ then $u(t_1)=1$ and

\begin{equation*} 6\,\hi+12\,\hj+12\,\hk=\vR'(t_1)=\vr'(1)\diff{u}{t}(t_1) =\big(3\,\hi+6\,\hj+6\,\hk\big)\diff{u}{t}(t_1) \end{equation*}

which implies that $\diff{u}{t}(t_1)=2\text{.}$

(d) By the product and chain rules,

\begin{equation*} \vR'(t)=\vr'\big(u(t)\big)\diff{u}{t}\implies \vR''(t)=\vr''\big(u(t)\big)\Big(\diff{u}{t}\Big)^2 +\vr'\big(u(t)\big)\difftwo{u}{t} \end{equation*}

In particular,

\begin{align*} 27\,\hi+30\,\hj+6\,\hk &=\vR''(t_1)=\vr''(1)\Big(\diff{u}{t}(t_1)\Big)^2 +\vr'\big(1\big)\difftwo{u}{t}(t_1)\\ &=\big(6\,\hi+6\,\hj\big)2^2+\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \end{align*}

Simplifying

\begin{equation*} 3\,\hi+6\,\hj+6\,\hk =\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \implies \difftwo{u}{t}(t_1)=1 \end{equation*}

1.6.2.24. (✳).

Solution.

(a) According to Newton,

\begin{equation*} m\vr''(t) = \vF(t)\qquad\text{so that}\qquad \vr''(t) = -3t\,\hi + \sin t\,\hj + 2e^{2t}\,\hk \end{equation*}

Integrating once gives

\begin{gather*} \vr'(t) = -3\frac{t^2}{2}\,\hi - \cos t\,\hj + e^{2t}\,\hk +\vc \end{gather*}

for some constant vector $\vc\text{.}$ We are told that $\vr'(0)=\vv_0 =\frac{\pi^2}{2}\,\hi\text{.}$ This forces $\vc=\frac{\pi^2}{2}\,\hi+\hj-\hk$ so that

\begin{gather*} \vr'(t) = \left(\frac{\pi^2}{2}-\frac{3t^2}{2}\right)\,\hi +(1- \cos t)\,\hj + \big(e^{2t}-1\big)\,\hk \end{gather*}

Integrating a second time gives

\begin{gather*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk + \vc \end{gather*}

for some (other) constant vector $\vc\text{.}$ We are told that $\vr(0)=\vr_0 =\frac{1}{2}\,\hk\text{.}$ This forces $\vc=\vZero$ so that

\begin{gather*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk \end{gather*}

(b) The particle is in the plane $x=0$ when

\begin{gather*} 0=\left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right) =\frac{t}{2}(\pi^2-t^2) \iff t=0, \pm\pi \end{gather*}

So the desired time is $t=\pi\text{.}$

\begin{align*} \vr'(\pi) &= \left(\frac{\pi^2}{2}-\frac{3\pi^2}{2}\right)\,\hi +(1- \cos\pi)\,\hj + \big(e^{2\pi}-1\big)\,\hk\\ &= -\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk \end{align*}

1.6.2.25. (✳).

Solution.

(a) Parametrize $C$ by $x\text{.}$ Since $y=x^2$ and $z=\frac{2}{3}x^3\text{,}$

\begin{align*} \vr(x)&=x\,\hi+x^2\,\hj+\frac{2}{3}x^3\,\hk\cr \vr'(x)&=\hi+2x\,\hj+2x^2\,\hk\cr \vr''(x)&=2\,\hj+4x\,\hk\cr \diff{s}{x} &=|\vr'(x)| =\sqrt{1+4x^2+4x^4}=1+2x^2\cr \end{align*}

and

\begin{equation*} \int_C \ \dee{s} =\int_0^3 \diff{s}{x}\ \dee{x} =\int_0^3 \big(1+2x^2\big)\ \dee{x} ={\Big[x+\frac{2}{3}x^3\Big]}_0^3 =21 \end{equation*}

(b) The particle travelled a distance of 21 units in $\frac{7}{2}$ time units. This corresponds to a speed of $\frac{21}{7/2}=6\text{.}$

\begin{equation*} \vR(t)=\vr\big(x(t)\big)\implies \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t} \end{equation*}

By parts (a) and (b) and the chain rule

\begin{equation*} 6=\diff{s}{t}=\diff{s}{x}\diff{x}{t}=(1+2x^2)\diff{x}{t} \implies \diff{x}{t}=\frac{6}{1+2x^2} \end{equation*}

In particular, the particle is at $\big(1,1,\frac{2}{3}\big)$ at $x=1\text{.}$ At this time $\diff{x}{t}=\frac{6}{1+2\times 1}=2$ and

\begin{equation*} \vR'=\vr'\big(1\big)\diff{x}{t} =\big(\hi+2\,\hj+2\,\hk\big)2 =2\hi+4\,\hj+4\,\hk \end{equation*}

(d) By the product and chain rules,

\begin{equation*} \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}\implies \vR''(t)=\vr''\big(x(t)\big){\Big(\diff{x}{t}\Big)}^2 +\vr'\big(x(t)\big)\difftwo{x}{t} \end{equation*}

Applying $\diff{}{t}$ to $6=\big(1+2x(t)^2\big)\diff{x}{t}(t)$ gives

\begin{equation*} 0=4x{\Big(\diff{x}{t}\Big)}^2+(1+2x^2)\difftwo{x}{t} \end{equation*}

In particular, when $x=1$ and $\diff{x}{t}=2\text{,}$ $0=4\times 1\big(2\big)^2+(3)\difftwo{x}{t}$ gives $\difftwo{x}{t}=-\frac{16}{3}$ and

\begin{equation*} \vR''=\big(2\,\hj+4\,\hk\big)\big(2\big)^2 -\big(\hi+2\,\hj+2\,\hk\big)\frac{16}{3} = -\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big) \end{equation*}

1.6.2.26.

Solution.

The question is already set up as an $xy$--plane, with the camera at the origin, so the vector in the direction the camera is pointing is $(x(t),y(t))\text{.}$ Let $\theta$ be the angle the camera makes with the positive $x$-axis (due east). The camera, the object, and the due-east direction (positive $x$-axis) make a right triangle.

\begin{align*} \tan\theta&=\frac{y}{x}\\ \end{align*}

Differentiating implicitly with respect to $t\text{:}$

\begin{align*} \sec^2\theta\,\diff{\theta}{t}&=\frac{xy'-yx'}{x^2}\\ \diff{\theta}{t}&=\cos^2\theta\left(\frac{xy'-yx'}{x^2}\right)=\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2\left(\frac{xy'-yx'}{x^2}\right)\\ &=\frac{xy'-yx'}{{x^2+y^2}} \end{align*}

1.6.2.27.

Solution.

Define $\vu(t)=e^{\alpha t}\frac{d\vr}{dt}(t)\text{.}$ Then

\begin{align*} \frac{d\vu}{dt}(t)&=\alpha e^{\alpha t}\frac{d\vr}{dt}(t) +e^{\alpha t}\frac{d^2\vr}{dt^2}(t)\cr &=\alpha e^{\alpha t}\frac{d\vr}{dt}(t) -ge^{\alpha t}\hk -\alpha e^{\alpha t}\frac{d\vr}{dt}(t)\cr &=-ge^{\alpha t}\hk\cr \end{align*}

Integrating both sides of this equation from $t=0$ to $t=T$ gives

\begin{align*} \vu(T)-\vu(0)&=-g\frac{e^{\alpha T}-1}{\alpha}\hk\\ \implies \vu(T)&=\vu(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk =\frac{d\vr}{dt}(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk\\ &=\vv_0-g\frac{e^{\alpha T}-1}{\alpha}\hk \end{align*}

Substituting in $\vu(T)=e^{\alpha t}\frac{d\vr}{dt}(T)$ and multiplying through by $e^{-\alpha T}$

\begin{equation*} \frac{d\vr}{dt}(T) =e^{-\alpha T}\vv_0-g\frac{1-e^{-\alpha T}}{\alpha}\hk \end{equation*}

Integrating both sides of this equation from $T=0$ to $T=t$ gives

\begin{align*} &\vr(t)-\vr(0) =\frac{e^{-\alpha t}-1}{-\alpha}\vv_0-g\frac{ t}{\alpha}\hk +g\frac{e^{-\alpha t}-1}{-\alpha^2}\hk\\ &\implies \vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk \end{align*}

1.6.2.28. (✳).

Solution.

(a) By definition,

\begin{equation*} \vv'(t)=\va(t)=\llt \cos t,\sin t,0\rgt \implies \vv(t)=\llt \sin t+c_1, -\cos t+c_2, c_3\rgt \end{equation*}

for some constants $c_1\text{,}$ $c_2\text{,}$ $c_3\text{.}$ To satisfy $\vv(0)=\llt 0,-1,1\rgt\text{,}$ we need $c_1=0\text{,}$ $c_2=0$ and $c_3=1\text{.}$ So $\vv(t)=\llt \sin t, -\cos t, 1\rgt\text{.}$ Similarly,

\begin{align*} \vr'(t)=\vv(t)&=\llt \sin t,-\cos t,1\rgt\\ &\implies \vr(t)=\llt -\cos t+d_1, -\sin t+d_2, t+d_3\rgt \end{align*}

for some constants $d_1\text{,}$ $d_2\text{,}$ $d_3\text{.}$ To satisfy $\vr(0)=\llt -1,0,0\rgt\text{,}$ we need $d_1=0\text{,}$ $d_2=0$ and $d_3=0\text{.}$ So $\vr(t)=\llt -\cos t, -\sin t, t\rgt\text{.}$

(b) To test for orthogonality, we compute the dot product

\begin{align*} \vv(t)\cdot\va(t)&=\llt \sin t, -\cos t, 1\rgt\cdot\llt \cos t,\sin t,0\rgt\\ &=\sin t\cos t-\cos t \sin t+1\times 0=0 \end{align*}

so $\vv(t)\perp\va(t)$ for all $t\text{.}$

(c) At $t=-\frac{\pi}{2}$ the particle is at $\vr\big(-\frac{\pi}{2}\big)=\llt 0,1,-\frac{\pi}{2}\rgt$ and has velocity $\vv\big(-\frac{\pi}{2}\big)=\llt -1,0,1\rgt\text{.}$ So the tangent line must pass through $\llt 0,1,-\frac{\pi}{2}\rgt$ and have direction vector $\llt -1,0,1\rgt\text{.}$ Here is a vector parametric equation for the tangent line.

\begin{equation*} \vr(u)=\llt 0,1,-\frac{\pi}{2}\rgt+u\llt -1,0,1\rgt \end{equation*}

(d) True. Look at the path followed by the particle from the top so that we only see $x$ and $y$ coordinates. The path we see (call this the projected path) is $x(t)=-\cos t\text{,}$ $y(t)=-\sin t\text{,}$ which is a circle of radius one centred on the origin. Any tangent line to any circle always remains outside the circle. So no tangent line to the projected path can pass through the $(0,0)\text{.}$ So no tangent line to the path followed by the particle can pass through the $z$--axis and, in particular, through $(0,0,0)\text{.}$

1.6.2.29. (✳).

Solution.

(a) Since

\begin{equation*} x(t)^2+y(t)^2=t^2\cos^2\big(\tfrac{\pi t}{2}\big) +t^2\sin^2\big(\tfrac{\pi t}{2}\big)=t^2\qquad\text{and}\qquad z(t)^2=t^2 \end{equation*}

are the same, the path of the particle lies on the cone $z^2=x^2+y^2\text{.}$

(b) By definition,

\begin{align*} \text{velocity}&=\vr'(t)\\ &= \big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]\hi +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]\hj +\hk\\ \text{speed}&=|\vr'(t)|\\ &=\sqrt{\big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]^2 +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]^2 +1^2}\\ &= \Big[\cos^2\big(\tfrac{\pi t}{2}\big) -2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\sin^2\big(\tfrac{\pi t}{2}\big)\\ &\hskip.2in +\sin^2\big(\tfrac{\pi t}{2}\big) +2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\cos^2\big(\tfrac{\pi t}{2}\big) +1\Big]^{1/2}\\ & =\sqrt{2+\frac{\pi^2 t^2}{4}} \end{align*}

(c) At $t=1\text{,}$ the particle is at $\vr(1)=(0,1,1)$ and has velocity $\vr'(1)=\llt -\frac{\pi}{2},1,1\rgt\text{.}$ So for $t\ge 1\text{,}$ the particle is at

\begin{equation*} \llt x,y,z\rgt = \llt 0,1,1\rgt+(t-1)\llt -\frac{\pi}{2},1,1\rgt \end{equation*}

This is also a vector parametric equation for the line.

(d) Assume that the particle’s speed remains constant as it flies along $L\text{.}$ Then the $x$-coordinate of the particle at time $t$ (for $t\ge 1$) is $-\frac{\pi}{2}(t-1)\text{.}$ This takes the value $-1$ when $t-1=\frac{2}{\pi}\text{.}$ So the particle hits $x=-1\text{,}$ $\frac{2}{\pi}$ seconds after it flew off the cone.

1.6.2.30. (✳).

Solution.

(a) The tangent vectors to the two curves are

\begin{equation*} \vr'_1(t)=\llt 1, 2t, 3t^2\rgt\qquad \vr'_2(t)=\llt-\sin t, \cos t, 1\rgt \end{equation*}

Both curves pass through $P$ at $t=0$ and then the tangent vectors are

\begin{equation*} \vr'_1(0)=\llt 1, 0, 0\rgt\qquad \vr'_2(0)=\llt 0, 1, 1\rgt \end{equation*}

So the angle of intersection, $\theta\text{,}$ is determined by

\begin{align*} \vr'_1(0)\cdot\vr'_2(0)=|\vr'_1(0)|\,|\vr'_2(0)|\,\cos\theta &\implies \llt 1, 0, 0\rgt\cdot\llt 0, 1, 1\rgt=1\cdot\sqrt{2}\cdot\cos\theta\\ &\implies\cos\theta=0\implies \theta=90^\circ \end{align*}

(b) Our strategy will be to

find a vector $\vv$ whose tail is on one line and whose head is on the other line and then
find a vector $\vn$ that is perpendicular to both lines.
Then, if we denote by $\theta$ the angle between $\vv$ and $\vn\text{,}$ the distance between the two lines is $|\vv|\cos\theta = \frac{|\vv\cdot\vn|}{|\vn|}$

Here we go

So the first step is to find a $\vv\text{.}$
- One point on the line $\vr(t)=\llt t, -1+2t, 1+3t\rgt$ is $\vr(0)=\llt 0,-1,1\rgt\text{.}$
- $(x,y,z)$ is on the other line if and only if $x+y-z=4$ and $2x-z=4\text{.}$ In particular, if $z=0$ then $x+y=4$ and $2x=4$ so that $x=2$ and $y=2\text{.}$
- So the vector $\vv=\llt 2-0\,,\,2-(-1)\,,\,0-1\rgt=\llt 2,3,-1\rgt$ has its head on one line and its tail on the other line.
Next we find a vector $\vn$ that is perpendicular to both lines.
- First we find a direction vector for the line $x+y-z=4\text{,}$ $2x-z=4\text{.}$ We already know that $x=y=2\text{,}$ $z=0$ is on that line. We can find a second point on that line by choosing, for example, $z=2$ and then solving $x+y=6\text{,}$ $2x=6$ to get $x=3\text{,}$ $y=3\text{.}$ So one direction vector for the line $x+y-z=4\text{,}$ $2x-z=4$ is $\vd_1=\llt3-2\,,\,3-2\,,\,2-0 \rgt = \llt 1,1,2\rgt\text{.}$
- A second way to get a direction vector for the line $x+y-z=4\text{,}$ $2x-z=4$ is to observe that $\llt 1,1,-1\rgt$ is normal to $x+y-z=4$ and so is perpendicular to the line and $\llt 2,0,-1\rgt$ is normal to $2x-z=4$ and so is also perpendicular to the line. So $\llt 1,1,-1\rgt\times\llt 2,0,-1\rgt$ is a direction vector for the line.
- A direction vector for the line $\vr(t)=\llt t, -1+2t, 1+3t\rgt$ is $\vd_2=\vr'(t)=\llt 1,2,3\rgt\text{.}$
- So
  \begin{equation*} \vn=\vd_2\times\vd_1 =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{matrix}\right] = \hi + \hj -\hk \end{equation*}
  is perpendicular to both lines.

The distance between the two lines is then

\begin{equation*} |\vv|\cos\theta = \frac{|\vv\cdot\vn|}{|\vn|} =\frac{\llt 2,3,-1\rgt \cdot\llt 1,1,-1\rgt}{|\llt 1,1,-1\rgt|} =\frac{6}{\sqrt{3}} =2\sqrt{3} \end{equation*}

1.7 Sketching Surfaces in 3d
1.7.2 Exercises

1.7.2.1. (✳).

Solution.

(a) $\varphi =\frac{\pi}{3}$ is a surface of constant (spherical coordinate) $\varphi\text{.}$ So it is a cone with vertex at the origin. We can express $\varphi=\frac{\pi}{3}$ in cartesian coordinates by observing that $0\le\varphi\le\frac{\pi}{2}$ so that $z\ge 0\text{,}$ and

\begin{align*} \varphi=\frac{\pi}{3} &\iff \tan\varphi =\frac{\sqrt{3}}{2} \iff \rho\sin\varphi =\frac{\sqrt{3}}{2}\rho\cos\varphi\\ &\iff \sqrt{x^2+y^2}=\frac{\sqrt{3}}{2} z \end{align*}

So the picture that corresponds to (a) is (C).

(b) As $r$ and $\theta$ are cylindrical coordinates

\begin{align*} r=2\cos\theta &\iff r^2=2 r\cos\theta \iff x^2+y^2 = 2x\\ &\iff (x-1)^2+y^2=1 \end{align*}

There is no $z$ appearing in $(x-1)^2+y^2=1\text{.}$ So every constant $z$ cross--section of $(x-1)^2+y^2=1$ is a (horizontal) circle of radius $1$ centred on the line $x=1\text{,}$ $y=0\text{.}$ It is a cylinder of radius $1$ centred on the line $x=1\text{,}$ $y=0\text{.}$ So the picture that corresponds to (b) is (F).

(c) Each constant $z$ cross--section of $x^2+y^2=z^2+1$ is a (horizontal) circle centred on the $z$--axis. The radius of the circle is $1$ when $z=0$ and grows as $z$ moves away from $z=0\text{.}$ So $x^2+y^2=z^2+1$ consists of a bunch of (horizontal) circles stacked on top of each other, with the radius increasing with $|z|\text{.}$ It is a hyperboloid of one sheet. The picture that corresponds to (c) is (D).

(d) Every point of $y=x^2+z^2$ has $y\ge 0\text{.}$ Only (B) has that property. We can also observe that every constant $y$ cross--section is a circle centred on $x=z=0\text{.}$ The radius of the circle is zero when $y=0$ and increases as $y$ increases. The surface $y=x^2+z^2$ is a paraboloid. The picture that corresponds to (d) is (B).

(e) As $\rho$ and $\varphi$ are spherical coordinates

\begin{align*} \rho=2\cos\varphi &\iff \rho^2=2\rho\cos\varphi \iff x^2+y^2+z^2=2z\\ &\iff x^2+y^2 + (z-1)^2=1 \end{align*}

This is the sphere of radius $1$ centred on $(0,0,1)\text{.}$ The picture that corresponds to (e) is (A).

(f) The only possibility left is that the picture that corresponds to (f) is (E).

1.7.2.2.

Solution.

Each solution below consists of three sketchs.

In the first sketch, we just redraw the given level curves with the $x$- and $y$-axes reoriented so that the sketch looks like we are high on the $z$-axis looking down at the $xy$-plane.
In the second sketch, we lift up each level curve $f(x,y)=C$ and draw it in the horizontal plane $z=C\text{.}$ That is we draw
\begin{align*} &\Set{(x,y,z)}{f(x,y)=C,\ z=C,\ x\ge 0, y\ge 0}\\ & =\Set{(x,y,z)}{z=f(x,y),\ z=C,\ x\ge 0, y\ge 0} \end{align*}
Finally, in the third sketch, we draw the part of graph $z=f(x,y)$ in the first octant, just by “filling in the gaps in the second sketch”.

(a)

(b)

1.7.2.3.

Solution.

We first add into the sketch of the graph the horizontal planes $z=C\text{,}$ for $C=3\text{,}$ $2\text{,}$ $1\text{,}$ $0.5\text{,}$ $0.25\text{.}$

To reduce clutter, for each $C\text{,}$ we have drawn in only

the (gray) intersection of the horizontal plane $z=C$ with the $yz$--plane, i.e. with the vertical plane $x=0\text{,}$ and
the (blue) intersection of the horizontal plane $z=C$ with the graph $z=f(x,y)\text{.}$

We have also omitted the label for the plane $z=0.25\text{.}$

The intersection of the plane $z=C$ with the graph $z=f(x,y)$ is line

\begin{equation*} \Set{(x,y,z)}{z=f(x,y),\ z=C} = \Set{(x,y,z)}{f(x,y)=C,\ z=C} \end{equation*}

Drawing this line (which is parallel to the $x$-axis) in the $xy$-plane, rather than in the plane $z=C\text{,}$ gives a level curve. Doing this for each of $C=3\text{,}$ $2\text{,}$ $1\text{,}$ $0.5\text{,}$ $0.25$ gives five level curves.

1.7.2.4.

Solution.

(a) For each fixed $c \gt 0\text{,}$ the level curve $x^2+2y^2=c$ is the ellipse centred on the origin with $x$ semi axis $\sqrt{c}$ and $y$ semi axis $\sqrt{c/2}\text{.}$ If $c=0\text{,}$ the level curve $x^2+2y^2=c=0$ is the single point $(0,0)\text{.}$

(b) For each fixed $c\ne 0\text{,}$ the level curve $xy=c$ is a hyperbola centred on the origin with asymptotes the $x$- and $y$-axes. If $c \gt 0\text{,}$ any $x$ and $y$ obeying $xy=c \gt 0$ are of the same sign. So the hyperbola is contained in the first and third quadrants. If $c \lt 0\text{,}$ any $x$ and $y$ obeying $xy=c \gt 0$ are of opposite sign. So the hyperbola is contained in the second and fourth quadrants. If $c=0\text{,}$ the level curve $xy=c=0$ is the single point $(0,0)\text{.}$

(c) For each fixed $c\ne 0\text{,}$ the level curve $xe^{-y}=c$ is the logarithmic curve $y=-\ln\frac{c}{x}\text{.}$ Note that, for $c \gt 0\text{,}$ the curve

is restricted to $x \gt 0\text{,}$ so that $\frac{c}{x} \gt 0$ and $\ln \frac{c}{x}$ is defined, and that
as $x\rightarrow 0^+\text{,}$ $y$ goes to $-\infty\text{,}$ while
as $x\rightarrow +\infty\text{,}$ $y$ goes to $+\infty\text{,}$ and
the curve crosses the $x$-axis (i.e. has $y=0$) when $x=c\text{.}$

and for $c \lt 0\text{,}$ the curve

is restricted to $x \lt 0\text{,}$ so that $\frac{c}{x} \gt 0$ and $\ln \frac{c}{x}$ is defined, and that
as $x\rightarrow 0^-\text{,}$ $y$ goes to $-\infty\text{,}$ while
as $x\rightarrow -\infty\text{,}$ $y$ goes to $+\infty\text{,}$ and
the curve crosses the $x$-axis (i.e. has $y=0$) when $x=c\text{.}$

If $c=0\text{,}$ the level curve $xe^{-y}=c=0$ is the $y$-axis, $x=0\text{.}$

1.7.2.5. (✳).

Solution.

If $C=0\text{,}$ the level curve $f=C=0$ is just the line $y=0\text{.}$ If $C\ne 0$ (of either sign), we may rewrite the equation, $f(x,y)=\frac{2y}{x^2+y^2}=C\text{,}$ of the level curve $f=C$ as

\begin{equation*} x^2-\frac{2}{C}y+y^2=0 \iff x^2+\left(y-\frac{1}{C}\right)^2 =\frac{1}{C^2} \end{equation*}

which is the equation of the circle of radius $\frac{1}{|C|}$ centred on $\left(0\,,\,\frac{1}{C}\right)\text{.}$

Remark. To be picky, the function $f(x,y)=\frac{2y}{x^2+y^2}$ is not defined at $(x,y)=(0,0)\text{.}$ The question should have either specified that the domain of $f$ excludes $(0,0)$ or have specified a value for $f(0,0)\text{.}$ In fact, it is impossible to assign a value to $f(0,0)$ in such a way that $f(x,y)$ is continuous at $(0,0)\text{,}$ because $\lim_{x\rightarrow 0}f(x,0)=0$ while $\lim_{y\rightarrow 0}f(0,|y|)=\infty\text{.}$ So it makes more sense to have the domain of $f$ being $\bbbr^2$ with the point $(0,0)$ removed. That’s why there is a little hole at the origin in the above sketch.

1.7.2.6. (✳).

Solution.

Observe that, for any constant $C\text{,}$ the curve $-x^2+4y^2=C$ is the level curve $f=e^C\text{.}$

If $C=0\text{,}$ then $-x^2+4y^2=C=0$ is the pair of lines $y=\pm\frac{x}{2}\text{.}$
If $C \gt 0\text{,}$ then $-x^2+4y^2=C \gt 0$ is the hyperbola $y=\pm\frac{1}{2}\sqrt{C+x^2}\text{.}$
If $C \lt 0\text{,}$ then $-x^2+4y^2=C \lt 0$ is the hyperbola $x=\pm\sqrt{|C|+4y^2}\text{.}$

1.7.2.7. (✳).

Solution.

(a) We can rewrite the equation as

\begin{equation*} x^2 + y^2 = (z-1)^2 - 1 \end{equation*}

The right hand side is negative for $|z-1| \lt 1\text{,}$ i.e. for $0 \lt z \lt 2\text{.}$ So no point on the surface has $0 \lt z \lt 2\text{.}$ For any fixed $z\text{,}$ outside that range, the curve $x^2 + y^2 = (z-1)^2 - 1$ is the circle of radius $\sqrt{(z-1)^2 - 1}$ centred on the $z$--axis. That radius is $0$ when $z=0,2$ and increases as $z$ moves away from $z=0,2\text{.}$ For very large $|z|\text{,}$ the radius increases roughly linearly with $|z|\text{.}$ Here is a sketch of some level curves.

(b) The surface consists of two stacks of circles. One stack starts with radius $0$ at $z=2\text{.}$ The radius increases as $z$ increases. The other stack starts with radius $0$ at $z=0\text{.}$ The radius increases as $z$ decreases. This surface is a hyperboloid of two sheets. Here are two sketchs. The sketch on the left is of the part of the surface in the first octant. The sketch on the right of the full surface.

1.7.2.8. (✳).

Solution.

For each fixed $z\text{,}$ $4x^2+y^2=1+z^2$ is an ellipse. So the surface consists of a stack of ellipses one on top of the other. The semi axes are $\frac{1}{2}\sqrt{1+z^2}$ and $\sqrt{1+z^2}\text{.}$ These are smallest when $z=0$ (i.e. for the ellipse in the $xy$-plane) and increase as $|z|$ increases. The intersection of the surface with the $xz$-plane (i.e. with the plane $y=0$) is the hyperbola $4x^2-z^2=1$ and the intersection with the $yz$-pane (i.e. with the plane $x=0$) is the hyperbola $y^2-z^2=1\text{.}$ Here are two sketches of the surface. The sketch on the left only shows the part of the surface in the first octant (with axes).

1.7.2.9.

Solution.

(a) If $c \gt 0\text{,}$ $f(x,y,z)=c\text{,}$ i.e. $x^2+y^2+z^2=c\text{,}$ is the sphere of radius $\sqrt{c}$ centered at the origin. If $c=0\text{,}$ $f(x,y,z)=c$ is just the origin. If $c \lt 0\text{,}$ no $(x,y,z)$ satisfies $f(x,y,z)=c\text{.}$

(b) $f(x,y,z)=c\text{,}$ i.e. $x+2y+3z=c\text{,}$ is the plane normal to $(1,2,3)$ passing through $(c,0,0)\text{.}$

(c) If $c \gt 0\text{,}$ $f(x,y,z)=c\text{,}$ i.e. $x^2+y^2=c\text{,}$ is the cylinder parallel to the $z$-axis whose cross-section is a circle of radius $\sqrt{c}$ that is parallel to the $xy$-plane and is centered on the $z$-axis. If $c=0\text{,}$ $f(x,y,z)=c$ is the $z$-axis. If $c \lt 0\text{,}$ no $(x,y,z)$ satisfies $f(x,y,z)=c\text{.}$

1.7.2.10.

Solution.

(a) The graph is $z=\sin x$ with $(x,y)$ running over $0\le x\le 2\pi\text{,}$ $0\le y\le 1\text{.}$ For each fixed $y_0$ between $0$ and $1\text{,}$ the intersection of this graph with the vertical plane $y=y_0$ is the same sin graph $z=\sin x$ with $x$ running from $0$ to $2\pi\text{.}$ So the whole graph is just a bunch of 2-d sin graphs stacked side-by-side. This gives the graph on the left below.

(b) The graph is $z=\sqrt{x^2+y^2}\text{.}$ For each fixed $z_0\ge 0\text{,}$ the intersection of this graph with the horizontal plane $z=z_0$ is the circle $\sqrt{x^2+y^2}=z_0\text{.}$ This circle is centred on the $z$-axis and has radius $z_0\text{.}$ So the graph is the upper half of a cone. It is the sketch on the right above.

(c) The graph is $z=|x|+|y|\text{.}$ For each fixed $z_0\ge 0\text{,}$ the intersection of this graph with the horizontal plane $z=z_0$ is the square $|x|+|y|=z_0\text{.}$ The side of the square with $x,y\ge 0$ is the straight line $x+y=z_0\text{.}$ The side of the square with $x\ge 0$ and $y\le 0$ is the straight line $x-y=z_0$ and so on. The four corners of the square are $(\pm z_0,0,z_0)$ and $(0, \pm z_0,z_0)\text{.}$ So the graph is a stack of squares. It is an upside down four-sided pyramid. The part of the pyramid in the first octant (that is, $x,y,z\ge 0$) is the sketch below.

1.7.2.11.

Solution.

(a) For each fixed $z_0\text{,}$ the $z=z_0$ cross-section (parallel to the $xy$-plane) of this surface is an ellipse centered on the origin with one semiaxis of length 2 along the $x$-axis and one semiaxis of length 4 along the $y$-axis. So this is an elliptic cylinder parallel to the $z$-axis. Here is a sketch of the part of the surface above the $xy$--plane.

(b) This is a plane through $(4,0,0)\text{,}$ $(0,4,0)$ and $(0,0,2)\text{.}$ Here is a sketch of the part of the plane in the first octant.

(c) For each fixed $x_0\text{,}$ the $x=x_0$ cross-section parallel to the $yz$-plane is an ellipse with semiaxes $3\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $y$-axis and $2\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $z$-axis. As you move out along the $x$-axis, away from $x=0\text{,}$ the ellipses grow at a rate proportional to $\sqrt{1+\frac{x^2}{16}}\text{,}$ which for large $x$ is approximately $\frac{|x|}{4}\text{.}$ This is called a hyperboloid of one sheet. Its

(d) For each fixed $y_0\text{,}$ the $y=x_0$ cross-section (parallel to the $xz$-plane) is a circle of radius $|y|$ centred on the $y$-axis. When $y_0=0$ the radius is $0\text{.}$ As you move further from the $xz$-plane, in either direction, i.e. as $|y_0|$ increases, the radius grows linearly. The full surface consists of a bunch of these circles stacked sideways. This is a circular cone centred on the $y$-axis.

(e) This is an ellipsoid centered on the origin with semiaxes $3\text{,}$ $\sqrt{12}=2\sqrt{3}$ and $3$ along the $x\text{,}$ $y$ and $z$-axes, respectively.

(f) Completing three squares, we have that $x^2+y^2+z^2+4x-by+9z-b=0$ if and only if $(x+2)^2+\big(y-\frac{b}{2}\big)^2+\big(z+\frac{9}{2}\big)^2 =b+4+\frac{b^2}{4}+\frac{81}{4}\text{.}$ This is a sphere of radius $r_b=\frac{1}{2}\sqrt{b^2+4b+97}$ centered on $\frac{1}{2}(-4,b,-9)\text{.}$

(g) There are no points on the surface with $x \lt 0\text{.}$ For each fixed $x_0 \gt 0$ the cross-section $x=x_0$ parallel to the $yz$-plane is an ellipse centred on the $x$--axis with semiaxes $\sqrt{x_0}$ in the $y$-axis direction and $\frac{3}{2}\sqrt{x_0}$ in the $z$--axis direction. As you increase $x_0\text{,}$ i.e. move out along the $x$-axis, the ellipses grow at a rate proportional to $\sqrt{x_0}\text{.}$ This is an elliptic paraboloid with axis the $x$-axis.

(h) This is called a parabolic cylinder. For any fixed $y_0\text{,}$ the $y=y_0$ cross-section (parallel to the $xz$-plane) is the upward opening parabola $z=x^2$ which has vertex on the $y$-axis.

1.7.2.12.

Solution.

Since the level curves are circles centred at the origin (in the $xy$-plane), when $z$ is a constant, the equation will have the form $x^2+y^2=c$ for some constant. That is, our equation looks like

\begin{equation*} x^2+y^2=g(z), \end{equation*}

where $g(z)$ is a function depending only on $z\text{.}$

Because our cross-sections are so nicely symmetric, we know the intersection of the figure with the left side of the $yz$-plane as well: $z=3(-y-1)=-3(y+1)$ (when $z\ge0$) and $z=-3(-y-1)=3(y+1)$ (when $z \lt 0$). Below is the intersection of our surface with the $yz$ plane.

Setting $x=0\text{,}$ our equation becomes $y^2=g(z)\text{.}$ Looking at the right side of the $yz$ plane, this should lead to: $\left.\begin{cases} z=3(y-1) &\mbox{if } z\geq 0,\ y\ge 1\\ z=-3(y-1) &\mbox{if } z \lt 0,\ y\ge 1 \end{cases}\right\} \text{.}$ That is:

\begin{align*} |z|&=3(y-1)\\ \frac{|z|}{3}+1&=y\\ \left(\frac{|z|}{3}+1\right)^2&=y^2 &(*) \end{align*}

A quick check: when we squared both sides of the equation in $(*)\text{,}$ we added another solution, $\frac{|z|}{3}+1=-y\text{.}$ Let’s make sure we haven’t diverged from our diagram.

\begin{align*} &&& \left(\frac{|z|}{3}+1\right)^2=y^2\\ &\Leftrightarrow&& \underbrace{\frac{|z|}{3}+1}_{\text{positive}}=\pm y\\ &\Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y \gt 0\\ \frac{|z|}{3}+1 =-y & y \lt 0 \end{cases}\\ & \Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y\ge1\\ \frac{|z|}{3}+1 =-y & y\le-1 \end{cases}\\ &\Leftrightarrow&& \begin{cases} |z| =3(y-1) & y\ge1\\ |z|=-3(y+1) & y\le-1 \end{cases}\\ &\Leftrightarrow&& \begin{cases} z =\pm \underbrace{3(y-1)}_{\text{positive}} & y\ge1 \\ z=\pm \underbrace{3(y+1)}_{\text{negative}} & y\le-1 \end{cases}\\ &\Leftrightarrow&& \begin{cases} z =3(y-1) & y\ge1,\ z\ge0 \\ z =-3(y-1) & y\ge1,\ z\le0 \\ z=-3(y+1) & y\le-1,\ z \ge 0 \\ z=3(y+1) & y\le-1,\ z\le 0 \end{cases} \end{align*}

This matches our diagram eactly. So, all together, the equation of the surface is

\begin{equation*} x^2+y^2=\left( \frac{|z|}{3}+1\right)^2 \end{equation*}

2 Partial Derivatives
2.1 Limits
2.1.2 Exercises

2.1.2.1.

Solution.

In general, this is false. Consider $f(x,y)=12-(1-10x)^2-(1-10y)^2\text{.}$

$\lim\limits_{(x,y)\to(0,0)}f(x,y)=12-1-1=10$ (the function is continuous)
$\displaystyle f(0.1,0.1)=12-(1-1)^2-(1-1)^2=12$
$\displaystyle f(0.2,0.2)=12-(1-2)^2-(1-2)^2=10$

We often (somewhat lazily) interpret the limit “$\lim\limits_{(x,y)\to(0,0)}f(x,y)=10$” to mean that, as $(x,y)$ gets closer and closer to the origin, $f(x,y)$ gets closer and closer to 10. This isn’t exactly what the definition means, though. The definition tells us that, we can guarantee that $f(x,y)$ be very close to 10 by choosing $(x,y)$ very close to $(0,0)\text{.}$

The function $f(x,y)$ can also be very close to 10 for some $(x,y)$’s that are not close to $(0,0)\text{.}$ Moreover, we don’t know how close to $(0,0)$ we have to be in order for $f(x,y)$ to be “very close” to 10.

2.1.2.2.

Solution.

The function we’re taking the limit of has its input as the position of the particle, and its output the size of the particle. So, $f(x,y)$ gives the size of particles found at position $(x,y)\text{.}$ In the definition, we write $\mathbf x = (x,y)\text{.}$ So, $\mathbf x$ is the position in the basin the particle was taken from.
Our claim deals with particles very close to where the millstone hits the basin, so $\mathbf a$ is the position in the basin where the millstone hits.
$\mathbf L$ is the limit of the function: in this case, 50 $\mu$m.

2.1.2.3.

Solution.

By inspection, when $y=0\text{,}$ then $f(x,y)=1$ as long as $x \neq 0\text{.}$ So, if we follow the $x$-axis in towards the origin, $f(x,y)=1$ along this route.
Also by inspection, when $x=0\text{,}$ then $f(x,y)=0$ as long as $y \neq 0\text{.}$ So, if we follow the $y$-axis in towards the origin, $f(x,y)=0$ along this route.
Since two different directions give us different values as we approach the origin, $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ does not exist.

2.1.2.4.

Solution.

Since $x=r\cos\theta$ and $y=r\sin\theta\text{,}$ we have that
\begin{equation*} f=x^2-y^2=r^2\cos^2\theta-r^2\sin^2\theta=r^2\cos(2\theta) \end{equation*}
When $r=1\text{,}$ $f=\cos(2\theta)\text{.}$ So, $f(x,y)$ runs between $-1$ and $1\text{.}$ It smallest value is $-1$ and its largest value is $+1\text{.}$
The distance from $(x,y)$ to the origin is $r$ (for $r\ge0)\text{.}$ So, at a distance $r\text{,}$ our function is $r^2\cos(2\theta)\text{.}$ Then $f(x,y)$ runs over the interval $[-r^2,r^2]\text{.}$ It smallest value is $-r^2$ and its largest value is $+r^2\text{.}$
Using our answer to the last part, we have that $|f|\le r^2\text{.}$ So for $0 \lt r \lt \sqrt\epsilon\text{,}$ we necessarily have that $|f(x,y)| \lt \epsilon$ whenever the distance from $(x,y)$ to the origin is at most $r\text{.}$
For every $\epsilon \gt 0\text{,}$ if we choose $(x,y)$ to be sufficiently close to $(0,0)$ (in particular, within a distance $r \lt \sqrt\epsilon$), then $f(x,y)$ is within distance $\epsilon$ of $0\text{.}$ By Definition 2.1.2, we have that $\lim\limits_{(x,y)\to(0,0)}f(x,y)=0\text{.}$

2.1.2.5.

Solution.

By Theorem 2.1.6, $f(x,y)$ is continuous over its domain. The domain of a polynomial is everywhere; in this case, $\mathbb R^2\text{.}$ So, $f(x,y)$ is continuous at $(a,b)\text{.}$ By the definition of continuity, $\lim\limits_{(x,y)\to(a,b)}f(x,y)=f(a,b)\text{.}$

2.1.2.6.

Solution.

(a) $\ds\lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2$

(b) Switching to polar coordinates,

\begin{align*} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2} &=\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \frac{r\cos\theta}{r^2} =\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \frac{\cos\theta}{r} \end{align*}

which does not exist, since, for example,

if $\theta=0\text{,}$ then
\begin{equation*} \lim_{\Atop{r\rightarrow0^+}{\theta=0}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{1}{r} =+\infty \end{equation*}
while if $\theta=\pi\text{,}$ then
\begin{equation*} \lim_{\Atop{r\rightarrow0^+}{\theta=\pi}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{-1}{r} =-\infty \end{equation*}

\begin{align*} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2} &=\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \frac{r^2\cos^2\theta}{r^2} =\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \cos^2\theta \end{align*}

which does not exist, since, for example,

if $\theta=0\text{,}$ then
\begin{equation*} \lim_{\Atop{r\rightarrow0^+}{\theta=0}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 1 = 1 \end{equation*}
while if $\theta=\frac{\pi}{2}\text{,}$ then
\begin{equation*} \lim_{\Atop{r\rightarrow0^+}{\theta=\pi/2}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 0 =0 \end{equation*}

(d) Switching to polar coordinates,

\begin{align*} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2} &=\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \frac{r^3\cos^3\theta}{r^2} =\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ r\cos^3\theta =0 \end{align*}

since $|\cos\theta|\le 1$ for all $\theta\text{.}$

(e) Switching to polar coordinates,

\begin{align*} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4} &=\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ \frac{r^2\cos^2\theta\ r^2\sin^2\theta} {r^2\cos^2\theta+r^4\sin^4\theta}\\ &=\lim_{\Atop{r\rightarrow0^+}{0\le\theta \lt 2\pi}}\ r^2\sin^2\theta \frac{\cos^2\theta}{\cos^2\theta+r^2\sin^4\theta}\\ &= 0 \end{align*}

Here, we used that

\begin{align*} \left|\sin^2\theta\frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\right| &\le \frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\\ &\le \left.\begin{cases} \frac{\cos^2\theta} {\cos^2\theta}&\text{ if }\cos\theta\ne 0 \\ 0 &\text { if }\cos\theta =0 \end{cases} \right\}\\ &\le 1 \end{align*}

for all $r \gt 0\text{.}$

(f) To start, observe that

\begin{gather*} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] \end{gather*}

We may evaluate $\ds\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right]$ by l’Hôpital’s rule or by using the definition of the derivative to give

\begin{equation*} \lim_{x\rightarrow 0}\ \frac{\sin x}{x} =\lim_{x\rightarrow 0}\ \frac{\sin x-\sin 0}{x-0} =\diff{}{x}\sin x\bigg|_{x=0} =\cos x\Big|_{x=0}=1 \end{equation*}

Similarly, we may evaluate $\ds\left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right]$ by l’Hôpital’s rule or by using the definition of the derivative to give

\begin{equation*} \lim_{y\rightarrow 0}\ \frac{e^y-1}{y} =\lim_{y\rightarrow 0}\ \frac{e^y-e^0}{y-0} =\diff{}{y}e^y\bigg|_{y=0} =e^y\Big|_{y=0}=1 \end{equation*}

So all together

2.1.2.7. (✳).

Solution.

(a) In polar coordinates, $x=r\cos\theta\text{,}$ $y=r\sin\theta\text{,}$ so that

\begin{align*} \frac{x^8+y^8}{x^4+y^4} &=\frac{r^8\cos^8\theta+r^8\sin^8\theta}{r^4\cos^4\theta+r^4\sin^4\theta} =r^4\frac{\cos^8\theta+\sin^8\theta}{\cos^4\theta+\sin^4\theta} \end{align*}

\begin{align*} \frac{\cos^8\theta+\sin^8\theta}{\cos^4\theta+\sin^4\theta} &\le \frac{\cos^8\theta+2\cos^4\theta\sin^4\theta+\sin^8\theta} {\cos^4\theta+\sin^4\theta} =\frac{\big(\cos^4\theta+\sin^4\theta\big)^2}{\cos^4\theta+\sin^4\theta}\\ &=\cos^4\theta+\sin^4\theta \le 2 \end{align*}

we have

\begin{equation*} 0\le \frac{x^8+y^8}{x^4+y^4}\le 2r^4 \end{equation*}

As $\ds\lim_{(x,y)\to (0,0)}2r^4=0\text{,}$ the squeeze theorem yields $\ds\lim_{(x,y)\to(0,0)}\frac{x^8+y^8}{x^4+y^4}=0\text{.}$

(b) In polar coordinates

\begin{align*} \frac{xy^5}{x^8+y^{10}} &=\frac{r^6\cos\theta\,\sin^5\theta}{r^8\cos^8\theta+r^{10}\sin^{10}\theta} =\frac{1}{r^2}\frac{\cos\theta\,\sin^5\theta}{\cos^8\theta+r^2\sin^{10}\theta} \end{align*}

As $(x,y)\to (0,0)$ the first fraction $\frac{1}{r^2}\to\infty$ but the second factor can take many different values. For example, if we send $(x,y)$ towards the origin along the $y$--axis, i.e. with $\theta=\pm\frac{\pi}{2}\text{,}$

\begin{gather*} \lim_{\Atop {(x,y)\to(0,0)}{x=0}}\frac{xy^5}{x^8+y^{10}} =\lim_{y\to 0} \frac{0}{y^{10}}=0 \end{gather*}

but if we send $(x,y)$ towards the origin along the line $y=x\text{,}$ i.e. with $\theta=\frac{\pi}{4},\frac{5\pi}{4}\text{,}$

\begin{gather*} \lim_{\Atop {(x,y)\to(0,0)}{y=x} }\frac{xy^5}{x^8+y^{10}} =\lim_{x\to 0} \frac{x^6}{x^8+x^{10}} =\lim_{x\to 0} \frac{1}{x^2}\frac{1}{1+x^2} =+\infty \end{gather*}

and if we send $(x,y)$ towards the origin along the line $y=-x\text{,}$ i.e. with $\theta=-\frac{\pi}{4},\frac{3\pi}{4}\text{,}$

\begin{gather*} \lim_{\Atop {(x,y)\to(0,0)}{y=-x} }\frac{xy^5}{x^8+y^{10}} =\lim_{x\to 0} \frac{-x^6}{x^8+x^{10}} =\lim_{x\to 0}- \frac{1}{x^2}\frac{1}{1+x^2} =-\infty \end{gather*}

So $\frac{xy^5}{x^8+y^{10}}$ does not approach a single value as $(x,y)\to(0,0)$ and the limit does not exist.

2.1.2.8. (✳).

Solution.

(a) In polar coordinates

\begin{equation*} \frac{x^3-y^3}{x^2+y^2}=\frac{r^3\cos^3\theta-r^3\sin^3\theta}{r^2} =r\cos^3\theta-r\sin^3\theta \end{equation*}

Since

\begin{equation*} \big|r\cos^3\theta-r\sin^3\theta\big|\le 2r \end{equation*}

and $2r\rightarrow 0$ as $r\rightarrow 0\text{,}$ the limit exists and is $0\text{.}$

(b) The limit as we approach $(0,0)$ along the $x$-axis is

\begin{gather*} \lim_{t\rightarrow 0}\frac{x^2-y^4}{x^2+y^4}\bigg|_{(x,y)=(t,0)} =\lim_{t\rightarrow 0}\frac{t^2-0^4}{t^2+0^4} =1 \end{gather*}

On the other hand the limit as we approach $(0,0)$ along the $y$-axis is

\begin{gather*} \lim_{t\rightarrow 0}\frac{x^2-y^4}{x^2+y^4}\bigg|_{(x,y)=(0,t)} =\lim_{t\rightarrow 0}\frac{0^2-t^4}{0^2+t^4} =-1 \end{gather*}

These are different, so the limit as $(x,y)\rightarrow 0$ does not exist.

We can gain a more detailed understanding of the behaviour of $\frac{x^2-y^4}{x^2+y^4}$ near the origin by switching to polar coordinates.

\begin{equation*} \frac{x^2-y^4}{x^2+y^4} =\frac{r^2\cos^2\theta-r^4\sin^4\theta}{r^2\cos^2\theta+r^4\sin^4\theta} =\frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta} \end{equation*}

Now fix any $\theta$ and let $r\rightarrow 0$ (so that we are approaching the origin along the ray that makes an angle $\theta$ with the positive $x$-axis). If $\cos\theta\ne 0$ (i.e. the ray is not part of the $y$-axis)

\begin{gather*} \lim_{r\rightarrow 0} \frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta} =\frac{\cos^2\theta}{\cos^2\theta} =1 \end{gather*}

But if $\cos\theta= 0$ (i.e. the ray is part of the $y$-axis)

\begin{gather*} \lim_{r\rightarrow 0} \frac{\cos^2\theta-r^2\sin^4\theta}{\cos^2\theta+r^2\sin^4\theta} =\lim_{r\rightarrow 0} \frac{-r^2\sin^4\theta}{r^2\sin^4\theta} =\frac{-\sin^4\theta}{\sin^4\theta} =-1 \end{gather*}

2.1.2.9. (✳).

Solution.

(a) In polar coordinates $x=r\cos\theta\text{,}$ $y=r\sin\theta$

\begin{align*} &\frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}\\ &\hskip0.5in=\frac{2r^2\cos^2\theta + r^3\cos^2\theta\sin\theta - r^3\cos\theta\sin^2\theta + 2r^2\sin^2\theta}{r^2}\\ &\hskip0.5in=2+ r\big[\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big] \end{align*}

\begin{equation*} r\big|\cos^2\theta\sin\theta - \sin^2\theta\cos\theta \big| \le 2r \rightarrow 0\text{ as }r\rightarrow 0 \end{equation*}

we have

\begin{equation*} \lim_{(x,y)\rightarrow(0,0)} \frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}=2 \end{equation*}

(b) Since

\begin{gather*} \frac{x^2y^2 -2 x^2y + x^2} {(x^2 + y^2-2y+1)^2} =\frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} \end{gather*}

and, in polar coordinates centred on $(0,1)\text{,}$ $x=r\cos\theta\text{,}$ $y=1+r\sin\theta\text{,}$

\begin{equation*} \frac{x^2(y-1)^2} {\big[x^2 + (y-1)^2\big]^2} =\frac{r^4\cos^2\theta\sin^2\theta}{r^4} =\cos^2\theta\sin^2\theta \end{equation*}

we have that the limit does not exist. For example, if we send $(x,y)$ to $(0,1)$ along the line $y=1\text{,}$ so that $\theta=0\text{,}$ we get the limit $0\text{,}$ while if we send $(x,y)$ to $(0,1)$ along the line $y=x+1\text{,}$ so that $\theta=\frac{\pi}{4}\text{,}$ we get the limit $\frac{1}{4}\text{.}$

2.1.2.10.

Solution.

(a) We have

\begin{align*} \lim_{r\rightarrow 0^+}f(r\cos\theta,r\sin\theta) &=\lim_{r\rightarrow 0^+} \frac{(r\cos\theta)^2(r\sin\theta)}{(r\cos\theta)^4+(r\sin\theta)^2}\\ &=\lim_{r\rightarrow 0^+}r\ \frac{\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\\ &=\lim_{r\rightarrow 0^+}r\ \lim_{r\rightarrow 0^+}\frac{\cos^2\theta\sin\theta} {r^2\cos^4\theta+\sin^2\theta} \end{align*}

Observe that, if $\sin\theta=0\text{,}$ then

\begin{equation*} \frac{\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}=0 \end{equation*}

for all $r\ne 0\text{.}$ If $\sin\theta\ne 0\text{,}$

\begin{align*} \lim_{r\rightarrow 0^+} \frac{\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta} &=\frac{\cos^2\theta\sin\theta}{\sin^2\theta} =\frac{\cos^2\theta}{\sin\theta} \end{align*}

So the limit $\ds\lim_{r\rightarrow 0^+} \frac{\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}$ exists (and is finite) for all fixed $\theta$ and

\begin{equation*} \lim\limits_{r\rightarrow 0^+}f(r\cos\theta,r\sin\theta)=0 \end{equation*}

(b) We have

\begin{equation*} \lim_{x\rightarrow 0}f(x,x^2) =\lim_{x\rightarrow 0}\frac{x^2x^2}{x^4+{(x^2)}^2} =\lim_{x\rightarrow 0}\frac{x^4}{2x^4} =\frac{1}{2} \end{equation*}

(c) Note that in part (a) we showed that as $(x,y)$ approaches $(0,0)$ along any straight line, $f(x,y)$ approaches the limit zero. In part (b) we have just shown that as $(x,y)$ approaches $(0,0)$ along the parabola $y=x^2\text{,}$ $f(x,y)$ approaches the limit $\half\text{,}$ not zero. So $f(x,y)$ takes values very close to $0\text{,}$ for some $(x,y)$’s that are really near $(0,0)$ and also takes values very close to $\frac{1}{2}\text{,}$ for other $(x,y)$’s that are really near $(0,0)\text{.}$ There is no single number, $L\text{,}$ with the property that $f(x,y)$ is really close to $L$ for all $(x,y)$ that are really close to $(0,0)\text{.}$ So the limit does not exist.

2.1.2.11. (✳).

Solution.

(a) Since, in polar coordinates,

\begin{equation*} \frac{xy}{x^2+y^2}=\frac{r^2\cos\theta\sin\theta}{r^2} =\cos\theta\sin\theta \end{equation*}

we have that the limit does not exist. For example,

if we send $(x,y)$ to $(0,0)$ along the positive $x$-axis, so that $\theta=0\text{,}$ we get the limit $\sin\theta\cos\theta\big|_{\theta=0}=0\text{,}$
while if we send $(x,y)$ to $(0,0)$ along the line $y=x$ in the first quadrant, so that $\theta=\frac{\pi}{4}\text{,}$ we get the limit $\sin\theta\cos\theta\big|_{\theta=\pi/4}=\frac{1}{2}\text{.}$

(b) This limit does not exist, since if it were to exist the limit

\begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2} &=\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{\sin(xy)}\ \frac{\sin(xy)}{x^2+y^2}\\ &=\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{\sin(xy)}\ \lim_{(x,y)\rightarrow(0,0)}\frac{\sin(xy)}{x^2+y^2} \end{align*}

would also exist. (Recall that $\ds\lim_{t\rightarrow 0}\tfrac{\sin t}{t} =1\text{.}$)

\begin{align*} \lim_{(x,y)\rightarrow(-1,1)}\big[x^2+2xy^2+y^4\big] &=(-1)^2+2(-1)(1)^2+(1)^4=0\\ \lim_{(x,y)\rightarrow(-1,1)}\big[1+y^4\big] &=1+(1)^4=2 \end{align*}

and the second limit is nonzero,

\begin{equation*} \lim_{(x,y)\rightarrow(-1,1)}\frac{x^2+2xy^2+y^4}{1+y^4}=\frac{0}{2}=0 \end{equation*}

(d) Since the limit along the positive $x$-axis

\begin{equation*} \lim_{\Atop{t\rightarrow 0}{t \gt 0}}|y|^x\Big|_{(x,y)=(t,0)} =\lim_{\Atop{t\rightarrow 0}{t \gt 0}}0^t =\lim_{\Atop{t\rightarrow 0}{t \gt 0}}0 =0 \end{equation*}

and the limit along the $y$-axis

\begin{equation*} \lim_{t\rightarrow 0}|y|^x\Big|_{(x,y)=(0,t)} =\lim_{t\rightarrow 0}|t|^0 =\lim_{t\rightarrow 0}1 =1 \end{equation*}

are different, the limit as $(x,y)\rightarrow 0$ does not exist.

2.1.2.12.

Solution.

(a) Let $a$ be any nonzero constant. When $y=x+\frac{x^2}{a}$ and $x\ne 0\text{,}$

\begin{equation*} \frac{x^2}{y-x} =\frac{x^2}{x^2/a} =a \end{equation*}

So the limit along the curve $y=x+\frac{x^2}{a}$ is

\begin{equation*} \lim_{t\rightarrow 0}\frac{x^2}{y-x}\Big|_{(x,y)=(t,t+t^2/a)} =\lim_{t\rightarrow 0}a =a \end{equation*}

In particular, the limit along the curve $y=x+x^2\text{,}$ which is $1\text{,}$ and the limit along the curve $y=x-x^2\text{,}$ which is $-1\text{,}$ are different. So the limit as $(x,y)\rightarrow 0$ does not exist.

(b) Let $a$ be any nonzero constant. When $y=x+\frac{x^8}{a}$ and $x\ne 0\text{,}$

\begin{equation*} \frac{x^8}{y-x} =\frac{x^8}{x^8/a} =a \end{equation*}

So the limit along the curve $y=x+\frac{x^8}{a}$ is

\begin{equation*} \lim_{t\rightarrow 0}\frac{x^8}{y-x}\Big|_{(x,y)=(t,t+t^8/a)} =\lim_{t\rightarrow 0}a =a \end{equation*}

In particular, the limit along the curve $y=x+x^8\text{,}$ which is $1\text{,}$ and the limit along the curve $y=x-x^8\text{,}$ which is $-1\text{,}$ are different. So the limit as $(x,y)\rightarrow 0$ does not exist.

2.2 Partial Derivatives
2.2.2 Exercises

2.2.2.1.

Solution.

By definition
\begin{equation*} \pdiff{f}{x}(0,0) =\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h} \end{equation*}
One approximation to this is
\begin{equation*} \pdiff{f}{x}(0,0) \approx \left.\frac{f(h,0)-f(0,0)}{h}\right|_{h=0.1} =\frac{1.10517-1}{0.1} =1.0517 \end{equation*}
Another approximation to this is
\begin{equation*} \pdiff{f}{x}(0,0) \approx \left.\frac{f(h,0)-f(0,0)}{h}\right|_{h=0.01} =\frac{1.01005-1}{0.01} =1.005 \end{equation*}
By definition
\begin{equation*} \pdiff{f}{y}(0,0) =\lim_{h\rightarrow 0}\frac{f(0,h)-f(0,0)}{h} \end{equation*}
One approximation to this is
\begin{equation*} \pdiff{f}{y}(0,0) \approx \left.\frac{f(0,h)-f(0,0)}{h}\right|_{h=-0.1} =\frac{0.99500-1}{-0.1} =0.0500 \end{equation*}
Another approximation to this is
\begin{equation*} \pdiff{f}{y}(0,0) \approx \left.\frac{f(0,h)-f(0,0)}{h}\right|_{h=-0.01} =\frac{0.99995-1}{-0.01} =.0050 \end{equation*}
To take the partial derivative with respect to $x$ at $(0,0)\text{,}$ we set $y=0\text{,}$ differentiate with respect to $x$ and then set $x=0\text{.}$ So
\begin{equation*} \pdiff{f}{x}(0,0) = \left.\diff{}{x} e^x\cos 0\right|_{x=0} =\left.e^x\right|_{x=0}=1 \end{equation*}
To take the partial derivative with respect to $y$ at $(0,0)\text{,}$ we set $x=0\text{,}$ differentiate with respect to $y$ and then set $y=0\text{.}$ So
\begin{equation*} \pdiff{f}{y}(0,0) = \left.\diff{}{y} e^0\cos y\right|_{y=0} =\left.\sin y\right|_{y=0} =0 \end{equation*}

2.2.2.2.

Solution.

If $f_y(0,0) \lt 0\text{,}$ then $f(0,y)$ decreases as $y$ increases from $0\text{.}$ Thus moving in the positive $y$ direction takes you downhill. This means you aren’t at the lowest point in a valley, since you can still move downhill. On the other hand, as $f_y(0,0) \lt 0\text{,}$ $f(0,y)$ also decreases as $y$ increases towards $0$ from slightly negative values. Thus if you move in the negative $y$-direction from $y=0\text{,}$ your height $z$ will increase. So you are not at a locally highest point—you’re not at a summit.

2.2.2.3. (✳).

Solution.

(a) By definition

\begin{align*} \pdiff{f}{x}(0,0) &=\lim_{\De x\rightarrow 0}\frac{f(\De x,0)-f(0,0)}{\De x}\\ &=\lim_{\De x\rightarrow 0}\frac{\frac{(\De x^2)(0)}{\De x^2+0^2}-0}{\De x}\\ &=0 \end{align*}

(b) By definition

\begin{align*} \pdiff{f}{y}(0,0) &=\lim_{\De y\rightarrow 0}\frac{f(0,\De y)-f(0,0)}{\De y}\\ &=\lim_{\De y\rightarrow 0}\frac{\frac{(0^2)(\De y)}{0^2+\De y^2}-0}{\De y}\\ &=0 \end{align*}

\begin{align*} \diff{}{t} f(t,t)\Big|_{t=0} &=\lim_{t\rightarrow 0}\frac{f(t,t)-f(0,0)}{t}\\ &=\lim_{t\rightarrow 0}\frac{\frac{(t^2)(t)}{t^2+t^2}-0}{t}\\ &=\lim_{t\rightarrow 0}\frac{t/2}{t}\\ &=\frac{1}{2} \end{align*}

2.2.2.4.

Solution.

(a)

\begin{align*} f_x(x,y,z)&=3x^2y^4z^5 & f_x(0,-1,-1)&=0\\ f_y(x,y,z)&=4x^3y^3z^5 & f_y(0,-1,-1)&=0\\ f_z(x,y,z)&=5x^3y^4z^4 & f_z(0,-1,-1)&=0 \end{align*}

(b)

\begin{align*} w_x(x,y,z)&=\frac{yz e^{xyz}}{1+e^{xyz}} & w_x(2,0,-1)&=0\\ w_y(x,y,z)&=\frac{xz e^{xyz}}{1+e^{xyz}} & w_y(2,0,-1)&=-1\\ w_z(x,y,z)&=\frac{xy e^{xyz}}{1+e^{xyz}} & w_z(2,0,-1)&=0 \end{align*}

(c)

\begin{align*} f_x(x,y)&=-\frac{x}{(x^2+y^2)^{3/2}} & f_x(-3,4)&=\frac{3}{125}\\ f_y(x,y)&=-\frac{y}{(x^2+y^2)^{3/2}} & f_y(-3,4)&=-\frac{4}{125} \end{align*}

2.2.2.5.

Solution.

By the quotient rule

\begin{alignat*}{2} \pdiff{z}{x}(x,y) &=\frac{(1)(x-y)-(x+y)(1)}{(x-y)^2} &&=\frac{-2y}{(x-y)^2}\\ \pdiff{z}{y}(x,y) &=\frac{(1)(x-y)-(x+y)(-1)}{(x-y)^2} &&=\frac{2x}{(x-y)^2} \end{alignat*}

Hence

\begin{equation*} x\pdiff{z}{x}(x,y)+y\pdiff{z}{y}(x,y) =\frac{-2xy+2yx}{(x-y)^2} =0 \end{equation*}

2.2.2.6. (✳).

Solution.

(a) We are told that $z(x,y)$ obeys

\begin{gather*} z(x,y)\, y - y + x = \ln\big(xy\,z(x,y)\big) \tag{$*$} \end{gather*}

for all $(x,y)$ (near $(-1,-2)$). Differentiating $(*)$ with respect to $x$ gives

\begin{gather*} y\,\pdiff{z}{x}(x,y) + 1 = \frac{1}{x} + \frac{\pdiff{z}{x}(x,y)}{z(x,y)} \implies \pdiff{z}{x}(x,y) = \frac{\frac{1}{x}-1}{y-\frac{1}{z(x,y)}} \end{gather*}

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $xz\text{,}$

\begin{gather*} \pdiff{z}{x} = \frac{z-xz}{xyz-x} = \frac{z(1-x)}{x(yz-1)} \end{gather*}

Differentiating $(*)$ with respect to $y$ gives

\begin{align*} &z(x,y)+y\,\pdiff{z}{y}(x,y) - 1 = \frac{1}{y} + \frac{\pdiff{z}{y}(x,y)}{z(x,y)}\\ &\hskip1in\implies \pdiff{z}{y}(x,y) = \frac{\frac{1}{y}+1-z(x,y)}{y-\frac{1}{z(x,y)}} \end{align*}

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $yz\text{,}$

\begin{gather*} \pdiff{z}{y} = \frac{z+yz-yz^2}{y^2z-y} = \frac{z(1+y-yz)}{y(yz-1)} \end{gather*}

(b) When $(x,y,z) = (-1, -2, 1/2)\text{,}$

\begin{align*} \pdiff{z}{x}(-1,-2) &= \left.\frac{\frac{1}{x}-1}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-1}-1}{-2-2} =\frac{1}{2}\\ \pdiff{z}{y}(-1,-2) &= \left.\frac{\frac{1}{y}+1-z}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-2}+1-\frac{1}{2}}{-2-2} =0 \end{align*}

2.2.2.7. (✳).

Solution.

We are told that the four variables $T\text{,}$ $U\text{,}$ $V\text{,}$ $W$ obey the single equation $(TU-V)^2 \ln(W-UV) = \ln 2\text{.}$ So they are not all independent variables. Roughly speaking, we can treat any three of them as independent variables and solve the given equation for the fourth as a function of the three chosen independent variables.

We are first asked to find $\pdiff{U}{T}\text{.}$ This implicitly tells to treat $T\text{,}$ $V$ and $W$ as independent variables and to view $U$ as a function $U(T,V,W)$ that obeys

\begin{equation*} \big(T\, U(T,V,W)-V\big)^2 \ln\big(W-U(T,V,W)\,V\big) = \ln 2 \tag{E1} \end{equation*}

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)\text{.}$ Differentiating (E1) with respect to $T$ gives

\begin{align*} &2\big(T\, U(T,V,W)\!-\!V\big)\! \left[ U(T,V,W)\!+\!T\,\pdiff{U}{T}(T,V,W)\right] \ln\big(W\!-\!U(T,V,W)\,V\big)\\ &\hskip0.5in -\big(T\, U(T,V,W)-V\big)^2 \frac{1}{W-U(T,V,W)\,V}\pdiff{U}{T}(T,V,W)\,V = 0 \end{align*}

In particular, for $(T, U, V, W)=(1, 1, 2, 4)\text{,}$

\begin{align*} &2\big((1)(1)-2\big) \left[ 1 +(1)\pdiff{U}{T}(1,2,4)\right] \ln\big(4-(1)(2)\big)\\ &\hskip1in -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)}\pdiff{U}{T}(1,2,4)\,(2) = 0 \end{align*}

This simplifies to

\begin{align*} &-2\left[ 1 +\pdiff{U}{T}(1,2,4)\right] \ln(2) -\pdiff{U}{T}(1,2,4)=0\\ &\hskip1.0in\implies \pdiff{U}{T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)} \end{align*}

We are then asked to find $\pdiff{T}{V}\text{.}$ This implicitly tells to treat $U\text{,}$ $V$ and $W$ as independent variables and to view $T$ as a function $T(U,V,W)$ that obeys

\begin{equation*} \big(T(U,V,W)\, U-V\big)^2 \ln\big(W-U\,V\big) = \ln 2 \tag{E2} \end{equation*}

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)\text{.}$ Differentiating (E2) with respect to $V$ gives

\begin{align*} &2\big(T(U,V,W)\, U-V\big)\ \left[\pdiff{T}{V}(U,V,W)\ U-1\right] \ln\big(W-U\,V\big)\\ &\hskip2.5in -\big(T(U,V,W)\, U-V\big)^2 \frac{U}{W-U\,V} = 0 \end{align*}

In particular, for $(T, U, V, W)=(1, 1, 2, 4)\text{,}$

\begin{align*} &2\big((1)(1)-2\big) \left[ (1)\pdiff{T}{V}(1,2,4)-1\right] \ln\big(4-(1)(2)\big)\\ &\hskip2.5in -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)} = 0 \end{align*}

This simplifies to

\begin{gather*} -2\left[\pdiff{T}{V}(1,2,4)-1\right] \ln(2) -\frac{1}{2}=0 \implies \pdiff{T}{V}(1,2,4) = 1 -\frac{1}{4\ln(2)} \end{gather*}

2.2.2.8. (✳).

Solution.

The function

\begin{align*} u(\rho, r,\theta) &= \big[\rho r\cos\theta\big]^2 +\big[\rho r\sin\theta\big] \rho r\\ &=\rho^2 r^2\cos^2\theta +\rho^2 r^2\sin\theta \end{align*}

\begin{align*} \pdiff{u}{r}(\rho, r,\theta) &=2 \rho^2 r\cos^2\theta +2 \rho^2 r\sin\theta \end{align*}

and

\begin{align*} \pdiff{u}{r}(2, 3,\pi/2) &=2 (2^2) (3) (0)^2 +2 (2^2) (3) (1) =24 \end{align*}

2.2.2.9.

Solution.

By definition

\begin{align*} f_x(x_0,y_0) \amp =\lim_{\De x\rightarrow 0}\frac{f(x_0+\De x,y_0)-f(x_0,y_0)}{\De x}\\ f_y(x_0,y_0) \amp =\lim_{\De y\rightarrow 0}\frac{f(x_0,y_0+\De y)-f(x_0,y_0)}{\De y} \end{align*}

Setting $x_0=y_0=0\text{,}$

\begin{align*} f_x(0,0)&=\lim_{\De x\rightarrow 0}\frac{f(\De x,0)-f(0,0)}{\De x} =\lim_{\De x\rightarrow 0}\frac{f(\De x,0)}{\De x}\\ &=\lim_{\De x\rightarrow 0}\frac{((\De x)^2-2\times0^2)/(\De x-0)}{\De x}\\ &=\lim_{\De x\rightarrow 0}1 =1\\ f_y(0,0)&=\lim_{\De y\rightarrow 0}\frac{f(0,\De y)-f(0,0)}{\De y} =\lim_{\De y\rightarrow 0}\frac{f(0,\De y)}{\De y}\\ &=\lim_{\De y\rightarrow 0}\frac{(0^2-2(\De y)^2)/(0-\De y)}{\De y}\\ &=\lim_{\De y\rightarrow 0}2 =2 \end{align*}

2.2.2.10.

Solution.

As $z(x,y)=f(x^2+y^2)$

\begin{align*} \pdiff{z}{x}(x,y)&=2xf'(x^2+y^2)\\ \pdiff{z}{y}(x,y)&=2yf'(x^2+y^2) \end{align*}

by the (ordinary single variable) chain rule. So

\begin{equation*} y\pdiff{z}{x}-x\pdiff{z}{y} =y(2x)f'(x^2+y^2)-x(2y)f'(x^2+y^2)=0 \end{equation*}

and the differential equation is always satisfied, assuming that $f$ is differentiable, so that the chain rule applies.

2.2.2.11.

Solution.

By definition

\begin{align*} \pdiff{f}{x}(0,0) &=\lim_{\De x\rightarrow 0}\frac{f(\De x,0)-f(0,0)}{\De x}\\ &=\lim_{\De x\rightarrow 0}\frac{\frac{(\De x+2\times 0)^2}{\De x+0}-0}{\De x}\\ &=\lim_{\De x\rightarrow 0}\frac{\De x}{\De x}\\ &=1 \end{align*}

and

\begin{align*} \pdiff{f}{y}(0,0) &=\lim_{\De y\rightarrow 0}\frac{f(0,\De y)-f(0,0)}{\De y}\\ &=\lim_{\De y\rightarrow 0}\frac{\frac{(0+2\De y)^2}{0+\De y}-0}{\De y}\\ &=\lim_{\De y\rightarrow 0}\frac{4\De y}{\De y}\\ &=4 \end{align*}

(b) $f(x,y)$ is not continuous at $(0,0)\text{,}$ even though both partial derivatives exist there. To see this, make a change of coordinates from $(x,y)$ to $(X,y)$ with $X=x+y$ (the denominator). Of course, $(x,y)\rightarrow (0,0)$ if and only if $(X,y)\rightarrow (0,0)\text{.}$ Now watch what happens when $(X,y)\rightarrow(0,0)$ with $X$ a lot smaller than $y\text{.}$ For example, $X=ay^2\text{.}$ Then

\begin{gather*} \frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(ay^2+y)^2}{ay^2} =\frac{(1+ay)^2}{a}\rightarrow\frac{1}{a} \end{gather*}

This depends on $a\text{.}$ So approaching $(0,0)$ along different paths gives different limits. (You can see the same effect without changing coordinates by sending $(x,y)\rightarrow (0,0)$ with $x=-y+ay^2\text{.}$) Even more dramatically, watch what happens when $(X,y)\rightarrow(0,0)$ with $X=y^3\text{.}$ Then

\begin{equation*} \frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(y^3+y)^2}{y^3} =\frac{{(1+y^2)}^2}{y}\rightarrow\pm\infty \end{equation*}

2.2.2.12.

Solution 1.

Let’s start by finding an equation for this surface. Every level curve is a horizontal circle of radius one, so the equation should be of the form

\begin{equation*} (x-f_1)^2+(y-f_2)^2=1 \end{equation*}

where $f_1$ and $f_2$ are functions depending only on $z\text{.}$ Since the centre of the circle at height $z$ is at position $x=0\text{,}$ $y=z\text{,}$ we see that the equation of our surface is

\begin{equation*} x^2+(y-z)^2=1 \end{equation*}

The height of the surface at the point $(x,y)$ is the $z(x,y)$ found by solving that equation. That is,

\begin{equation*} x^2+\big(y-z(x,y)\big)^2=1 \tag{$*$} \end{equation*}

We differentiate this equation implicitly to find $z_x(x,y)$ and $z_y(x,y)$ at the desired point $(x,y)= (0,-1)\text{.}$ First, differentiating $(*)$ with respect to $y$ gives

\begin{align*} 0+2\big(y-z(x,y)\big)\big(1- z_y(x,y)\big)&=0\\ 2(-1-0)\big(1-z_y(0,-1)\big)&=0& & \mbox{ at } (0,-1,0) \end{align*}

so that the slope looking in the positive $y$ direction is $z_y(0,-1)=1\text{.}$ Similarly, differentiating $(*)$ with respect to $x$ gives

\begin{align*} 2x+2\big(y-z(x,y)\big)\cdot\big(0-z_x(x,y)\big)&=0\\ 2x&=2\big(y-z(x,y)\big)\cdot z_x(x,y)\\ z_x(x,y)&=\frac{x}{y-z(x,y)}\\ z_x(0,-1)&=0 \hskip1in\mbox{ at } (0,-1,0) \end{align*}

The slope looking in the positive $x$ direction is $z_x(0,-1)=0\text{.}$

Solution 2.

Standing at $(0,-1,0)$ and looking in the positive $y$ direction, the surface follows the straight line that

passes through the point $(0,-1,0)\text{,}$ and
is parallel to the central line $z=y, x=0$ of the cylinder.

Shifting the central line one unit in the $y$-direction, we get the line $z=y+1\text{,}$ $x=0\text{.}$ (As a check, notice that $(0,-1,0)$ is indeed on $z=y+1\text{,}$ $x=0\text{.}$) The slope of this line is 1.

Standing at $(0,-1,0)$ and looking in the positive $x$ direction, the surface follows the circle $x^2+y^2=1\text{,}$ $z=0\text{,}$ which is the intersection of the cylinder with the $xy$-plane. As we move along that circle our $z$ coordinate stays fixed at $0\text{.}$ So the slope in that direction is 0.

2.3 Higher Order Derivatives
2.3.3 Exercises

2.3.3.1.

Solution.

We have to derive a bunch of equalities.

Fix any real number $x$ and set $g(y,z)=f_x(x,y,z)\text{.}$ By (Clairaut’s) Theorem 2.3.4 $g_{yz}(y,z)=g_{zy}(y,z)\text{,}$ so
\begin{equation*} f_{xyz}(x,y,z) = g_{yz}(y,z) =g_{zy}(y,z) = f_{xzy}(x,y,z) \end{equation*}
For every fixed real number $z\text{,}$ (Clairaut’s) Theorem 2.3.4 gives $f_{xy}(x,y,z)=f_{yx}(x,y,z)\text{.}$ So
\begin{equation*} f_{xyz}(x,y,z) = \pdiff{}{z} f_{xy}(x,y,z)= \pdiff{}{z} f_{yx}(x,y,z) =f_{yxz}(x,y,z) \end{equation*}
So far, we have
\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z) \end{equation*}
Fix any real number $y$ and set $g(x,z)=f_y(x,y,z)\text{.}$ By (Clairaut’s) Theorem 2.3.4 $g_{xz}(x,z)=g_{zx}(x,z)\text{.}$ So
\begin{equation*} f_{yxz}(x,y,z) = g_{xz}(x,z) =g_{zx}(x,z) = f_{yzx}(x,y,z) \end{equation*}
So far, we have
\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z)= f_{yzx}(x,y,z) \end{equation*}
For every fixed real number $y\text{,}$ (Clairaut’s) Theorem 2.3.4 gives $f_{xz}(x,y,z)=f_{zx}(x,y,z)\text{.}$ So
\begin{equation*} f_{xzy}(x,y,z) = \pdiff{}{y} f_{xz}(x,y,z)= \pdiff{}{y} f_{zx}(x,y,z) =f_{zxy}(x,y,z) \end{equation*}
So far, we have
\begin{align*} f_{xyz}(x,y,z) &= f_{xzy}(x,y,z)=f_{yxz}(x,y,z)\\ &= f_{yzx}(x,y,z)=f_{zxy}(x,y,z) \end{align*}
Fix any real number $z$ and set $g(x,y)=f_z(x,y,z)\text{.}$ By (Clairaut’s) Theorem 2.3.4 $g_{xy}(x,y)=g_{yx}(x,y)\text{.}$ So
\begin{equation*} f_{zxy}(x,y,z) = g_{xy}(x,y) =g_{yx}(x,y) = f_{zxy}(x,y,z) \end{equation*}
We now have all of
\begin{align*} f_{xyz}(x,y,z) &= f_{xzy}(x,y,z)=f_{yxz}(x,y,z)\\ &= f_{yzx}(x,y,z)=f_{zxy}(x,y,z) = f_{zxy}(x,y,z) \end{align*}

2.3.3.2.

Solution.

No such $f(x,y)$ exists, because if it were to exist, then we would have that $f_{xy}(x,y)=f_{yx}(x,y)\text{.}$ But

\begin{align*} f_{xy}(x,y)&=\pdiff{}{y}f_x(x,y)=\pdiff{}{y}e^y=e^y\\ f_{yx}(x,y)&=\pdiff{}{x}f_y(x,y)=\pdiff{}{x}e^x=e^x \end{align*}

are not equal.

2.3.3.3.

Solution.

(a) We have

\begin{align*} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3\\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{align*}

(b) We have

\begin{alignat*}{2} f_x(x,y) &= y^2e^{xy^2} \\ f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2}\\ f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2}\quad & f_{xyy}(x,y) &= \big(2\!+\!4xy^2\!+\!6xy^2\!+\!4x^2y^4\big)e^{xy^2}\\ & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{alignat*}

\begin{align*} \pdiff{f}{u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2}\\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3}\\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{align*}

In particular

\begin{align*} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{align*}

2.3.3.4.

Solution.

Let $f(x,y)=\sqrt{x^2+5y^2}\text{.}$ Then

\begin{align*} f_x&=\frac{x}{\sqrt{x^2+5y^2}} \\ f_{xx}&=\frac{1}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(x)(2x)}{(x^2+5y^2)^{3/2}} & f_{xy}&=-\frac{1}{2}\frac{(x)(10y)}{(x^2+5y^2)^{3/2}}\\ f_y&=\frac{5y}{\sqrt{x^2+5y^2}} \\ f_{yy}&=\frac{5}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(5y)(10y)}{(x^2+5y^2)^{3/2}}& f_{yx}&=-\frac{1}{2}\frac{(5y)(2x)}{(x^2+5y^2)^{3/2}} \end{align*}

Simplifying, and in particular using that $\frac{1}{\sqrt{x^2+5y^2}} =\frac{x^2+5y^2}{(x^2+5y^2)^{3/2}}\text{,}$

\begin{align*} f_{xx}&=\frac{5y^2}{(x^2+5y^2)^{3/2}}\qquad f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}\\ f_{yy}&=\frac{5x^2}{(x^2+5y^2)^{3/2}} \end{align*}

2.3.3.5.

Solution.

(a) As $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z\text{,}$ we have $f_z(x,y,z) = 0$ and hence

\begin{equation*} f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0 \end{equation*}

(b) Write $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)\text{,}$ $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ and $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)\text{.}$ Then

As $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z\text{,}$ we have $u_z(x,y,z) = 0$ and hence $u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0$
As $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ is independent of $y\text{,}$ we have $v_y(x,y,z) = 0$ and hence $v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0$
As $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ is independent of $x\text{,}$ we have $w_x(x,y,z) = 0$ and hence $w_{xyz}(x,y,z) =0$

As $f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)\text{,}$ we have

\begin{equation*} f_{xyz}(x,y,z)=u_{xyz}(x,y,z)+v_{xyz}(x,y,z)+w_{xyz}(x,y,z)=0 \end{equation*}

(c) In the course of evaluating $f_{xx}(x,0,0)\text{,}$ both $y$ and $z$ are held fixed at $0\text{.}$ Thus, if we set $g(x) = f(x,0,0)\text{,}$ then $f_{xx}(x,0,0)=g''(x)\text{.}$ Now

\begin{equation*} g(x) = f(x,0,0) = \arctan\big(e^{\sqrt{xyz}}\big)\Big|_{y=z=0} =\arctan(1) =\frac{\pi}{4} \end{equation*}

for all $x\text{.}$ So $g'(x)=0$ and $g''(x)=0$ for all $x\text{.}$ In particular,

\begin{equation*} f_{xx}(1,0,0) = g''(1) = 0 \end{equation*}

2.3.3.6. (✳).

Solution.

(a) The first order derivatives are

\begin{equation*} f_r(r,\theta)=mr^{m-1}\cos m\theta\qquad f_\theta(r,\theta)=-mr^m\sin m\theta \end{equation*}

The second order derivatives are

\begin{align*} f_{rr}(r,\theta)&=m(m-1)r^{m-2}\cos m\theta \qquad f_{r\theta}(r,\theta)=-m^2r^{m-1}\sin m\theta \\ f_{\theta\theta}(r,\theta)&=-m^2r^m\cos m\theta \end{align*}

so that

\begin{equation*} f_{rr}(1,0)=m(m-1),\ f_{r\theta}(1,0)=0,\ f_{\theta\theta}(1,0)=-m^2 \end{equation*}

(b) By part (a), the expression

\begin{align*} f_{rr}+\frac{\la}{r}f_r+\frac{1}{r^2}f_{\theta\theta} &=m(m-1)r^{m-2}\cos m\theta+\la mr^{m-2}\cos m\theta\\ &\hskip1in-m^2r^{m-2}\cos m\theta \end{align*}

vanishes for all $r$ and $\theta$ if and only if

\begin{equation*} m(m-1)+\la m-m^2=0\iff m(\la-1)=0\iff \la=1 \end{equation*}

2.3.3.7.

Solution.

\begin{align*} u_t(x,y,z,t) &=-\frac{3}{2}\frac{1}{t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\\ &\hskip1in +\frac{1}{4\al\,t^{7/2}}(x^2+y^2+z^2) e^{-(x^2+y^2+z^2)/(4\al t)}\\ u_x(x,y,z,t) &=-\frac{x}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\\ u_{xx}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{x^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\\ u_{yy}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{y^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\\ u_{zz}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{z^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \end{align*}

we have

\begin{align*} \al\big(u_{xx}\!+\!u_{yy}\!+\!u_{zz} \big) &=-\frac{3}{2\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \!+\!\frac{x^2\!+\!y^2\!+\!z^2}{4\al\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)}\\ &=u_t \end{align*}

2.4 The Chain Rule
2.4.5 Exercises

2.4.5.1.

Solution.

(c) We’ll start with part (c) and follow the procedure given in §2.4.1. We are to compute the derivative of $h(x,y,z)=f\big(u(x,y,z),v(x,y),w(x)\big)$ with respect to $x\text{.}$ For this function, the template of Step 2 in §2.4.1 is

\begin{equation*} \pdiff{h}{x}=\frac{\partial f}{ }\frac{ }{\partial x} \end{equation*}

Note that

The function $h$ appears once in the numerator on the left. The function $f\text{,}$ from which $h$ is constructed by a change of variables, appears once in the numerator on the right.
The variable, $x\text{,}$ in the denominator on the left appears once in the denominator on the right.

Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $u\text{,}$ $v$ and $w\text{,}$ it may only be differentiated with respect to $u\text{,}$ $v$ and $w\text{.}$ So we add together three copies of our template — one for each of $u\text{,}$ $v$ and $w\text{:}$

\begin{gather*} \pdiff{h}{x}=\pdiff{f}{u}\pdiff{u}{x} +\pdiff{f}{v}\pdiff{v}{x} +\pdiff{f}{w}\diff{w}{x} \end{gather*}

Since $w$ is a function of only one variable, we use the ordinary derivative symbol $\diff{w}{x}\text{,}$ rather than the partial derivative symbol $\pdiff{w}{x}$ in the third copy. Finally we put in the only functional dependence that makes sense. The left hand side is a function of $x\text{,}$ $y$ and $z\text{,}$ because $h$ is a function of $x\text{,}$ $y$ and $z\text{.}$ Hence the right hand side must also be a function of $x\text{,}$ $y$ and $z\text{.}$ As $f$ is a function of $u\text{,}$ $v$ and $w\text{,}$ this is achieved by evaluating $f$ at $u=u(x,y,z)\text{,}$ $v=v(x,y)$ and $w=w(x)\text{.}$

\begin{align*} \pdiff{h}{x}(x,y,z) &=\pdiff{f}{u}\big(u(x,y,z),v(x,y),w(x)\big) \pdiff{u}{x}(x,y,z)\\ &\hskip0.5in +\pdiff{f}{v}\big(u(x,y,z),v(x,y),w(x)\big) \pdiff{v}{x}(x,y)\\ &\hskip1in +\pdiff{f}{w}\big(u(x,y,z),v(x,y),w(x)\big) \diff{w}{x}(x) \end{align*}

(a) We again follow the procedure given in §2.4.1. We are to compute the derivative of $h(x,y)=f\big(x,u(x,y)\big)$ with respect to $x\text{.}$ For this function, the template of Step 2 in §2.4.1 is

\begin{equation*} \pdiff{h}{x}=\frac{\partial f}{ }\frac{ }{\partial x} \end{equation*}

Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $x$ and $u\text{,}$ it may only be differentiated with respect to $x\text{,}$ and $u\text{.}$ So we add together two copies of our template — one for $x$ and one for $u\text{:}$

\begin{gather*} \pdiff{h}{x}=\pdiff{f}{x}\diff{x}{x} +\pdiff{f}{u}\pdiff{u}{x} \end{gather*}

In $\diff{x}{x}$ we are to differentiate the (explicit) function $x$ (i.e. the function $F(x)=x$) with respect to $x\text{.}$ The answer is of course $1\text{.}$ So

\begin{gather*} \pdiff{h}{x}=\pdiff{f}{x} +\pdiff{f}{u}\pdiff{u}{x} \end{gather*}

Finally we put in the only functional depedence that makes sense. The left hand side is a function of $x\text{,}$ and $y\text{,}$ because $h$ is a function of $x$ and $y\text{.}$ Hence the right hand side must also be a function of $x$ and $y\text{.}$ As $f$ is a function of $x\text{,}$ $u\text{,}$ this is achieved by evaluating $f$ at $u=u(x,y)\text{.}$

\begin{gather*} \pdiff{h}{x}(x,y) =\pdiff{f}{x}\big(x,u(x,y)\big) +\pdiff{f}{u}\big(x,u(x,y)\big) \pdiff{u}{x}(x,y) \end{gather*}

(b) Yet again we follow the procedure given in §2.4.1. We are to compute the derivative of $h(x)=f\big(x,u(x),v(x)\big)$ with respect to $x\text{.}$ For this function, the template of Step 2 in §2.4.1 is

\begin{equation*} \diff{h}{x}=\frac{\partial f}{ }\frac{ }{\partial x} \end{equation*}

(As $h$ is function of only one variable, we use the ordinary derivative symbol $\diff{h}{x}$ on the left hand side.) Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $x\text{,}$ $u$ and $v\text{,}$ it may only be differentiated with respect to $x\text{,}$ $u$ and $v\text{.}$ So we add together three copies of our template — one for each of $x\text{,}$ $u$ and $v\text{:}$

\begin{align*} \diff{h}{x}&=\pdiff{f}{x}\diff{x}{x} +\pdiff{f}{u}\diff{u}{x} +\pdiff{f}{v}\diff{v}{x}\\ &=\pdiff{f}{x} +\pdiff{f}{u}\diff{u}{x} +\pdiff{f}{v}\diff{v}{x} \end{align*}

Finally we put in the only functional depedence that makes sense.

\begin{align*} \diff{h}{x}(x) &=\pdiff{f}{x}\big(x,u(x),v(x)\big) +\pdiff{f}{u}\big(x,u(x),v(x)\big) \diff{u}{x}(x)\\ &\hskip1in+\pdiff{f}{v}\big(x,u(x),v(x)\big) \diff{v}{x}(x) \end{align*}

2.4.5.2.

Solution.

To visualize, in a simplified setting, the situation from Example 2.4.10, note that $w'(x)$ is the rate of change of $z$ as we slide along the blue line, while $f_x(x,y)$ is the change of $z$ as we slide along the orange line.

In the partial derivative $f_x(x,y)\approx \frac{\Delta f}{\Delta x}\text{,}$ we let $x$ change, while $y$ stays the same. Necessarily, that forces $f$ to change as well. Starting at point $P_0\text{,}$ if we move $x$ but keep $y$ fixed, we end up at $P_2\text{.}$ According to the labels on the diagram, $\Delta x$ is $x_2-x_1\text{,}$ and $\Delta f$ is $z_2-z_1\text{.}$

The function $w(x)$ is a constant function, so we expect $w'(x)=0\text{.}$ In the approximation $\diff{w}{x}\approx \frac{\Delta w}{\Delta x}\text{,}$ we let $x$ change, but $w$ stays the same. Necessarily, to stay on the surface, this forces $y$ to change. Starting at point $P_0\text{,}$ if we move $x$ but keep $z=f(x,y)$ fixed, we end up at $P_1\text{.}$ According to the labels on the diagram, $\Delta x$ is $x_2-x_1$ again, and $\Delta w=z_1-z_1=0\text{.}$

To compare the two situations, note the first case has $\Delta y =0$ while the second case has $\Delta f =0\text{.}$

2.4.5.3. (✳).

Solution.

We are told in the statement of the question that $w(t)= f\big(x(t),y(t),t\big)\text{.}$ Applying the chain rule to $w(t)= f\big(x(t),y(t),t\big),$ by following the procedure given in §2.4.1, gives

\begin{align*} \diff{w}{t}(t) &=\pdiff{f}{x}\big(x(t),y(t),t\big)\diff{x}{t}(t) +\pdiff{f}{y}\big(x(t),y(t),t\big)\diff{y}{t}(t)\\ &\hskip1in+\pdiff{f}{t}\big(x(t),y(t),t\big)\diff{t}{t}\\ &=\pdiff{f}{x}\big(x(t),y(t),t\big)\diff{x}{t}(t) +\pdiff{f}{y}\big(x(t),y(t),t\big)\diff{y}{t}(t)\\ &\hskip1in+\pdiff{f}{t}\big(x(t),y(t),t\big) \end{align*}

Substituting in the values given in the question

\begin{equation*} \diff{w}{t} =2\times 1 -3\times 2 +5 =1 \end{equation*}

On the other hand, we are told explicitly in the question that $f_t$ is $5\text{.}$ The reason that $f_t$ and $\diff{w}{t}$ are different is that

$f_t$ gives the rate of change of $f(x,y,t)$ as $t$ varies while $x$ and $y$ are held fixed, but
$\diff{w}{t}$ gives the rate of change of $f\big(x(t),y(t),t\big)\text{.}$ For the latter all of $x=x(t)\text{,}$ $y=y(t)$ and $t$ are changing at once.

2.4.5.4.

Solution.

The basic assumption is that the three quantites $x\text{,}$ $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0\text{,}$ which can be solved to

determine $x$ as a function of $y$ and $z$ (say $x=f(y,z)$) and can alternatively be solved to
determine $y$ as a function of $x$ and $z$ (say $y=g(x,z)$) and can alternatively be solved to
determine $z$ as a function of $x$ and $y$ (say $z=h(x,y)$).

As an example, if $F(x,y,z) = xyz-1\text{,}$ then

$F(x,y,z)=xyz-1=0$ implies that $x=\frac{1}{yz}=f(y,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $y=\frac{1}{xz}=g(x,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $z=\frac{1}{xy}=h(x,y)$

In general, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

\begin{equation*} F\big(f(y,z),y,z\big)=0 \tag{$*$} \end{equation*}

for all $y$ and $z\text{.}$ Set $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)\text{.}$ Applying the chain rule to $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ (with $y$ and $z$ independent variables) gives

\begin{align*} \pdiff{\mathcal{F}}{z}(y,z) &= \pdiff{F}{x}\big(f(y,z),y,z\big) \pdiff{f}{z}(y,z) +\pdiff{F}{z}\big(f(y,z),y,z\big) \end{align*}

The equation $(*)$ says that $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ for all $y$ and $z\text{.}$ So differentiating the equation $(*)$ with respect to $z$ gives

\begin{align*} \pdiff{\mathcal{F}}{z}(y,z)&=\pdiff{F}{x}\big(f(y,z),y,z\big) \pdiff{f}{z}(y,z) +\pdiff{F}{z}\big(f(y,z),y,z\big)=0\\ &\implies \pdiff{f}{z}(y,z) =-\frac{\pdiff{F}{z}\big(f(y,z),y,z\big)} {\pdiff{F}{x}\big(f(y,z),y,z\big)} \end{align*}

for all $y$ and $z\text{.}$ Similarly, differentiating $F\big(x,g(x,z),z\big)=0$ with respect to $x$ and $F\big(x,y,h(x,y)\big)=0$ with respect to $y$ gives

\begin{gather*} \pdiff{g}{x}(x,z) =-\frac{\pdiff{F}{x}\big(x,g(x,z),z\big)} {\pdiff{F}{y}\big(x,g(x,z),z\big)}\qquad \pdiff{h}{y}(x,y) =-\frac{\pdiff{F}{y}\big(x,y,h(x,y)\big)} {\pdiff{F}{z}\big(x,y,h(x,y)\big)} \end{gather*}

If $(x,y,z)$ is any point satisfying $F(x,y,z)=0$ (so that $x=f(y,z)$ and $y=g(x,z)$ and $z=h(x,y)$), then

\begin{gather*} \pdiff{f}{z}(y,z) =-\frac{\pdiff{F}{z}\big(x,y,z\big)} {\pdiff{F}{x}\big(x,y,z\big)}\qquad \pdiff{g}{x}(x,z) =-\frac{\pdiff{F}{x}\big(x,y,z\big)} {\pdiff{F}{y}\big(x,y,z\big)}\\ \pdiff{h}{y}(x,y) =-\frac{\pdiff{F}{y}\big(x,y,z\big)} {\pdiff{F}{z}\big(x,y,z\big)} \end{gather*}

and

\begin{align*} \pdiff{f}{z}(y,z)\ \pdiff{g}{x}(x,z)\ \pdiff{h}{y}(x,y) &=-\frac{\pdiff{F}{z}\big(x,y,z\big)} {\pdiff{F}{x}\big(x,y,z\big)}\ \frac{\pdiff{F}{x}\big(x,y,z\big)} {\pdiff{F}{y}\big(x,y,z\big)}\ \frac{\pdiff{F}{y}\big(x,y,z\big)} {\pdiff{F}{z}\big(x,y,z\big)}\\ &=-1 \end{align*}

2.4.5.5.

Solution.

The problem is that $\pdiff{w}{x}$ is used to represent two completely different functions in the same equation. The careful way to write the equation is the following. Let $f(x,y,z)$ and $g(x,y)$ be continuously differentiable functions and define $w(x,y)=f\big(x,y,g(x,y)\big)\text{.}$ By the chain rule,

\begin{align*} \pdiff{w}{x}(x,y) &=\pdiff{f}{x}\big(x,y,g(x,y)\big) \pdiff{x}{x} +\pdiff{f}{y}\big(x,y,g(x,y)\big) \pdiff{y}{x}\\ &\hskip1in +\pdiff{f}{z}\big(x,y,g(x,y)\big) \pdiff{g}{x}(x,y)\\ &=\pdiff{f}{x}\big(x,y,g(x,y)\big) +\pdiff{f}{z}\big(x,y,g(x,y)\big) \pdiff{g}{x}(x,y) \end{align*}

While $w(x,y)=f\big(x,y,g(x,y)\big)\text{,}$ it is not true that $\pdiff{w}{x}(x,y) =\pdiff{f}{x}\big(x,y,g(x,y)\big)\text{.}$ For example, take $f(x,y,z)=x-z$ and $g(x,y)=x\text{.}$ Then $w(x,y)=f\big(x,y,g(x,y)\big)=x-g(x,y)=0$ for all $(x,y)\text{,}$ so that $\pdiff{w}{x}(x,y)=0$ while $\pdiff{f}{x}(x,y,z)=1$ for all $(x,y,z)\text{.}$

2.4.5.6.

Solution.

Method 1: Since $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st\text{,}$ $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$ we can write out $w(s,t)$ explicitly:

\begin{align*} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{and}\qquad w_t(s,t)=s^2(2t) \end{align*}

Method 2: The question specifies that $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st\text{,}$ $y(s,t)=s\cos t$ and $z(s,t)=s\sin t\text{.}$ That is, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ with $W(x,y,z)= x^2+y^2+z^2\text{.}$ Applying the chain rule to $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ and noting that $\pdiff{W}{x}=2x\text{,}$ $\pdiff{W}{y}=2y\text{,}$ $\pdiff{W}{z}=2z\text{,}$ gives

\begin{align*} \pdiff{w}{s}(s,t) &= \pdiff{W}{x}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{x}{s}(s,t)\\ &\hskip1in + \pdiff{W}{y}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{y}{s}(s,t)\\ &\hskip1in + \pdiff{W}{z}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{z}{s}(s,t)\\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t)\\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \pdiff{w}{t}(s,t) &= \pdiff{W}{x}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{x}{t}(s,t)\\ &\hskip1in + \pdiff{W}{y}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{y}{t}(s,t)\\ &\hskip1in + \pdiff{W}{z}\big(x(s,t), y(s,t), z(s,t)\big)\pdiff{z}{t}(s,t)\\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{align*}

2.4.5.7.

Solution.

By definition,

\begin{equation*} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\pdiff{}{x}\left[\pdiff{}{y}\left(\pdiff{}{y}f(2x+3y,xy)\right) \right] \end{equation*}

We’ll compute the derivatives from the inside out. Let’s call $F(x,y)=f(2x+3y,xy)$ so that the innermost derivative is $G(x,y)=\pdiff{}{y}f(2x+3y,xy)=\pdiff{}{y}F(x,y)\text{.}$ By the chain rule

\begin{align*} G(x,y)&=\pdiff{}{y}F(x,y)\\ &=f_1(2x+3y,xy)\pdiff{}{y}(2x+3y)+f_2(2x+3y,xy)\pdiff{}{y}(xy)\\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{align*}

Here the subscript $1$ means take the partial derivative of $f$ with respect to the first argument while holding the second argument fixed, and the subscript $2$ means take the partial derivative of $f$ with respect to the second argument while holding the first argument fixed. Next call the middle derivative $H(x,y)=\pdiff{}{y}\left(\pdiff{}{y}f(2x+3y,xy)\right)$ so that

\begin{align*} H(x,y) &= \pdiff{}{y} G(x,y)\\ &=\pdiff{}{y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big)\\ &=3\pdiff{}{y}\Big(f_1(2x+3y,xy)\Big) +x\pdiff{}{y}\Big(f_2(2x+3y,xy)\Big) \end{align*}

By the chain rule (twice),

\begin{align*} \pdiff{}{y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\pdiff{}{y}(2x+3y)\\ &\hskip1in+f_{12}(2x+3y,xy)\pdiff{}{y}(xy)\\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy)\\ \pdiff{}{y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\pdiff{}{y}(2x+3y)\\ &\hskip1in+f_{22}(2x+3y,xy)\pdiff{}{y}(xy)\\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{align*}

so that

\begin{align*} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big)\\ &\hskip1.5in +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big)\\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{align*}

In the last equality we used that $f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy)\text{.}$ The notation $f_{21}$ means first differentiate with respect to the second argument and then differentiate with respect to the first argument. For example, if $f(x,y)=e^{2y}\sin x\text{,}$ then

\begin{gather*} f_{21}(x,y) =\pdiff{}{x} \Big[\pdiff{}{y}\big(e^{2y}\sin x\big)\Big] =\pdiff{}{x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{gather*}

Finally, we get to

\begin{align*} &\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\pdiff{}{x} H(x,y)\\ &\hskip0.4in =\pdiff{}{x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &\hskip0.4in=9\pdiff{}{x}\Big(f_{11}(2x+3y,xy)\Big)\\ &\hskip0.7in +6f_{12}(2x+3y,xy) +6x\pdiff{}{x}\Big(f_{12}(2x+3y,xy)\Big)\\ &\hskip0.7in +2xf_{22}(2x+3y,xy) +x^2\pdiff{}{x}\Big(f_{22}(2x+3y,xy)\Big) \end{align*}

By three applications of the chain rule

\begin{align*} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\ &\hskip0.2in +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\ &\hskip0.2in +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112}\\ &\hskip0.5in +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{align*}

All functions on the right hand side have arguments $(2x+3y,xy)\text{.}$

2.4.5.8.

Solution.

The given function is

\begin{equation*} g(s,t)=f(2s+3t,3s-2t) \end{equation*}

The first order derivatives are

\begin{align*} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{align*}

The second order derivatives are

\begin{align*} g_{ss}(s,t) &=\pdiff{}{s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)\\ &\hskip1in+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\pdiff{}{t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)\\ &\hskip1in-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\pdiff{}{t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)\\ &\hskip1in+4f_{22}(2s+3t,3s-2t) \end{align*}

Here $f_1$ denotes the partial derivative of $f$ with respect to its first argument, $f_{12}$ is the result of first taking one partial derivative of $f$ with respect to its first argument and then taking a partial derivative with respect to its second argument, and so on.

2.4.5.9. (✳).

Solution.

By the chain rule,

\begin{align*} \pdiff{g}{s}(s,t) &=\pdiff{}{s} f (s - t, s + t)\\ &= \pdiff{f}{x}\big(s-t\,,\,s+t\big)\pdiff{}{s}\big(s-t\big) +\pdiff{f}{y}\big(s-t\,,\,s+t\big)\pdiff{}{s}\big(s+t\big)\\ &= \pdiff{f}{x}\big(s-t\,,\,s+t\big) +\pdiff{f}{y}\big(s-t\,,\,s+t\big)\\ \frac{\partial^2 g}{\partial s^2}(s,t) &=\color{blue}{\pdiff{}{s}\left[\pdiff{f}{x}\big(s-t\,,\,s+t\big)\right]} +\color{red}{\pdiff{}{s}\left[\pdiff{f}{y}\big(s-t\,,\,s+t\big)\right]}\\ &=\color{blue}{ \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)}\\ &\hskip0.2in +\color{red}{\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)}\\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s\!-\!t,s\!+\!t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s\!-\!t,s\!+\!t\big) +\frac{\partial^2 f}{\partial y^2}\big(s\!-\!t,s\!+\!t\big)\right\} \end{align*}

and

\begin{align*} \pdiff{g}{t}(s,t)&=\pdiff{}{t} f (s - t, s + t)\\ &= \pdiff{f}{x}\big(s-t\,,\,s+t\big)\pdiff{}{t}\big(s-t\big) +\pdiff{f}{y}\big(s-t\,,\,s+t\big)\pdiff{}{t}\big(s+t\big)\\ &= -\pdiff{f}{x}\big(s-t\,,\,s+t\big) +\pdiff{f}{y}\big(s-t\,,\,s+t\big)\\ \frac{\partial^2 g}{\partial t^2}(s,t) &=-\color{blue}{\pdiff{}{t}\left[\pdiff{f}{x}\big(s-t\,,\,s+t\big)\right]} +\color{red}{\pdiff{}{t}\left[\pdiff{f}{y}\big(s-t\,,\,s+t\big)\right]}\\ &= -\color{blue}{\Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big]}\\ &\hskip0.2in +\color{red}{\Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big]}\\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s\!-\!t,s\!+\!t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s\!-\!t,s\!+\!t\big) +\frac{\partial^2 f}{\partial y^2}\big(s\!-\!t,s\!+\!t\big)\right\} \end{align*}

Suppressing the arguments

\begin{align*} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\}\\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right]\\ &=0 \end{align*}

as desired.

2.4.5.10. (✳).

Solution.

The notation in the statement of this question is horrendous — the symbol $z$ is used with two different meanings in one equation. On the left hand side, it is a function of $x$ and $y\text{,}$ and on the right hand side, it is a function of $s$ and $t\text{.}$ Unfortunately that abuse of notation is also very common. Let us undo the notation conflict by renaming the function of $s$ and $t$ to $F(s,t)\text{.}$ That is,

\begin{equation*} F(s,t) = f\big(2s+t\,,\,s-t\big) \end{equation*}

In this new notation, we are being asked to find $a\text{,}$ $b$ and $c$ so that

\begin{align*} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{align*}

with the arguments on the right hand side being $(s,t)$ and the arguments on the left hand side being $\big(2s+t\,,\,s-t\big)\text{.}$

By the chain rule,

\begin{align*} \pdiff{F}{s}(s,t)&= \pdiff{f}{x}\big(2s+t\,,\,s-t\big)\pdiff{}{s}(2s+t) +\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\pdiff{}{s}(s-t)\\ &= 2\pdiff{f}{x}\big(2s+t\,,\,s-t\big) +\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\\ \frac{\partial^2 F}{\partial s^2}(s,t) &=\color{blue}{2\pdiff{}{s}\left[\pdiff{f}{x}\big(2s+t\,,\,s-t\big)\right]} +\color{red}{\pdiff{}{s}\left[\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\right]}\\ &= \color{blue}{4\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) + 2\frac{\partial^2 f}{\partial y\partial x} \big(2s+t\,,\,s-t\big)}\\ &\hskip0.2in +\color{red}{2\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big)} \end{align*}

and

\begin{align*} \pdiff{F}{t}(s,t)&= \pdiff{f}{x}\big(2s+t\,,\,s-t\big)\pdiff{}{t}(2s+t) +\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\pdiff{}{t}(s-t)\\ &= \pdiff{f}{x}\big(2s+t\,,\,s-t\big) -\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\\ \frac{\partial^2 F}{\partial t^2}(s,t) &=\color{blue}{\pdiff{}{t}\left[\pdiff{f}{x}\big(2s+t\,,\,s-t\big)\right]} \color{red}{-\pdiff{}{t}\left[\pdiff{f}{y}\big(2s+t\,,\,s-t\big)\right]}\\ &= \color{blue}{\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) -\frac{\partial^2 f}{\partial y\partial x}\big(2s+t\,,\,s-t\big)}\\ &\hskip0.2in \color{red}{-\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) } \end{align*}

Suppressing the arguments

\begin{align*} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{align*}

Finally, translating back into the (horrendous) notation of the question

\begin{align*} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{align*}

so that $a=5$ and $b=c=2\text{.}$

2.4.5.11. (✳).

Solution.

Let $u(x,y) = x^2 - y^2$ , and $v(x,y) = 2xy\text{.}$ Then $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)\text{.}$ By the chain rule

\begin{align*} &\pdiff{}{y}F(x^2 - y^2 , 2xy) =\pdiff{}{y}F(u(x,y) , v(xy))\\ &\hskip0.25in=F_u(u(x,y) , v(xy))\pdiff{u}{y}(x,y) + F_v(u(x,y) , v(xy))\pdiff{v}{y}(x,y)\\ &\hskip0.25in= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x)\\ &\frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) =\pdiff{}{x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\}\\ &\hskip0.25in=\color{blue}{-2y\pdiff{}{x}\left[ F_u(x^2 - y^2 , 2xy)\right]} +2 F_v(x^2 - y^2 , 2xy)\\ &\hskip1.0in +\color{red}{2x \pdiff{}{x}\left[F_v(x^2 - y^2 , 2xy)\right]}\\ &\hskip0.25in= \color{blue}{-4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy)} +2 F_v(x^2 - y^2 , 2xy)\\ &\hskip1.0in + \color{red}{4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy)}\\ &\hskip0.25in= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy)\\ &\hskip1.0in +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy)\\ &\hskip1.0in +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{align*}

2.4.5.12. (✳).

Solution.

For any (differentiable) function $F\text{,}$ we have, by the chain and product rules,

\begin{align*} \pdiff{u}{x}(x,y)&= \pdiff{}{x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\pdiff{}{x}\Big[F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F'\big(xe^{-y^2}\big)\pdiff{}{x}\Big(xe^{-y^2}\Big)\\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2}\\ \pdiff{u}{y}(x,y)&= \pdiff{}{y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\pdiff{}{y}\Big[F\big(xe^{-y^2}\big)\Big]\\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \pdiff{}{y}\Big(xe^{-y^2}\Big)\\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{align*}

(a) In particular, when $F(z)=\ln(z)\text{,}$ $F'(z)=\frac{1}{z}$ and

\begin{align*} \pdiff{u}{x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \pdiff{u}{y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y\\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{align*}

(b) In general

\begin{align*} 2xy\pdiff{u}{x} + \pdiff{u}{y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big)\\ &\hskip1in + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2}\\ &=e^y\, F\big(xe^{-y^2}\big)\\ &=u \end{align*}

2.4.5.13. (✳).

Solution.

By the chain rule,

\begin{align*} \pdiff{h}{t}(s,t) &=\pdiff{}{t}\big[f(2s + 3t)\big] + \pdiff{}{t}\big[g(s - 6t)\big]\\ &=f'(2s + 3t)\pdiff{}{t}(2s+3t) + g(s - 6t)\pdiff{}{t}(s-6t)\\ &= 3f'(2s + 3t) - 6g'(s - 6t)\\ \frac{\partial^2 h}{\partial t^2}(s,t) &=\color{blue}{3\pdiff{}{t}\big[f''(2s + 3t)\big]} \color{red}{- 6\pdiff{}{t}\big[ g''(s - 6t) \big]}\\ &=\color{blue}{3\,f''(2s + 3t)\pdiff{}{t}(2s+3t)} \color{red}{- 6\,g''(s - 6t) \pdiff{}{t}(s-6t)}\\ &=\color{blue}{9f''(2s+3t)} +\color{red}{36g''(s-6t)} \end{align*}

In particular

\begin{gather*} \frac{\partial^2 h}{\partial t^2}(2,1) =9f''(7) +36g''(-4) =9(-2) +36(-1) =-54 \end{gather*}

2.4.5.14. (✳).

Solution.

We’ll first compute the first order partial derivatives of $w(x,y,z)\text{.}$ Write $u(x,y,z)=xz$ and $v(x,y,z)=yz$ so that $w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big)\text{.}$ By the chain rule,

\begin{align*} &\pdiff{w}{x}(x,y,z) = \pdiff{}{x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\pdiff{f}{u}\big(u(x,y,z), v(x,y,z)\big)\pdiff{u}{x}(x,y,z)\\ &\hskip1in +\pdiff{f}{v}\big(u(x,y,z), v(x,y,z)\big)\pdiff{v}{x}(x,y,z)\\ &=z\pdiff{f}{u}(xz, yz)\\ &\pdiff{w}{y}(x,y,z) = \pdiff{}{y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\pdiff{f}{u}\big(u(x,y,z), v(x,y,z)\big)\pdiff{u}{y}(x,y,z)\\ &\hskip1in +\pdiff{f}{v}\big(u(x,y,z), v(x,y,z)\big)\pdiff{v}{y}(x,y,z)\\ &=z\pdiff{f}{v}(xz, yz)\\ &\pdiff{w}{z}(x,y,z) = \pdiff{}{z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\pdiff{f}{u}\big(u(x,y,z), v(x,y,z)\big)\pdiff{u}{z}(x,y,z)\\ &\hskip1in +\pdiff{f}{v}\big(u(x,y,z), v(x,y,z)\big)\pdiff{v}{z}(x,y,z)\\ &=x\pdiff{f}{u}(xz, yz) + y\pdiff{f}{v}(xz, yz) \end{align*}

\begin{align*} x\pdiff{w}{x} + y\pdiff{w}{y} &=xz\pdiff{f}{u}(xz, yz) + yz\pdiff{f}{v}(xz, yz)\\ &=z\left[x\pdiff{f}{u}(xz, yz) + y\pdiff{f}{v}(xz, yz)\right]\\ &=z\pdiff{w}{z} \end{align*}

as desired.

2.4.5.15. (✳).

Solution.

By definition $z(r,t) = f(r\cos t\,,\,r\sin t)\text{.}$

(a) By the chain rule

\begin{align*} \pdiff{z}{t}(r,t) &= \pdiff{}{t}\Big[f(r\cos t\,,\,r\sin t)\Big]\\ &= \pdiff{f}{x}(r\cos t\,,\,r\sin t)\pdiff{}{t}(r\cos t) +\pdiff{f}{y}(r\cos t\,,\,r\sin t)\pdiff{}{t}(r\sin t)\\ &= -r\sin t\ \pdiff{f}{x}(r\cos t\,,\,r\sin t) +r\cos t\ \pdiff{f}{y}(r\cos t\,,\,r\sin t) \end{align*}

(b) By linearity, the product rule and the chain rule

\begin{align*} &\frac{\partial^2 z}{\partial t^2}(r,t) = -\pdiff{}{t}\left[r\sin t \pdiff{f}{x}(r\cos t,r\sin t)\right] \!+\!\pdiff{}{t}\left[r\cos t \pdiff{f}{y}(r\cos t,r\sin t)\right]\\ &\hskip0.1in= -r\cos t\ \pdiff{f}{x}(r\cos t\,,\,r\sin t) \color{blue}{ -r\sin t\ \pdiff{}{t}\left[\pdiff{f}{x}(r\cos t\,,\,r\sin t)\right]}\\ &\hskip0.6in -r\sin t\ \pdiff{f}{y}(r\cos t\,,\,r\sin t) \color{red}{ +r\cos t\ \pdiff{}{t}\left[\pdiff{f}{y}(r\cos t\,,\,r\sin t)\right]}\\ &\hskip0.1in= -r\cos t \ \pdiff{f}{x}(r\cos t\,,\,r\sin t)\\ &\hskip0.6in \color{blue}{ +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t)}\\ &\hskip0.1in\phantom{=}-r\sin t \ \pdiff{f}{y}(r\cos t\,,\,r\sin t)\\ &\hskip0.6in \color{red}{-r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t)}\\ &\hskip0.1in= -r\cos t \ \pdiff{f}{x} -r\sin t \ \pdiff{f}{y}\\ &\hskip0.6in +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{align*}

with all of the partial derivatives of $f$ evaluated at $(r\cos t\,,\,r\sin t)\text{.}$

2.4.5.16. (✳).

Solution.

Write $w(t) = z\big(x(t),y(t)\big) = f\big(x(t),y(t)\big)$ with $x(t)=2t^2\text{,}$ $y(t)=t^3\text{.}$ We are to compute $\difftwo{w}{t}(1)\text{.}$ By the chain rule

\begin{align*} \diff{w}{t}(t)&= \diff{}{t} f\big(x(t),y(t)\big)\\ &= f_x(x(t)\,,\,y(t))\,\diff{x}{t}(t) +f_y(x(t)\,,\,y(t))\,\diff{y}{t}(t)\\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{align*}

By linearity, the product rule, and the chain rule,

\begin{align*} &\difftwo{}{t} f\big(x(t),y(t)\big) = \diff{}{t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\diff{}{t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right]\\ &\hskip0.3in= 4\,f_x\big(x(t),y(t)\big) + \color{blue}{4t \diff{}{t}\left[f_x\big(x(t),y(t)\big) \right] }\\ &\hskip0.3in\phantom{=} +6t\,f_y\big(x(t),y(t)\big) +\color{red}{3t^2\diff{}{t}\left[f_y\big(x(t),y(t)\big) \right]}\\ &\hskip0.3in= 4\,f_x(2t^2\,,\,t^3) + \color{blue}{4t\Big[f_{xx}\big(x(t),y(t)\big)\,\diff{x}{t}(t) + f_{xy}\big(x(t),y(t)\big)\,\diff{y}{t}(t)\Big]}\\ &\hskip0.3in\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + \color{red}{3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\diff{x}{t}(t) + f_{yy}\big(x(t),y(t)\big)\,\diff{y}{t}(t)\Big]}\\ &\hskip0.3in= 4\,f_x(2t^2\,,\,t^3) + \color{blue}{16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3)}\\ &\hskip0.3in\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + \color{red}{12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3)} \end{align*}

In particular, when $t=1\text{,}$ and since $f_{xy}(2, 1)=f_{yx}(2, 1)\text{,}$

\begin{align*} \left.\difftwo{}{t} f\big(x(t),y(t)\big) \right|_{t=1} &= 4\,(5) +\color{blue}{ 16\,(2) + 12\,(1)}\\ &\phantom{=} + 6\,(-2) + \color{red}{12\,(1) + 9\,(-4)}\\ &=28 \end{align*}

2.4.5.17. (✳).

Solution.

By the chain rule

\begin{align*} &\pdiff{G}{t}(\ga,s,t) = \pdiff{}{t} \Big[F (\gamma + s, \gamma-s, At)\Big]\\ &=\pdiff{F}{x}(\gamma + s, \gamma-s, At)\, \pdiff{}{t}(\gamma + s) +\pdiff{F}{y}(\gamma + s, \gamma-s, At)\, \pdiff{}{t}(\gamma - s)\\ &\hskip1in +\pdiff{F}{z}(\gamma + s, \gamma-s, At)\,\pdiff{}{t}(At)\\ &= A\,\pdiff{F}{z}(\gamma + s, \gamma-s, At) \end{align*}

and

\begin{align*} &\pdiff{G}{\ga}(\ga,s,t) = \pdiff{}{\ga} \Big[F (\gamma + s, \gamma-s, At)\Big]\\ &=\pdiff{F}{x}(\gamma + s, \gamma-s, At)\, \pdiff{}{\ga}(\gamma + s)\\ &\hskip1in +\pdiff{F}{y}(\gamma + s, \gamma-s, At)\, \pdiff{}{\ga}(\gamma - s)\\ &\hskip1in +\pdiff{F}{z}(\gamma + s, \gamma-s, At)\,\pdiff{}{\ga}(At)\\ &= \pdiff{F}{x}(\gamma + s, \gamma-s, At) +\pdiff{F}{y}(\gamma + s, \gamma-s, At) \tag{E1} \end{align*}

and

\begin{align*} &\pdiff{G}{s}(\ga,s,t) = \pdiff{}{s} \Big[F (\gamma + s, \gamma-s, At)\Big]\\ &=\pdiff{F}{x}(\gamma + s, \gamma-s, At)\, \pdiff{}{s}(\gamma + s)\\ &\hskip1in +\pdiff{F}{y}(\gamma + s, \gamma-s, At)\, \pdiff{}{s}(\gamma - s)\\ &\hskip1in +\pdiff{F}{z}(\gamma + s, \gamma-s, At)\,\pdiff{}{s}(At)\\ &= \pdiff{F}{x}(\gamma + s, \gamma-s, At) -\pdiff{F}{y}(\gamma + s, \gamma-s, At) \tag{E2} \end{align*}

We can evaluate the second derivatives by applying the chain rule to the four terms on the right hand sides of

\begin{align*} \frac{\partial^2G}{\partial\ga^2}(\ga,s,t) &=\pdiff{}{\ga}\Big[\pdiff{G}{\ga}(\ga, s, t)\Big]\\ & =\color{red}{\pdiff{}{\ga}\Big[\pdiff{F}{x}(\gamma + s, \gamma-s, At)\Big]} +\color{orange}{\pdiff{}{\ga}\Big[\pdiff{F}{y}(\gamma + s, \gamma-s, At) \Big]}\\ \frac{\partial^2G}{\partial s^2}(\ga,s,t) &=\pdiff{}{s}\Big[\pdiff{G}{s}(\ga, s, t)\Big]\\ & =\color{blue}{\pdiff{}{s}\Big[\pdiff{F}{x}(\gamma + s, \gamma-s, At)\Big]} -\color{violet}{\pdiff{}{s}\Big[\pdiff{F}{y}(\gamma + s, \gamma-s, At) \Big]} \end{align*}

Alternatively, we can observe that replacing $F$ by $\pdiff{F}{x}$ in (E1) and (E2) gives

\begin{align*} &\color{red}{\pdiff{}{\ga}\Big[\pdiff{F}{x}(\gamma + s, \gamma-s, At)\Big]}\\ &\hskip0.5in=\color{red}{\frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)}\\ &\color{blue}{\pdiff{}{s}\Big[\pdiff{F}{x}(\gamma + s, \gamma-s, At) \Big]}\\ &\hskip0.5in=\color{blue}{\frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)} \end{align*}

replacing $F$ by $\pdiff{F}{y}$ in (E1) and (E2) gives

\begin{align*} &\color{orange}{\pdiff{}{\ga}\Big[\pdiff{F}{y}(\gamma + s, \gamma-s,At)\Big]}\\ &\hskip0.5in=\color{orange}{\frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s,At)}\\ &\color{violet}{\pdiff{}{s}\Big[\pdiff{F}{y}(\gamma + s, \gamma-s, At) \Big]}\\ &\hskip0.5in=\color{violet}{\frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)} \end{align*}

Consequently

\begin{align*} \frac{\partial^2G}{\partial\ga^2}(\ga,s,t) &= \color{red}{\frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)}\\ &\phantom{=}+ \color{orange}{\frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)}\\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)\\ &\phantom{=} +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{align*}

and

\begin{align*} \frac{\partial^2G}{\partial s^2}(\ga,s,t) &= \color{blue}{\frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)}\\ &\phantom{=}-\Big[ \color{violet}{\frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)}\Big]\\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At)\\ &\phantom{=} +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{align*}

So, suppressing the arguments,

\begin{align*} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \pdiff{F}{z} =2\pdiff{F}{z}-A \pdiff{F}{z} =0 \end{align*}

if $A=2\text{.}$

2.4.5.18. (✳).

Solution.

By the chain rule

\begin{gather*} \pdiff{g}{s}(s,t) = \pdiff{}{s}\big[f(as-bt)\big] =f'(as-bt)\pdiff{}{s}(as-bt) = af'(as-bt) \end{gather*}

In particular

\begin{gather*} \pdiff{g}{s}(b,a) = af'(ab-ba) =af'(0) =10 a \end{gather*}

2.4.5.19. (✳).

Solution.

We are told that the function $z(x,y)$ obeys

\begin{equation*} f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$} \end{equation*}

for all $x$ and $y\text{.}$ By the chain rule,

\begin{align*} &\pdiff{}{x}\big[f\big(x\,z(x,y), y\,z(x,y)\big)\big]\\ &=f_u\big(x\,z(x,y), y\,z(x,y)\big)\pdiff{}{x}\big[x\,z(x,y)\big]\\ &\hskip1in + f_v\big(x\,z(x,y), y\,z(x,y)\big)\pdiff{}{x}\big[y\,z(x,y)\big]\\ &=f_u\big(x\,z(x,y), y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big]\\ &\hskip1in + f_v\big(x\,z(x,y), y\,z(x,y)\big)\,y\,z_x(x,y)\\ &\pdiff{}{y}\big[f\big(x\,z(x,y), y\,z(x,y)\big)\big]\\ &=f_u\big(x\,z(x,y), y\,z(x,y)\big)\pdiff{}{y}\big[x\,z(x,y)\big]\\ &\hskip1in + f_v\big(x\,z(x,y), y\,z(x,y)\big)\pdiff{}{y}\big[y\,z(x,y)\big]\\ &=f_u\big(x\,z(x,y), y\,z(x,y)\big)\, x\,z_y(x,y)\\ &\hskip1in + f_v\big(x\,z(x,y), y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{align*}

so differentiating $(*)$ with respect to $x$ and with respect to $y$ gives

\begin{align*} &f_u\big(x z(x,y),yz(x,y)\big)\big[z(x,y)+xz_x(x,y)\big] + f_v\big(xz(x,y), yz(x,y)\big)y z_x(x,y) \\ &\hskip4in=0\\ &f_u\big(x\,z(x,y), yz(x,y)\big) xz_y(x,y) + f_v\big(xz(x,y), yz(x,y)\big) \big[z(x,y)+yz_y(x,y)\big] \\ &\hskip4in=0 \end{align*}

or, leaving out the arguments,

\begin{align*} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0\\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{align*}

Solving the first equation for $z_x$ and the second for $z_y$ gives

\begin{align*} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v}\\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{align*}

so that

\begin{gather*} x\pdiff{z}{x}+y\pdiff{z}{y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{gather*}

as desired.

Remark: This is of course under the assumption that $x\,f_u+y\,f_v$ is nonzero. That is equivalent, by the chain rule, to the assumption that $\pdiff{}{z}\big[f(xz,yz)\big]$ is non zero. That, in turn, is almost, but not quite, equivalent to the statement that $f(xz,yz)=0$ is can be solved for $z$ as a function of $x$ and $y\text{.}$

2.4.5.20. (✳).

Solution.

(a) By the chain rule

\begin{align*} w_s(s,t) &=\pdiff{}{s}\big[u(2s + 3t, 3s - 2t)\big]\\ &= u_x(2s + 3t, 3s - 2t)\pdiff{}{s}\big(2s + 3t\big)\\ &\hskip0.25in +u_y(2s + 3t, 3s - 2t)\pdiff{}{s}\big(3s - 2t\big)\\ &= 2\,u_x(2s + 3t, 3s - 2t) +3\,u_y(2s + 3t, 3s - 2t) \end{align*}

and

\begin{align*} w_{ss}(s,t) &=\color{blue}{2\pdiff{}{s}\big[u_x(2s + 3t, 3s - 2t)\big]} + \color{red}{3\pdiff{}{s}\big[u_y(2s + 3t, 3s - 2t)\big]}\\ &=\color{blue}{\big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big]}\\ &\hskip1.0in +\color{red}{\big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big]}\\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t)\\ &\hskip1in +9\,u_{yy}(2s + 3t, 3s - 2t) \end{align*}

(b) Again by the chain rule

\begin{align*} w_t(s,t) &=\pdiff{}{t}\big[u(2s + 3t, 3s - 2t)\big]\\ &= u_x(2s + 3t, 3s - 2t)\pdiff{}{t}\big(2s + 3t\big)\\ &\hskip1in +u_y(2s + 3t, 3s - 2t)\pdiff{}{t}\big(3s - 2t\big)\\ &= 3\,u_x(2s + 3t, 3s - 2t) -2\, u_y(2s + 3t, 3s - 2t) \end{align*}

and

\begin{align*} w_{tt}(s,t) &=\color{blue}{3\pdiff{}{t}\big[u_x(2s + 3t, 3s - 2t)\big]} -\color{red}{2\pdiff{}{t}\big[u_y(2s + 3t, 3s - 2t)\big]}\\ &=\color{blue}{\big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big]}\\ &\hskip1.0in -\color{red}{\big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big]}\\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t)\\ &\hskip1in +4\,u_{yy}(2s + 3t, 3s - 2t) \end{align*}

Consquently, for any constant $A\text{,}$

\begin{align*} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{align*}

Given that $u_{xx} + u_{yy}=0\text{,}$ this will be zero, as desired, if $A=-1\text{.}$ (Then $(4-9A)=(9-4A)=13\text{.}$)

2.4.5.21. (✳).

Solution.

This question uses bad (but standard) notation, in that the one symbol $f$ is used for two different functions, namely $f(x,y)$ and $f(r,\theta)=f(x,y)\big|_{x=r\cos\theta,\,y=r\sin\theta}\text{.}$ Let us undo this notation conflict by renaming the function of $r$ and $\theta$ to $F(r,\theta)\text{.}$ That is,

\begin{equation*} F(r,\theta) = f\big(r\cos\theta\,,\,r\sin\theta\big) \end{equation*}

Similarly, rename $g\text{,}$ viewed as a function of $r$ and $\theta\text{,}$ to $G(r,\theta)\text{.}$ That is,

\begin{equation*} G(r,\theta) = g\big(r\cos\theta\,,\,r\sin\theta\big) \end{equation*}

In this new notation, we are being asked

in part (a) to find $F_\theta\text{,}$ $F_r$ and $F_{r\theta}$ in terms of $r\text{,}$ $\theta\text{,}$ $f_x$ and $f_y\text{,}$ and
in part (b) to express $F_r$ and $F_\theta$ in terms of $r\text{,}$ $\theta$ and $G_r\text{,}$ $G_\theta\text{.}$

(a) By the chain rule

\begin{align*} F_\theta(r,\theta) &=\pdiff{}{\theta}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big]\\ &= f_x\big(r\cos\theta\,,\,r\sin\theta\big) \ \pdiff{}{\theta}\big(r\cos\theta\big)\\ &\hskip1in +f_y\big(r\cos\theta\,,\,r\sin\theta\big) \ \pdiff{}{\theta}\big(r\sin\theta\big)\\ &=-r\sin\theta\, f_x\big(r\cos\theta\,,\,r\sin\theta\big) +r\cos\theta\, f_y\big(r\cos\theta\,,\,r\sin\theta\big) \tag{E1}\\ F_r(r,\theta) &=\pdiff{}{r}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big]\\ &= f_x\big(r\cos\theta\,,\,r\sin\theta\big) \ \pdiff{}{r}\big(r\cos\theta\big)\\ &\hskip1in +f_y\big(r\cos\theta\,,\,r\sin\theta\big) \ \pdiff{}{r}\big(r\sin\theta\big)\\ &= \cos\theta\, f_x\big(r\cos\theta\,,\,r\sin\theta\big) +\sin\theta\, f_y\big(r\cos\theta\,,\,r\sin\theta\big) \tag{E2} \end{align*}

and

\begin{align*} &F_{r\theta}(r,\theta) = \pdiff{}{\theta}\big[F_r(r,\theta)\big]\\ &=\pdiff{}{\theta}\Big[\cos\theta\, f_x\big(r\cos\theta\,,\,r\sin\theta\big) +\sin\theta\, f_y\big(r\cos\theta\,,\,r\sin\theta\big)\Big]\\ &=-\sin\theta\ f_x\big(r\cos\theta\,,\,r\sin\theta\big) +\cos\theta\, \color{blue}{ \pdiff{}{\theta} \Big[f_x\big(r\cos\theta\,,\,r\sin\theta\big)\Big]}\\ &\hskip0.2in +\cos\theta\ f_y\big(r\cos\theta\,,\,r\sin\theta\big) +\sin\theta\, \color{red}{ \pdiff{}{\theta}\Big[f_y\big(r\cos\theta\,,\,r\sin\theta\big)\Big]}\\ &=-\sin\theta\ f_x\big(r\cos\theta\,,\,r\sin\theta\big)\\ &\hskip0.4in +\cos\theta \color{blue}{ \Big[f_{xx}\big(r\cos\theta\,,\,r\sin\theta\big)\,(-r\sin\theta) +f_{xy}\big(r\cos\theta\,,\,r\sin\theta\big)\,(r\cos\theta)\Big]}\\ &\hskip0.2in +\cos\theta\ f_y\big(r\cos\theta\,,\,r\sin\theta\big)\\ &\hskip0.4in +\sin\theta \color{red}{ \Big[f_{yx}\big(r\cos\theta\,,\,r\sin\theta\big)\,(-r\sin\theta) +f_{yy}\big(r\cos\theta\,,\,r\sin\theta\big)\,(r\cos\theta)\Big]}\\ &=-\sin\theta\, f_x +\cos\theta\, f_y\\ &\hskip0.4in -r\sin\theta\cos\theta\, f_{xx} +r[\cos^2\theta-\sin^2\theta]\, f_{xy} +r\sin\theta\cos\theta\, f_{yy} \end{align*}

with the arguments of $f_x\text{,}$ $f_y\text{,}$ $f_{xx}\text{,}$ $f_{xy}$ and $f_{yy}$ all being $\big(r\cos\theta\,,\,r\sin\theta\big)\text{.}$

(b) Replacing $f$ by $g$ in (E1) gives

\begin{align*} G_\theta(r,\theta) &=\pdiff{}{\theta}\big[g\big(r\cos\theta\,,\,r\sin\theta\big)\big]\\ &=-r\sin\theta\, g_x\big(r\cos\theta\,,\,r\sin\theta\big) +r\cos\theta\, g_y\big(r\cos\theta\,,\,r\sin\theta\big)\\ &=-r\sin\theta\, f_y\big(r\cos\theta\,,\,r\sin\theta\big) -r\cos\theta\, f_x\big(r\cos\theta\,,\,r\sin\theta\big)\\ &=-r \pdiff{}{r}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big] \qquad\text{by (E2)} \end{align*}

Replacing $f$ by $g$ in (E2) gives

\begin{align*} G_r(r,\theta) &=\pdiff{}{r}\big[g\big(r\cos\theta\,,\,r\sin\theta\big)\big]\\ &= \cos\theta\, g_x\big(r\cos\theta\,,\,r\sin\theta\big) +\sin\theta\, g_y\big(r\cos\theta\,,\,r\sin\theta\big)\\ &= \cos\theta\, f_y\big(r\cos\theta\,,\,r\sin\theta\big) -\sin\theta\, f_x\big(r\cos\theta\,,\,r\sin\theta\big)\\ &=\frac{1}{r} \pdiff{}{\theta}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big] \qquad\text{by (E1)} \end{align*}

\begin{align*} \pdiff{}{r}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big] &=-\frac{1}{r} \pdiff{}{\theta}\big[g\big(r\cos\theta\,,\,r\sin\theta\big)\big]\\ \pdiff{}{\theta}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big] &=r\pdiff{}{r}\big[g\big(r\cos\theta\,,\,r\sin\theta\big)\big] \end{align*}

2.4.5.22. (✳).

Solution.

By the chain rule

\begin{align*} \pdiff{z}{s}(s,t) &=\pdiff{}{s}f\big(g(s,t),h(s,t)\big)\\ &=\pdiff{f}{x}\big(g(s,t),h(s,t)\big) \pdiff{g}{s}(s,t) +\pdiff{f}{y}\big(g(s,t),h(s,t)\big) \pdiff{h}{s}(s,t)\\ \pdiff{z}{t}(s,t) &=\pdiff{}{t}f\big(g(s,t),h(s,t)\big)\\ &=\pdiff{f}{x}\big(g(s,t),h(s,t)\big) \pdiff{g}{t}(s,t) +\pdiff{f}{y}\big(g(s,t),h(s,t)\big) \pdiff{h}{t}(s,t) \end{align*}

In particular

\begin{align*} \pdiff{z}{s}(1,2) &=\pdiff{f}{x}\big(g(1,2),h(1,2)\big) \pdiff{g}{s}(1,2) +\pdiff{f}{y}\big(g(1,2),h(1,2)\big) \pdiff{h}{s}(1,2)\\ &=\pdiff{f}{x}(3,6) \pdiff{g}{s}(1,2) +\pdiff{f}{y}(3,6)) \pdiff{h}{s}(1,2)\\ &=7\times(-1)+8\times(-5) =-47\\ \pdiff{z}{t}(1,2) &=\pdiff{f}{x}\big(g(1,2),h(1,2)\big) \pdiff{g}{t}(1,2) +\pdiff{f}{y}\big(g(1,2),h(1,2)\big) \pdiff{h}{t}(1,2)\\ &=7\times4+8\times 10 =108 \end{align*}

Hence $\vnabla z(1,2)=\llt -47,108\rgt\text{.}$

2.4.5.23. (✳).

Solution.

(a) By the product and chain rules

\begin{align*} w_x(x,y) &= \pdiff{}{x}\big[e^{-y}f(x\!-\!y)\big] = e^{-y} \pdiff{}{x}\big[f(x\!-\!y)\big] = e^{-y} f'(x-y)\pdiff{}{x}(x\!-\!y)\\ &= e^{-y}f'(x-y)\\ % w_y(x,y) &= \pdiff{}{y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \pdiff{}{y}\big[f(x-y)\big]\\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\pdiff{}{y}(x-y)\\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{align*}

Hence

\begin{align*} w+\pdiff{w}{x}+\pdiff{w}{y} &=e^{-y}f(x\!-\!y)+e^{-y}f'(x\!-\!y)-e^{-y}f(x\!-\!y)-e^{-y}f'(x\!-\!y)\\ &=0 \end{align*}

as desired.

(b) Think of $x=u^3-3uv^2\text{,}$ $y=3u^2v-v^3$ as two equations in the two unknowns $u\text{,}$ $v$ with $x\text{,}$ $y$ just being given parameters. The question implicitly tells us that those two equations can be solved for $u\text{,}$ $v$ in terms of $x\text{,}$ $y\text{,}$ at least near $(u,v)=(2,1)\text{,}$ $(x,y)=(2,11)\text{.}$ That is, the question implicitly tells us that the functions $u(x,y)$ and $v(x,y)$ are determined by

\begin{equation*} x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3 \end{equation*}

Applying $\pdiff{}{x}$ to both sides of the equation $x=u(x,y)^3-3u(x,y)v(x,y)^2$ gives

\begin{align*} 1&=3\,u(x,y)^2\,\pdiff{u}{x}(x,y) -3\pdiff{u}{x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\pdiff{v}{x}(x,y) \end{align*}

Then applying $\pdiff{}{x}$ to both sides of $y=3u(x,y)^2v(x,y)-v(x,y)^3$ gives

\begin{align*} 0&=6\,u(x,y)\,\pdiff{u}{x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\pdiff{v}{x}(x,y) -3\,v(x,y)^2\,\pdiff{v}{x}(x,y) \end{align*}

Substituting in $x=2\text{,}$ $y=11\text{,}$ $u=2\text{,}$ $v=1$ gives

\begin{alignat*}{2} 1&=12\pdiff{u}{x}(2,11) \!-\!3\pdiff{u}{x}(2,11) \!-\!12\pdiff{v}{x}(2,11) &&=9\pdiff{u}{x}(2,11) \!-\!12\pdiff{v}{x}(2,11)\\ 0&=12\pdiff{u}{x}(2,11) \!+\!12\pdiff{v}{x}(2,11) \!-\!3\pdiff{v}{x}(2,11) &&=12\pdiff{u}{x}(2,11) \!+\!9\pdiff{v}{x}(2,11) \end{alignat*}

From the second equation $\pdiff{v}{x}(2,11) =-\frac{4}{3}\pdiff{u}{x}(2,11)\text{.}$ Substituting into the first equation gives

\begin{equation*} 1=9\pdiff{u}{x}(2,11) -12\left[-\frac{4}{3}\pdiff{u}{x}(2,11)\right] = 25\pdiff{u}{x}(2,11) \end{equation*}

so that $\pdiff{u}{x}(2,11)=\frac{1}{25}$ and $\pdiff{v}{x}(2,11)=-\frac{4}{75}\text{.}$ The question also tells us that $z(x,y)=u(x,y)^2-v(x,y)^2\text{.}$ Hence

\begin{align*} &\pdiff{z}{x}(x,y) =2u(x,y)\pdiff{u}{x}(x,y) -2v(x,y)\pdiff{v}{x}(x,y)\\ \implies & \pdiff{z}{x}(2,11) =4\pdiff{u}{x}(2,11) -2\pdiff{v}{x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{align*}

2.4.5.24. (✳).

Solution.

(a) We are told that

\begin{equation*} x(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u \end{equation*}

Applying $\pdiff{}{u}$ to both equations gives

\begin{align*} 2x(u,v)\pdiff{x}{u}(u,v) -\pdiff{y}{u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0\\ 2x(u,v)\pdiff{x}{u}(u,v) +2y(u,v)\pdiff{y}{u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{align*}

Setting $u=\frac{\pi}{2}\text{,}$ $v=0\text{,}$ $x\big(\frac{\pi}{2},0\big)=1\text{,}$ $y\big(\frac{\pi}{2},0\big)=1$ gives

\begin{align*} 2\pdiff{x}{u}\left(\frac{\pi}{2},0\right) -\pdiff{y}{u}\left(\frac{\pi}{2},0\right)&=0\\ 2\pdiff{x}{u}\left(\frac{\pi}{2},0\right) +2\pdiff{y}{u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{align*}

Substituting $\pdiff{y}{u}\big(\frac{\pi}{2},0\big)=2\pdiff{x}{u}\big(\frac{\pi}{2},0\big)\text{,}$ from the first equation, into the second equation gives $6\pdiff{x}{u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi}$ so that $\pdiff{x}{u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi}$ and $\pdiff{y}{u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}\text{.}$

(b) We are told that $z(u,v)=x(u,v)^4+y(u,v)^4\text{.}$ So

\begin{align*} \pdiff{z}{u}(u,v) &=4x(u,v)^3\ \pdiff{x}{u}(u,v) +4y(u,v)^3\ \pdiff{y}{u}(u,v) \end{align*}

Substituting in $u=\frac{\pi}{2}\text{,}$ $v=0\text{,}$ $x\big(\frac{\pi}{2},0\big)=1\text{,}$ $y\big(\frac{\pi}{2},0\big)=1$ and using the results of part (a),

\begin{align*} \pdiff{z}{u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \pdiff{x}{u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \pdiff{y}{u}\!\left(\frac{\pi}{2},0\right)\\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right)\\ &=\frac{8}{\pi} \end{align*}

2.4.5.25. (✳).

Solution.

This question uses bad (but standard) notation, in that the one symbol $f$ is used for two different functions, namely $f(u,v)$ and $f(x,y)=f(u,v)\big|_{u=x+y,v=x-y}\text{.}$ A better wording is

[] Let $f(u,v)$ and $F(x,y)$ be differentiable functions such that $F(x,y)=f(x+y,x-y)\text{.}$ Find a constant, $\al\text{,}$ such that
\begin{equation*} F_x(x,y)^2+F_y(x,y)^2=\al\big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\} \end{equation*}

By the chain rule

\begin{align*} \pdiff{F}{x}(x,y) &=f_u(x+y,x-y)\pdiff{}{x}(x+y) +f_v(x+y,x-y)\pdiff{}{x}(x-y)\\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \pdiff{F}{y}(x,y) &=f_u(x+y,x-y)\pdiff{}{y}(x+y) +f_v(x+y,x-y)\pdiff{}{y}(x-y)\\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{align*}

Hence

\begin{align*} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2\\ &\hskip1in +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{align*}

So $\al=2$ does the job.

2.4.5.26.

Solution.

Recall that $u(x,t)=v\big(\xi(x,t),\eta(x,t)\big)\text{.}$ By the chain rule

\begin{align*} \pdiff{u}{x}(x,t) &= \pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{x} +\pdiff{v}{\eta}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{x}\\ &= \pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big) +\pdiff{v}{\eta}\big(\xi(x,t),\eta(x,t)\big)\\ \pdiff{u}{t}(x,t) &= \pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{t} +\pdiff{v}{\eta}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{t}\\ &= -c\pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big) +c\pdiff{v}{\eta}\big(\xi(x,t),\eta(x,t)\big) \end{align*}

Again by the chain rule

\begin{align*} \frac{\partial^2 u}{\partial x^2}(x,t) &=\color{blue}{ \pdiff{}{x} \Big[\pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big)\Big]} +\color{red}{\pdiff{}{x} \Big[\pdiff{v}{\eta}\big(\xi(x,t),\eta(x,t)\big)\Big]}\cr &=\color{blue}{ \frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{x} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{x}}\\ &\hskip0.2in +\color{red}{\frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{x} +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{x}}\\ &=\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) +2\frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big)\\ &\hskip0.2in +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{align*}

and

\begin{align*} \frac{\partial^2 u}{\partial t^2}(x,t) &=-c \color{blue}{\pdiff{}{t} \Big[\pdiff{v}{\xi}\big(\xi(x,t),\eta(x,t)\big)\Big]} +c\color{red}{\pdiff{}{t} \Big[ \frac{\partial v}{\partial\eta} \big(\xi(x,t),\eta(x,t)\big)\Big]}\\ &=-c\color{blue}{\Big[\frac{\partial^2 v}{\partial \xi^2} \big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{t} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{t}\Big]}\\ &\hskip0.2in +c\color{red}{\Big[\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) \pdiff{\xi}{t} +\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \pdiff{\eta}{t}\Big]}\\ &=c^2\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) -2c^2\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)\\ &\hskip0.2in +c^2\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{align*}

so that

\begin{equation*} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t) =4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) \end{equation*}

Hence

\begin{align*} &\frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 \text{ for all }(x,t)\\ &\iff 4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)=0\hbox{ for all }(x,t)\\ &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta) \end{align*}

(b) Now $\frac{\partial^2 v\,}{\partial\xi\partial\eta}\big(\xi,\eta\big) =\frac{\partial}{\partial\xi} \Big[\frac{\partial v}{\partial\eta}\Big]=0\text{.}$ Temporarily rename $\pdiff{v}{\eta}=w\text{.}$ The equation $\pdiff{w}{\xi}(\xi,\eta)=0$ says that, for each fixed $\eta\text{,}$ $w(\xi,\eta)$ is a constant. The value of the constant may depend on $\eta\text{.}$ That is, $\pdiff{v}{\eta}(\xi,\eta)=w(\xi,\eta) =H(\eta)\text{,}$ for some function $H\text{.}$ (As a check, observe that $\pdiff{}{\xi}H(\eta)=0\text{.}$) So the derivative of $v$ with respect to $\eta\text{,}$ (viewing $\xi$ as a constant) is $H(\eta)\text{.}$

Let $G(\eta)$ be any function whose derivative is $H(\eta)$ (i.e. an indefinite integral of $H(\eta)$). Then $\pdiff{}{\eta}\big[v(\xi,\eta)-G(\eta)]=H(\eta)-H(\eta)=0\text{.}$ This is the case if and only if, for each fixed $\xi\text{,}$ $v(\xi,\eta)-G(\xi,\eta)$ is a constant, independent of $\eta\text{.}$ That is, if and only if

\begin{equation*} v(\xi,\eta)-G(\eta)=F(\xi) \end{equation*}

for some function $F\text{.}$ Hence

\begin{align*} \frac{\partial^2 u}{\partial x^2}(x,t) &-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 \iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta)\\ &\iff v(\xi,\eta)=F(\xi)+G(\eta)\text{ for some functions $F$ and $G$}\\ &\iff u(x,t)=v\big(\xi(x,t),\eta(x,t)\big) =F\big(\xi(x,t)\big)+G\big(\eta(x,t)\big)\\ &\phantom{\iff u(x,t)\,}=F(x-ct)+G(x+ct) \end{align*}

(c) We’ll give the interpretation of $F(x-ct)\text{.}$ The case $G(x+ct)$ is similar. Suppose that $u(x,t)=F(x-ct)\text{.}$ Think of $u(x,t)$ as the height of water at position $x$ and time $t\text{.}$ Pick any number $z\text{.}$ All points $(x,t)$ in space time for which $x-ct=z$ have the same value of $u\text{,}$ namely $F(z)\text{.}$ So if you move so that your position is $x=z+ct$ (i.e. you move the right with speed $c$) you always see the same wave height. Thus $F(x-ct)$ represents a wave moving to the right with speed $c\text{.}$

Similarly, $G(x+ct)$ represents a wave moving to the left with speed $c\text{.}$

2.4.5.27.

Solution.

(a) We are told to evaluate $\pdiff{y}{z}\text{.}$ So $y$ has to be a function of $z$ and possibly some other variables. We are also told that $x\text{,}$ $y\text{,}$ and $z$ are related by the single equation $e^{yz}-x^2 z \ln y = \pi\text{.}$ So we are to think of $x$ and $z$ as being independent variables and think of $y(x,z)$ as being determined by solving $e^{yz}-x^2 z \ln y = \pi$ for $y$ as a function of $x$ and $z\text{.}$ That is, the function $y(x,z)$ obeys

\begin{equation*} e^{y(x,z)\,z}-x^2 z \ln y(x,z) = \pi \end{equation*}

for all $x$ and $z\text{.}$ Applying $\pdiff{}{z}$ to both sides of this equation gives

\begin{align*} &\left[y(x,z)+z\pdiff{y}{z}(x,z)\right]e^{y(x,z)\,z}-x^2 \ln y(x,z) -x^2 z\frac{1}{y(x,z)}\pdiff{y}{z}(x,z) = 0\\ & \implies \pdiff{y}{z}(x,z) =\frac{x^2 \ln y(x,z)-y(x,z)e^{y(x,z)\,z}}{ze^{y(x,z)\,z}-\frac{x^2 z}{y(x,z)}} \end{align*}

(b) We are told to evaluate $\diff{y}{x}\text{.}$ So $y$ has to be a function of the single variable $x\text{.}$ We are also told that $x$ and $y$ are related by $F(x,y,x^2-y^2)=0\text{.}$ So the function $y(x)$ has to obey

\begin{equation*} F\big(x,y(x),x^2-y(x)^2\big)=0 \end{equation*}

for all $x\text{.}$ Applying $\diff{}{x}$ to both sides of that equation and using the chain rule gives

\begin{align*} &F_1\big(x,y(x),x^2-y(x)^2\big)\,\ \diff{x}{x} +F_2\big(x,y(x),x^2-y(x)^2\big)\ \diff{y}{x}(x)\\ &\hskip1in +F_3\big(x,y(x),x^2-y(x)^2\big)\,\diff{}{x}\left[x^2-y(x)^2\right]= 0\\ \implies &F_1\big(x,y(x),x^2-y(x)^2\big) +F_2\big(x,y(x),x^2-y(x)^2\big)\ \diff{y}{x}(x)\\ &\hskip1in +F_3\big(x,y(x),x^2-y(x)^2\big)\,\left[2x-2y(x)\diff{y}{x}(x)\right]= 0\\ \implies & \diff{y}{x}(x) =-\frac{F_1\big(x,y(x),x^2-y(x)^2\big)+2x\,F_3\big(x,y(x),x^2-y(x)^2\big)} {F_2\big(x,y(x),x^2-y(x)^2\big)-2y(x)\,F_3\big(x,y(x),x^2-y(x)^2\big)} \end{align*}

(c) The hard part of this question is figuring out what it is that we are to compute. We are asked to find some derivative of some function $y\text{.}$ But what function? Four variables appear in this question. Namely $x\text{,}$ $y\text{,}$ $u$ and $v\text{.}$ But we are not free to assign arbitrary values to all four of them. They have to be related by the two equations $xyuv=1$ and $x+y+u+v=0\text{.}$ If we assign values to any two of $x\text{,}$ $y\text{,}$ $u$ and $v\text{,}$ the values of the other two are to be determined by solving $xyuv=1\text{,}$ $x+y+u+v=0\text{.}$ That is, we may choose any two of $x\text{,}$ $y\text{,}$ $u$ and $v$ to be independent variables (i.e. variables that may be assigned any value). Then the other two variables are functions of those independent variables that are determined by solving the given equations.

We are told to evaluate $\left(\pdiff{y}{x}\right)_u\text{.}$ According to Definition 2.2.2, $\left(\pdiff{y}{x}\right)_u$ is the partial derivative of $y$ with respect to $x$ with $u$ being held fixed. So $x$ and $u$ have to be independent variables and $y$ has to be a function of $x$ and $u\text{.}$ The fourth variable $v$ also has to be a function of $x$ and $u\text{.}$ The functions $y(x,u)$ and $v(x,u)$ must obey

\begin{align*} x\,y(x,u)\,u\,v(x,u)&=1\\ x+y(x,u)+u+v(x,u)&=0 \end{align*}

for all $x$ and $u\text{.}$ Applying $\pdiff{}{x}$ to both sides of both of these equations gives

\begin{align*} y\,u\,v\ +\ x\,\pdiff{y}{x}\,u\,v\ +\ x\,y\,u\,\pdiff{v}{x}&=0\\ 1+\pdiff{y}{x}+0+\pdiff{v}{x}&=0 \end{align*}

Substituting, $\pdiff{v}{x}=-1-\pdiff{y}{x}\text{,}$ from the second equation, into the first equation gives

\begin{gather*} y\,u\,v\ +\ x\,\pdiff{y}{x}\,u\,v-x\,y\,u \left(1+\pdiff{y}{x}\right)=0 \end{gather*}

Now $u$ cannot be $0$ because $x\,y(x,u)\,u\,v(x,u)=1\text{.}$ So

\begin{align*} &y\,v\ +\ x\,\pdiff{y}{x}\,v-x\,y\left(1+\pdiff{y}{x}\right)=0\\ &\implies \left(\pdiff{y}{x}\right)_u\!\!(x,u) =\frac{y(x,u)\,v(x,u)-x\,y(x,u)}{x\,y(x,u)-x\,v(x,u)} \end{align*}

2.5 Tangent Planes and Normal Lines
2.5.3 Exercises

2.5.3.1.

Solution.

Write $F(x,y,z) = x^2+y^2+(z-1)^2-1$ and $G(x,y,z) = x^2+y^2+(z+1)^2-1\text{.}$ Let $S_1$ denote the surface $F(x,y,z)=0$ and $S_2$ denote the surface $G(x,y,z)=0\text{.}$ First note that $F(0,0,0)=G(0,0,0)=0$ so that the point $(0,0,0)$ lies on both $S_1$ and $S_2\text{.}$ The gradients of $F$ and $G$ are

\begin{align*} \vnabla F(x,y,z) &=\llt\pdiff{F}{x}(x,y,z)\,,\, \pdiff{F}{y}(x,y,z)\,,\, \pdiff{F}{z}(x,y,z)\rgt\\ &=\llt 2x\,,\,2y\,,\,2(z-1)\rgt\\ \vnabla G(x,y,z) &=\llt\pdiff{G}{x}(x,y,z)\,,\, \pdiff{G}{y}(x,y,z)\,,\, \pdiff{G}{z}(x,y,z)\rgt\\ &=\llt 2x\,,\,2y\,,\,2(z+1)\rgt \end{align*}

In particular,

\begin{equation*} \vnabla F(0,0,0)=\llt 0,0,-2\rgt\qquad \vnabla G(0,0,0)=\llt 0,0,2\rgt \end{equation*}

so that the vector $\hk=-\frac{1}{2}\vnabla F(0,0,0) =\frac{1}{2}\vnabla G(0,0,0)$ is normal to both surfaces at $(0,0,0)\text{.}$ So the tangent plane to both $S_1$ and $S_2$ at $(0,0,0)$ is

\begin{equation*} \hk\cdot\llt x-0,y-0,z-0\rgt=0\qquad\text{or}\qquad z=0 \end{equation*}

Denote by $P$ the plane $z=0\text{.}$ Thus $S_1$ is tangent to $P$ at $(0,0,0)$ and $P$ is tangent to $S_2$ at $(0,0,0)\text{.}$ So it is reasonable to say that $S_1$ and $S_2$ are tangent at $(0,0,0)\text{.}$

2.5.3.2.

Solution.

Denote by $S$ the surface $G(x,y,z)=0$ and by $C$ the parametrized curve $\vr(t)=\big(x(t),y(t),z(t)\big)\text{.}$ To start, we’ll find the tangent plane to $S$ at $\vr_0$ and the tangent line to $C$ at $\vr_0\text{.}$

The tangent vector to $C$ at $\vr_0$ is $\llt x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \rgt\text{,}$ so the parametric equations for the tangent line to $C$ at $\vr_0$ are
\begin{equation*} x-x_0 = t x'(t_0)\qquad y-y_0 = t y'(t_0)\qquad z-z_0 = t z'(t_0) \tag{$E_1$} \end{equation*}
The gradient $\llt\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\rgt$ is a normal vector to the surface $S$ at $(x_0,y_0,z_0)\text{.}$ So the tangent plane to the surface $S$ at $(x_0,y_0,z_0)$ is
\begin{equation*} \llt\pdiff{G}{x}, \pdiff{G}{y}, \pdiff{G}{z}\rgt \cdot \llt x-x_0,y-y_0,z-z_0\rgt = 0 \end{equation*}
with the derivatives of $G$ evaluated at $\big(x_0,y_0,z_0\big)\text{,}$ or
\begin{align*} \pdiff{G}{x}\big( x_0,y_0,z_0\big)(x-x_0) &+\pdiff{G}{y}\big( x_0,y_0,z_0\big)(y-y_0)\\ &+\pdiff{G}{z}\big( x_0,y_0,z_0\big)(z-z_0) = 0 \tag{$E_2$} \end{align*}

Next, we’ll show that the tangent vector $\llt x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \rgt$ to $C$ at $\vr_0$ and the normal vector $\llt\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\rgt$ to $S$ at $\vr_0$ are perpendicular to each other. To do so, we observe that, for every $t\text{,}$ the point $\big(x(t),y(t),z(t)\big)$ lies on the surface $G(x,y,z)=0$ and so obeys

\begin{gather*} G\big(x(t),y(t),z(t)\big) =0 \end{gather*}

Differentiating this equation with respect to $t$ gives, by the chain rule,

\begin{align*} 0&= \diff{}{t}G\big(x(t),y(t),z(t)\big)\\ &=\pdiff{G}{x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\pdiff{G}{y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t)\\ &\hskip0.5in +\pdiff{G}{z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{align*}

Then setting $t=t_0$ gives

\begin{align*} \pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) &+\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0)\\ &+\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$} \end{align*}

Finally, we are in a position to show that if $(x,y,z)$ is any point on the tangent line to $C$ at $\vr_0\text{,}$ then $(x,y,z)$ is also on the tangent plane to $S$ at $\vr_0\text{.}$ As $(x,y,z)$ is on the tangent line to $C$ at $\vr_0$ then there is a $t$ such that, by $(E_1)\text{,}$

\begin{align*} &\pdiff{G}{x}\big(x_0,y_0,z_0\big)\color{blue}{\{x\!-\!x_0\}} +\pdiff{G}{y}\big(x_0,y_0,z_0\big)\color{blue}{\{y\!-\!y_0\}} +\pdiff{G}{z}\big(x_0,y_0,z_0\big)\color{blue}{\{z\!-\!z_0\}}\\ &=\pdiff{G}{x}\big(x_0,y_0,z_0\big) \color{blue}{\big\{t x'(t_0)\big\}} +\pdiff{G}{y}\big(x_0,y_0,z_0\big) \color{blue}{\big\{ty'(t_0)\big\}}\\ &\hskip1in +\pdiff{G}{z}\big(x_0,y_0,z_0\big) \color{blue}{\big\{tz'(t_0)\big\}}\\ &=\color{blue}{t}\left[\pdiff{G}{x}\big(x_0,y_0,z_0\big) \color{blue}{x'(t_0)} +\pdiff{G}{y}\big(x_0,y_0,z_0\big) \color{blue}{y'(t_0)} +\pdiff{G}{z}\big(x_0,y_0,z_0\big)\ \color{blue}{z'(t_0)} \right]\\ &=0 \end{align*}

by $(E_3)\text{.}$ That is, $(x,y,z)$ obeys the equation, $(E_2)\text{,}$ of the tangent plane to $S$ at $\vr_0$ and so is on that tangent plane. So the tangent line to $C$ at $\vr_0$ is contained in the tangent plane to $S$ at $\vr_0\text{.}$

2.5.3.3.

Solution.

Use $S_1$ to denote the surface $F(x,y,z)=0\text{,}$ $S_2$ to denote the surface $G(x,y,z)=0$ and $C$ to denote the curve of intersection of $S_1$ and $S_2\text{.}$

Since $C$ is contained in $S_1\text{,}$ the tangent line to $C$ at $(x_0,y_0,z_0)$ is contained in the tangent plane to $S_1$ at $(x_0,y_0,z_0)\text{,}$ by Q[2.5.3.2]. In particular, any tangent vector, $\vt\text{,}$ to $C$ at $(x_0,y_0,z_0)$ must be perpendicular to $\vnabla F(x_0,y_0,z_0)\text{,}$ the normal vector to $S_1$ at $(x_0,y_0,z_0)\text{.}$ (See Theorem 2.5.1.)
Since $C$ is contained in $S_2\text{,}$ the tangent line to $C$ at $(x_0,y_0,z_0)$ is contained in the tangent plane to $S_2$ at $(x_0,y_0,z_0)\text{,}$ by Q[2.5.3.2]. In particular, any tangent vector, $\vt\text{,}$ to $C$ at $(x_0,y_0,z_0)$ must be perpendicular to $\vnabla G(x_0,y_0,z_0)\text{,}$ the normal vector to $S_2$ at $(x_0,y_0,z_0)\text{.}$

So any tangent vector to $C$ at $(x_0,y_0,z_0)$ must be perpendiular to both $\vnabla F(x_0,y_0,z_0)$ and $\vnabla G(x_0,y_0,z_0)\text{.}$ One such tangent vector is

\begin{gather*} \vt = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{gather*}

(Because the vectors $\vnabla F(x_0,y_0,z_0)$ and $\vnabla G(x_0,y_0,z_0)$ are nonzero and not parallel, $\vt$ is nonzero.) So the normal plane in question passes through $(x_0,y_0,z_0)$ and has normal vector $\vn=\vt\text{.}$ Consquently, the normal plane is

\begin{align*} &\vn\cdot\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =0\\ &\qquad\text{where } \vn=\vt=\vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{align*}

2.5.3.4.

Solution.

Use $S_1$ to denote the surface $z=f(x,y)\text{,}$ $S_2$ to denote the surface $z=g(x,y)$ and $C$ to denote the curve of intersection of $S_1$ and $S_2\text{.}$

Since $C$ is contained in $S_1\text{,}$ the tangent line to $C$ at $(x_0,y_0,z_0)$ is contained in the tangent plane to $S_1$ at $(x_0,y_0,z_0)\text{,}$ by Q[2.5.3.2]. In particular, any tangent vector, $\vt\text{,}$ to $C$ at $(x_0,y_0,z_0)$ must be perpendicular to $-f_x(x_0,y_0)\,\hi -f_y(x_0,y_0)\,\hj+\hk\text{,}$ the normal vector to $S_1$ at $(x_0,y_0,z_0)\text{.}$ (See Theorem 2.5.1.)
Since $C$ is contained in $S_2\text{,}$ the tangent line to $C$ at $(x_0,y_0,z_0)$ is contained in the tangent plane to $S_2$ at $(x_0,y_0,z_0)\text{,}$ by Q[2.5.3.2]. In particular, any tangent vector, $\vt\text{,}$ to $C$ at $(x_0,y_0,z_0)$ must be perpendicular to $-g_x(x_0,y_0)\,\hi -g_y(x_0,y_0)\,\hj+\hk\text{,}$ the normal vector to $S_2$ at $(x_0,y_0,z_0)\text{.}$

So any tangent vector to $C$ at $(x_0,y_0,z_0)$ must be perpendicular to both of the vectors $-f_x(x_0,y_0)\,\hi-f_y(x_0,y_0)\,\hj+\hk$ and $-g_x(x_0,y_0)\,\hi -g_y(x_0,y_0)\,\hj+\hk\text{.}$ One such tangent vector is

\begin{align*} &\vt = \big[-f_x(x_0,y_0)\,\hi - f_y(x_0,y_0)\,\hj+\hk\big]\times \big[-g_x(x_0,y_0)\,\hi - g_y(x_0,y_0)\,\hj+\hk\big]\\ &=\det\left[\begin{matrix} \hi & \hj & \hk\\ -f_x(x_0,y_0) & -f_y(x_0,y_0) & 1\\ -g_x(x_0,y_0) & -g_y(x_0,y_0) & 1 \end{matrix}\right]\\ &=\big\lt g_y(x_0,y_0)-f_y(x_0,y_0)\,,\, f_x(x_0,y_0)-g_x(x_0,y_0)\,,\,\\ &\hskip2in f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\big\gt \end{align*}

So the tangent line in question passes through $(x_0,y_0,z_0)$ and has direction vector $\vd=\vt\text{.}$ Consquently, the tangent line is

\begin{equation*} \llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt = t\,\vd \end{equation*}

\begin{align*} x&=x_0+t\big[g_y(x_0,y_0)-f_y(x_0,y_0)\big]\\ y&=y_0+t\big[f_x(x_0,y_0)-g_x(x_0,y_0)\big]\\ z&=z_0+ t\big[f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\big] \end{align*}

2.5.3.5. (✳).

Solution.

We are going to use Theorem 2.5.1. To do so, we need the first order derivatives of $f(x,y)$ at $(x,y)=(-1,1)\text{.}$ So we find them first.

\begin{alignat*}{2} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9}\\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{alignat*}

The tangent plane is

\begin{align*} z&=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1)\\ & =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1)\\ &=\frac{2}{9}-\frac{2}{9}x-\frac{1}{9}y \end{align*}

or $2x+y+9z=2\text{.}$

2.5.3.6. (✳).

Solution.

The equation of the given surface is of the form $G(x,y,z)=9$ with $G(x,y,z) =\frac{27}{\sqrt{x^2+y^2+z^2+3}}\text{.}$ So, by Theorem 2.5.5, a normal vector to the surface at $(2,1,1)$ is

\begin{align*} \vnabla G(2,1,1) &=-\frac{1}{2}\ \frac{27}{(x^2+y^2+z^2+3)^{3/2}}\big(2x\,,\,2y\,,\,2z\big) \bigg|_{(x,y,z)=(2,1,1)}\\ &=-\llt 2\,,\,1\,,\,1\rgt \end{align*}

and the equation of the tangent plane is

\begin{equation*} -\llt 2\,,\,1\,,\,1\rgt\cdot \llt x-2\,,\,y-1\,,\,z-1\rgt=0\qquad\text{or}\qquad 2x+y+z = 6 \end{equation*}

2.5.3.7.

Solution.

(a) The specified graph is $z=f(x,y)=x^2-y^2$ or $F(x,y,z)=x^2-y^2-z=0\text{.}$ Observe that $f(-2,1)=3\text{.}$ The vector

\begin{align*} \vnabla F(-2,1,3) &= \llt F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\rgt\Big|_{(x,y,z)=(-2,1,3)}\\ &= \llt 2x,-2y,-1\rgt\Big|_{(x,y,z)=(-2,1,3)}\\ &= \llt -4,-2,-1\rgt \end{align*}

is a normal vector to the graph at $(-2,1,3)\text{.}$ So the tangent plane is

\begin{equation*} -4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3 \end{equation*}

and the normal line is

\begin{equation*} \llt x,y,z\rgt=\llt -2,1,3\rgt+t\llt 4,2,1\rgt \end{equation*}

(b) The specified graph is $z=f(x,y)=e^{xy}$ or $F(x,y,z)=e^{xy}-z=0\text{.}$ Observe that $f(2,0)=1\text{.}$ The vector

\begin{align*} \vnabla F(2,0,1) &= \llt F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\rgt\Big|_{(x,y,z)=(2,0,1)}\\ &= \llt ye^{xy},xe^{xy},-1\rgt\Big|_{(x,y,z)=(2,0,1)}\\ &= \llt 0,2,-1\rgt \end{align*}

is a normal vector to the graph at $(2,0,1)\text{.}$ So the tangent plane is

\begin{equation*} 0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1 \end{equation*}

and the normal line is

\begin{equation*} \llt x,y,z\rgt=\llt 2,0,1\rgt+t\llt 0,2,-1\rgt \end{equation*}

2.5.3.8. (✳).

Solution.

We may use $G(x,y,z) = xyz^2 + y^2 z^3 - 3 - x^2 = 0$ as an equation for the surface. Note that $(-1,1,2)$ really is on the surface since

\begin{gather*} G(-1,1,2) = (-1)(1)(2)^2 + (1)^2 (2)^3 - 3 - (-1)^2 = -4 + 8 - 3 - 1 =0 \end{gather*}

By Theorem 2.5.5, since

\begin{alignat*}{2} G_x(x,y,z)&=yz^2 -2x \qquad & G_x(-1,1,2)&=6\\ G_y(x,y,z)&=xz^2 +2yz^3 \qquad & G_y(-1,1,2)&=12\\ G_z(x,y,z)&=2xyz+3y^2z^2 \qquad & G_z(-1,1,2)&=8 \end{alignat*}

one normal vector to the surface at $(-1,1,2)$ is $\vnabla G(-1,1,2) = \llt 6\,,\,12\,,\,8\rgt$ and an equation of the tangent plane to the surface at $(-1,1,2)$ is

\begin{gather*} \llt 6\,,\,12\,,\,8\rgt \cdot \llt x+1\,,\,y-1\,,\,z-2\rgt = 0\qquad\text{or}\qquad 6x+12 y+ 8z = 22 \end{gather*}

\begin{equation*} z = -\frac{3}{4} x- \frac{3}{2} y +\frac{11}{4} \end{equation*}

2.5.3.9. (✳).

Solution.

(a) The surface is $G(x,y,z)=z-x^2+2xy-y^2=0\text{.}$ When $x=a$ and $y=2a$ and $(x,y,z)$ is on the surface, we have $z= a^2-2(a)(2a) +(2a)^2=a^2\text{.}$ So, by Theorem 2.5.5, a normal vector to this surface at $(a,2a,a^2)$ is

\begin{gather*} \vnabla G(a,2a,a^2) = \llt -2x+2y\,,\,2x-2y\,,\,1\rgt\Big|_{(x,y,z)=(a,2a,a^2)} = \llt 2a\,,\,-2a\,,\,1\rgt \end{gather*}

and the equation of the tangent plane is

\begin{gather*} \llt 2a\,,\,-2a\,,\,1\rgt\cdot\llt x-a\,,\,y-2a\,,\,z-a^2\rgt =0 \quad\text{or}\quad 2ax -2ay +z = -a^2 \end{gather*}

(b) The two planes are parallel when their two normal vectors, namely $\llt 2a\,,\,-2a\,,\,1\rgt$ and $\llt 1\,,\,-1\,,\,1\rgt\text{,}$ are parallel. This is the case if and only if $a=\frac{1}{2}\text{.}$

2.5.3.10. (✳).

Solution.

The first order partial derivatives of $f$ are

\begin{alignat*}{2} f_x(x,y) & = -\frac{4xy}{{(x^2+y^2)}^2}\quad & f_x(-1,2) & = \frac{8}{25}\\ f_y(x,y) & = \frac{2}{x^2+y^2}-\frac{4y^2}{{(x^2+y^2)}^2}\quad & f_y(-1,2) & = \frac{2}{5}-\frac{16}{25} =-\frac{6}{25} \end{alignat*}

So, by Theorem 2.5.1, a normal vector to the surface at $(x,y)=(-1,2)$ is $\llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt\text{.}$ As $f(-1,2)= \frac{4}{5}\text{,}$ the tangent plane is

\begin{align*} &\llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt\cdot\llt x+1\,,\,y-2\,,\, z -\frac{4}{5}\rgt=0\\ & \quad \text{or}\quad \frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5} \end{align*}

and the normal line is

\begin{gather*} \llt x,y,z\rgt = \llt -1,2,\frac{4}{5}\rgt +t \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt \end{gather*}

2.5.3.11. (✳).

Solution.

A normal vector to the surface $x^2 + 9y^2 + 4z^2 = 17$ at the point $(x,y,z)$ is $\llt 2x\,,\, 18y\,,\,8z\rgt\text{.}$ A normal vector to the plane $x - 8z = 0$ is $\llt 1\,,\,0\,,\,-8\rgt\text{.}$ So we want $\llt 2x\,,\, 18y\,,\,8z\rgt$ to be parallel to $\llt 1\,,\,0\,,\,-8\rgt\text{,}$ i.e. to be a nonzero constant times $\llt 1\,,\,0\,,\,-8\rgt\text{.}$ This is the case whenever $y=0$ and $z=-2x$ with $x\ne 0\text{.}$ In addition, we want $(x,y,z)$ to lie on the surface $x^2 + 9y^2 + 4z^2 = 17\text{.}$ So we want $y=0\text{,}$ $z=-2x$ and

\begin{gather*} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{gather*}

So the allowed points are $\pm(1,0,-2)\text{.}$

2.5.3.12. (✳).

Solution.

The equation of $S$ is of the form $G(x,y,z) = x^2 + 2y^2 + 2y-z = 1\text{.}$ So one normal vector to $S$ at the point $(x_0,y_0,z_0)$ is

\begin{equation*} \vnabla G(x_0,y_0,z_0) = 2x_0\,\hi + (4y_0+2)\,\hj -\hk \end{equation*}

and the normal line to $S$ at $(x_0,y_0,z_0)$ is

\begin{equation*} (x,y,z) = (x_0,y_0,z_0) +t\llt 2x_0\,,\,4y_0+2\,,\, -1\rgt \end{equation*}

For this normal line to pass through the origin, there must be a $t$ with

\begin{gather*} (0,0,0) = (x_0,y_0,z_0) +t\llt 2x_0\,,\,4y_0+2\,,\, -1\rgt \end{gather*}

\begin{align*} x_0 + 2x_0\,t & =0 \tag{E1}\\ y_0 +(4y_0+2)t &=0 \tag{E2}\\ z_0 -t &=0 \tag{E3} \end{align*}

Equation (E3) forces $t=z_0\text{.}$ Substituting this into equations (E1) and (E2) gives

\begin{align*} x_0(1+2z_0) & =0 \tag{E1}\\ y_0 +(4y_0+2)z_0 &=0 \tag{E2} \end{align*}

The question specifies that $x_0\ne 0\text{,}$ so (E1) forces $z_0=-\frac{1}{2}\text{.}$ Substituting $z_0=-\frac{1}{2}$ into (E2) gives

\begin{equation*} -y_0-1=0 \implies y_0=-1 \end{equation*}

Finally $x_0$ is determined by the requirement that $(x_0,y_0,z_0)$ must lie on $S$ and so must obey

\begin{align*} z_0 = x_0^2 + 2y_0^2 + 2y_0 - 1 &\implies -\frac{1}{2} = x_0^2 + 2(-1)^2 +2(-1)-1\\ &\implies x_0^2 = \frac{1}{2} \end{align*}

So the allowed points $P$ are $\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)$ and $\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)\text{.}$

2.5.3.13. (✳).

Solution.

Let $(x_0,y_0,z_0)$ be a point on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0\text{.}$ A normal vector to the plane $2x-y+z=0$ is $\llt 2,-1,1\rgt\text{.}$ Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and $\vnabla G(x,y,z) = \llt 8x,2y,-2z\rgt\text{,}$ a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is $\vnabla G(x_0,y_0,z_0)=\llt 8x_0,2y_0,-2z_0\rgt\text{.}$ So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

\begin{align*} &\llt 8x_0,2y_0,-2z_0\rgt =t\llt 2,-1,1\rgt \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\half,-1,-1\big) \end{align*}

2.5.3.14.

Solution.

One vector normal to the surface $F(x,y,z)=4x^2+9y^2-z^2=0$ at $(2,1,-5)$ is

\begin{gather*} \vnabla F(2,1,-5) = \llt 8x,18y,-2z\rgt\Big|_{(2,1,-5)}=\llt 16,18,10\rgt \end{gather*}

One vector normal to the surface $G(x,y,z)=6x+3y+2z=5$ at $(2,1,-5)$ is

\begin{gather*} \vnabla G(2,1,-5) = \llt 6,3,2\rgt \end{gather*}

Now

The curve lies in the surface $z^2=4x^2+9y^2\text{.}$ So the tangent vector to the curve at $(2,1,-5)$ is perpendicular to the normal vector $\frac{1}{2}\llt 16,18,10\rgt=\llt 8,9,5\rgt\text{.}$
The curve also lies in the surface $6x+3y+2z=5\text{.}$ So the tangent vector to the curve at $(2,1,-5)$ is also perpendicular to the normal vector $\llt 6,3,2\rgt\text{.}$
So the tangent vector to the curve at $(2,1,-5)$ is parallel to
\begin{align*} \llt 8,9,5\rgt\times \llt 6,3,2\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 8 & 9 & 5\\ 6 & 3 & 2 \end{matrix}\right] =\llt 3,14,-30\rgt \end{align*}

The desired vectors are

\begin{equation*} \pm\sqrt{3}\frac{\llt 3,14,-30\rgt}{|\llt 3,14,-30\rgt|} =\pm\sqrt{\frac{3}{1105}}\llt 3,14,-30\rgt \end{equation*}

2.5.3.15.

Solution.

Let $(x_0,y_0,z_0)$ be any point on the surface. A vector normal to the surface at $(x_0,y_0,z_0)$ is

\begin{align*} &\vnabla\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)}\\ &\hskip0.3in =\llt y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\rgt \end{align*}

The tangent plane to the surface at $(x_0,y_0,z_0)$ is horizontal if and only if this vector is vertical, which is the case if and only if its $x$- and $y$-components are zero, which in turn is the case if and only if

\begin{align*} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{align*}

The values of $z_0$ at these points are $0\text{,}$ $e^{-1}\text{,}$ $-e^{-1}\text{,}$ $-e^{-1}$ and $e^{-1}\text{,}$ respectively. So the horizontal tangent planes are $z=0\text{,}$ $z=e^{-1}$ and $z=-e^{-1}\text{.}$ At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}\text{,}$ respectively.

2.5.3.16. (✳).

Solution.

(a) A normal vector to the surface at $(0,2,1)$ is

\begin{align*} &\vnabla\big(xy-2x+yz+x^2+y^2+z^3-7\big)\big|_{(0,2,1)}\\ &\hskip0.5in=\llt y-2+2x\,,\,x+z+2y\,,\,y+3z^2\rgt\big|_{(0,2,1)}\\ &\hskip0.5in=\llt 0, 5, 5\rgt \end{align*}

So the tangent plane is

\begin{gather*} 0(x-0)+5(y-2)+5(z-1)=0\text{ or }y+z=3 \end{gather*}

The vector parametric equations for the normal line are

\begin{equation*} \vr(t)=\llt 0,2,1\rgt+t\llt 0,5,5\rgt \end{equation*}

(b) Differentiating

\begin{equation*} xy-2x+y\,z(x,y)+x^2+y^2+z(x,y)^3=7 \end{equation*}

gives

\begin{align*} y-2+y\,z_x(x,y)+2x+3z(x,y)^2z_x(x,y)&=0\\ &\hskip-1in\implies z_x(x,y)=\frac{2-2x-y}{y+3z(x,y)^2}\\ x+z(x,y)+y\,z_y(x,y)+2y+3z(x,y)^2z_y(x,y)&=0\\ &\hskip-1in\implies z_y(x,y)=-\frac{x+2y+z(x,y)}{y+3z(x,y)^2} \end{align*}

In particular, at $(0,2,1)\text{,}$ $z_y(0,2)=-\frac{4+1}{2+3}=-1\text{.}$

\begin{align*} &z_{xy}(x,y) =-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2}\big(1+6z(x,y)z_y(x,y)\big)\\ &=-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2} \left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right) \end{align*}

As an alternate solution, we could also differentiate $z_y$ with respect to $x\text{.}$ This gives

\begin{align*} z_{yx}(x,y) &=-\frac{1+z_x(x,y)}{y+3z(x,y)^2} +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)z_x(x,y)\\ &=-\frac{1}{y+3z(x,y)^2}\left(1+\frac{2-2x-y}{y+3z(x,y)^2}\right)\\ &\hskip0.5in +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)\frac{2-2x-y}{y+3z(x,y)^2} \end{align*}

2.5.3.17. (✳).

Solution.

(a) A vector perpendicular to $x^2+z^2=10$ at $(1,1,3)$ is

\begin{equation*} \vnabla(x^2+z^2)\big|_{(1,1,3)} =(2x\hi+2z\hk)\big|_{(1,1,3)} =2\hi+6\hk\hbox{ or } \frac{1}{2} \llt 2,0,6\rgt=\llt 1,0,3\rgt \end{equation*}

(b) A vector perpendicular to $y^2+z^2=10$ at $(1,1,3)$ is

\begin{equation*} \vnabla(y^2+z^2)\big|_{(1,1,3)} =(2y\hj+2z\hk)\big|_{(1,1,3)} =2\hj+6\hk\hbox{ or }\frac{1}{2} \llt 0,2,6\rgt=\llt 0,1,3\rgt \end{equation*}

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both $(1,0,3)$ and $(0,1,3)\text{.}$ One such vector is

\begin{equation*} \llt 0,1,3\rgt\times\llt1,0,3\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =\llt 3,3,-1\rgt \end{equation*}

(c) The specified tangent line passes through $(1,1,3)$ and has direction vector $\llt 3,3,-1\rgt$ and so has vector parametric equation

\begin{equation*} \vr(t)=\llt 1,1,3\rgt+t\llt 3,3,-1\rgt \end{equation*}

2.5.3.18. (✳).

Solution.

$\vr(t)=\llt x(t)\,,\,y(t)\,,\,z(t)\rgt$ intersects $z^3 + xyz -2 = 0$ when

\begin{align*} z(t)^3+x(t)\,y(t)\,z(t)-2=0&\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0\\ &\iff 2t^6=2\\ &\iff t=1 \end{align*}

since $t$ is required to be positive. The direction vector for the curve at $t=1$ is

\begin{equation*} \vr'(1)=3\,\hi+\hj+2\,\hk \end{equation*}

A normal vector for the surface at $\vr(1)=\llt 1,1,1\rgt$ is

\begin{equation*} \vnabla(z^3+xyz)\big|_{(1,1,1)}=[yz\hi+xz\hj+(3z^2+xy)\hk]_{(1,1,1)} =\hi+\hj+4\hk \end{equation*}

The angle $\theta$ between the curve and the normal vector to the surface is determined by

\begin{align*} \big|\llt 3,1,2\rgt\big|\,\big|\llt 1,1,4\rgt\big|\cos\theta =\llt 3,1,2\rgt \cdot\llt 1,1,4\rgt &\iff \sqrt{14}\sqrt{18}\cos\theta=12\\ &\iff \sqrt{7\times 36}\cos\theta=12\\ &\iff \cos\theta=\frac{2}{\sqrt{7}}\\ &\iff \theta=40.89^\circ \end{align*}

The angle between the curve and the surface is $90-40.89=49.11^\circ$ (to two decimal places).

2.5.3.19.

Solution.

Let $(x,y,z)$ be any point on the paraboloid $z=x^2+y^2\text{.}$ The square of the distance from $(1,1,0)$ to this point is

\begin{align*} D(x,y)&=(x-1)^2+(y-1)^2+z^2\\ &=(x-1)^2+(y-1)^2+{(x^2+y^2)}^2 \end{align*}

We wish to minimize $D(x,y)\text{.}$ That is, to find the lowest point on the graph $z=D(x,y)\text{.}$ At this lowest point, the tangent plane to $z=D(x,y)$ is horizontal. So at the minimum, the normal vector to $z=D(x,y)$ has $x$ and $y$ components zero. So

\begin{alignat*}{2} 0&=\pdiff{D}{x}(x,y)&&= 2(x-1)+2(x^2+y^2)(2x)\\ 0&=\pdiff{D}{y}(x,y)&&= 2(y-1)+2(x^2+y^2)(2y) \end{alignat*}

By symmetry (or multiplying the first equation by $y\text{,}$ multiplying the second equation by $x$ and subtracting) the solution will have $x=y$ with

\begin{equation*} 0=2(x-1)+2(x^2+x^2)(2x)=8x^3+2x-2 \end{equation*}

Observe that the value of $8x^3+2x-2=2(4x^3+x-1)$ at $x=\frac{1}{2}$ is $0\text{.}$ (See Appendix A.16 of the CLP-2 text for some useful tricks that can help you guess roots of polynomials with integer coefficients.) So $\big(x-\frac{1}{2}\big)$ is a factor of

\begin{equation*} 4x^3+x-1 ={\textstyle 4\big(x^3+\frac{x}{4}-\frac{1}{4}\big) =4\big(x-\frac{1}{2}\big)\big(x^2+\half x+\half\big)} \end{equation*}

and the minimizing $(x,y)$ obeys $x=y$ and

\begin{equation*} 0=8x^3+2x-2 =8\big(x-\half\big)\big(x^2+\half x+\half\big)=0 \end{equation*}

By the quadratic root formula, $x^2+\half x+\half$ has no real roots, so the only solution is $x=y=\half\text{,}$ $z=\big(\half\big)^2+\big(\half\big)^2=\half$ and the distance is $\sqrt{\big(\half-1\big)^2+\big(\half-1\big)^2 +\big(\half\big)^2}=\frac{\sqrt{3}}{2}\text{.}$

2.6 Linear Approximations and Error
2.6.3 Exercises

2.6.3.1.

Solution.

(a) The first order partial derivatives of $P(x,y)$ at $x=x_0$ and $y=y_0$ are

\begin{equation*} P_x(x_0,y_0) = m x_0^{m-1} y_0^n\qquad P_y(x_0,y_0) = n x_0^m y_0^{n-1} \end{equation*}

So, by (2.6.1), the linear approximation is

\begin{align*} P(x_0+\De x,y_0+\De y) &\approx P(x_0,y_0) + P_x(x_0,y_0)\,\De x + P_y(x_0,y_0)\,\De y\\ &\approx P(x_0,y_0) + mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y \end{align*}

(b) By part (a)

\begin{align*} \frac{P(x_0+\De x,y_0+\De y)-P(x_0,y_0)}{P(x_0,y_0)} &\approx \frac{mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y}{x_0^my_0^n}\\ &=m\frac{\De x}{x_0} + n\frac{\De y}{y_0} \end{align*}

Hence

\begin{align*} P_\% &\approx 100\left|m\frac{\De x}{x_0} + n\frac{\De y}{y_0}\right|\\ &\le |m| 100\left|\frac{\De x}{x_0}\right| + |n| 100\left|\frac{\De y}{y_0}\right|\\ &\le |m| x_\% + |n| y_\% \end{align*}

Warning. The answer $m\,x_\% +n\,y_\%\text{,}$ without absolute values on $m$ and $n\text{,}$ can be seriously wrong. As an example, suppose that $m=1\text{,}$ $n=-1\text{,}$ $x_0=y_0=1\text{,}$ $\De x = 0.05$ and $\De y=-0.05\text{.}$ Then

\begin{align*} P_\% &\approx 100\left|m\frac{\De x}{x_0} + n\frac{\De y}{y_0}\right|\\ &= 100\left|(1)\frac{0.05}{1} + (-1)\frac{-0.05}{1}\right|\\ & = 10\% \end{align*}

while

\begin{align*} m\,x_\% +n\,y_\% &= m\, 100\left|\frac{\De x}{x_0}\right| + n\, 100\left|\frac{\De y}{y_0}\right|\\ &= (1) 100\left|\frac{0.05}{1}\right| + (-1) 100\left|\frac{-0.05}{1}\right|\\ &= 0 \end{align*}

The point is that $m$ and $n$ being of opposite sign does not guarantee that there is a cancelation between the two terms of $m\frac{\De x}{x_0} + n\frac{\De y}{y_0}\text{,}$ because $\frac{\De x}{x_0}$ and $\frac{\De y}{y_0}$ can also be of opposite sign.

2.6.3.2.

Solution.

We used that $\diff{}{\theta}\sin\theta = \cos\theta\text{.}$ That is true only if $\theta$ is given in radians, not degrees. (See Lemma 2.8.3 and Warning 3.4.23 in the CLP-1 text.) So we have to convert $2^\circ$ to radians, which is $2\times\frac{\pi}{180}=\frac{\pi}{90}\text{.}$ The correct computation is

\begin{align*} Y\big(0.9,\tfrac{\pi}{90}\big) &= Y\big(1-0.1\,,\, 0+\tfrac{\pi}{90}\big) \approx 0\ +\ (0)\,(-0.1)\ +\ (1)\big(\tfrac{\pi}{90}\big)\\ &=\tfrac{\pi}{90} \approx 0.035 \end{align*}

Just out of general interest, $0.9\sin\tfrac{\pi}{90} =0.0314$ to four decimal places.

2.6.3.3.

Solution.

Apply the linear approximation $f(0.01,1.05)\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05)\text{,}$ with

\begin{align*} f(x,y)&=\sin(\pi xy+\ln y) & f(0,1)&=\sin 0=0\\ f_x(x,y)&=\pi y\cos(\pi xy+\ln y) & f_x(0,1)&=\pi\cos 0=\pi\\ f_y(x,y)&=\left(\pi x+\frac{1}{y}\right)\cos(\pi xy+\ln y) & f_y(0,1)&=\cos 0=1 \end{align*}

This gives

\begin{align*} f(0.01,1.05)&\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05)\\ &=0+\pi(0.01)+1(0.05)\\ &=0.01\,\pi + 0.05 \approx 0.0814 \end{align*}

2.6.3.4. (✳).

Solution.

We are going to need the first order derivatives of $f(x,y)$ at $(x,y)=(-1,1)\text{.}$ So we find them first.

The linear approximation to $f(x,y)$ about $(-1,1)$ is

\begin{align*} f(x,y)&\approx f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1)\\ & =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \end{align*}

In particular

\begin{gather*} f(-0.9,1.1)\approx \frac{1}{3} -\frac{2}{9}\,(0.1) -\frac{1}{9}\,(0.1) =\frac{27}{90}=0.3 \end{gather*}

2.6.3.5.

Solution.

Let the four numbers be $x_1\text{,}$ $x_2\text{,}$ $x_3$ and $x_4\text{.}$ Let the four rounded numbers be $x_1+\veps_1\text{,}$ $x_2+\veps_2\text{,}$ $x_3+\veps_3$ and $x_4+\veps_4\text{.}$ Then $0\le x_1,x_2,x_3,x_4\le 50$ and $|\veps_1|,|\veps_2|,|\veps_3|,|\veps_4|\le 0.05\text{.}$ If $P(x_1,x_2,x_3,x_4)=x_1x_2x_3x_4\text{,}$ then the error in the product introduced by rounding is, using the four variable variant of the linear approximation (2.6.2),

\begin{align*} &\big|P(x_1+\veps_1,x_2+\veps_2,x_3+\veps_3,x_4+\veps_4)-P(x_1,x_2,x_3,x_4)\big|\\ &\approx\Big|\pdiff{P}{x_1}(x_1,x_2,x_3,x_4)\veps_1 +\pdiff{P}{x_2}(x_1,x_2,x_3,x_4)\veps_2\\ &\hskip0.5in+\pdiff{P}{x_3}(x_1,x_2,x_3,x_4)\veps_3 +\pdiff{P}{x_4}(x_1,x_2,x_3,x_4)\veps_4\Big|\\ &=\big|x_2x_3x_4\veps_1 +x_1x_3x_4\veps_2 +x_1x_2x_4\veps_3 +x_1x_2x_3\veps_4\big|\\ &\le 4\times 50\times 50\times 50\times 0.05 =25000 \end{align*}

2.6.3.6. (✳).

Solution.

Denote by $x$ and $y$ the lengths of sides with $x=3\pm 0.1$ and $y=4\pm 0.2\text{.}$ Then the length of the hypotenuse is $f(x,y)=\sqrt{x^2+y^2}\text{.}$ Note that

\begin{alignat*}{2} f(x,y)&=\sqrt{x^2+y^2} & f(3,4)&=5\\ f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}\qquad & f_x(3,4)&=\frac{3}{5}\\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}} & f_y(3,4)&=\frac{4}{5} \end{alignat*}

By the linear approximation

\begin{align*} f(x,y)&\approx f(3,4) + f_x(3,4)\,(x-3) + f_y(3,4)\,(y-4)\\ & =5 +\frac{3}{5}\,(x-3) +\frac{4}{5}\,(y-4) \end{align*}

So the approximate maximum error in calculating the length of the hypotenuse is

\begin{gather*} \frac{3}{5}\,(0.1) +\frac{4}{5}\,(0.2) =\frac{1.1}{5} = 0.22 \end{gather*}

2.6.3.7. (✳).

Solution.

The function $R(R_1,R_2)$ is defined implictly by

\begin{equation*} \frac{1}{R(R_1,R_2)} =\frac{1}{R_1}+\frac{1}{R_2} \tag{$*$} \end{equation*}

In particular

\begin{equation*} \frac{1}{R(2,8)} = \frac{1}{2}+\frac{1}{8} = \frac{5}{8} \implies R(2,8) = \frac{8}{5} \end{equation*}

We wish to use the linear approximation

\begin{equation*} R(R_1,R_2) \approx R(2,8) + \pdiff{R}{R_1}(2,8)\,(R_1-2) + \pdiff{R}{R_2}(2,8)\,(R_2-8) \end{equation*}

To do so, we need the partial derivatives $\pdiff{R}{R_1}(2,8)$ and $\pdiff{R}{R_2}(2,8)\text{.}$ To find them, we differentiate $(*)$ with respect to $R_1$ and $R_2\text{:}$

\begin{align*} -\frac{1}{R(R_1,R_2)^2}\pdiff{R}{R_1}(R_1,R_2) &= -\frac{1}{R_1^2}\\ -\frac{1}{R(R_1,R_2)^2}\pdiff{R}{R_2}(R_1,R_2) &= -\frac{1}{R_2^2} \end{align*}

Setting $R_1=2$ and $R_2=8$ gives

\begin{alignat*}{3} -\frac{1}{(8/5)^2}\pdiff{R}{R_1}(2,8) &= -\frac{1}{4} &\quad\implies\quad & &\pdiff{R}{R_1}(2,8) &=\frac{16}{25}\\ -\frac{1}{(8/5)^2}\pdiff{R}{R_2}(2,8) &= -\frac{1}{64} &\quad\implies\quad & &\pdiff{R}{R_2}(2,8) &=\frac{1}{25} \end{alignat*}

So the specified change in $R$ is

\begin{gather*} R(1.9,8.1)-R(2,8) \approx \frac{16}{25} (-0.1) + \frac{1}{25}(0.1) =-\frac{15}{250} = -0.06 \end{gather*}

2.6.3.8.

Solution.

First, we compute the values of the partial derivatives of $R(R_1,R_2,R_3)$ at the measured values of $R_1\text{,}$ $R_2\text{,}$ $R_3\text{.}$ Applying $\pdiff{}{R_i}\text{,}$ with $i=1,2,3$ to both sides of the defining equation

\begin{equation*} \frac{1}{R(R_1,R_2,R_3)}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \end{equation*}

for $R(R_1,R_2,R_3)$ gives

\begin{align*} -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_1}(R_1,R_2,R_3) &=-\frac{1}{R_1^2}\\ -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_2}(R_1,R_2,R_3) &=-\frac{1}{R_2^2}\\ -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_3}(R_1,R_2,R_3) &=-\frac{1}{R_3^2} \end{align*}

When $R_1=25\Om\text{,}$ $R_2=40\Om$ and $R_3=50\Om$

\begin{align*} \frac{1}{R(25,40,50)}&=\frac{1}{25}+\frac{1}{40}+\frac{1}{50} =\frac{8+5+4}{200}\\ &\implies R(25,40,50)=\frac{200}{17}=11.765 \end{align*}

Substituting in these values of $R_1\text{,}$ $R_2\text{,}$ $R_3$ and $R\text{,}$

\begin{alignat*}{2} \pdiff{R}{R_1}(25,40,50) &=\frac{R(25,40,50)^2}{25^2}&=\frac{64}{17^2}&=0.221\\ \pdiff{R}{R_2}(25,40,50) &=\frac{R(25,40,50)^2}{40^2}&=\frac{25}{17^2}&=0.0865\\ \pdiff{R}{R_3}(25,40,50) &=\frac{R(25,40,50)^2}{50^2}&=\frac{16}{17^2}&=0.0554 \end{alignat*}

If the absolute errors in measuring $R_1\text{,}$ $R_2$ and $R_3$ are denoted $\veps_1\text{,}$ $\veps_2$ and $\veps_3\text{,}$ respectively, then, , using the linear approximation (2.6.2), the corresponding error $E$ in $R$ is

\begin{align*} E &= R(25+\veps_1,40+\veps_2,50+\veps_3)-R(25,40,50)\\ &\approx \pdiff{R}{R_1}(25,40,50)\veps_1 +\pdiff{R}{R_2}(25,40,50)\veps_2 +\pdiff{R}{R_3}(25,40,50)\veps_3 \end{align*}

and obeys

\begin{align*} |E|&\le \frac{64}{17^2}|\veps_1|+\frac{25}{17^2}|\veps_2| +\frac{16}{17^2}|\veps_3|\\ \text{or}\quad|E|&\le 0.221|\veps_1|+0.0865|\veps_2|+0.0554|\veps_3| \end{align*}

We are told that the percentage error in each measurement is no more that $0.5\%\text{.}$ So

\begin{align*} |\veps_1|&\le \frac{0.5}{100}25=\frac{1}{8}=0.125\\ |\veps_2|&\le \frac{0.5}{100}40=\frac{1}{5}=0.2\\ |\veps_3|&\le \frac{0.5}{100}50=\frac{1}{4}=0.25 \end{align*}

so that

\begin{align*} |E|&\le \frac{8}{17^2}+\frac{5}{17^2}+\frac{4}{17^2} =\frac{1}{17}\\ \text{or}\quad|E|&\le 0.221\times0.125+0.0865\times0.2+0.0554\times0.25 =0.059 \end{align*}

2.6.3.9.

Solution.

By the linear approximation

\begin{gather*} \De S\approx \pdiff{S}{A}(20,12)\,\De A +\pdiff{S}{W}(20,12)\,\De W \end{gather*}

with $S(A,W)=\frac{A}{A-W} =1+\frac{W}{A-W}\text{.}$ So

\begin{align*} S(A,W)&=\frac{A}{A-W} & S(20,12)&=\frac{20}{8}=\frac{5}{2}\\ S_A(A,W)&=-\frac{W}{(A-W)^2} & S_A(20,12)&=-\frac{12}{8^2}=-\frac{3}{16}\\ S_W(A,W)&=\frac{A}{(A-W)^2} & S_W(20,12)&=\frac{20}{8^2}=\frac{5}{16} \end{align*}

For any given $\De A$ and $\De W\text{,}$ the percentage error is

\begin{equation*} \left|100\frac{\De S}{S}\right| =\left|100\frac{2}{5}\Big(-\frac{3}{16}\De A+\frac{5}{16}\De W\Big)\right| \end{equation*}

We are told that $|\De A|\le 0.01$ and $|\De W|\le 0.02\text{.}$ To maximize $\big|100\frac{2}{5}\big(-\frac{3}{16}\De A+\frac{5}{16}\De W\big)\big|$ take $\De A=- 0.01$ and $\De W=+ 0.02\text{.}$ So the maximum percentage error is

\begin{equation*} 100\frac{2}{5}\left[-\frac{3}{16}(- 0.01)+\frac{5}{16}(0.02)\right] = \frac{2}{5}\times \frac{13}{16} =\frac{13}{40}=0.325\% \end{equation*}

2.6.3.10. (✳).

Solution.

The linear approximation to $P(s,r)$ at $(2,2)$ is

\begin{gather*} P(s,r)\approx P(2,2) +P_s(2,2)\,(s-2) + P_r(2,2)\,(r-2) \end{gather*}

\begin{align*} P(2,2) &= (2)(2)\big[4(2)^2-(2)^2-2\big] = 40 \qquad\text{(which we don't actually need) }\\ P_s(2,2) &= \Big[12s^2r-r^3-2r\Big]_{s=r=2} = 84\\ P_r(2,2) &= \Big[4s^3-3sr^2-2s\Big]_{s=r=2} = 4 \end{align*}

the linear approximation is

\begin{gather*} P(s,r)\approx 40 +84\,(s-2) + 4\,(r-2) \end{gather*}

Under method 1, the maximum error in $P$ will have magnitude at most (approximately)

\begin{gather*} 84(0.01) + 4(0.1) = 1.24 \end{gather*}

Under method 2, the maximum error in $P$ will have magnitude at most (approximately)

\begin{gather*} 84(0.02) + 4(0.02) = 1.76 \end{gather*}

Method 1 is better.

2.6.3.11.

Solution.

Using the four variable variant of the linear approximation (2.6.2),

\begin{align*} \De S&\approx \pdiff{S}{p}\De p +\pdiff{S}{\ell}\De \ell +\pdiff{S}{w}\De w +\pdiff{S}{h}\De h\\ & =C\frac{p\ell^4}{w h^3}\left[\frac{\De p}{p}+4\frac{\De\ell}{\ell} -\frac{\De w}{w}-3\frac{\De h}{h} \right] \end{align*}

When $w\approx0.1$ and $h\approx0.2\text{,}$

\begin{equation*} \frac{\De w}{w}\approx 10\De w\qquad 3\frac{\De h}{h}\approx 15\De h \end{equation*}

So a change in height by $\De h=\veps$ produces a change in sag of about $\De S = 15\veps$ times $-C\frac{p\ell^4}{w h^3}\text{,}$ while a change $\De w$ in width by the same $\veps$ produces a change in sag of about $\De S=10\veps$ times the same $-C\frac{p\ell^4}{w h^3}\text{.}$ The sag is more sensitive to $\De h\text{.}$

2.6.3.12. (✳).

Solution.

The first order partial derivatives of $f$ are

The linear approximation of $f(x,y)$ about $(-1,2)$ is

\begin{align*} f(x,y)&\approx f(-1,2) + f_x(-1,2)\,(x+1) + f_y(-1,2)\,(y-2)\\ &=\frac{4}{5} +\frac{8}{25}\,(x+1) - \frac{6}{25}\,(y-2) \end{align*}

In particular, for $x=-0.8$ and $y=2.1\text{,}$

\begin{align*} f(-0.8,2.1)&\approx \frac{4}{5} +\frac{8}{25}\,(0.2) - \frac{6}{25}\,(0.1)\\ &=0.84 \end{align*}

2.6.3.13. (✳).

Solution.

(a) The function $f(x,y)$ obeys

\begin{equation*} xy\,f(x,y) + x + y^2 + f(x,y)^3 =0 \tag{$*$} \end{equation*}

for all $x$ and $y$ (sufficiently close to $(-1,1)$). Differentiating $(*)$ with respect to $x$ gives

\begin{align*} &y\,f(x,y) +xy\,f_x(x,y) + 1 + 3f(x,y)^2 f_x(x,y) = 0\\ &\hskip1in\implies f_x(x,y) = -\frac{y\,f(x,y) +1}{3f(x,y)^2 +xy} \end{align*}

Without knowing $f(x,y)$ explicitly, there’s not much that we can do with this.

(b) $f(-1,1)$ obeys

\begin{equation*} (-1)(1)\,f(-1,1)\! +\! (-1) \!+\! (1)^2 \!+\! f(-1,1)^3 =0 \iff f(-1,1)^3 \!-\!f(-1,1) =0 \end{equation*}

Since $f(-1, 1) \lt 0$ we may divide this equation by $f(-1, 1) \lt 0\text{,}$ giving $f(-1,1)^2 - 1=0\text{.}$ Since $f(-1, 1) \lt 0\text{,}$ we must have $f(-1, 1)=-1\text{.}$ By part (a)

\begin{gather*} f_x(-1, 1) = -\frac{(1)\,f(-1, 1) +1}{3f(-1, 1)^2 +(-1)(1)} = 0 \end{gather*}

To get the linear approximation, we still need $f_y(-1,1)\text{.}$ Differentiating $(*)$ with respect to $y$ gives

\begin{equation*} x\,f(x,y) + xy\,f_y(x,y) + 2y + 3f(x,y)^2 f_y(x,y) =0 \end{equation*}

Then setting $x=-1\text{,}$ $y=1$ and $f(-1,1)=-1$ gives

\begin{align*} &(-1)\,(-1) + (-1)(1)\,f_y(-1,1) + 2(1) + 3(-1)^2 f_y(-1,1) =0\\ &\hskip1in\implies f_y(-1,1) =-\frac{3}{2} \end{align*}

So the linear approximation is

\begin{equation*} f(x,y) \approx f(-1,1) + f_x(-1,1)\,(x\!+\!1) + f_y(-1,1)\,(y\!-\!1) = -1 -\frac{3}{2} (y\!-\!1) \end{equation*}

\begin{gather*} f(-1.02,0.97) \approx -1 -\frac{3}{2} (0.97-1) =-0.955 \end{gather*}

2.6.3.14. (✳).

Solution.

By definition, the differential at $x=a\text{,}$ $y=b$ is

\begin{equation*} f_x(a,b)\,\dee{x} + f_y(a,b)\,\dee{y} \end{equation*}

so we have to determine the partial derivatives $f_x(a,b)$ and $f_y(a,b)\text{.}$ We are told that

\begin{equation*} e^{f(x,y)} + y\,f(x,y) = x + y \end{equation*}

for all $x$ and $y\text{.}$ Differentiating this equation with respect to $x$ and with respect to $y$ gives, by the chain rule,

\begin{gather*} e^{f(x,y)}f_x(x,y) + y\,f_x(x,y) = 1\\ e^{f(x,y)}f_y(x,y) + f(x,y) +y\,f_y(x,y) = 1 \end{gather*}

Solving the first equation for $f_x$ and the second for $f_y$ gives

\begin{align*} f_x(x,y) &= \frac{1}{e^{f(x,y)}+y}\\ f_y(x,y) &= \frac{1-f(x,y)}{e^{f(x,y)}+y} \end{align*}

So the differential at $x=a\text{,}$ $y=b$ is

\begin{equation*} \frac{\dee{x}}{e^{f(a,b)}+b} + \frac{1-f(a,b)}{e^{f(a,b)}+b} \,\dee{y} \end{equation*}

Since we can’t solve explicitly for $f(a,b)$ for general $a$ and $b\text{.}$ There’s not much more that we can do with this.

(b) In particular, when $a=1$ and $b=0\text{,}$ we have

\begin{equation*} e^{f(1,0)} + 0\,f(1,0) = 1 + 0 \implies e^{f(1,0)} = 1 \implies f(1,0)=0 \end{equation*}

and the linear approximation simpifies to

\begin{equation*} f\big(1+\dee{x}\,,\,\dee{y}\big) \approx f(1,0) + \frac{\dee{x}}{e^{f(1,0)}+0} + \frac{1-f(1,0)}{e^{f(1,0)}+0} \,\dee{y} = \dee{x} +\dee{y} \end{equation*}

Choosing $\dee{x} = -0.01$ and $\dee{y} = 0.01\text{,}$ we have

\begin{equation*} f\big(0.99\,,\,0.01\big) \approx -0.01 + 0.01 =0 \end{equation*}

2.6.3.15. (✳).

Solution.

Let $C(A,B,\theta)=\sqrt{A^2+B^2-2AB\cos\theta}\text{.}$ Then $C\big(3,4,\frac{\pi}{2}\big)=5\text{.}$ Differentiating $C^2=A^2+B^2-2AB\cos\theta$ gives

\begin{alignat*}{8} 2C\frac{\partial C}{\partial A}(A,B,\theta)&=2A-2B\cos\theta &\quad&\implies\quad & 10\frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)&=6\cr 2C\frac{\partial C}{\partial B}(A,B,\theta)&=2B-2A\cos\theta & &\implies & 10\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)&=8\cr 2C\frac{\partial C}{\partial \theta}(A,B,\theta)&=2AB\sin\theta & &\implies & 10\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)&=24\cr \end{alignat*}

Hence the approximate maximum error in the computed value of $C$ is

\begin{align*} |\De C|&\approx \left| \frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)\De A +\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)\De B +\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)\De\theta\right|\\ &\le(0.6)(0.1)+(0.8)(0.1)+(2.4)\frac{\pi}{180}\cr &=\frac{\pi}{75}+0.14\le 0.182 \end{align*}

2.6.3.16. (✳).

Solution.

Substituting $(x_0,y_0)=(3,4)$ and $(x,y)=(3.02,3.96)$ into

\begin{equation*} f(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) \end{equation*}

gives

\begin{align*} f(3.02,3.96)&\approx f(3,4)+0.02f_x(3,4)-0.04f_y(3,4)\\ &=60+0.02\left(20+\frac{36}{5}\right)-0.04\left(15+\frac{48}{5}\right)\\ &=59.560 \end{align*}

since

\begin{equation*} f_x(x,y)=y\sqrt{x^2+y^2}+\frac{x^2y}{\sqrt{x^2+y^2}}\qquad f_y(x,y)=x\sqrt{x^2+y^2}+\frac{xy^2}{\sqrt{x^2+y^2}} \end{equation*}

2.6.3.17. (✳).

Solution.

The volume of a cylinder of diameter $d$ and height $h$ is $V(d,h)=\pi\big(\frac{d}{2}\big)^2h\text{.}$ The wording of the question is a bit ambiguous in that it does not specify if the given dimensions are inside dimensions or outside dimensions. Assume that they are outside dimensions. Then the volume of the can, including the metal, is $V(8,12)$ and the volume of the interior, excluding the metal, is

\begin{align*} &V(8-2\times0.04\,,\,12-2\times 0.04)\\ &\hskip0.5in\approx V(8,12) +V_d(8,12)(-2\times 0.04) +V_h(8,12)(-2\times 0.04)\\ &\hskip0.5in=V(8,12) +\frac{1}{2}\pi\times 8\times 12\times(-2\times 0.04) +\pi\left(\frac{8}{2}\right)^2\!\!(-2\times 0.04)\\ &\hskip0.5in=V(8,12) -\pi\times 128\times 0.04 \end{align*}

So the volume of metal is approximately $\pi\times 128\times 0.04= 5.12\pi\approx 16.1$cc. (To this level of approximation, it doesn’t matter whether the dimensions are inside or outside dimensions.)

2.6.3.18. (✳).

Solution.

(a) The function $z(x,y)$ obeys

\begin{equation*} z(x,y)^3-z(x,y)+2xy-y^2=0 \end{equation*}

for all $(x,y)$ near $(2,4)\text{.}$ Differentiating with respect to $x$ and $y$

\begin{align*} 3z(x,y)^2\pdiff{z}{x}(x,y) -\pdiff{z}{x}(x,y)+2y&=0\\ 3z(x,y)^2\pdiff{z}{y}(x,y) -\pdiff{z}{y}(x,y)+2x-2y&=0 \end{align*}

Substituting in $x=2\text{,}$ $y=4$ and $z(2,4)=1$ gives

\begin{alignat*}{3} 3\pdiff{z}{x}(2,4) -\pdiff{z}{x}(2,4)+8&=0 & &\iff & \pdiff{z}{x}(2,4) & = -4\\ 3\pdiff{z}{y}(2,4) -\pdiff{z}{y}(2,4)+4-8&=0 & &\iff & \pdiff{z}{y}(2,4) & = 2 \end{alignat*}

The linear approximation is

\begin{align*} z(x,y)&\approx z(2,4)+z_x(2,4)(x-2)+z_y(2,4)(y-4)\\ &= 1-4(x-2)+2(y-4)\\ &=1-4x+2y \end{align*}

(b) Substituting in $x=2.02$ and $y=3.96$ gives

\begin{gather*} z(2.02,3.96)\approx 1-4\times0.02+2\times(-0.04) =0.84 \end{gather*}

2.6.3.19. (✳).

Solution.

(a) We are told that

\begin{equation*} z(x,y)^3 - xy\,z(x,y)^2 - 4x = 0 \end{equation*}

for all $(x,y)$ (sufficiently near $(1,1)$). Differentiating this equation with respect to $x$ gives

\begin{align*} & 3z(x,y)^2\,\pdiff{z}{x}(x,y) -y\, z(x,y)^2 -2xy\,z(x,y)\pdiff{z}{x}(x,y) - 4=0\\ & \implies \pdiff{z}{x} = \frac{4+yz^2}{3z^2-2xyz} \end{align*}

and differentiating with respect to $y$ gives

\begin{align*} & 3z(x,y)^2\,\pdiff{z}{y}(x,y) -x\, z(x,y)^2 -2xy\,z(x,y)\pdiff{z}{y}(x,y) =0\\ & \implies \pdiff{z}{y} = \frac{xz^2}{3z^2-2xyz} \end{align*}

(b) When $(x,y,z)=(1,1,2)\text{,}$

\begin{align*} \pdiff{z}{x}(1,1) &= \frac{4+(1)(2)^2}{3(2)^2-2(1)(1)(2)} = 1\\ \pdiff{z}{y}(1,1) &= \frac{(1)(2)^2}{3(2)^2-2(1)(1)(2)} = \frac{1}{2} \end{align*}

\begin{align*} z(x,y) &\approx z(1,1) + z_x(1,1)\,(x-1) + z_y\,(1,1)\,(y-1)\\ &= 2 + (x-1) +\frac{1}{2}(y-1) \end{align*}

So errors of $\pm 0.03$ in $x$ and $\pm 0.02$ in $y$ leads of errors of about

\begin{equation*} \pm\left[0.03 + \frac{1}{2}(0.02)\right] =\pm 0.04 \end{equation*}

in $z\text{.}$

(d) By the chain rule

\begin{align*} \diff{}{\theta} z\big(x(\theta),y(\theta)\big) &= z_x\big(x(\theta),y(\theta)\big)\,x'(\theta) +z_y\big(x(\theta),y(\theta)\big)\,y'(\theta)\\ &=-z_x\big(1 + \cos\theta,\sin\theta\big)\,\sin\theta +z_y\big(1 + \cos\theta,\sin\theta\big)\,\cos\theta \end{align*}

At $A\text{,}$ $x=2\text{,}$ $y=0\text{,}$ $z=2$ (since $z^3-(2)(0)z^2-4(2)=0$) and $\theta=0\text{,}$ so that

\begin{align*} \pdiff{z}{x}(2,0) &= \frac{4+(0)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{1}{3}\\ \pdiff{z}{y}(2,0) &= \frac{(2)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{2}{3} \end{align*}

and

\begin{align*} \diff{z}{\theta} &=-\frac{1}{3}\sin(0) +\frac{2}{3}\cos(0) =\frac{2}{3} \end{align*}

At $B\text{,}$ $x=1\text{,}$ $y=1\text{,}$ $z=2$ and $\theta=\frac{\pi}{2}\text{,}$ so that, by part (b),

\begin{gather*} \pdiff{z}{x}(1,1) = 1\qquad \pdiff{z}{y}(1,1) = \frac{1}{2} \end{gather*}

and

\begin{align*} \diff{z}{\theta} &=-\sin\frac{\pi}{2} +\frac{1}{2}\cos\frac{\pi}{2} = -1 \end{align*}

2.6.3.20. (✳).

Solution.

We are going to need the first order partial derivatives of $f(x,y)=ye^{-x}$ at $(x,y)=(1,e)\text{.}$ Here they are.

\begin{align*} f_x(x,y)&= -ye^{-x} & f_x(1,e)&=-e\,e^{-1}=-1\\ f_y(x,y)&=e^{-x} & f_y(1,e)&= e^{-1} \end{align*}

(a) The linear approximation to $f(x,y)$ at $(x,y)=(1,e)$ is

\begin{align*} f(x,y) &\approx f(1,e) + f_x(1,e)\,(x-1) +f_y(1,e)\,(y-e)\\ &= 1 -(x-1) +e^{-1}(y-e) \end{align*}

The maximum error is then approximately

\begin{gather*} -1(-0.1) +e^{-1}(0.1) =\frac{1+e^{-1}}{10} \end{gather*}

(b) The equation of the graph is $g(x,y,z) = f(x,y) -z =0\text{.}$ Any vector that is a nonzero constant times

\begin{gather*} \vnabla g(1,e,1) =\llt f_x(1,e)\,,\,f_y(1,e)\,,\,-1\rgt =\llt -1\,,\,e^{-1}\,,\,-1\rgt \end{gather*}

is perpendicular to $g=0$ at $(1,e,1)\text{.}$

2.6.3.21. (✳).

Solution.

(a) We are told that for all $x,y$ (with $(x,y,z)$ near $(2,-1/2,1)$), the function $z(x,y)$ obeys

\begin{gather*} z(x,y)^4 -xy^2 z(x,y)^2 +y=0 \tag{$*$} \end{gather*}

Differentiating $(*)$ with respect to $x$ gives

\begin{align*} &4z(x,y)^3\pdiff{z}{x}(x,y) - y^2z(x,y)^2 -2xy^2z(x,y)\pdiff{z}{x}(x,y) =0\\ &\implies \pdiff{z}{x}(x,y) = \frac{y^2 z(x,y)^2}{4z(x,y)^3-2xy^2z(x,y)} \end{align*}

Similarly, differentiating this equation with respect to $y$ gives

\begin{align*} &4z(x,y)^3\pdiff{z}{y}(x,y) - 2xy\,z(x,y)^2 -2xy^2z(x,y)\pdiff{z}{y}(x,y) +1=0\\ &\implies \pdiff{z}{y}(x,y) = \frac{2xy\, z(x,y)^2-1}{4z(x,y)^3-2xy^2z(x,y)} \end{align*}

(b) Substituting $(x, y, z) = (2, -1/2, 1)$ into the results of part (a) gives

\begin{align*} \pdiff{z}{x}(2,-1/2) &= \frac{1/4}{4-1} =\frac{1}{12}\\ \pdiff{z}{y}(2,-1/2) &= \frac{-2-1}{4-1}=-1 \end{align*}

\begin{align*} f(x,y) &\approx f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2)\\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{align*}

In particular

\begin{gather*} f(1.94,-0.4) \approx 1 -\frac{0.06}{12}-0.1 \end{gather*}

so that

\begin{gather*} f(1.94,-0.4) - 1 \approx -0.105 \end{gather*}

(d) The tangent plane is

\begin{align*} z&=f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2)\\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{align*}

\begin{gather*} \frac{x}{12} -y -z = -\frac{1}{3} \end{gather*}

2.6.3.22. (✳).

Solution.

(a) The linear approximation to $f(x,y)$ at $(1,3)$ is

\begin{align*} f(x,y) &\approx f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3)\\ & = 1 + 3(x-1) -2(y-3) \end{align*}

So the change is $z$ is approximately

\begin{gather*} 3(1.2-1) -2(2.6-3) = 1.4 \end{gather*}

(b) The equation of the tangent plane is

\begin{gather*} z = f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3) = 1 + 3(x-1) -2(y-3) \end{gather*}

\begin{equation*} 3x-2y-z = -4 \end{equation*}

2.6.3.23. (✳).

Solution.

Think of the volume as being the function $V(p,T)$ of pressure and temperature that is determined implicitly (at least for $p\approx 1\text{,}$ $T\approx 5$ and $V\approx 2$) by the equation

\begin{equation*} \big(pV(p,T)^2 + 16\big)\big(V(p,T) - 1\big) = T V(p,T)^2 \tag{$*$} \end{equation*}

To determine the approximate change in $V\text{,}$ we will use the linear approximation to $V(p,T)$ at $p=1\text{,}$ $T=5\text{.}$ So we will need the partial derivatives $V_p(1,5)$ and $V_T(1,5)\text{.}$ As the equation $(*)$ is valid for all $p$ near $1$ and $T$ near 5, we may differentiate $(*)$ with respect to $p\text{,}$ giving

\begin{align*} &\big(V^2 +2 pV V_p\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_p = 2T V V_p \end{align*}

and we may also differentiate $(*)$ with respect to $T\text{,}$ giving

\begin{align*} &\big(2 pV V_T\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_T = V^2 + 2T V V_T \end{align*}

In particular, when $p=1\text{,}$ $V=2\text{,}$ $T=5\text{,}$

\begin{align*} \big(4 +4 V_p(1,5)\big)\big(2 - 1\big) + \big(4 + 16\big)V_p(1,5) &= 20 V_p(1,5)\\ &\hskip-0.5in\implies V_p(1,5)=-1\\ 4 V_T(1,5)\big(2 - 1\big) + \big(4 + 16\big)V_T(1,5) &= 4 + 20 V_T(1,5)\\ &\hskip-0.5in\implies V_T(1,5)=1 \end{align*}

so that the change in $V$ is

\begin{gather*} V(1.2\,,\,5.3)-V(1,5) \approx V_p(1,5)\,(0.2) + V_T(1,5)\,(0.3) = -0.2+0.3 = 0.1 \end{gather*}

2.6.3.24. (✳).

Solution.

Since

\begin{align*} f_x(2,1) & = -2x e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = -4\\ f_y(2,1) & = \phantom{-}8y e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = 8 \end{align*}

The tangent plane to $z=f(x,y)$ at $(2,1)$ is

\begin{align*} z &= f(2,1) +f_x(2,1)\,(x-2) +f_y(2,1)\,(y-1) = 1 -4 (x-2) +8(y-1)\\ &= 1-4x +8y \end{align*}

and the tangent plane approximation to the value of $f(1.99, 1.01)$ is

\begin{gather*} f(1.99, 1.01) \approx 1 -4 (1.99-2) +8(1.01-1) = 1.12 \end{gather*}

2.6.3.25. (✳).

Solution.

(a) The linear approximation to $f(x,y)$ at $(a,b)$ is

\begin{align*} f(x,y) &\approx f(a,b) + f_x(a,b)\,(x-a) + f_y(a,b)\,(y-b)\\ &= \ln(4a^2 + b^2) + \frac{8a}{4a^2+b^2}\, (x-a) + \frac{2b}{4a^2+b^2}\, (y-b) \end{align*}

In particular, for $a=0$ and $b=1\text{,}$

\begin{align*} f(x,y) &\approx 2\, (y-1) \end{align*}

and, for $x=0.1$ and $y=1.2\text{,}$

\begin{align*} f(0.1,1.2) &\approx 0.4 \end{align*}

(b) The point $(a,b,c)$ is on the surface $z=f(x,y)$ if and only if

\begin{equation*} c = f(a,b) = \ln(4a^2 + b^2) \end{equation*}

Note that this forces $4a^2 + b^2$ to be nonzero. The tangent plane to the surface $z = f (x, y)$ at the point $(a,b,c)$ is parallel to the plane $2x + 2y - z = 3$ if and only if $\llt 2\,,\,2\,,\,-1\rgt$ is a normal vector for the tangent plane. That is, there is a nonzero number $t$ such that

\begin{gather*} \llt 2\,,\,2\,,\,-1\rgt =t\llt f_x(a,b)\,,\,f_y(a,b)\,,\ -1\rgt =t\llt \frac{8a}{4a^2+b^2} \,,\, \frac{2b}{4a^2+b^2} \,,\,-1\rgt \end{gather*}

For the $z$--coordinates to be equal, $t$ must be $1\text{.}$ Then, for the $x$-- and $y$--coordinates to be equal, we need

\begin{align*} \frac{8a}{4a^2+b^2} &= 2\\ \frac{2b}{4a^2+b^2} &= 2 \end{align*}

Note that these equations force both $a$ and $b$ to be nonzero. Dividing these equations gives $\frac{8a}{2b}=1$ and hence $b=4a\text{.}$ Substituting $b=4a$ into either of the two equations gives

\begin{gather*} \frac{8a}{20a^2}=2 \implies a=\frac{1}{5} \end{gather*}

So $a=\frac{1}{5}\text{,}$ $b=\frac{4}{5}$ and

\begin{gather*} c=\ln\left(\frac{4}{5^2}+\frac{4^2}{5^2}\right) =\ln\frac{4}{5} \end{gather*}

2.6.3.26. (✳).

Solution.

(a) The surface has equation $G(x,y,z) = x^2 z^3 + y \sin(\pi x) + y^2 =0\text{.}$ So a normal vector to the surface at $(1,1-1)$ is

\begin{align*} &\vnabla G(1,1,-1)\\ &\hskip0.25in= \big[ \big(2x z^3 +\pi y\cos(\pi x)\big)\hi +\big(\sin(\pi x) +2y\big)\hj +3z^2 x^2\,\hk\big]_{(x,y,z)=(1,1,-1)}\\ &\hskip0.25in= \big(-2-\pi\big)\hi +2\,\hj +3\,\hk \end{align*}

So the equation of the tangent plane is

\begin{align*} &\big(-2-\pi\big)(x-1) + 2(y-1) +3 (z+1) =0\\ &\text{or}\quad -(2+\pi)x +2y +3z = -\pi-3 \end{align*}

(b) The functions $z(x,y)$ obeys

\begin{gather*} x^2 z(x,y)^3 +y\sin(\pi x) +y^2 = 0 \end{gather*}

for all $x$ and $y\text{.}$ Differentiating this equation with respect to $x$ gives

\begin{gather*} 2x z(x,y)^3 + 3 x^2 z(x,y)^2 \pdiff{z}{x}(x,y) + \pi y \cos(\pi x) =0 \end{gather*}

Evaluating at $(1,1,-1)$ gives

\begin{gather*} -2 + 3 \pdiff{z}{x}(1,1) -\pi =0 \implies \pdiff{z}{x}(1,1) = \frac{\pi+2}{3} \end{gather*}

\begin{gather*} z(x,1) \approx z(1,1) +\pdiff{z}{x}(1,1)\ (x-1) \end{gather*}

gives

\begin{gather*} z(0.97,1) \approx -1 +\frac{\pi+2}{3}\ (-0.03) = -1 -\frac{\pi+2}{100} = -\frac{\pi+102}{100} \end{gather*}

2.6.3.27. (✳).

Solution.

(a) The function $F(y,z)$ obeys $F(y,z)^4+y^4+z^4+F(y,z)yz=17$ for all $y$ and $z$ near $y=1\text{,}$ $z=2\text{.}$ Applying the derivatives $\pdiff{}{y}$ and $\pdiff{}{z}$ to this equation gives

\begin{align*} 4F(y,z)^3F_y(y,z)+4y^3+F_y(y,z)yz +F(y,z)z&=0\\ 4F(y,z)^3F_z(y,z)+4z^3+F_z(y,z)yz +F(y,z)y&=0 \end{align*}

Substiututing $F(1,2)=0\text{,}$ $y=1$ and $z=2$ gives

\begin{alignat*}{3} 4+2F_y(1,2)&=0 & &\ \implies\ & &F_y(1,2)=-2\\ 32+2F_z(1,2)&=0 & &\ \implies\ & & F_z(1,2)=-16 \end{alignat*}

(b) Using the tangent plane to $x=F(y,z)$ at $y=1$ and $z=2\text{,}$ which is

\begin{equation*} x \approx F(1,2) +F_y(1,2)\,(y-1) +F_z(1,2)\,(z-2) \end{equation*}

with $y=1.01$ and $z=1.98$ gives

\begin{align*} x=F(1.01, 1.98) &\approx F(1,2)+F_y(1,2)(1.01-1)+F_z(1,2)(1.98-2)\\ &=0-2(.01)-16(-0.02) =0.3 \end{align*}

2.7 Directional Derivatives and the Gradient
2.7.2 Exercises

2.7.2.1. (✳).

Solution.

The partial derivatives, at a general point $(x,y,z)$ and also at the point of interest $(0,1,1)\text{,}$ are

\begin{alignat*}{2} f_x(x,y,z)&=yz e^{xyz}\qquad & f_x(0,1,1)&= 1\\ f_y(x,y,z)&=xz e^{xyz}\qquad & f_y(0,1,1)&= 0\\ f_z(x,y,z)&=xy e^{xyz}\qquad & f_z(0,1,1)&= 0 \end{alignat*}

So $\vnabla f(0,1,1) =\llt 1,0,0\rgt$ and the specified directional derivative is

\begin{gather*} D_{\frac{\llt 0,1,1\rgt}{\sqrt{2}}}f(0,1,1) =\llt 1,0,0\rgt\cdot\frac{\llt 0,1,1\rgt}{\sqrt{2}} =0 \end{gather*}

2.7.2.2. (✳).

Solution.

In two dimensions, write $g(x,y) = y^2+\sin(xy)\text{.}$ Then

\begin{gather*} \vnabla g =\llt g_x\,,\,g_y\rgt =\llt y\cos(xy)\,,\, 2y + x\cos(xy) \rgt \end{gather*}

In three dimensions, write $g(x,y,z) = y^2+\sin(xy)\text{.}$ Then

\begin{gather*} \vnabla g =\llt g_x\,,\,g_y\,,\,g_z\rgt =\llt y\cos(xy)\,,\, 2y + x\cos(xy)\,,\, 0 \rgt \end{gather*}

2.7.2.3.

Solution.

(a) The gradient of $f$ is $\vnabla f(x,y)=\llt 3,-4\rgt\text{.}$ So the specified rate of change is

\begin{equation*} \llt 3,-4\rgt\cdot\frac{\llt -2,0\rgt}{|\llt -2,0\rgt|}=-3 \end{equation*}

(b) The gradient of $f$ is $\vnabla f(x,y,z)=\llt -x^{-2},-y^{-2},-z^{-2}\rgt\text{.}$ In particular, the gradient of $f$ at the point $(2,-3,4)$ is $\vnabla f(2,-3,4)=\llt -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\rgt\text{.}$ So the specified rate of change is

\begin{equation*} \llt -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\rgt\cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}}= -\frac{61}{144\sqrt{3}}\approx-0.2446 \end{equation*}

2.7.2.4.

Solution.

The gradient of $f(x,y)$ is $\vnabla f(x,y)=\llt y,x\rgt\text{.}$ In particular, the gradient of $f$ at the point $(2,0)$ is $\vnabla f(2,0)=\llt 0,2 \rgt\text{.}$ So the rate of change in the direction that makes angle $\theta$ with respect to the $x$-axis, that is, in the direction $\llt \cos\theta,\sin\theta\rgt$ is

\begin{equation*} \llt \cos\theta,\sin\theta\rgt \cdot\vnabla f(2,0) =\llt \cos\theta,\sin\theta\rgt \cdot\llt 0, 2\rgt =2\sin\theta \end{equation*}

(a) To get a rate $-1\text{,}$ we need

\begin{equation*} \sin\theta=-\frac{1}{2} \implies \theta=-30^\circ,\ -150^\circ \end{equation*}

So the desired directions are

\begin{equation*} \llt \cos\theta,\sin\theta\rgt =\llt\pm\frac{\sqrt{3}}{2} ,-\frac{1}{2} \rgt \end{equation*}

(b) To get a rate $-2\text{,}$ we need

\begin{equation*} \sin\theta=-1 \implies \theta=-90^\circ \end{equation*}

So the desired direction is

\begin{equation*} \llt \cos\theta,\sin\theta\rgt =\llt 0 ,-1 \rgt \end{equation*}

\begin{equation*} \sin\theta=-\frac{3}{2} \end{equation*}

No $\theta$ obeys this, since $-1\le\sin\theta\le 1$ for all $\theta\text{.}$ So no direction works!

2.7.2.5.

Solution.

Denote $\vnabla f(a,b)=\llt \al,\be\rgt\text{.}$ We are told that

\begin{alignat*}{3} \llt\al,\be\rgt\cdot\llt\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rgt&=3\sqrt{2} & &\quad\text{or}\quad & \al+\be&=6\\ \llt\al,\be\rgt\cdot\llt\frac{3}{5},-\frac{4}{5}\rgt&=5 & &\quad\text{or}\quad & 3\al-4\be&=25 \end{alignat*}

Adding 4 times the first equation to the second equation gives $7\al=49\text{.}$ Substituting $\al=7$ into the first equation gives $\be=-1\text{.}$ So $\vnabla f(a,b)=\llt 7,-1\rgt\text{.}$

2.7.2.6. (✳).

Solution.

Use a coordinate system with the positive $y$-axis pointing north, with the positive $x$-axis pointing east and with our current location being $x=y=0\text{.}$ Denote by $z(x,y)$ the elevation of the earth’s surface at $(x,y)\text{.}$ We are told that

\begin{align*} \vnabla z(0,0)\cdot(-\hj) &= 4\\ \vnabla z(0,0)\cdot\left(\frac{\hi-\hj}{\sqrt{2}}\right) &= \sqrt{2} \end{align*}

The first equation implies that $z_y(0,0)=-4$ and the second equation implies that

\begin{gather*} \frac{z_x(0,0)-z_y(0,0)}{\sqrt{2}}=\sqrt{2} \implies z_x(0,0)=z_y(0,0)+2 = -2 \end{gather*}

So the slope in the eastern direction is

\begin{gather*} \vnabla z(0,0)\cdot\hi = z_x(0,0) = -2 \end{gather*}

2.7.2.7. (✳).

Solution.

(a) Use $\vnabla f(P)$ to denote the gradient vector of $f$ at $P\text{.}$ We are told that

directional derivative of $f$ at $P$ is a maximum in the direction $2\hi - \hj + \hk\text{,}$ which implies that $\vnabla f(P)$ is parallel to $2\hi - \hj + \hk\text{,}$ and
the magnitude of the directional derivative in that direction is $3\sqrt{6}\text{,}$ which implies that $|\vnabla f(P)|=3\sqrt{6}\text{.}$

\begin{gather*} \vnabla f(P) = 3\sqrt{6} \frac{2\hi - \hj + \hk}{|2\hi - \hj + \hk|} = 6\hi - 3\hj + 3\hk \end{gather*}

(b) The directional derivative of $f$ at $P$ in the direction $\hi+\hj$ is

\begin{equation*} \vnabla f(P) \cdot\frac{\hi+\hj}{|\hi+\hj|} =\frac{1}{\sqrt{2}}\big(6\hi - 3\hj + 3\hk\big)\cdot\big(\hi+\hj\big) =\frac{3}{\sqrt{2}} \end{equation*}

2.7.2.8. (✳).

Solution.

(a) The gradient of $f$ at $(x,y)=(2,1)$ is

\begin{gather*} \vnabla f(2,1) = \llt -y^2\,,\,-2xy \rgt\Big|_{(x,y)=(2,1)} = \llt -1\,,\, -4\rgt \end{gather*}

So the path of steepest ascent is in the direction $-\frac{1}{\sqrt{17}}\llt 1\,,\, 4\rgt\text{,}$ which is a little west of south. The slope is

\begin{equation*} |\vnabla f(2,1)| = |\llt -1\,,\, -4\rgt| = \sqrt{17} \end{equation*}

(b) The directional derivative in the north direction is

\begin{equation*} D_{\llt 0,1\rgt}f(2,1) = \vnabla f(2,1)\cdot\llt 0,1\rgt = \llt -1\,,\, -4\rgt \cdot \llt 0,1\rgt = -4 \end{equation*}

So the hiker descends with slope $|-4|=4\text{.}$

(c) To contour, i.e. remain at the same height, the hiker should walk in a direction perpendicular to $\vnabla f(2,1)= \llt -1\,,\, -4\rgt\text{.}$ Two unit vectors perpendicular to $\llt -1\,,\, -4\rgt$ are $\pm\frac{1}{\sqrt{17}}\llt 4, -1\rgt\text{.}$

2.7.2.9.

Solution.

The gradient of $h(x,y)=1000-2x^2-3y^2$ is $\vnabla h(x,y)=(-4x,-6y)\text{.}$ This gradient (which points in the direction of steepest ascent) must be parallel to the tangent to $y=ax^b$ at all points on $y=ax^b\text{.}$ A tangent to $y=ax^b$ is $\llt 1,\diff{y}{x}\rgt =\llt 1, abx^{b-1}\rgt\text{.}$

\begin{equation*} \llt -4x,-6y\rgt \parallel \llt 1, abx^{b-1}\rgt \implies \frac{abx^{b-1}}{1}=\frac{-6y}{-4x} \implies \frac{3}{2}y=abx^b \end{equation*}

This is true at all points on $y=ax^b$ if and only if $b=\frac{3}{2}\text{.}$ As $(1,1)$ must also be on $y=ax^b\text{,}$ we need $1=a1^b\text{,}$ which forces $a=1\text{,}$ $b=\frac{3}{2}\text{.}$ Here is a contour map showing the hiking trail.

2.7.2.10. (✳).

Solution.

(a) The temperature gradient at $(3,2,1)$ is

\begin{gather*} \vnabla T(3,2,1) = \llt 4x\,,\,2y\,,\,-2z\rgt\Big|_{(x,y,z)=(3,2,1)} = \llt 12\,,\,4\,,\,-2\rgt \end{gather*}

She wishes to fly in a direction that is perpendicular to $\vnabla T(3,2,1)\text{.}$ That is, she wishes to fly in a direction $\llt a\,,\,b\,,\,c\rgt$ that obeys

\begin{equation*} 0 = \llt 12\,,\,4\,,\,-2\rgt\cdot \llt a\,,\,b\,,\,c\rgt = 12a+4b-2c \end{equation*}

Any nonzero $\llt a\,,\,b\,,\,c\rgt$ that obeys $12a+4b-2c=0$ is an allowed direction. Four allowed unit vectors are $\pm\frac{\llt 0\,,\,1\,,\,2\rgt}{\sqrt{5}}$ and $\pm\frac{\llt 1\,,\,-3\,,\,0\rgt}{\sqrt{10}}\text{.}$

(b) No they need not be the same. Four different explicit directions were given in part (a).

(c) To cool down as quickly as possible, she should move in the direction opposite to the temperature gradient. A unit vector in that direction is $-\frac{\llt 6\,,\,2\,,\,-1\rgt}{\sqrt{41}}\text{.}$

2.7.2.11. (✳).

Solution.

The temperature gradient at $(2,1,3)$ is

\begin{gather*} \vnabla T(2,1,3) = \llt 2x\,,\,z\,,\,y \rgt\Big|_{(x,y,z)=(2,1,3)} = \llt 4\,,\,3\,,\,1 \rgt \end{gather*}

(a) The bird is flying in the direction $\llt 4-2\,,\,3-1\,,\,4-3 \rgt =\llt 2\,,\,2\,,\,1\rgt$ at speed $2$ and so has velocity $\vv=2\frac{\llt 2\,,\,2\,,\,1\rgt}{|\llt 2\,,\,2\,,\,1\rgt|} =\frac{2}{3} \llt 2\,,\,2\,,\,1\rgt \text{.}$ The rate of change of air temperature experienced by the bird at that instant is

\begin{gather*} \vnabla T(2,1,3) \cdot\vv =\frac{2}{3} \llt 4\,,\,3\,,\,1 \rgt \cdot \llt 2\,,\,2\,,\,1\rgt = 10 \end{gather*}

(b) To maintain constant altitude (while not being stationary), the bird’s direction of travel has to be of the form $\llt a\,,\,b\,,\,0\rgt\text{,}$ for some constants $a$ and $b\text{,}$ not both zero. To keep the air temperature fixed, its direction of travel has to be perpendicular to $\vnabla T(2,1,3)= \llt 4\,,\,3\,,\,1 \rgt\text{.}$ So $a$ and $b$ have to obey

\begin{gather*} 0 = \llt a\,,\,b\,,\,0\rgt \cdot \llt 4\,,\,3\,,\,1 \rgt = 4a+3b \iff b=-\frac{4}{3}a \end{gather*}

and the direction of travel has to be a nonzero constant times $\llt 3\,,-4\,,0\rgt\text{.}$ The two such unit vectors are $\pm \frac{1}{5}\llt 3\,,-4\,,0\rgt\text{.}$

2.7.2.12. (✳).

Solution.

We are going to need, in both parts of this question, the gradient of $f(x,y)$ at $(x,y)=\left(1, -\frac{4}{3}\right)\text{.}$ So we find it first.

\begin{alignat*}{2} f_x(x,y)&=4x+3y\qquad & f_x(1,-4/3)&=0\\ f_y(x,y)&=3x+2y\qquad & f_y(1,-4/3)&=\frac{1}{3} \end{alignat*}

so $\vnabla f\left(1, -\frac{4}{3}\right) =\llt 0,\frac{1}{3}\rgt\text{.}$

(a) The maximum rate of change of $f$ at $P$ is

\begin{gather*} \left|\vnabla f\left(1, -\tfrac{4}{3}\right)\right| =\left|\llt 0,\tfrac{1}{3}\rgt\right| =\tfrac{1}{3} \end{gather*}

(b) If $\llt a,b\rgt$ is a unit vector, the directional derivative of $f$ at $P$ in the direction $\llt a,b\rgt$ is

\begin{gather*} D_{\llt a,b\rgt}f\left(1, -\tfrac{4}{3}\right) =\vnabla f\left(1, -\tfrac{4}{3}\right)\cdot \llt a,b\rgt =\llt 0,\tfrac{1}{3}\rgt\cdot \llt a,b\rgt =\tfrac{b}{3} \end{gather*}

So we need $\frac{b}{3}=\frac{1}{5}$ and hence $b=\frac{3}{5}\text{.}$ For $\llt a,b\rgt$ to be a unit vector, we also need

\begin{gather*} a^2+b^2=1 \iff a^2=1-b^2=1-\frac{3^2}{5^2}=\frac{16}{25} \iff a=\pm\frac{4}{5} \end{gather*}

So the allowed directions are $\llt \pm\frac{4}{5}\,,\,\frac{3}{5}\rgt\text{.}$

2.7.2.13. (✳).

Solution.

The slope of $y=x^2$ at $(1,1)$ is $\diff{}{x}x^2\Big|_{x=1}=2\text{.}$ So a unit vector in the bug’s direction of motion is $\frac{\llt 1,2\rgt}{\sqrt{5}}$ and the bug’s velocity vector is $\vv=0.01\frac{\llt 1,2\rgt}{\sqrt{5}}\text{.}$

The temperature gradient at $(1,1)$ is

\begin{gather*} \vnabla T(1,1) = \llt 2xy e^{x^2}\,,\,e^{x^2}\rgt\Big|_{(x.y)=(1,1)} =\llt 2e\,,\,e\rgt \end{gather*}

and the rate of change of $T$ (per unit time) that the bug feels as it passes through the point $(1, 1)$ is

\begin{gather*} \vnabla T(1,1)\cdot \vv =\frac{0.01}{\sqrt{5}} \llt 2e\,,\,e\rgt\cdot \llt 1,2\rgt =\frac{0.04 e}{\sqrt{5}} \end{gather*}

2.7.2.14. (✳).

Solution.

(a) We are to find the directional derivative in the direction

\begin{equation*} \llt 0-3\,,\,1-2\,,\,2-1\rgt = \llt -3\,,\,-1\,,\,1\rgt \end{equation*}

As the gradient of $F$ is

\begin{gather*} \vnabla F(x,y,z) = \llt y-1\,,\,x-2y\,,\,2z \rgt \end{gather*}

the directional derivative is

\begin{align*} D_{\frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|}}F(3,2,1) &=\vnabla F(3,2,1)\cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|}\\ &= \llt 2-1\,,\,3-2(2)\,,\,2(1) \rgt \cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|}\\ &=\llt 1\,,\,-1\,,\,2\rgt \cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{\sqrt{11}}\\ &=0 \end{align*}

(b) The temperature decreases most rapidly in the direction opposite the gradient. A unit vector in that direction is

\begin{gather*} -\frac{\vnabla F(3,2,1)}{|\vnabla F(3,2,1)|} = -\frac{\llt 1\,,\,-1\,,\,2 \rgt}{|\llt 1\,,\,-1\,,\,2 \rgt|} = \frac{1}{\sqrt{6}}\llt -1\,,\,1\,,\,-2 \rgt \end{gather*}

\begin{gather*} \vv=\llt x'(0)\,,\,y'(0)\,,\,z'(0)\rgt =\llt 3e^t\,,\,-2\sin t\,,\,\frac{1}{2\sqrt{1+t}}\rgt\Big|_{t=0} =\llt 3\,,\,0\,,\,\frac{1}{2}\rgt \end{gather*}

So the rate of change of temperature with respect to $t$ at $t=0$ is

\begin{gather*} \vnabla F(3,2,1)\cdot \vv =\llt 1\,,\,-1\,,\,2 \rgt\cdot \llt 3\,,\,0\,,\,\frac{1}{2}\rgt =4 \end{gather*}

(d) For $\hi+5\hj+a\hk$ to be tangent to the level surface $F(x, y, z) = 3$ at $(3, 2, 1)\text{,}$ $\hi+5\hj+a\hk$ must be perpendicular to $\vnabla F(3,2,1)\text{.}$ So

\begin{gather*} 0=\llt 1\,,\,5\,,\,a \rgt \cdot \llt 1\,,\,-1\,,\,2 \rgt = -4+2a \end{gather*}

So $a=2\text{.}$

2.7.2.15. (✳).

Solution.

(a) The first order partial derivatives of $f$ and $g$ are

\begin{align*} \pdiff{f}{x}(x,y,z)&= 2e^{-(x^2 +y^2 +z^2)} -2x(2x + y)e^{-(x^2 +y^2 +z^2)}\\ &\hskip2in\implies \pdiff{f}{x}(0,1,-1)=2e^{-2}\\ \pdiff{f}{y}(x,y,z)&= e^{-(x^2 +y^2 +z^2)} -2y(2x + y)e^{-(x^2 +y^2 +z^2)}\\ &\hskip2in\implies \pdiff{f}{y}(0,1,-1)=-e^{-2}\\ \pdiff{f}{z}(x,y,z)&= -2z(2x + y)e^{-(x^2 +y^2 +z^2)}\\ &\hskip2in\implies \pdiff{f}{z}(0,1,-1)=2e^{-2}\\ \pdiff{g}{x}(x,y,z)&= z\\ &\hskip2in\implies \pdiff{g}{x}(0,1,-1)=-1\\ \pdiff{g}{y}(x,y,z)&= 2y+z\\ &\hskip2in\implies \pdiff{g}{y}(0,1,-1)= 1\\ \pdiff{g}{z}(x,y,z)&= x+y+2z\\ &\hskip2in\implies \pdiff{g}{z}(0,1,-1)=-1 \end{align*}

so that gradients are

\begin{equation*} \vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt\qquad \vnabla g(0,1,-1) =\llt -1,1,-1\rgt \end{equation*}

(b) The bird’s velocity is the vector of length $6$ in the direction of $\vnabla f(0,1,-1)\text{,}$ which is

\begin{equation*} \vv= 6\frac{\llt 2,-1,2\rgt}{|\llt 2,-1,2\rgt|} =\llt 4,-2,4\rgt \end{equation*}

The rate of change of $g$ (per unit time) seen by the bird is

\begin{gather*} \vnabla g(0,1,-1) \cdot \vv =\llt -1,1,-1\rgt \cdot \llt 4,-2,4\rgt = -10 \end{gather*}

(c) The direction of flight for the bat has to be perpendicular to both $\vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt$ and $\vnabla g(0,1,-1) =\llt -1,1,-1\rgt\text{.}$ Any vector which is a non zero constant times

\begin{align*} \llt 2,-1,2\rgt \times \llt -1,1,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 2 & -1 & 2\\ -1 & 1 & -1 \end{matrix}\right] =\llt -1 \,,\, 0 \,,\, 1 \rgt \end{align*}

is perpendicular to both $\vnabla f(0,1,-1)$ and $\vnabla g(0,1,-1)\text{.}$ In addition, the direction of flight for the bat must have a positive $z$-component. So any vector which is a (strictly) positive constant times $\llt -1 \,,\, 0 \,,\, 1 \rgt$ is fine.

2.7.2.16. (✳).

Solution.

(a) Let’s use $\vv$ to denote the bee’s velocity vector at time $t=2\text{.}$

The bee’s direction of motion is tangent to the curve. That tangent is perpendicular to both the normal vector to $3z + x^2 + y^2 = 2$ at $(1,1,0)\text{,}$ which is
\begin{equation*} \llt 2x\,,\,2y\,,\,3\rgt\Big|_{(x,y,z)=(1,1,0)} = \llt 2\,,\,2\,,\,3\rgt \end{equation*}
and the normal vector to $z = x^2 - y^2$ at $(1,1,0)\text{,}$ which is
\begin{equation*} \llt 2x\,,\,-2y\,,\,-1\rgt\Big|_{(x,y,z)=(1,1,0)} = \llt 2\,,\,-2\,,\,-1\rgt \end{equation*}
So $\vv$ has to be some constant times
\begin{align*} \llt 2\,,\,2\,,\,3\rgt \times \llt 2\,,\,-2\,,\,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk\\ 2 & 2 & 3\\ 2 & -2 & -1 \end{matrix}\right] =\llt 4 \,,\, 8 \,,\, -8 \rgt \end{align*}
or, equivalently, some constant times $\llt 1 \,,\, 2 \,,\, -2 \rgt\text{.}$
Since the $z$-component of $\vv$ has to be positive, $\vv$ has to be a positive constant times $\llt -1 \,,\, -2 \,,\, 2 \rgt\text{.}$
Since the speed has to be $6\text{,}$ $\vv$ has to have length $6\text{.}$

As $|\llt -1 \,,\, -2 \,,\, 2 \rgt|=3$

\begin{equation*} \vv = 2\llt -1 \,,\, -2 \,,\, 2 \rgt =\llt -2 \,,\, -4 \,,\, 4 \rgt \end{equation*}

(b) Solution 1: Suppose that the bee is at $\big(x(t),y(t),z(t)\big)$ at time $t\text{.}$ Then the temperature that the bee feels at time $t$ is

\begin{gather*} T\big(x(t),y(t),z(t),t\big) = x(t) y(t) -3x(t) +2y(t) t +z(t) \end{gather*}

Then the rate of change of temperature (per unit time) felt by the bee at time $t=2$ is

\begin{align*} &\diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2}\\ \hskip1in&=x'(2)y(2) + x(2)y'(2) -3x'(2) +2y'(2)2+2y(2) +z'(2) \end{align*}

Recalling that, at time $t=2\text{,}$ the bee is at $(1,1,0)$ and has velocity $\llt -2 \,,\, -4 \,,\, 4 \rgt$

\begin{align*} &\diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2}\\ &\hskip1in=(-2)(1) + (1)(-4) -3(-2) +2(-4)2+2(1) +4\\ &\hskip1in=-10 \end{align*}

(b) Solution 2: Suppose that the bee is at $\big(x(t),y(t),z(t)\big)$ at time $t\text{.}$ Then the temperature that the bee feels at time $t$ is

\begin{gather*} T\big(x(t),y(t),z(t),t\big) \end{gather*}

By the chain rule, the rate of change of temperature (per unit time) felt by the bee at time $t=2$ is

\begin{align*} &\diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2}\\ &\hskip0.5in=\bigg[\pdiff{T}{x}\big(x(t),y(t),z(t),t\big)\,x'(t) +\pdiff{T}{y}\big(x(t),y(t),z(t),t\big)\,y'(t)\\ &\hskip1.0in +\pdiff{T}{z}\big(x(t),y(t),z(t),t\big)\,z'(t) +\pdiff{T}{t}\big(x(t),y(t),z(t),t\big)\bigg]_{t=2} \end{align*}

Recalling that $T = xy - 3x+2yt+z\text{,}$ we have

\begin{align*} &\diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2}\\ &\hskip1in=[y(2)-3]x'(2) + [x(2)+2\times 2]y'(2) +z'(2) +2y(2) \end{align*}

Also recalling that, at time $t=2\text{,}$ the bee is at $(1,1,0)$ and has velocity $\llt -2 \,,\, -4 \,,\, 4 \rgt$

\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=[-2](-2) + [5](-4) +4 +2\\ &=-10 \end{align*}

2.7.2.17. (✳).

Solution.

(a) We are to find the rate of change of $T(x,y,z)$ at $(1, 2, -1)$ in the direction $\llt 1, 1, 0\rgt - \llt 1, 2, -1\rgt = \llt 0,-1,1 \rgt\text{.}$ That rate of change (per unit distance) is the directional derivative

\begin{gather*} D_{\frac{\llt 0,-1,1 \rgt}{\sqrt{2}}} T(1,2,-1) =\vnabla T(1,2,-1)\cdot \frac{\llt 0,-1,1 \rgt}{\sqrt{2}} \end{gather*}

\begin{align*} \pdiff{T}{x}(x,y,z)&= -20 x\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{x}(1,2,-1)&= -20\, e^{-9}\\ \pdiff{T}{y}(x,y,z)&= -10 y\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{y}(1,2,-1)&= -20\, e^{-9}\\ \pdiff{T}{z}(x,y,z)&= -30 z\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{z}(1,2,-1)&= 30\, e^{-9} \end{align*}

the directional derivative

\begin{align*} D_{\frac{\llt 0,-1,1 \rgt}{\sqrt{2}}} T(1,2,-1) &=e^{-9}\llt -20,-20,30\rgt\cdot \frac{\llt 0,-1,1 \rgt}{\sqrt{2}}\\ &=\frac{50}{\sqrt{2}}e^{-9} =25\,\sqrt{2}\,e^{-9} \end{align*}

(b) The direction of maximum rate of decrease is $-\vnabla T(1,2,-1)\text{.}$ A unit vector in that direction is $\frac{\llt 2,2,-3\rgt}{\sqrt{17}}\text{.}$

(c) The maximum rate of decrease at $P$ is $-|\vnabla T(1,2,-1)|=-10 e^{-9} |\llt -2,-2,3\rgt| = -10\sqrt{17} e^{-9}\text{.}$

2.7.2.18. (✳).

Solution.

Denote by $\llt a\,,\,b\,,\,c\rgt$ the gradient of the function $f$ at $P\text{.}$ We are told

\begin{align*} \llt a\,,\,b\,,\,c\rgt \cdot \llt 1\,,\,0\,,\,0\rgt &= 2\\ \llt a\,,\,b\,,\,c\rgt \cdot \frac{1}{\sqrt{2}}\llt 1\,,\,1\,,\,0\rgt &= -\sqrt{2}\\ \llt a\,,\,b\,,\,c\rgt \cdot \frac{1}{\sqrt{3}}\llt 1\,,\,1\,,\,1\rgt &= -\frac{5}{\sqrt{3}} \end{align*}

Simplifying

\begin{align*} a&=2\\ a+b&=-2\\ a+b+c&=-5 \end{align*}

From these equations we read off, in order, $a=2\text{,}$ $b=-4$ and $c=-3\text{.}$ The function $f$ has maximum rate of change at $P$ in the direction if the gradient of $f\text{.}$ The unit vector in that direction is

\begin{equation*} \frac{\llt 2\,,\,-4\,,\,-3\rgt}{|\llt 2\,,\,-4\,,\,-3\rgt|} =\frac{\llt 2\,,\,-4\,,\,-3\rgt}{\sqrt{29}} \end{equation*}

The maximum rate of change is the magnitude of the gradient, which is $\sqrt{29}\text{.}$

2.7.2.19. (✳).

Solution.

We are told that the direction of fastest increase for the function $f(x,y)$ at the origin is given by the vector $\llt 1, 2\rgt\text{.}$ This implies that $\vnabla f(0,0)$ is parallel to $\llt 1, 2\rgt\text{.}$ This in turn implies that $\llt 1, 2\rgt$ is normal to the level curve of $f(x,y)$ that passes through the origin. So $\llt 2, -1\rgt\text{,}$ being perpendicular to $\llt 1, 2\rgt\text{,}$ is tangent to the level curve of $f(x,y)$ that passes through the origin. The unit vectors that are parallel to $\llt 2, -1\rgt$ are $\pm\frac{1}{\sqrt{5}}\llt 2, -1\rgt\text{.}$

2.7.2.20. (✳).

Solution.

Write $h(x,y) = 1000 -0.02\,x^2-0.01 y^2$ so that the hill is $z=h(x,y)\text{.}$

(a) The direction of steepest ascent at $(0,100,900)$ is the direction of maximum rate of increase of $h(x,y)$ at $(0,100)$ which is $\vnabla h(0,100) = \llt 0\,,\, -0.01(2)(100)\rgt = \llt 0\,,\,-2\rgt\text{.}$ In compass directions that is South.

(b) The slope of the hill there is

\begin{gather*} \vnabla h(0,100)\cdot\llt 0,-1\rgt =-\pdiff{h}{y}(0,100) = 2 \end{gather*}

(c) Denote by $\big(x(t),y(t),z(t)\big)$ your position at time $t$ and suppose that you are at $(0,100,900)$ at time $t=0\text{.}$ Then we know

$z(t) = 1000 -0.02\,x(t)^2-0.01 y(t)^2\text{,}$ so that $z'(t) = -0.04\,x(t)x'(t)-0.02 y(t)y'(t)\text{,}$ since you are on the hill and
$x'(0)=0$ and $y'(0) \gt 0$ since you are going in the direction of steepest descent and
$x'(0)^2+y'(0)^2+z'(0)^2=25$ since you are moving at speed $5\text{.}$

Since $x(0)$ and $y(0)=100\text{,}$ we have $z'(0)= -0.02(100)y'(0) = -2y'(0)\text{.}$ So

\begin{align*} &25 = x'(0)^2+y'(0)^2+z'(0)^2 =5\ y'(0)^2\\ &\hskip1in\implies y'(0) = \sqrt{5}\\ &\hskip1in\implies \llt x'(0)\,,\,y'(0)\,,\,z'(0) \rgt =\llt 0\,,\, \sqrt{5}\,,\, -2\sqrt{5}\rgt \end{align*}

and your rate of change of altitude is

\begin{align*} \diff{}{t} h\big(x(t)\,,\,y(t)\big)\Big|_{t=0} &=\vnabla h(0,100)\cdot \llt x'(0)\,,\,y'(0) \rgt =\llt 0\,,\,-2\rgt \cdot \llt 0\,,\, \sqrt{5}\rgt\\ &=-2\sqrt{5} \end{align*}

2.7.2.21. (✳).

Solution.

Reading through the question as a whole we see that we will need

for part (a), the gradient of $PT$ at $(2t, t^2 - 1, \cos t)\Big|_{t=0} =(0,-1,1)$
for part (b), the gradients of both $P$ and $T$ at $(0,-1,1)$ and
for part (c), the gradient of $T$ at $(0,-1,1)$ and the gradient of $S=z^3+xz+y^2$ at $(0,-1,1)$ (to get the normal vector to the surface at that point).

So, by way of preparation, let’s compute all of these gradients.

\begin{align*} \vnabla P(x,y,z) &= \frac{2x}{1+z^2}\hi + \frac{4y}{1+z^2}\hj -\frac{(x^2+2y^2)2z}{{(1+z^2)}^2}\hk\\ &\hskip1in\implies \vnabla P(0,-1,1) = -2\,\hj - \hk\\ \vnabla T(x,y,z) &= y\,\hi + x\,\hj -2z\,\hk\\ &\hskip1in\implies\vnabla T(0,-1,1) = -\hi - 2\hk\\ \vnabla S(x,y,z) &= z\,\hi + 2y\,\hj +(x+3z^2)\,\hk\\ &\hskip1in\implies\vnabla S(0,-1,1) = \hi - 2\hj +3\,\hk \end{align*}

To get the gradient of $PT$ we use the product rule

\begin{equation*} \vnabla(PT)(x,y,z) =T(x,y,z)\,\vnabla P(x,y,z) +P(x,y,z)\,\vnabla T(x,y,z) \end{equation*}

so that

\begin{align*} \vnabla(PT)(0,-1,1) &=T(0,-1,1)\,\vnabla P(0,-1,1) +P(0,-1,1)\,\vnabla T(0,-1,1)\\ &=(5+0-1) \big(-2\,\hj - \hk\big) + \frac{0+2}{1+1}\big(-\hi - 2\hk\big)\\ &= -\hi -8\,\hj -6\,\hk \end{align*}

(a) Since $\diff{}{t}(PT)^2 = 2(PT)\diff{}{t}(PT)\text{,}$ and the velocity vector of the plane at time $0$ is

\begin{equation*} \diff{}{t}\llt 2t, t^2 - 1, \cos t\rgt\Big|_{t=0} =\llt 2, 2t, -\sin t\rgt\Big|_{t=0} =\llt 2,0,0\rgt \end{equation*}

we have

\begin{align*} \diff{}{t}(PT)^2\Big|_{t=0} &= 2\,P(0,-1,1)\,T(0,-1,1)\ \vnabla(PT)(0,-1,1)\cdot \llt 2,0,0\rgt\\ &= 2\ \frac{0+2}{1+1}\ (5+0-1)\llt -1,-8,-6\rgt \cdot \llt 2,0,0\rgt\\ &= -16 \end{align*}

(b) The direction should be perpendicular to $\vnabla P(0,-1,1)$ (to keep $P$ constant) and should also be perpendicular to $\vnabla T(0,-1,1)$ (to keep $T$ constant). So any nonzero constant times

\begin{align*} \pm \vnabla P(0,-1,1) \times \vnabla T(0,-1,1) &=\pm \llt 0 \,,\, -2 \,,\, -1\rgt \times \llt -1\,,\,0 \,,\,- 2\rgt\\ &=\pm \det\left[\begin{matrix} \hi & \hj & \hk\\ 0 & 2 & 1\\ 1 & 0 & 2 \end{matrix}\right]\\ &=\pm \llt 4\,,\, 1 \,,\, -2\rgt \end{align*}

are allowed directions.

(c) We want the direction to be as close as possible to $\vnabla T(0,-1,1) =\llt -1 \,,\, 0 \,,\, -2\rgt$ while still being tangent to the surface, i.e. being perpendicular to the normal vector $\vnabla S(0,-1,1)=\llt 1 \,,\, -2 \,,\, 3\rgt\text{.}$ We can get that optimal direction by subtracting from $\vnabla T(0,-1,1)$ the projection of $\vnabla T(0,-1,1)$ onto the normal vector.

The projection of $\vnabla T(0,-1,1)$ onto the normal vector $\vnabla S(0,-1,1)$ is

\begin{align*} \text{proj}_{\vnabla S(0,-1,1)}\vnabla T(0,-1,1) &=\frac{\vnabla T(0,-1,1)\cdot \vnabla S(0,-1,1)} {|\vnabla S(0,-1,1)|^2}\vnabla S(0,-1,1)\\ &=\frac{\llt -1\,,\,0 \,,\,- 2\rgt\cdot \llt 1\,,\, -2 \,,\,3\rgt} {|\llt 1\,,\, -2 \,,\,3\rgt|^2}\llt 1\,,\, -2 \,,\,3\rgt\\ &=\frac{-7}{14}\llt 1\,,\, -2 \,,\,3\rgt \end{align*}

So the optimal direction is

\begin{align*} \vd&=\vnabla T(0,-1,1) - \text{proj}_{\vnabla S(0,-1,1)}\vnabla T(0,-1,1)\\ &=\llt -1\,,\,0 \,,\,- 2\rgt -\frac{-7}{14}\llt 1\,,\, -2 \,,\,3\rgt\\ &= \llt -\frac{1}{2}\,,\,-1 \,,\,- \frac{1}{2}\rgt \end{align*}

So any positive non zero multiple of $-\llt 1\,,\, 2\,,\,1\rgt$ will do. Note, as a check, that $-\llt 1\,,\, 2\,,\,1\rgt$ has dot product zero, i.e. is perpendicular to, $\vnabla S(0,-1,1)=\llt 1 \,,\, -2 \,,\, 3\rgt\text{.}$

2.7.2.22. (✳).

Solution.

Write $\vnabla f(a,b,c) =\llt F,G,H\rgt\text{.}$ We are told that

\begin{align*} D_\vu f&=\frac{1}{\sqrt{6}} \llt 1, 1, 2\rgt \cdot \llt F,G,H\rgt = 0\\ D_\vv f&=\frac{1}{\sqrt{3}} \llt 1, -1, -1\rgt \cdot \llt F,G,H\rgt = 0\\ D_\vw f&=\frac{1}{\sqrt{3}} \llt 1, 1, 1\rgt \cdot \llt F,G,H\rgt = 4 \end{align*}

so that

\begin{align*} F + G + 2H &= 0 \tag{E1}\\ F - G - H &= 0 \tag{E2}\\ F + G + H &= 4\sqrt{3} \tag{E3} \end{align*}

Adding (E2) and (E3) gives $2F=4\sqrt{3}$ or $F=2\sqrt{3}\text{.}$ Substituting $F=2\sqrt{3}$ into (E1) and (E2) gives

\begin{align*} G + 2H &= -2\sqrt{3} \tag{E1}\\ - G - H &= -2\sqrt{3} \tag{E2} \end{align*}

Adding (E1) and (E2) gives $H=-4\sqrt{3}$ and substituting $H=-4\sqrt{3}$ back into (E2) gives $G=6\sqrt{3}\text{.}$ All together

\begin{equation*} \vnabla f(a,b,c) =\sqrt{3} \llt 2,6,-4\rgt \end{equation*}

2.7.2.23. (✳).

Solution.

(a) The expression $\lim_{t\rightarrow 0}\frac{f((1,1)+t\vu)-f(1,1)}{t}$ is the directional derivative of $f$ at $(1,1)$ in the direction $\vu\text{,}$ which is $D_\vu f(1,1)=\vnabla f(1,1)\cdot\vu\text{.}$ This is mazimized when $\vu$ is parallel to $\vnabla f(1,1)\text{.}$ Since

\begin{equation*} f_x(x,y)=2xy^2e^{-x-y}-x^2y^2e^{-x-y}\qquad f_y(x,y)=2x^2ye^{-x-y}-x^2y^2e^{-x-y} \end{equation*}

we have

\begin{equation*} \vnabla f(1,1)=e^{-2}\llt 1,1\rgt \end{equation*}

so that the desired unit vector $\vu$ is $\frac{1}{\sqrt{2}}\llt 1,1\rgt\text{.}$

(b) In order to remain at elevation $e^{-2}\text{,}$ the ant must move so that $D_\vu f(1,1)=0\text{.}$ This is the case if $\vu\perp\vnabla f(1,1)\text{.}$ For example, we can take $\vu=\llt 1,-1\rgt\text{.}$ When the ant moves in this direction, while remaining on the surface of the hill, its vertical component of velocity is zero. So $\vv=c\llt 1,-1,0\rgt$ for any nonzero constant $c\text{.}$

(c) In order to maximize its instantaneous rate of level increase, the ant must choose the $x$ and $y$ coordinates of its velocity vector in the same direction as $\vnabla f(1,1)\text{.}$ Namely $\vu=c\llt 1,1\rgt$ for any $c \gt 0\text{.}$ To make $\vu$ a unit vector, we choose $c=\frac{1}{\sqrt{2}}\text{.}$ The corresponding value of the $z$ coordinate of its velocity vector is the rate of change of $f$ per unit horizontal distance travelled, which is the directional derivative

\begin{equation*} D_\vu f(1,1)=\vnabla f(1,1)\cdot\vu=e^{-2}\llt 1,1\rgt \cdot\llt c,c\rgt =2ce^{-2} \end{equation*}

So $\vv=\frac{1}{\sqrt{2}}\llt 1,1,2e^{-2}\rgt\text{.}$ Any positive multiple of this vector is also a correct answer.

2.7.2.24. (✳).

Solution.

(a) The direction of motion at $s=1$ is given by the tangent vector

\begin{equation*} \vr'(s)=\llt s^2,2,2s\rgt\big|_{s=1}=\llt 1,2,2\rgt \end{equation*}

Since the length of the velocity vector must be $3\text{,}$

\begin{equation*} \text{velocity}=\vv=3\frac{\llt 1,2,2\rgt}{|\llt 1,2,2\rgt|}=\llt 1,2,2\rgt \end{equation*}

(b) The rate of change of temperature per unit distance felt by the sparrow at $s=1$ is $\vnabla T\big(\frac{1}{3},2,1\big)\cdot\frac{\vv}{|\vv|}\text{.}$ The rate of change of temperature per unit time felt by the sparrow at $s=1$ is

\begin{align*} \vnabla T\left(\frac{1}{3},2,1\right)\cdot\frac{\vv}{|\vv|}\ |\vv| &=\vnabla T\left(\frac{1}{3},2,1\right)\cdot\vv =\vv\cdot\llt 6x,y,4z\rgt\Big|_{({1\over3},2,1\big)}\\ &=\llt 1,2,2\rgt\cdot\llt 2,2,4\rgt=14^\circ/{\rm s} \end{align*}

(c) The temperature decreases at maximum rate in the direction opposite the temperature gradient, which is (any positive constant times) $-\llt 2,2,4\rgt\text{.}$

(d) The eagle is moving at right angles to the direction of motion of the sparrow, which is $\llt 1,2,2\rgt\text{.}$ As the eagle is also moving in a direction for which the temperature remains constant, it must be moving perpendicularly to the temperature gradient, $\llt 2,2,4\rgt\text{.}$ So the direction of the eagle must be (a posiitve constant times) one of

\begin{align*} \pm\llt 1,2,2\rgt \times\llt 2,2,4\rgt =\pm\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 2 & 2\\ 2 & 2 & 4 \end{matrix}\right] =\pm\llt 4,0,-2\rgt \end{align*}

or equivalently, any positive constant times $\pm\llt 2,0,-1\rgt\text{.}$

2.7.2.25. (✳).

Solution.

(a) The moth is moving the direction of the temperature gradient at $(3,4,0)\text{,}$ which is

\begin{equation*} \vnabla T(3,4,0) =-200\frac{2x\hi+2y\hj+2z\hk}{{(1+x^2+y^2+z^2)}^2}\bigg|_{(3,4,0)} =-400\frac{3\hi+4\hj}{26^2} \end{equation*}

Since the speed of the moth is $1 {\rm m/s}$ its velocity vector is a vector of length one in direction $-\frac{400}{26^2}\llt 3,4,0\rgt$ and hence is $\vv=-\frac{\llt 3,4,0\rgt}{|\llt 3,4,0\rgt|} =-\llt\frac{3}{5},\frac{4}{5},0\rgt\text{.}$

(b) The rate of change of temperature (per unit time) the moth feels at that time is

\begin{equation*} \vnabla T(3,4,0)\cdot\vv =\frac{400}{26^2}\llt 3,4,0\rgt\cdot \llt\frac{3}{5},\frac{4}{5},0\rgt =\frac{400\times25}{26^2\times 5} =\frac{500}{169}\approx 2.96^\circ/{\rm s} \end{equation*}

2.7.2.26. (✳).

Solution.

(a) We are told that $T(x,y,z)=\frac{k}{|\llt x,y,z\rgt|} =\frac{k}{\sqrt{x^2+y^2+z^2}}$ for some constant $k$ and that

\begin{equation*} 120=T(1,2,2)=\frac{k}{|\llt 1,2,2\rgt|} \implies k=120\times\sqrt{1+2^2+2^2} = 360 \end{equation*}

(b) The (unit) direction from $(1,2,2)$ to $(2,1,3)$ is $\vd=\frac{\llt 2,1,3\rgt-\llt 1,2,2\rgt}{|\llt 2,1,3\rgt-\llt 1,2,2\rgt|} =\frac{\llt 1,-1,1\rgt}{|\llt 1,-1,1\rgt|}=\frac{1}{\sqrt{3}}\llt 1,-1,1\rgt\text{.}$ The desired rate of change of temperature is

\begin{align*} D_{\vd} T(1,2,2) &=\vnabla T(1,2,2)\cdot\vd =-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}}\Big|_{\llt 1,2,2\rgt}\cdot\vd\\ &=-360\frac{\llt 1,2,2\rgt}{27}\cdot\frac{\llt 1,-1,1\rgt}{\sqrt{3}} =-\frac{40}{3\sqrt{3}}\approx-7.70 \end{align*}

degrees per unit distance.

(c) At $(x,y,z)\text{,}$ the direction of greatest increase is in the direction of the temperature gradient at $(x,y,z)\text{,}$ which is $\vnabla T(x,y,z) =-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}}$ and which points opposite to the radius vector. That is, it points towards the origin. This argument only fails at $(x,y,z)=(0,0,0)\text{,}$ where the gradient, and indeed $T(x,y,z)\text{,}$ is not defined.

2.7.2.27. (✳).

Solution.

(a) The shoreline is $f(x,y)=0$ or $x^2+4x+4y^2=32$ or $(x+2)^2+4y^2=36\text{,}$ which is an ellipse centred on $(-2,0)$ with semiaxes $6$ in the $x$-direction and $3$ in the $y$-direction.

(b,c) The gradient of $f$ at $(-1,1)$ is

\begin{equation*} \vnabla f(-1,1)=\big[(-2x-4)\,\hi-8y\,\hj\big]_{(-1,1)}=-2\,\hi-8\,\hj \end{equation*}

To remain at constant depth, he should swim perpendicular to the depth gradient. So he should swim in direction $\pm\frac{1}{\sqrt{17}}\llt 4,-1\rgt\text{.}$ To increase his depth as rapidly as possible, he should swim in the direction of the depth gradient, which is $-\frac{1}{\sqrt{17}}\llt 1,4\rgt\text{.}$

2.7.2.28.

Solution.

(a) The curve on which the temperature is $T_0$ is $x^2-2y^2=T_0\text{.}$ If $T_0=0\text{,}$ this is the pair of straight lines $y=\pm\frac{x}{\sqrt{2}}\text{.}$ If $T_0 \gt 0\text{,}$ it is a hyperbola on which $x^2=2y^2+T_0\ge T_0\text{.}$ If $T_0 \lt 0\text{,}$ it is a hyperbola on which $2y^2=x^2-T_0\ge |T_0|\text{.}$ Here is a sketch which show the isotherms $T=0,\ 1,\ -1$ as well as the branch of the $T=2$ isotherm that contains the ant’s location $(2,-1)\text{.}$

Note that the temperature gradient is $\vnabla T(x,y)=\llt 2x,-4y\rgt\text{.}$ In particular, the temperature gradient at $(2,-1)$ is $\vnabla T(2,-1)=\llt 4,4\rgt\text{.}$

(b) To achieve maximum rate of cooling, the ant should move in the direction opposite the temperature gradient at $(2,-1)\text{.}$ So the direction of maximum rate of cooling is

\begin{equation*} -\frac{\llt 4,4\rgt}{4\sqrt{2}} =\frac{\llt -1,-1\rgt}{\sqrt{2}} \end{equation*}

(c) If the ant moves in the direction of part (b), its rate of cooling per unit distance is $|\vnabla T(2,-1)|=|\llt 4,4\rgt| = 4\sqrt{2}\text{.}$ It the ant is moving at speed $v\text{,}$ its rate of cooling per unit time is $4\sqrt{2}\,v\text{.}$

(d) If the ant moves from $(2,-1)$ in direction $\llt -1,-2\rgt$ its temperature increases at the rate

\begin{equation*} D_{\frac{\llt -1,-2\rgt}{\sqrt{5}}} T(2,-1) = \llt 4,4\rgt\cdot \frac{\llt -1,-2\rgt}{\sqrt{5}} = -\frac{12}{\sqrt{5}} \end{equation*}

per unit distance. So, if the ant is moving at speed $v\text{,}$ its rate of decrease of temperature per unit time is $\frac{12}{\sqrt{5}}\,v$

(e) Suppose that the ant moves along the curve $y=y(x)\text{.}$ For the ant to always experience maximum rate of cooling (or maximum rate of heating), the tangent to this curve must be parallel to $\vnabla T(x,y)$ at every point of the curve. A tangent to the curve at $(x,y)$ is $\llt 1,\diff{y}{x}(x)\rgt\text{.}$ This is parallel to $\vnabla T(x,y)=\llt 2x,-4y\rgt$ when

\begin{gather*} \frac{\diff{y}{x}}{1}=\frac{-4y}{2x} \implies \frac{dy}{y}=-2\frac{dx}{x} \implies \ln y=-2\ln x+C \implies y=C'x^{-2} \end{gather*}

To pass through $(2,-1)\text{,}$ we need $C'=-4\text{,}$ so $y=-\frac{4}{x^2}\text{.}$

2.7.2.29. (✳).

Solution.

The first order partial derivatives of $f\text{,}$ both at a general point $(x,y,z)$ and at the point $(1, 0, \pi/2)\text{,}$ are

\begin{alignat*}{2} f_x(x,y,z)&= 2x\qquad & f_x(1, 0, \pi/2)&= 2\\ f_y(x,y,z)&= -z\sin(yz)\qquad & f_y(1, 0, \pi/2)&= 0\\ f_z(x,y,z)&= -y\sin(yz)\qquad & f_z(1, 0, \pi/2)&= 0 \end{alignat*}

(a) The rate of increase of $f$ is largest in the direction of $\vnabla f(1, 0, \pi/2)=\llt 2,0,0\rgt\text{.}$ A unit vector in that direction is $\hi\text{.}$

(b) The gradient vector $\vnabla f(1, 0, \pi/2)=\llt 2,0,0\rgt$ is a normal vector to the surface $f=1$ at $(1, 0, \pi/2)\text{.}$ So the specified tangent plane is

\begin{gather*} \llt 2,0,0\rgt \cdot \llt x-1\,,\,y-0\,,\,z-\pi/2\rgt=0\qquad\text{or}\qquad x=1 \end{gather*}

(c) The vector from the point $(0,1,0)$ to the point $(1,1,0)\text{,}$ on $T\text{,}$ is $\llt 1,0,0\rgt\text{,}$ which is perpendicular to $T\text{.}$ So $(1,1,0)$ is the point on $T$ nearest $(0,1,0)$ and the distance from $(0,1,0)$ to $T$ is $|\llt 1,0,0\rgt|=1\text{.}$

(d) The vector $\llt 1,0,1\rgt$ is perpendicular to the plane $x+z=0\text{.}$ So the angle between the planes $T$ and $x+z=0$ is the same as the angle $\theta$ between the vectors $\llt 1,0,0\rgt$ and $\llt 1,0,1\rgt\text{,}$ which obeys

\begin{align*} & |\llt 1,0,0\rgt| \ |\llt 1,0,1\rgt|\ \cos\theta =|\llt 1,0,0\rgt\cdot\llt 1,0,1\rgt| =1\\ &\implies \cos\theta=\frac{1}{\sqrt{2}} \implies \theta=\frac{\pi}{4} \end{align*}

2.7.2.30. (✳).

Solution.

(a) We are being asked for the directional derivative of $T$ in the direction of the unit vector from $P=(2,1,1)$ to $Q=(3,2,2)\text{,}$ which is $\frac{\llt 1,1,1\rgt}{\sqrt{3}}\text{.}$ That directional derivative is

\begin{gather*} \vnabla T(P)\cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}} =\llt 1,2,3\rgt \cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}} =2\sqrt{3} \end{gather*}

(b) The linear approximation to $T$ at $P$ is

\begin{align*} T(2+\De x\,,\,1+\De y\,,\,1+\De z) &\approx T(P) + T_x(P)\,\De x + T_y(P)\,\De y + T_z(P)\,\De z\\ &= 5 +\De x +2\,\De y +3\,\De z \end{align*}

Applying this with $\De x = -0.1\text{,}$ $\De y = 0\text{,}$ $\De z=0.2$ gives

\begin{align*} T(1.9\,,\,1\,,\,1.2) &\approx 5 +(-0.1) +2\,(0) +3\,(0.2) =5.5 \end{align*}

(c) For the rate of change of $T$ to be zero, the direction of motion must be perpendicular to $\vnabla T(P) = \llt 1,2,3\rgt\text{.}$ For the rate of change of $S$ to also be zero, the direction of motion must also be perpendicular to $\vnabla S(P) = \llt 1,0,1\rgt\text{.}$ The vector

\begin{align*} \llt 1,2,3\rgt \times \llt 1,0,1\rgt &=\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 2 & 3\\ 1 & 0 & 1\end{matrix}\right] =\llt 2,2,-2\rgt \end{align*}

is perpendicular to both $\vnabla T(P)$ and $\vnabla S(P)\text{.}$ So the desired unit vectors are $\pm\frac{\llt 1,1,-1\rgt}{\sqrt{3}}\text{.}$

2.7.2.31. (✳).

Solution.

We are going to need the gradients of both $F$ and $G$ at $(0,1,2)\text{.}$ So we compute

\begin{align*} \pdiff{F}{x}(x,y,z)&=y^2+z & \pdiff{F}{y}(x,y,z)&=2xy & \pdiff{F}{z}(x,y,z)&=3z^2+x\\ \pdiff{G}{x}(x,y,z)&=3 & \pdiff{G}{y}(x,y,z)&=-1 & \pdiff{G}{z}(x,y,z)&=4 \end{align*}

and then

\begin{equation*} \vnabla F(0,1,2) = \llt 3,0,12 \rgt\qquad \vnabla G(0,1,2) = \llt 3,-1,4 \rgt \end{equation*}

(a) The linear approximation to $F$ at $(0,1,2)$ is

\begin{align*} F(x,y,z) &\approx F(0,1,2)\!+\!F_x(0,1,2)\,x\!+\! F_y(0,1,2)\,(y\!-\!1)\!+\!F_z(0,1,2)\,(z\!-\!2)\\ &=8+ 3 x + 12 (z-2) \end{align*}

In particular

\begin{align*} F(0.1\,,\,0.9\,,\,1.8) - F(0,1,2) &\approx 3(0.1) + 12(-0.2) =-2.1 \end{align*}

(b) The direction along which $G$ increases most rapidly at $P$ is $\vnabla G(0,1,2) = \llt 3,-1,4 \rgt\text{.}$ The directional derivative of $F$ in that direction is

\begin{gather*} D_{\frac{\llt 3,-1,4\rgt}{\sqrt{26}}}F(0,1,2) = \vnabla F(0,1,2) \cdot \frac{\llt 3,-1,4\rgt}{\sqrt{26}} = \llt 3,0,12 \rgt\cdot \frac{\llt 3,-1,4\rgt}{\sqrt{26}} \gt 0 \end{gather*}

So $F$ increases.

(c) For the rate of change of $F$ to be zero, $\llt a\,,\,b\,,\,c\rgt$ must be perpendicular to $\vnabla F(0,1,2) = \llt 3,0,12 \rgt\text{.}$

For the rate of change of $G$ to be zero, $\llt a\,,\,b\,,\,c\rgt$ must be perpendicular to $\vnabla G(0,1,2) = \llt 3,-1,4 \rgt\text{.}$

So any nonzero constant times

\begin{align*} \det\left[\begin{matrix} \hi & \hj & \hk\\ 3 & 0 & 12\\ 3 & -1 & 4 \end{matrix}\right] =\llt 12 \,,\, 24 \,,\, -3 \rgt =3 \llt 4 \,,\, 8 \,,\, -1 \rgt \end{align*}

is an allowed direction.

2.7.2.32. (✳).

Solution.

(a) Since

\begin{equation*} z=-\frac{100}{x^2+2x+4y^2+11}=-\frac{100}{(x+1)^2+4y^2+10} \end{equation*}

the bottom of the crater is at $x=-1\text{,}$ $y=0$ (where the denominator is a minimum) and the contours (level curves) are ellipses having equations $(x+1)^2+4y^2=C\text{.}$ In the sketch below, the filled dot represents the bottom of the crater and the open dot represents the car park. The contours sketched are (from inside out) $z=-7.5, -5, -2.5, -1\text{.}$ Note that the trail crosses the contour lines at right angles.

(b) The trail is to be parallel to

\begin{equation*} \vnabla z = \frac{100}{{(x^2+2x+4y^2+11)}^2}(2x+2,8y) \end{equation*}

At the car park $\vnabla z(4,5)\parallel \llt 10, 40\rgt\parallel \llt 1,4\rgt\text{.}$ To move towards the bottom of the crater, we should leave in the direction $-\llt 1,4\rgt\text{.}$

2.7.2.33. (✳).

Solution.

We have

\begin{equation*} \vnabla h(x,y)= -200 e^{-(x^2+2y^2)}\llt x,2y\rgt \end{equation*}

and, in particular,

\begin{equation*} \vnabla h(3,2)= -200 e^{-17}\llt 3,4\rgt \end{equation*}

(a) At $(3,2)$ the dune slopes downward the most steeply in the direction opposite $\vnabla h(3,2)\text{,}$ which is (any positive multiple of) $\llt 3,4\rgt\text{.}$

(b) The rate is $D_{\hj} h(3,2)=\vnabla h(3,2)\cdot\hj=-800 e^{-17}\text{.}$

(c) To remain at the same height, you should walk perpendicular to $\vnabla h(3,2)\text{.}$ So you should walk in one of the directions $\pm\big(\frac{4}{5},-\frac{3}{5}\big)\text{.}$

(d) Suppose that you are walking along a steepest descent curve. Then the direction from $(x,y)$ to $(x+\dee{x}, y+\dee{y})\text{,}$ with $(\dee{x},\dee{y})$ infinitesmal, must be opposite to $\vnabla h(x,y)= -200 e^{-(x^2+2y^2)}(x,2y)\text{.}$ Thus $(\dee{x},\dee{y})$ must be parallel to $(x,2y)$ so that the slope

\begin{gather*} \diff{y}{x}=\frac{2y}{x} \implies \frac{\dee{y}}{y}=2\frac{\dee{x}}{x} \implies \ln y=2\ln x+C \end{gather*}

We must choose $C$ to obey $\ln 2=2\ln 3+C$ in order to pass through the point $(3,2)\text{.}$ Thus $C=\ln\frac{2}{9}$ and the curve is $\ln y=2\ln x+\ln\frac{2}{9}$ or $y=\frac{2}{9}x^2\text{.}$

2.7.2.34. (✳).

Solution.

(a) Denote $\vnabla f(1,2)= \llt a,b\rgt\text{.}$ We are told that

\begin{alignat*}{3} D_\vu f(1,2)&=\vu\cdot(a,b)&&=\frac{3}{5}a+\frac{4}{5}b&&=10\\ D_\vv f(1,2)&=\vv\cdot(a,b)&&=\frac{3}{5}a-\frac{4}{5}b&&=2 \end{alignat*}

Adding these two equations gives $\frac{6}{5}a=12\text{,}$ which forces $a=10\text{,}$ and subtracting the two equations gives $\frac{8}{5}b=8\text{,}$ which forces $b=5\text{,}$ as desired.

(b) The rate of change of $f$ at $(1,2)$ in the direction of the vector $\hi+2\hj$ is

\begin{gather*} \frac{\hi+2\hj}{|\hi+2\hj|}\cdot \vnabla f(1,2) =\frac{1}{\sqrt{5}}\llt 1,2\rgt\cdot\llt 10,5\rgt =4\sqrt{5}\approx 8.944 \end{gather*}

\begin{align*} f\big(x_0+\De x\,,\,y_0+\De y\big) &\approx f\big(x_0\,,\,y_0\big) + \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x + \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y \end{align*}

with $x_0=1\text{,}$ $\De x=0.01\text{,}$ $y_0=2\text{,}$ and $\De y=0.05\text{,}$ gives

\begin{align*} f(1.01,2.05) &\approx f(1,2)+f_x(1,2)\times(1.01-1)+f_y(1,2)\times(2.05-2)\\ &=7+10\times0.01+5\times0.05\\ &=7.35 \end{align*}

2.8 A First Look at Partial Differential Equations
2.8.3 Exercises

2.8.3.1.

Solution.

We start by evaluating $u_t(x,t)$ and $u_{xx}(x,t)+u(x,t)$ when $u(x,t)= e^{-t-x^2}\text{.}$

\begin{align*} u(x,t)\amp= e^{-t-x^2}\\ u_t(x,t)\amp= -e^{-t-x^2}\\ u_x(x,t)\amp= -2x e^{-t-x^2}\\ u_{xx}(x,t)\amp= -2 e^{-t-x^2} +4x^2 e^{-t-x^2} \end{align*}

\begin{align*} u_{xx}(x,t)+u(x,t) \amp= \big[-2 e^{-t-x^2} +4x^2 e^{-t-x^2}\big] +e^{-t-x^2} \\ \amp= \big[4x^2-1]e^{-t-x^2} \end{align*}

For this to equal $g(x)\, u_t(x,t) = -g(x)\, e^{-t-x^2}\text{,}$ we need $g(x) = 1-4x^2\text{.}$

2.8.3.2.

Solution.

(a) Fix any $y_0$ and set $v(x)=u(x,y_0)\text{.}$ Then

\begin{equation*} \diff{v}{x}(x) =\pdiff{u}{x}(x,y_0) = 0 \end{equation*}

So, for each fixed $y_0\text{,}$ $v(x)=u(x,y_0)\text{,}$ which is a function of $x\text{,}$ has to be a constant. The constant may be different for each different choice of $y_0\text{.}$ So $u(x,y_0) = C(y_0)$ with $C(y_0)$ depending only on $y_0\text{,}$ not on $x\text{.}$ Or, renaming $y_0$ back to $y\text{,}$ $u(x,y)=C(y)$ with $C(y)$ being any function of the single variable $y\text{.}$

(b) Fix any $y_0$ and set $v(x)=u(x,y_0)\text{.}$ Then

\begin{equation*} \diff{v}{x}(x) =\pdiff{u}{x}(x,y_0) = f(x) \end{equation*}

In words, $v(x)$ has to have derivative $f(x)\text{,}$ i.e. be an antiderivative of $f(x)\text{.}$ So if $F(x)$ is any function whose derivative is $f(x)\text{,}$ i.e. if $F(x)$ is any antiderivative of $f(x)\text{,}$ then, for each fixed $y_0\text{,}$ $v(x)=u(x,y_0) = F(x) +C\text{,}$ with $C$ being a constant. The constant may be different for each different choice of $y_0\text{.}$ So $u(x,y_0) = F(x)+ C(y_0)$ with $C(y_0)$ depending only on $y_0\text{,}$ not on $x\text{.}$ Or, renaming $y_0$ back to $y\text{,}$ $u(x,y)=F(x) + C(y)$ with $F(x)$ being any antiderivative of $f(x)$ and $C(y)$ being any function of the single variable $y\text{.}$

2.8.3.3.

Solution.

(a) If $u(x,y) = x^3-3xy^2\text{,}$ then

\begin{alignat*}{2} u_x\amp= 3x^2-3y^2 \qquad\amp u_{xx}\amp= 6x \\ u_y\amp= -6xy \qquad\amp u_{yy}\amp= -6x \end{alignat*}

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6x=0$ and $x^3-3xy^2$ is harmonic.

(b) If $u(x,y) = x^3-y^3\text{,}$ then

\begin{alignat*}{2} u_x\amp= 3x^2 \qquad\amp u_{xx}\amp= 6x \\ u_y\amp= -3y^2 \qquad\amp u_{yy}\amp= -6y \end{alignat*}

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6y$ is not identically zero and $x^3-y^3$ is not harmonic.

\begin{alignat*}{2} u_x\amp= \cos(x)\,\cos(y) \qquad\amp u_{xx}\amp= -\sin(x)\,\cos(y) \\ u_y\amp= -\sin(x)\,\sin(y) \qquad\amp u_{yy}\amp= -\sin(x)\,\cos(y) \end{alignat*}

So $u_{xx}(x,y)+u_{yy}(x,y)=-2\sin(x)\,\cos(y)$ is not identically zero and $\sin(x)\,\cos(y)$ is not harmonic.

(d) If $u(x,y) = e^{7x}\,\cos(7y)\text{,}$ then

\begin{alignat*}{2} u_x\amp= 7\,e^{7x}\,\cos(7y) \qquad\amp u_{xx}\amp= 49\,e^{7x}\,\cos(7y) \\ u_y\amp= -7\,e^{7x}\ \sin(7y) \qquad\amp u_{yy}\amp= -49\,e^{7x}\,\cos(7y) \end{alignat*}

So $u_{xx}(x,y)+u_{yy}(x,y)=49\,e^{7x}\,\cos(7y)-49\,e^{7x}\,\cos(7y)=0$ and $e^{7x}\,\cos(7y)$ is harmonic.

(e) If $u(x,y) = \ln(x^2+y^2)\text{,}$ then

\begin{alignat*}{2} u_x\amp= \frac{2x}{x^2+y^2} \qquad\amp u_{xx}\amp= \frac{2}{x^2+y^2}- \frac{4x^2}{{(x^2+y^2)}^2}\\ u_y\amp= \frac{2y}{x^2+y^2} \qquad\amp u_{yy}\amp= \frac{2}{x^2+y^2}- \frac{4y^2}{{(x^2+y^2)}^2} \end{alignat*}

\begin{align*} u_{xx}(x,y)+u_{yy}(x,y)\amp=\frac{2}{x^2+y^2}- \frac{4x^2}{{(x^2+y^2)}^2} +\frac{2}{x^2+y^2}- \frac{4y^2}{{(x^2+y^2)}^2} \\ \amp=\frac{4}{x^2+y^2} - 4\frac{x^2+y^2}{{(x^2+y^2)}^2} \\ \amp=\frac{4}{x^2+y^2}-\frac{4}{x^2+y^2} \\ \amp=0 \end{align*}

and $\ln(x^2+y^2)$ is harmonic.

2.8.3.4. (✳).

Solution.

We evaluate both sides of the given PDE with $u=u(x,t) = e^{t+ax} + e^{t-ax}\text{.}$ Since

\begin{alignat*}{2} u(x,t) \amp= e^{t+ax} + e^{t-ax} \\ u_t(x,t) \amp = e^{t+ax} + e^{t-ax} \\ u_x(x,t) \amp= ae^{t+ax} - ae^{t-ax}\qquad \amp u_{xx}(x,t) \amp= a^2e^{t+ax} + a^2e^{t-ax} \end{alignat*}

the left hand side of the PDE is

\begin{equation*} 5u_t = 5e^{t+ax} + 5e^{t-ax} \end{equation*}

and the right hand side of the PDE is

\begin{align*} u_{xx} + u \amp=\big(a^2e^{t+ax} + a^2e^{t-ax}\big)+\big(e^{t+ax} + e^{t-ax}\big) \\ \amp= (a^2+1)e^{t+ax} + (a^2+1)e^{t-ax} \end{align*}

The left and right hand sides are equal if and only if

\begin{equation*} 5=(a^2+1) \iff a^2=4 \iff a=\pm 2 \end{equation*}

2.8.3.5.

Solution.

We evaluate $u_{xx}+u_{yy}+u_{zz}$ with $u = u(x,y,z) = e^{3x+4y}\sin(az)\text{.}$ Since

\begin{alignat*}{2} u(x,y,z) \amp= e^{3x+4y}\sin(az) \\ u_x(x,y,z) \amp = 3\,e^{3x+4y}\sin(az)\qquad \amp u_{xx}(x,y,z) \amp= 9\,e^{3x+4y}\sin(az) \\ u_y(x,y,z) \amp= 4\,e^{3x+4y}\sin(az)\qquad \amp u_{yy}(x,y,z) \amp= 16\,e^{3x+4y}\sin(az)\\ u_z(x,y,z) \amp= a\,e^{3x+4y}\cos(az)\qquad \amp u_{zz}(x,y,z) \amp= -a^2\,e^{3x+4y}\sin(az) \end{alignat*}

We have

\begin{equation*} u_{xx}+u_{yy}+u_{zz} = \big(9+16-a^2) e^{3x+4y}\sin(az) \end{equation*}

This is zero (for all $x\text{,}$ $y\text{,}$ $z$) if and only if

\begin{equation*} a^2=9+16=25 \iff a=\pm 5 \end{equation*}

2.8.3.6.

Solution.

We evaluate both sides of the given PDE with $u=u(x,t)=\sin(at)\,\cos(bx)\text{.}$ Since

\begin{alignat*}{2} u(x,t) \amp= \sin(at)\,\cos(bx) \\ u_t(x,t) \amp = a\,\cos(at)\,\cos(bx)\qquad \amp u_{tt}(x,t) \amp= -a^2\,\sin(at)\,\cos(bx) \\ u_x(x,t) \amp= -b\,\sin(at)\,\sin(bx)\qquad \amp u_{xx}(x,t) \amp= -b^2\sin(at)\,\cos(bx) \end{alignat*}

the left hand side of the PDE is

\begin{equation*} u_{tt} = -a^2\,\sin(at)\,\cos(bx) \end{equation*}

and the right hand side of the PDE is

\begin{equation*} u_{xx} =-b^2\,\sin(at)\,\cos(bx) \end{equation*}

The left and right hand sides are equal if and only if

\begin{equation*} a^2=b^2 \iff a=\pm b \end{equation*}

2.8.3.7.

Solution.

We simply evaluate the two terms on the left hand side when $z=z(x,y)=F\big(x^2+y^2\big)\text{.}$ By the chain rule,

\begin{align*} y\pdiff{z}{x}\amp= y\pdiff{}{x}F\big(x^2+y^2\big) =yF'\big(x^2+y^2\big) \pdiff{}{x}\,\big(x^2+y^2\big) =yF'\big(x^2+y^2\big)\left(2x\right) \\ \amp=2xy\,F'\big(x^2+y^2\big)\\ x\pdiff{z}{y}\amp= x\pdiff{}{y}F\big(x^2+y^2\big) =xF'\big(x^2+y^2\big) \pdiff{}{y}\,\big(x^2+y^2\big) =xF'\big(x^2+y^2\big)\left(2y\right) \\ \amp=2xyF'\big(x^2+y^2\big) \end{align*}

\begin{equation*} y\pdiff{z}{x} - x\pdiff{z}{y} =2xy\,F'\big(x^2+y^2\big) - 2xy\,F'\big(x^2+y^2\big) =0 \end{equation*}

and $z(x,y)=F\big(x^2+y^2\big)$ really does solve the PDE $y\pdiff{z}{x} - x\pdiff{z}{y}=0$ for any differentiable function $F\text{.}$

2.8.3.8.

Solution.

We evaluate both sides of the given PDE with $u=u(x,t)=f(t)\,\cos(2x)\text{.}$ Since

\begin{alignat*}{2} u(x,t) \amp= f(t)\,\cos(2x) \\ u_t(x,t) \amp = f'(t)\,\cos(2x)\\ u_x(x,t) \amp= -2\,f(t)\,\sin(2x)\qquad \amp u_{xx}(x,t) \amp= -4\,f(t)\,\cos(2x) \end{alignat*}

the left hand side of the PDE is

\begin{equation*} u_t(x,t) = f'(t)\,\cos(2x) \end{equation*}

and the right hand side of the PDE is

\begin{equation*} u_{xx}(x,t) = -4\,f(t)\,\cos(2x) \end{equation*}

The left and right hand sides are equal if and only if

\begin{equation*} f'(t)=-4 f(t) \end{equation*}

This is the type of ordinary differential equation that we studied in Section 3.3, on exponential growth and decay, in the CLP-1 text. We found in Theorem 3.3.2 there that the general solution to this ODE is $f(t) = Ce^{-4t}$ with $C$ being an arbitrary constant.

2.8.3.9.

Solution.

Let $u_1(x,t)$ and $u_2(x,t)$ obey $\frac{\partial^2}{\partial t^2}u_1(x,t) =\frac{\partial^2}{\partial x^2}u_1(x,t)$ and $\frac{\partial^2}{\partial t^2}u_2(x,t) =\frac{\partial^2}{\partial x^2}u_2(x,t)\text{.}$ Then $u(x,t)=a_1u_1(x,t)+a_2u_2(t,x)$ obeys

\begin{align*} u_{tt}(x,t) \amp= \frac{\partial^2}{\partial t^2} \big[a_1u_1(x,t)+a_2u_2(t,x)\big] \\ \amp=a_1\frac{\partial^2}{\partial t^2}u_1(x,t) + a_2\frac{\partial^2}{\partial t^2}u_2(x,t) \\ \amp=a_1\frac{\partial^2}{\partial x^2}u_1(x,t) + a_2\frac{\partial^2}{\partial x^2}u_2(x,t) \\ \amp= \frac{\partial^2}{\partial x^2}\big[a_1u_1(x,t)+a_2u_2(t,x)\big] \\ \amp=u_{xx}(x,t) \end{align*}

as desired.

2.8.3.10.

Solution.

We evaluate $u_{xx}+u_{yy}$ with $u(x,y)=v(ax+by\,,\,cx+dy)\text{.}$ Since, by the chain rule,

\begin{align*} u(x,y) \amp= v(ax+by\,,\,cx+dy)\\ u_x(x,y) \amp= a\,v_x(ax+by\,,\,cx+dy)+c\,v_y(ax+by\,,\,cx+dy) \\ u_y(x,y) \amp= b\,v_x(ax+by\,,\,cx+dy)+d\,v_y(ax+by\,,\,cx+dy) \\ u_{xx}(x,y) \amp= a^2\,v_{xx}(ax+by\,,\,cx+dy)+ ac\,v_{xy}(ax+by\,,\,cx+dy) \\ \amp\hskip0.5in + ca\,v_{yx}(ax+by\,,\,cx+dy) +c^2\,v_{yy}(ax+by\,,\,cx+dy) \\ u_{yy}(x,y) \amp= b^2\,v_{xx}(ax+by\,,\,cx+dy)+ bd\,v_{xy}(ax+by\,,\,cx+dy) \\ \amp\hskip0.5in + db\,v_{yx}(ax+by\,,\,cx+dy) +d^2\,v_{yy}(ax+by\,,\,cx+dy) \end{align*}

we have

\begin{align*} u_{xx}+u_{yy} \amp=(a^2+b^2) v_{xx}(ax+by\,,\,cx+dy) +(c^2+d^2) v_{yy}(ax+by\,,\,cx+dy) \\ \amp\hskip0.5in +2(ac+bd) v_{xy}(ax+by\,,\,cx+dy) \end{align*}

If $a^2+b^2=c^2+d^2\text{,}$ i.e. if $\llt a,b\rgt$ and $\llt c,d\rgt$ have the same length, then the first line of the right hand side is zero, since $v_{xx}+v_{yy}=0\text{.}$
If $\llt a,b\rgt\cdot\llt c,d\rgt=ac+bd=0\text{,}$ i.e. if $\llt a,b\rgt$ and $\llt c,d\rgt$ are mutally perpendicular, then the second line of the right hand side is zero.

So if $\llt a,b\rgt$ and $\llt c,d\rgt$ have the same length and are mutally perpendicular, then $u_{xx}+u_{yy}=0\text{.}$ The missing word is “perpendicular”.

2.8.3.11.

Solution.

In preparation for substituting into the PDE, we compute $u_{xx}\text{,}$ $u_{yy}$ and $u_{zz}\text{.}$

\begin{align*} u(x,y,z) \amp= r(x,y,z)^n = \big(x^2+y^2+z^2\big)^{n/2}\\ u_x(x,y,z)\amp= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \pdiff{}{x}\big(x^2+y^2+z^2\big) \\ \amp= nx\ \big(x^2+y^2+z^2\big)^{n/2-1}\\ u_{xx}(x,y,z)\amp= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nx\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2x) \\ \amp=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)x^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_y(x,y,z)\amp= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \pdiff{}{y}\big(x^2+y^2+z^2\big)\\ \amp= ny\ \big(x^2+y^2+z^2\big)^{n/2-1}\\ u_{yy}(x,y,z)\amp= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + ny\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2y) \\ \amp=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)y^2 \big(x^2+y^2+z^2\big)^{n/2-2} \\ u_z(x,y,z)\amp= \frac{n}{2}\,\big(x^2+y^2+z^2\big)^{n/2-1}\ \pdiff{}{z}\big(x^2+y^2+z^2\big) \\ \amp= nz\ \big(x^2+y^2+z^2\big)^{n/2-1}\\ u_{zz}(x,y,z)\amp= n\ \big(x^2+y^2+z^2\big)^{n/2-1} + nz\ (n/2-1) \big(x^2+y^2+z^2\big)^{n/2-2} (2z) \\ \amp=n\ \big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)z^2 \big(x^2+y^2+z^2\big)^{n/2-2} ) \end{align*}

\begin{align*} u_{xx}+u_{yy}+u_{zz} \amp=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ (x^2+y^2+z^2) \big(x^2+y^2+z^2\big)^{n/2-2} \\ \amp=3n\big(x^2+y^2+z^2\big)^{n/2-1} +n(n-2)\ \big(x^2+y^2+z^2\big)^{n/2-1} \\ \amp=[3n+n^2-2n]\ \big(x^2+y^2+z^2\big)^{n/2-1} \end{align*}

This is zero if and only if

\begin{equation*} n+n^2=n(1+n)=0 \iff n=0,-1 \end{equation*}

2.8.3.12.

Solution.

(a) Substituting $u(x,t)= X(x)\,T(t)$ into the given PDE yields

\begin{equation*} X(x)\,T'(t) = u_t=x\,u_x=x\, X'(x)\,T(t) \end{equation*}

Then dividing both sides by $X(x)\,T(t)$ gives

\begin{equation*} \frac{T'(t)}{T(t)} = x\,\frac{X'(x)}{X(x)} \end{equation*}

as desired.

(b) The left hand side $\frac{T'(t)}{T(t)}$ is independent of $x\text{,}$ and the right hand side $x\,\frac{X'(x)}{X(x)}$ is independent of $t\text{.}$ The left and right hand sides are equal to each other, so both are independent of both $t$ and $x\text{,}$ i.e. are constant. If we call the constant $\la\text{,}$ then

\begin{align*} \frac{T'(t)}{T(t)} \amp= x\,\frac{X'(x)}{X(x)}=\la \\ \implies T'(t)\amp=\la\,T(t),\qquad X'(x)=\frac{\la}{x} X(x) \end{align*}

(c) The equation $T'(t)=\la\,T(t)$ is the type of ordinary differential equation that we studied in Section 3.3, on exponential growth and decay, in the CLP-1 text. We found in Theorem 3.3.2 there that the general solution to this ODE is $T(t) = Ce^{\la t}$ with $C$ being an arbitrary constant, which we require to be positive to make $T>0\text{.}$ The equation $X'(x)=\frac{\la}{x} X(x)$ is a separable ODE. We studied such ODE’s in Section 2.4 in the CLP-2 text. To solve it, we divide across by $X(x)\text{,}$ giving

\begin{alignat*}{2} \frac{X'(x)}{X(x)} = \frac{\la}{x} \amp\implies \diff{}{x} \ln X(x) = \frac{\la}{x} \qquad\amp\amp\text{assuming }X,x>0 \\ \amp\implies \ln X(x) = \la\ln x+K'\qquad\amp\amp\text{with }K'\text{ constant} \\ \amp\implies X(x) = K x^\la \qquad\amp\amp\text{with }K=e^{K'}>0\text{ constant} \end{alignat*}

\begin{equation*} u(x,t)= X(x)\,T(t) = D\,e^{\la t}\,x^\la\qquad \text{with }D=CK>0\text{ a constant} \end{equation*}

solves the PDE $u_t=xu_x$ for $x>0\text{.}$

2.8.3.13.

Solution.

By the chain rule,

\begin{align*} \diff{}{t}u\big(X(t),Y(t)\big) \amp=u_x\big(X(t),Y(t)\big)\,\diff{X}{t}(t) +u_y\big(X(t),Y(t)\big)\,\diff{Y}{t}(t) \\ \amp=\al\big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\be\big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) \end{align*}

But evaluating $\al(x,y)\,u_x(x,y) +\be(x,y)\,u_y(x,y)=0$ at $x=X(t)\text{,}$ $y=Y(t)$ gives

\begin{equation*} \al\big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\be\big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) =0 \end{equation*}

\begin{equation*} \diff{}{t}u\big(X(t),Y(t)\big)=0 \end{equation*}

2.8.3.14.

Solution.

(a) Suppose that $u(x,y)$ obeys the PDE

\begin{equation*} 3u_x(x,y) + 6u_y(x,y)=u(x,y) \end{equation*}

Define $v(X,Y) = u(X, Y+2X)\text{.}$ Then, by the chain rule,

\begin{align*} v_X(X,Y)\amp=\pdiff{}{X}\big[ u(X, Y+2X)\big] \\ \amp=u_x(X, Y+2X) +2u_y(X, Y+2X) \\ \amp=\frac{1}{3}\big\{3u_x(X, Y+2X) + 6u_y(X, Y+2X)\big\} \\ \amp=\frac{1}{3} u(X, Y+2X) \\ \amp=\frac{1}{3} v(X,Y) \end{align*}

(b) Define $v(X,Y) = u(X, Xe^Y)\text{.}$ Then, by the chain rule,

\begin{align*} v_X(X,Y)\amp=\pdiff{}{X}\big[ u(X, Xe^Y)\big] \\ \amp=u_x(X, Xe^Y) +e^Yu_y(X, Xe^Y) \end{align*}

Now notice that if $xu_x(x,y) + yu_y(x,y)=u(x,y)\text{,}$ then, evaluating at $x=X$ and $y=Xe^Y$ gives

\begin{equation*} Xu_x(X, Xe^Y) + Xe^Yu_y(X, Xe^Y)=u(X, Xe^Y) \end{equation*}

\begin{align*} v_X(X,Y) \amp=\frac{1}{X}\big\{X u_x(X, Xe^Y) +X e^Y u_y(X, Xe^Y) \big\} \\ \amp=\frac{1}{X} u(X, Xe^Y) \\ \amp=\frac{1}{X} v(X,Y) \end{align*}

2.9 Maximum and Minimum Values
2.9.4 Exercises

2.9.4.1. (✳).

Solution.

a) (i) $\vnabla f$ is zero at critical points. The point $T$ is a local maximum and the point $U$ is a saddle point. The remaining points $P\text{,}$ $R\text{,}$ $S\text{,}$ are not critical points.

(a) (ii) Only $U$ is a saddle point.

(a) (iii) We have $f_y(x,y) \gt 0$ if $f$ increases as you move vertically upward through $(x,y)\text{.}$ Looking at the diagram, we see

\begin{align*} f_y(P) &\lt 0& f_y(Q) &\lt 0& f_y(R) &=0\\ f_y(S) &\gt 0& f_y(T) &=0& f_y(U) &=0 \end{align*}

So only $S$ works.

(a) (iv) The directional derivative of $f$ in the direction $\llt 0,-1\rgt$ is $\vnabla f\cdot \llt 0,-1\rgt =-f_y\text{.}$ It is negative if and only if $f_y \gt 0\text{.}$ So, again, only $S$ works.

(b) (i) The function $z=F(x,2)$ is increasing at $x=1\text{,}$ because the $y=2.0$ graph in the diagram has positive slope at $x=1\text{.}$ So $F_x(1,2) \gt 0\text{.}$

(b) (ii) The function $z=F(x,2)$ is also increasing (though slowly) at $x=2\text{,}$ because the $y=2.0$ graph in the diagram has positive slope at $x=2\text{.}$ So $F_x(2,2) \gt 0\text{.}$ So $F$ does not have a critical point at $(2,2)\text{.}$

(b) (iii) From the diagram the looks like $F_x(1,1.9) \gt F_x(1,2.0) \gt F_x(1,2.1)\text{.}$ That is, it looks like the slope of the $y=1.9$ graph at $x=1$ is larger than the slope of the $y=2.0$ graph at $x=1\text{,}$ which in turn is larger than the slope of the $y=2.1$ graph at $x=1\text{.}$ So it looks like $F_x(1,y)$ decreases as $y$ increases through $y=2\text{,}$ and consequently $F_{xy}(1,2) \lt 0\text{.}$

2.9.4.2.

Solution.

The height $\sqrt{x^2+y^2}$ at $(x,y)$ is the distance from $(x,y)$ to $(0,0)\text{.}$ So the minimum height is zero at $(0,0,0)\text{.}$ The surface is a cone. The cone has a point at $(0,0,0)$ and the derivatives $z_x$ and $z_y$ do not exist there. The maximum height is achieved when $(x,y)$ is as far as possible from $(0,0)\text{.}$ The highest points are at $(\pm 1,\pm 1,\sqrt{2})\text{.}$ There $z_x$ and $z_y$ exist but are not zero. These points would not be the highest points if it were not for the restriction $|x|,|y|\le 1\text{.}$

2.9.4.3.

Solution.

Define $f(t)=g(\va+t\vd)$ and determine $t_0$ by $\vx_0=\va+t_0\vd\text{.}$ Then $f'(t)=\vnabla g(\va+t\vd)\cdot\vd\text{.}$ To see this, write $\va=\llt a_1,a_2,a_3\rgt$ and $\vd=\llt d_1,d_2,d_3\rgt\text{.}$ Then

\begin{equation*} f(t)=g(a_1+td_1,a_2+td_2,a_3+td_3) \end{equation*}

So, by the chain rule,

\begin{align*} f'(t) &= \pdiff{g}{x}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_1\\ &\hskip1in +\pdiff{g}{y}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_2\\ &\hskip1in +\pdiff{g}{z}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_3\\ &=\vnabla g(\va+t\vd)\cdot\vd \end{align*}

Then $\vx_0$ is a local max or min of the restriction of $g$ to the specified line if and only if $t_0$ is a local max or min of $f(t)\text{.}$ If so, $f'(t_0)$ necessarily vanishes. So if $\vx_0$ is a local max or min of the restriction of $g$ to the specified line, then $\vnabla g(\vx_0)\cdot\vd=0\text{,}$ i.e. $\vnabla g(\vx_0)\perp\vd\text{,}$ and $\vx_0=\va+t_0\vd$ for some $t_0\text{.}$ The second condition is to ensure that $x_0$ lies on the line.

2.9.4.4. (✳).

Solution.

(a)

The level curve $z=0$ is $y^2-x^2=0\text{,}$ which is the pair of $45^\circ$ lines $y=\pm x\text{.}$
When $C \gt 0\text{,}$ the level curve $z=C^4$ is ${(y^2-x^2)}^2=C^4\text{,}$ which is the pair of hyperbolae $y^2-x^2=C^2\text{,}$ $y^2-x^2=-C^2$ or
\begin{equation*} y=\pm\sqrt{x^2+C^2}\qquad x=\pm\sqrt{y^2+C^2} \end{equation*}
The hyperbola $y^2-x^2=C^2$ crosses the $y$--axis (i.e. the line $x=0$) at $(0,\pm C)\text{.}$ The hyperbola $y^2-x^2=-C^2$ crosses the $x$--axis (i.e. the line $y=0$) at $(\pm C,0)\text{.}$

Here is a sketch showing the level curves $z=0\text{,}$ $z=1$ (i.e. $C=1$), and $z=16$ (i.e. $C=2$).

(b) As $f_x(x,y)=-4x(y^2-x^2)$ and $f_y(x,y)=4y(y^2-x^2)\text{,}$ we have $f_x(0,0) = f_y(0,0) = 0$ so that $(0,0)$ is a critical point. Note that

$f(0,0)=0\text{,}$
$f(x,y)\ge 0$ for all $x$ and $y\text{.}$

So $(0,0)$ is a local (and also absolute) minimum.

\begin{alignat*}{2} f_{xx}(x,y) &=-4y^2+12x^2\qquad& f_{xx}(x,y)&=0\\ f_{yy}(x,y) &= 12y^2-4x^2\qquad& f_{yy}(x,y)&=0\\ f_{xy}(x,y) &=-8xy\qquad& f_{xx}(x,y)&=0 \end{alignat*}

As $f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2=0 \text{,}$ the Second Derivative Test (Theorem 2.9.16) tells us absolutely nothing.

2.9.4.5. (✳).

Solution.

Write $f(x,y) = x^2 +cxy +y^2\text{.}$ Then

\begin{align*} f_x(x,y)&= 2x+cy & f_x(0,0) = 0\\ f_y(x,y)&= cx+2y & f_y(0,0) = 0\\ f_{xx}(x,y) &= 2\\ f_{xy}(x,y) &= c\\ f_{yy}(x,y) &= 2 \end{align*}

As $f_x(0,0)=f_y(0,0)=0\text{,}$ we have that $(0,0)$ is always a critical point for $f\text{.}$ According to the Second Derivative Test, $(0,0)$ is also a saddle point for $f$ if

\begin{gather*} f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 \lt 0 \iff 4-c^2 \lt 0 \iff |c| \gt 2 \end{gather*}

As a remark, the Second Derivative Test provides no information when the expression $f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 =0\text{,}$ i.e. when $c=\pm 2\text{.}$ But when $c=\pm 2\text{,}$

\begin{equation*} f(x,y) = x^2 \pm 2xy + y^2 =(x\pm y)^2 \end{equation*}

and $f$ has a local minimum, not a saddle point, at $(0,0)\text{.}$

2.9.4.6. (✳).

Solution.

To find the critical points we will need the gradient of $f\text{,}$ and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of $f$ up to order two. Here they are.

\begin{alignat*}{3} f&=x^3 - y^3 - 2xy + 6\\ f_x&=3x^2-2y & f_{xx}&=6x \qquad & f_{xy}&= -2\\ f_y&=-3y^2-2x \qquad & f_{yy}&=-6y\qquad & f_{yx}&= -2 \end{alignat*}

The critical points are the solutions of

\begin{equation*} f_x=3x^2-2y=0 \qquad f_y=-3y^2-2x = 0 \end{equation*}

Substituting $y=\frac{3}{2}x^2\text{,}$ from the first equation, into the second equation gives

\begin{align*} -3\left(\frac{3}{2}x^2\right)^2-2x =0 &\iff -2x\left(\frac{3^3}{2^3}x^3+1\right)=0\\ &\iff x=0,\ -\frac{2}{3} \end{align*}

So there are two critical points: $(0,0)\text{,}$ $\left(-\frac{2}{3},\frac{2}{3}\right)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$0\times 0 -(-2)^2 \lt 0$		saddle point
$\left(-\frac{2}{3},\frac{2}{3}\right)$	$(-4)\times (-4)-(-2)^2 \gt 0$	$-4$	local max

2.9.4.7. (✳).

Solution.

\begin{alignat*}{3} f&=x^3 + x^2y + xy^2 - 9x\\ f_x&=3x^2+2xy+y^2-9 \qquad & f_{xx}&=6x+2y \qquad & f_{xy}&= 2x+2y\\ f_y&=x^2+2xy & f_{yy}&=2x\qquad & f_{yx}&= 2x+2y \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{alignat*}{2} f_x&=3x^2+2xy+y^2-9&&=0 \tag{E1}\\ f_y&=x(x+2y) &&= 0 \tag{E2} \end{alignat*}

Equation (E2) is satisfied if at least one of $x=0\text{,}$ $x=-2y\text{.}$

If $x=0\text{,}$ equation (E1) reduces to $y^2-9=0\text{,}$ which is satisfied if $y=\pm 3\text{.}$
If $x=-2y\text{,}$ equation (E1) reduces to
\begin{gather*} 0=3(-2y)^2+2(-2y)y+y^2-9=9y^2-9 \end{gather*}
which is satisfied if $y=\pm 1\text{.}$

So there are four critical points: $(0,3)\text{,}$ $(0,-3)\text{,}$ $(-2,1)$ and $(2,-1)\text{.}$ The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,3)$	$(6)\times (0)-(6)^2 \lt 0$		saddle point
$(0,-3)$	$(-6)\times (0)-(-6)^2 \lt 0$		saddle point
$(-2,1)$	$(-10)\times (-4)-(-2)^2 \gt 0$	$-10$	local max
$(2,-1)$	$(10)\times (4)-(2)^2 \gt 0$	$10$	local min

2.9.4.8. (✳).

Solution.

The region of interest is

\begin{equation*} D=\Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5} \end{equation*}

First observe that, on the boundary of this region, at least one of $x\text{,}$ $y$ and $z$ is zero. So $f(x,y,z)=x^2 y^2 z$ is zero on the boundary. As $f$ takes values which are strictly bigger than zero at all points of $D$ that are not on the boundary, the minimum value of $f$ is $0$ on

\begin{align*} \partial D &= \big\{\ (x,y,z)\ \big|\ x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5,\\ &\hskip2.5in \text{at least one of }x,y,z\text{ zero}\ \big\} \end{align*}

The maximum value of $f$ will be taken at a critical point. On $D$

\begin{align*} f &= x^2 y^2 (5-2x-y) =5x^2 y^2 - 2x^3 y^2 -x^2 y^3 \end{align*}

So the critical points are the solutions of

\begin{align*} 0&=f_x(x,y) = 10xy^2 -6x^2y^2 -2xy^3\\ 0&=f_y(x,y) = 10x^2y -4x^3y -3x^2y^2 \end{align*}

or, dividing by the first equation by $xy^2$ and the second equation by $x^2y\text{,}$ (recall that $x,y\ne 0$)

\begin{alignat*}{2} 10 -6x -2y &=0 &\qquad\text{or}\qquad 3x+y&=5\\ 10 -4x -3y &=0 &\qquad\text{or}\qquad 4x+3y&=10 \end{alignat*}

Substituting $y=5-3x\text{,}$ from the first equation, into the second equation gives

\begin{equation*} 4x+3(5-3x)=10 \implies -5x +15 =10 \implies x=1,\ y=5-3(1)=2 \end{equation*}

So the maximum value of $f$ is $(1)^2(2)^2(5-2-2)=4$ at $(1,2,1)\text{.}$

2.9.4.9.

Solution.

\begin{alignat*}{3} f&=x^2+y^2+x^2y+4\\ f_x&=2x+2xy\qquad & f_{xx}&=2+2y\qquad & f_{xy}&= 2x\\ f_y&=2y+x^2 & f_{yy}&=2 \end{alignat*}

The critical points are the solutions of

\begin{alignat*}{3} & & &f_x=0 & &f_y=0\\ &\iff\quad & &2x(1+y)=0 & &2y+x^2=0\\ &\iff & &x=0\text{ or }y=-1\qquad & &2y+x^2=0 \end{alignat*}

When $x=0\text{,}$ $y$ must be $0\text{.}$ When $y=-1\text{,}$ $x^2$ must be $2\text{.}$ So, there are three critical points: $(0,0)\text{,}$ $\big(\pm\sqrt{2},-1\big)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$2\times 2-0^2 \gt 0$	$2 \gt 0$	local min
$(\sqrt{2},-1)$	$0\times 2-(2\sqrt{2})^2 \lt 0$		saddle point
$(-\sqrt{2},-1)$	$0\times 2-(-2\sqrt{2})^2 \lt 0$		saddle point

2.9.4.10. (✳).

Solution.

\begin{alignat*}{3} f&=x^3 + x^2 - 2xy + y^2 - x\\ f_x&=3x^2+2x-2y-1 \qquad & f_{xx}&=6x+2 \qquad & f_{xy}&= -2\\ f_y&=-2x+2y & f_{yy}&=2\qquad & f_{yx}&= -2 \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{alignat*}{2} f_x&=3x^2+2x-2y-1&&=0 \tag{E1}\\ f_y&=-2x+2y &&= 0 \tag{E2} \end{alignat*}

Substituting $y=x\text{,}$ from (E2), into (E1) gives

\begin{gather*} 3x^2-1=0 \iff x=\pm\frac{1}{\sqrt{3}}=0 \end{gather*}

So there are two critical points: $\pm\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)$	$(2\sqrt{3}+2)\times (2)-(-2)^2 \gt 0$	$2\sqrt{3}+2 \gt 0$	local min
$-\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)$	$(-2\sqrt{3}+2)\times (2)-(-2)^2 \lt 0$		saddle point

2.9.4.11. (✳).

Solution.

To find the critical points we will need the gradient of $f$ and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of $f$ up to order two. Here they are.

\begin{alignat*}{3} f&=x^3 + xy^2 - 3x^2 - 4y^2 + 4\\ f_x&=3x^2+y^2-6x\qquad & f_{xx}&=6x-6 \qquad & f_{xy}&= 2y\\ f_y&=2xy -8y & f_{yy}&=2x-8\qquad & f_{yx}&= 2y \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{equation*} f_x=3x^2+y^2-6x=0 \qquad f_y=2(x-4)y = 0 \end{equation*}

The second equation is satisfied if at least one of $x=4\text{,}$ $y=0$ are satisfied.

If $x=4\text{,}$ the first equation reduces to $y^2=-24\text{,}$ which has no real solutions.
If $y=0\text{,}$ the first equation reduces to $3x(x-2)=0\text{,}$ which is satisfied if either $x=0$ or $x=2\text{.}$

So there are two critical points: $(0,0)\text{,}$ $(2,0)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times(-8)-(0)^2 \gt 0$	$-6$	local max
$(2,0)$	$6\times(-4)-(0)^2 \lt 0$		saddle point

2.9.4.12.

Solution.

The specified function and its first order derivatives are

\begin{gather*} f(x,y)=xy-x^3y^2\qquad f_x(x,y)=y-3x^2y^2\qquad f_y(x,y)=x-2x^3y \end{gather*}

First, we find the critical points.

\begin{alignat*}{5} f_x&=0 & &\quad\iff\quad &y(1-3x^2y)&=0 & &\quad\iff\quad & &y=0 \text{ or } 3x^2y=1\\ f_y&=0 & &\quad\iff\quad &x(1-2x^2y)&=0 & &\quad\iff\quad & &x=0 \text{ or } 2x^2y=1 \end{alignat*}
- If $y=0\text{,}$ we cannot have $2x^2y=1\text{,}$ so we must have $x=0\text{.}$
- If $3x^2y=1\text{,}$ we cannot have $x=0\text{,}$ so we must have $2x^2y=1\text{.}$ Dividing gives $1=\frac{3x^2y}{2x^2y}=\frac{3}{2}$ which is impossible.
So the only critical point in the square is $(0,0)\text{.}$ There $f=0\text{.}$
Next, we look at the part of the boundary with $x=0\text{.}$ There $f=0\text{.}$
Next, we look at the part of the boundary with $y=0\text{.}$ There $f=0\text{.}$
Next, we look at the part of the boundary with $x=1\text{.}$ There $f=y-y^2\text{.}$ As $\diff{}{y}(y-y^2)=1-2y\text{,}$ the max and min of $y-y^2$ for $0\le y\le 1$ must occur either at $y=0\text{,}$ where $f=0\text{,}$ or at $y=\half\text{,}$ where $f=\frac{1}{4}\text{,}$ or at $y=1\text{,}$ where $f=0\text{.}$
Next, we look at the part of the boundary with $y=1\text{.}$ There $f=x-x^3\text{.}$ As $\diff{}{x}(x-x^3)=1-3x^2\text{,}$ the max and min of $x-x^3$ for $0\le x\le 1$ must occur either at $x=0\text{,}$ where $f=0\text{,}$ or at $x=\frac{1}{\sqrt{3}}\text{,}$ where $f=\frac{2}{3\sqrt{3}}\text{,}$ or at $x=1\text{,}$ where $f=0\text{.}$

All together, we have the following candidates for max and min.

point	$(0,0)$	$x=0$	$y=0$	$(1,0)$	$(1,\half)$	$(1,1)$	$(0,1)$	$(\frac{1}{\sqrt{3}},1)$	$(1,1)$
value of $f$	$0$	$0$	$0$	$0$	$\frac{1}{4}$	$0$	$0$	$\frac{2}{3\sqrt{3}}$	$0$
	min	min	min	min		min	min	max	min

The largest and smallest values of $f$ in this table are

\begin{equation*} \text{min}=0\qquad \text{max}=\frac{2}{3\sqrt{3}}\approx0.385 \end{equation*}

2.9.4.13.

Solution.

The specified temperature and its first order derivatives are

\begin{align*} T(x,y)&=(x+y)e^{-x^2-y^2}\\ T_x(x,y)&=(1-2x^2-2xy)e^{-x^2-y^2}\\ T_y(x,y)&=(1-2xy-2y^2)e^{-x^2-y^2} \end{align*}

First, we find the critical points.
\begin{alignat*}{3} T_x&=0 & &\quad\iff\quad & 2x(x+y)&=1\\ T_y&=0 & &\quad\iff\quad & 2y(x+y)&=1 \end{alignat*}
As $x+y$ may not vanish, this forces $x=y$ and then $x=y=\pm\half\text{.}$ So the only critical points are $(\half,\half)$ and $(-\half,-\half)\text{.}$
On the boundary $x=\cos\theta$ and $y=\sin\theta\text{,}$ so $T=(\cos\theta+\sin\theta)e^{-1}\text{.}$ This is a periodic function and so takes its max and min at zeroes of $\diff{T}{\theta}=\big(-\sin\theta+\cos\theta\big)e^{-1}\text{.}$ That is, when $\sin\theta=\cos\theta\text{,}$ which forces $\sin\theta=\cos\theta=\pm\frac{1}{\sqrt{2}}\text{.}$

All together, we have the following candidates for max and min.

point	$(\half,\half)$	$(-\half,-\half)$	$(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$	$(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
value of $T$	$\frac{1}{\sqrt{e}}\approx0.61$	$-\frac{1}{\sqrt{e}}$	$\frac{\sqrt{2}}{e}\approx0.52$	$-\frac{\sqrt{2}}{e}$
	max	min

The largest and smallest values of $T$ in this table are

\begin{equation*} \text{min}=-\frac{1}{\sqrt{e}}\qquad \text{max}=\frac{1}{\sqrt{e}} \end{equation*}

2.9.4.14. (✳).

Solution.

Both of the functions $f(x, y) = \sqrt{x^2 + y^2}$ (i.e. (ii)) and $f(x, y) = x^2 + y^2$ (i.e. (iv)) are invariant under rotations around the $(0,0)\text{.}$ So their level curves are circles centred on the origin. In polar coordinates $\sqrt{x^2 + y^2}$ is $r\text{.}$ So the sketched level curves of the function in (ii) are $r = 0, 0.1, 0.2, \ldots , 1.9, 2\text{.}$ They are equally spaced. So at this point, we know that the third picture goes with (iv) and the fourth picture goes with (ii).

Notice that the lines $x=y\text{,}$ $x=-y$ and $y=0$ are all level curves of the function $f(x, y) = y(x + y)(x - y) + 1$ (i.e. of (iii)) with $f=1\text{.}$ So the first picture goes with (iii). And the second picture goes with (i).

Here are the pictures with critical points marked on them. There are saddle points where level curves cross and there are local max’s or min’s at “bull’s eyes”.

(i)

(ii)

(iii)

(iv)

2.9.4.15. (✳).

Solution.

\begin{alignat*}{3} f&=x^3+3xy+3y^2-6x-3y-6\\ f_x&=3x^2+3y-6 & f_{xx}&=6x \qquad & f_{xy}&= 3\\ f_y&=3x+6y-3 \qquad & f_{yy}&=6\qquad & f_{yx}&= 3 \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{alignat*}{2} f_x&=3x^2+3y-6&&=0 \tag{E1}\\ f_y&=3x\ +6y-3 &&= 0 \tag{E2} \end{alignat*}

Subtracting equation (E2) from twice equation (E1) gives

\begin{gather*} 6x^2-3x-9=0 \iff (2x-3)(3x+3)=0 \end{gather*}

So we must have either $x=\frac{3}{2}$ or $x=-1\text{.}$

If $x=\frac{3}{2}\text{,}$ (E2) reduces to $\frac{9}{2}+6y-3=0$ so $y=-\frac{1}{4}\text{.}$
If $x=-1\text{,}$ (E2) reduces to $-3+6y-3=0$ so $y=1\text{.}$

So there are two critical points: $\big(\frac{3}{2},-\frac{1}{4}\big)$ and $(-1,1)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$\big(\frac{3}{2},-\frac{1}{4}\big)$	$(9)\times (6)-(3)^2 \gt 0$	$9$	local min
$(-1,1)$	$(-6)\times (6)-(3)^2 \lt 0$		saddle point

2.9.4.16. (✳).

Solution.

(a) To find the critical points we will need the gradient of $h$ and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of $f$ up to order two. Here they are.

\begin{alignat*}{3} h&=y(4 - x^2 - y^2)\\ h_x&=-2xy & h_{xx}&=-2y \qquad & h_{xy}&= -2x\\ h_y&=4-x^2-3y^2 \qquad & h_{yy}&=-6y\qquad & h_{yx}&= -2x \end{alignat*}

(Of course, $h_{xy}$ and $h_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{equation*} h_x=-2xy=0 \qquad h_y=4-x^2-3y^2 = 0 \end{equation*}

The first equation is satisfied if at least one of $x=0\text{,}$ $y=0$ are satisfied.

If $x=0\text{,}$ the second equation reduces to $4-3y^2=0\text{,}$ which is satisfied if $y=\pm\frac{2}{\sqrt{3}}\text{.}$
If $y=0\text{,}$ the second equation reduces to $4-x^2=0$ which is satisfied if $x=\pm 2\text{.}$

So there are four critical points: $\left(0,\frac{2}{\sqrt{3}}\right)\text{,}$ $\left(0,-\frac{2}{\sqrt{3}}\right)\text{,}$ $(2,0)\text{,}$ $(-2,0)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$h_{xx}h_{yy}-h_{xy}^2$	$h_{xx}$	type
$\left(0,\frac{2}{\sqrt{3}}\right)$	$\left(\frac{-4}{\sqrt{3}}\right)\times \left(-\frac{12}{\sqrt{3}}\right)-(0)^2 \gt 0$	$\frac{-4}{\sqrt{3}}$	local max
$\left(0,-\frac{2}{\sqrt{3}}\right)$	$\left(\frac{4}{\sqrt{3}}\right)\times \left(\frac{12}{\sqrt{3}}\right)-(0)^2 \gt 0$	$\frac{4}{\sqrt{3}}$	local min
$(2,0)$	$0\times 0-(-4)^2 \lt 0$		saddle point
$(-2,0)$	$0\times 0-(4)^2 \lt 0$		saddle point

(b) The absolute max and min can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle $x^2+y^2=1\text{.}$

Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of $h\text{.}$ We found all of the critical points of $h$ in part (a). Since $2 \gt 1$ and $\frac{2}{\sqrt{3}} \gt 1$ none of the critical points are in the disk.
At each point of $x^2+y^2=1$ we have $h(x,y)=3y$ with $-1\le y\le 1\text{.}$ Clearly the maximum value is $3$ (at $(0,1)$) and the minimum value is $-3$ (at $(0,-1)$).

So all together, the maximum and minimum values of $h(x,y)$ in $x^2+y^2\le 1$ are $3$ (at $(0,1)$) and $-3$ (at $(0,-1)$), respectively.

2.9.4.17. (✳).

Solution.

The maximum and minimum must either occur at a critical point or on the boundary of $R\text{.}$

The critical points are the solutions of
\begin{align*} 0&=f_x(x,y) = 2-2x\\ 0&=f_y(x,y) = -8y \end{align*}
So the only critical point is $(1,0)\text{.}$
On the side $x=-1\text{,}$ $-1\le y\le 1$ of the boundary of $R$
\begin{equation*} f(-1,y) = 2-4y^2 \end{equation*}
This function decreases as $|y|$ increases. So its maximum value on $-1\le y\le 1$ is achieved at $y=0$ and its minimum value is achieved at $y=\pm 1\text{.}$
On the side $x=3\text{,}$ $-1\le y\le 1$ of the boundary of $R$
\begin{equation*} f(3,y) = 2-4y^2 \end{equation*}
This function decreases as $|y|$ increases. So its maximum value on $-1\le y\le 1$ is achieved at $y=0$ and its minimum value is achieved at $y=\pm 1\text{.}$
On both sides $y=\pm 1\text{,}$ $-1\le x\le 3$ of the boundary of $R$
\begin{equation*} f(x,\pm 1) = 1+2x-x^2 = 2 -(x-1)^2 \end{equation*}
This function decreases as $|x-1|$ increases. So its maximum value on $-1\le x\le 3$ is achieved at $x=1$ and its minimum value is achieved at $x= 3$ and $x=-1$ (both of whom are a distance $2$ from $x=1$).

So we have the following candidates for the locations of the min and max

point	$(1,0)$	$(-1,0)$	$(1,\pm 1)$	$(-1,\pm 1)$	$(3,0)$	$(3,\pm 1)$
value of $f$	$6$	$2$	$2$	$-2$	$2$	$-2$
	max			min		min

So the minimum is $-2$ and the maximum is $6\text{.}$

2.9.4.18. (✳).

Solution.

Since $\vnabla h = \llt -4\,,\,-2 \rgt$ is never zero, $h$ has no critical points and the minimum of $h$ on the disk $x^2+y^2\le 1$ must be taken on the boundary, $x^2+y^2=1\text{,}$ of the disk. To find the minimum on the boundary, we parametrize $x^2+y^2\le 1$ by $x=\cos\theta\text{,}$ $y=\sin\theta$ and find the minimum of

\begin{gather*} H(\theta) = -4\cos\theta -2\sin\theta +6 \end{gather*}

Since

\begin{gather*} 0=H'(\theta) = 4\sin\theta -2\cos\theta \implies x=\cos\theta = 2\sin\theta =2y \end{gather*}

\begin{gather*} 1=x^2+y^2 = 4y^2 +y^2 =5 y^2 \implies y = \pm\frac{1}{\sqrt{5}},\ x = \pm\frac{2}{\sqrt{5}} \end{gather*}

At these two points

\begin{gather*} h = 6-4x-2y = 6 - 10y =6 \mp \frac{10}{\sqrt{5}} =6 \mp 2\sqrt{5} \end{gather*}

The minimum is $6-2\sqrt{5}\text{.}$

2.9.4.19. (✳).

Solution.

(a) Thinking a little way ahead, to find the critical points we will need the gradient of $f$ and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of $f$ up to order two. Here they are.

\begin{alignat*}{3} f&=xy(x + y - 3)\\ f_x&=2xy+y^2-3y & f_{xx}&=2y \qquad & f_{xy}&= 2x+2y-3\\ f_y&=x^2+2xy-3x \qquad & f_{yy}&=2x\qquad & f_{yx}&= 2x+2y-3 \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{equation*} f_x=y(2x+y-3)=0 \qquad f_y=x(x+2y-3) = 0 \end{equation*}

The first equation is satisfied if at least one of $y=0\text{,}$ $y=3-2x$ are satisfied.

If $y=0\text{,}$ the second equation reduces to $x(x-3)=0\text{,}$ which is satisfied if either $x=0$ or $x=3\text{.}$
If $y=3-2x\text{,}$ the second equation reduces to $x(x+6-4x-3)=x(3-3x)=0$ which is satisfied if $x=0$ or $x=1\text{.}$

So there are four critical points: $(0,0)\text{,}$ $(3,0)\text{,}$ $(0,3)\text{,}$ $(1,1)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$0\times 0-(-3)^2 \lt 0$		saddle point
$(3,0)$	$0\times 6-(3)^2 \lt 0$		saddle point
$(0,3)$	$6\times 0-(3)^2 \lt 0$		saddle point
$(1,1)$	$2\times 2-(1)^2 \gt 0$	2	local min

(b) The absolute max and min can occur either in the interior of the triangle or on the boundary of the triangle. The boundary of the triangle consists of the three line segments.

\begin{align*} L_1 &= \Set{(x,y)}{x=0,\ 0\le y\le 8}\\ L_2 &= \Set{(x,y)}{y=0,\ 0\le x\le 8}\\ L_3 &= \Set{(x,y)}{x+y=8,\ 0\le x\le 8} \end{align*}

Any absolute max or min in the interior of the triangle must also be a local max or min and, since there are no singular points, must also be a critical point of $f\text{.}$ We found all of the critical points of $f$ in part (a). Only one of them, namely $(1,1)$ is in the interior of the triangle. (The other three critical points are all on the boundary of the triangle.) We have $f(1,1) = -1\text{.}$
At each point of $L_1$ we have $x=0$ and so $f(x,y)=0\text{.}$
At each point of $L_2$ we have $y=0$ and so $f(x,y)=0\text{.}$
At each point of $L_3$ we have $f(x,y)=x(8-x)(5)=40x-5x^2=5[8x-x^2]$ with $0\le x\le 8\text{.}$ As $\diff{}{x}\big(40x-5x^2\big)= 40-10x\text{,}$ the max and min of $40x-5x^2$ on $0\le x\le 8$ must be one of $5\big[8x-x^2\big]_{x=0}=0$ or $5\big[8x-x^2\big]_{x=8}=0$ or $5\big[8x-x^2\big]_{x=4}=80\text{.}$

So all together, we have the following candidates for max and min, with the max and min indicated.

point(s)	$(1,1)$	$L_1$	$L_2$	$(0,8)$	$(8,0)$	$(4,4)$
value of $f$	$-1$	$0$	$0$	$0$	$0$	$80$
	min					max

2.9.4.20. (✳).

Solution.

Thinking a little way ahead, to find the critical points we will need the gradient of $f\text{,}$ and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of $f$ up to order two. Here they are.

\begin{alignat*}{3} f&=3x^2 y + y^3 - 3x^2 - 3y^2 + 4\\ f_x&=6xy-6x & f_{xx}&=6y-6 \qquad & f_{xy}&= 6x\\ f_y&=3x^2+3y^2-6y \qquad & f_{yy}&=6y-6\qquad & f_{yx}&= 6x \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{equation*} f_x=6x(y-1)=0 \qquad f_y=3x^2+3y^2-6y = 0 \end{equation*}

The first equation is satisfied if at least one of $x=0\text{,}$ $y=1$ are satisfied.

If $x=0\text{,}$ the second equation reduces to $3y^2-6y=0\text{,}$ which is satisfied if either $y=0$ or $y=2\text{.}$
If $y=1\text{,}$ the second equation reduces to $3x^2 -3=0$ which is satisfied if $x=\pm 1\text{.}$

So there are four critical points: $(0,0)\text{,}$ $(0,2)\text{,}$ $(1,1)\text{,}$ $(-1,1)\text{.}$

The classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times (-6)-(0)^2 \gt 0$	$-6$	local max
$(0,2)$	$6\times 6-(0)^2 \gt 0$	6	local min
$(1,1)$	$0\times 0-(6)^2 \lt 0$		saddle point
$(-1,1)$	$0\times 0-(-6)^2 \lt 0$		saddle point

2.9.4.21. (✳).

Solution.

(a) Since

\begin{alignat*}{4} f&=2x^3 - 6xy + y^2 +4y\\ f_x&=6x^2-6y & f_{xx}&=12x\qquad && f_{xy}&= -6\\ f_y&=-6x+2y+4\qquad & f_{yy}&=2 \end{alignat*}

the critical points are the solutions of

\begin{alignat*}{3} & & &f_x=0\qquad & &f_y=0\\ &\iff\qquad& &y=x^2\qquad & &y-3x+2=0\\ &\iff& &y=x^2 & &x^2-3x+2=0\\ &\iff& &y=x^2 & &x=1\text{ or }2 \end{alignat*}

So, there are two critical points: $(1,1),\ (2,4)\text{.}$

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(1,1)$	$12\times 2-(-6)^2 \lt 0$		saddle point
$(2,4)$	$24\times 2-(-6)^2 \gt 0$	$24$	local min

(b) There are no critical points in the interior of the allowed region, so both the maximum and the minimum occur only on the boundary. The boundary consists of the line segments (i) $x=1\text{,}$ $0\le y\le 1\text{,}$ (ii) $y=1\text{,}$ $0\le x\le 1$ and (iii) $y=1-x\text{,}$ $0\le x\le 1\text{.}$

First, we look at the part of the boundary with $x=1\text{.}$ There $f=y^2-2y+2\text{.}$ As $\diff{}{y}(y^2-2y+2)=2y-2$ vanishes only at $y=1\text{,}$ the max and min of $y^2-2y+2$ for $0\le y\le 1$ must occur either at $y=0\text{,}$ where $f=2\text{,}$ or at $y=1\text{,}$ where $f=1\text{.}$
Next, we look at the part of the boundary with $y=1\text{.}$ There $f=2x^3-6x+5\text{.}$ As $\diff{}{x}(2x^3-6x+5)=6x^2-6\text{,}$ the max and min of $2x^3-6x+5$ for $0\le x\le 1$ must occur either at $x=0\text{,}$ where $f=5\text{,}$ or at $x=1\text{,}$ where $f=1\text{.}$
Next, we look at the part of the boundary with $y=1-x\text{.}$ There
\begin{gather*} f=2x^3-6x(1-x)+(1-x)^2+4(1-x)=2x^3+7x^2-12x+5 \end{gather*}
As
\begin{align*} \diff{}{x}(2x^3+7x^2-12x+5)&=6x^2+14x-12 =2\big(3x^2+7x-6\big)\\ &=2(3x-2)(x+3) \end{align*}
the max and min of $2x^3+7x^2-12x+5$ for $0\le x\le 1$ must occur either at $x=0\text{,}$ where $f=5\text{,}$ or at $x=1\text{,}$ where $f=2\text{,}$ or at $x=\frac{2}{3}\text{,}$ where
\begin{gather*} f=2(\frac{8}{27})-6(\frac{2}{3})(\frac{1}{3})+\frac{1}{9}+\frac{4}{3} =\frac{16-36+3+36}{27} =\frac{19}{27} \end{gather*}

So all together, we have the following candidates for max and min, with the max and min indicated.

point	$(1,0)$	$(1,1)$	$(0,1)$	$\big(\frac{2}{3},\frac{1}{3}\big)$
value of $f$	$2$	$1$	$5$	$\frac{19}{27}$
			max	min

2.9.4.22. (✳).

Solution.

We have

\begin{alignat*}{3} f(x,y)&=x^4+y^4-4xy+2\quad & f_x(x,y)&=4x^3-4y\quad & f_{xx}(x,y)&=12x^2\\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2\\ & & & &f_{xy}(x,y)&=-4 \end{alignat*}

At a critical point

\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3\\ &\iff x=x^9\text{ and }y=x^3\\ &\iff x(x^8-1)=0,\ y=x^3\\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{align*}

Here is a table giving the classification of each of the three critical points.

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$0\times 0-(-4)^2 \lt 0$		saddle point
$(1,1)$	$12\times 12-(-4)^2 \gt 0$	12	local min
$(-1,-1)$	$12\times 12-(-4)^2 \gt 0$	12	local min

2.9.4.23. (✳).

Solution.

(a) We have

\begin{alignat*}{2} f(x,y)&=xy(x+2y-6)\\ f_x(x,y)&=2xy+2y^2-6y\qquad & f_{xx}(x,y)&=2y\\ f_y(x,y)&=x^2+4xy-6x\qquad & f_{yy}(x,y)&=4x\\ & & f_{xy}(x,y)&=2x+4y-6 \end{alignat*}

At a critical point

\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff 2y(x+y-3)=0\text{ and }x(x+4y-6)=0\\ &\hskip-0.5in\iff \{y=0\text{ or }x+y=3\}\text{ and }\{x=0\text{ or }x+4y=6\}\\ &\hskip-0.5in\iff \{x=y=0\}\text{ or }\{y=0,\ x+4y=6\}\\ &\text{ or }\{x+y=3,\ x=0\} \text{ or }\{x+y=3,\ x+4y=6\}\\ &\hskip-0.5in\iff (x,y)=(0,0)\text{ or }(6,0)\text{ or }(0,3)\text{ or }(2,1) \end{align*}

Here is a table giving the classification of each of the four critical points.

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$0\times 0-(-6)^2 \lt 0$		saddle point
$(6,0)$	$0\times 24-6^2 \lt 0$		saddle point
$(0,3)$	$6\times 0-6^2 \lt 0$		saddle point
$(2,1)$	$2\times 8-2^2 \gt 0$	2	local min

(b) Observe that $xy=4$ and $x+2y=6$ intersect when $x=6-2y$ and

\begin{align*} (6-2y)y=4 &\iff 2y^2-6y+4=0 \iff 2(y-1)(y-2)=0\\ &\iff (x,y) = (4,1)\text{ or }(2,2) \end{align*}

The shaded region in the sketch below is $D\text{.}$

None of the critical points are in $D\text{.}$ So the max and min must occur at either $(2,2)$ or $(4,1)$ or on $xy=4\text{,}$ $2 \lt x \lt 4$ (in which case $F(x)=f\big(x,\frac{4}{x}\big)=4\big(x+\frac{8}{x}-6)$ obeys $F'(x)=4-\frac{32}{x^2}=0\iff x=\pm2\sqrt{2}$) or on $x+2y=6\text{,}$ $2 \lt x \lt 4$ (in which case $f(x,y)$ is identically zero). So the min and max must occur at one of

$(x,y)$	$f(x,y)$
$(2,2)$	$2\times 2(2+2\times2-6)=0$
$(4,1)$	$4\times 1(4+2\times 1-6)=0$
$(2\sqrt{2},\sqrt{2})$	$4(2\sqrt{2}+2\sqrt{2}-6) \lt 0$

The maximum value is $0$ and the minimum value is $4(4\sqrt{2}-6) \approx -1.37\text{.}$

2.9.4.24. (✳).

Solution.

We have

\begin{alignat*}{3} f(x,y)&=x^4+y^4-4xy\qquad & f_x(x,y)&=4x^3-4y\qquad & f_{xx}(x,y)&=12x^2\\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2\\ & & & &f_{xy}(x,y)&=-4 \end{alignat*}

At a critical point

\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \iff x=x^9\text{ and }y=x^3\\ &\iff x(x^8-1)=0,\ y=x^3\\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{align*}

Here is a table giving the classification of each of the three critical points.

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$0\times 0-(-4)^2 \lt 0$		saddle point
$(1,1)$	$12\times 12-(-4)^2 \gt 0$	12	local min
$(-1,-1)$	$12\times 12-(-4)^2 \gt 0$	12	local min

2.9.4.25. (✳).

Solution.

The coldest point must be either on the boundary of the plate or in the interior of the plate.

On the semi--circular part of the boundary $0\le y\le 2$ and $x^2+y^2=4$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln 5-y\text{.}$ The smallest value of $\ln 5-y$ is taken when $y$ is as large as possible, i.e. when $y=2\text{,}$ and is $\ln 5 -2\approx -0.391\text{.}$
On the flat part of the boundary, $y=0$ and $-2\le x\le 2$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln\big(1+x^2\big)\text{.}$ The smallest value of $\ln\big(1+x^2\big)$ is taken when $x$ is as small as possible, i.e. when $x=0\text{,}$ and is $0\text{.}$
If the coldest point is in the interior of the plate, it must be at a critical point of $T(x,y)\text{.}$ Since
\begin{gather*} T_x(x,y)=\frac{2x}{1+x^2+y^2}\qquad T_y(x,y)=\frac{2y}{1+x^2+y^2}-1 \end{gather*}
a critical point must have $x=0$ and $\frac{2y}{1+x^2+y^2}-1=0\text{,}$ which is the case if and only if $x=0$ and $2y-1-y^2=0\text{.}$ So the only critical point is $x=0,\ y=1\text{,}$ where $T=\ln 2-1\approx -0.307\text{.}$

Since $-0.391 \lt -0.307 \lt 0\text{,}$ the coldest temperture is $-0.391$ and the coldest point is $(0,2)\text{.}$

2.9.4.26. (✳).

Solution.

We have

\begin{alignat*}{3} f(x,y)&=x^3+xy^2-x\qquad & f_x(x,y)&=3x^2+y^2-1\qquad & f_{xx}(x,y)&=6x\\ & & f_y(x,y)&=2xy & f_{yy}(x,y)&=2x\\ & & & & f_{xy}(x,y)&=2y \end{alignat*}

At a critical point

\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff xy=0\text{ and }3x^2+y^2=1\\ &\hskip-0.75in\iff \{x=0\text{ or }y=0\}\text{ and }3x^2+y^2=1\\ &\hskip-0.75in\iff (x,y)=(0,1)\text{ or }(0,-1)\text{ or }\left(\frac{1}{\sqrt{3}},0\right) \text{ or }\left(-\frac{1}{\sqrt{3}},0\right) \end{align*}

Here is a table giving the classification of each of the four critical points.

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,1)$	$0\times 0-2^2 \lt 0$		saddle point
$(0,-1)$	$0\times 0-(-2)^2 \lt 0$		saddle point
$\big(\frac{1}{\sqrt{3}},0\big)$	$2\sqrt{3}\times\frac{2}{\sqrt{3}}-0^2 \gt 0$	$2\sqrt{3}$	local min
$\big(-\frac{1}{\sqrt{3}},0\big)$	$-2\sqrt{3}\times \big(-\frac{2}{\sqrt{3}}\big)-0^2 \gt 0$	$-2\sqrt{3}$	local max

2.9.4.27. (✳).

Solution.

(a) We have

\begin{alignat*}{3} g(x,y)&=x^2-10y-y^2 \qquad& g_x(x,y)&=2x \qquad& g_{xx}(x,y)&=2\\ & & g_y(x,y)&=-10-2y\qquad & g_{yy}(x,y)&=-2\\ & & & &g_{xy}(x,y)&=0 \end{alignat*}

At a critical point

\begin{align*} g_x(x,y)=g_y(x,y)=0 &\iff 2x=0\text{ and }-10-2y=0\\ &\iff (x,y)=(0,-5) \end{align*}

Since $g_{xx}(0,-5)g_{yy}(0,-5)-g_{xy}(0,-5)^2=2\times(-2)-0^2 \lt 0\text{,}$ the critical point is a saddle point.

(b) The extrema must be either on the boundary of the region or in the interior of the region.

On the semi-elliptical part of the boundary $-2\le y\le 0$ and $x^2+4y^2=16$ so that $g=x^2-10y-y^2=16-10y-5y^2=21-5(y+1)^2\text{.}$ This has a minimum value of 16 (at $y=0,-2$) and a maximum value of 21 (at $y=-1$). You could also come to this conclusion by checking the critical point of $16-10y-5y^2$ (i.e. solving $\diff{}{y}(16-10y-5y^2)=0$) and checking the end points of the allowed interval (namely $y=0$ and $y=-2$).
On the flat part of the boundary $y=0$ and $-4\le x\le 4$ so that $g=x^2\text{.}$ The smallest value is taken when $x=0$ and is $0$ and the largest value is taken when $x=\pm 4$ and is $16\text{.}$
If an extremum is in the interior of the plate, it must be at a critical point of $g(x,y)\text{.}$ The only critical point is not in the prescribed region.

Here is a table giving all candidates for extrema:

$(x,y)$	$g(x,y)$
$(0,-2)$	$16$
$(\pm 4,0)$	$16$
$(\pm \sqrt{12},-1)$	$21$
$(0,0)$	$0$

From the table the smallest value of $g$ is $0$ at $(0,0)$ and the largest value is $21$ at $(\pm 2\sqrt{3},-1)\text{.}$

2.9.4.28. (✳).

Solution.

We have

\begin{alignat*}{2} f(x,y)&=x^3-3xy^2-3x^2-3y^2\\ f_x(x,y)&=3x^2-3y^2-6x\qquad & f_{xx}(x,y)&=6x-6\\ f_y(x,y)&=-6xy-6y & f_{yy}(x,y)&=-6x-6\\ & &f_{xy}(x,y)&=-6y \end{alignat*}

At a critical point

\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff 3(x^2-y^2-2x)=0\text{ and }-6y(x+1)=0\\ &\hskip-0.5in\iff \{x=-1\text{ or }y=0\}\text{ and }x^2-y^2-2x=0\\ &\hskip-0.5in\iff (x,y)=(-1,\sqrt{3})\text{ or }(-1,-\sqrt{3})\text{ or }(0,0) \text{ or }(2,0) \end{align*}

Here is a table giving the classification of each of the four critical points.

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times(-6)-0^2 \gt 0$	$-6$	local max
$(2,0)$	$6\times(-18)-0^2 \lt 0$		saddle point
$(-1,\sqrt{3})$	$(-12)\times0-(-6\sqrt{3})^2 \lt 0$		saddle point
$(-1,-\sqrt{3})$	$(-12)\times 0-(6\sqrt{3})^2 \lt 0$		saddle point

2.9.4.29. (✳).

Solution.

The maximum must be either on the boundary of $D$ or in the interior of $D\text{.}$

On the circular part of the boundary $r=2\text{,}$ $0\le\theta\le\frac{\pi}{2}$ (in polar coordinates) so that $f=r^2\cos\theta\sin\theta e^{-r^2/2}=2\sin(2\theta)e^{-2}\text{.}$ This has a maximum value of $2e^{-2}$ at $\theta=\frac{\pi}{4}$ or $x=y=\sqrt{2}\text{.}$
On the two flat parts of the boundary $x=0$ or $y=0$ so that $f=0\text{.}$
If the maximum is in the interior of $D\text{,}$ it must be at a critical point of $f(x,y)\text{.}$ Since
\begin{gather*} f_x(x,y)=e^{-(x^2 + y^2) / 2}\big[y-x^2y\big]\qquad f_y(x,y)=e^{-(x^2 + y^2) / 2}\big[x-xy^2\big] \end{gather*}
$(x,y)$ is a critical point if and only if
\begin{align*} &y(1-x^2)=0\text{ and }x(1-y^2)=0\\ &\iff \{y=0\text{ or }x=1\text{ or }x=-1\}\text{ and } \{x=0\text{ or }y=1\text{ or }y=-1\} \end{align*}
There are two critical points with $x,y\ge 0\text{,}$ namely $(0,0)$ and $(1,1)\text{.}$ The first of these is on the boundary of $D$ and the second is in the interior of $D\text{.}$

Here is a table giving all candidates for the maximum:

$(x,y)$	$g(x,y)$
$(\sqrt{2},\sqrt{2})$	$2e^{-2}\approx0.271$
$(x,0)$	$0$
$(0,y)$	$0$
$(1,1)$	$e^{-1}\approx0.368$

Since $e \gt 2\text{,}$ we have that $2e^{-2}=e^{-1}\frac{2}{e} \lt e^{-1}$ and the largest value is $e^{-1}\text{.}$

2.9.4.30.

Solution.

Suppose that the bends are made a distance $x$ from the ends of the fence and that the bends are through an angle $\theta\text{.}$ Here is a sketch of the enclosure.

It consists of a rectangle, with side lengths $100-2x$ and $x\sin\theta\text{,}$ together with two triangles, each of height $x\sin\theta$ and base length $x\cos\theta\text{.}$ So the enclosure has area

\begin{align*} A(x,\theta)&=(100-2x)x\sin\theta+2\cdot\half\cdot x\sin\theta\cdot x\cos\theta\\ &=(100x-2x^2)\sin\theta+\half x^2\sin(2\theta) \end{align*}

The maximize the area, we need to solve

\begin{align*} 0=A_x&=(100-4x)\sin\theta+x\sin(2\theta) \\ &\hskip1in\implies\quad (100-4x)+2x\cos\theta=0\\ 0=A_\theta&=(100x-2x^2)\cos\theta+x^2\cos(2\theta)\\ &\hskip0.5in\implies\quad (100-2x)\cos\theta+x\cos(2\theta)=0 \end{align*}

Here we have used that the fence of maximum area cannot have $\sin\theta=0$ or $x=0\text{,}$ because in either of these two cases, the area enclosed will be zero. The first equation forces $\cos\theta=-\frac{100-4x}{2x}$ and hence $\cos(2\theta)=2\cos^2\theta-1=\frac{(100-4x)^2}{2x^2}-1\text{.}$ Substituting these into the second equation gives

\begin{alignat*}{2} & & -(100-2x)\frac{100-4x}{2x}+x\Big[\frac{(100-4x)^2}{2x^2}-1\Big]&=0\\ &\implies & -(100-2x)(100-4x)+(100-4x)^2-2x^2&=0\\ &\implies & 6x^2-200x&=0\\ &\implies & x=\frac{100}{3} \quad\cos\theta=-\frac{-100/3}{200/3}=\frac{1}{2}\quad \theta&=60^\circ\\ & & A= \left(100\frac{100}{3}-2\frac{100^2}{3^2}\right) \frac{\sqrt{3}}{2}+\frac{1}{2} \frac{100^2}{3^2}\frac{\sqrt{3}}{2} &=\frac{2500}{\sqrt{3}} \end{alignat*}

2.9.4.31.

Solution.

Suppose that the box has side lengths $x\text{,}$ $y$ and $z\text{.}$ Here is a sketch.

Because the box has to have volume $V$ we need that $V=xyz\text{.}$ We wish to minimize the area $A=xy+2yz+2xz$ of the four sides and bottom. Substituting in $z=\frac{V}{xy}\text{,}$

\begin{align*} A&=xy+2\frac{V}{x}+2\frac{V}{y}\\ A_x&=y-2\frac{V}{x^2}\\ A_y&=x-2\frac{V}{y^2} \end{align*}

To minimize, we want $A_x=A_y=0\text{,}$ which is the case when $yx^2=2V,\ xy^2=2V\text{.}$ This forces $yx^2=xy^2\text{.}$ Since $V=xyz$ is nonzero, neither $x$ nor $y$ may be zero. So $x=y=(2V)^{1/3}\text{,}$ $z=2^{-2/3}V^{1/3}\text{.}$

2.9.4.32. (✳).

Solution.

(a) The maximum and minimum can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle $x^2+y^2=4\text{.}$

Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of $h\text{.}$ Since $T_x=-8x$ and $T_y=-2y\text{,}$ the only critical point is $(x,y)=(0,0)\text{,}$ where $T=20\text{.}$ Since $4x^2+y^2\ge 0\text{,}$ we have $T(x,y)=20-4x^2-y^2\le 20\text{.}$ So the maximum value of $T$ (even in $\bbbr^2$) is $20\text{.}$
At each point of $x^2+y^2=4$ we have $T(x,y)=20-4x^2-y^2=20 -4x^2-(4-x^2)=16-3x^2$ with $-2\le x\le 2\text{.}$ So $T$ is a minimum when $x^2$ is a maximum. Thus the minimum value of $T$ on the disk is $16-3(\pm 2)^2=4\text{.}$

So all together, the maximum and minimum values of $T(x,y)$ in $x^2+y^2\le 4$ are $20$ (at $(0,0)$) and $4$ (at $(\pm 2,0)$), respectively.

(b) To increase its temperature as quickly as possible, the ant should move in the direction of the temperature gradient $\vnabla T(1,1)=\llt -8x,-2y\rgt\Big|_{(x,y)=(1,1)}=\llt -8,-2\rgt\text{.}$ A unit vector in that direction is $\frac{1}{\sqrt{17}}\llt -4,-1\rgt\text{.}$

\begin{gather*} \vnabla T(1,1)\cdot\llt -2,-1\rgt =\llt -8,-2\rgt\cdot\llt -2,-1\rgt =18 \end{gather*}

(d) We are being asked to find the $(x,y)=(x,2-x^2)$ which maximizes

\begin{gather*} T\big(x,2-x^2\big) =20 -4x^2-\big(2-x^2\big)^2 = 16-x^4 \end{gather*}

The maximum of $16-x^4$ is obviously $16$ at $x=0\text{.}$ So the ant should go to $\big(0,2-0^2\big)=(0,2)\text{.}$

2.9.4.33. (✳).

Solution.

\begin{alignat*}{3} f&=3kx^2 y + y^3 - 3x^2 - 3y^2 + 4\\ f_x&=6kxy-6x & f_{xx}&=6ky-6 \qquad & f_{xy}&= 6kx\\ f_y&=3kx^2+3y^2-6y \qquad & f_{yy}&=6y-6\qquad & f_{yx}&= 6kx \end{alignat*}

(Of course, $f_{xy}$ and $f_{yx}$ have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

\begin{equation*} f_x=6x(ky-1)=0 \qquad f_y=3kx^2+3y^2-6y = 0 \end{equation*}

The first equation is satisfied if at least one of $x=0\text{,}$ $y=\frac{1}{k}$ are satisfied. (Recall that $k \gt 0\text{.}$)

If $x=0\text{,}$ the second equation reduces to $3y(y-2)=0\text{,}$ which is satisfied if either $y=0$ or $y=2\text{.}$
If $y=\frac{1}{k}\text{,}$ the second equation reduces to $3kx^2+\frac{3}{k^2}-\frac{6}{k}=3kx^2+\frac{3}{k^2}(1-2k)=0\text{.}$

Case $k \lt \frac{1}{2}\text{:}$ If $k \lt \frac{1}{2}\text{,}$ then $\frac{3}{k^2}(1-2k) \gt 0$ and the equation $3kx^2+\frac{3}{k^2}(1-2k)=0$ has no real solutions. In this case there are two critical points: $(0,0)\text{,}$ $(0,2)$ and the classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times(-6)-(0)^2 \gt 0$	$-6$	local max
$(0,2)$	$(12k-6)\times 6-(0)^2 \lt 0$		saddle point

Case $k=\frac{1}{2}\text{:}$ If $k=\frac{1}{2}\text{,}$ then $\frac{3}{k^2}(1-2k)=0$ and the equation $3kx^2+\frac{3}{k^2}(1-2k)=0$ reduces to $3kx^2=0$ which has as its only solution $x=0\text{.}$ We have already seen this third critical point, $x=0\text{,}$ $y=\frac{1}{k}=2\text{.}$ So there are again two critical points: $(0,0)\text{,}$ $(0,2)$ and the classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times(-6)-(0)^2 \gt 0$	$-6$	local max
$(0,2)$	$(12k-6)\times 6-(0)^2=0$		unknown

Case $k \gt \frac{1}{2}\text{:}$ If $k \gt \frac{1}{2}\text{,}$ then $\frac{3}{k^2}(1-2k) \lt 0$ and the equation $3kx^2+\frac{3}{k^2}(1-2k)=0$ reduces to $3kx^2=\frac{3}{k^2}(2k-1)$ which has two solutions, namely $x=\pm\sqrt{\frac{1}{k^3}(2k-1)}\text{.}$ So there are four critical points: $(0,0)\text{,}$ $(0,2)\text{,}$ $\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)$ and $\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)$ and the classification is

$\Atop{\text{critical}}{\text{point}}$	$f_{xx}f_{yy}-f_{xy}^2$	$f_{xx}$	type
$(0,0)$	$(-6)\times(-6)-(0)^2 \gt 0$	$-6$	local max
$(0,2)$	$(12k-6)\times 6-(0)^2 \gt 0$	$12k-6 \gt 0$	local min
$\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)$	$(6-6)\times (\frac{6}{k}-6)-( \gt 0)^2 \lt 0$		saddle point
$\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)$	$(6-6)\times (\frac{6}{k}-6)-( \lt 0)^2 \lt 0$		saddle point

2.9.4.34. (✳).

Solution.

(a) For $x,y \gt 0\text{,}$

\begin{align*} f_x&=2-\frac{1}{x^2y}=0\iff y=\frac{1}{2x^2}\cr f_y&=4-\frac{1}{xy^2}=0 \end{align*}

Substituting $y=\frac{1}{2x^2}\text{,}$ from the first equation, into the second gives $4-4x^3=0$ which forces $x=1\text{,}$ $y=\half\text{.}$ At $x=1\text{,}$ $y=\half\text{,}$

\begin{equation*} f\big(1,\half\big)=2+2+2=6 \end{equation*}

(b) The second derivatives are

\begin{equation*} f_{xx}(x,y)=\frac{2}{x^3y}\qquad f_{xy}(x,y)=\frac{1}{x^2y^2}\qquad f_{yy}(x,y)=\frac{2}{xy^3} \end{equation*}

In particular

\begin{equation*} f_{xx}\big(1,\half\big)=4\qquad f_{xy}\big(1,\half\big)=4\qquad f_{yy}\big(1,\half\big)=16 \end{equation*}

Since $f_{xx}\big(1,\half\big)f_{yy}\big(1,\half\big)- f_{xy}\big(1,\half\big)^2=4\times 16-4^2=48 \gt 0$ and $f_{xx}\big(1,\half\big)=4 \gt 0\text{,}$ the point $\big(1,\half\big)$ is a local minimum.

(c) As $x$ or $y$ tends to infinity (with the other at least zero), $2x+4y$ tends to $+\infty\text{.}$ As $(x,y)$ tends to any point on the first quadrant part of the $x$- and $y$--axes, $\frac{1}{xy}$ tends to $+\infty\text{.}$ Hence as $x$ or $y$ tends to the boundary of the first quadrant (counting infinity as part of the boundary), $f(x,y)$ tends to $+\infty\text{.}$ As a result $\big(1,\half\big)$ is a global (and not just local) minimum for $f$ in the first quadrant. Hence $f(x,y)\ge f\big(1,\half\big)=6$ for all $x,y \gt 0\text{.}$

2.9.4.35.

Solution.

We wish to choose $m$ and $b$ so as to minimize the (square of the) rms error $E(m,b)=\sum\limits_{i=1}^n (mx_i+b-y_i)^2\text{.}$

\begin{alignat*}{3} 0&=\pdiff{E}{m}&&=\smsum_{i=1}^n 2(mx_i+b-y_i)x_i &&=m\Big[\smsum_{i=1}^n 2x^2_i\Big]+b\Big[\smsum_{i=1}^n 2x_i\Big] -\Big[\smsum_{i=1}^n 2x_iy_i\Big]\\ 0&=\pdiff{E}{b}&&=\smsum_{i=1}^n 2(mx_i+b-y_i) &&=m\Big[\smsum_{i=1}^n 2x_i\Big]+b\Big[\smsum_{i=1}^n 2\Big] -\Big[\smsum_{i=1}^n 2y_i\Big] \end{alignat*}

There are a lot of symbols in those two equations. But remember that only two of them, namely $m$ and $b\text{,}$ are unknowns. All of the $x_i$’s and $y_i$’s are given data. We can make the equations look a lot less imposing if we define $S_x=\smsum_{i=1}^n x_i\text{,}$ $S_y=\smsum_{i=1}^n y_i\text{,}$ $S_{x^2}=\smsum_{i=1}^n x^2_i$ and $S_{xy}=\smsum_{i=1}^n x_iy_i\text{.}$ In terms of this notation, the two equations are (after dividing by two)

\begin{align*} S_{x^2}\, m+S_x\, b&=S_{xy} \tag{1}\\ S_{x}\,m+n\,b&=S_{y} \tag{2} \end{align*}

This is a system of two linear equations in two unknowns. One way

²⁵

This procedure is probably not the most efficient one. But it has the advantage that it always works, it does not require any ingenuity on the part of the solver, and it generalizes easily to larger linear systems of equations.

to solve them, is to use one of the two equations to solve for one of the two unknowns in terms of the other unknown. For example, equation (2) gives that

\begin{equation*} b=\frac{1}{n}\big(S_y-S_x\,m\big) \end{equation*}

If we now substitute this into equation (1) we get

\begin{equation*} S_{x^2}\, m+\frac{S_x}{n}\big(S_y-S_x\,m\big)=S_{xy} \implies \left(S_{x^2}-\frac{S_x^2}{n}\right)m = S_{xy} -\frac{S_xS_y}{n} \end{equation*}

which is a single equation in the single unkown $m\text{.}$ We can easily solve it for $m\text{.}$ It tells us that

\begin{equation*} m=\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2} \end{equation*}

Then substituting this back into $b=\frac{1}{n}\big(S_y-S_x\,m\big)$ gives us

\begin{equation*} b=\frac{S_y}{n} -\frac{S_x}{n}\left(\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2}\right) =\frac{S_yS_{x^2}-S_xS_{xy}}{nS_{x^2} -S_x^2} \end{equation*}

2.10 Lagrange Multipliers
2.10.2 Exercises

2.10.2.1. (✳).

Solution.

(a) $f(x, y) = x^2 +y^2$ is the square of the distance from the point $(x,y)$ to the origin. There are points on the curve $xy=1$ that have either $x$ or $y$ arbitrarily large and so whose distance from the origin is arbitrarily large. So $f$ has no maximum on the curve. On the other hand $f$ will have a minimum, achieved at the points of $xy=1$ that are closest to the origin.

(b) On the curve $xy=1$ we have $y=\frac{1}{x}$ and hence $f=x^2+\frac{1}{x^2}\text{.}$ As

\begin{gather*} \diff{}{x}\left(x^2+\frac{1}{x^2}\right) =2x-\frac{2}{x^3} =\frac{2}{x^3}(x^4-1) \end{gather*}

and as no point of the curve has $x=0\text{,}$ the minimum is achieved when $x=\pm 1\text{.}$ So the minima are at $\pm (1,1)\text{,}$ where $f$ takes the value $2\text{.}$

2.10.2.2.

Solution.

(a) As you leave $(x_0,y_0,z_0)$ walking in the direction $\vd\ne\vZero\text{,}$ $f$ has to be decreasing, or at least not increasing, because $f$ takes its largest value on $S$ at $(x_0,y_0,z_0)\text{.}$ So the directional derivative

\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}\le 0 \tag{E1} \end{equation*}

As you leave $(x_0,y_0,z_0)$ walking in the direction $-\vd\ne\vZero\text{,}$ $f$ also has to be decreasing, or at least not increasing, because $f$ still takes its largest value on $S$ at $(x_0,y_0,z_0)\text{.}$ So the directional derivative

\begin{equation*} D_{-\vd/|\vd|}f(x_0,y_0,z_0)=-\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}\le 0 \tag{E2} \end{equation*}

(E1) and (E2) can both be true only if the directional derivative

\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|} = 0 \end{equation*}

(b) By Definition 2.7.5, the directional derivative is

\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|} \end{equation*}

As $(x_0,y_0,z_0)$ is a local maximum for $f$ on $S\text{,}$ the method of Lagrange multipliers, Theorem 2.10.2, gives that $\vnabla f(x_0,y_0,z_0) =\la\vnabla g(x_0,y_0,z_0)$ for some $\la\text{.}$
By Theorem 2.5.5, the vector $\vnabla g(x_0,y_0,z_0)$ is perpendicular to the surface $S$ at $(x_0,y_0,z_0)\text{,}$ and, in particular, is perpendicular to the vector $\vd\text{,}$ which after all is tangent to the surface $S$ at $(x_0,y_0,z_0)\text{.}$

So $\vnabla g(x_0,y_0,z_0)\cdot\vd=0$ and the directional derivative

\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}=0 \end{equation*}

2.10.2.3.

Solution.

We are to find the maximum and minimum of $f(x,y,z)=x+y-z$ subject to the constraint $g(x,y,z) = x^2+y^2+z^2 -1=0\text{.}$ According to the method of Lagrange multipliers, we need to find all solutions to

\begin{alignat*}{5} f_x = 1 = 2\la x &= \la g_x \ &&\implies\ && x=\frac{1}{2\la} \tag{E1}\\ f_y = 1 = 2\la y &= \la g_y \ &&\implies\ && y=\frac{1}{2\la}\tag{E2}\\ f_z = -1 = 2\la z &= \la g_z \ &&\implies\ && z=-\frac{1}{2\la} \tag{E3}\\ x^2+y^2+z^2&=1 \ &&\implies\ &&3\left(\frac{1}{2\la}\right)^2=1 \ &&\implies\ & \la&=\pm\frac{\sqrt{3}}{2}\tag{E4} \end{alignat*}

Thus the critical points are $\big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\text{,}$ where $f=-\sqrt{3}$ and $\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)\text{,}$ where $f=\sqrt{3}\text{.}$ So, the max is $f=\sqrt{3}$ and the min is $f=-\sqrt{3}\text{.}$

2.10.2.4.

Solution.

The ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ passes through the point $(1,2,1)$ if and only if $\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=1\text{.}$ We are to minimize $f(a,b,c)=\frac{4}{3}\pi abc$ subject to the constraint that $g(a,b,c) = \frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2} -1=0\text{.}$ According to the method of Lagrange multipliers, we need to find all solutions to

\begin{alignat*}{3} f_a = \frac{4}{3}\pi bc = -\frac{2\la}{a^3} &= \la g_a \quad&&\implies\quad && \frac{3}{2\pi}\la=-a^3bc \tag{E1}\\ f_b = \frac{4}{3}\pi ac = - \frac{8\la}{b^3} &= \la g_b \quad&&\implies\quad && \frac{3}{2\pi}\la=-\frac{1}{4}ab^3c \tag{E2}\\ f_c = \frac{4}{3}\pi ab = -\frac{2\la}{c^3} &= \la g_c \quad&&\implies\quad && \frac{3}{2\pi}\la=-abc^3 \tag{E3}\\ \frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}&=1 \tag{E4} \end{alignat*}

The equations $-\frac{3}{2\pi}\la=a^3bc=\frac{1}{4}ab^3c$ force $b=2a$ (since we want $a,b,c \gt 0$). The equations $-\frac{3}{2\pi}\la=a^3bc=abc^3$ force $a=c\text{.}$ Hence, by (E4),

\begin{gather*} 1=\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=\frac{3}{a^2} \implies a=c=\sqrt{3},\ b=2\sqrt{3} \end{gather*}

2.10.2.5. (✳).

Solution.

So we are to minimize $f(x,y) = x^2+y^2 $ subject to the constraint $g(x,y) = x^2 y -1=0\text{.}$ According to the method of Lagrange multipliers, we need to find all solutions to

\begin{align*} f_x = 2x &=2 \la xy = \la g_x \tag{E1}\\ f_y = 2y &= \la x^2 = \la g_y \tag{E2}\\ x^2y&=1 \tag{E3} \end{align*}

Equation (E1), $2x(1-\la y)=0\text{,}$ gives that either $x=0$ or $\la y=1\text{.}$
But substituting $x=0$ in (E3) gives $0=1\text{,}$ which is impossible.
Also note that $\la=0$ is impossible, since substituting $\la=0$ in (E1) and (E2) gives $x=y=0\text{,}$ which violates (E3).
So $y=\frac{1}{\la}\text{.}$
Substituting $y=\frac{1}{\la}$ into (E2) gives $\frac{2}{\la} = \la x^2$ or $x^2=\frac{2}{\la^2}\text{.}$ So $x=\pm\frac{\sqrt{2}}{\la}\text{.}$
Substituting $y=\frac{1}{\la}\text{,}$ $x=\pm\frac{\sqrt{2}}{\la}$ into (E3) gives $\frac{2}{\la^3} =1$ or $\la^3 = 2$ or $\la= \root{3}\of{2}\text{.}$
$\la= 2^{1/3}$ gives $x=\pm 2^{\frac{1}{2}-\frac{1}{3}}=\pm 2^{\frac{1}{6}}$ and $y= 2^{-\frac{1}{3}}\text{.}$

So the two critical points are $\big(2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big)$ and $\big(-2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big)\text{.}$ For both of these critical points,

\begin{equation*} x^2+y^2= 2^{\frac{1}{3}} + 2^{-\frac{2}{3}} = 2^{\frac{1}{3}} + \frac{1}{2}2^{\frac{1}{3}} =\frac{3}{2}\root{3}\of{2} =\frac{3}{\sqrt[3]{4}} \end{equation*}

2.10.2.6. (✳).

Solution.

Let $r$ and $h$ denote the radius and height, respectively, of the cylinder. We can always choose our coordinate system so that the axis of the cylinder is parallel to the $z$--axis.

If the axis of the cylinder does not lie exactly on the $z$--axis, we can enlarge the cylinder sideways. (See the first figure below. It shows the $y=0$ cross--section of the cylinder.) So we can assume that the axis of the cylinder lies on the $z$--axis
If the top and/or the bottom of the cylinder does not touch the sphere $x^2+y^2+z^2=1\text{,}$ we can enlarge the cylinder vertically. (See the second figure below.)
So we may assume that the cylinder is
\begin{equation*} \Set{(x,y,z)}{x^2+y^2\le r^2,\ -h/2\le z\le h/2} \end{equation*}
with $r^2+(h/2)^2=1\text{.}$ See the third figure below.

So we are to maximize the volume, $f(r,h) = \pi r^2 h\text{,}$ of the cylinder subject to the constraint $g(r,h) = r^2+ \frac{h^2}{4} -1=0\text{.}$ According to the method of Lagrange multipliers, we need to find all solutions to

\begin{align*} f_r = 2\pi r h &=2 \la r = \la g_r \tag{E1}\\ f_h = \pi r^2 &= \la \frac{h}{2} = \la g_h \tag{E2}\\ r^2+ \frac{h^2}{4}&=1 \tag{E3} \end{align*}

Equation (E1), $2r(\pi h-\la)=0\text{,}$ gives that either $r=0$ or $\la=\pi h\text{.}$ Clearly $r=0$ cannot give the maximum volume, so $\la=\pi h\text{.}$ Substituting $\la=\pi h$ into (E2) gives

\begin{equation*} \pi r^2 = \frac{1}{2}\pi h^2 \implies r^2=\frac{h^2}{2} \end{equation*}

Substituting $r^2=\frac{h^2}{2}$ into (E3) gives

\begin{gather*} \frac{h^2}{2} + \frac{h^2}{4} =1 \implies h^2 =\frac{4}{3} \end{gather*}

Clearly both $r$ and $h$ have to be positive, so $h=\frac{2}{\sqrt{3}}$ and $r=\sqrt{\frac{2}{3}}\text{.}$

2.10.2.7. (✳).

Solution.

For this problem the objective function is $f(x,y) = xy$ and the constraint function is $g(x,y)=x^2 + 2y^2 - 1\text{.}$ To apply the method of Lagrange multipliers we need $\vnabla f$ and $\vnabla g\text{.}$ So we start by computing the first order derivatives of these functions.

\begin{equation*} f_x=y\qquad f_y=x\qquad g_x=2x\qquad g_y=4y \end{equation*}

So, according to the method of Lagrange multipliers, we need to find all solutions to

\begin{align*} y&=\la (2x) \tag{E1}\\ x&=\la (4y) \tag{E2}\\ x^2+2y^2-1&=0 \tag{E3} \end{align*}

First observe that none of $x\text{,}$ $y\text{,}$ $\la$ can be zero, because if at least one of them is zero, then (E1) and (E2) force $x=y=0\text{,}$ which violates (E3). Dividing (E1) by (E2) gives $\frac{y}{x} = \frac{x}{2y}$ so that $x^2=2y^2$ or $x=\pm \sqrt{2}\,y\text{.}$ Then (E3) gives

\begin{gather*} 2y^2+2y^2=1 \iff y=\pm\frac{1}{2} \end{gather*}

The method of Lagrange multipliers, Theorem 2.10.2, gives that the only possible locations of the maximum and minimum of the function $f$ are $\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{2}\right)\text{.}$ So the maximum and minimum values of $f$ are $\frac{1}{2\sqrt{2}}$ and $-\frac{1}{2\sqrt{2}}\text{,}$ respectively.

2.10.2.8. (✳).

Solution.

This is a constrained optimization problem with the objective function being $f(x,y) = x^2 + y^2$ and the constraint function being $g(x,y) =x^4 + y^4 - 1\text{.}$ By Theorem 2.10.2, any minimum or maximum $(x,y)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2x &=4 \la x^3 = \la g_x \tag{E1}\\ f_y = 2y &=4 \la y^3 = \la g_y \tag{E2}\\ x^4 + y^4 &= 1 \tag{E3} \end{align*}

for some real number $\la\text{.}$ By equation (E1), $2x(1-2\la x^2)=0\text{,}$ which is obeyed if and only if at least one of $x=0\text{,}$ $2\la x^2=1$ is obeyed. Similarly, by equation (E2), $2y(1-2\la y^2)=0\text{,}$ which is obeyed if and only if at least one of $y=0\text{,}$ $2\la y^2=1$ is obeyed.

If $x=0\text{,}$ (E3) reduces to $y^4=1$ or $y=\pm 1\text{.}$ At both $\big(0,\pm 1\big)$ we have $f\big(0,\pm1\big)=1\text{.}$
If $y=0\text{,}$ (E3) reduces to $x^4=1$ or $x=\pm 1\text{.}$ At both $\big(\pm 1,0\big)$ we have $f\big(\pm1,0\big)=1\text{.}$
If both $x$ and $y$ are nonzero, we have $x^2=\frac{1}{2\lambda}=y^2\text{.}$ Then (E3) reduces to
\begin{gather*} 2x^4=1 \end{gather*}
so that $x^2=y^2=\frac{1}{\sqrt{2}}$ and $x=\pm 2^{-1/4}\text{,}$ $y=\pm 2^{-1/4}\text{.}$ At all four of these points, we have $f=\sqrt{2}\text{.}$

So the minimum value of $f$ on $x^4+y^4=1$ is $1$ and the maximum value of $f$ on $x^4+y^4=1$ is $\sqrt{2}\text{.}$

2.10.2.9. (✳).

Solution.

The function $f(x,y,z)=(x-1)^2+(y+2)^2+(z-1)^2$ gives the square of the distance from the point $(x,y,z)$ to the point $(1,-2,1)\text{.}$ So it suffices to find the $(x,y,z)$ which minimizes $f(x,y,z)=(x-1)^2+(y+2)^2+(z-1)^2$ subject to the constraint $g(x,y,z) = z^2 + x^2 + y^2 - 2y - 10=0\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2(x-1) &= 2 \la x = \la g_x \tag{E1}\\ f_y = 2(y+2) &=2 \la (y-1) = \la g_y \tag{E2}\\ f_z = 2(z-1) &= 2\la z = \la g_z \tag{E3}\\ z^2 + x^2 + y^2 - 2y &= 10 \tag{E4} \end{align*}

for some real number $\la\text{.}$ Now

\begin{align*} \text{(E1)} &\implies x=\frac{1}{1-\la}\\ \text{(E2)} &\implies y=-\frac{2+\la}{1-\la}\\ \text{(E3)} &\implies z=\frac{1}{1-\la} \end{align*}

(Note that $\la$ cannot be $1\text{,}$ because if it were (E1) would reduce to $-2=0\text{.}$) Substituting these into (E4), and using that

\begin{equation*} y-2=-\frac{2+\la}{1-\la} -\frac{2-2\la}{1-\la}=-\frac{4-\la}{1-\la} \end{equation*}

gives

\begin{align*} &\frac{1}{{(1-\la)}^2}+\frac{1}{{(1-\la)}^2} +\frac{2+\la}{1-\la}\ \frac{4-\la}{1-\la} = 10\\ \iff & 2+(2+\la)(4-\la) = 10 (1-\la)^2\\ \iff & 11\la^2 -22\la =0\\ \iff & \la=0\text{ or }\la=2 \end{align*}

When $\la=0\text{,}$ we have $(x,y,z) = (1,-2,1)$ (nasty!), which gives distance zero and so is certainly the closest point. When $\la=2\text{,}$ we have $(x,y,z) = (-1,4,-1)\text{,}$ which does not give distance zero and so is certainly the farthest point.

2.10.2.10. (✳).

Solution.

We are to maximize and minimize $f(x,y,z)=x^2 + y^2 -\frac{1}{20} z^2$ subject to the constraints $g(x,y,z)=x + 2y + z - 10=0$ and $h(x,y,z) = x^2 + y^2 - z=0\text{.}$ By Theorem 2.10.8, any local minimum or maximum $(x,y,z)$ must obey the double Lagrange multiplier equations

\begin{align*} f_x = 2x &=\la + 2 \mu x = \la g_x +\mu h_x\tag{E1}\\ f_y = 2y &=2\la + 2 \mu y = \la g_y +\mu h_y\tag{E2}\\ f_z = -\frac{z}{10} &=\la - \mu = \la g_z +\mu h_z\tag{E3}\\ x + 2y + z &= 10\tag{E4}\\ x^2 + y^2 - z &= 0 \tag{E5} \end{align*}

for some real numbers $\la$ and $\mu\text{.}$

Equation (E1) gives $2(1-\mu)x=\la$ and equation (E2) gives $(1-\mu)y=\la\text{.}$ So

\begin{gather*} 2(1-\mu)x=(1-\mu)y \implies (1-\mu)(2x-y)=0 \end{gather*}

So at least one of $\mu=1$ and $y=2x$ must be true.

If $\mu=1\text{,}$ equations (E1) and (E2) both reduce to $\la=0$ and then the remaining equations reduce to
\begin{align*} -\frac{z}{10} &=-1\tag{E3}\\ x + 2y + z &= 10\tag{E4}\\ x^2 + y^2 - z &= 0 \tag{E5} \end{align*}
Then (E3) implies $z=10\text{,}$ and (E4) in turn implies $x+2y+10=10$ so that $x=-2y\text{.}$ Finally, substituting $z=10$ and $x=-2y$ into (E5) gives
\begin{gather*} 4y^2+y^2-10=0 \iff 5y^2=10 \iff y=\pm\sqrt{2} \end{gather*}
If $y=2x\text{,}$ equations (E4) and (E5) reduce to
\begin{align*} 5x + z &= 10\tag{E4}\\ 5x^2 - z &= 0 \tag{E5} \end{align*}
Substituting $z=5x^2\text{,}$ from (E5), into (E4) gives
\begin{gather*} 5x^2+5x-10=0 \iff x^2+x-2=0 \iff (x+2)(x-1)=0 \end{gather*}
So we have either $x=-2\text{,}$ $y=2x=-4\text{,}$ $z=5x^2=20$ or $x=1\text{,}$ $y=2x=2\text{,}$ $z=5x^2=5\text{.}$ (In both cases, we could now solve (E1) and (E3) for $\la$ and $\mu\text{,}$ but we don’t care what the values of $\la$ and $\mu$ are.)

So we have the following candidates for the locations of the min and max

point	$(-2\sqrt{2},\sqrt{2}, 10)$	$(2\sqrt{2},-\sqrt{2}, 10)$	$(-2,-4,20) $	$(1,2,5) $
value of $f$	$8+2-5$	$8+2-5$	$4+16-20$	$1+4-\frac{25}{20}$
	max	max	min

So the maximum is $5$ and the minimum is $0\text{.}$

2.10.2.11. (✳).

Solution.

The function $f(x,y,z)=x^2+y^2+z^2$ gives the square of the distance from the point $(x,y,z)$ to the origin. So it suffices to find the $(x,y,z)$ (in the first octant) which minimizes $f(x,y,z)=x^2+y^2+z^2$ subject to the constraint $g(x,y,z) = x^3y^2z -6\sqrt{3}=0\text{.}$ To start, we’ll find the minimizers in all of $\bbbr^3\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2x &= 3 \la x^2y^2z = \la g_x \tag{E1}\\ f_y = 2y &=2 \la x^3 yz = \la g_y \tag{E2}\\ f_z = 2z &= \la x^3y^2 = \la g_z \tag{E3}\\ x^3y^2z &= 6\sqrt{3} \tag{E4} \end{align*}

for some real number $\la\text{.}$

Multiplying (E1) by $2x\text{,}$ (E2) by $3y\text{,}$ and (E3) by $6z$ gives

\begin{align*} 4x^2 &= 6 \la x^3y^2z \tag{E1'}\\ 6y^2 &= 6 \la x^3y^2z \tag{E2'}\\ 12z^2&= 6 \la x^3y^2z \tag{E3'} \end{align*}

The three right hand sides are all identical. So the three left hand sides must all be equal.

\begin{equation*} 4x^2=6y^2=12z^2 \iff x=\pm\sqrt{3}\, z,\ y=\pm\sqrt{2}\, z \end{equation*}

Equation (E4) forces $x$ and $z$ to have the same sign. So we must have $x=\sqrt{3}\,z$ and $y=\pm \sqrt{2}\,z\text{.}$ Substituting this into (E4) gives

\begin{gather*} \big(\sqrt{3}\,z\big)^3 \big(\pm \sqrt{2}\,z\big)^2 z=6\sqrt{3} \iff z^6=1 \iff z=\pm 1 \end{gather*}

So our minimizer (in all of $\bbbr^3$) must be one of $\big(\sqrt{3}\,,\,\pm\sqrt{2}\,,\,1\big)$ or $\big(-\sqrt{3}\,,\,\pm\sqrt{2}\,,\,-1\big)\text{.}$ All of these points give exactly the same value of $f$ (namely $3+2+1=6$). That is all four points are a distance $\sqrt{6}$ from the origin and all other points on $x^3y^2z=6\sqrt{3}$ have distance from the origin strictly greater than $\sqrt{6}\text{.}$ So the first octant point on $x^3y^2z=6\sqrt{3}$ that is closest to the origin is $\big(\sqrt{3}\,,\,\sqrt{2}\,,\,1\big)\text{.}$

2.10.2.12. (✳).

Solution.

This is a constrained optimization problem with the objective function being

\begin{equation*} f(x,y,z) = xyz \end{equation*}

and the constraint function being

\begin{equation*} G(x,y,z) =x^2 + xy + y^2 + 3z^2 - 9 \end{equation*}

By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = yz &= \la (2x+y) = \la G_x \tag{E1}\\ f_y = xz &= \la (2y+x) = \la G_y \tag{E2}\\ f_z = xy &= 6\la z = \la G_z \tag{E3}\\ x^2 + xy + y^2 + 3z^2 &= 9 \tag{E4} \end{align*}

for some real number $\la\text{.}$

If $\la=0\text{,}$ then, by (E1), $yz=0$ so that $f(x,y,z)=xyz=0\text{.}$ This cannot possibly be the maximum value of $f$ because there are points $(x,y,z)$ on $g(x,y,z)=9$ (for example $x=y=1\text{,}$ $z=\sqrt{2}$) with $f(x,y,z) \gt 0\text{.}$
If $\la\ne 0\text{,}$ then multiplying (E1) by $x\text{,}$ (E2) by $y\text{,}$ and (E3) by $z$ gives

\begin{align*} &xyz = \la (2x^2+xy) = \la(2y^2 +xy) =6\la z^2\\ &\hskip1in\implies 2x^2+xy =2y^2 +xy =6z^2\\ &\hskip1in\implies x=\pm y,\ z^2=\frac{1}{6}(2x^2+xy) \end{align*}
- If $x=y\text{,}$ then $z^2=\frac{x^2}{2}$ and, by (E4)
  \begin{align*} x^2+x^2+x^2 +\frac{3}{2}x^2=9 &\implies x^2=2\\ &\implies x=y=\pm \sqrt{2},\ z=\pm 1 \end{align*}
  For these points
  \begin{equation*} f(x,y,z)=2z=\begin{cases} 2&\text{if }z=1 \\ -2&\text{if }z=-1 \end{cases} \end{equation*}
- If $x=-y\text{,}$ then $z^2=\frac{x^2}{6}$ and, by (E4)
  \begin{align*} x^2-x^2+x^2 +\frac{x^2}{2}=9 &\implies x^2=6\\ &\implies x=-y=\pm \sqrt{6},\ z=\pm 1 \end{align*}
  For these points
  \begin{equation*} f(x,y,z)=-6z=\begin{cases} -6&\text{if }z=1 \\ 6&\text{if }z=-1 \end{cases} \end{equation*}

So the maximum is $6$ and is achieved at $\big(\sqrt{6}\,,\,-\sqrt{6}\,,\,-1\big)$ and $\big(-\sqrt{6}\,,\,\sqrt{6}\,,\,-1\big)\text{.}$

2.10.2.13. (✳).

Solution.

In order for a sphere of radius $r$ centred on the origin to be enclosed in the ellipsoid, every point of the ellipsoid must be at least a distance $r$ from the origin. So the largest allowed $r$ is the distance from the origin to the nearest point on the ellipsoid.

We have to minimize $f(x,y,z)=x^2+y^2+z^2$ subject to the constraint $g(x,y,z) = 2(x+1)^2 + y^2 + 2(z-1)^2 -8\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2x &=4 \la (x+1) = \la g_x \tag{E1}\\ f_y = 2y &=2 \la y = \la g_y \tag{E2}\\ f_z = 2z &=4 \la (z-1) = \la g_z \tag{E3}\\ 2(x+1)^2 + y^2 + 2(z-1)^2 &= 8 \tag{E4} \end{align*}

for some real number $\la\text{.}$

By equation (E2), $2y(1-\la)=0\text{,}$ which is obeyed if and only if at least one of $y=0\text{,}$ $\la=1$ is obeyed.

If $y=0\text{,}$ the remaining equations reduce to
\begin{align*} x &=2 \la (x+1) \tag{E1}\\ z &=2 \la (z-1) \tag{E3}\\ (x+1)^2 + (z-1)^2 &= 4 \tag{E4} \end{align*}
Note that $2\la$ cannot be $1$ — if it were, (E1) would reduce to $0=1\text{.}$ So equation (E1) gives
\begin{gather*} x = \frac{2\la}{1-2\la}\qquad\text{or}\qquad x+1 = \frac{1}{1-2\la} \end{gather*}
Equation (E3) gives
\begin{gather*} z = -\frac{2\la}{1-2\la}\qquad\text{or}\qquad z-1 = -\frac{1}{1-2\la} \end{gather*}
Substituting $x+1 = \frac{1}{1-2\la}$ and $z-1 = -\frac{1}{1-2\la}$ into (E4) gives
\begin{align*} \frac{1}{(1-2\la)^2} + \frac{1}{(1-2\la)^2} =4 &\iff \frac{1}{(1-2\la)^2} = 2\\ &\iff \frac{1}{1-2\la} =\pm\sqrt{2} \end{align*}
So we now have two candidates for the location of the max and min, namely $(x,y,z) = \big(-1 + \sqrt{2}, 0, 1-\sqrt{2}\big)$ and $(x,y,z) = \big(-1 - \sqrt{2}, 0, 1+\sqrt{2}\big)\text{.}$
If $\la=1\text{,}$ the remaining equations reduce to
\begin{align*} x &=2 (x+1) \tag{E1}\\ z &=2 (z-1) \tag{E3}\\ 2(x+1)^2 + y^2 + 2(z-1)^2 &= 8 \tag{E4} \end{align*}
Equation (E1) gives $x=-2$ and equation (E3) gives $z=2\text{.}$ Substituting these into (E4) gives
\begin{gather*} 2+y^2+2=8 \iff y^2=4 \iff y=\pm 2 \end{gather*}

So we have the following candidates for the locations of the min and max

point	$\big(-1 + \sqrt{2}, 0, 1-\sqrt{2}\big)$	$\big(-1 - \sqrt{2}, 0, 1+\sqrt{2}\big)$	$(-2,2,2)$	$(-2,-2,2) $
value of $f$	$2\big(3-2\sqrt{2}\big)$	$2\big(3+2\sqrt{2}\big)$	$12$	$12$
	min		max	max

Recalling that $f(x,y,z)$ is the square of the distance from $(x,y,z)$ to the origin, the maximum allowed radius for the enclosed sphere is $\sqrt{6-4\sqrt{2}}\approx 0.59\text{.}$

2.10.2.14. (✳).

Solution.

(a) We are to maximize $f(x,y,z)=z$ subject to the constraints $g(x,y,z)=x+y+z-2=0$ and $h(x,y,z) = x^2 + y^2 + z^2 -2=0\text{.}$ By Theorem 2.10.8, any local minimum or maximum $(x,y,z)$ must obey the double Lagrange multiplier equations

\begin{align*} f_x = 0 &=\la + 2 \mu x = \la g_x +\mu h_x\tag{E1}\\ f_y = 0 &=\la + 2 \mu y = \la g_y +\mu h_y\tag{E2}\\ f_z = 1 &=\la + 2 \mu z = \la g_z +\mu h_z\tag{E3}\\ x + y + z &= 2\tag{E4}\\ x^2 + y^2 + z^2 &= 2 \tag{E5} \end{align*}

for some real numbers $\la$ and $\mu\text{.}$ Subtracting (E2) from (E1) gives $2\mu(x-y)=0\text{.}$ So at least one of $\mu=0$ and $y=x$ must be true.

If $\mu=0\text{,}$ equations (E1) and (E3) reduce to $\la=0$ and $\la=1\text{,}$ which is impossible. So $\mu\ne 0\text{.}$
If $y=x\text{,}$ equations (E2) through (E5) reduce to
\begin{align*} \la + 2 \mu x &= 0 \tag{E2}\\ \la + 2 \mu z &= 1\tag{E3}\\ 2x + z &= 2\tag{E4}\\ 2x^2 + z^2 &= 2 \tag{E5} \end{align*}
By (E4), $x=\frac{2-z}{2}\text{.}$ Substituting this into (E5) gives
\begin{align*} 2\frac{(2-z)^2}{4} +z^2 =2 &\iff (2-z)^2 +2z^2 = 4 \iff 3z^2-4z=0\\ &\iff z = 0,\ \frac{4}{3} \end{align*}

The maximum $z$ is thus $\frac{4}{3}\text{.}$

(b) Presumably the “lowest point” is the point with the minimal $z$--coordinate. By our work in part (a), we have that the minimal value of $z$ on $C$ is $0\text{.}$ We have also already seen in part (a) that $y=x\text{.}$ When $z=0\text{,}$ (E4) reduces to $2x=2\text{.}$ So the desired point is $(1,1) \text{.}$

2.10.2.15. (✳).

Solution.

(a) This is a constrained optimization problem with the objective function being $f(x,y,z) = (x - 2)^2 + (y + 2)^2 + (z - 4)^2$ and the constraint function being $g(x,y,z) =x^2 + y^2 + z^2 - 6\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2(x-2) &=2 \la x = \la g_x \tag{E1}\\ f_y = 2(y+2) &=2 \la y = \la g_y \tag{E2}\\ f_z = 2(z-4) &=2 \la z = \la g_z \tag{E3}\\ x^2 + y^2 + z^2 &= 6 \tag{E4} \end{align*}

for some real number $\la\text{.}$ Simplifying

\begin{align*} x-2 &= \la x \tag{E1}\\ y+2 &= \la y \tag{E2}\\ z-4 &= \la z \tag{E2}\\ x^2 + y^2 + z^2 &= 6 \tag{E4} \end{align*}

Note that we cannot have $\la=1\text{,}$ because then (E1) would reduce to $-2=0\text{.}$ Substituting $x=\frac{2}{1-\la}\text{,}$ from (E1), and $y=\frac{-2}{1-\la}\text{,}$ from (E2), and $z=\frac{4}{1-\la}\text{,}$ from (E3), into (E4) gives

\begin{align*} \frac{4}{(1-\la)^2} + \frac{4}{(1-\la)^2} + \frac{16}{(1-\la)^2} =6 &\iff (1-\la)^2=4\\ &\iff 1-\la =\pm 2 \end{align*}

and hence

\begin{equation*} (x,y,z) = \pm \frac{(2,-2,4)}{2}= \pm (1,-1,2) \end{equation*}

So we have the following candidates for the locations of the min and max

point	$(1,-1,2)$	$-(1,-1,2)$
value of $f$	$6$	$54$
	min	max

So the minimum is $6$ and the maximum is $54\text{.}$

(b) $f(x,y,z)$ is the square of the distance from $(x,y,z)$ to $(2,-2,4)\text{.}$ So the point on the sphere $x^2 + y^2 + z^2 = 6$ that is farthest from the point $(2, -2, 4)$ is the point from part (a) that maximizes $f\text{,}$ which is $(-1,1,-2)\text{.}$

2.10.2.16. (✳).

Solution.

(a) This is a constrained optimization problem with the objective function being $f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2$ and the constraint function being $g(x,y,z) =x^2 + y^2 + z^2 - 1\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 2(x-2) &=2 \la x = \la g_x \tag{E1}\\ f_y = 2(y-1) &=2 \la y = \la g_y \tag{E2}\\ f_z = 2z &=2 \la z = \la g_z \tag{E3}\\ x^2 + y^2 + z^2 &= 1 \tag{E4} \end{align*}

for some real number $\la\text{.}$ By equation (E3), $2z(1-\la)=0\text{,}$ which is obeyed if and only if at least one of $z=0\text{,}$ $\la=1$ is obeyed.

If $z=0$ and $\la\ne 1\text{,}$ the remaining equations reduce to
\begin{align*} x-2 &= \la x \tag{E1}\\ y-1 &= \la y \tag{E2}\\ x^2 + y^2 &= 1 \tag{E4} \end{align*}
Substituting $x=\frac{2}{1-\la}\text{,}$ from (E1), and $y=\frac{1}{1-\la}\text{,}$ from (E2), into (E3) gives
\begin{gather*} \frac{4}{(1-\la)^2} + \frac{1}{(1-\la)^2} =1 \iff (1-\la)^2=5 \iff 1-\la =\pm\sqrt{5} \end{gather*}
and hence
\begin{equation*} (x,y,z) = \pm\frac{1}{\sqrt{5}}(2,1,0) \end{equation*}
To aid in the evaluation of $f(x,y,z)$ at these points note that, at these points,
\begin{align*} &x-2=\la x = \frac{2\la}{1-\la},\qquad y-1=\la y = \frac{\la}{1-\la}\\ &\implies f(x,y,z) =\frac{4\la^2}{(1\!-\!\la)^2}\!+\!\frac{\la^2}{(1\!-\!\la)^2} =\frac{5\la^2}{(1\!-\!\la)^2} =\la^2 =\big(1\mp\sqrt{5}\big)^2 \end{align*}
If $\la=1\text{,}$ the remaining equations reduce to
\begin{align*} x-2 &=x \tag{E1}\\ y -1 &=y \tag{E2}\\ x^2 +y^2 + z^2 &= 1 \tag{E4} \end{align*}
Since $-2\ne 0$ and $-1\ne 0\text{,}$ neither (E1) nor (E2) has any solution.

So we have the following candidates for the locations of the min and max

point	$\frac{1}{\sqrt{5}}(2,1,0)$	$-\frac{1}{\sqrt{5}}(2,1,0)$
value of $f$	$\big(1-\sqrt{5}\big)^2$	$\big(1+\sqrt{5}\big)^2$
	min	max

So the minimum is $\big(\sqrt{5}-1\big)^2=6-2\sqrt{5}\text{.}$

(b) The function $f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2$ is the square of the distance from the point $(x,y,z)$ to the point $(2,1,0)\text{.}$ So the minimum of $f$ subject to the constraint $x^2+y^2+z^2=1$ is the square of the distance from $(2,1,0)$ to the point on the sphere $x^2+y^2+z^2=1$ that is nearest $(2,1,0)\text{.}$

2.10.2.17. (✳).

Solution.

For this problem the objective function is $f(x,y,z) = (x + z)e^y$ and the constraint function is $g(x,y,z)=x^2 + y^2 + z^2 -6\text{.}$ To apply the method of Lagrange multipliers we need $\vnabla f$ and $\vnabla g\text{.}$ So we start by computing the first order derivatives of these functions.

\begin{equation*} f_x=e^y\qquad f_y=(x+z)e^y\qquad f_z=e^y\qquad g_x=2x\qquad g_y=2y\qquad g_z=2z \end{equation*}

So, according to the method of Lagrange multipliers, we need to find all solutions to

\begin{align*} e^y&=\la (2x) \tag{E1}\\ (x+z)e^y&=\la (2y) \tag{E2}\\ e^y&=\la (2z) \tag{E3}\\ x^2+y^2+z^2-6&=0 \tag{E4} \end{align*}

First notice that, since $e^y\ne 0\text{,}$ equation (E1) guarantees that $\la\ne 0$ and $x\ne 0$ and equation (E3) guarantees that $z\ne 0$ too.

So dividing (E1) by (E3) gives $\frac{x}{z}=1$ and hence $x=z\text{.}$
Then subbing $x=z$ into (E2) gives $2z e^y = \la(2y)\text{.}$ Dividing this equation by (E3) gives $2z=\frac{y}{z}$ or $y=2z^2\text{.}$
Then subbing $x=z$ and $y=2z^2$ into (E4) gives
\begin{align*} z^2+4z^4+z^2-6=0 &\iff 4z^4 +2z^2 -6 = 0\\ &\iff (2z^2+3)(2z^2-2) =0 \end{align*}
As $2z^2+3 \gt 0\text{,}$ we must have $2z^2-2=0$ or $z=\pm 1\text{.}$

Recalling that $x=z$ and $y=2z^2\text{,}$ the method of Lagrange multipliers, Theorem 2.10.2, gives that the only possible locations of the maximum and minimum of the function $f$ are $(1,2,1)$ and $(-1,2,-1)\text{.}$ To complete the problem, we only have to compute $f$ at those points.

point	$(1,2,1)$	$(-1,2,-1)$
value of $f$	$2e^2$	$-2e^2$
	max	min

Hence the maximum value of $(x + z)e^y$ on $x^2 + y^2 + z^2 = 6$ is $2e^2$ and the minimum value is $-2 e^2\text{.}$

2.10.2.18. (✳).

Solution.

Let $(x,y)$ be a point on $2x^2 + 4xy + 5y^2 = 30\text{.}$ We wish to maximize and minimize $x^2+y^2$ subject to $2x^2 + 4xy + 5y^2 = 30\text{.}$ Define $L(x,y,\la)=x^2+y^2-\la(2x^2 + 4xy + 5y^2 - 30)\text{.}$ Then

\begin{alignat*}{3} 0&=L_x=2x-\la (4x+4y)\qquad&&\implies\qquad &(1-2\la)x-2\la y&=0\tag{1}\\ 0&=L_y=2y-\la (4x+10y)\qquad&&\implies& -2\la x+(1-5\la)y&=0\tag{2}\\ 0&=L_\la=2x^2 + 4xy + 5y^2 - 30 \end{alignat*}

Note that $\la$ cannot be zero because if it is, (1) forces $x=0$ and (2) forces $y=0\text{,}$ but $(0,0)$ is not on the ellipse. So equation (1) gives $y=\frac{1-2\la}{2\la}x\text{.}$ Substituting this into equation (2) gives $-2\la x+\frac{(1-5\la)(1-2\la)}{2\la}x=0\text{.}$ To get a nonzero $(x,y)$ we need

\begin{align*} &-2\la +\frac{(1-5\la)(1-2\la)}{2\la}=0\\ &\iff 0=-4\la^2+(1-5\la)(1-2\la) =6\la^2-7\la+1=(6\la-1)(\la-1) \end{align*}

So $\la$ must be either $1$ or $\frac{1}{6}\text{.}$ Substituting these into either (1) or (2) gives

\begin{alignat*}{5} \la&=1&&\implies -x-2y&=0 &&\implies x&=-2y &&\implies 8y^2-8y^2+5y^2=30\\ &&&&&&&&&\implies y=\pm \sqrt{6}\\ \la&=\frac{1}{6}&&\implies \frac{2}{3} x-\frac{1}{3} y&=0 &&\implies y&=2x &&\implies 2x^2+8x^2+20x^2=30\\ &&&&&&&&&\implies x=\pm 1 \end{alignat*}

The farthest points are $\pm\sqrt{6}(-2,1)\text{.}$ The nearest points are $\pm(1,2)\text{.}$

2.10.2.19.

Solution.

Let $(x,y)$ be a point on $3x^2-2xy+3y^2=4\text{.}$ This point is at the end of a major axis when it maximizes its distance from the centre, $(0,0)\text{,}$ of the ellipse. It is at the end of a minor axis when it minimizes its distance from $(0,0)\text{.}$ So we wish to maximize and minimize $f(x,y)=x^2+y^2$ subject to the constraint $g(x,y)=3x^2-2xy+3y^2-4=0\text{.}$ According to the method of Lagrange multipliers, we need to find all solutions to

\begin{alignat*}{3} f_x = 2x = \la (6x-2y)\phantom{-} &= \la g_x \quad&&\implies\quad & (1-3\la)x+\la y&=0 \tag{E1}\\ f_y = 2y = \la (-2x+6y) &= \la g_y \quad&&\implies\quad & \la x+(1-3\la)y&=0\tag{E2}\\ 3x^2-2xy+3y^2&=4 \tag{E3} \end{alignat*}

To start, let’s concentrate on the first two equations. Pretend for a couple of minutes, that we already know the value of $\la$ and are trying to find $x$ and $y\text{.}$ The system of equations $(1-3\la)x+\la y=0\text{,}$ $\la x+(1-3\la)y=0$ has one obvious solution. Namely $x=y=0\text{.}$ But this solution is not acceptable because it does not satisfy the equation of the ellipse. If you have already taken a linear algebra course, you know that a system of two linear homogeneous equations in two unknowns has a nonzero solution if and only if the determinant of the matrix of coefficients is zero. (You use this when you find eigenvalues and eigenvectors.) For the equations of interest, this is

\begin{align*} &\det\left[\begin{matrix}1-3\la&\la\\ \la&1-3\la\end{matrix}\right] =(1-3\la)^2-\la^2 =(1-2\la)(1-4\la)=0\\ &\hskip3in\implies\la=\frac{1}{2},\frac{1}{4} \end{align*}

Even if you have not already taken a linear algebra course, you also come to this conclusion directly when you try to solve the equations. Note that $\la$ cannot be zero because if it is, (E1) forces $x=0$ and (E2) forces $y=0\text{.}$ So equation (E1) gives $y=-\frac{1-3\la}{\la}x\text{.}$ Substituting this into equation (E2) gives $\la x-\frac{(1-3\la)^2}{\la}x=0\text{.}$ To get a nonzero $(x,y)$ we need

\begin{equation*} \la -\frac{(1-3\la)^2}{\la}=0\iff \la^2-(1-3\la)^2=0 \end{equation*}

By either of these two methods, we now know that $\la$ must be either $\frac{1}{2}$ or $\frac{1}{4}\text{.}$ Substituting these into either (E1) or (E2) and then using (E3) gives

\begin{alignat*}{6} \la&=\frac{1}{2}&&\implies& -\frac{1}{2} x+\frac{1}{2} y&=0 &&\implies\ & x&=y &&\implies 3x^2-2x^2+3x^2=4\\ &&&&&&&&&& &\implies\ x=\pm 1\\ \la&=\frac{1}{4}&&\implies& \frac{1}{4} x+\frac{1}{4} y&=0 &&\implies\ & x&=-y &&\implies 3x^2+2x^2+3x^2=4\\ &&&&&&&&&& &\implies\ x=\pm \frac{1}{\sqrt{2}} \end{alignat*}

The ends of the minor axes are $\pm\big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\big)\text{.}$ The ends of the major axes are $\pm(1,1)\text{.}$

2.10.2.20. (✳).

Solution.

Let the box have dimensions $x\times y\times z\text{.}$ Use units of money so that the sides and bottom cost one unit per square meter and the top costs two units per square meter. Then the top costs $2xy\text{,}$ the bottom costs $xy$ and the four sides cost $2xz+2yz\text{.}$ We are to find the $x\text{,}$ $y$ and $z$ that minimize the cost $f(x,y,z)=2xy +xy +2xz+2yz$ subject to the constraint that $g(x,y,z)=xyz-96=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{2} f_x&=3y+2z&&=\la yz=\la g_x\\ f_y&=3x+2z&&=\la xz=\la g_y\\ f_z&=2x+2y&&=\la xy=\la g_z\\ &\ \ \ xyz-96&&=0 \end{alignat*}

Multiplying the first equation by $x\text{,}$ the second equation by $y$ and the third equation by $z$ and then substituting in $xyz=96$ gives

\begin{align*} 3xy+2xz&=96\la\\ 3xy+2yz&=96\la\\ 2xz+2yz&=96\la \end{align*}

Subtracting the second equation from the first gives $2z(x-y)=0\text{.}$ Since $z=0$ is impossible, we must have $x=y\text{.}$ Substituting this in,

\begin{equation*} 3x^2+2xz=96\la\qquad 4xz=96\la \end{equation*}

Subtracting,

\begin{align*} 3x^2-2xz=0&\implies z=\frac{3}{2}x \implies 96=xyz=\frac{3}{2}x^3 \implies x^3=64\\ &\implies x=y=4,\ z=6\ \text{meters} \end{align*}

2.10.2.21. (✳).

Solution.

We are to find the $x\text{,}$ $y$ and $z$ that minimize the temperature $T(x,y,z)=40xy^2z$ subject to the constraint that $g(x,y,z)=x^2+y^2+z^2-1=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{2} T_x&=40y^2z&&=\la(2x)=\la g_x\\ T_y&=80xyz&&=\la(2y)=\la g_y\\ T_z&=40xy^2&&=\la(2z)=\la g_z\\ &\hskip-0.5in x^2+y^2+z^2-1&&=0 \end{alignat*}

Multiplying the first equation by $x\text{,}$ the second equation by $y/2$ and the third equation by $z$ gives

\begin{align*} 40xy^2z&=2x^2\la\\ 40xy^2z&=y^2\la\\ 40xy^2z&=2z^2\la \end{align*}

Hence we must have

\begin{equation*} 2x^2\la=y^2\la=2z^2\la \end{equation*}

If $\la=0\text{,}$ then $40y^2z=0,\ 80xyz=0,\ 40xy^2=0$ which is possible only if at least one of $x,y,z$ is zero so that $T(x,y,z)=0\text{.}$
If $\la\ne 0\text{,}$ then
\begin{align*} 2x^2=y^2=2z^2 &\implies 1=x^2+y^2+z^2=x^2+2x^2+x^2=4x^2\\ &\implies x=\pm\half,\ y^2=\half,\ z=\pm \half\\ &\implies T=40\big(\pm\half)\half\big(\pm\half)=\pm 5 \end{align*}

(The sign of $x$ and $z$ need not be the same.) So the hottest temperature is $+5$ and the coldest temperature is $-5\text{.}$

2.10.2.22. (✳).

Solution.

The optimal box will have vertices $(\pm x,\pm y, 0)\text{,}$ $(\pm x,\pm y,\ z)$ with $x,y,z \gt 0$ and $z=48-4x^2-3y^2\text{.}$ (If the lower vertices are not in the $xy$--plane, the volume of the box can be increased by lowering the bottom of the box to the $xy$--plane. If any of the four upper vertices are not on the hemisphere, the volume of the box can be increased by moving the upper vertices outwards to the hemisphere.) The volume of this box will be $(2x)(2y)z\text{.}$ So we are to find the $x\text{,}$ $y$ and $z$ that maximize the volume $f(x,y,z)=4xyz$ subject to the constraint that $g(x,y,z)=48-4x^2-3y^2-z =0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{7} f_x&=4yz=-8\la x&&=\la g_x\cr f_y&=4xz=-6\la y&&=\la g_y\cr f_z&=4xy=-\la&&=\la g_z\cr &\hskip-0.2in 48-4x^2-3y^2-z&&=0 \end{alignat*}

Multiplying the first equation by $x\text{,}$ the second equation by $y$ and the third equation by $z$ gives

\begin{align*} 4xyz&=-8\la x^2\\ 4xyz&=-6\la y^2\\ 4xyz&=-\la z \end{align*}

This forces $8\la x^2=6\la y^2=\la z\text{.}$ Since $\la$ cannot be zero (because that would force $4xyz=0$), this in turn gives $8 x^2=6 y^2= z\text{.}$ Substituting in to the fourth equation gives

\begin{equation*} 48 -\frac{z}{2}-\frac{z}{2}-z=0\implies 2z=48\implies z=24,\ 8x^2=24,\ 6y^2=24 \end{equation*}

The dimensions of the box of biggest volume are $2x=2\sqrt{3}$ by $2y=4$ by $z=24\text{.}$

2.10.2.23. (✳).

Solution.

Use units of money for which cardboard costs one unit per square meter. Then, if the bin has dimensions $x\times y\times z\text{,}$ it costs $3xy+2xz+2yz\text{.}$ We are to find the $x\text{,}$ $y$ and $z$ that minimize the cost $f(x,y,z)=3xy+2xz+2yz$ subject to the constraint that $g(x,y,z)=xyz-12=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{2} f_x&=3y+2z&&=\la yz=\la g_x\\ f_y&=3x+2z&&=\la xz=\la g_y\\ f_z&=2x+2y&&=\la xy=\la g_z\\ &\ \ \ xyz-12&&=0 \end{alignat*}

Multiplying the first equation by $x\text{,}$ the second equation by $y$ and the third equation by $z$ and then substituting in $xyz=12$ gives

\begin{align*} 3xy+2xz&=12\la\\ 3xy+2yz&=12\la\\ 2xz+2yz&=12\la \end{align*}

Subtracting the second equation from the first gives $2z(x-y)=0\text{.}$ Since $z=0$ is impossible, we must have $x=y\text{.}$ Substituting this in

\begin{equation*} 3x^2+2xz=12\la\qquad 4xz=12\la \end{equation*}

Subtracting

\begin{align*} 3x^2-2xz=0\implies z=\frac{3}{2}x &\implies 12=xyz=\frac{3}{2}x^3 \implies x^3=8\\ &\implies x=y=2,\ z=3\ {\rm meters} \end{align*}

2.10.2.24. (✳).

Solution.

If the box has dimensions $x\times y\times z\text{,}$ it costs $24xy+16xz+16yz\text{.}$ We are to find the $x\text{,}$ $y$ and $z$ that minimize the cost $f(x,y,z)=24xy+16xz+16yz$ subject to the constraint that $g(x,y,z)=xyz-4=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{2} f_x&=24y+16z&&=\la yz=\la g_x\\ f_y&=24x+16z&&=\la xz=\la g_y\\ f_z&=16x+16y&&=\la xy=\la g_z\\ &\ \ \ \ \ \ xyz-4&&=0 \end{alignat*}

Multiplying the first equation by $x\text{,}$ the second equation by $y$ and the third equation by $z$ and then substituting in $xyz=4$ gives

\begin{align*} 24xy+16xz&=4\la\\ 24xy+16yz&=4\la\\ 16xz+16yz&=4\la \end{align*}

Subtracting the second equation from the first gives $16z(x-y)=0\text{.}$ Since $z=0$ is impossible, we must have $x=y\text{.}$ Subbing this in

\begin{equation*} 24x^2+16xz=4\la\qquad 32xz=4\la \end{equation*}

Subtracting

\begin{align*} 24x^2-16xz=0&\implies z=\frac{3}{2}x \implies 4=xyz=\frac{3}{2}x^3 \implies x^3=\frac{8}{3}\\ &\implies x=y=\frac{2}{\root 3\of 3},\ z=3^{2/3}{\rm metres} \end{align*}

2.10.2.25. (✳).

Solution.

The vertices of the pyramid are $(0,0,0)\text{,}$ $\big(\frac{1}{a},0,0\big)\text{,}$ $\big(0, \frac{1}{b},0\big)$ and $\big(0,0,\frac{1}{c}\big)\text{.}$ So the base of the pyramid is a triangle of area $\half\frac{1}{a}\frac{1}{b}$ and the height of the pyramid is $\frac{1}{c}\text{.}$ So the volume of the pyramid is $\frac{1}{6abc}\text{.}$ The plane passes through $(1,2,3)$ if and only if $a+2b+3c=1\text{.}$ Thus we are to find the $a\text{,}$ $b$ and $c$ that maximize the volume $f(a,b,c)=\frac{1}{6abc}$ subject to the constraint that $g(a,b,c)=a+2b+3c-1=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the maximizing $a\text{,}$ $b\text{,}$ $c$ must obey

\begin{alignat*}{4} f_a&=-\frac{1}{6a^2bc}&&=\la&=\la g_a\quad &&\iff\quad 6\la a^2bc&=-1\\ f_b&=-\frac{1}{6ab^2c}&&=2\la&=\la g_b\quad&&\iff\quad 6\la ab^2c&=-\half\\ f_c&=-\frac{1}{6abc^2}&&=3\la&=\la g_c\quad &&\iff\quad 6\la abc^2&=-\frac{1}{3}\\ &a+2b+3c&&=1 \end{alignat*}

Dividing the first two equations gives $\frac{a}{b}=2$ and dividing the first equation by the third gives $\frac{a}{c}=3\text{.}$ Substituting $b=\half a$ and $c=\frac{1}{3} a$ in to the final equation gives

\begin{equation*} a+2b+3c=3a=1 \implies a=\frac{1}{3},\ b=\frac{1}{6},\ c=\frac{1}{9} \end{equation*}

and the maximum volume is $\frac{3\times 6\times 9}{6}=27\text{.}$

2.10.2.26. (✳).

Solution.

We’ll find the minimum distance$^2$ and then take the square root. That is, we’ll find the minimum of $f(x,y,z)=x^2+y^2+z^2$ subject to the constraints $g(x,y,z)=x-z-4=0$ and $h(x,y,z) = x + y + z -3=0\text{.}$ By Theorem 2.10.8, any local minimum or maximum $(x,y,z)$ must obey the double Lagrange multiplier equations

\begin{align*} f_x = 2x &=\la + \mu = \la g_x +\mu h_x\tag{E1}\\ f_y = 2y &= \mu = \la g_y +\mu h_y\tag{E2}\\ f_z = 2z &=-\la + \mu = \la g_z +\mu h_z\tag{E3}\\ x - z &= 4\tag{E4}\\ x + y + z &= 3 \tag{E5} \end{align*}

for some real numbers $\la$ and $\mu\text{.}$ Adding (E1)and (E3) and then subtracting 2 times (E2) gives

\begin{equation*} 2x-4y+2z=0\qquad\text{or}\qquad x-2y+z=0 \tag{E6} \end{equation*}

Substituting $x=4+z$ (from (E4)) into (E5) and (E6) gives

\begin{align*} y+2z&=-1 \tag{E5'}\\ -2y+2z&=-4 \tag{E6'} \end{align*}

Substituing $y=-1-2z$ (from (E5’)) into (E6’) gives

\begin{gather*} 6z =-6 \implies z=-1 \implies y=-1-2(-1)=1 \implies x=4+(-1)=3 \end{gather*}

So the closest point is $(3,1,-1)$ and the minimum distance is $\sqrt{3^2+1^2+(-1)^2}=\sqrt{11}\text{.}$

2.10.2.27. (✳).

Solution 1.

This is a constrained optimization problem with objective function $f(x,y,z) = 6 x +y^2 +xz$ and constraint function $g(x,y,z) =x^2+y^2+z^2-36\text{.}$ By Theorem 2.10.2, any local minimum or maximum $(x,y,z)$ must obey the Lagrange multiplier equations

\begin{align*} f_x = 6+z &=2 \la x = \la g_x \tag{E1}\\ f_y = 2y &=2 \la y = \la g_y \tag{E2}\\ f_z = x &=2 \la z = \la g_z \tag{E3}\\ x^2 + y^2 + z^2 &= 36 \tag{E4} \end{align*}

for some real number $\la\text{.}$ By equation (E2), $y(1-\la)=0\text{,}$ which is obeyed if and only if at least one of $y=0\text{,}$ $\la=1$ is obeyed.

If $y=0\text{,}$ the remaining equations reduce to
\begin{align*} 6+z &=2 \la x \tag{E1}\\ x &=2 \la z \tag{E3}\\ x^2 + z^2 &= 36 \tag{E4} \end{align*}
Substituting (E3) into (E1) gives $6 + z = 4\la^ 2 z\text{,}$ which forces $4\la^2\ne 1$ (since $6\ne 0$) and gives $z = \frac{6}{4\la^2-1}$ and then $x=\frac{12\la}{4\la^2-1}\text{.}$ Substituting this into (E4) gives
\begin{align*} \frac{144\la^2}{{(4\la^2-1)}^2} +\frac{36}{{(4\la^2-1)}^2}&=36\\ \frac{4\la^2}{{(4\la^2-1)}^2} +\frac{1}{{(4\la^2-1)}^2}&=1\\ 4\la^2+1 &= {(4\la^2-1)}^2 \end{align*}
Write $\mu=4\la^2\text{.}$ Then this last equation is
\begin{align*} \mu+1 = \mu^2 -2\mu +1 &\iff \mu^2-3\mu =0\\ &\iff \mu=0,3 \end{align*}
When $\mu=0\text{,}$ we have $z=\frac{6}{\mu-1}=-6$ and $x=0$ (by (E4)). When $\mu=3\text{,}$ we have $z=\frac{6}{\mu-1}=3$ and then $x=\pm \sqrt{27} =\pm 3\sqrt{3}$ (by (E4)).
If $\la=1\text{,}$ the remaining equations reduce to
\begin{align*} 6+z &=2 x \tag{E1}\\ x &=2 z \tag{E3}\\ x^2 +y^2 + z^2 &= 36 \tag{E4} \end{align*}
Substituting (E3) into (E1) gives $6+z=4z$ and hence $z=2\text{.}$ Then (E3) gives $x=4$ and (E4) gives $4^2 + y^2 + 2^2 =36$ or $y^2=16$ or $y=\pm 4\text{.}$

So we have the following candidates for the locations of the min and max

point	$(0,0, -6)$	$(3\sqrt{3},0,3)$	$(-3\sqrt{3},0,3)$	$(4,4,2) $	$(4,-4,2) $
value of $f$	$0$	$27\sqrt{3}$	$-27\sqrt{3}$	$48$	$48$
			min	max	max

Solution 2.

On the sphere we have $y^2=36-x^2-z^2$ and hence $f= 36 + 6x +xz -x^2-z^2$ and $x^2+z^2\le 36\text{.}$ So it suffices to find the max and min of $h(x,z)= 36 + 6x +xz -x^2-z^2$ on the disk $D=\Set{(x,z)}{x^2+z^2\le 36}\text{.}$

If a max or min occurs at an interior point $(x,z)$ of $D\text{,}$ then $(x,z)$ must be a critical point of $h$ and hence must obey
\begin{align*} h_x = 6+z-2x&=0\\ h_z = x-2z=0 \end{align*}
Substituting $x=2z$ into the first equation gives $6-3z=0$ and hence $z=2$ and $x=4\text{.}$
If a max or min occurs a point $(x,z)$ on the boundary of $D\text{,}$ we have $x^2+z^2=36$ and hence $x=\pm\sqrt{36-z^2}$ and $h=6x+zx=\pm(6+z)\sqrt{36-z^2}$ with $-6\le z\le 6\text{.}$ So the max or min can occur either when $z=-6$ or $z=+6$ or at a $z$ obeying
\begin{gather*} 0=\diff{}{z}\big[(6+z)\sqrt{36-z^2}\big] =\sqrt{36-z^2} - \frac{z(6+z)}{\sqrt{36-z^2}} \end{gather*}
or equivalently
\begin{align*} 36-z^2-z(6+z)&=0\\ 2z^2 +6z -36 &=0\\ z^2 +3z -18 &=0\\ (z+6)(z-3)&=0 \end{align*}
So the max or min can occur either when $z=\pm 6$ or $z=3\text{.}$

So we have the following candidates for the locations of the min and max

point	$(0,0, \pm 6)$	$(3\sqrt{3},0,3)$	$(-3\sqrt{3},0,3)$	$(4,4,2) $	$(4,-4,2) $
value of $f$	$0$	$27\sqrt{3}$	$-27\sqrt{3}$	$48$	$48$
			min	max	max

2.10.2.28. (✳).

Solution.

By way of preparation, we have

\begin{gather*} \pdiff{T}{x}(x,y) = 2x\,e^y\qquad \pdiff{T}{y}(x,y) = e^y\big(x^2+y^2+2y\big) \end{gather*}

(a-i) For this problem the objective function is $T(x,y) = e^y\big(x^2+y^2\big)$ and the constraint function is $g(x,y)=x^2 + y^2 - 100\text{.}$ According to the method of Lagrange multipliers, Theorem 2.10.2, we need to find all solutions to

\begin{align*} T_x = 2x\,e^y &=\la (2x) = \la g_x \tag{E1}\\ T_y = e^y\big(x^2+y^2+2y\big) &=\la (2y) = \la g_y \tag{E2}\\ x^2+y^2&=100 \tag{E3} \end{align*}

(a-ii) According to equation (E1), $2x(e^y-\la)=0\text{.}$ This condition is satisfied if and only if at least one of $x=0\text{,}$ $\la=e^y$ is obeyed.

If $x=0\text{,}$ then equation (E3) reduces to $y^2=100\text{,}$ which is obeyed if $y=\pm 10\text{.}$ Equation (E2) then gives the corresponding values for $\la\text{,}$ which we don’t need.
If $\la=e^y\text{,}$ then equation (E2) reduces to
\begin{equation*} e^y\big(x^2+y^2+2y\big) = (2y)e^y \iff e^y\big(x^2+y^2\big)=0 \end{equation*}
which conflicts with (E3). So we can’t have $\la=e^y\text{.}$

So the only possible locations of the maximum and minimum of the function $T$ are $(0,10)$ and $(0,-10)\text{.}$ To complete the problem, we only have to compute $T$ at those points.

point	$(0,10)$	$(0,-10)$
value of $T$	$100 e^{10}$	$100 e^{-10}$
	max	min

Hence the maximum value of $T(x,y) = e^y\big(x^2+y^2\big)$ on $x^2 + y^2 = 100$ is $100 e^{10}$ at $(0,10)$ and the minimum value is $100 e^{-10}$ at $(0,-10)\text{.}$

We remark that, on $x^2+y^2=100\text{,}$ the objective function $T(x,y) = e^y\big(x^2+y^2\big) = 100 e^y\text{.}$ So of course the maximum value of $T$ is achieved when $y$ is a maximum, i.e. when $y=10\text{,}$ and the minimum value of $T$ is achieved when $y$ is a minimum, i.e. when $y=-10\text{.}$

(b-i) By definition, the point $(x,y)$ is a critical point of $T(x,y)$ if ane only if

\begin{align*} T_x = 2x\,e^y &=0 \tag{E1}\\ T_y = e^y\big(x^2+y^2+2y\big) &=0 \tag{E2} \end{align*}

(b-ii) Equation (E1) forces $x=0\text{.}$ When $x=0\text{,}$ equation (E2) reduces to

\begin{gather*} e^y\big(y^2+2y\big) =0 \iff y(y+2)=0 \iff y=0\text{ or }y=-2 \end{gather*}

So there are two critical points, namely $(0,0)$ and $(0,-2)\text{.}$

(c) Note that $T(x,y) = e^y\big(x^2+y^2\big)\ge 0$ on all of $\bbbr^2\text{.}$ As $T(x,y)=0$ only at $(0,0)\text{,}$ it is obvious that $(0,0)$ is the coolest point.

In case you didn’t notice that, here is a more conventional solution.

The coolest point on the solid disc $x^2+y^2\le 100$ must either be on the boundary, $x^2+y^2= 100\text{,}$ of the disc or be in the interior, $x^2+y^2 \lt 100\text{,}$ of the disc.

In part (a-ii) we found that the coolest point on the boundary is $(0,-10)\text{,}$ where $T=100 e^{-10}\text{.}$

If the coolest point is in the interior, it must be a critical point and so must be either $(0,0)\text{,}$ where $T=0\text{,}$ or $(0,-2)\text{,}$ where $T= 4e^{-2}\text{.}$

So the coolest point is $(0,0)\text{.}$

2.10.2.29. (✳).

Solution.

(a) A normal vector to $F(x,y,z)=4x^2+4y^2+z^2=96$ at $(x_0,y_0,z_0)$ is $\vnabla F(x_0,y_0,z_0)=\llt 8x_0,8y_0,2z_0\rgt\text{.}$ (Note that this normal vector is never the zero vector because $(0,0,0)$ is not on the surface.) So the tangent plane to $4x^2+4y^2+z^2=96$ at $(x_0,y_0,z_0)$ is

\begin{align*} &8x_0(x-x_0)+8y_0(y-y_0)+2z_0(z-z_0)=0\\ &\text{or}\quad 8x_0x+8y_0y+2z_0z=8x_0^2+8y_0^2+2z_0^2 \end{align*}

This plane is of the form $x+y+z=c$ if and only if $8x_0=8y_0=2z_0\text{.}$ A point $(x_0,y_0,z_0)$ with $8x_0=8y_0=2z_0$ is on the surface $4x^2+4y^2+z^2=96$ if and only if

\begin{align*} 4x_0^2+4y_0^2+z_0^2=4x_0^2+4x_0^2+(4x_0)^2=96 \iff 24 x_0^2=96 &\iff x_0^2=4\\ &\iff x_0=\pm 2 \end{align*}

When $x_0=\pm 2\text{,}$ we have $y_0=\pm 2$ and $z_0=\pm 8$ (upper signs go together and lower signs go together) so that the tangent plane $8x_0x+8y_0y+2z_0z=8x_0^2+8y_0^2+2z_0^2$ is

\begin{align*} 8(\pm 2)x+8(\pm 2)y+2(\pm 8)z&=8(\pm 2)^2+8(\pm 2)^2+2(\pm 8)^2\\ &\quad\text{or}\quad \pm x \pm y \pm z=2+2+8\\ &\quad\text{or}\quad x + y + z=\mp 12\\ &\implies c=\pm 12 \end{align*}

(b) We are to find the $x\text{,}$ $y$ and $z$ that minimize or maximize $f(x,y,z)=x+y+z$ subject to the constraint that $g(x,y,z)=4x^2+4y^2+z^2-96=0\text{.}$ By the method of Lagrange multipliers (Theorem 2.10.2), the minimizing/maximizing $x\text{,}$ $y\text{,}$ $z$ must obey

\begin{alignat*}{2} f_x&=1=\la(8x)&&=\la g_x\\ f_y&=1=\la(8y)&&=\la g_y\\ f_z&=1=\la(2z)&&=\la g_z\\ &\hskip-0.6in 4x^2+4y^2+z^2-96&&=0 \end{alignat*}

The first three equations give

\begin{equation*} x=\frac{1}{8\la}\qquad y=\frac{1}{8\la}\qquad z=\frac{1}{2\la} \qquad\text{with }\la\ne 0 \end{equation*}

Substituting this into the fourth equation gives

\begin{align*} 4\left(\frac{1}{8\la}\right)^2+4\left(\frac{1}{8\la}\right)^2 +\left(\frac{1}{2\la}\right)^2=96 &\iff \left(\frac{1}{16}+\frac{1}{16}+\frac{1}{4}\right)\frac{1}{\la^2}=96\\ &\iff \la^2=\frac{3}{8}\frac{1}{96}=\frac{1}{8\times 32}\\ &\iff \la=\pm \frac{1}{16} \end{align*}

Hence $x=\pm 2\text{,}$ $y=\pm 2$ and $z=\pm 8$ so that the largest and smallest values of $x+y+z$ on $4x^2+4y^2+z^2-96$ are $\pm 2\pm2 \pm 8$ or $\pm 12\text{.}$

(c) The level surfaces of $x+y+z$ are planes with equation of the form $x+y+z=c\text{.}$ To find the largest (smallest) value of $x+y+z$ on $4x^2+4y^2+z^2=96$ we keep increasing (decreasing) $c$ until we get to the largest (smallest) value of $c$ for which the plane $x+y+z=c$ intersects $4x^2+4y^2+z^2=96\text{.}$ For this value of $c\text{,}$ $x+y+z=c$ is tangent to $4x^2+4y^2+z^2=96\text{.}$

2.10.2.30.

Solution.

Note that if $(x,y)$ obeys $g(x,y)=xy-1=0\text{,}$ then $x$ is necessarily nonzero. So we may assume that $x\ne 0\text{.}$ Then

\begin{align*} &\text{There is a $\la$ such that $(x,y,\la)$ obeys (E1)}\\ & \iff \text{there is a $\la$ such that } f_x(x,y)=\la g_x(x,y),\ f_y(x,y)=\la g_y(x,y),\\ &\hskip2.0in g(x,y)=0\\ & \iff \text{there is a $\la$ such that } f_x(x,y)=\la y,\quad f_y(x,y)=\la x,\quad xy=1\\ & \iff \text{there is a $\la$ such that } \frac{1}{y}f_x(x,y)= \frac{1}{x} f_y(x,y)=\la ,\quad xy=1\\ & \iff \frac{1}{y}f_x(x,y)= \frac{1}{x} f_y(x,y) ,\quad xy=1\\ & \iff xf_x\Big(\!x,\frac{1}{x}\Big)= \frac{1}{x} f_y\Big(\!x,\frac{1}{x}\Big) ,\quad y=\frac{1}{x}\\ & \iff F'(x) = \diff{}{x} f\Big(\!x,\frac{1}{x}\Big) = f_x\Big(\!x,\frac{1}{x}\Big) -\frac{1}{x^2}f_y\Big(\!x,\frac{1}{x}\Big) =0,\quad y=\frac{1}{x} \end{align*}

3 Multiple Integrals
3.1 Double Integrals
3.1.7 Exercises

3.1.7.1.

Solution.

(a) The given double integral $\int_{-1}^3\int_{-4}^1 \dee{y}\,\dee{x}=\dblInt_R \dee{x}\,\dee{y}$ where

\begin{equation*} R=\Set{(x,y)}{ -1\le x\le 3,\ -4\le y\le1} \end{equation*}

and so the integral is the area of a rectangle with sides of lengths $4$ and $5\text{.}$ Thus $\int_{-1}^3\int_{-4}^1 \dee{y}\,\dee{x}=4\times 5=20\text{.}$

(b) The given double integral $\ds\int_0^2\int_0^{\sqrt{4-y^2}} \dee{x}\,\dee{y} =\ds\dblInt_R \dee{x}\,\dee{y}$ where

\begin{align*} R&=\Set{(x,y)}{ 0\le y\le 2,\ 0\le x\le\sqrt{4-y^2}}\\ &=\Set{(x,y)}{ x\ge 0,\ y\ge 0,\ y\le 2,\ x^2+y^2\le 4} \end{align*}

So $R$ is the first quadrant part of the circular disk of radius $2$ centred on $(0,0)\text{.}$ The area of the full disk is $\pi\,2^2=4\pi\text{.}$ The given integral is one quarter of that, which is $\pi\text{.}$

(c) The given double integral $\ds\int_{-3}^3\int_0^{\sqrt{9-y^2}}\sqrt{9-x^2-y^2}\ \dee{x}\,\dee{y} =\ds\dblInt_{\cR} z(x,y)\ \dee{x}\,\dee{y}$ where $z(x,y)=\sqrt{9-x^2-y^2}$ and

\begin{align*} \cR&=\Set{(x,y)}{ -3\le y\le 3,\ 0\le x\le\sqrt{9-y^2}}\\ &=\Set{(x,y)}{ x\ge 0,\ -3\le y\le 3,\ \ x^2+y^2\le 9} \end{align*}

So $\cR$ is the right half of the circular disk of radius $3$ centred on $(0,0)\text{.}$ By Equation (3.1.9), the given integral is the volume of the solid

\begin{align*} \cV&=\Big\{\,(x,y,z)\,\Big|\,(x,y)\in\cR,\ 0\le z\le \sqrt{9-x^2-y^2}\,\Big\}\\ &=\Big\{\,(x,y,z)\,\Big|\,(x,y)\in\cR,\ z\ge 0,\ x^2+y^2+z^2\le 9\,\Big\} \end{align*}

Thus $\cV$ is the one quarter of the spherical ball of radius $3$ and centre $(0,0,0)$ with $x\ge 0$ and $z\ge 0\text{.}$ So

\begin{equation*} \int_{-3}^3\int_0^{\sqrt{9-y^2}}\sqrt{9-x^2-y^2}\ \dee{x}\,\dee{y} =\frac{1}{4}\Big(\frac{4}{3}\pi 3^3\Big) =9\pi \end{equation*}

3.1.7.2.

Solution.

(a) The integral with respect to $x$ treats $y$ as a constant. So

\begin{align*} \int_0^3 f(x,y)\,\dee{x} &= \int_0^3 12 x^2y^3\,\dee{x} = \Big[4x^3y^3\Big]_{x=0}^{x=3} = 108 y^3 \end{align*}

(b) The integral with respect to $y$ treats $x$ as a constant. So

\begin{align*} \int_0^2 f(x,y)\,\dee{y} &= \int_0^2 12 x^2y^3\,\dee{y} = \Big[3x^2y^4\Big]_{y=0}^{y=2} = 48x^2 \end{align*}

\begin{align*} \int_0^2\int_0^3 f(x,y)\,\dee{x}\,\dee{y} &=\int_0^2\left[\int_0^3 f(x,y)\,\dee{x}\right]\dee{y} =\int_0^2 108y^3\,\dee{y}\\ &=\Big[27y^4\Big]_{y=0}^{y=2} =27\times 16 = 432 \end{align*}

(d) By part (b)

\begin{align*} \int_0^3\int_0^2 f(x,y)\,\dee{y}\,\dee{x} &=\int_0^3\left[\int_0^2 f(x,y)\,\dee{y}\right]\dee{x} =\int_0^3 48x^2\,\dee{y}\\ &=\Big[16x^3\Big]_{x=0}^{x=3} =16\times 27 = 432 \end{align*}

(e) This time

\begin{align*} \int_0^3\int_0^2 f(x,y)\,\dee{x}\,\dee{y} &=\int_0^3\left[\int_0^2 12 x^2y^3\,\dee{x}\right]\dee{y} =\int_0^3\left[4 x^3y^3\right]_0^2\dee{y}\\ &=\int_0^3 32 y^3\dee{y} =\Big[8y^4\Big]_0^3 =8\times 81 = 648 \end{align*}

3.1.7.3.

Solution.

The following figures show the domains of integration for the integrals in this problem.

(a)

(b)

(c)

(d)

(e)

(f)

\begin{align*} \dblInt_R (x^2+y^2)\,\dee{x}\,\dee{y} &=\int_0^a \dee{x}\int_0^b\dee{y}\ (x^2+y^2) \tag{a}\\ &=\int_0^a \dee{x}\,\left(x^2b+\frac{1}{3}b^3\right)\\ &=\frac{1}{3}\big(a^3b+ab^3\big)\\ \dblInt_T (x-3y)\,\dee{x}\,\dee{y} &=\int_0^a \dee{x}\int_0^{b(1-{x\over a})}\dee{y}\,(x-3y) \tag{b}\\ &=\int_0^a \dee{x}\,\left[bx\left(1-\frac{x}{a}\right) -\frac{3}{2}b^2\left(1-\frac{x}{a}\right)^2\right] \notag\\ &=\left[\frac{b}{2}x^2-\frac{b}{3a}x^3 +\frac{a}{2}b^2\left(1-\frac{x}{a}\right)^3\right]_0^a \notag\\ &=\frac{a^2b}{2}-\frac{a^2b}{3}-\frac{ab^2}{2} =\frac{a^2b}{6}-\frac{ab^2}{2}\\ \dblInt_R xy^2\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{x^2}^{\sqrt{x}}\dee{y}\ xy^2 \tag{c}\\ &=\frac{1}{3}\int_0^1 \dee{x}\ x\big(x^{3/2}-x^6 \big)\\ &=\frac{1}{3}\left(\frac{2}{7}-\frac{1}{8}\right) =\frac{3}{56}\\ \dblInt_D x\cos y\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{0}^{1-x^2}\dee{y}\ x\cos y \tag{d}\\ &=\int_0^1 \dee{x}\ x\sin(1-x^2)\\ &=\frac{1}{2}\Big[\cos(1-x^2)\Big]_0^1 =\frac{1}{2}(1-\cos 1)\\ \dblInt_R \frac{x}{y}e^y\,\dee{x}\,\dee{y} &=\int_0^1 \dee{y}\int_{y}^{\sqrt{y}}\dee{x}\ \frac{x}{y}e^y =\int_0^1 \dee{y}\ \frac{y-y^2}{2y}e^y \tag{e}\\ &=\frac{1}{2}\int_0^1 \dee{y}\ (1-y)e^y\\ &=\frac{1}{2}\Big[-ye^y+2e^y\Big]_0^1 =\frac{1}{2}(e-2) \notag\\ \dblInt_T \frac{xy}{1+x^4}\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{x}^{1}\dee{y}\ \frac{xy}{1+x^4} \tag{f}\\ &=\frac{1}{2}\int_0^1 \dee{x}\ \frac{x(1-x^2)}{1+x^4}\\ &=\frac{1}{4}\int_0^1 \dee{t}\ \frac{1-t}{1+t^2}\hbox{ where } t=x^2 \notag\\ &=\frac{1}{4}\left[\arctan t-\frac{1}{2}\ln(1+t^2)\right]_0^1\\ &=\frac{1}{4}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right) \end{align*}

3.1.7.4.

Solution.

The following figures show the domains of integration for the integrals in this problem.

(a)

(b)

(c)

\begin{align*} \int_0^2 \dee{x}\int_1^{e^x} \dee{y} &=\int_0^2 \dee{x}\ \big[e^x-1\big] =\big[e^x-x\big]_0^2 =e^2-3 \tag{a}\\ \int_1^{e^2} \dee{y}\int_{\ln y}^2 \dee{x} &=\int_1^{e^2} \dee{y}\ \big[2-\ln y\big] =\big[2y-y\ln y+y\big]_1^{e^2}\\ &=e^2-3\\ \int_0^{\sqrt{2}}\dee{y}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}}\dee{x}\ y &=\int_0^{\sqrt{2}}\dee{y}\ 2y\sqrt{4-2y^2} \tag{b}\\ &=-\frac{1}{3}\Big[{(4-2y^2)}^{3/2}\Big]_0^{\sqrt{2}} =\frac{8}{3}\\ \int_{-2}^{2} \dee{x}\int_0^{\sqrt{2-{x^2\over 2}}} \dee{y}\ y &=\int_{-2}^{2} \dee{x}\ \left[1-\frac{x^2}{4}\right] =2\int_0^{2} \dee{x}\ \left[1-\frac{x^2}{4}\right]\\ &=2\left[x-\frac{x^3}{12}\right]_0^{2} =\frac{8}{3}\\ \int_{-2}^1 \dee{x}\int_{x^2+4x}^{3x+2}\dee{y} &=\int_{-2}^1 \dee{x}\ \big[-x^2-x+2\big] \tag{c}\\ &=\left[-\frac{x^3}{3}-\frac{x^2}{2}+2x\right]_{-2}^1 =\frac{9}{2}\\ \int_{-4}^{5} \dee{y}\int_{{y-2\over 3}}^{-2+\sqrt{4+y}} \dee{x} &=\int_{-4}^{5} \dee{y}\ \left[\!-\frac{4}{3}-\frac{y}{3}+\sqrt{4+y}\right]\\ &=\left[-\frac{4y}{3}-\frac{y^2}{6}+\frac{2}{3}{(4+y)}^{3\over 2}\right]_{-4}^{5} =\frac{9}{2} \end{align*}

In part (c), we used that the equation $y=x^2+4x$ is equivalent to $y+4=(x+2)^2$ and hence to $x=-2\pm\sqrt{y+4}\text{.}$

3.1.7.5. (✳).

Solution.

In the given integrals

$y$ runs for $0$ to $2\text{,}$ and
for each fixed $y$ between $0$ and $1\text{,}$ $x$ runs from $0$ to $y$ and
for each fixed $y$ between $1$ and $2\text{,}$ $x$ runs from $0$ to $2-y$

The figure on the left below contains a sketch of that region together with the generic horizontal slices that were used to set up the given integrals.

To reverse the order of integration, we switch to vertical, rather than horizontal, slices, as in the figure on the right above. Looking at that figure, we see that

$x$ runs for $0$ to $1\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $x$ to $2-x\text{.}$

So the desired integral is

\begin{gather*} \int_{x=0}^{x=1}\int_{y=x}^{y=2-x} f(x,y)\ \dee{y}\,\dee{x} \end{gather*}

3.1.7.6. (✳).

Solution.

(a) In the given integral

$x$ runs from $0$ to $1$ and
for each fixed $x$ between $0$ and $1\text{,}$ $y$ runs from $x$ to $1$

So the domain of integration is

\begin{equation*} D = \Set{(x,y)}{0\le x\le 1,\ x\le y\le 1} \end{equation*}

It is sketched in the figure on the left below.

(b) The given integral decomposed the domain of integration into vertical strips like the blue strip in the figure on the right above. To reverse the order of integration, we instead use horizontal strips. Looking at the pink strip in the figure on the right above, we see that this entails

having $y$ run from $0$ to $1$ and
for each fixed $y$ between $0$ and $1\text{,}$ having $x$ run from $0$ to $y$

This gives

\begin{equation*} \int_0^1\dee{y}\int_0^y \dee{x}\ e^{x/y}=\int_0^1\dee{y}\ \Big[ye^{x/y}\Big]_0^y =\int_0^1\dee{y}\ y(e-1) =\frac{1}{2} (e-1) \end{equation*}

3.1.7.7. (✳).

Solution.

(a) On $R$

$y$ runs from $1$ to $4$ (from $1$ to $\sqrt{2}$ in the first integral and from $\sqrt{2}$ to $4$ in the second).
For each fixed $y$ between $1$ and $\sqrt{2}\text{,}$ $x$ runs from $\frac{1}{y}$ to $\sqrt{y}$ and
for each fixed $y$ between $\sqrt{2}$ and $4\text{,}$ $x$ runs from $\frac{y}{2}$ to $\sqrt{y}\text{.}$

The figure on the left below is a sketch of $R\text{,}$ together with generic horizontal strips as were used in setting up the integral.

(b) To reverse the order of integration we use vertical strips as in the figure on the right above. Looking at that figure, we see that, on $R\text{,}$

$x$ runs from $1/\sqrt{2}$ to $2\text{.}$
For each fixed $x$ between $1/\sqrt{2}$ and $1\text{,}$ $y$ runs from $\frac{1}{x}$ to $2x$ and
for each fixed $x$ between $1$ and $2\text{,}$ $y$ runs from $x^2$ to $2x\text{.}$

\begin{gather*} I= \int_{1/\sqrt{2}}^1 \int_{1/x}^{2x} f(x,y)\,\dee{y}\,\dee{x} +\int_1^2 \int_{x^2}^{2x} f(x,y)\,\dee{y}\,\dee{x} \end{gather*}

\begin{align*} I &= \int_1^{\sqrt{2}} \int_{1/y}^{\sqrt{y}} \frac{x}{y}\,\dee{x}\,\dee{y} +\int_{\sqrt{2}}^4 \int_{y/2}^{\sqrt{y}} \frac{x}{y}\,\dee{x}\,\dee{y}\\ &=\int_1^{\sqrt{2}} \frac{1}{y}\left[\frac{y}{2}-\frac{1}{2y^2}\right] \,\dee{y} +\int_{\sqrt{2}}^4 \frac{1}{y}\left[\frac{y}{2}-\frac{y^2}{8}\right] \,\dee{y}\\ &=\left[\frac{y}{2}+\frac{1}{4y^2}\right]_1^{\sqrt{2}} +\left[\frac{y}{2}-\frac{y^2}{16}\right]_{\sqrt{2}}^4\\ &=\frac{1}{\sqrt{2}} +\frac{1}{8}-\frac{1}{2}-\frac{1}{4} +2 -1 -\frac{1}{\sqrt{2}} +\frac{1}{8}\\ &=\frac{1}{2} \end{align*}

3.1.7.8. (✳).

Solution.

(a) When $f(x,y)=x\text{,}$

\begin{align*} \int_{x=-1}^{x=3}\left[\int_{y=x^2}^{y=2x+3} x \dee{y}\right] \dee{x} &= \int_{x=-1}^{x=3}\left[x(2x+3-x^2) \right] \dee{x}\\ &=\left[\frac{2x^3}{3}+\frac{3x^2}{2}-\frac{x^4}{4}\right]_{-1}^3\\ &=18+\frac{27}{2}-\frac{81}{4} +\frac{2}{3}-\frac{3}{2}+\frac{1}{4}\\ &=18+12-20+\frac{2}{3} =\frac{32}{3} \end{align*}

(b) On the region $E$

$x$ runs from $-1$ to $3$ and
for each $x$ in that range, $y$ runs from $x^2$ to $2x+3$

Here are two sketches of $E\text{,}$ with the left one including a generic vertical strip as was used in setting up the given integral.

(c) To reverse the order of integration we use horizontal strips as in the figure on the right above. Looking at that figure, we see that, on the region $E\text{,}$

$y$ runs from $0$ to $9$ and
for each $y$ between $0$ and $1\text{,}$ $x$ runs from $-\sqrt{y}$ to $\sqrt{y}$
for each $y$ between $1$ and $9\text{,}$ $x$ runs from $(y-3)/2$ to $\sqrt{y}$

\begin{gather*} I = \int_0^1\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ x +\int_1^9\dee{y} \int_{(y-3)/2}^{\sqrt{y}}\dee{x}\ x \end{gather*}

3.1.7.9. (✳).

Solution.

The antiderivative of the function $\sin(y^2)$ cannot be expressed in terms of familiar functions. So we do not want the inside integral to be over $y\text{.}$ So we’ll use horizontal slices as in the figure

On the domain of integration

$y$ runs from $0$ to $4\text{,}$ and
for each fixed $y$ in that range, $x$ runs from $-y$ to $y/2$

The given integral

\begin{align*} \dblInt_D \sin(y^2)\ \dee{A} &=\int_0^4\dee{y}\int_{-y}^{y/2}\dee{x}\ \sin(y^2)\\ &=\int_0^4 \dee{y}\ \frac{3}{2}y\,\sin(y^2)\\ &=\left[-\frac{3}{4}\cos(y^2)\right]_0^4\\ &=\frac{3}{4}\big[1-\cos(16)\big] \end{align*}

3.1.7.10. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $0$ to $1$ and
for each fixed $y$ in that range, $x$ runs from $\sqrt{y}$ to $1\text{.}$

The figure on the left below is a sketch of that domain, together with a generic horizontal strip as was used in setting up the integral.

(b) The inside integral, $\int_{\sqrt{y}}^1 \frac{\sin(\pi x^2)}{x}\ \dee{x}\text{,}$ in the given form of $I$ looks really nasty. So let’s try exchanging the order of integration. Looking at the figure on the right above, we see that, on the domain of integration,

$x$ runs from $0$ to $1$ and
for each fixed $x$ in that range, $y$ runs from $0$ to $x^2\text{.}$

\begin{align*} I &= \int_0^1\dee{x}\int_0^{x^2}\dee{y}\ \frac{\sin(\pi x^2)}{x}\\ &=\int_0^1\dee{x}\ x\sin(\pi x^2)\\ &=\left[-\frac{\cos(\pi x^2)}{2 \pi}\right]_0^1 \qquad\text{(Looks pretty rigged!)}\\ &=\frac{1}{\pi} \end{align*}

3.1.7.11. (✳).

Solution.

(a) Let’s call the triangle $\cT\text{.}$ Here are two sketches of $\cT\text{,}$ one including a generic vertical strip and one including a generic horizontal strip. Notice that the equation of the line through $(0,0)$ and $(1,1)$ is $y=x\text{.}$

First, we’ll set up the integral using vertical strips. Looking at the figure on the left above, we see that, on $\cT\text{,}$

$x$ runs from $0$ to $1$ and
for each $x$ in that range, $y$ runs from $x$ to $1\text{.}$

So the integral

\begin{gather*} I=\int_0^1\dee{x}\int_x^1\dee{y}\ y^2\sin xy \end{gather*}

Next, we’ll set up the integral using horizontal strips. Looking at the figure on the right above, we see that, on $\cT\text{,}$

$y$ runs from $0$ to $1$ and
for each $y$ in that range, $x$ runs from $0$ to $y\text{.}$

So the integral

\begin{gather*} I=\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy \end{gather*}

(b) To evaluate the inside integral, $\int_x^1\dee{y}\ y^2\sin xy\text{,}$ of the vertical strip version, will require two integration by parts to get rid of the $y^2\text{.}$ So we’ll use the horizontal strip version.

\begin{align*} I&=\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy\\ &=\int_0^1\dee{y}\ \Big[-y\cos xy\Big]_0^y\\ &=\int_0^1\dee{y}\ \big[y-y\cos y^2\big]\\ &=\left[\frac{y^2}{2}-\frac{\sin y^2}{2}\right]_0^1 \qquad\text{(Look's pretty rigged!)}\\ &= \frac{1-\sin 1}{2} \end{align*}

3.1.7.12. (✳).

Solution.

If we call the triangular base region $\cT\text{,}$ then the volume is

\begin{gather*} V=\dblInt_\cT f(x,y)\ \dee{A} = \dblInt_\cT e^{-x^2}\ \dee{x}\,\dee{y} \end{gather*}

If we set up the integral using horizontal slices, so that the inside integral is the $x$--integral, there will be a big problem — the integrand $e^{-x^2}$ does not have an obvious anti--derivative. (In fact its antiderivative cannot be expressed in terms of familiar functions.) So let’s try vertical slices as in the sketch

Looking at that sketch we see that

$x$ runs from $0$ to $1\text{,}$ and
for each $x$ in that range, $y$ runs from $0$ to $x\text{.}$

So the integral is

\begin{align*} V&=\int_0^1\dee{x}\int_0^x\dee{y}\ e^{-x^2}\\ &=\int_0^1 \dee{x}\ xe^{-x^2}\\ &=\left[-\frac{1}{2}e^{-x^2}\right]_0^1\\ &=\frac{1-e^{-1}}{2} \end{align*}

3.1.7.13. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $0$ to $1$ and
for each $y$ in that range $x$ runs from $y$ to $2-y\text{.}$ So the left hand side of the domain is the line $x=y$ and the right hand side of the domain is $x=2-y\text{.}$

The figure on the left below is a sketch of that domain, together with a generic horizontal strip as was used in setting up the integral.

(b) To reverse the order of integration we use vertical, rather than horizontal, strips. Looking at the figure on the right above, we see that, in the domain of integration

$x$ runs from $0$ to $2$ and
for each $x$ between $0$ and $1\text{,}$ $y$ runs from $0$ to $x\text{,}$ while
for each $x$ between $1$ and $2\text{,}$ $y$ runs from $0$ to $2-x\text{.}$

So the integral

\begin{gather*} I=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x} \end{gather*}

\begin{align*} I&=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x}\\ &=\frac{1}{2}\int_0^1\dee{x}\ x +\frac{1}{2}\int_1^2\dee{x}\ \frac{(2-x)^2}{x}\\ &=\frac{1}{4} +\frac{1}{2}\int_1^2\dee{x}\ \left(\frac{4}{x}-4+x\right)\\ &=\frac{1}{4} +\frac{1}{2}\left[4\ln 2 -4 + \frac{4-1}{2}\right]\\ &=2\ln 2 -1 \end{align*}

3.1.7.14. (✳).

Solution.

(a) On the domain of integration,

$x$ runs from $0$ to $1\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $\sqrt{x}$ to $1\text{.}$ We may rewrite $y=\sqrt{x}$ as $x=y^2\text{,}$ which is a rightward opening parabola.

Here are two sketches of the domain of integration, which we call $D\text{.}$ The left hand sketch also shows a vertical slice, as was used in setting up the integral.

(b) The inside integral, $\int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}\text{,}$ of the given integral looks pretty nasty. So let’s reverse the order of integration, by using horizontal, rather than vertical, slices. Looking at the figure on the right above, we see that

$y$ runs from $0$ to $1\text{,}$ and
for each fixed $y$ in that range $x$ runs from $0$ to $y^2\text{.}$

\begin{align*} I &= \int_0^1\dee{y} \int_0^{y^2}\dee{x}\ \sqrt{1+y^3}\\ &=\int_0^1\dee{y} \ y^2\sqrt{1+y^3}\\ &=\int_1^2\frac{\dee{u}}{3}\ \sqrt{u} \qquad\text{with } u=1+y^3,\ \dee{u}=3y^2\,\dee{y}. \text{ Looks pretty rigged!}\\ &=\frac{1}{3} \left[\frac{u^{3/2}}{3/2}\right]_1^2\\ &=\frac{2\big(2\sqrt{2}-1\big)}{9} \end{align*}

3.1.7.15. (✳).

Solution.

(a) Observe that the parabola $y^2=x$ and the line $y=x-2$ meet when $x=y+2$ and

\begin{equation*} y^2=y+2 \iff y^2-y-2=0 \iff (y-2)(y+1)=0 \end{equation*}

So the points of intersection of $x=y^2$ and $y=x-2$ are $(1,-1)$ and $(4,2)\text{.}$ Here is a sketch of $D\text{.}$

To evaluate $J\text{,}$ we’ll use horizontal slices as in the figure above. (If we were to use vertical slices we would have to split the integral in two, with $0\le x\le 1$ in one part and $1\le x\le 4$ in the other.) From the figure, we see that, on $D\text{,}$

$y$ runs from $-1$ to $2$ and
for each fixed $y$ in that range, $x$ runs from $y^2$ to $y+2\text{.}$

Hence

\begin{align*} J &=\dblInt_D 3y\ \dee{A} =\int_{-1}^2\dee{y}\int_{y^2}^{y+2}\dee{x}\ 3y\\ &=3\int_{-1}^2\dee{y}\ y(y+2-y^2)\\ &=3\left[\frac{y^3}{3}+y^2-\frac{y^4}{4}\right]_{-1}^2\\ &=3\left[\frac{8}{3}+4-4+\frac{1}{3}-1+\frac{1}{4}\right]\\ &=\frac{27}{4} \end{align*}

(b) On the domain of integration,

$x$ runs from $0$ to $4$ and
for each fixed $x$ in that range, $y$ runs from $\frac{1}{2}\sqrt{x}$ to $1\text{.}$

The figure on the left below is a sketch of that domain, together with a generic vertical strip as was used in setting up the integral.

The inside integral, over $y\text{,}$ looks pretty nasty because $e^{y^3}$ does not have an obvious antiderivative. So let’s reverse the order of integration. That is, let’s use horizontal, rather than vertical, strips. From the figure on the right above, we see that, on the domain of integration

$y$ runs from $0$ to $1$ and
for each fixed $y$ in that range, $x$ runs from $0$ to $4y^2\text{.}$

\begin{align*} I &= \int_0^1\dee{y}\int_0^{4y^2}\dee{x}\ e^{y^3}\\ &= \int_0^1\dee{y}\ 4y^2 e^{y^3}\\ &= \frac{4}{3}\int_0^1\dee{u}\ e^u \qquad\text{with } u=y^3,\ \dee{u}=3y^2\,\dee{y}\qquad \text{(Looks rigged!)}\\ &=\frac{4}{3}\big[e-1\big] \end{align*}

3.1.7.16. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $-4$ to $0$ and
for each $y$ in that range, $x$ runs from $\sqrt{-y}$ (when $y=-x^2$) to $2\text{.}$

The figure on the left below provides a sketch of the domain of integration. It also shows the generic horizontal slice that was used to set up the given iterated integral.

(b) The inside integral, $\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}$ looks nasty. So let’s reverse the order of integration and use vertical, rather than horizontal, slices. From the figure on the right above, on the domain of integration,

$x$ runs from $0$ to $2$ and
for each $x$ in that range, $y$ runs from $-x^2$ to $0\text{.}$

So the integral

\begin{align*} \int_{-4}^0\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}\,\dee{y} &=\int_0^2\dee{x}\int_{-x^2}^0\dee{y}\ \cos(x^3)\\ &=\int_0^2\dee{x}\ x^2\ \cos(x^3) =\left[\frac{\sin(x^3)}{3}\right]_0^2\\ &=\frac{\sin(8)}{3} \end{align*}

3.1.7.17. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $0$ to $4$ and
for each $y$ in the range $0\le y\le 1\text{,}$ $x$ runs from $-\sqrt{y}$ to $\sqrt{y}$ and
for each $y$ in the range $1\le y\le 4\text{,}$ $x$ runs from $y-2$ to $\sqrt{y}\text{.}$

Both figures below provide sketches of the domain of integration.

To reverse the order of integration observe, from the figure on the right above that, on the domain of integration,

$x$ runs from $-1$ to $2$ and
for each $x$ in that range, $y$ runs from $x^2$ to $x+2\text{.}$

So the integral

\begin{equation*} I = \int_{-1}^2\int_{x^2}^{x+2} f(x,y)\ \dee{y}\,\dee{x} \end{equation*}

(b) We’ll use the integral with the order of integration reversed that we found in part (a). When $f(x,y)=\frac{e^x}{2-x}$

\begin{align*} I &= \int_{-1}^2\int_{x^2}^{x+2} \frac{e^x}{2-x}\ \dee{y}\,\dee{x}\\ &= \int_{-1}^2 (x+2-x^2)\frac{e^x}{2-x}\ \dee{x} = -\int_{-1}^2 (x-2)(x+1)\frac{e^x}{2-x}\ \dee{x}\\ &= \int_{-1}^2 (x+1) e^x\ \dee{x}\\ &= \Big[xe^x\Big]_{-1}^2\\ &= 2e^2 + \frac{1}{e} \end{align*}

3.1.7.18. (✳).

Solution.

On the domain of integration

$y$ runs from $0$ to $4\text{.}$ In inequalities, $0\le y\le 4\text{.}$
For each fixed $y$ in that range, $x$ runs from $\sqrt{y}$ to $\sqrt{8-y}\text{.}$ In inequalities, that is $\sqrt{y}\le x\le \sqrt{8-y}\text{,}$ or $y\le x^2\le 8-y\text{.}$

Here are two sketchs of the domain of integration.

(b) To reverse the order we observe, from the figure on the right above, that, on the domain of integration,

$x$ runs from $0$ to $\sqrt{8}\text{.}$ In inequalities, $0\le x\le \sqrt{8}\text{.}$
For each fixed $x$ between $0$ and $2\text{,}$ $y$ runs from $0$ to $x^2\text{.}$ In inequalities, that is $0\le y\le x^2\text{.}$
For each fixed $x$ between $2$ and $\sqrt{8}\text{,}$ $y$ runs from $0$ to $8-x^2\text{.}$ In inequalities, that is $0\le y\le 8-x^2\text{.}$

So the integral is

\begin{equation*} \int_0^2\int_0^{x^2} f(x,y)\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} f(x,y)\,\dee{y}\,\dee{x} \end{equation*}

\begin{align*} &\int_0^2\int_0^{x^2} \frac{1}{(1+y)^2}\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} \frac{1}{(1+y)^2}\,\dee{y}\,\dee{x}\\ &=-\int_0^2 \left[\frac{1}{1+y}\right]_0^{x^2}\,\dee{x} -\int_2^{\sqrt{8}} \left[\frac{1}{1+y}\right]_0^{8-x^2}\,\dee{x}\\ &=\int_0^2 \left[1-\frac{1}{1+x^2}\right]\,\dee{x} +\int_2^{\sqrt{8}} \left[1-\frac{1}{9-x^2}\right]\,\dee{x}\\ &=\sqrt{8}-\arctan x\bigg|_0^2 -\frac{1}{6}\int_2^{\sqrt{8}} \left[\frac{1}{3+x}+\frac{1}{3-x}\right]\,\dee{x}\\ &=\sqrt{8}-\arctan 2 -\frac{1}{6}\Big[\ln(3+x)-\ln(3-x)\Big]_2^{\sqrt{8}}\\ &=\sqrt{8}-\arctan 2 -\frac{1}{6}\left[\ln\frac{3+\sqrt{8}}{3-\sqrt{8}} -\ln 5\right] \end{align*}

3.1.7.19. (✳).

Solution.

The antiderivative of the function $e^{-y^2}$ cannot be expressed in terms of elementary functions. So the inside integral $\int_{-2}^{2x} e^{y^2}\ \dee{y}$ cannot be evaluated using standard calculus 2 techniques. The trick for dealing with this integral is to reverse the order of integration. On the domain of integration

$x$ runs from $-1$ to $0\text{.}$ In inequalities, $-1\le x\le 0\text{.}$
For each fixed $x$ in that range, $y$ runs from $-2$ to $2x\text{.}$ In inequalities, $-2\le y\le 2x\text{.}$

The domain of integration, namely

\begin{equation*} \Set{(x,y)}{-1\le x\le 0,\ -2\le y\le 2x} \end{equation*}

is sketched in the figure on the left below.

Looking at the figure on the right above, we see that we can also express the domain of integration as

\begin{equation*} \Set{(x,y)}{-2\le y\le 0,\ y/2\le x\le 0} \end{equation*}

So the integral

\begin{align*} \int_{-1}^0 \int_{-2}^{2x} e^{y^2}\ \dee{y}\,\dee{x} &=\int_{-2}^0 \int_{y/2}^{0} e^{y^2}\ \dee{x}\,\dee{y}\\ &=-\frac{1}{2}\int_{-2}^0 y e^{y^2}\ \dee{y}\\ &=-\frac{1}{2}\left[\frac{1}{2}e^{y^2}\right]_{-2}^0\\ &=\frac{1}{4}\big[e^4-1\big] \end{align*}

3.1.7.20. (✳).

Solution.

We first have to get a picture of the domain of integration. The first integral has domain of integration

\begin{equation*} \Set{(x,y)}{0\le x\le 2,\ 0\le y\le x} \end{equation*}

and the second integral has domain of integration

\begin{equation*} \Set{(x,y)}{2\le x\le 6,\ 0\le y\le \sqrt{6-x}} \end{equation*}

Here is a sketch. The domain of integration for the first integral is the shaded triangular region to the left of $x=2$ and the domain of integration for the second integral is the shaded region to the right of $x=2\text{.}$

To exchange the order of integration, we use horizontal slices as in the figure below.

The bottom slice has $y=0$ and the top slice has $y=2\text{.}$ On the slice at height $y\text{,}$ $x$ runs from $y$ to $6-y^2\text{.}$ So

\begin{equation*} I = \int_0^2 \int_y^{6-y^2} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}

3.1.7.21. (✳).

Solution.

(a), (b) Looking at the figure on the left below, we see that we can write the domain

\begin{equation*} D = \Set{(x,y)}{0\le y\le 1,\ -\sqrt{1-y}\le x\le\sqrt{1-y}} \end{equation*}

\begin{equation*} \dblInt_D f(x,y)\ \dee{A} =\int_0^1\dee{y} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}\dee{x}\ f(x,y) =\int_0^1 \int_{-\sqrt{1-y}}^{\sqrt{1-y}} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}

Looking at the figure on the right above, we see that we can write the domain

\begin{equation*} D = \Set{(x,y)}{-1\le x\le 1,\ 0\le y\le 1-x^2} \end{equation*}

\begin{equation*} \dblInt_D f(x,y)\ \dee{A} =\int_{-1}^1\dee{x} \int_0^{1-x^2}\dee{y}\ f(x,y) =\int_{-1}^1 \int_0^{1-x^2} f(x,y)\ \dee{y}\,\dee{x} \end{equation*}

\begin{align*} \dblInt_D e^{x-(x^3/3)}\ \dee{A} &=\int_{-1}^1\dee{x} \int_0^{1-x^2}\dee{y}\ e^{x-(x^3/3)}\\ &=\int_{-1}^1 (1-x^2) e^{x-(x^3/3)}\ \dee{x}\\ &=\int_{-2/3}^{2/3} e^u\ \dee{u} \qquad\text{with } u=x -\frac{x^3}{3},\ \dee{u} = \big(1-x^2)\,\dee{x}\\ &= e^{2/3}-e^{-2/3} \end{align*}

3.1.7.22. (✳).

Solution.

(a) On the domain of integration,

$x$ runs from $0$ to $1$ and
for each fixed $x$ in that range, $y$ runs from $x^2$ to $1\text{.}$

The figure on the left below is a sketch of that domain, together with a generic vertical strip as was used in setting up the integral.

(b) As it stands, the inside integral, over $y\text{,}$ looks pretty nasty because $\sin(y^3)$ does not have an obvious antiderivative. So let’s reverse the order of integration. The given integral was set up using vertical strips. So, to reverse the order of integration, we use horizontal strips as in the figure on the right above. Looking at that figure we see that, on the domain of integration,

$y$ runs from $0$ to $1$ and
for each fixed $y$ in that range, $x$ runs from $0$ to $\sqrt{y}\text{.}$

\begin{align*} I&=\int_0^1\dee{y} \int_0^{\sqrt{y}}\dee{x}\ x^3\ \sin(y^3)\\ &=\int_0^1\dee{y}\ \sin(y^3)\left[\frac{x^4}{4}\right]_0^{\sqrt{y}}\\ &=\frac{1}{4}\int_0^1\dee{y}\ y^2\sin(y^3)\\ &=\frac{1}{4}\left[-\frac{\cos(y^3)}{3}\right]_0^1\\ &=\frac{1-\cos(1)}{12} \end{align*}

3.1.7.23. (✳).

Solution.

(a) The solid is the set of all $(x,y,z)$ obeying $0\le x\le 3\text{,}$ $0\le y\le 3$ and $0\le z\le 6-xy\text{.}$ The base of this region is the set of all $(x,y)$ for which there is a $z$ such that $(x,y,z)$ is in the solid. So the base is the set of all $(x,y)$ obeying $0\le x\le 3\text{,}$ $0\le y\le 3$ and $6-xy\ge 0\text{,}$ i.e. $xy\le 6\text{.}$ This region is sketched in the figure on the left below.

(b) We’ll deompose the base region into vertical strips as in the figure on the right above. Observe that the line $y=3$ intersects the curve $xy=6$ at the point $(2,3)$ and that on the base

$x$ runs from $0$ to $3$ and that
for each fixed $x$ between $0$ and $2\text{,}$ $y$ runs from $0$ to $3\text{,}$ while
for each fixed $x$ between $2$ and $3\text{,}$ $y$ runs from $0$ to $6/x$

and that, for each $(x,y)$ in the base, $z$ runs from $0$ to $6-xy\text{.}$ So the

\begin{align*} \text{Volume}&=\int_0^2\dee{x}\int_0^3\dee{y}\ (6-xy)+ \int_2^3\dee{x}\int_0^{6/x} \dee{y}\ (6-xy)\\ &=\int_0^2\dee{x}\ \left[6y-\frac{1}{2} xy^2\right]_0^3 +\int_2^3\dee{x}\ \left[6y-\frac{1}{2} xy^2\right]_0^{6/x}\\ &=\int_0^2\dee{x}\ \left[18-\frac{9}{2}x\right] +\int_2^3\dee{x}\ \left[\frac{36}{x}-\frac{18}{x}\right]\\ &=\left[18x-\frac{9}{4}x^2\right]_0^2+\Big[18\ln x\Big]_2^3 =27+18\ln\frac{3}{2}\approx 34.30 \end{align*}

3.1.7.24. (✳).

Solution.

In the given integral

$x$ runs from $-2$ to $2$ and
for each fixed $x$ between $-2$ and $2\text{,}$ $y$ runs from $x^2$ to $4$

So the domain of integration is

\begin{equation*} D = \Set{(x,y)}{-2\le x\le 2,\ x^2\le y\le 4} \end{equation*}

This is sketched below.

The inside integral, $\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}\text{,}$ in the given integral looks really nasty. So let’s try exchanging the order of integration. The given integral was formed by decomposing the domain of integration $D$ into horizontal strips, like the blue strip in the figure above. To exchange the order of integration we instead decompose the domain of integration $D$ into vertical strips, like the pink strip in the figure above. To do so, we observe that, on $D\text{,}$

$y$ runs from $0$ to $4$ and
for each fixed $y$ between $0$ and $4\text{,}$ $x$ runs from $-\sqrt{y}$ to $\sqrt{y}\text{.}$

That is, we reexpress the domain of integration as

\begin{equation*} D = \Set{(x,y)}{0\le y\le 4,\ -\sqrt{y}\le x\le \sqrt{y}} \end{equation*}

and the given integral as

\begin{align*} \int_{-2}^2\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}\,\dee{x} &=\int_0^4\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ \cos\big(y^{3/2}\big)\cr &=\int_0^4\dee{y} \ 2\sqrt{y}\cos\big(y^{3/2}\big)\cr &=\frac{4}{3}\int_0^8\dee{t}\ \cos t\quad\hbox{where } t=y^{3/2},\ \dee{t}=\frac{3}{2}\sqrt{y}\ \dee{y}\cr &=\frac{4}{3}\sin t\Big|_0^8 =\frac{4}{3}\sin 8\approx 1.319 \end{align*}

3.1.7.25. (✳).

Solution.

(a) We may rewrite the equation $x^2+y^2=2y$ of the cylinder as $x^2+(y-1)^2=1\text{.}$ We are (in part (c)) to find the volume of the set

\begin{equation*} V=\Set{(x,y,z)}{x^2+(y-1)^2\le 1,\ 0\le z\le 8+2xy} \end{equation*}

When we look at this solid from far above (so that we can’t see $z$) we see the set of points $(x,y)$ that obey $x^2+(y-1)^2\le 1$ and $8+2xy\ge 0$ (so that there is at least one allowed $z$ for that $(x,y)$). All points in $x^2+(y-1)^2\le 1$ have $-1\le x\le 1$ and $0\le y\le 2$ and hence $-2\le xy\le 2$ and $8+2xy\ge 0\text{.}$ So the domain of integration consists of the full disk

\begin{equation*} D = \Set{(x,y)}{x^2+(y-1)^2\le 1} \end{equation*}

The volume is

\begin{equation*} I=\dblInt_D (8+2xy)\ \dee{x}\dee{y} \end{equation*}

(b) We can express the double integral over $D$ as iterated integrals by decomposing $D$ into horizontal strips, like the pink strip in the figure below, and also by decomposing $D$ into blue strips, like the blue strip in the figure below.

For horizontal strips, we use that, on $D$

$y$ runs from $0$ to $2$ and,
for each fixed $y$ between $0$ and $2\text{,}$ $x$ runs from $-\sqrt{2y-y^2}$ to $\sqrt{2y-y^2}$

so that

\begin{equation*} D=\Set{(x,y)}{0\le y\le 2,\ -\sqrt{2y-y^2}\le x\le \sqrt{2y-y^2}} \end{equation*}

For vertical strips, we use that, on $D$

$x$ runs from $-1$ to $1$ and,
for each fixed $x$ between $-1$ and $1\text{,}$ $y$ runs from $1-\sqrt{1-x^2}$ to $1+\sqrt{1-x^2}$

so that

\begin{equation*} D=\Set{(x,y)}{-1\le x\le 1,\ 1-\sqrt{1-x^2}\le y\le 1+\sqrt{1-x^2}} \end{equation*}

Thus

\begin{align*} I&=\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ (8+2xy)\\ &=\int_{-1}^1 \dee{x}\int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\dee{y}\ (8+2xy) \end{align*}

\begin{align*} \text{Volume}&=8\pi+\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ 2xy =8\pi+2\int_0^2 \dee{y}\ y\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ x\\ &=8\pi \end{align*}

because $\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ x=0$ for all $y\text{,}$ because the integrand is odd and the domain of integration is even.

3.1.7.26. (✳).

Solution.

In the given integral

$y$ runs from $0$ to $9$ and
for each fixed $y$ between $0$ and $9\text{,}$ $x$ runs from $\sqrt{y}$ to $3$

So the domain of integration is

\begin{equation*} D = \Set{(x,y)}{0\le y\le 9,\ \sqrt{y}\le x\le 3} \end{equation*}

This is sketched below.

The inside integral, $\int_{\sqrt{y}}^3\sin\big(\pi x^3\big)\ \dee{x}\text{,}$ in the given integral looks really nasty. So let’s try exchanging the order of integration. The given integral was formed by decomposing the domain of integration $D$ into horizontal strips, like the blue strip in the figure above. To exchange the order of integration we instead decompose the domain of integration $D$ into vertical strips, like the pink strip in the figure above. To do so, we observe that, on $D\text{,}$

$x$ runs from $0$ to $3$ and
for each fixed $x$ between $0$ and $3\text{,}$ $y$ runs from $0$ to $x^2\text{.}$

That is, we reexpress the domain of integration as

\begin{equation*} D = \Set{(x,y)}{0\le x\le 3,\ 0\le y\le x^2} \end{equation*}

and the given integral as

\begin{align*} \int_0^9\int_{\sqrt{y}}^3\sin(\pi x^3)\ \dee{x}\dee{y} &=\int_0^3\dee{x} \int_0^{x^2}\dee{y}\ \sin(\pi x^3) \cr &=\int_0^3\dee{x} \ x^2\sin(\pi x^3)\cr &=\frac{1}{3\pi}\int_0^{27\pi}\dee{t}\ \sin t\quad\hbox{where } t=\pi x^3,\ \dee{t}=3\pi x^2\,\dee{x}\cr &=-\frac{1}{3\pi}\cos t\Big|_0^{27\pi}=-\frac{1}{3\pi}\cos t\Big|_0^\pi =\frac{2}{3\pi}\approx 0.212 \end{align*}

3.1.7.27. (✳).

Solution.

(a) In the given integral

$x$ runs from $0$ to $1\text{,}$ and
for each fixed $x$ between $0$ and $1\text{,}$ $y$ runs from $-\sqrt{x}$ to $\sqrt{x}\text{.}$

So the region

\begin{equation*} R=\Set{(x,y)}{0\le x\le 1,\ -\sqrt{x}\le y\le \sqrt{x}} \end{equation*}

It is sketched below.

(b) The given integral was formed by decomposing the domain of integration $R$ into vertical strips, like the pink strip in the figure above. To exchange the order of integration we instead decompose the domain of integration $R$ into horizontal strips, like the blue strip in the figure above. To do so, we observe that, on $R\text{,}$

$y$ runs from $-1$ to $1\text{,}$ and
for each fixed $y$ between $-1$ and $1\text{,}$ $x$ runs from $y^2$ to $1\text{.}$

\begin{equation*} I=\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y} \end{equation*}

(c) The easy way to evaluate $I$ is to observe that, since $\sin\big(y^3-3y\big)$ is odd under $y\rightarrow -y\text{,}$ the integral

\begin{equation*} \int_{-\sqrt{x}}^{\sqrt{x}} \sin\big(y^3-3y\big)\,\dee{y}=0 \end{equation*}

for all $x\text{.}$ Hence $I=0\text{.}$ The hard way is

\begin{align*} I&=\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y}\\ &=\int_{-1}^1 (1-y^2) \sin\big(y^3-3y\big)\ \dee{y}\\ &=\int_2^{-2} \sin t\ \frac{\dee{t}}{-3} \qquad\hbox{ where }t=y^3-3y,\ \dee{t} =3(y^2-1)\,\dee{y}\\ &=\frac{1}{3}\cos t\ \Big|_2^{-2}=0 \end{align*}

again, since $\cos$ is even.

3.1.7.28. (✳).

Solution.

The parabola $y^2=2x+6$ and the line $y=x-1$ meet when $x=y+1$ with $y^2=2(y+1)+6$ or $y^2-2y-8=(y-4)(y+2)=0\text{.}$ So they meet at $(-1,-2)$ and $(5,4)\text{.}$ The domain of integration is sketched below.

On this domain

$y$ runs from $-2$ to $4\text{,}$ and
for each fixed $y$ between $-2$ and $4\text{,}$ $x$ runs from $\frac{y^2}{2}-3$ to $y+1\text{.}$

So the integral is

\begin{align*} &\int_{-2}^4\dee{y}\int_{y^2/2-3}^{y+1}\dee{x}\ xy =\int_{-2}^4\dee{y}\ \frac{1}{2} x^2y\bigg|_{y^2/2-3}^{y+1}\\ &\hskip0.5in=\frac{1}{2}\int_{-2}^4\dee{y}\ \left[y^3+2y^2+y-\frac{1}{4}y^5+3y^3-9y\right]\\ &\hskip0.5in=\frac{1}{2}\int_{-2}^4\dee{y}\ \left[-8y+2y^2+4y^3-\frac{1}{4}y^5\right]\\ &\hskip0.5in=\left[-2y^2+\frac{1}{3}y^3+\frac{1}{2}y^4-\frac{1}{48}y^6\right]_{-2}^4\\ &\hskip0.5in=-2(16-4)+\frac{1}{3}(64+8)+\frac{1}{2}(256-16) -\frac{1}{48}(4096-64)\cr &\hskip0.5in=-24+24+120-84=36 \end{align*}

3.1.7.29.

Solution.

Looking down from the top, we see the cylinder $x^2+2y^2\le 8\text{.}$ That gives the base region. The top of the solid, above any fixed $(x,y)$ in the base region, is at $z=8-x$ (this is always positive because $x$ never gets bigger than $\sqrt{8}$) . The bottom of the solid, below any fixed $(x,y)$ in the base region, is at $z=y-4$ (this is always negative because $y$ is always smaller than $2$). So the height of the solid at any $(x,y)$ is

\begin{equation*} z_{\rm top}-z_{\rm bottom} =(8-x)-(y-4)=12-x-y \end{equation*}

The volume is

\begin{equation*} \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ (12-x-y) \end{equation*}

Recall, from Theorem 1.2.11 in the CLP-2 text, that if $f(x)$ is an odd function (meaning that $f(-x)=-f(x)$ for all $x$), then $\int_{-a}^a f(x)\ \dee{x}=0$ (because the two integrals $\int_0^a f(x)\ \dee{x}$ and $\int_{-a}^0 f(x)\ \dee{x}$ have the same magnitude but opposite signs). Applying this twice gives

\begin{equation*} \int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ x=0\text{ and } \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ y =\int_{-2}^{2}\dee{y}\ 2y\sqrt{8-2y^2}=0 \end{equation*}

since $x$ and $y\sqrt{8-2y^2}$ are both odd. Thus

\begin{equation*} \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ (-x-y)=0 \implies \text{Volume} = \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ 12 \end{equation*}

so that the volume is just 12 times the area of the ellipse $x^2+2y^2=8\text{,}$ which is

\begin{equation*} 12\big(\pi\,\sqrt{8}\,2\big)= 48\sqrt{2}\,\pi \end{equation*}

3.2 Double Integrals in Polar Coordinates
3.2.5 Exercises

3.2.5.1.

Solution.

The first hand sketch below contains the points, $(x_1,y_1)\text{,}$ $(x_3,y_3)\text{,}$ $(x_5,y_5)\text{,}$ that are on the axes. The second hand sketch below contains the points, $(x_2,y_2)\text{,}$ $(x_4,y_4)\text{,}$ that are not on the axes.

Recall that the polar coordinates $r\text{,}$ $\theta$ are related to the cartesian coordinates $x\text{,}$ $y\text{,}$ by $x=r\cos\theta\text{,}$ $y=r\sin\theta\text{.}$ So $r=\sqrt{x^2+y^2}$ and $\tan\theta=\frac{y}{x}$ (assuming that $x\ne 0$ and $r\gt 0$) and

\begin{alignat*}{2} (x_1,y_1) &= (3,0) &&\implies r_1=3,\ \tan\theta_1=0\\ & &&\implies \theta_1 =0 \text{ as $(x_1,y_1)$ is on the positive $x$-axis}\\ (x_2,y_2) &= (1,1) &&\implies r_2=\sqrt{2},\ \tan\theta_2=1\\ & &&\implies \theta_2=\frac{\pi}{4} \text{ as $(x_2,y_2)$ is in the first octant}\\ (x_3,y_3) &= (0,1) &&\implies r_3=1,\ \cos\theta_3=0\\ & &&\implies \theta_3=\frac{\pi}{2} \text{ as $(x_3,y_3)$ is on the positive $y$-axis}\\ (x_4,y_4) &= (-1,1) &&\implies r_4=\sqrt{2},\ \tan\theta_4=-1\\ & &&\implies \theta_4=\frac{3\pi}{4} \text{ as $(x_4,y_4)$ is in the third octant}\\ (x_5,y_5) &= (-2,0) &&\implies r_5=2,\ \tan\theta_5=0\\ & &&\implies \theta_5 =\pi \text{ as $(x_5,y_5)$ is on the negative $x$-axis} \end{alignat*}

3.2.5.2.

Solution.

In this solution, we’ll supress the subscripts. That is, we’ll write $r$ in place of $r_i$ and $\theta$ in place of $\theta_i\text{.}$ Note that the distance from the point $\big(r\cos\theta\,,\,r\sin\theta\big)$ to the origin is

\begin{equation*} \sqrt{r^2\cos^2\theta + r^2\sin^2\theta} =\sqrt{r^2} =|r| \end{equation*}

Thus $r$ can be either the distance to the origin or minus the distance to the origin.

(a) The distance from $(-2,0)$ to the origin is $2\text{.}$ So either $r=2$ or $r=-2\text{.}$

If $r=2\text{,}$ then $\theta$ must obey
\begin{align*} (-2,0) = \big(2\cos\theta\,,\,2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1\\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=-1\\ &\iff \theta= n\pi,\ n\text{ odd integer } \end{align*}
If $r=-2\text{,}$ then $\theta$ must obey
\begin{align*} (-2,0) = \big(-2\cos\theta\,,\,-2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1\\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=1\\ &\iff \theta= n\pi,\ n\text{ even integer } \end{align*}

In particular, $\big(r=-2\,,\,\theta= 0\big)$ has $r\lt 0$ and $0\le\theta\lt 2\pi\text{.}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\pi\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\pi\text{,}$ $r \lt 0\text{.}$ In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=0\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=0\text{,}$ $r \lt 0\text{.}$

(b) The distance from $(1,1)$ to the origin is $\sqrt{2}\text{.}$ So either $r=\sqrt{2}$ or $r=-\sqrt{2}\text{.}$

If $r=\sqrt{2}\text{,}$ then $\theta$ must obey
\begin{align*} (1,1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=\frac{1}{\sqrt{2}}\\ &\iff \theta= \frac{\pi}{4} + 2n\pi,\ n\text{ integer } \end{align*}
If $r=-\sqrt{2}\text{,}$ then $\theta$ must obey
\begin{align*} (1,1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-\frac{1}{\sqrt{2}}\\ &\iff \theta= \frac{5\pi}{4} + 2n\pi,\ n\text{ integer } \end{align*}

In particular, $\big(r=-\sqrt{2}\,,\, \theta= \frac{5\pi}{4}\big)$ has $r\lt 0$ and $0\le\theta\lt 2\pi\text{.}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}\text{,}$ $r \lt 0\text{.}$ In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}\text{,}$ $r \lt 0\text{.}$

(c) The distance from $(-1,-1)$ to the origin is $\sqrt{2}\text{.}$ So either $r=\sqrt{2}$ or $r=-\sqrt{2}\text{.}$

If $r=\sqrt{2}\text{,}$ then $\theta$ must obey
\begin{align*} (-1,-1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-\frac{1}{\sqrt{2}}\\ &\iff \theta= \frac{5\pi}{4} + 2n\pi,\ n\text{ integer } \end{align*}
If $r=-\sqrt{2}\text{,}$ then $\theta$ must obey
\begin{align*} (-1,-1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=\frac{1}{\sqrt{2}}\\ &\iff \theta= \frac{\pi}{4} + 2n\pi,\ n\text{ integer } \end{align*}

In particular, $\big(r=-\sqrt{2}\,,\, \theta= \frac{\pi}{4}\big)$ has $r\lt 0$ and $0\le\theta\lt 2\pi\text{.}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}\text{,}$ $r \lt 0\text{.}$ In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}\text{,}$ $r \lt 0\text{.}$

(d) The distance from $(3,0)$ to the origin is $3\text{.}$ So either $r=3$ or $r=-3\text{.}$

If $r=3\text{,}$ then $\theta$ must obey
\begin{align*} (3,0) = \big(3\,\cos\theta\,,\,3\,\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1\\ &\iff \theta= 0 + 2n\pi,\ n\text{ integer } \end{align*}
If $r=-3\text{,}$ then $\theta$ must obey
\begin{align*} (3,0) = \big(-3\,\cos\theta\,,\,-3\,\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1\\ &\iff \theta= \pi + 2n\pi,\ n\text{ integer } \end{align*}

In particular, $\big(r=-3\,,\, \theta= \pi\big)$ has $r\lt 0$ and $0\le\theta\lt 2\pi\text{.}$

(e) The distance from $(0,1)$ to the origin is $1\text{.}$ So either $r=1$ or $r=-1\text{.}$

If $r=1\text{,}$ then $\theta$ must obey
\begin{align*} (0,1) = \big(\cos\theta\,,\,\sin\theta\big) &\iff \cos\theta=0,\ \sin\theta=1\\ &\iff \theta= \frac{\pi}{2} + 2n\pi,\ n\text{ integer } \end{align*}
If $r=-1\text{,}$ then $\theta$ must obey
\begin{align*} (0,1) = \big(-\cos\theta\,,\,-\sin\theta\big) &\iff \cos\theta=0,\ \sin\theta=-1\\ &\iff \theta= \frac{3\pi}{2} + 2n\pi,\ n\text{ integer } \end{align*}

In particular, $\big(r=-1\,,\, \theta= \frac{3\pi}{2}\big)$ has $r\lt 0$ and $0\le\theta\lt 2\pi\text{.}$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{3\pi}{2}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{3\pi}{2}\text{,}$ $r \lt 0\text{.}$ In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{2}\text{,}$ $r \gt 0$ and the orange half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{2}\text{,}$ $r \lt 0\text{.}$

3.2.5.3.

Solution.

(a) The lengths are

\begin{align*} |\he_r(\theta)| &= \sqrt{\cos^2\theta+\sin^2\theta} = 1\\ |\he_\theta(\theta)| &= \sqrt{(-\sin\theta)^2+\cos^2\theta} = 1 \end{align*}

\begin{equation*} \he_r(\theta) \cdot \he_\theta(\theta) = (\cos\theta)(-\sin\theta) +(\sin\theta)(\cos\theta)=0 \end{equation*}

the two vectors are perpendicular and the angle between them is $\frac{\pi}{2}\text{.}$ The cross product is

\begin{align*} \he_r(\theta) \times \he_\theta(\theta) &=\det\left[\begin{matrix} \hi & \hj & \hk\\ \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0 \end{matrix}\right] =\hk \end{align*}

(b) Note that for $\theta$ determined by $x=r\cos\theta\text{,}$ $y=r\sin\theta\text{,}$

the vector $\he_r(\theta)$ is a unit vector in the same direction as the vector from $(0,0)$ to $(x,y)$ and
the vector $\he_\theta(\theta)$ is a unit vector that is perpendicular to $\he_r(\theta)\text{.}$
The $y$-component of $\he_\theta(\theta)$ has the same sign as the $x$-component of $\he_r(\theta)\text{.}$ The $x$-component of $\he_\theta(\theta)$ has opposite sign to that of the $y$-component of $\he_r(\theta)\text{.}$

Here is a sketch of $(x_i,y_i)\text{,}$ $\he_r(\theta_i)\text{,}$ $\he_\theta(\theta_i)$ for $i =1,3,5$ (the points on the axes)

and here is a sketch (to a different scale) of $(x_i,y_i)\text{,}$ $\he_r(\theta_i)\text{,}$ $\he_\theta(\theta_i)$ for $i =2,4$ (the points off the axes).

3.2.5.4.

Solution.

Here is a sketch of $\llt a, b\rgt$ and $\llt A, B\rgt\text{.}$

(a) From the sketch,

\begin{align*} a&=r\cos\theta\\ b&=r\sin\theta \end{align*}

(b) The length of the vector $\llt A, B\rgt$ is again $r$ and the angle between $\llt A, B\rgt$ and the $x$-axis is $\theta +\varphi\text{.}$ So

\begin{alignat*}{3} A&=r\cos(\theta+\varphi) &&=r\cos\theta\cos\varphi-r\sin\theta\sin\varphi &&=a\cos\varphi-b\sin\varphi\\ B&=r\sin(\theta+\varphi) &&=r\sin\theta\cos\varphi+r\cos\theta\sin\varphi &&=b\cos\varphi+a\sin\varphi \end{alignat*}

3.2.5.5.

Solution.

(a) The region

\begin{equation*} \cR=\Set{(x,y)}{0\le x^2+y^2\le 4,\ 0\le y\le x} \end{equation*}

In polar coordinates,

the circle $x^2+y^2=4$ becomes $r^2=4$ or $r=2$ and
the line $y=x$ becomes $r\sin\theta=r\cos\theta$ or $\tan\theta=1$ or $\theta=\frac{\pi}{4}\text{.}$

Thus the domain of integration is

\begin{equation*} \cR=\Set{(r\cos\theta,r\sin\theta)}{0\le r\le 2,\ 0\le\theta\le\tfrac{\pi}{4}} \end{equation*}

On this domain,

$\theta$ runs from $0$ to $\frac{\pi}{4}\text{.}$
For each fixed $\theta$ in that range, $r$ runs from $0$ to $2\text{,}$ as in the figure on the left below.

In polar coordinates $\dee{x}\,\dee{y} = r\,\dee{r}\,\dee{\theta}\text{,}$ so that

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\frac{\pi}{4}}\dee{\theta} \int_0^2\dee{r}\ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

Alternatively, on $\cR\text{,}$

$r$ runs from $0$ to $2\text{.}$
For each fixed $r$ in that range, $\theta$ runs from $0$ to $\frac{\pi}{4}\text{,}$ as in the figure on the right above.

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^2\dee{r} \int_0^{\frac{\pi}{4}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

(b) The region

\begin{equation*} \cR=\Set{(x,y)}{1\le x^2+y^2\le 4,\ x\ge 0,\ y\ge 0} \end{equation*}

In polar coordinates,

the circle $x^2+y^2=1$ becomes $r^2=1$ or $r=1$ and
the circle $x^2+y^2=4$ becomes $r^2=4$ or $r=2$ and
the positive $x$-axis, $x\ge0\text{,}$ $y=0\text{,}$ becomes $\theta=0$ and
the positive $y$-axis, $x=0\text{,}$ $y\ge0\text{,}$ becomes $\theta=\frac{\pi}{2}\text{.}$

Thus the domain of integration is

\begin{equation*} \cR=\Set{(r\cos\theta,r\sin\theta)}{1\le r\le 2,\ 0\le\theta\le\tfrac{\pi}{2}} \end{equation*}

On this domain,

$\theta$ runs from $0$ to $\frac{\pi}{2}\text{.}$
For each fixed $\theta$ in that range, $r$ runs from $1$ to $2\text{,}$ as in the figure on the left below.

In polar coordinates $\dee{x}\,\dee{y} = r\,\dee{r}\,\dee{\theta}\text{,}$ so that

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\frac{\pi}{2}}\dee{\theta} \int_1^2\dee{r}\ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

Alternatively, on $\cR\text{,}$

$r$ runs from $1$ to $2\text{.}$
For each fixed $r$ in that range, $\theta$ runs from $0$ to $\frac{\pi}{2}\text{,}$ as in the figure on the right above.

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_1^2\dee{r} \int_0^{\frac{\pi}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

\begin{equation*} \cR=\Set{(x,y)}{(x-1)^2+y^2\le 1,\ y \ge 0} \end{equation*}

In polar coordinates, the circle $(x-1)^2+y^2= 1\text{,}$ or $x^2-2x+y^2=0\text{,}$ is $r^2-2r\cos\theta=0$ or $r=2\cos\theta\text{.}$ Note that, on $r=2\cos\theta\text{,}$

when $\theta=0\text{,}$ $r=2$ and
as $\theta$ increases from $0$ towards $\frac{\pi}{2}\text{,}$ $r$ decreases but remains strictly bigger than $0$ (look at the figure below), until
when $\theta=\frac{\pi}{2}\text{,}$ $r=0\text{.}$

Thus the domain of integration is

\begin{equation*} \cR=\Set{(r\cos\theta,r\sin\theta)}{0\le\theta\le\tfrac{\pi}{2},\ 0\le r\le 2\cos\theta} \end{equation*}

On this domain,

$\theta$ runs from $0$ to $\frac{\pi}{2}\text{.}$
For each fixed $\theta$ in that range, $r$ runs from $0$ to $2\cos\theta\text{,}$ as in the figure on the left below.

In polar coordinates $\dee{x}\,\dee{y} = r\,\dee{r}\,\dee{\theta}\text{,}$ so that

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^{\frac{\pi}{2}}\dee{\theta} \int_0^{2\cos\theta}\dee{r}\ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

Alternatively, on $\cR\text{,}$

$r$ runs from $0$ (at the point $(0,0)$) to $2$ (at the point $(2,0)$).
For each fixed $r$ in that range, $\theta$ runs from $0$ to $\arccos\frac{r}{2}$ (which was gotten by solving $r=2\cos\theta$ for $\theta$ as a function of $r$), as in the figure on the right above.

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_0^2\dee{r} \int_0^{\arccos\frac{r}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

(d) The region

\begin{equation*} \cR=\Set{(x,y)}{0\le y\le 2,\ 0\le x\le y} \end{equation*}

In polar coordinates,

the line $y=2$ becomes $r\sin\theta=2$ and
the positive $y$-axis, $x=0\text{,}$ $y\ge0\text{,}$ becomes $\theta=\frac{\pi}{2}$ and
the line $y=x$ becomes $r\sin\theta=r\cos\theta$ or $\tan\theta=1$ or $\theta=\frac{\pi}{4}\text{.}$

Thus the domain of integration is

\begin{equation*} \cR=\Set{(r\cos\theta,r\sin\theta)}{\tfrac{\pi}{4}\le\theta\le\tfrac{\pi}{2},\ 0\le r\sin\theta\le 2} \end{equation*}

On this domain,

$\theta$ runs from $\frac{\pi}{4}$ to $\frac{\pi}{2}\text{.}$
For each fixed $\theta$ in that range, $r$ runs from $0$ to $\frac{2}{\sin\theta}\text{,}$ as in the first figure below.

In polar coordinates $\dee{x}\,\dee{y} = r\,\dee{r}\,\dee{\theta}\text{,}$ so that

\begin{equation*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} =\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dee{\theta} \int_0^{\frac{2}{\sin\theta}}\dee{r}\ r\ f(r\cos\theta,r\sin\theta) \end{equation*}

Alternatively, on $\cR\text{,}$

$r$ runs from $0$ (at the point $(0,0)$) to $2\sqrt{2}$ (at the point $(2,2)$).
For each fixed $r$ between $0$ and $2\text{,}$ $\theta$ runs from $\frac{\pi}{4}$ to $\frac{\pi}{2}\text{,}$ as in the second figure above.
For each fixed $r$ between $2$ and $2\sqrt{2}\text{,}$ $\theta$ runs from $\frac{\pi}{4}$ to $\arcsin\frac{2}{r}$ (which was gotten by solving $r\sin\theta=2$ for $\theta$ as a function of $r$), as in the third figure above.

\begin{align*} \dblInt_\cR f(x,y)\,\dee{x}\,\dee{y} &=\int_0^2\dee{r} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta)\\ &\hskip1in+\int_2^{2\sqrt{2}}\dee{r} \int_{\frac{\pi}{4}}^{\arcsin\frac{2}{r}}\dee{\theta} \ r\ f(r\cos\theta,r\sin\theta) \end{align*}

3.2.5.6.

Solution.

(a) Let $D$ denote the domain of integration. The symbols $\displaystyle \int_1^2\dee{r} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dee{\theta}$ say that, on $D\text{,}$

$r$ runs from $1$ to $2$ and
for each $r$ in that range, $\theta$ runs from $-\frac{\pi}{4}$ to $\frac{\pi}{4}\text{.}$

In Cartesian coordinates

$r=1$ is the circle $x^2+y^2=1$ and
$r=2$ is the circle $x^2+y^2=4$ and
$\theta=\frac{\pi}{4}$ is the ray $y=x\text{,}$ $x\ge 0$ and
$\theta=-\frac{\pi}{4}$ is the ray $y=-x\text{,}$ $x\ge 0\text{.}$

\begin{gather*} D=\Set{(x,y)}{1\le x^2+y^2\le 4,\ -x\le y\le x,\ x\ge 0} \end{gather*}

Here are two sketches. $D$ is the shaded region in the sketch on the right.

(b) Let $D$ denote the domain of integration. The symbols $\int_0^{\frac{\pi}{4}}\dee{\theta} \int_0^{\frac{2}{\sin\theta+\cos\theta}}\dee{r}$ say that, on $D\text{,}$

$\theta$ runs from $0$ to $\frac{\pi}{4}$ and
for each $\theta$ in that range, $r$ runs from $0$ to $\frac{2}{\sin\theta+\cos\theta}\text{.}$

In Cartesian coordinates

$\theta=0$ is the positive $x$-axis and
$\theta=\frac{\pi}{4}$ is the ray $y=x\text{,}$ $x\ge 0$ and
$r=\frac{2}{\sin\theta+\cos\theta}\text{,}$ or equivalently $r\cos\theta+r\sin\theta=2\text{,}$ is the line $x+y=2\text{.}$

Looking at the sketch on the left below, we see that, since the lines $y=x$ and $x+y=2$ cross at $(1,1)\text{,}$

\begin{gather*} D=\Set{(x,y)}{0\le y\le 1,\ y\le x\le 2-y} \end{gather*}

$D$ is the shaded region in the sketch on the right.

(c) Let $D$ denote the domain of integration. The symbols $\int_0^{2\pi}\dee{\theta} \int_0^{\frac{3}{\sqrt{\cos^2\theta+9\sin^2\theta}}}\dee{r}$ say that, on $D\text{,}$

$\theta$ runs all the way from $0$ to $2\pi$ and
for each $\theta\text{,}$ $r$ runs from $0$ to $\frac{3}{\sqrt{\cos^2\theta+9\sin^2\theta}}\text{.}$

In Cartesian coordinates

$r=\frac{3}{\sqrt{\cos^2\theta+9\sin^2\theta}}\text{,}$ or equivalently $r^2\cos^2\theta+9r^2\sin^2\theta=9\text{,}$ is the ellipse $x^2+9y^2=9\text{.}$

So $D$ is the interior of the ellipse $x^2+9y^2=9$ and $D$ is the shaded region in the lower sketch.

3.2.5.7.

Solution.

(a) In polar coordinates, the domain of integration, $x^2+y^2\le a^2\text{,}$ $0\le y\le \sqrt{3}x\text{,}$ becomes

\begin{equation*} 0\le r\le a,\ 0\le r\sin\theta\le\sqrt{3}r\cos\theta\quad\text{ or }\quad 0\le r\le a,\ 0\le\theta\le \arctan\sqrt{3}=\frac{\pi}{3} \end{equation*}

The integral is

\begin{align*} \dblInt_S (x+y)\dee{x}\,\dee{y} &=\int_0^a \dee{r}\int_0^{{\pi\over 3}}\dee{\theta}\ r\,(r\cos\theta+r\sin\theta)\\ &=\int_0^a \dee{r}\ r^2\Big[\sin\theta-\cos\theta\Big]_0^{{\pi\over 3}} =\frac{a^3}{3}\left[\frac{\sqrt{3}}{2}-\frac{1}{2}+1\right]\\ &=\frac{a^3}{6}\big[\sqrt{3}+1\big] \end{align*}

(b) In polar coordinates, the domain of integration, $x^2+y^2\le 2\text{,}$ $x\ge 1\text{,}$

becomes

\begin{equation*} r\le \sqrt{2},\ r\cos\theta\ge 1\qquad\text{ or }\qquad \frac{1}{\cos\theta}\le r\le\sqrt{2} \end{equation*}

For $\frac{1}{\cos\theta}\le r\le\sqrt{2}$ to be nonempty, we need $\cos\theta\le \frac{1}{\sqrt{2}}$ or $|\theta|\le\frac{\pi}{4}\text{.}$ By symmetry under $y\rightarrow -y\text{,}$ the integral is

\begin{align*} \dblInt_Sx\ \dee{x}\,\dee{y} &=2\int_0^{{\pi\over 4}}\dee{\theta}\int_{1\over\cos\theta}^{\sqrt{2}} \dee{r}\ r\,(r\cos\theta)\\ &=2\int_0^{{\pi\over 4}}\dee{\theta}\ \cos\theta\ \frac{r^3}{3}\bigg|_{1\over\cos\theta}^{\sqrt{2}} =\frac{2}{3}\int_0^{{\pi\over 4}}\dee{\theta}\ \Big[2^{3/2}\cos\theta-\frac{1}{\cos^2\theta}\Big]\\ &=\frac{2}{3} \Big[2^{3/2}\sin\theta-\tan\theta\Big]_0^{{\pi\over 4}} =\frac{2}{3} \big[2^{3/2}\frac{1}{\sqrt{2}}-1\big] =\frac{2}{3} \end{align*}

(c) In polar coordinates, the triangle with vertices $(0,0), (1,0)$ and $(1,1)$ has sides $\theta=0\text{,}$ $\theta=\frac{\pi}{4}$ and $r=\frac{1}{\cos\theta}$ (which is the polar coordinates version of $x=1$). The integral is

\begin{align*} \dblInt_T (x^2+y^2)\ \dee{x}\,\dee{y} &=\int_0^{{\pi\over 4}}\dee{\theta}\int_0^{1\over\cos\theta} \dee{r}\ r(r^2)\\ &=\int_0^{{\pi\over 4}}\dee{\theta}\ \frac{r^4}{4}\bigg|_0^{1\over\cos\theta} =\frac{1}{4}\int_0^{{\pi\over 4}}\dee{\theta}\ \frac{1}{\cos^4\theta}\\ &=\frac{1}{4}\int_0^{{\pi\over 4}}\dee{\theta}\ \sec^4\theta\\ &=\frac{1}{4}\int_0^{{\pi\over 4}}\dee{\theta}\ \sec^2\theta\big(1+\tan^2\theta\big)\\ &=\frac{1}{4}\int_0^1d t\ \big(1+t^2\big)\text{ where }t=\tan\theta\\ &=\frac{1}{4}\left[t+\frac{t^3}{3}\right]_0^1 =\frac{1}{4}\ \frac{4}{3} =\frac{1}{3} \end{align*}

(d) In polar coordinates, the domain of integration, $x^2+y^2\le 1\text{,}$ becomes $0\le r\le 1\text{,}$ $0\le\theta\le 2\pi\text{.}$ So

\begin{align*} \dblInt_{x^2+y^2\le 1}\hskip-5pt \ln(x^2+y^2)\,\dee{x}\,\dee{y} &=\int_0^{2\pi}\hskip-3pt \dee{\theta}\int_0^1\dee{r}\ r\ln r^2 =2\pi\int_0^1\dee{r}\ r\ln r^2\\ &=\pi\int_0^1 \dee{s}\ \ln s\text{ where } s=r^2\\ &=\pi\Big[s\ln s-s\Big]_0^1 =-\pi \end{align*}

To be picky, $\ln s$ tends to $-\infty$ as $s$ tends to $0\text{.}$ So $\int_0^1\dee{s}\,\ln s$ is an improper integral. The careful way to evaluate it is

\begin{align*} \int_0^1 \dee{s}\ \ln s &=\lim_{\veps\to 0^+}\int_\veps^1 \dee{s}\ \ln s =\lim_{\veps\to 0^+}\Big[s\ln s-s\Big]_\veps^1\\ &=\lim_{\veps\to 0^+}\Big[-1-\veps\ln\veps+\veps\Big] =-1 \end{align*}

That $\lim\limits_{\veps\to 0^+}\veps\ln\veps = 0$ was shown in Example 3.7.15 of the CLP-1 text.

3.2.5.8.

Solution.

The top surface $x^2+y^2+z^2=2$ meets the bottom surface $z=x^2+y^2$ when $z$ obeys $x^2+y^2=z=2-z^2\text{.}$ That is, when $0=z^2+z-2=(z-1)(z+2)\text{.}$ The root $z=-2$ is inconsistent with $z=x^2+y^2\ge 0\text{.}$ So the top and bottom surfaces meet at the circle $z=1\text{,}$ $x^2+y^2=1\text{.}$

In polar coordinates, the top surface is $z^2=2-r^2\text{,}$ or equivalently $z=\sqrt{2-r^2}\text{,}$ and the bottom surface is $z=r^2\text{.}$ So the height of the volume above the point with polar coordinates $(r,\theta)$ is $\sqrt{2-r^2}-r^2$ and

\begin{align*} \text{Volume}&=\int_0^1\dee{r}\int_0^{2\pi} \dee{\theta}\ r\,\big[\sqrt{2-r^2}-r^2\big] =2\pi \int_0^1\dee{r}\ r\,\big[\sqrt{2-r^2}-r^2\big]\\ &=2\pi \left[-\frac{1}{3}{(2-r^2)}^{3/2}-\frac{r^4}{4}\right]_0^1 =2\pi \left[-\frac{1}{3}-\frac{1}{4}+\frac{1}{3}2^{3/2}\right]\\ &=\pi \left[\frac{4}{3}\sqrt{2}-\frac{7}{6}\right] \approx 2.26 \end{align*}

In Cartesian coordinates

\begin{align*} \text{Volume}&=4\int_0^1\dee{x}\int_0^{\sqrt{1-x^2}} \dee{y}\ \big[\sqrt{2-x^2-y^2}-x^2-y^2\big] \end{align*}

The $y$ integral can be done using the substitution $y=\sqrt{2-x^2}\cos t\text{,}$ but it is easier to use polar coordinates.

3.2.5.9.

Solution.

For this region $x$ and $y$ run over the interior of the cylinder $x^2+(y-a)^2=a^2\text{.}$ For each $(x,y)$ inside the cylinder, $z$ runs from $-\sqrt{x^2+y^2}$ to $\sqrt{x^2+y^2}\text{.}$ As $x^2+(y-a)^2=a^2$ if and only if $x^2+y^2-2ay=0\text{,}$ the cylinder has equation $r^2=2ar\sin\theta\text{,}$ or equivalently, $r=2a\sin\theta\text{,}$ in polar coordinates.

Thus $(r,\theta)$ runs over $0\le\theta\le\pi,\ 0\le r\le 2a\sin\theta$ and for each $(r,\theta)$ in this region $z$ runs from $-r$ to $r\text{.}$ By symmetry under $x\rightarrow -x\text{,}$ the volume is

\begin{align*} \text{Volume} &=2\int_0^{{\pi\over 2}}\dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r\big[r-(-r)\big] =4\int_0^{{\pi\over 2}}\dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r^2\\ &=\frac{4}{3}\int_0^{{\pi\over 2}}\!\!\dee{\theta}\ (2a\sin\theta)^3\\ &=\frac{32}{3}a^3\int_0^{{\pi\over 2}}\!\!\dee{\theta}\ \sin\theta(1-\cos^2\theta)\\ &=-\frac{32}{3}a^3\int_1^0 dt\ (1-t^2)\text{ where } t=\cos\theta\\ &=-\frac{32}{3}a^3\left[t-\frac{t^3}{3}\right]_1^0 =\frac{64}{9}a^3 \end{align*}

3.2.5.10.

Solution.

The figure below shows the top view of the specified solid. $(x,y)$ runs over the interior of the circle $x^2+y^2=2ax\text{.}$ For each fixed $(x,y)$ in this disk, $z$ runs from $-\sqrt{2ax}$ to $+\sqrt{2ax}\text{.}$ In polar coordinates, the circle is $r^2=2ar\cos\theta$ or $r=2a\cos\theta\text{.}$

The solid is symmetric under $y\rightarrow -y$ and $z\rightarrow -z\text{,}$ so we can restrict to $y\ge 0\text{,}$ $z\ge 0$ and multiply by 4. The volume is

\begin{align*} \text{Volume} &=4\int_0^{\pi\over 2} \dee{\theta}\int_0^{2a\cos\theta} \dee{r}\ r\sqrt{2ar\cos\theta}\\ &=4\int_0^{\pi\over 2} \dee{\theta}\ \sqrt{2a\cos\theta}\ \frac{2}{5}\ r^{5/2}\bigg|_0^{2a\cos\theta}\\ &=\frac{8}{5}\int_0^{\pi\over 2} \dee{\theta}\ \big(2a\cos\theta\big)^3 =\frac{64}{5}a^3\int_0^{\pi\over 2} \dee{\theta}\ \cos\theta\big(1-\sin^2\theta\big)\\ &=\frac{64}{5}a^3\int_0^1 dt\ \big(1-t^2\big) \qquad\text{ where } t=\sin\theta\\ &=\frac{64}{5}a^3\left[t-\frac{t^3}{3}\right]_0^1 =\frac{128}{15}a^3 \end{align*}

3.2.5.11. (✳).

Solution.

(a)

The equation $x^2 + y^2 \le 2y$ is equivalent to the equation $x^2 + (y-1)^2=1\text{,}$ which is the equation of the cylinder whose $z=z_0$ cross--section is the horizontal circle of radius $1\text{,}$ centred on $x=0\text{,}$ $y=1\text{,}$ $z=z_0\text{.}$ The part of this cylinder in the first octant is sketched in the first figure below.
$z \le \sqrt{x^2 + y^2}$ is the equation of the cone with vertex $(0,0,0)\text{,}$ and axis the positive $z$--axis, whose radius at height $z=2$ is $2\text{.}$ The part of this cone in the first octant is sketched in the second figure below.

The region $E$ is the part of the cylinder that is above the $xy$--plane (since $z\ge 0$) outside the cone (since $z\le\sqrt{x^2+y^2}$). The part of $E$ that is in the first octant is outlined in red in the figure below. Both $x^2 + y^2 \le 2y$ and $0 \le z \le \sqrt{x^2 + y^2}$ are invariant under $x\rightarrow -x\text{.}$ So $E$ is also invariant under $x\rightarrow -x\text{.}$ That is, $E$ is symmetric about the $yz$--plane and contains, in the octant $x\le 0\text{,}$ $y\ge 0\text{,}$ $z\ge 0\text{,}$ a mirror image of the first octant part of $E\text{.}$

(b) In polar coordinates, $x^2 + y^2 \le 2y$ becomes

\begin{equation*} r^2\le 2r\sin\theta \iff r\le 2\sin\theta \end{equation*}

Let us denote by $D$ the base region of the part of $E$ in the first octant (i.e. the shaded region in the figure above). Think of $D$ as being part of the $xy$--plane. In polar coordinates, on $D$

$\theta$ runs from $0$ to $\frac{\pi}{2}\text{.}$ (Recall that $D$ is contained in the first quadrant.)
For each $\theta$ in that range, $r$ runs from $0$ to $2\sin\theta\text{.}$

Because

in polar coordinates $\dee{A} = r\,\dee{r}\,\dee{\theta}\text{,}$ and
the height of $E$ above each point $(x,y)$ in $D$ is $\sqrt{x^2+y^2}\text{,}$ or, in polar coordinates, $r\text{,}$ and
the volume of $E$ is twice the volume of the part of $E$ in the first octant,

we have

\begin{align*} \text{Volume}(E) &= 2\int_0^{\frac{\pi}{2}}\dee{\theta} \int_0^{2\sin\theta}\dee{r}\ r^2\\ &=\frac{16}{3} \int_0^{\frac{\pi}{2}}\dee{\theta}\ \sin^3\theta =\frac{16}{3} \int_0^{\frac{\pi}{2}}\dee{\theta}\ \sin\theta\ \big(1-\cos^2\theta\big)\\ &=-\frac{16}{3} \int_1^0\dee{u}\ \big(1-u^2\big)\qquad \text{with }u=\cos\theta,\ \dee{u}=-\sin\theta\ \dee{\theta}\\ &=\frac{16}{3}\left[1-\frac{1}{3}\right]\\ &=\frac{32}{9} \end{align*}

3.2.5.12. (✳).

Solution.

On the domain of integration

$x$ runs for $0$ to $2\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $0$ to $\sqrt{4-x^2}\text{.}$ The equation $y=\sqrt{4-x^2}$ is equivalent to $x^2+y^2=4\text{,}$ $y\ge 0\text{.}$

This domain is sketched in the figure on the left below.

Considering that

the integrand, ${(x^2+y^2)}^{\frac{3}{2}}\text{,}$ is invariant under rotations about the origin and
the outer curve, $x^2+y^2=4\text{,}$ is invariant under rotations about the origin

we’ll use polar coordinates. In polar coordinates,

the outer curve, $x^2+y^2=4\text{,}$ is $r=2\text{,}$ and
the integrand, ${(x^2+y^2)}^{\frac{3}{2}}$ is $r^3\text{,}$ and
$\displaystyle \dee{A}=r\,\dee{r}\,\dee{\theta}$

Looking at the figure on the right above, we see that the given integral is, in polar coordinates,

\begin{align*} \int_0^{\pi/2}\dee{\theta}\int_0^2\dee{r}\ r (r^3) &=\frac{\pi}{2}\ \frac{2^5}{5} =\frac{16\pi}{5} \end{align*}

3.2.5.13. (✳).

Solution.

(a) The region $\cL$ is sketched in the figure on the left below.

(b) In polar coordinates

the circle $x^2+y^2=2$ is $r^2=2$ or $r=\sqrt{2}\text{,}$ and
the circle $x^2+y^2=4$ is $r^2=4$ or $r=2\text{,}$ and
the line $y=x$ is $r\sin\theta = r\cos\theta\text{,}$ or $\tan\theta =1 \text{,}$ or (for the part in the first quadrant) $\theta=\frac{\pi}{4}\text{,}$ and
the positive $x$--axis ($y=0\text{,}$ $x\ge 0$) is $\theta=0$

Looking at the figure on the right above, we see that, in $\cL\text{,}$

$\theta$ runs from $0$ to $\frac{\pi}{4}\text{,}$ and
for each fixed $\theta$ in that range, $r$ runs from $\sqrt{2}$ to $2\text{.}$
$\dee{A}$ is $r\,\dee{r}\,\dee{\theta}$

\begin{gather*} M = \int_0^{\pi/4}\dee{\theta}\int_{\sqrt{2}}^2\dee{r}\ r\, \rho(r\cos\theta\,,\,r\sin\theta) \end{gather*}

\begin{gather*} \rho = \frac{2xy}{x^2+y^2} = \frac{2r^2\cos\theta\,\sin\theta}{r^2} =\sin(2\theta) \end{gather*}

we have

\begin{align*} M &= \int_0^{\pi/4}\dee{\theta}\int_{\sqrt{2}}^2\dee{r}\ r\, \sin(2\theta)\\ &=\left[\int_0^{\pi/4}\sin(2\theta)\ \dee{\theta}\right] \left[\int_{\sqrt{2}}^2 r\ \dee{r}\right]\\ &=\left[-\frac{1}{2}\cos(2\theta)\right]_0^{\pi/4} \left[ \frac{r^2}{2}\right]_{\sqrt{2}}^2 =\frac{1}{2}\ \frac{4-2}{2}\\ &=\frac{1}{2} \end{align*}

3.2.5.14. (✳).

Solution.

We’ll use polar coordinates. The domain of integration is

\begin{equation*} \bbbr^2=\Set{(r\cos\theta\,,\,r\sin\theta)}{0\le r \lt \infty,\ 0\le\theta\le 2\pi} \end{equation*}

The given integral is improper, so we’ll start by integrating $r$ from $0$ to an arbitrary $R \gt 0\text{,}$ and then we’ll take the limit $R\rightarrow\infty\text{.}$ In polar coordinates, the integrand $\frac{1}{{(1+x^2+y^2)}^2}=\frac{1}{{(1+r^2)}^2}\text{,}$ and $\dee{A}=r\,\dee{r}\,\dee{\theta}\text{,}$ so

\begin{align*} \dblInt_{\bbbr^2} \frac{1}{{(1+x^2+y^2)}^2}\ \dee{A} &=\lim_{R\rightarrow\infty} \int_0^{2\pi}\dee{\theta}\int_0^R\dee{r} \frac{r}{{(1+r^2)}^2}\\ &=\lim_{R\rightarrow\infty} \int_0^{2\pi}\dee{\theta}\ \left[-\frac{1}{2(1+r^2)}\right]_0^R\\ &=\lim_{R\rightarrow\infty} 2\pi \left[\frac{1}{2}-\frac{1}{2(1+R^2)}\right]\\ &=\pi \end{align*}

3.2.5.15. (✳).

Solution.

Let’s switch to polar coordinates. In polar coordinates, the circle $x^2+y^2=2$ is $r=\sqrt{2}$ and the line $y=x$ is $\theta=\frac{\pi}{4}\text{.}$

In polar coordinates $\dee{A} = r\,\dee{r}\,\dee{\theta}\text{,}$ so the integral

\begin{align*} \dblInt_D y\sqrt{x^2+y^2}\,\dee{A} &=\int_0^{\pi/4}\dee{\theta}\int_0^{\sqrt{2}}\dee{r}\ r\ \overbrace{r\sin\theta}^{y}\ \overbrace{r}^{\sqrt{x^2+y^2}}\\ &=\int_0^{\pi/4}\dee{\theta}\ \sin\theta\ \left[\frac{r^4}{4}\right]_0^{\sqrt{2}}\\ &=\Big[-\cos\theta\Big]_0^{\pi/4}\\ &= 1-\frac{1}{\sqrt{2}} \end{align*}

3.2.5.16. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $0$ to $1\text{.}$ In inequalities, $0\le y\le 1\text{.}$
For each fixed $y$ in that range, $x$ runs from $\sqrt{3}\,y$ to $\sqrt{4-y^2}\text{.}$ In inequalities, that is $\sqrt{3}\,y\le x\le \sqrt{4-y^2}\text{.}$ Note that the inequalities $x\le \sqrt{4-y^2}\text{,}$ $x\ge 0$ are equivalent to $x^2+y^2\le 4\text{,}$ $x\ge 0\text{.}$

Note that the line $x=\sqrt{3}\, y$ and the circle $x^2+y^2\le 4$ intersect when $3y^2+y^2=4\text{,}$ i.e. $y=\pm 1\text{.}$ Here is a sketch.

(b) In polar coordinates, the circle $x^2+y^2= 4$ is $r=2$ and the line $x=\sqrt{3}\, y\text{,}$ i.e. $\frac{y}{x} =\frac{1}{\sqrt{3}}\text{,}$ is $\tan\theta=\frac{1}{\sqrt{3}}$ or $\theta=\frac{\pi}{6}\text{.}$ As $\dee{x}\,\dee{y} = r\,\dee{r}\,\dee{\theta}\text{,}$ the domain of integration is

\begin{gather*} \Set{(r\cos\theta,r\sin\theta)}{0\le \theta\le\frac{\pi}{6},\ 0\le r\le 2} \end{gather*}

and

\begin{align*} \int_0^1 \int_{\sqrt{3}y}^{\sqrt{4-y^2}} \ln\big(1+x^2+y^2)\ \dee{x}\,\dee{y} &= \int_0^2\dee{r}\int_0^{\pi/6}\dee{\theta}\ r\,\ln(1+r^2)\\ &= \frac{\pi}{6}\int_0^2\dee{r}\ r\,\ln(1+r^2)\\ &=\frac{\pi}{12}\int_1^5 \dee{u} \ln(u)\\ &\hskip0.5in\text{with }u=1+r^2,\ \dee{u}=2r\,\dee{r}\\ &=\frac{\pi}{12}\Big[u\ln(u)-u\Big]_1^5\\ &=\frac{\pi}{12}\big[5\ln(5)-4\big] \end{align*}

3.2.5.17. (✳).

Solution.

Here is a sketch of $D\text{.}$

We’ll use polar coordinates. In polar coordinates the circle $x^2+y^2=16$ is $r=4$ and the line $x=2$ is $r\cos\theta =2\text{.}$ So

\begin{gather*} D = \left\{(r\cos\theta\,,\,r\sin\theta)\ \left|\ -\frac{\pi}{3}\le\theta\le\frac{\pi}{3},\ \frac{2}{\cos\theta}\le r\le 4\ \right.\right\} \end{gather*}

and, as $\dee{A} = r\,\dee{r}\,\dee{\theta}\text{,}$ the specified integral is

\begin{align*} \dblInt_D\big(x^2+y^2\big)^{-3/2}\ \dee{A} &= \int_{-\pi/3}^{\pi/3} \dee{\theta}\int_{2/\cos\theta}^4\dee{r}\ r\frac{1}{r^3}\\ &= \int_{-\pi/3}^{\pi/3} \dee{\theta}\ \left[-\frac{1}{r}\right]_{2/\cos\theta}^4\\ &= \int_{-\pi/3}^{\pi/3} \dee{\theta}\ \left[\frac{\cos\theta}{2}-\frac{1}{4}\right]\\ &= \left[\frac{\sin\theta}{2}-\frac{\theta}{4}\right]_{-\pi/3}^{\pi/3}\\ &=\frac{\sqrt{3}}{2}-\frac{\pi}{6} \end{align*}

3.2.5.18. (✳).

Solution.

(a) The inequality $x^2 + y^2 \le 2x$ is equivalent to $(x-1)^2 + y^2 \le 1$ and says that $(x,y)$ is to be inside the disk of radius $1$ centred on $(1,0)\text{.}$ Here is a sketch.

In polar coordinates, $x=r\cos\theta\text{,}$ $y=r\sin\theta$ so that the line $y=x$ is $\theta=\frac{\pi}{4}$ and the circle $x^2+y^2=2x$ is

\begin{equation*} r^2=2r\cos\theta \qquad\text{or}\qquad r=2\cos\theta \end{equation*}

Consequently

\begin{equation*} D = \Set{(r\cos\theta\,,\,r\sin\theta)} {-\frac{\pi}{2}\le\theta\le\frac{\pi}{4},\ 0\le r\le 2\cos\theta} \end{equation*}

(b) The solid has height $z=r$ above the point in $D$ with polar coordinates $r\text{,}$ $\theta\text{.}$ So the

\begin{align*} \text{Volume} &= \dblInt_D r\ \dee{A}=\dblInt_D r^2\ \dee{r}\, \dee{\theta} =\int_{-\pi/2}^{\pi/4}\dee{\theta}\int_0^{2\cos\theta} \dee{r}\ r^2\\ &= \frac{8}{3}\int_{-\pi/2}^{\pi/4}\dee{\theta}\ \cos^3\theta = \frac{8}{3}\int_{-\pi/2}^{\pi/4}\dee{\theta}\ \cos\theta\big[1-\sin^2\theta\big]\\ &=\frac{8}{3} \left[\sin\theta -\frac{\sin^3\theta}{3}\right]_{-\pi/2}^{\pi/4}\\ &=\frac{8}{3}\left[\left(\frac{1}{\sqrt{2}} -\frac{1}{6\sqrt{2}}\right) -\left(-1 + \frac{1}{3}\right) \right]\\ &=\frac{40}{18\sqrt{2}} +\frac{16}{9} \end{align*}

3.2.5.19. (✳).

Solution.

We’ll use polar coordinates. In $D$

$\theta$ runs from $0$ to $\frac{\pi}{2}$ and
for each fixed $\theta$ between $0$ and $\frac{\pi}{2}\text{,}$ $r$ runs from $1$ to $1+\cos(\theta)\text{.}$

So the area of $D$ is

\begin{align*} \text{area}&=A=\int_0^{\pi/2} \dee{\theta}\int_1^{1+\cos\theta} \dee{r}\ r =\int_0^{\pi/2} \dee{\theta}\ \frac{1}{2} r^2\bigg|_1^{1+\cos\theta}\\ &=\int_0^{\pi/2} \dee{\theta}\ \left[\frac{1}{2} \cos^2\theta+\cos\theta\right] \end{align*}

We are interested in the average value of $r$ on $D\text{,}$ which is

\begin{align*} \text{ave dist}&=\frac{1}{A}\int_0^{\pi/2} \dee{\theta}\int_1^{1+\cos\theta} \dee{r}\ r^2 =\frac{1}{A}\int_0^{\pi/2} \dee{\theta}\ \frac{1}{3} r^3\bigg|_1^{1+\cos\theta}\\ &=\frac{1}{A}\int_0^{\pi/2} \dee{\theta}\ \left[\frac{1}{3} \cos^3\theta +\cos^2\theta+\cos\theta\right] \end{align*}

Now we evaluate the integrals of the various powers of cosine.

\begin{align*} \int_0^{\pi/2}\cos\theta\ \dee{\theta}&=\sin\theta\,\bigg|_0^{\pi/2}=1\\ \int_0^{\pi/2}\cos^2\theta\ \dee{\theta}&=\frac{\cos\theta\sin\theta}{2}\,\bigg|_0^{\pi/2} +\frac{1}{2}\int _0^{\pi/2}\dee{\theta}=\frac{\pi}{4}\\ \int_0^{\pi/2}\cos^3\theta\ \dee{\theta}&=\frac{\cos^2\theta\sin\theta}{3}\,\bigg|_0^{\pi/2} +\frac{2}{3}\int _0^{\pi/2}\cos\theta\ \dee{\theta}=\frac{2}{3} \end{align*}

So $A=\frac{\pi}{8}+1$ and

\begin{gather*} \text{ave dist}=\frac{8}{\pi+8}\left[\frac{2}{9}+\frac{\pi}{4}+1\right] =2\frac{\pi+44/9}{\pi+8}\approx 1.442 \end{gather*}

3.2.5.20. (✳).

Solution.

(a) Observe that

the condition $x^2+y^2\le 1$ restricts $G$ to the interior of the circle of radius $1$ centred on the origin, and
the conditions $0\le x\le 2y$ restricts $G$ to $x\ge 0\text{,}$ $y\ge 0\text{,}$ i.e. to the first quadrant, and
the conditions $x\le 2y$ and $y\le 2x$ restrict $\frac{x}{2}\le y\le 2x\text{.}$ So $G$ lies below the (steep) line $y=2x$ and lies above the (not steep) line $y=\frac{x}{2}\text{.}$

Here is a sketch of $G$

(b) Observe that the line $y=2x$ crosses the circle $x^2+y^2=1$ at a point $(x,y)$ obeying

\begin{gather*} x^2+(2x)^2=x^2+y^2=1 \implies 5x^2=1 \end{gather*}

and that the line $x=2y$ crosses the circle $x^2+y^2=1$ at a point $(x,y)$ obeying

\begin{gather*} (2y)^2+y^2=x^2+y^2=1 \implies 5y^2=1 \end{gather*}

So the intersection point of $y=2x$ and $x^2+y^2=1$ in the first octant is $\left(\frac{1}{\sqrt{5}}\,,\,\frac{2}{\sqrt{5}}\right)$ and the intersection point of $x=2y$ and $x^2+y^2=1$ in the first octant is $\left(\frac{2}{\sqrt{5}}\,,\,\frac{1}{\sqrt{5}}\right)\text{.}$

We’ll set up the iterated integral using horizontal strips as in the sketch

Looking at that sketch, we see that, on $G\text{,}$

$y$ runs from $0$ to $\frac{2}{\sqrt{5}}\text{,}$ and
for each fixed $y$ between $0$ and $\frac{1}{\sqrt{5}}\text{,}$ $x$ runs from $\frac{y}{2}$ to $2y\text{,}$ and
for each fixed $y$ between $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ $x$ runs from $\frac{y}{2}$ to $\sqrt{1-y^2}\text{.}$

\begin{align*} \dblInt_G f(x,y)\ \dee{A} &=\int_0^{\frac{1}{\sqrt{5}}}\dee{y}\int_{y/2}^{2y}\dee{x}\ f(x,y) +\int_{\frac{1}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\dee{y}\int_{y/2}^{\sqrt{1-y^2}} \dee{x}\ f(x,y) \end{align*}

(b) In polar coordinates

the equation $x^2+y^2=1$ becomes $r=1\text{,}$ and
the equation $y=x/2$ becomes $r\sin\theta = \frac{r}{2}\cos\theta$ or $\tan\theta =\frac{1}{2}\text{,}$ and
the equation $y=2x$ becomes $r\sin\theta = 2r\cos\theta$ or $\tan\theta =2\text{.}$

Looking at the sketch

we see that, on $G\text{,}$

$\theta$ runs from $\arctan\frac{1}{2}$ to $\arctan 2\text{,}$ and
for each fixed $\theta$ in that range, $r$ runs from $0$ to $1\text{.}$

As $\dee{A}=r\,\dee{r}\,\dee{\theta}\text{,}$ and $x=r\cos\theta\text{,}$ $y=r\sin\theta\text{,}$

\begin{align*} \dblInt_G f(x,y)\ \dee{A} &=\int_{\arctan\frac{1}{2}}^{\arctan 2}\dee{\theta} \int_0^1\dee{r}\ r\,f(r\cos\theta,r\sin\theta) \end{align*}

3.2.5.21. (✳).

Solution.

(a) On the domain of integration

$y$ runs from $0$ to $\sqrt{2}$ and
for each $y$ in that range, $x$ runs from $y$ to $\sqrt{4-y^2}\text{.}$ We can rewrite $x=\sqrt{4-y^2}$ in the more familiar form $x^2+y^2=4\text{,}$ $x\ge 0\text{.}$

The figure on the left below provides a sketch of the domain of integration. It also shows the generic horizontal slice that was used to set up the given iterated integral.

(b) To reverse the order of integration observe, we use vertical, rather than horizontal slices. From the figure on the right above that, on the domain of integration,

$x$ runs from $0$ to $2$ and
for each $x$ in the range $0\le x\le\sqrt{2}\text{,}$ $y$ runs from $0$ to $x\text{.}$
for each $x$ in the range $\sqrt{2}\le x\le 2\text{,}$ $y$ runs from $0$ to $\sqrt{4-x^2}\text{.}$

So the integral

\begin{equation*} J = \int_0^{\sqrt{2}}\int_0^x \frac{y}{x} e^{x^2+y^2}\ \dee{y}\,\dee{x} + \int_{\sqrt{2}}^2\int_0^{\sqrt{4-x^2}} \frac{y}{x} e^{x^2+y^2}\ \dee{y}\,\dee{x} \end{equation*}

(c) In polar coordinates, the line $y=x$ is $\theta=\frac{\pi}{4}\text{,}$ the circle $x^2+y^2=4$ is $r=2\text{,}$ and $\dee{x}\,\dee{y}=r\,\dee{r}\,\dee{\theta}\text{.}$ So

\begin{align*} J&=\int_0^{\pi/4}\dee{\theta}\int_0^2\dee{r}\ r \overbrace{\frac{r\sin\theta}{r\cos\theta}}^{\frac{y}{x}}e^{r^2}\\ &=\int_0^{\pi/4}\dee{\theta}\ \frac{\sin\theta}{\cos\theta} \left[\frac{1}{2} e^{r^2}\right]_0^2\\ &=-\frac{1}{2}\left[e^4-1\right] \int_1^{1/\sqrt{2}}\dee{u}\ \frac{1}{u} \qquad\text{with }u=\cos\theta,\ \dee{u}=-\sin\theta\,\dee{\theta}\\ &=-\frac{1}{2}\left[e^4-1\right]\ \Big[\ln|u|\Big]_1^{1/\sqrt{2}}\\ &= \frac{1}{4}\left[e^4-1\right]\ln 2 \end{align*}

3.2.5.22.

Solution.

The paraboloid hits the $xy$--plane at $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1\text{.}$

\begin{align*} \text{Volume} &=\int_0^a \dee{x}\int_0^{b\sqrt{1-{x^2\over a^2}}} \dee{y}\ \left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\\ &=b\int_0^a \dee{x}\int_0^{\sqrt{1-{x^2\over a^2}}} \dee{v}\ \left(1-\frac{x^2}{a^2}-v^2\right) \qquad\text{ where } y=bv \end{align*}

Think of this integral as being of the form

\begin{equation*} \displaystyle b\int_0^a \dee{x}\ g(x)\quad \text{with}\quad \displaystyle g(x)=\int_0^{\sqrt{1-{x^2\over a^2}}} \dee{v}\ \left(1-\frac{x^2}{a^2}-v^2\right) \end{equation*}

Then, substituting $x=au\text{,}$

\begin{align*} \text{Volume} &=ab\int_0^1 \dee{u}\int_0^{\sqrt{1-u^2}} \dee{v}\ \big(1-u^2-v^2\big)\\ &=ab\dblInt_{u^2+v^2\le 1\atop u,v\ge 0} \dee{u} \dee{v}\ \big(1-u^2-v^2\big) \end{align*}

Now switch to polar coordinates using $u=r\cos\theta\text{,}$ $v=r\sin\theta\text{.}$

\begin{align*} \text{Volume} &=ab\int_0^1 \dee{r}\int_0^{\pi\over 2} \dee{\theta}\ r\big(1-r^2\big) =ab\ \frac{\pi}{2}\ \left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 =\frac{\pi}{8}ab \end{align*}

3.2.5.23.

Solution.

Let $r(z)$ be the radius of the urn at height $z$ above its middle. Because the bounding surface of the urn is parabolic, $r(z)$ must be a quadratic function of $z$ that varies between $3$ at $z=0$ and $2$ at $z=\pm 6\text{.}$ That is, $r(z)$ must be of the form $r(z)=az^2+bz+c\text{.}$ The condition that $r(0)=3$ tells us that $c=3\text{.}$ The conditions that $r(\pm 6)=2$ tells us that

\begin{align*} 6^2a + 6b + 3 &=2\\ 6^2a - 6b + 3 &=2 \end{align*}

So $b=0$ and $6^2a=-1$ so that $a=-\frac{1}{6^2}\text{.}$ All together $r(z)=3-\big(\frac{z}{6}\big)^2\text{.}$

Slice the urn into horzontal slices, with the slice at height $z$ a disk of radius $r(z)$ and thickness $\dee{z}$ and hence of volume $\pi r(z)^2\dee{z}\text{.}$ The volume to height $z_0$ is

\begin{align*} V(z) &=\int_{-6}^{z_0} \dee{z}\ \pi r(z)^2 =\int_{-6}^{z_0} \dee{z}\ \pi\left[3-\frac{z^2}{36}\right]^2 =\pi\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{z_0} \end{align*}

We are told that the mark is to be at the 6 cup level and that the urn holds 24 cups. So the mark is to be at the height $z_0$ for which the volume, $V(z_0)\text{,}$ is one quarter of the total volume, $V(6)\text{.}$ That is, we are to choose $z_0$ so that $V(z_0)=\frac{1}{4} V(6)$ or

\begin{align*} \pi\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{z_0} &=\frac{\pi}{4}\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{6}\\ &=\frac{\pi}{2}\left[9\times 6-\frac{6^3}{18}+\frac{6^5}{5\times36^2}\right] \end{align*}

\begin{equation*} 9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{5\times36^2} =-\frac{1}{2}\left[9\times 6-\frac{6^3}{18}+\frac{6^5}{5\times36^2}\right] =-21.60 \end{equation*}

Since $\Big[9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{6480}\Big]_{z_0=-2.495}=-21.61$ and $\Big[9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{6480}\Big]_{z_0=-2.490}=-21.57\text{,}$ there is a solution $z_0=-2.49$ (to two decimal places). The mark should be about 3.5’’ above the bottom.

3.2.5.24. (✳).

Solution.

(a) In polar coordinates, the base region $x^2+y^2\le 9$ is $r\le 3\text{,}$ $0\le\theta\le 2\pi\text{.}$ So the

\begin{align*} \text{Volume} &=\dblInt_{x^2+y^2\le 9} e^{x^2+y^2}\ \dee{x}\dee{y} =\int_0^3 \dee{r}\int_0^{2\pi}\dee{\theta}\ r e^{r^2} =2\pi\int_0^3\dee{r}\ r e^{r^2}\\ &=\pi e^{r^2}\Big|_0^3\\ &=\pi\big(e^9-1\big)\approx 25,453 \end{align*}

(b) The two integrals have domains

\begin{gather*} \Set{(x,y)}{0\le y\le 1,\ 0\le x\le y}\qquad \Set{(x,y)}{1\le y\le 2,\ 0\le x\le 2-y} \end{gather*}

The union of those two domains (as well as horizontal strips that were used in setting up the two given integrals) is sketched in the figure on the left below.

To reverse the order of integration, we decompose the domain using vertical strips as in the figure on the right above. As

$x$ runs from $0$ to $1$ and
for each fixed $x$ between $0$ and $1\text{,}$ $y$ runs from $x$ to $2-x\text{.}$

we have that the

\begin{equation*} \text{Volume}=\int_0^1 \dee{x}\int_x^{2-x}\dee{y}\ e^{x^2+y^2} \end{equation*}

3.3 Applications of Double Integrals
3.3.4 Exercises

3.3.4.1.

Solution.

(a) $\dblInt_D x\ \dee{x}\,\dee{y}=0$ because $x$ is odd under $x\rightarrow-x\text{,}$ i.e. under reflection about the $y$--axis, while the domain of integration is symmetric about the $y$--axis. $\dblInt_D 3\ \dee{x}\,\dee{y}$ is the three times the area of a half disc of radius $2\text{.}$ So, $\dblInt_D(x+3)\dee{x}\,\dee{y}=3\times \half\times\pi 2^2=6\pi\text{.}$

(s) $\dblInt_R x\ \dee{x}\,\dee{y}/\dblInt_R \dee{x}\,\dee{y}$ is the average value of $x$ in the rectangle $R\text{,}$ namely $\frac{a}{2}\text{.}$ Similarly, $\dblInt_R y\ \dee{x}\,\dee{y}/\dblInt_R \dee{x}\,\dee{y}$ is the average value of $y$ in the rectangle $R\text{,}$ namely $\frac{b}{2}\text{.}$ $\dblInt_R \dee{x}\,\dee{y}$ is area of the rectangle $R\text{,}$ namely $ab\text{.}$ So,

$\dblInt_R x\ \dee{x}\,\dee{y} =\frac{a}{2} \dblInt_R \dee{x}\,\dee{y} =\frac{a}{2} ab$ and
$\displaystyle \dblInt_R y\ \dee{x}\,\dee{y} =\frac{b}{2} \dblInt_R \dee{x}\,\dee{y} =\frac{b}{2} ab$

and $\dblInt_R (x+y)\dee{x}\,\dee{y}=\half ab (a+b)\text{.}$

3.3.4.2. (✳).

Solution.

Here is a sketch of $D\text{.}$

By definition, the centre of mass is $(\bar x, \bar y)\text{,}$ with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$--coordinates, respectively, over $D\text{.}$ That is,

\begin{gather*} \bar x = \frac{\dblInt_D x\ \rho(x,y)\ \dee{A}}{\dblInt_D \rho(x,y)\ \dee{A}} \qquad \bar y = \frac{\dblInt_D y\ \rho(x,y)\ \dee{A}}{\dblInt_D \rho(x,y)\ \dee{A}} \end{gather*}

By symmetry under reflection in the $y$--axis, we have $\bar x=0\text{.}$ So we just have to determine $\bar y\text{.}$ We’ll evaluate the integrals using vertical strips as in the figure above. Looking at that figure, we see that

$x$ runs from $-1$ to $1\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $x^2$ to $1\text{.}$

So the denominator is

\begin{align*} \dblInt_D \rho(x,y)\ \dee{A} &= \int_{-1}^1\dee{x}\int_{x^2}^1\dee{y}\ \overbrace{y}^{\rho(x,y)}\\ &= \frac{1}{2}\int_{-1}^1 \dee{x}\ (1-x^4) =\int_0^1 \dee{x}\ (1-x^4)\\ &=\frac{4}{5} \end{align*}

and the numerator of $\bar y$ is

\begin{align*} \dblInt_D y\,\rho(x,y)\ \dee{A} &= \int_{-1}^1\dee{x}\int_{x^2}^1\dee{y}\ y\overbrace{y}^{\rho(x,y)}\\ &= \frac{1}{3}\int_{-1}^1 \dee{x}\ (1-x^6) =\frac{2}{3}\int_0^1 \dee{x}\ (1-x^6)\\ &=\frac{2}{3}\ \frac{6}{7} = \frac{4}{7} \end{align*}

All together, $\bar x=0$ and

\begin{gather*} \bar y = \frac{\frac{4}{7}} {\frac{4}{5}} = \frac{5}{7} \end{gather*}

3.3.4.3. (✳).

Solution.

(a) Here is a sketch of $R\text{.}$

(b) Considering that

$\rho(x,y)$ is invariant under rotations about the origin and
the outer curve $x^2+y^2=4$ is invariant under rotations about the origin and
the given hint involves a $\theta$ integral

we’ll use polar coordinates.

Observe that the line $x=1$ and the circle $x^2+y^2=4$ intersect when

\begin{equation*} 1+y^2=4 \iff y=\pm\sqrt{3} \end{equation*}

and that the polar coordinates of the point $(x,y)=\big(1,\sqrt{3}\big)$ are $r=\sqrt{x^2+y^2}=2$ and $\theta=\arctan\frac{y}{x}=\arctan \sqrt{3} =\frac{\pi}{3}\text{.}$ Looking at the sketch

we see that, on $R\text{,}$

$\theta$ runs from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$ and
for each fixed $\theta$ in that range, $r$ runs from $\frac{1}{\cos\theta} =\sec\theta$ to $2\text{.}$
In polar coordinates, $\dee{A}=r\,\dee{r}\,\dee{\theta}\text{,}$ and
the density $\rho =\frac{1}{\sqrt{x^2+y^2}} =\frac{1}{r}$

So the mass is

\begin{align*} M&=\dblInt_R \rho(x,y)\ \dee{A} =\int_{-\pi/3}^{\pi/3}\dee{\theta}\int_{\sec\theta}^2\dee{r}\ \frac{r}{r} =\int_{-\pi/3}^{\pi/3}\dee{\theta}\ \big[2-\sec\theta\big]\\ &=2\int_0^{\pi/3}\dee{\theta}\ \big[2-\sec\theta\big]\\ &= 2\Big[2\theta - \ln\big(\sec\theta+\tan\theta\big)\Big]_0^{\pi/3}\\ &= 2\left[\frac{2\pi}{3} - \ln\big(2+\sqrt{3}\big) + \ln\big(1+0\big)\right]\\ &= \frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big) \end{align*}

(c) By definition, the centre of mass is $(\bar x, \bar y)\text{,}$ with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$--coordinates, respectively, over $R\text{.}$ That is,

\begin{gather*} \bar x = \frac{\dblInt_R x\ \rho(x,y)\ \dee{A}}{\dblInt_R \rho(x,y)\ \dee{A}} \qquad \bar y = \frac{\dblInt_R y\ \rho(x,y)\ \dee{A}}{\dblInt_R \rho(x,y)\ \dee{A}} \end{gather*}

By symmetry under reflection in the $x$--axis, we have $\bar y=0\text{.}$ So we just have to determine $\bar x\text{.}$ The numerator is

\begin{align*} \dblInt_R x\ \rho(x,y)\ \dee{A} &=\int_{-\pi/3}^{\pi/3}\dee{\theta}\int_{\sec\theta}^2\dee{r}\ \frac{r}{r}\ \overbrace{r\cos\theta}^{x}\\ &=\frac{1}{2}\int_{-\pi/3}^{\pi/3}\dee{\theta}\ \big[4-\sec^2\theta\big] \cos\theta =\int_0^{\pi/3}\dee{\theta}\ \big[4\cos\theta-\sec\theta\big]\\ &= \Big[4\sin\theta - \ln\big(\sec\theta+\tan\theta\big)\Big]_0^{\pi/3}\\ &= \left[4\frac{\sqrt{3}}{2} - \ln\big(2+\sqrt{3}\big) + \ln\big(1+0\big)\right]\\ &= 2\sqrt{3}- \ln\big(2+\sqrt{3}\big) \end{align*}

All together, $\bar y=0$ and

\begin{gather*} \bar x = \frac{2\sqrt{3}- \ln\big(2+\sqrt{3}\big)} {\frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big)} \approx 1.38 \end{gather*}

3.3.4.4. (✳).

Solution.

Let’s call the plate $\cP\text{.}$ By definition, the $x$--coordinate of its centre of mass is

\begin{gather*} \bar x = \frac{\dblInt_\cP x\ \dee{A}}{\dblInt_\cP\dee{A}} \end{gather*}

Here is a sketch of the plate.

The cardiod is given to us in polar coordinates, so let’s evaluate the integrals in polar coordinates. Looking at the sketch above, we see that, on $\cP\text{,}$

$\theta$ runs from $0$ to $\pi/2$ and
for each fixed $\theta$ in that range, $r$ runs from $0$ to $1+\sin\theta\text{.}$
In polar coordinates $\dee{A} = r\,\dee{r}\,\dee{\theta}$

So the two integrals of interest are

\begin{align*} \dblInt_\cP\dee{A} &=\int_0^{\pi/2}\dee{\theta}\int_0^{1+\sin\theta}\dee{r}\ r\\ &=\frac{1}{2} \int_0^{\pi/2}\dee{\theta}\ \big(1+2\sin\theta +\sin^2\theta\big)\\ &=\frac{1}{2}\frac{\pi}{2} +\Big[-\cos\theta\Big]_0^{\pi/2} +\frac{1}{2} \int_0^{\pi/2}\dee{\theta}\ \frac{1-\cos(2\theta)}{2}\\ &=\frac{\pi}{4} + 1 +\frac{1}{4}\left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{\pi/2}\\ &=\frac{3\pi}{8} + 1 \end{align*}

and

\begin{align*} \dblInt_\cP x\ \dee{A} &=\int_0^{\pi/2}\dee{\theta}\int_0^{1+\sin\theta}\dee{r}\ r \overbrace{(r\cos\theta)}^{x}\\ &=\frac{1}{3} \int_0^{\pi/2}\dee{\theta}\ \big(1+\sin\theta\big)^3\cos\theta\\ &=\frac{1}{3}\int_1^2\dee{u}\ u^3\qquad\text{with }u=1+\sin\theta,\ \dee{u} = \cos\theta\,\dee{\theta}\\ &=\frac{1}{12}\big[2^4-1^4\big]\\ &=\frac{5}{4} \end{align*}

All together

\begin{gather*} \bar x = \frac{\frac{5}{4}}{\frac{3\pi}{8} + 1} =\frac{10}{3\pi+8} \approx 0.57 \end{gather*}

For an efficient, sneaky, way to evaluate $\int_0^{\pi/2}\sin^2\theta\ \dee{\theta}\text{,}$ see Remark 3.3.5.

3.3.4.5. (✳).

Solution.

Call the plate $P\text{.}$ By definition, the centre of mass is $(\bar x, \bar y)\text{,}$ with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$--coordinates, respectively, over $P\text{.}$ That is,

\begin{gather*} \bar x = \frac{\dblInt_P x\ \rho(x,y)\ \dee{A}}{\dblInt_P \rho(x,y)\ \dee{A}} \qquad \bar y = \frac{\dblInt_P y\ \rho(x,y)\ \dee{A}}{\dblInt_P \rho(x,y)\ \dee{A}} \end{gather*}

with $\rho(x,y)=k\text{.}$ Here is a sketch of $P\text{.}$

By symmetry under reflection in the line $y=x\text{,}$ we have $\bar y=\bar x\text{.}$ So we just have to determine

\begin{equation*} \bar x = \frac{\dblInt_P x\ \dee{A}}{\dblInt_P \dee{A}} \end{equation*}

The denominator is just one quarter of the area of circular disk of radius $1\text{.}$ That is, $\dblInt_P \dee{A}=\frac{\pi}{4}\text{.}$ We’ll evaluate the numerator using polar coordinates as in the figure above. Looking at that figure, we see that

$\theta$ runs from $0$ to $\frac{\pi}{2}\text{,}$ and
for each fixed $\theta$ in that range, $r$ runs from $0$ to $1\text{.}$

As $\dee{A}=r\,\dee{r}\,\dee{\theta}\text{,}$ and $x=r\cos\theta\text{,}$ the numerator

\begin{align*} \dblInt_P x\ \dee{A} &=\int_0^{\pi/2}\dee{\theta} \int_0^1\dee{r}\ r\overbrace{r\cos\theta}^{x} =\left[\int_0^{\pi/2}\dee{\theta}\ \cos\theta\right] \left[ \int_0^1\dee{r}\ r^2\right]\\ &=\Big[\sin\theta\Big]_0^{\pi/2} \left[\frac{r^3}{3}\right]_0^1\\ &=\frac{1}{3} \end{align*}

All together

\begin{gather*} \bar x = \bar y = \frac{1/3}{\pi/4} =\frac{4}{3\pi} \end{gather*}

3.3.4.6. (✳).

Solution.

Here is a sketch of $R\text{.}$

Note that

the equation of the straight line through $(2,0)$ and $(0,2)$ is $y=2-x\text{,}$ or $x=2-y\text{.}$ (As a check note that both points $(2,0)$ and $(0,2)$ are on $x=2-y\text{.}$
The equation of the straight line through $(1,0)$ and $(0,2)$ is $y=2-2x\text{,}$ or $x=\frac{2-y}{2}\text{.}$ (As a check note that both points $(0,2)$ and $(1,0)$ are on $x=\frac{2-y}{2}\text{.}$

By definition, the $y$--coordinate of the center of mass of $R$ is the weighted average of $y$ over $R\text{,}$ which is

\begin{equation*} \bar y =\frac{\dblInt_R y\,\rho(x,y)\,\dee{A}}{\dblInt_R \rho(x,y)\,\dee{A}} =\frac{\dblInt_R y^3\,\dee{A}}{\dblInt_R y^2\,\dee{A}} \end{equation*}

On $R\text{,}$

$y$ runs from $0$ to $2\text{.}$ That is, $0\le y\le 2\text{.}$
For each fixed $y$ in that range, $x$ runs from $\frac{2-y}{2}$ to $2-y\text{.}$ In inequalities, that is $\frac{2-y}{2}\le x\le 2-y\text{.}$

Thus

\begin{equation*} R = \left\{\ (x,y)\ \left|\ 0\le y\le 2,\ \frac{2-y}{2}\le x\le 2-y \right.\right\} \end{equation*}

For both $n=2$ and $n=3\text{,}$ we have

\begin{align*} \dblInt_R y^n\,\dee{A} &=\int_0^2\dee{y} \int_{\frac{2-y}{2}}^{2-y}\dee{x}\ y^n\\ &=\int_0^2\dee{y} \ y^n\frac{2-y}{2}\\ &=\frac{1}{2}\left[\frac{2y^{n+1}}{n+1}-\frac{y^{n+2}}{n+2}\right]_0^2\\ &=\frac{1}{2}\left[\frac{2^{n+2}}{n+1}-\frac{2^{n+2}}{n+2}\right]\\ &=\frac{2^{n+1}}{(n+1)(n+2)} \end{align*}

\begin{gather*} \bar y =\frac{\dblInt_R y^3\,\dee{A}}{\dblInt_R y^2\,\dee{A}} =\frac{ \frac{2^4}{(4)(5)} }{ \frac{2^{3}}{(3)(4)} } =\frac{6}{5} \end{gather*}

3.3.4.7. (✳).

Solution.

By the definition given in the statement with $(a,b)=(0,0)\text{,}$ the average is

\begin{gather*} \frac{1}{A(D)}\dblInt_D \sqrt{x^2+y^2}\ \dee{x}\,\dee{y} \end{gather*}

The denominator $A(D) = \pi\text{.}$ We’ll use polar coordinates to evaluate the numerator.

\begin{align*} \dblInt_D \sqrt{x^2+y^2}\ \dee{x}\,\dee{y} &=\int_0^{2\pi}\dee{\theta}\int_0^1\dee{r} \ r\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}\\ &=\int_0^{2\pi}\dee{\theta}\int_0^1\dee{r}\ r^2 =\int_0^{2\pi}\dee{\theta}\ \frac{1}{3}\\ &=\frac{2\pi}{3} \end{align*}

So the average is

\begin{equation*} \frac{\frac{2\pi}{3}}{\pi}=\frac{2}{3} \end{equation*}

3.3.4.8. (✳).

Solution.

Note that $x^2+y^2=x$ is equivalent to $\left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4}\text{,}$ which is the circle of radius $\frac{1}{2}$ centred on $\left(\frac{1}{2},0\right)\text{.}$ Let’s call the crescent $\cC$ and write

\begin{align*} D &= \Set{(x,y)}{x^2+y^2\le 1}\\ H &= \Set{(x,y)}{\left(x-\tfrac{1}{2}\right)^2+y^2\le\tfrac{1}{4}} \end{align*}

so that

\begin{equation*} \cC= D\setminus H \end{equation*}

meaning that $\cC$ is the disk $D$ with the “hole” $H$ removed. Here is a sketch.

(a) As $D$ is a disk of radius $1\text{,}$ it has area $\pi\text{.}$ As $H$ is a disk of radius $\frac{1}{2}\text{,}$ it has area $\frac{\pi}{4}\text{.}$ As $\cC$ has density $1\text{,}$

\begin{align*} \text{Mass}(\cC) &= \dblInt_\cC \dee{A} = \dblInt_D\dee{A} -\dblInt_H\dee{A}\\ &=\pi - \frac{\pi}{4}\\ &=\frac{3\pi}{4} \end{align*}

(b) Recall that, by definition, the $x$--coordinate of the centre of mass of $\cC$ is the average value of $x$ over $\cC\text{,}$ which is

\begin{gather*} \bar x = \frac{\dblInt_\cC x\,\dee{A}}{\dblInt_\cC \dee{A}} \end{gather*}

We have already found that $\dblInt_\cC \dee{A}=\frac{3\pi}{4}\text{.}$ So we have to determine the numerator

\begin{equation*} \dblInt_\cC x\,\dee{A} = \dblInt_D x\,\dee{A} - \dblInt_H x\,\dee{A} \end{equation*}

As $x$ is an odd function and $D$ is invariant under $x\rightarrow -x\text{,}$ $\dblInt_D x\,\dee{A}=0\text{.}$ So we just have to determine $\dblInt_H x\,\dee{A}\text{.}$ To do so we’ll work in polar coordinates, so that $\dee{A} = r\,\dee{r}\,\dee{\theta}\text{.}$ In polar coordinates $x^2 + y^2 = x$ is $r^2 =r\cos\theta$ or $r=\cos\theta\text{.}$ So, looking at the figure above (just before the solution to part (a)), on the domain of integration,

$\theta$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}\text{.}$
For each fixed $\theta$ in that range, $r$ runs from $0$ to $\cos\theta\text{.}$

So the integral is

\begin{align*} \dblInt_H x\,\dee{A} &= \int_{-\pi/2}^{\pi/2}\dee{\theta}\int_0^{\cos\theta}\dee{r}\ r\overbrace{(r\cos\theta)}^{x}\\ &=\int_{-\pi/2}^{\pi/2}\dee{\theta} \ \frac{\cos^4\theta}{3}\\ &= \frac{\pi}{8} \end{align*}

So all together

\begin{gather*} \bar x = \frac{\dblInt_\cC x\,\dee{A}}{\dblInt_\cC \dee{A}} = \frac{\dblInt_D x\,\dee{A} - \dblInt_H x\,\dee{A}}{\dblInt_\cC \dee{A}} =\frac{0-\frac{\pi}{8}}{\frac{3\pi}{4}} =-\frac{1}{6} \end{gather*}

3.3.4.9. (✳).

Solution.

The domain is pictured below.

The two circles intersect when $x^2+y^2=2$ and

\begin{align*} x^2+(y-1)^2=2-y^2+(y-1)^2=1 &\iff -2y+3=1\\ &\iff y=1 \end{align*}

and $x=\pm 1\text{.}$ In polar coordinates $x^2+y^2=2$ is $r=\sqrt{2}$ and $x^2+(y-1)^2=x^2+y^2-2y+1=1$ is $r^2-2r\sin\theta=0$ or $r=2\sin\theta\text{.}$ The two curves intersect when $r=\sqrt{2}$ and$\sqrt{2}=2\sin\theta$ so that $\theta=\frac{\pi}{4}$ or $\frac{3}{4}\pi\text{.}$ So

\begin{equation*} D=\Set{(r\cos\theta,r\sin\theta)}{\tfrac{1}{4}\pi\le\theta\le\tfrac{3}{4}\pi,\ \sqrt{2}\le r\le 2\sin\theta} \end{equation*}

and, as the density is $\frac{2}{r}\text{,}$

\begin{align*} \text{mass} &=\int_{\pi/4}^{3\pi/4}\dee{\theta}\int^{2\sin\theta}_{\sqrt{2}}\dee{r}\ r\frac{2}{r} =2\int_{\pi/4}^{3\pi/4}\dee{\theta}\ \big[2\sin\theta -\sqrt{2}\,\big]\\ &=4\int_{\pi/4}^{\pi/2}\dee{\theta}\ \big[2\sin\theta -\sqrt{2}\,\big] =4\Big[-2\cos\theta -\sqrt{2}\theta\Big] _{\pi/4}^{\pi/2}\\ &=4\sqrt{2} -\sqrt{2}\pi\approx 1.214 \end{align*}

3.3.4.10. (✳).

Solution.

(a) The side of the triangle from $(-a,0)$ to $(0,c)$ is straight line that passes through those two points. As $y=0$ when $x=-a\text{,}$ the line must have an equation of the form $y=K(x+a)$ for some constant $K\text{.}$ Since $y=c$ when $x=0\text{,}$ the constant $K=\frac{c}{a}\text{.}$ So that the equation is $y=\frac{c}{a}(x+a)\text{.}$ has equation $cx-ay=-ac\text{.}$ Similarly the side of the triangle from $(b,0)$ to $(0,c)$ has equation $y=\frac{c}{b}(b-x)\text{.}$ The triangle has area $A=\frac{1}{2}(a+b)c\text{.}$ It has centre of mass $(\bar x,\bar y)$ with

\begin{gather*} \bar x=\frac{1}{A}\dblInt_T x\ \dee{x}\dee{y}\qquad \bar y=\frac{1}{A}\dblInt_T y\ \dee{x}\dee{y} \end{gather*}

To evaluate the integrals we’ll decompose the triangle into vertical strips as in the figure

\begin{align*} \bar x&=\frac{1}{A}\dblInt_T x\ \dee{x}\dee{y}\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \dee{x}\int_0^{c+{c\over a}x}\dee{y}\ x +\int_0^b \dee{x}\int_0^{c-{c\over b}x}\dee{y}\ x\bigg)\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \dee{x}\ x\left(c+\frac{c}{a}x\right) +\int_0^b \dee{x}\ x\left(c-\frac{c}{b}x\right)\bigg)\\ &=\frac{1}{A}\left(\left[\frac{1}{2} cx^2+\frac{c}{3a}x^3\right]_{-a}^0 +\left[\frac{1}{2} cx^2-\frac{c}{3b}x^3\right]_0^b\right)\\ &=2\frac{\frac{1}{2} c(b^2-a^2)+\frac{c}{3}(a^2-b^2)}{(a+b)c} =\frac{1}{3}(b-a)\\ \bar y&=\frac{1}{A}\dblInt_T y\ \dee{x}\dee{y}\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \dee{x}\int_0^{c+{c\over a}x}\dee{y}\ y +\int_0^b \dee{x}\int_0^{c-{c\over b}x}\dee{y}\ y\bigg)\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \dee{x}\ \frac{1}{2}\left(c+\frac{c}{a}x\right)^2 +\int_0^b \dee{x}\ \frac{1}{2}\left(c-\frac{c}{b}x\right)^2\bigg)\\ &=\frac{1}{A}\left(\frac{a}{6c}\left[c+\frac{c}{a}x\right]^3\bigg|_{-a}^0 -\frac{b}{6c}\left(c-\frac{c}{b}x\right)^3\bigg|_0^b\, \right)\\ &=2\frac{\frac{ac^2}{6}+\frac{bc^2}{6}}{(a+b)c} =\frac{c}{3} \end{align*}

(b) The midpoint of the side opposite $(-a,0)$ is $\frac{1}{2}\big[(b,0)+(0,c)\big]=\frac{1}{2}(b,c)\text{.}$ The vector from $(-a,0)$ to $\frac{1}{2}(b,c)$ is $\frac{1}{2}\llt b,c\rgt-\llt-a,0\rgt =\llt a+\frac{b}{2},\frac{c}{2}\rgt\text{.}$ So the line joining these two points has vector parametric equation

\begin{equation*} \vr(t)=\llt -a,0\rgt+t\llt a+\frac{1}{2} b\,,\,\frac{1}{2} c\rgt \end{equation*}

The point $(\bar x,\bar y)$ lies on this line since

\begin{equation*} \vr\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a)\,,\,\frac{c}{3}\right) =(\bar x,\bar y) \end{equation*}

Similarly, the midpoint of the side opposite $(b,0)$ is $\frac{1}{2}(-a,c)\text{.}$ The line joining these two points has vector parametric equation

\begin{equation*} \vr(t)=\llt b,0\rgt +t\llt-b-\frac{1}{2} a\,,\,\frac{1}{2} c\rgt \end{equation*}

The point $(\bar x,\bar y)$ lies on this line too, since

\begin{equation*} \vr\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a),\frac{c}{3}\right) =(\bar x,\bar y) \end{equation*}

It is not really necessary to check that $(\bar x,\bar y)$ lies on the third median, but let’s do it anyway. The midpoint of the side opposite $(0,c)$ is $\frac{1}{2}(b-a,0)\text{.}$ The line joining these two points has vector parametric equation

\begin{equation*} \vr(t)=\llt 0,c\rgt+t\llt\frac{b}{2}-\frac{a}{2},-c\rgt \end{equation*}

The point $(\bar x,\bar y)$ lies on this median too, since

\begin{equation*} \vr\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a),\frac{c}{3}\right) =(\bar x,\bar y) \end{equation*}

3.4 Surface Area

Exercises

3.4.1.

Solution.

(a) $S$ is the part of the plane $z=y\,\tan\theta$ that lies above the rectangle in the $xy$-plane with vertices $(0,0)\text{,}$ $(a,0)\text{,}$ $(0,b)\text{,}$ $(a,b)\text{.}$ So $S$ is the rectangle with vertices $(0,0,0)\text{,}$ $(a,0,0)\text{,}$ $(0,b,b\tan\theta)\text{,}$ $(a,b,b\tan\theta)\text{.}$ So it has side lengths

\begin{align*} |\llt a,0,0\rgt -\llt 0,0,0\rgt| &=a\\ |\llt 0,b,b\tan\theta\rgt -\llt 0,0,0\rgt| &= \sqrt{b^2+b^2\tan^2\theta} \end{align*}

and hence area $ab\sqrt{1+\tan^2\theta}=ab\sec\theta\text{.}$

(b) $S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = y\,\tan\theta$ and with $(x,y)$ running over

\begin{equation*} \cD =\Set{(x,y)}{0\le x\le a,\ 0\le y\le b} \end{equation*}

Hence by Theorem 3.4.2

\begin{align*} \text{Area}(S)&=\dblInt_\cD \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &=\int_0^a\dee{x}\int_0^b\dee{y}\ \sqrt{1+0^2+\tan^2\theta}\\ &=ab\sqrt{1+\tan^2\theta}=ab\sec\theta \end{align*}

3.4.2.

Solution.

$S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = \frac{d-ax-by}{c}$ and with $(x,y)$ running over $D\text{.}$ Hence by Theorem 3.4.2

\begin{align*} \text{Area}(S)&=\dblInt_D \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &=\dblInt_D\ \sqrt{1+\frac{a^2}{c^2}+\frac{b^2}{c^2}}\\ &=\frac{\sqrt{a^2+b^2+c^2}}{c} A(D) \end{align*}

3.4.3.

Solution.

Note that all three vertices $(a,0,0)\text{,}$ $(0,b,0)$ and $(0,0,c)$ lie on the plane $\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1\text{.}$ So the triangle is part of that plane.

Method 1. $S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = c\left(1-\frac{x}{a}-\frac{y}{b}\right)$ and with $(x,y)$ running over the triangle $T_{xy}$ in the $xy$-plane with vertices $(0,0,0)$ $(a,0,0)$ and $(0,b,0)\text{.}$ Hence by part a of Theorem 3.4.2

\begin{align*} \text{Area}(S)&=\dblInt_{T_{xy}} \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &=\dblInt_{T_{xy}}\ \sqrt{1+\frac{c^2}{a^2}+\frac{c^2}{b^2}}\ \dee{x}\,\dee{y}\\ &=\sqrt{1+\frac{c^2}{a^2}+\frac{c^2}{b^2}}\ A(T_{xy}) \end{align*}

where $A(T_{xy})$ is the area of $T_{xy}\text{.}$ Since the triangle $T_{xy}$ has base $a$ and height $b$ (see the figure below), it has area $\frac{1}{2}ab\text{.}$ So

\begin{equation*} \text{Area}(S)=\frac{1}{2}\sqrt{1+\frac{c^2}{a^2}+\frac{c^2}{b^2}}\ ab =\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2} \end{equation*}

Method 2. $S$ is the part of the surface $x=g(y,z)$ with $g(y,z) = a\left(1-\frac{y}{b}-\frac{z}{c}\right)$ and with $(y,z)$ running over the triangle $T_{yz}$ in the $yz$-plane with vertices $(0,0,0)$ $(0,b,0)$ and $(0,0,c)\text{.}$ Hence by part b of Theorem 3.4.2

\begin{align*} \text{Area}(S)&=\dblInt_{T_{yz}} \sqrt{1+g_y(y,z)^2+g_z(y,z)^2}\ \dee{y}\,\dee{z}\\ &=\dblInt_{T_{yz}}\ \sqrt{1+\frac{a^2}{b^2}+\frac{a^2}{c^2}}\ \dee{y}\,\dee{z}\\ &=\sqrt{1+\frac{a^2}{b^2}+\frac{a^2}{c^2}}\ A(T_{yz}) \end{align*}

where $A(T_{yz})$ is the area of $T_{yz}\text{.}$ Since $T_{yz}$ has base $b$ and height $c\text{,}$ it has area $\frac{1}{2}bc\text{.}$ So

\begin{equation*} \text{Area}(S)=\frac{1}{2}\sqrt{1+\frac{a^2}{b^2}+\frac{a^2}{c^2}}\ bc =\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2} \end{equation*}

Method 3. $S$ is the part of the surface $y=h(x,z)$ with $h(x,z) = b\left(1-\frac{x}{a}-\frac{z}{c}\right)$ and with $(x,z)$ running over the triangle $T_{xz}$ in the $xz$-plane with vertices $(0,0,0)$ $(a,0,0)$ and $(0,0,c)\text{.}$ Hence by part c of Theorem 3.4.2

\begin{align*} \text{Area}(S)&=\dblInt_{T_{xz}} \sqrt{1+h_x(x,z)^2+h_z(x,z)^2}\ \dee{x}\,\dee{z}\\ &=\dblInt_{T_{xz}}\ \sqrt{1+\frac{b^2}{a^2}+\frac{b^2}{c^2}}\ \dee{x}\,\dee{z}\\ &=\sqrt{1+\frac{b^2}{a^2}+\frac{b^2}{c^2}}\ A(T_{xz}) \end{align*}

where $A(T_{xz})$ is the area of $T_{xz}\text{.}$ Since $T_{xz}$ has base $a$ and height $c\text{,}$ it has area $\frac{1}{2}ac\text{.}$ So

\begin{equation*} \text{Area}(S)=\frac{1}{2}\sqrt{1+\frac{b^2}{a^2}+\frac{b^2}{c^2}}\ bc =\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2} \end{equation*}

(b) We have already seen in the solution to part (a) that

\begin{equation*} \text{Area}(T_{xy})=\frac{ab}{2}\qquad \text{Area}(T_{xz})=\frac{ac}{2}\qquad \text{Area}(T_{yz})=\frac{bc}{2}\qquad \end{equation*}

Hence

\begin{align*} \text{Area}(S) &=\sqrt{\frac{a^2b^2}{4}+\frac{a^2c^2}{4}+\frac{b^2c^2}{4}}\\ &=\sqrt{\text{Area}(T_{xy})^2 +\text{Area}(T_{xz})^2 +\text{Area}(T_{yz})^2 } \end{align*}

3.4.4. (✳).

Solution.

For the surface $z=f(x,y)=y^{3/2}\text{,}$

\begin{equation*} \dee{S}=\sqrt{1+f_x^2+f_y^2}\ \dee{x}\dee{y} =\sqrt{1+\Big(\frac{3}{2}\sqrt{y}\Big)^2}\ \dee{x}\dee{y} =\sqrt{1+\frac{9}{4}y}\ \dee{x}\dee{y} \end{equation*}

by Theorem 3.4.2.a, So the area is

\begin{align*} \int_0^1\dee{x}\int_0^1\dee{y}\ \sqrt{1+\frac{9}{4}y} &=\int_0^1\dee{x}\ \frac{8}{27}\Big[\Big(1+\frac{9}{4}y\Big)^{3/2}\Big]_0^1\\ &=\int_0^1\dee{x}\ \frac{8}{27}\Big[\Big(\frac{13}{4}\Big)^{3/2}-1\Big]\\ &=\frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right] \end{align*}

3.4.5. (✳).

Solution.

First observe that any point $(x,y,z)$ on the paraboliod lies above the $xy$-plane if and only if

\begin{equation*} 0\le z = a^2-x^2-y^2 \iff x^2+y^2\le a^2 \end{equation*}

That is, if and only if $(x,y)$ lies in the circular disk of radius $a$ centred on the origin. The equation of the paraboloid is of the form $z=f(x,y)$ with $f(x,y)=a^2-x^2-y^2\text{.}$ So, by Theorem 3.4.2.a,

\begin{align*} \text{Surface area} &= \dblInt_{x^2+y^2\le a^2}\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &= \dblInt_{x^2+y^2\le a^2}\sqrt{1+4x^2+4y^2}\ \dee{x}\,\dee{y} \end{align*}

Switching to polar coordinates,

\begin{align*} \text{Surface area} &= \int_0^a\dee{r}\int_0^{2\pi}\dee{\theta}\ r\sqrt{1+4r^2}\\ &= 2\pi \int_0^a\dee{r}\ r\sqrt{1+4r^2}\\ &= 2\pi \int_1^{1+4a^2}\frac{\dee{s}}{8}\ \sqrt{s}\qquad \text{with } s=1+4r^2, \dee{s}=8r\,\dee{r}\\ &=\frac{\pi}{4}\ \frac{2}{3}s^{3/2}\bigg|_{s=1}^{s=1+4a^2}\\ &=\frac{\pi}{6}\big[{(1+4a^2)}^{3/2}-1\big] \end{align*}

3.4.6. (✳).

Solution.

First observe that any point $(x,y,z)$ on the cone lies between the planes $z=2$ and $z=3$ if and only if $4\le x^2+y^2\le 9\text{.}$

The equation of the cone can be rewritten in the form $z=f(x,y)$ with $f(x,y)=\sqrt{x^2+y^2}\text{.}$ Note that

\begin{gather*} f_x(x,y)=\frac{x}{\sqrt{x^2+y^2}}\qquad f_y(x,y)=\frac{y}{\sqrt{x^2+y^2}} \end{gather*}

So, by Theorem 3.4.2.a,

\begin{align*} \text{Surface area} &= \dblInt_{4\le x^2+y^2\le 9}\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &= \dblInt_{4\le x^2+y^2\le 9} \sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}\ \dee{x}\,\dee{y}\\ &=\sqrt{2} \dblInt_{4\le x^2+y^2\le 9} \dee{x}\,\dee{y} \end{align*}

Now the domain of integration is a circular washer with outside radius $3$ and inside radius $2$ and hence of area $\pi(3^2-2^2)=5\pi\text{.}$ So the surface area is $5\sqrt{2}\pi\text{.}$

3.4.7. (✳).

Solution.

The equation of the surface is of the form $z=f(x,y)$ with $f(x,y)=\frac{2}{3}\big(x^{3/2} + y^{3/2}\big)\text{.}$ Note that

\begin{gather*} f_x(x,y)=\sqrt{x}\qquad f_y(x,y)=\sqrt{y} \end{gather*}

So, by Theorem 3.4.2.a,

\begin{align*} \text{Surface area} &= \int_0^1\dee{x}\int_0^1\dee{y}\ \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\\ &= \int_0^1\dee{x}\int_0^1\dee{y}\ \sqrt{1+x+y}\\ &= \int_0^1\dee{x}\ \Big[\frac{2}{3}(1+x+y)^{3/2}\Big]_{y=0}^{y=1}\\ &= \frac{2}{3}\int_0^1\dee{x}\ \big[(2+x)^{3/2}-(1+x)^{3/2}\big]\\ &= \frac{2}{3}\ \frac{2}{5}\Big[(2+x)^{5/2}-(1+x)^{5/2}\Big]_{x=0}^{x=1}\\ &= \frac{4}{15} \big[3^{5/2}-2^{5/2}-2^{5/2}+1^{5/2}\big]\\ &= \frac{4}{15}\big[9\sqrt{3}-8\sqrt{2}+1\big] \end{align*}

3.4.8. (✳).

Solution.

(a) By Theorem 3.4.2.a, $F(x,y) = \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\text{.}$

(b) (i) The “dimple” to be painted is part of the upper sphere $x^2+y^2+\big(z-2\sqrt{3}\big)^2=4\text{.}$ It is on the bottom half of the sphere and so has equation $z=f(x,y)=2\sqrt{3}-\sqrt{4-x^2-y^2}\text{.}$ Note that

\begin{gather*} f_x(x,y) = \frac{x}{\sqrt{4-x^2-y^2}}\qquad f_y(x,y) = \frac{y}{\sqrt{4-x^2-y^2}} \end{gather*}

The point on the dimple with the largest value of $x$ is $(1,0,\sqrt{3})\text{.}$ (It is marked by a dot in the figure above.) The dimple is invariant under rotations around the $z$--axis and so has $(x,y)$ running over $x^2+y^2\le 1\text{.}$ So, by Theorem 3.4.2.a,

\begin{align*} \text{Surface area} &= \dblInt_{x^2+y^2\le 1}\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ \dee{x}\,\dee{y}\\ &= \dblInt_{x^2+y^2\le 1}\sqrt{1+\frac{x^2}{4-x^2-y^2} +\frac{y^2}{4-x^2-y^2}}\ \dee{x}\,\dee{y}\\ &= \dblInt_{x^2+y^2\le 1}\frac{2}{\sqrt{4-x^2-y^2}}\ \dee{x}\,\dee{y} \end{align*}

Switching to polar coordinates,

\begin{align*} \text{Surface area} &= \int_0^{2\pi}\dee{\theta}\int_0^1\dee{r}\ \frac{2r}{\sqrt{4-r^2}} \end{align*}

(b) (ii) Observe that if we flip the dimple up by reflecting it in the plane $z=\sqrt{3}\text{,}$ as in the figure below, the “Death Star” becomes a perfect ball of radius $2\text{.}$

The area of the pink dimple in the figure above is identical to the area of the blue cap in that figure. So the total surface area of the Death Star is exactly the surface area of a sphere of radius $a=2$ and so (see Example 3.4.5) is $4\pi a^2 = 4 \pi 2^2=16\pi\text{.}$

3.4.9.

Solution.

The equation of the half of the cone with $x\ge 0$ can be rewritten in the form $x=g(y,z)$ with $g(y,z)=\sqrt{3}\sqrt{y^2+z^2}\text{.}$ Note that

\begin{equation*} g_y(y,z)=\frac{\sqrt{3}y}{\sqrt{y^2+z^2}}\quad g_z(y,z)=\frac{\sqrt{3}z}{\sqrt{y^2+z^2}} \end{equation*}

so that, by Theorem 3.4.2,

\begin{align*} \dee{S}&=\sqrt{1+g_y(y,z)^2+g_z(y,z)^2}\,\dee{y}\,\dee{z}\\ &=\sqrt{1+\frac{3y^2}{y^2+z^2}+\frac{3z^2}{y^2+z^2}}\,\dee{y}\,\dee{z}\\ &=2\,\dee{y}\,\dee{z} \end{align*}

A point $(x,y,z)$ on $x=\sqrt{3}\sqrt{y^2+z^2}$ has $\sqrt{3}\le x\le 2\sqrt{3}$ if and only if $1\le\sqrt{y^2+z^2}\le 2\text{.}$ So

\begin{align*} &\text{Area}= \dblInt_{1\le \sqrt{y^2+z^2}\le 2}2\,\dee{y}\,\dee{z} \\ &\hskip0.25in=2\,\Big[\text{area of }\Set{(y,z)}{\sqrt{y^2+z^2}\le 2}- \text{area of }\Set{(y,z)}{\sqrt{y^2+z^2}\le 1}\Big]\\ &\hskip0.25in=2\,\big[\pi 2^2-\pi 1^2\big] =6\pi\approx 18.85 \end{align*}

3.4.10. (✳).

Solution.

We are to find the surface area of part of a hemisphere. On the hemisphere

\begin{align*} z=f(x,y)&=\sqrt{a^2-x^2-y^2}\\ f_x(x,y)&=-\frac{x}{\sqrt{a^2-x^2-y^2}}\qquad f_y(x,y)=-\frac{y}{\sqrt{a^2-x^2-y^2}} \end{align*}

so that

\begin{align*} \dee{S}&=\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,\dee{x}\,\dee{y}\\ &=\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}\,\dee{x}\,\dee{y}\\ &=\sqrt{\frac{a^2}{a^2-x^2-y^2}}\,\dee{x}\,\dee{y} \end{align*}

In polar coordinates, this is $\dee{S}=\frac{a}{\sqrt{a^2-r^2}}\,r\,\dee{r}\,\dee{\theta}\text{.}$ We are to find the surface area of the part of the hemisphere that is inside the cylinder, $x^2-ax+y^2=0\text{,}$ which is polar coordinates is becomes $r^2-ar\cos\theta=0$ or $r=a\cos\theta\text{.}$ The top half of the domain of integration is sketched below.

So the

\begin{align*} {\rm Surface\ Area} &= 2\int_0^{\pi/2}\dee{\theta}\int_0^{a\cos\theta}\dee{r}\ r \frac{a}{\sqrt{a^2-r^2}}\\ &= 2a\int_0^{\pi/2}\dee{\theta}\ \Big[-\sqrt{a^2-r^2}\,\Big]_0^{a\cos\theta}\\ &= 2a\int_0^{\pi/2}\dee{\theta}\ \big[a-a\sin\theta\big]\\ &= 2a^2\Big[\theta+\cos\theta\Big]_0^{\pi/2} =a^2[\pi-2] \end{align*}

3.5 Triple Integrals

Exercises

3.5.1.

Solution.

$\dblInt_R \sqrt{b^2-y^2}\,\dee{x}\,\dee{y} =\tripInt_V \dee{x}\,\dee{y}\,\dee{z}\text{,}$ where

\begin{align*} V&=\Set{(x,y,z)} {0\le z\le\sqrt{b^2-y^2},\ 0\le x\le a,\ 0\le y\le b}\\ &=\Set{(x,y,z)} {y^2+z^2\le b^2,\ 0\le x\le a,\ y\ge 0,\ z\ge 0} \end{align*}

Now $y^2+z^2\le b^2$ is a cylinder of radius $b$ centered on the $x$--axis and the part of $y^2+z^2\le b^2\text{,}$ with $y\ge0,\ z\ge0$ is one quarter of this cylinder. It has cross--sectional area $\frac{1}{4}\pi b^2\text{.}$ $V$ is the part of this quarter--cylinder with $0\le x\le a\text{.}$ It has length $a$ and cross--sectional area $\frac{1}{4}\pi b^2\text{.}$ So, $\dblInt_R \sqrt{b^2-y^2}\,\dee{x}\,\dee{y}=\frac{1}{4}\pi ab^2\text{.}$

3.5.2. (✳).

Solution.

The mass is

\begin{align*} \int_0^1\dee{x} \int_0^2\dee{y}\int_0^3\dee{z}\ x &= 6\int_0^1 \dee{x}\ x =3 \end{align*}

3.5.3.

Solution.

The domain of integration is

\begin{equation*} V=\Set{(x,y,z)}{x,y,z\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}+\tfrac{z}{c}\le 1} \end{equation*}

In $V\text{,}$ $\tfrac{z}{c}\le 1-\tfrac{x}{a}-\tfrac{y}{b}$ and $x,y\ge0\text{,}$ so the biggest value of $z$ in $V$ is achieved when $x=y=0$ and is $c\text{.}$ Thus, in $V\text{,}$ $z$ runs from $0$ to $c\text{.}$
For each fixed $0\le z\le c\text{,}$ $(x,y)$ takes all values in
\begin{equation*} D_z=\Set{(x,y)}{x,y\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}\le 1-\tfrac{z}{c}} \end{equation*}
The biggest value of $y$ on $D_z$ is achieved when $x=0$ and is $b\left(1-\tfrac{z}{c}\right)\text{.}$ Thus, on $D_z\text{,}$ $y$ runs from $0$ to $b\left(1-\tfrac{z}{c}\right)\text{.}$
For each fixed $0\le z\le c$ and $0\le y \le b\left(1-\tfrac{z}{c}\right)\text{,}$ $x$ runs over
\begin{equation*} D_{y,z}=\Set{x}{0\le x \le a\left(1-\tfrac{y}{b}-\tfrac{z}{c}\right)} \end{equation*}
This is pictured in the second figure below.

So the specified integral is

\begin{align*} \tripInt_R x\ \dee{V} &=\int_0^c \dee{z} \dblInt_{D_z} \dee{x}\,\dee{y}\ x = \int_0^c \dee{z}\int_0^{b(1-{z\over c})}\dee{y}\int_{D_{y,z}}\dee{x}\ x\\ &=\int_0^c \dee{z}\int_0^{b(1-{z\over c})}\dee{y} \int_0^{a(1-{y\over b}-{z\over c})} \dee{x}\ x\\ &=\int_0^c \dee{z}\int_0^{b(1-{z\over c})} \dee{y}\ \frac{a^2}{2} \left(1-\frac{y}{b}-\frac{z}{c}\right)^2\\ &=\int_0^c \dee{z}\ \left[-\frac{a^2b}{6} \left(1-\frac{y}{b}-\frac{z}{c}\right)^3 \right]_0^{b(1-{z\over c})}\\ &=\int_0^c \dee{z}\ \frac{a^2b}{6}\left(1-\frac{z}{c}\right)^3\\ &=\left[-\frac{a^2bc}{24}\left(1-\frac{z}{c}\right)^4 \right]_0^c =\frac{a^2bc}{24} \end{align*}

3.5.4.

Solution.

The domain of integration is

\begin{equation*} R = \Set{(x,y,z)}{0\le x,y,z\le 1,\ z\ge 1-y,\ z\le2-x-y} \end{equation*}

In the figure on the below, the more darkly shaded region is part of $z=1-y$ and the more lightly shaded region is part of $z=2-x-y\text{.}$

In $R\text{,}$ $z$ runs from $0$ (for example $(0,1,0)$ is in $R$) to $1$ (for example $(0,0,1)$ is in $R$).
For each fixed $0\le z\le 1\text{,}$ $(x,y)$ runs over

\begin{equation*} D_z = \Set{(x,y)}{0\le x,y\le 1,\ y\ge 1-z,\ x+y\le2-z} \end{equation*}

Here is a sketch of a top view of $D_z\text{.}$

On $D_z\text{,}$ $y$ runs from $1-z$ to $1\text{.}$
For each fixed $0\le z\le 1$ and $1-z\le y\le 1\text{,}$ $x$ runs from $0$ to $2-y-z\text{.}$

So the specified integral is

\begin{align*} \tripInt_R y\ \dee{V} &=\int_0^1\dee{z}\dblInt_{D_z} \dee{x}\dee{y}\ y =\int_0^1 \dee{z}\int_{1-z}^1 \dee{y}\int_0^{2-y-z} \dee{x}\ y\\ &=\int_0^1 \dee{z}\int_{1-z}^1 \dee{y}\ y(2-y-z)\\ &=-\int_0^1 \dee{z}\int_z^0 du\ (1-u)(1+u-z)\qquad\text{ where } u=1-y\\ &=\int_0^1 \dee{z}\int_0^z du\ (1-u^2-z+uz)\\ &=\int_0^1 \dee{z}\ \big(z-\frac{z^3}{3}-z^2+\frac{z^3}{2})\\ &=\frac{1}{2}-\frac{1}{12}-\frac{1}{3}+\frac{1}{8} =\frac{5}{24} \end{align*}

3.5.5.

Solution.

(a) The domain of integration is

\begin{align*} V &= \Set{(x,y,z)}{0\le z\le 1,\ 0\le y\le 1-z,\ 0\le x\le 1-z}\\ &= \Set{(x,y,z)}{ x,y,z\ge 0,\ x+z\le 1,\ y+z\le 1 } \end{align*}

This is sketched in the figure below. The front face is $x+z=1$ and the lightly shaded right face is $y+z=1\text{.}$

In $V\text{,}$

$x$ takes all values between $0$ and $1\text{.}$
For each fixed $0\le x\le 1\text{,}$ $(y,z)$ takes all values in

\begin{equation*} D_x=\Set{(y,z)}{y,z\ge 0,\ z\le 1-x,\ y+z\le 1} \end{equation*}

Here is a sketch of $D_x\text{.}$
Looking at the sketch above, we see that, on $D_x\text{,}$ $y$ runs from $0$ to $1$ and
- for each fixed $y$ between $0$ and $x\text{,}$ $z$ runs from $0$ to $1-x$ and
- for each fixed $y$ between $x$ and $1\text{,}$ $z$ runs from $0$ to $1-y$

So the integral is, in the new order,

\begin{align*} \tripInt_V f(x,y,z)\ \dee{V} &= \int_0^1\dee{x}\dblInt_{D_x} \dee{y}\,\dee{z}\ f(x,y,z)\\ &= \int_0^1\dee{x}\int_0^{x}\dee{y}\int_0^{1-x}\hskip-10pt \dee{z}\hskip3pt f(x,y,z)\\ &\hskip0.5in+\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{1-y}\hskip-10pt \dee{z}\hskip3pt f(x,y,z) \end{align*}

(b) The domain of integration is

\begin{align*} V &= \Set{(x,y,z)}{0\le z\le 1,\ \sqrt{z}\le y\le 1,\ 0\le x\le y}\\ &= \Set{(x,y,z)}{0\le z\le y^2,\ 0\le x\le y\le 1} \end{align*}

In this region, $x$ takes all values between $0$ and $1\text{.}$ For each fixed $x$ between $0$ and $1\text{,}$ $(y,z)$ takes all values in

\begin{equation*} D_x=\Set{(y,z)}{0\le z\le y^2,\ x\le y\le 1} \end{equation*}

Here is a sketch of $D_x\text{.}$

In the new order, the integral is

\begin{equation*} \int_0^1\dee{x}\dblInt_{D_x} \dee{y}\,\dee{z}\ f(x,y,z) =\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{y^2}\!\! \dee{z}\ f(x,y,z) \end{equation*}

3.5.6. (✳).

Solution.

(a) In the domain of integration for the given integral

$y$ runs from $-1$ to $1\text{,}$ and
for each fixed $y$ in that range $z$ runs from $0$ to $1-y^2\text{,}$ and
for each fixed $y$ and $z$ as above, $x$ runs from $0$ to $2-y-z\text{.}$

That is,

\begin{equation*} E = \Set{(x,y,z)}{-1\le y\le 1,\ 0\le z\le 1-y^2,\ 0\le x\le 2-y-z} \end{equation*}

Each constant $x$ cross--section of the surface $z=1-y^2$ is an upside down parabola. So the surface $z=1-y^2$ consists of a bunch of copies of the parabola $z=1-y^2$ stacked front to back. The top figure below provides a sketch of $z=1-y^2\text{.}$
The surface $x = 2-y-z\text{,}$ or equivalently, $x+y+z=2$ is a plane. It passes through the points $(2,0,0)\text{,}$ $(0,2,0)$ and $(0,0,2)\text{.}$ It is sketched in the bottom figure below. We know that our domain of integration extends to $y=-1\text{,}$ so we have chosen to include in the sketch the part of the plane in $x\ge 0\text{,}$ $y\ge-1\text{,}$ $z\ge 0\text{.}$

The domain $E$ is constructed by using the plane $x+y+z=2$ to chop the front off of the “tunnel” $0\le z\le 1-y^2\text{.}$ It is outlined in red in the figure below.

(b) We are to change the order of integration so that the outside integral is over $y$ (the same as the given integral), the middle integral is over $x\text{,}$ and the inside integral is over over $z\text{.}$

We still have $y$ running from $-1$ to $1\text{.}$
For each fixed $y$ in that range, $(x,z)$ runs over
\begin{equation*} E_y=\Set{(x,z)}{0\le z\le 1-y^2,\ 0\le x+z\le 2-y} \end{equation*}
The biggest value of $x$ in $E_y$ is $2-y\text{.}$ It is achieved when $z=0\text{.}$ You can also see this in the figure below. The shaded region in that figure is $E_y\text{.}$
For each fixed $x$ and $y$ as above, $z$ runs over
\begin{equation*} E_{x,y} = \Set{z}{0\le z\le 1-y^2,\ 0\le z\le 2-x-y} \end{equation*}
That is, $z$ runs from $0$ to the smaller of $1-y^2$ and $2-x-y\text{.}$ Note that $1-y^2\le 2-x-y$ if and only if $x\le 1+y^2-y\text{.}$
So if $0\le x\le 1+y^2-y\text{,}$ $z$ runs from $0$ to $1-y^2$ and if $1+y^2-y\le x\le 2-y\text{,}$ $z$ runs from $0$ to $2-x-y\text{.}$

So the integral is

\begin{align*} &\int_{y=-1}^{y=1} \int_{x=0}^{x=1+y^2-y} \int_{z=0}^{z=1-y^2} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y}\\ &\hskip1in +\int_{y=-1}^{y=1} \int_{x=1+y^2-y}^{x=2-y} \int_{z=0}^{z=2-x-y} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \end{align*}

3.5.7. (✳).

Solution.

(a) In the given integral $J\text{,}$

$x$ runs from $0$ to $1\text{,}$
for each fixed $x$ in that range, $z$ runs from 0 to $1-\frac{x}{2}\text{,}$ and
for each fixed $x$ and $z$ as above, $y$ runs from $0$ to $4-2x-4z\text{.}$

\begin{gather*} E = \Set{(x,y,z)}{0\le x\le 1,\ 0\le z\le 1-\tfrac{x}{2},\ 0 \le y\le 4-2x-4z} \end{gather*}

Notice that the condition $y\le 4-2x-4z$ can be rewritten as $z\le 1-\frac{x}{2} -\frac{y}{4}\text{.}$ When $y\ge 0\text{,}$ this implies that $z\le 1-\frac{x}{2}\text{,}$ so that we can drop the condition $z\le 1-\frac{x}{2}$ from our description of $E\text{:}$

\begin{gather*} E = \Set{(x,y,z)}{0\le x\le 1,\ 0 \le y\le 4-2x-4z,\ z\ge 0} \end{gather*}

First, we figure out what $E$ looks like. The plane $2x+y+4z=4$ intersects the $x$--, $y$-- and $z$--axes at $(2,0,0)\text{,}$ $(0,4,0)$ and $(0,0,1)\text{,}$ respectively. That plane is shown in the first sketch below. The set of points $\Set{(x,y,z)}{x,y,z\ge 0,\ y\le 4-2x-4z}$ is outlined with heavy lines.

So it only remains to impose the condtion $x\le 1\text{,}$ which chops off the front bit of the tetrahedron. This is done in the second sketch above. Here is a cleaned up sketch of $E\text{.}$

(b) We are to reorder the integration so that the outside integral is over $y\text{,}$ the middle integral is over $x\text{,}$ and the inside integral is over $z\text{.}$ Looking at the figure below,

we see that

$y$ runs from $0$ to $4\text{,}$ and
for each fixed $y$ in that range, $(x,z)$ runs over
\begin{equation*} \Set{(x,z)}{0\le x\le 1,\ 2x+4z\le 4-y,\ z\ge 0} \end{equation*}
for each fixed $y$ between $0$ and $2$ (as in the left hand shaded bit in the figure above)
- $x$ runs from $0$ to $1\text{,}$ and then
- for each fixed $x$ in that range, $z$ runs from $0$ to $\frac{4-2x-y}{4}\text{.}$
for each fixed $y$ between $2$ and $4$ (as in the right hand shaded bit in the figure above)
- $x$ runs from $0$ to $\frac{4-y}{2}$ (the line of intersection of the plane $2x+y+4z=4$ and the $xy$--plane is $z=0\text{,}$ $2x+y=4$), and then
- for each fixed $x$ in that range, $z$ runs from $0$ to $\frac{4-2x-y}{4}\text{.}$

So the integral

\begin{align*} J &= \int_{y=0}^{y=2} \int_{x=0}^{x=1} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y}\\ &\hskip0.5in+\int_{y=2}^{y=4} \int_{x=0}^{x=\frac{4-y}{2}} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}

3.5.8. (✳).

Solution.

Let’s use $V$ to denote the domain of integration for the given integral. On $V$

$x$ runs from $0$ to $1\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $\sqrt{x}$ to $1\text{.}$ In particular $0\le y\le 1\text{.}$ We can rewrite $y=\sqrt{x}$ as $x=y^2$ (with $y\ge 0$).
For each fixed $x$ and $y$ as above, $z$ runs from $0$ to $1-y\text{.}$

\begin{align*} V &= \Set{(x,y,z)}{0\le x\le 1,\ \sqrt{x}\le y\le 1,\ 0\le z\le 1-y}\\ &= \Set{(x,y,z)}{x,z\ge 0,\ x\le 1,\ y\ge \sqrt{x},\ y\le 1,\ z\le 1-y} \end{align*}

Outside integral is with respect to $x$: We have already seen that $0\le x\le 1$ and that, for each fixed $x$ in that range, $(y,z)$ runs over

\begin{equation*} V_x =\Set{(y,z)}{\sqrt{x}\le y\le 1,\ 0\le z\le 1-y} \end{equation*}

Here are two sketches of $V_x\text{.}$ The sketch on the left shows a vertical strip as was used in setting up the integral given in the statement of this problem.

To reverse the order of the $y$-- and $z$--integrals we use horizontal strips as in the figure on the right above. Looking at that figure, we see that, on $V_x\text{,}$

$z$ runs from $0$ to $1-\sqrt{x}\text{,}$ and
for each fixed $z$ in that range, $y$ runs from $\sqrt{x}$ to $1-z\text{.}$

\begin{align*} I &= \int_0^1\dee{x} \int_0^{1-\sqrt{x}}\dee{z} \int_{\sqrt{x}}^{1-z}\dee{y} \ f(x,y,z)\\ &= \int_0^1 \int_0^{1-\sqrt{x}} \int_{\sqrt{x}}^{1-z} \ f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \end{align*}

Outside integral is with respect to $y$: Looking at the figures above we see that, for each $0\le x\le 1\text{,}$ $y$ runs from $\sqrt{x}$ to $1$ on $V_x\text{.}$ As $x$ runs from $0$ to $1$ in $V\text{,}$ we have that $\sqrt{x}$ also runs from $0$ to $1$ on $V\text{,}$ so that $y$ runs from $0$ to $1$ on $V\text{.}$ Reviewing the definition of $V\text{,}$ we see that, for each fixed $0\le y\le 1\text{,}$ $(x,z)$ runs over

\begin{equation*} V_y= \Set{(x,z)}{0\le x\le y^2,\ 0\le z\le 1-y} \end{equation*}

Here are two sketches of $V_y\text{.}$

Looking at the figure on the left (with the vertical strip), we see that, on $V_y\text{,}$

$x$ runs from $0$ to $y^2\text{,}$ and
for each fixed $x$ in that range, $z$ runs from $0$ to $1-y\text{.}$

\begin{equation*} I = \int_0^1 \dee{y} \int_0^{y^2}\dee{x} \int_0^{1-y}\dee{z}\ f(x,y,z) = \int_0^1 \int_0^{y^2}\int_0^{1-y}f(x,y,z)\ \dee{z}\,\dee{x}\, \dee{y} \end{equation*}

Looking at the figure on the right above (with the horizontal strip), we see that, on $V_y\text{,}$

$z$ runs from $0$ to $1-y\text{.}$
for each fixed $z$ in that range, $x$ runs from $0$ to $y^2\text{.}$

\begin{equation*} I = \int_0^1 \dee{y} \int_0^{1-y}\dee{z} \int_0^{y^2}\dee{x}\ f(x,y,z) = \int_0^1 \int_0^{1-y} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{z}\,\dee{y} \end{equation*}

Outside integral is with respect to $z$: Looking at the sketches of $V_x$ above we see that, for each $0\le x\le 1\text{,}$ $z$ runs from $0$ to $1-\sqrt{x}$ on $V_x\text{.}$ As $x$ runs from $0$ to $1$ in $V\text{,}$ $1-\sqrt{x}$ also runs between $0$ to $1$ on $V\text{,}$ so that $z$ runs from $0$ to $1$ on $V\text{.}$ Reviewing the definition of $V\text{,}$ we see that, for each fixed $0\le z\le 1\text{,}$ $(x,y)$ runs over

\begin{equation*} V_z= \Set{(x,y)}{0\le x\le y^2,\ \sqrt{x}\le y\le 1-z} \end{equation*}

Here are two sketches of $V_z\text{.}$

Looking at the figure on the left (with the vertical strip), we see that, on $V_z\text{,}$

$x$ runs from $0$ to $(1-z)^2\text{,}$ and
for each fixed $x$ in that range, $y$ runs from $\sqrt{x}$ to $1-z\text{.}$

\begin{align*} I &= \int_0^1 \dee{z} \int_0^{(1-z)^2}\dee{x} \int_{\sqrt{x}}^{1-z}\dee{y}\ f(x,y,z)\\ &= \int_0^1 \int_0^{(1-z)^2} \int_{\sqrt{x}}^{1-z}f(x,y,z)\ \dee{y}\,\dee{x}\, \dee{z} \end{align*}

Looking at the figure on the right above (with the horizontal strip), we see that, on $V_z\text{,}$

$y$ runs from $0$ to $1-z\text{.}$
for each fixed $y$ in that range, $x$ runs from $0$ to $y^2\text{.}$

\begin{equation*} I = \int_0^1 \dee{z} \int_0^{1-z}\dee{y} \int_0^{y^2}\dee{x}\ f(x,y,z) = \int_0^1 \int_0^{1-z} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{y}\,\dee{z} \end{equation*}

Summary: We have found that

\begin{alignat*}{2} I&= \int_0^1 \int_{\sqrt{x}}^1 \int_0^{1-y} f(x,y,z)\,\dee{z}\,\dee{y}\,\dee{x} &&=\int_0^1 \int_0^{1-\sqrt{x}}\!\! \int_{\sqrt{x}}^{1-z} \!\!f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x}\\ &=\int_0^1 \int_0^{y^2}\int_0^{1-y}f(x,y,z)\ \dee{z}\,\dee{x}\, \dee{y} &&=\int_0^1 \int_0^{1-y} \int_0^{y^2}\!\! f(x,y,z)\ \dee{x}\, \dee{z}\,\dee{y}\\ &=\int_0^1\!\int_0^{(1-z)^2}\!\! \int_{\sqrt{x}}^{1-z}\!\!f(x,y,z)\, \dee{y}\,\dee{x}\, \dee{z} &&=\int_0^1 \int_0^{1-z}\!\! \int_0^{y^2} f(x,y,z) \dee{x}\, \dee{y}\,\dee{z} \end{alignat*}

3.5.9. (✳).

Solution.

First we have to get some idea as to what $E$ looks like. Here is a sketch.

We are going to need the equation of the plane that contains the points $(-1,0,0)\text{,}$ $(0,-2,0)$ and $(0,0,3)\text{.}$ This plane does not contain the origin and so has an equation of the form $ax+by+cz=1\text{.}$

$(-1,0,0)$ lies on the plane $ax+by+cz=1$ if and only if $a(-1)+b(0)+c(0)=1\text{.}$ So $a=-1\text{.}$
$(0,-2,0)$ lies on the plane $ax+by+cz=1$ if and only if $a(0)+b(-2)+c(0)=1\text{.}$ So $b=-\frac{1}{2}\text{.}$
$(0,0,3)$ lies on the plane $ax+by+cz=1$ if and only if $a(0)+b(0)+c(3)=1\text{.}$ So $c=\frac{1}{3}\text{.}$

So the plane that contains the points $(-1,0,0)\text{,}$ $(0,-2,0)$ and $(0,0,3)$ is $-x-\frac{y}{2}+\frac{z}{3}=1\text{.}$

We can now get a detailed mathematical description of $E\text{.}$ A point $(x,y,z)$ is in $E$ if and only if

$(x,y,z)$ lies above the $xy$--plane, i.e. $z\ge 0\text{,}$ and
$(x,y,z)$ lies to the left of the $xz$--plane, i.e. $y\le 0\text{,}$ and
$(x,y,z)$ lies behind the $yz$--plane, i.e. $x\le 0\text{,}$ and
$(x,y,z)$ lies on the same side of the plane $-x-\frac{y}{2}+\frac{z}{3}=1$ as the origin. That is $-x-\frac{y}{2}+\frac{z}{3}\le 1\text{.}$ (Go ahead and check that $(0,0,0)$ obeys this inequality.)

\begin{equation*} E=\Set{(x,y,z)}{x\le 0,\ y\le 0,\ z\ge 0,\ -x-\tfrac{y}{2}+\tfrac{z}{3}\le 1} \end{equation*}

(a) Note that we want the outside integral to be the $x$--integral. On $E$

$x$ runs from $-1$ to $0$ and
for each fixed $x$ in that range $(y,z)$ runs over

\begin{gather*} E_x=\Set{(y,z)}{y\le 0,\ z\ge 0,\ -\tfrac{y}{2}+\tfrac{z}{3}\le 1+x} \end{gather*}

Here is a sketch of $E_x\text{.}$
On $E_x\text{,}$ $y$ runs from $-2(1+x)$ to $0$ and
for each fixed such $y\text{,}$ $z$ runs from $0$ to $3(1+x+y/2)$

\begin{gather*} I = \int_{x=-1}^{x=0}\int_{y=-2(1+x)}^{y=0}\int_{z=0}^{z=3(1+x+y/2)} f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x} \end{gather*}

(b) This time we want the outside integral to be the $z$--integral. Looking back at the sketch of $E\text{,}$ we see that, on $E\text{,}$

$z$ runs from $0$ to $3$ and
for each fixed $z$ in that range $(x,y)$ runs over

\begin{gather*} E_z=\Set{(x,y)}{x\le 0,\ y\le 0,\ -x-\tfrac{y}{2}\le 1-\tfrac{z}{3}} \end{gather*}

Here is a sketch of $E_z\text{.}$
On $E_z\text{,}$ $x$ runs from $-(1-z/3)$ to $0$ and
for each fixed such $x\text{,}$ $y$ runs from $-2(1+x-z/3)$ to $0$

\begin{gather*} I = \int_{z=0}^{z=3}\int_{x=-(1-z/3)}^{x=0}\int_{y=-2(1+x-z/3)}^{y=0} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{gather*}

3.5.10. (✳).

Solution.

The plane $x+y+z=1$ intersects the coordinate plane $z=0$ along the line $x+y=1\text{,}$ $z=0\text{.}$ So

\begin{align*} T&=\Set{(x,y,z)}{x\ge 0,\ y\ge0,\ x+y\le 1,\ 0\le z\le 1-x-y}\\ &=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 1-x,\ 0\le z\le 1-x-y} \end{align*}

and

\begin{align*} K&=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\int_0^{1-x-y}\dee{z}\ \frac{1}{ (1 + x + y + z)^4}\\ &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \left[-\frac{1}{ 3(1 + x + y + z)^3}\right]_{z=0}^{z=1-x-y}\\ &=\frac{1}{3}\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \left[\frac{1}{(1 + x + y)^3} - \frac{1}{2^3}\right]\\ &=\frac{1}{3}\int_0^1\dee{x}\ \left[-\frac{1}{2(1 + x + y)^2} - \frac{y}{2(4)}\right]_{y=0}^{y=1-x}\\ &=\frac{1}{6}\int_0^1\dee{x}\ \left[\frac{1}{(1 + x)^2}-\frac{1}{2^2}- \frac{1-x}{4}\right]\\ &=\frac{1}{6}\int_0^1\dee{x}\ \left[\frac{1}{(1 + x)^2} -\frac{1}{2} + \frac{x}{4}\right]\\ &=\frac{1}{6} \left[-\frac{1}{1 + x} -\frac{x}{2} + \frac{x^2}{8}\right]_{x=0}^{x=1}\\ &=\frac{1}{6} \left[1-\frac{1}{2} -\frac{1}{2} + \frac{1}{8}\right]\\ &=\frac{1}{48} \end{align*}

3.5.11. (✳).

Solution.

Note that the planes $z = x + y$ and $z = 2$ intersect along the line $x+y=2\text{,}$ $z=2\text{.}$

\begin{align*} E &=\Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ x+y\le 2,\ x+y\le z\le 2}\\ &= \Set{(x,y,z)}{0\le x\le 2,\ 0\le y\le 2-x,\ x+y\le z\le 2} \end{align*}

and the mass of $E$ is

\begin{align*} \tripInt_E \rho(x,y,z)\,\dee{V} &=\int_0^2\dee{x} \int_0^{2-x}\dee{y} \int_{x+y}^2\dee{z}\ z\\ &=\frac{1}{2}\int_0^2\dee{x} \int_0^{2-x}\dee{y} \ \big[4-(x+y)^2\big]\\ &=\frac{1}{2}\int_0^2\dee{x}\ \left[4(2-x)-\frac{\big(x+(2-x)\big)^3-x^3}{3}\right]\\ &=\frac{1}{2} \left[4(2)(2)-2(2)^2 -\frac{8}{3}(2)+\frac{2^4}{12} \right]\\ &=\frac{1}{2} \left[8 -\frac{16}{3}+\frac{4}{3} \right]\\ &=2 \end{align*}

3.5.12. (✳).

Solution.

First, we need to develop an understanding of what $E$ looks like. Here are sketches of the parabolic cylinder $y=x^2\text{,}$ on the left, and the plane $y+z=1\text{,}$ on the right.

$E$ is constructed by using the plane $y+z=1$ to chop the top off of the parabolic cylinder $y=x^2\text{.}$ Here is a sketch.

\begin{equation*} E=\Set{(x,y,z)}{0\le x\le 1,\ x^2\le y\le 1,\ 0\le z\le 1-y} \end{equation*}

and the integral

\begin{align*} \tripInt_E x\ \dee{V} &=\int_0^1\dee{x}\int_{x^2}^1\dee{y}\int_0^{1-y}\dee{z}\ x\\ &=\int_0^1\dee{x}\int_{x^2}^1\dee{y}\ x(1-y)\\ &=\int_0^1\dee{x}\ x\left[y-\frac{y^2}{2}\right]_{x^2}^1\\ &=\int_0^1\dee{x}\ \left[\frac{x}{2} -x^3+\frac{x^5}{2}\right]\\ &=\frac{1}{4}-\frac{1}{4}+\frac{1}{12}\\ &=\frac{1}{12} \end{align*}

3.5.13. (✳).

Solution.

First, we need to develop an understanding of what $E$ looks like. Here are sketches of the plane $x+y=1\text{,}$ on the left, and of the “tower” bounded by the coordinate planes $x=0\text{,}$ $y=0\text{,}$ $z=0$ and the plane $x+y=1\text{,}$ on the right.

Now here is the parabolic cylinder $z=y^2$ on the left. $E$ is constructed by using the parabolic cylinder $z=y^2$ to chop the top off of the tower $x\ge 0\text{,}$ $y\ge 0\text{,}$ $z\ge 0\text{,}$ $x+y\le 1\text{.}$ The figure on the right is a sketch.

\begin{equation*} E=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 1-x,\ 0\le z\le y^2} \end{equation*}

and the integral

\begin{align*} \tripInt_E z\ \dee{V} &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\int_0^{y^2}\dee{z}\ z\\ &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \frac{y^4}{2}\\ &=\int_0^1\dee{x}\ \frac{(1-x)^5}{10}\\ &=\left[-\frac{(1-x)^6}{60}\right]_0^1\\ &=\frac{1}{60} \end{align*}

3.5.14. (✳).

Solution.

The integral

\begin{align*} \tripInt_R yz^2 e^{-xyz}\ \dee{V} &=\int_0^3\dee{z}\int_0^2\dee{y}\int_0^1\dee{x}\ yz^2 e^{-xyz}\\ &=\int_0^3\dee{z}\int_0^2\dee{y}\ \Big[-z e^{-xyz}\Big]_{x=0}^{x=1}\\ & =\int_0^3\dee{z}\int_0^2\dee{y}\ \Big[z-z e^{-yz}\Big]\\ &=\int_0^3\dee{z}\ \Big[zy+ e^{-yz}\Big]_{y=0}^{y=2}\\ &=\int_0^3\dee{z}\ \Big[2z+ e^{-2z}-1\Big]\\ &=\left[z^2-\frac{1}{2} e^{-2z}-z\right]_0^3 =\frac{13}{2}-\frac{e^{-6}}{2} \end{align*}

3.5.15. (✳).

Solution.

(a) Each constant $y$ cross section of $z=1-x^2$ is an upside down parabola. So the surface is a bunch of upside down parabolas stacked side by side. The figure on the left below is a sketch of the part of the surface with $y\ge 0$ and $z\ge 0$ (both of which conditions will be required in part (b)).

(b) The figure on the right above is a sketch of the plane $y=z\text{.}$ It intersects the surface $z=1-x^2$ in the solid blue sloped parabolic curve in the figure below.

Observe that, on the curve $z=1-x^2\text{,}$ $z=y\text{,}$ we have $y=1-x^2\text{.}$ So that when one looks at the solid $E$ from high on the $z$--axis, one sees

\begin{equation*} \Set{(x,y)}{0\le y\le 1-x^2} \end{equation*}

The $y=1-x^2$ boundary of that region is the dashed blue line in the $xy$--plane in the figure above. So

\begin{equation*} E = \Set{(x,y,z)}{-1\le x\le 1,\ 0\le y\le 1-x^2,\ y\le z\le 1-x^2} \end{equation*}

and the integral

\begin{equation*} \tripInt_E f(x,y,z)\,\dee{V} =\int_{-1}^1\dee{x}\int_0^{1-x^2}\dee{y}\int_y^{1-x^2}\dee{z}\ f(x,y,z) \end{equation*}

3.5.16. (✳).

Solution.

In the integral $J\text{,}$

$x$ runs from $0$ to $1\text{.}$ In inequalities, $0\le x\le 1\text{.}$
Then, for each fixed $x$ in that range, $y$ runs from $0$ to $x\text{.}$ In inequalities, $0\le y\le x\text{.}$
Then, for each fixed $x$ and $y$ in those ranges, $z$ runs from $0$ to $y\text{.}$ In inequalities, $0\le z\le y\text{.}$

These inequalties can be combined into

\begin{equation*} 0\le z\le y\le x\le 1 \tag{$*$} \end{equation*}

We wish to reverse the order of integration so that the $z$--integral is on the outside, the $y$--integral is in the middle and the $x$--integral is on the inside.

The smallest $z$ compatible with $(*)$ is $z=0$ and the largest $z$ compatible with $(*)$ is $z=1$ (when $x=y=z=1$). So $0\le z\le 1\text{.}$
Then, for each fixed $z$ in that range, $(x,y)$ run over $z\le y\le x\le 1\text{.}$ In particular, the smallest allowed $y$ is $y=z$ and the largest allowed $y$ is $y=1$ (when $x=y=1$). So $z\le y\le 1\text{.}$
Then, for each fixed $y$ and $z$ in those ranges, $x$ runs over $y\le x\le 1\text{.}$

\begin{equation*} J = \int_0^1 \int_z^1\int_y^1 f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{equation*}

3.5.17. (✳).

Solution.

The hard part of this problem is figuring out what $E$ looks like. First here are separate sketches of the plane $x=3$ and the plane $z=2x$ followed by a sketch of the two planes together.

Next for the parabolic cylinder $z=y^2\text{.}$ It is a bunch of parabolas $z=y^2$ stacked side by side along the $x$--axis. Here is a sketch of the part of $z=y^2$ in the first octant.

Finally, here is a sketch of the part of $E$ in the first octant. $E$ does have a second half gotten from the sketch by reflecting it in the $xz$--plane, i.e. by replacing $y\rightarrow-y\text{.}$

The question doesn’t specify on which side of the three surfaces $E$ lies. When in doubt take the finite region bounded by the given surfaces. That’s what we have done.

\begin{equation*} E = \Set{(x,y,z)}{x\le 3,\ -\sqrt{6}\le y\le \sqrt{6},\ y^2\le z\le 2x} \end{equation*}

Order $\dee{z}\,\dee{x}\,\dee{y}\text{:}$ On $E\text{,}$ $y$ runs from $-\sqrt{6}$ to $\sqrt{6}\text{.}$ For each fixed $y$ in this range $(x,z)$ runs over $E_y=\Set{(x,z)}{x\le 3,\ y^2\le z\le 2x}\text{.}$ Here is a sketch of $E_y\text{.}$

From the sketch

\begin{equation*} E_y = \Set{(x,z)}{y^2/2\le x\le 3,\ y^2\le z\le 2x} \end{equation*}

and the integral is

\begin{equation*} \int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{x=y^2/2}^{x=3}\int_{z=y^2}^{z=2x} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{equation*}

Order $\dee{x}\,\dee{z}\,\dee{y}\text{:}$ Also from the sketch of $E_y$ above

\begin{equation*} E_y = \Set{(x,z)}{y^2\le z\le 6,\ z/2\le x\le 3} \end{equation*}

and the integral is

\begin{equation*} \int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{z=y^2}^{z=6}\int_{x=z/2}^{x=3} f(x,y,z)\ \dee{x}\,\dee{z}\,\dee{y} \end{equation*}

Order $\dee{y}\,\dee{x}\,\dee{z}\text{:}$ From the sketch of the part of $E$ in the first octant, we see that, on $E\text{,}$ $z$ runs from $0$ to $6\text{.}$ For each fixed $z$ in this range $(x,y)$ runs over

\begin{align*} E_z&=\Set{(x,y)}{x\le 3,\ -\sqrt{6}\le y\le \sqrt{6},\ y^2\le z\le 2x}\\ &=\Set{(x,y)}{z/2\le x\le 3,\ y^2\le z}\\ &=\Set{(x,y)}{z/2\le x\le 3,\ -\sqrt{z}\le y\le \sqrt{z} } \end{align*}

So the integral is

\begin{equation*} \int_{z=0}^{z=6}\int_{x=z/2}^{x=3}\int_{y=-\sqrt{z}}^{y=\sqrt{z}} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{equation*}

3.5.18. (✳).

Solution.

(a) The region $E$ is

\begin{align*} E &= \Set{(x,y,z)}{x^2+y^2\le 1,\ -1\le z\le y} \end{align*}

Here is are sketches, one without axes and one with axes, of the front half of $E\text{,}$ outlined in red.

The integral

\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{x^2+y^2\le 1}\dee{x}\,\dee{y}\int_{-1}^y\dee{z}\ f(x,y,z)\\ &=\int_{-1}^1\dee{y}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\dee{x} \int_{-1}^y\dee{z}\ f(x,y,z)\\ &=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_{-1}^y f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}

(b) Here is a sketch of (the front half of) a constant $z$ slice of $E\text{.}$

Note that

in $E\text{,}$ $z$ runs from $-1$ to $1\text{.}$
Once $z$ has been fixed, $x$ and $y$ must obey $x^2+y^2\le 1\text{,}$ $z\le y\le 1$

\begin{align*} E &= \Set{(x,y,z)}{-1\le z\le 1,\ z\le y\le 1,\ -\sqrt{1-y^2}\le x\le\sqrt{1-y^2}} \end{align*}

and

\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1\dee{z}\int_z^1\dee{y} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\dee{x}\ f(x,y,z)\\ &=\int_{-1}^1\int_z^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{align*}

Note that

in $E\text{,}$ $x$ runs from $-1$ to $1\text{.}$
Once $x$ has been fixed, $y$ and $z$ must obey
\begin{equation*} -\sqrt{1-x^2}\le y\le\sqrt{1-x^2}\qquad -1\le z\le y \end{equation*}

Here is a sketch.

Note that

$z$ runs from $-1$ to $\sqrt{1-x^2}\text{.}$
For each $z$ between $-1$ and $-\sqrt{1-x^2}\text{,}$ $y$ runs from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}\text{,}$ while
for each $z$ between $-\sqrt{1-x^2}$ and $\sqrt{1-x^2}\text{,}$ $y$ runs from $z$ to $\sqrt{1-x^2}\text{.}$

\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1\dee{x} \int_{-1}^{-\sqrt{1-x^2}}\dee{z} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\dee{y} \ f(x,y,z)\\ &\hskip1in +\int_{-1}^1\dee{x} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\dee{z} \int_z^{\sqrt{1-x^2}}\dee{y} \ f(x,y,z) \end{align*}

\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1 \int_{-1}^{-\sqrt{1-x^2}} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x}\\ &\hskip1in +\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_z^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \end{align*}

3.5.19. (✳).

Solution.

First, we need to develop an understanding of what $E$ looks like. Note that all of the equations $y=0\text{,}$ $y=2\text{,}$ $y+z=3$ and $z=x^2$ are invariant under $x\rightarrow -x\text{.}$ So $E$ is invariant under $x\rightarrow -x\text{,}$ i.e. is symmetric about the $yz$--plane. We’ll sketch the first octant (i.e. $x,y,z\ge 0$) part of $E\text{.}$ There is also a $x\le 0\text{,}$ $y\ge 0\text{,}$ $z\ge 0$ part.

Here are sketches of the plane $y=2\text{,}$ on top, the plane $y+z=3$ in the middle and of the “tunnel” bounded by the coordinate planes $x=0\text{,}$ $y=0\text{,}$ $z=0$ and the planes $y=2\text{,}$ $y+z=3\text{,}$ on the bottom.

Now here is the parabolic cylinder $z=x^2$ on the top. $E$ is constructed by using the parabolic cylinder $z=x^2$ to chop the front off of the tunnel $x\ge 0\text{,}$ $0\le y\le 2\text{,}$ $z\ge 0\text{,}$ $y+z\le 3\text{.}$ The figure on the bottom is a sketch.

\begin{equation*} E=\Set{(x,y,z)}{0\le y\le 2,\ x^2\le z\le 3-y} \end{equation*}

(a) On $E$

$y$ runs from $0$ to $2\text{.}$
For each fixed $y$ in that range, $(x,z)$ runs over $\Set{(x,z)}{x^2\le z\le 3-y}\text{.}$
In particular, the largest $x^2$ is $3-y$ (when $z=3-y$). So $x$ runs from $-\sqrt{3-y}$ to $\sqrt{3-y}\text{.}$
For fixed $y$ and $x$ as above, $z$ runs from $x^2$ to $3-y\text{.}$

so that

\begin{gather*} I=\tripInt_E f(x,y,z)\ \dee{V} = \int_0^2 \int_{-\sqrt{3-y}}^{\sqrt{3-y}} \int_{x^2}^{3-y} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{gather*}

(b) On $E$

$z$ runs from $0$ to $3\text{.}$
For each fixed $z$ in that range, $(x,y)$ runs over
\begin{align*} &\Set{(x,y)}{0\le y\le 2,\ x^2\le z\le 3-y}\\ &\hskip1in=\Set{(x,y)}{0\le y\le 2,\ y\le 3-z,\ x^2\le z} \end{align*}
In particular, $y$ runs from $0$ to the minimum of $2$ and $3-z\text{.}$
So if $0\le z\le 1$ (so that $3-z\ge 2$), $(x,y)$ runs over $\Set{(x,y)}{0\le y\le 2,\ x^2\le z}\text{,}$ while
if $1\le z\le 3\text{,}$ (so that $3-z\le 2$), $(x,y)$ runs over $\Set{(x,y)}{0\le y\le 3-z,\ x^2\le z}\text{,}$

so that

\begin{gather*} I= \int_0^1 \int_0^2 \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} +\int_1^3 \int_0^{3-z} \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{gather*}

$z$ runs from $0$ to $3\text{.}$
For each fixed $z$ in that range, $(x,y)$ runs over
\begin{equation*} \Set{(x,y)}{0\le y\le 2,\ x^2\le z\le 3-y} \end{equation*}
In particular, $y$ runs from $0$ to the minimum of $2$ and $3-z\text{.}$
So if $0\le z\le 1$ (so that $3-z\ge 2$), $(x,y)$ runs over $\Set{(x,y)}{0\le y\le 2,\ x^2\le z}\text{,}$ while
if $1\le z\le 3\text{,}$ (so that $3-z\le 2$), $(x,y)$ runs over $\Set{(x,y)}{0\le y\le 3-z,\ x^2\le z}\text{,}$

so that

\begin{gather*} I= \int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}}\int_0^2 f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} +\int_1^3 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^{3-z} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{gather*}

3.5.20. (✳).

Solution.

The cylinder $y^2+z^2=1$ is centred on the $x$ axis. The part of the cylinder in the first octant intersects the plane $z=0$ in the line $y= 1\text{,}$ intersects to plane $y=0$ in the line $z=1$ and intersects the plane $x=0$ in the quarter circle $y^2+z^2=1\text{,}$ $x=0\text{,}$ $y,z\ge 0\text{.}$ Here is a sketch of $E\text{.}$

Viewed from above, the region $E$ is bounded by the lines $x=0\text{,}$ $y=0\text{,}$ $x+y=2$ and $y=1\text{.}$ This base region is pictured below.

To set up the domain of integration, let’s decompose the base region into horizontal strips as in the figure above. On the base region

$y$ runs from $0$ to $1$ and
for each fixed $y$ between $0$ and $1\text{,}$ $x$ runs from $0$ to $2-y\text{.}$
For each fixed $(x,y)$ in the base region $z$ runs from $0$ to $\sqrt{1-y^2}$

\begin{equation*} E=\Set{(x,y,z)}{0\le y\le 1,\ 0\le x\le 2-y,\ 0\le z\le\sqrt{1-y^2} } \end{equation*}

and

\begin{align*} \tripInt_E z\,\dee{V}&=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\int_0^{\sqrt{1-y^2}}\dee{z}\ z\\ &=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\ \frac{1}{2} z^2\Big|_0^{\sqrt{1-y^2}}\\ &=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\ \frac{1}{2} (1-y^2)\\ &=\int_0^1 \dee{y}\ \frac{1}{2} (1-y^2)(2-y) =\frac{1}{2}\int_0^1 \dee{y}\ \big(2-y-2y^2+y^3\big)\\ &=\frac{1}{2}\left[2-\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right] =\frac{13}{24}\approx 0.5417 \end{align*}

3.5.21. (✳).

Solution.

The planes $x=1\text{,}$ $y=1\text{,}$ $z=1\text{,}$ and $x+y+z=2$ and the region $D$ are sketched below.

And here is a sketch of $D$ without the planes cluttering up the figure.

On $D$

$z$ runs from $0$ to $1$ and
for each fixed $z\text{,}$ between 0 and 1, $(x,y)$ runs over the triangle $T_z$ bounded by $x=1\text{,}$ $y=1$ and $x+y=2-z\text{.}$ Observe that when $z=0\text{,}$ this triangle is just a point (the bottom vertex of the tetrahedron). As $z$ increases, the triangle grows, reaching its maximum size when $z=1\text{.}$

Here is a sketch of $T_z\text{.}$

In setting up the domain of integration, we’ll decompose, for each $0\le z\le 1\text{,}$ $T_z$ into vertical strips as in the figure above. On $T_z$

$x$ runs from $1-z$ to $1$ and
for each fixed $x$ between $1-z$ and $1\text{,}$ $y$ runs from $2-x-z$ to $1$

so that

\begin{align*} \tripInt_D x\,\dee{V} &=\int_0^1\dee{z}\dblInt_{T_z}\dee{x}\,\dee{y}\ x =\int_0^1 \dee{z}\int_{1-z}^1 \dee{x}\int_{2-x-z}^1 \dee{y}\ x\\ &=\int_0^1 \dee{z}\int_{1-z}^1 \dee{x}\ x(x+z-1)\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}x^3+\frac{1}{2} x^2(z-1)\right]_{1-z}^1\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}+\frac{1}{2} (z-1)-\frac{1}{3}(1-z)^3 -\frac{1}{2}(1-z)^2(z-1)\right]\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}+\frac{1}{2} (z-1) -\frac{1}{6}(z-1)^3\right]\\ &=\left[\frac{1}{3}z+\frac{1}{4} (z-1)^2-\frac{1}{24}(z-1)^4\right]_0^1 =\frac{1}{3}-\frac{1}{4}+\frac{1}{24}=\frac{3}{24}\\ &=\frac{1}{8}=0.125 \end{align*}

3.5.22. (✳).

Solution.

(a) Here is a 3d sketch of the region. The coordinates of the labelled corners are

\begin{align*} a&=(0,0,1)& b&=(0,0,0)& c&=(1,0,0)\\ d&=(0,1,1)& f&=(0,2,0)& g&=(1,1,0) \end{align*}

(b) Here is a sketch of the side view of $T\text{,}$ looking down the $y$ axis.

We’ll set up the limits of integration by using it as the base region. We decompose the base region into vertical strips as in the figure above. On the base region

$x$ runs from $0$ to $1$ and
for each fixed $x$ between $0$ and $1\text{,}$ $z$ runs from $0$ to $1-x^2\text{.}$
In $T\text{,}$ for each fixed $(x,y)$ in the base region, $y$ runs from $0$ to $2-x-z\text{.}$

\begin{align*} \tripInt_T x\,\dee{V} &=\int_0^1\dee{x}\int_0^{1-x^2}\dee{z}\int_0^{2-x-z}\dee{y}\ x\\ &=\int_0^1\dee{x}\int_0^{1-x^2}\dee{z}\ (2-x-z)x\cr &=\int_0^1 \dee{x}\ \left[x(2-x)(1-x^2)-\frac{1}{2} x{(1-x^2)}^2\right]\\ &=\int_0^1 \dee{x}\ \left[2x-x^2-2x^3+x^4-\frac{1}{2} x+x^3 -\frac{1}{2} x^5\right]\\ &=\int_0^1 \dee{x}\ \left[\frac{3}{2}x-x^2-x^3+x^4-\frac{1}{2} x^5\right]\\ &=\left[\frac{3}{4}x^2-\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5 -\frac{1}{12}x^6\right]_0^1\\ &=\frac{3}{4}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{12} =\frac{17}{60} \end{align*}

3.6 Triple Integrals in Cylindrical Coordinates
3.6.4 Exercises

3.6.4.1.

Solution.

(a), (b) Since the cylindrical coordinate $r(x,y,z)$ of a point $(x,y,z)$ is the distance, $\sqrt{x^2+y^2}\text{,}$ from $(x,y,z)$ to the $z$-axis, the sets

\begin{align*} \Set{(x,y,z)}{r(x,y,z)=0} &=\Set{(x,y,z)}{x^2+y^2=0}\\ &=\Set{(x,y,z)}{x=y=0}\\ &=\text{the $z$-axis}\\ \Set{(x,y,z)}{r(x,y,z)=1} &=\Set{(x,y,z)}{x^2+y^2=1}\\ &=\text{the cylinder of radius $1$ centred on the $z$-axis} \end{align*}

(c), (d) Since the cylindrical coordinate $\theta(x,y,z)$ of a point $(x,y,z)$ is the angle between the positive $x$-axis and the line from $(0,0,0)$ to $(x,y,0)\text{,}$ the sets

\begin{align*} \Set{(x,y,z)}{\theta(x,y,z)=0} &=\text{the half of the $xz$-plane with } x \gt 0\\ \Set{(x,y,z)}{\theta(x,y,z)=\frac{\pi}{4}} &=\text{the half of the plane $y=x$ with } x \gt 0 \end{align*}

3.6.4.2.

Solution.

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

The cylindrical coordinate $r$ is the distance of the point from the $z$-axis. In particular all points with $r=0$ lie on the $z$-axis (for all values of $\theta$).
The cylindrical coordinate $z$ is the distance of the point from the $xy$-plane. In particular all points with $z=0$ lie on the $xy$-plane.

3.6.4.3.

Solution.

(a) When $\theta=0\text{,}$ $\sin\theta=0$ and $\cos\theta=1\text{,}$ so that the polar coordinates $r=1\text{,}$ $\theta=0\text{,}$ $z=0$ correspond to the Cartesian coordinates

\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = (1\times\cos 0,\ 1\times\sin 0,\ 0) =(1,0,0) \end{equation*}

(b) When $\theta=\frac{\pi}{4}\text{,}$ $\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}\text{,}$ so that the polar coordinates $r=1\text{,}$ $\theta=\frac{\pi}{4}\text{,}$ $z=0$ correspond to the Cartesian coordinates

\begin{align*} (x,y,z) &= (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 0\right)\\ &= \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) \end{align*}

(c) When $\theta=\frac{\pi}{2}\text{,}$ $\sin\theta=1$ and $\cos\theta=0\text{,}$ so that the polar coordinates $r=1\text{,}$ $\theta=\frac{\pi}{2}\text{,}$ $z=0$ correspond to the Cartesian coordinates

\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(1\times\cos \frac{\pi}{2},\ 1\times\sin \frac{\pi}{2},\ 0\right) = (0,1,0) \end{equation*}

(d) When $\theta=\pi\text{,}$ $\sin\theta=0$ and $\cos\theta=-1\text{,}$ so that the polar coordinates $r=0\text{,}$ $\theta=\pi\text{,}$ $z=1$ correspond to the Cartesian coordinates

\begin{equation*} (x,y,z) = (r\cos\theta,r\sin\theta,z) = \left(0\times\cos\pi,\ 0\times\sin\pi,\ 1\right) = (0,0,1) \end{equation*}

(e) When $\theta=\frac{\pi}{4}\text{,}$ $\sin\theta=\cos\theta=\frac{1}{\sqrt{2}}\text{,}$ so that the polar coordinates $r=1\text{,}$ $\theta=\frac{\pi}{4}\text{,}$ $z=1$ correspond to the Cartesian coordinates

\begin{align*} (x,y,z) &= (r\cos\theta,r\sin\theta,z)\\ &= \left(1\times\cos \frac{\pi}{4},\ 1\times\sin \frac{\pi}{4},\ 1\right)\\ &= \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right) \end{align*}

3.6.4.4.

Solution.

(a) The cylindrical coordinates must obey

\begin{equation*} 1=x=r\cos\theta\qquad 1=y=r\sin\theta\qquad 2=z \end{equation*}

So $z=2\text{,}$ $r=\sqrt{1^2+1^2}=\sqrt{2}$ and $\tan\theta=\frac{y}{x} =\frac{1}{1}=1\text{.}$ Recall that $\tan\left(\frac{\pi}{4}+k\pi\right)=1$ for all integers $k\text{.}$ As $(x,y)=(1,1)$ lies in the first quadrant, $0\le\theta\le\frac{\pi}{2}\text{.}$ So $\theta=\frac{\pi}{4}$ (plus possibly any integer multiple of $2\pi$).

(b) The cylindrical coordinates must obey

\begin{equation*} -1=x=r\cos\theta\qquad -1=y=r\sin\theta\qquad 2=z \end{equation*}

So $z=2\text{,}$ $r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$ and $\tan\theta=\frac{y}{x} =\frac{-1}{-1}=1\text{.}$ Recall that $\tan\left(\frac{\pi}{4}+k\pi\right)=1$ for all integers $k\text{.}$ As $(x,y)=(-1,-1)$ lies in the third quadrant, $\pi\le\theta\le\frac{3\pi}{2}\text{.}$ So $\theta=\frac{5\pi}{4}$ (plus possibly any integer multiple of $2\pi$).

\begin{equation*} -1=x=r\cos\theta\qquad \sqrt{3}=y=r\sin\theta\qquad 0=z \end{equation*}

So $z=0\text{,}$ $r=\sqrt{(-1)^2+\big(\sqrt{3}\big)^2}=2$ and $\tan\theta=\frac{y}{x} =\frac{\sqrt{3}}{-1}=-\sqrt{3}\text{.}$ Recall that $\tan\left(\frac{2\pi}{3}+k\pi\right)=-\sqrt{3}$ for all integers $k\text{.}$ As $(x,y)=(-1,\sqrt{3})$ lies in the second quadrant, $\frac{\pi}{2}\le\theta\le\pi\text{.}$ So $\theta=\frac{2\pi}{3}$ (plus possibly any integer multiple of $2\pi$).

(d) The cylindrical coordinates must obey

\begin{equation*} 0=x=r\cos\theta\qquad 0=y=r\sin\theta\qquad 1=z \end{equation*}

So $z=1\text{,}$ $r=\sqrt{0^2+0^2}=0$ and $\theta$ is completely arbitrary.

3.6.4.5.

Solution.

(a) As $x=r\cos\theta$ and $y=r\sin\theta\text{,}$

\begin{equation*} z=2xy \iff z=2r^2\cos\theta\sin\theta = r^2\sin(2\theta) \end{equation*}

(b) As $x=r\cos\theta$ and $y=r\sin\theta\text{,}$

\begin{equation*} x^2+y^2+z^2=1 \iff r^2\cos^2\theta+r^2\sin^2\theta + z^2 =1 \iff r^2+z^2=1 \end{equation*}

\begin{align*} (x-1)^2 + y^2 =1 &\iff (r\cos\theta-1)^2 + (r\sin\theta)^2 =1\\ &\iff r^2\cos^2\theta -2r\cos\theta +1 + r^2\sin^2\theta = 1\\ &\iff r^2=2r\cos\theta \iff r=2\cos\theta\text{ or }r=0\\ &\iff r=2\cos\theta \end{align*}

Note that the solution $r=0$ is included in $r=2\cos\theta$ — just choose $\theta=\frac{\pi}{2}\text{.}$

3.6.4.6.

Solution.

(a) In cylindrical coordinates, the cone $z=2a-\sqrt{x^2+y^2}$ is $z=2a-r$ and the cylinder $x^2+y^2=2ay$ is $r^2=2ar\sin\theta$ or $r=2a\sin\theta\text{.}$ The figures below show the parts of the cone, the cylinder and the intersection, respectively, that are in the first octant.

The specified region is

\begin{equation*} V=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)} {0\le\theta\le\pi,\ r\le 2a\sin\theta,\ 0\le z\le 2a-r} \end{equation*}

By symmetry under $x\rightarrow -x\text{,}$ the full volume is twice the volume in the first octant.

So the

\begin{align*} \text{Volume} &= 2\int_0^{{\pi\over 2}} \dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r \int_0^{2a -r}\dee{z}\\ &= 2\int_0^{{\pi\over 2}} \dee{\theta}\int_0^{2a\sin\theta}\dee{r}\ r(2a-r)\\ &= 2\int_0^{{\pi\over 2}} \dee{\theta}\ \left[4a^3\sin^2\theta-\frac{8a^3}{3}\sin^3\theta\right]\\ &= 8a^3\left[\int_0^{{\pi\over 2}} \dee{\theta}\ \frac{1-\cos(2\theta)}{2} +\frac{2}{3}\int^0_1 \dee{t}\ \big(1-t^2\big)\right] \quad\text{where } t=\cos\theta\\ &= 8a^3\left[\frac{\pi}{4}-\frac{2}{3}\left(1-\frac{1}{3}\right)\right] =a^3\big(2\pi-\frac{32}{9}\big) \end{align*}

For an efficient, sneaky, way to evaluate $\int_0^{{\pi\over 2}} \dee{\theta}\ \sin^2\theta\text{,}$ see Remark 3.3.5.

(b) The domain of integration is

\begin{equation*} V = \Set{(x,y,z)}{-x\le y\le \sqrt{3}x,\ 0\le z \le 1-x^2-y^2} \end{equation*}

Recall that in polar coordinates $\frac{y}{x}=\tan\theta\text{.}$ So the boundaries of the wedge $-x\le y\le \sqrt{3}x\text{,}$ or equivalently $-1\le\frac{y}{x}\le\sqrt{3}\text{,}$ correspond, in polar coordinates, to $\theta=\tan^{-1}(-1)=-\frac{\pi}{4}$ and $\theta=\tan^{-1}\sqrt{3}=\frac{\pi}{3}\text{.}$ In cylindrical coordinates, the paraboloid $z=1-x^2-y^2$ becomes $z=1-r^2\text{.}$ There are $z$’s that obey $0\le z\le 1-r^2$ if and only if $r\le 1\text{.}$ So, in cylindrical coordinates,

\begin{equation*} V = \Set{(r\cos\theta,r\sin\theta,z)} {-\tfrac{\pi}{4}\le \theta\le \tfrac{\pi}{3},\ 0\le r\le 1,\ 0\le z \le 1-r^2} \end{equation*}

and

\begin{align*} \text{Volume} &= \int_{-{\pi\over 4}}^{{\pi\over 3}} \dee{\theta}\int_0^1\dee{r}\ r \int_0^{1 -r^2}\dee{z} = \left(\frac{\pi}{3}+\frac{\pi}{4}\right)\int_0^1\dee{r}\ r(1-r^2)\\ &= \frac{7}{12}\pi\left(\frac{1}{2}-\frac{1}{4}\right) = \frac{7}{48}\pi \end{align*}

\begin{equation*} V = \Set{(x,y,z)}{x^2+y^2\le z\le 2y} \end{equation*}

There are $z$’s that obey $x^2+y^2\le z\le 2y$ if and only if

\begin{align*} x^2+y^2\le 2y &\iff x^2+y^2-2y \le 0 \iff x^2 + (y-1)^2 \le 1 \end{align*}

This disk is sketched in the figure

In cylindrical coordinates,

the bottom, $z=x^2+y^2\text{,}$ is $z=r^2\text{,}$
the top, $z=2y\text{,}$ is $z=2r\sin\theta\text{,}$ and
the disk $x^2+y^2\le 2y$ is $r^2\le 2r\sin\theta\text{,}$ or equivalently $r\le2\sin\theta\text{,}$

so that, looking at the figure above,

\begin{equation*} V = \Set{(r\cos\theta,r\sin\theta,z)}{0\le\theta\le\pi, \ 0\le r\le 2\sin\theta,\ r^2\le z\le 2r\sin\theta} \end{equation*}

By symmetry under $x\rightarrow -x\text{,}$ the full volume is twice the volume in the first octant so that

\begin{align*} \text{Volume} &= 2\int_{0}^{{\pi\over 2}} \dee{\theta}\int_0^{2\sin\theta}\dee{r}\ r\int_{r^2}^{2r\sin\theta}\dee{z}\\ &= 2\int_{0}^{{\pi\over 2}} \dee{\theta}\int_0^{2\sin\theta}\dee{r}\ r(2r\sin\theta-r^2)\\ &= 2\int_{0}^{{\pi\over 2}} \dee{\theta}\ \left(\frac{2^4}{3}-\frac{2^4}{4}\right)\sin^4\theta \end{align*}

To integrate

⁸

For a general discussion of trigonometric integrals see §1.8 in the CLP-2 text. In particular the integral $\int \cos^4 x\ \dee{x}$ is evaluated in Example 1.8.8 in the CLP-2 text.

$\sin^4\theta\text{,}$ we use the double angle formulae $\sin^2 x= \frac{1-\cos(2x)}{2}$ and $\cos^2 x= \frac{1+\cos(2x)}{2}$ to write

\begin{align*} \sin^4\theta &= \left[ \frac{1-\cos(2\theta)}{2} \right]^2\\ &= \frac{1}{4} - \frac{1}{2} \cos(2\theta) + \frac{1}{4}\cos^2(2\theta)\\ &= \frac{1}{4} - \frac{1}{2} \cos(2\theta) + \frac{1}{8}\left(1 + \cos(4\theta) \right)\\ &= \frac{3}{8} - \frac{1}{2} \cos(2\theta) + \frac{1}{8}\cos(4\theta) \end{align*}

\begin{align*} \text{Volume} &= 2\ \frac{2^4}{12} \left[ \frac{3}{8}\theta - \frac{1}{4} \sin(2\theta) + \frac{1}{32}\sin(4\theta) \right]_0^{\pi\over 2} = 2\ \frac{2^4}{12}\ \frac{3}{16}\pi =\frac{\pi}{2} \end{align*}

3.6.4.7. (✳).

Solution.

Note that the paraboloids $z=x^2+y^2$ and $z=2-x^2-y^2$ intersect when $z=x^2+y^2=1\text{.}$ We’ll use cylindrical coordinates. Then $x^2+y^2=r^2\text{,}$ $\dee{V}=r \ \dee{r}\,\dee{\theta}\,\dee{z}\text{,}$ and

\begin{equation*} E=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z))}{0\le r\le 1,\ r^2\le z\le 2-r^2,\ 0\le\theta\le 2\pi} \end{equation*}

so that

\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_0^1\dee{r}\int_{r^2}^{2-r^2}\dee{z}\int_0^{2\pi}\dee{\theta}\ r\ \overbrace{r^3}^{f}\\ &=2\pi \int_0^1\dee{r}\ r^4\big(2-r^2-r^2\big)\\ &=2\pi \left[2\frac{1^5}{5}-2\frac{1^7}{7}\right]\\ &=\frac{8\pi}{35} \end{align*}

3.6.4.8. (✳).

Solution.

Observe that both the sphere $x^2+y^2+z^2=2$ and the paraboloid $z=x^2+y^2$ are invariant under rotations around the $z$--axis. So $E$ is invariant under rotations around the $z$--axis and the centroid (centre of mass) of $E$ will lie on the $z$--axis. Thus $\bar x=\bar y=0$ and we just have to find

\begin{gather*} \bar z = \frac{\tripInt_E z\ \dee{V}}{\tripInt_E\dee{V}} \end{gather*}

The surfaces $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 2$ intersect when $z=x^2+y^2$ and

\begin{equation*} z+z^2=2 \iff z^2+z-2=0 \iff(z+2)(z-1)=0 \end{equation*}

Since $z=x^2+y^2\ge 0\text{,}$ the surfaces intersect on the circle $z=1\text{,}$ $x^2+y^2=2\text{.}$ So

\begin{equation*} E = \Set{(x,y,z)}{ x^2+y^2\le 1,\ x^2+y^2\le z\le \sqrt{2-x^2-y^2}} \end{equation*}

Here is a sketch of the $y=0$ cross section of E.

Let’s use cylindrical coordinates to do the two integrals. In cylindrical coordinates

$E = \Set{(r\cos\theta,r\sin\theta,z)}{ 0\le r\le 1,\ 0\le\theta\le 2\pi,\ r^2\le z\le \sqrt{2-r^2}}\text{,}$ and
$\dee{V}$ is $r\,\dee{r}\,\dee{\theta}\,\dee{z}$

so, for $n=0,1$ (we’ll try to do both integrals at the same time)

\begin{align*} \tripInt_E z^n\ \dee{V} &=\int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_{r^2}^{\sqrt{2-r^2}} \dee{z}\ r\ z^n\\ &=2\pi\int_0^1\dee{r} \ r\begin{cases} \sqrt{2-r^2}-r^2 & \text{if }n=0\\ \frac{1}{2}\big(2-r^2-r^4\big) & \text{if }n=1 \end{cases} \end{align*}

Since

\begin{align*} \int_0^1 \dee{r}\ r\sqrt{2-r^2} &=\left[-\frac{1}{3}(2-r^2)^{3/2}\right]_0^1 =\frac{1}{3}\big(2\sqrt{2}-1\big) \end{align*}

we have

\begin{align*} \tripInt_E z^n\ \dee{V} &=2\pi\left.\begin{cases} \frac{1}{3}(2\sqrt{2}-1)-\frac{1}{4} &\text{if }n=0\\ \frac{1}{2}-\frac{1}{8}-\frac{1}{12} &\text{if }n=1 \end{cases}\right\}\\ &=2\pi\begin{cases} \frac{2}{3}\sqrt{2}-\frac{7}{12} &\text{if }n=0\\ \frac{7}{24} &\text{if }n=1 \end{cases} \end{align*}

and $\bar x=\bar y=0$ and

\begin{gather*} \bar z =\frac{\tripInt_E z\ \dee{V}}{\tripInt_E\dee{V}} =\frac{\frac{7}{24}}{\frac{2}{3}\sqrt{2}-\frac{7}{12}} =\frac{7}{16\sqrt{2}-14} \approx 0.811 \end{gather*}

3.6.4.9. (✳).

Solution.

Note that both surfaces are invariant under rotations about the $z$--axis. Here is a sketch of the $y=0$ cross section of E.

The surfaces $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 6$ intersect when $z=x^2+y^2$ and

\begin{equation*} z+z^2=6 \iff z^2+z-6=0 \iff(z+3)(z-2)=0 \end{equation*}

Since $z=x^2+y^2\ge 0\text{,}$ the surfaces intersect on the circle $z=2\text{,}$ $x^2+y^2=2\text{.}$ So

\begin{equation*} E = \Set{(x,y,z)}{ x^2+y^2\le 2,\ x^2+y^2\le z\le \sqrt{6-x^2-y^2}} \end{equation*}

Let’s use cylindrical coordinates to do the integral. In cylindrical coordinates

$E = \Set{(r\cos\theta,r\sin\theta,z)}{ r\le \sqrt{2},\ 0\le\theta\le 2\pi,\ r^2\le z\le \sqrt{6-r^2}}\text{,}$ and
$\dee{V}$ is $r\,\dee{r}\,\dee{\theta}\,\dee{z}$

\begin{align*} \tripInt_E (x^2+y^2)\ \dee{V} &=\int_0^{\sqrt{2}}\dee{r} \int_0^{2\pi}\dee{\theta} \int_{r^2}^{\sqrt{6-r^2}} \dee{z}\ r\ r^2\\ &=2\pi\int_0^{\sqrt{2}}\dee{r} \ r^3\big(\sqrt{6-r^2}-r^2\big)\\ & =2\pi\int_0^{\sqrt{2}}\dee{r}\,r\ r^2\sqrt{6-r^2} - 2\pi\int_0^{\sqrt{2}}\dee{r} \ r^5\\ &=2\pi\int_6^4\frac{\dee{u}}{-2} \ (6-u)\sqrt{u} -2\pi\frac{2^3}{6}\\ &\hskip1in\qquad\text{with } u = 6-r^2,\ \dee{u} = -2r\,\dee{r}\\ &=-\pi\left[6\frac{u^{3/2}}{3/2}-\frac{u^{5/2}}{5/2}\right]_6^4-\frac{8\pi}{3}\\ &=-\pi\left[4\big(8-6\sqrt{6}\big)-\frac{2}{5}\big(32-36\sqrt{6}\big)\right] -\frac{8\pi}{3}\\ &=\pi\left[\frac{64}{5}-32-\frac{8}{3} +\left(24-\frac{72}{5}\right)\sqrt{6}\right]\\ &=\pi\left[\frac{48}{5}\sqrt{6}-\frac{328}{15}\right] \approx 1.65\pi \end{align*}

3.6.4.10. (✳).

Solution.

We’ll use cylindrical coordinates. In cylindrical coordinates

the sphere $x^2+y^2+z^2=a^2$ becomes $r^2+z^2=a^2$ and
the circular cylinder $x^2+y^2=ax$ (or equivalently $(x-a/2)^2+ y^2=a^2/4$) becomes $r^2=ar\cos\theta$ or $r=a\cos\theta\text{.}$

Here is a sketch of the top view of the solid.

The solid is

\begin{align*} \big\{\ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\,,\, & 0\le r\le a\cos\theta\,,\,\\ &\hskip0.25in -\sqrt{a^2-r^2}\le z\le \sqrt{a^2-r^2}\ \big\} \end{align*}

By symmetry, the volume of the specified solid is four times the volume of the solid

\begin{equation*} \Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{ 0\le\theta\le\frac{\pi}{2}\,,\, 0\le r\le a\cos\theta\,,\, 0\le z\le \sqrt{a^2-r^2}} \end{equation*}

Since $\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}\text{,}$ the volume of the solid is

\begin{align*} 4\int_0^{\pi/2}\dee{\theta} \int_0^{a\cos\theta} \dee{r} \int_0^{\sqrt{a^2-r^2}}\dee{z}\ r &=4\int_0^{\pi/2}\dee{\theta} \int_0^{a\cos\theta} \dee{r}\ r\sqrt{a^2-r^2}\\ &=-\frac{4}{3}\int_0^{\pi/2}\dee{\theta}\ \big(a^2-r^2\big)^{3/2}\Big|_0^{a\cos\theta}\\ &=\frac{4}{3}\int_0^{\pi/2}\dee{\theta}\ \Big[a^3-\big(a^2-a^2\cos^2\theta\big)^{3/2}\Big]\\ &=\frac{4a^3}{3}\int_0^{\pi/2}\dee{\theta}\ \big[1-\sin^3\theta\big]\\ &=\frac{4a^3}{3}\left[\theta - \frac{1}{12}\cos(3\theta) + \frac{3}{4}\cos\theta \right]_0^{\pi/2}\\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} + \frac{1}{12} - \frac{3}{4} \right]\\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right] \end{align*}

3.6.4.11. (✳).

Solution.

Note that the surfaces meet when $z=y^2=4-x^2$ and then $(x,y)$ runs over the circle $x^2+y^2=4\text{.}$ So the domain of integration is

\begin{equation*} E = \Set{(x,y,z)}{x^2+y^2\le 4,\ y^2\le z\le 4-x^2} \end{equation*}

Let’s switch to cylindrical coordinates. Then

\begin{align*} E = \big\{\ (r\cos\theta,r\sin\theta,z)\ \big|\ 0\le r\le 2,\ &0\le\theta\le 2\pi,\\ &\hskip0.25in r^2\sin^2\theta\le z\le 4-r^2\cos^2\theta\ \big\} \end{align*}

and, since $\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}\text{,}$

\begin{align*} \tripInt_E y^2\ \dee{V} &=\int_0^2\dee{r} \int_0^{2\pi} \dee{\theta} \int_{r^2\sin^2\theta}^{4-r^2\cos^2\theta}\dee{z}\ r\ \overbrace{r^2\sin^2\theta}^{y^2}\\ &=\int_0^2\dee{r} \int_0^{2\pi} \dee{\theta}\ r^3\sin^2\theta\big[4-r^2\cos^2\theta - r^2\sin^2\theta\big]\\ &=\int_0^2\dee{r}\ \big[4r^3-r^5\big] \int_0^{2\pi} \dee{\theta}\ \frac{1-\cos(2\theta)}{2}\\ &=\frac{1}{2}\int_0^2\dee{r}\ \big[4r^3-r^5\big]\ \left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{2\pi}\\ &=\pi\left[r^4-\frac{r^6}{6}\right]_0^2\\ &=\frac{16\pi}{3} \end{align*}

For an efficient, sneaky, way to evaluate $\int_0^{2\pi}\sin^2\theta\ \dee{\theta}\text{,}$ see Remark 3.3.5.

3.6.4.12.

Solution.

By symmetry, $\bar x=\bar y=\bar z\text{,}$ so it suffices to compute, for example, $\bar z\text{.}$ The mass of the body is the density, $\rho\text{,}$ times its volume, which is one eighth of the volume of a sphere. So

\begin{equation*} M=\frac{\rho}{8}\ \frac{4}{3}\pi a^3 \end{equation*}

In cylindrical coordinates, the equation of the spherical surface of the body is $r^2+z^2=a^2\text{.}$ The part of the body at height $z$ above the $xy$--plane is one quarter of a disk of radius $\sqrt{a^2-z^2}\text{.}$ The numerator of $\bar z$ is

\begin{align*} \tripInt_B z\rho\ \dee{V} &=\rho\int_0^a \dee{z}\int_0^{\pi/2}d\theta\int_0^{\sqrt{a^2-z^2}}dr\ r\ z\\ &=\rho\int_0^a \dee{z}\int_0^{\pi/2}d\theta\ z\ \frac{r^2}{2} \bigg|_0^{\sqrt{a^2-z^2}}\\ &=\frac{\rho}{2}\int_0^a \dee{z}\int_0^{\pi/2}d\theta\ z(a^2-z^2)\\ &=\frac{\pi}{4}\rho\int_0^a \dee{z}\ z(a^2-z^2)\\ &=\frac{\pi}{4}\rho\left[a^2\frac{z^2}{2}-\frac{z^4}{4}\right]_0^a =\frac{\pi}{16}\rho a^4 \end{align*}

Dividing by $M=\frac{\pi}{6}\rho a^3$ gives $\bar x=\bar y=\bar z=\frac{3}{8}a\text{.}$

3.6.4.13. (✳).

Solution.

(a) In cylindrical coordinates the equation of a sphere of radius 2 centred on the origin is $r^2+z^2=2^2\text{.}$ Since $\dee{V}=r\, \dee{r}\, \dee{\theta}\, \dee{z}$ and $\dee{m} =\frac{5}{\sqrt{3}}(z^2+1)r\, \dee{r}\, \dee{\theta}\, \dee{z}$ and the hole has radius $1/2\text{,}$ the integral is

\begin{gather*} \text{mass} = \int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r \end{gather*}

(b) By part (a)

\begin{align*} \text{mass} = &\int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r\\ &\hskip0.5in=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\int_0^{\sqrt{4-r^2}}\dee{z}\ (z^2+1)\\ &\hskip0.5in=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\left[\frac{z^3}{3}+z\right]_0^{\sqrt{4-r^2}}\\ &\hskip0.5in=4\pi\frac{5}{\sqrt{3}}\int_{1/2}^2\dee{r}\ r\left[\frac{1}{3}{(4-r^2)}^{3/2}+{(4-r^2)}^{1/2}\right] \end{align*}

Make the change of variables $s=4-r^2\text{,}$ $\dee{s}=-2r\,\dee{r}\text{.}$ This gives

\begin{align*} \text{mass} &= 4\pi\frac{5}{\sqrt{3}}\int_{15/4}^0\frac{\dee{s}}{-2}\ \left[\frac{1}{3}s^{3/2}+s^{1/2}\right] = -2\pi\frac{5}{\sqrt{3}} \left[\frac{2}{15}s^{5/2}+\frac{2}{3}s^{3/2} \right]_{15/4}^0\\ &= 2\pi\frac{5}{\sqrt{3}} \left[\frac{2}{15}\frac{15^{5/2}}{32} +\frac{2}{3}\frac{15^{3/2}}{8}\right]\\ &=2\pi\frac{5}{\sqrt{3}} \left[\frac{1}{16}+\frac{1}{12}\right]15^{3/2} =\frac{525}{24}\sqrt{5}\pi\approx 153.7\text{kg} \end{align*}

3.6.4.14. (✳).

Solution.

(a) The solid consists of all $(x,y,z)$ with

$(x,y)$ running over the disk $x^2+y^2\le 4$ and
for each fixed $(x,y)$ obeying $x^2+y^2\le 4\text{,}$ $z$ running from $0$ to $e^{-x^2-y^2}$

On the disk $x^2+y^2\le 4\text{,}$

$x$ runs from $-2$ to $2$ and
for each fixed $x$ obeying $-2\le x\le 2\text{,}$ $y$ runs from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$

\begin{equation*} \text{Volume}=\int_{-2}^2 \dee{x} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\dee{y}\int_0^{e^{-x^2-y^2}}\dee{z} \end{equation*}

(b) Switching to cylindrical coordinates

\begin{align*} \text{Volume}&=\int_0^2 \dee{r}\int_0^{2\pi}\dee{\theta} \int_0^{e^{-r^2}}\dee{z}\ r =\int_0^2 \dee{r}\int_0^{2\pi}\dee{\theta} \ r e^{-r^2} =\int_0^2 \dee{r}\ 2\pi\ r e^{-r^2}\\ &=-\pi e^{-r^2}\Big|_0^2 =\pi\big[1-e^{-4}\big]\approx 3.084 \end{align*}

3.6.4.15. (✳).

Solution.

The solid consists of the set of all points $(x,y,z)$ such that $x^2+y^2\le 4$ and $0\le z\le\frac{y}{2}\text{.}$ In particular $y\ge 0\text{.}$ When we look at the solid from above, we see all $(x,y)$ with $x^2+y^2\le 4$ and $y\ge 0\text{.}$ This is sketched in the figure on the left below.

We’ll use cylindrical coordinates. In the base region (the shaded region in the figure on the left above)

$\theta$ runs from $0$ to $\pi$ and
for each fixed $\theta$ between $0$ and $\pi\text{,}$ $r$ runs from $0$ to $2\text{.}$
For each fixed point $(x,y)=(r\cos\theta,r\sin\theta)$ in the base region, $z$ runs from $0$) to $\frac{y}{2}=\frac{r\sin\theta}{2}\text{.}$

So the volume is

\begin{align*} \int_0^\pi \dee{\theta}\int_0^2 \dee{r}\int_0^{r\sin\theta/2} dz\ r &=\int_0^\pi \dee{\theta}\int_0^2 \dee{r}\ \frac{1}{2} r^2\sin\theta =\int_0^\pi \dee{\theta}\ \frac{r^3}{6}\sin\theta\,\bigg|_0^2\\ &=\frac{4}{3}\int_0^\pi \dee{\theta}\ \sin\theta\\ &=-\frac{4}{3}\cos\theta\,\bigg|_0^\pi =\frac{8}{3} \end{align*}

3.6.4.16. (✳).

Solution.

(a) The direction of maximum rate of increase is $\vnabla\rho(1,0,-1)\text{.}$ As

\begin{align*} \pdiff{\rho}{x}(x,y,z) &= \frac{4x}{1+x^2+y^2} -\frac{2x(z+2x^2)}{{(1+x^2+y^2)}^2} \\ &\hskip0.5in \pdiff{\rho}{x}(1,0,-1) = \frac{4}{2} -\frac{2(-1+2)}{{(2)}^2} =\frac{3}{2}\\ \pdiff{\rho}{y}(x,y,z) &= -\frac{2y(z+2x^2)}{{(1+x^2+y^2)}^2} \\ &\hskip0.5in \pdiff{\rho}{y}(1,0,-1)= 0\\ \pdiff{\rho}{z}(x,y,z) &= \frac{1}{1+x^2+y^2} \\ &\hskip0.5in \pdiff{\rho}{z}(1,0,-1) = \frac{1}{2} \end{align*}

So $\vnabla\rho(1,0,-1) = \frac{1}{2}(3,0,1)\text{.}$ The unit vector in this direction is $\frac{1}{\sqrt{10}}(3,0,1)\text{.}$

(b) The region swept by the space craft is, in cylindrical coordinates,

\begin{equation*} V=\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le\theta\le 2\pi,\ 0\le r\le 1,\ 0\le z\le 2} \end{equation*}

and the amount of hydrogen collected is

\begin{align*} \tripInt_V \rho\ \dee{V} &=\tripInt_V\frac{z+2r^2\cos^2\theta}{1+r^2}r\dee{r}\,\dee{\theta}\,\dee{z}\\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta} \int_0^1\dee{r}\ \frac{zr+(2\cos^2\theta)r^3}{1+r^2}\\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta} \int_0^1\dee{r}\ \left[ z \frac{r}{1+r^2} +2r\cos^2\theta -\cos^2\theta\frac{2r}{1+r^2}\right]\\ &\hskip1.5in\text{since }\frac{r^3}{1+r^2} = \frac{r+r^3 - r}{1+r^2} =r-\frac{r}{1+r^2}\\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta}\ \left[\frac{z}{2}\ln(1+r^2)+r^2\cos^2\theta -\ln(1+r^2)\cos^2\theta\right]_0^1\\ &=\int_0^2\dee{z} \int_0^{2\pi} \dee{\theta}\ \left[\frac{\ln 2}{2} z+ \cos^2\theta -\ln(2)\cos^2\theta\right]\\ &=\int_0^2\dee{z} \ \left[(\pi\ln 2) z+ \pi-\pi\ln 2\right]\\ &= 2\pi\ln 2 +2\pi -2\pi\ln 2\\ &=2\pi \end{align*}

3.6.4.17.

Solution.

We may choose our coordinate axes so that the torus is constructed by rotating the circle $(x-b)^2+z^2=a^2$ (viewed as lying in the $xz$--plane) about the $z$--axis. On this circle, $x$ runs from $b-a$ to $b+a\text{.}$

In cylindrical coordinates, the torus has equation $(r-b)^2+z^2=a^2\text{.}$ (Recall that the cylindrical coordinate $r$ of a point is its distance from the $z$--axis.) On this torus,

$r$ runs from $b-a$ to $b+a\text{.}$
For each fixed $r\text{,}$ $z$ runs from $-\sqrt{a^2-(r-b)^2}$ to $\sqrt{a^2-(r-b)^2}\text{.}$

As the torus is symmetric about the $xy$--plane, its volume is twice that of the volume of the part with $z\ge 0\text{.}$

\begin{align*} \text{Volume} &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\int_{0}^{\sqrt{a^2-(r-b)^2}}\dee{z}\\ &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\sqrt{a^2-(r-b)^2}\\ &= 4\pi \int_{-a}^{a}\dee{s}\ (s+b)\sqrt{a^2-s^2} \qquad\text{ where } s=r-b \end{align*}

As $s\sqrt{a^2-s^2}$ is odd under $s\rightarrow -s\text{,}$ $\int_{-a}^{a}\dee{s}\ s\sqrt{a^2-s^2}=0\text{.}$ Also, $\int_{-a}^{a}\dee{s}\ \sqrt{a^2-s^2}$ is precisely the area of the top half of a circle of radius $a\text{.}$ So

\begin{equation*} \text{Volume }= 4b\pi \int_{-a}^{a}\dee{s}\ \sqrt{a^2-s^2} =2\pi^2a^2b \end{equation*}

So the mass density of the torus is $\frac{M}{2\pi^2a^2b}$ and $\dee{m} = \frac{M}{2\pi^2a^2b}\,\dee{V} =\frac{M}{2\pi^2a^2b}\,r\,\dee{r}\,d\theta\,\dee{z}$ and

\begin{align*} \text{moment of inertia} &= 2\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r\int_{0}^{\sqrt{a^2-(r-b)^2}}\dee{z}\ \frac{M}{2\pi^2a^2b} r^2\\ &= \frac{M}{\pi^2a^2b}\int_{0}^{2\pi} d\theta\int_{b-a}^{b+a}\dee{r}\ r^3\sqrt{a^2-(r-b)^2}\\ &= \frac{2M}{\pi a^2b} \int_{-a}^{a}\dee{s}\ (s+b)^3\sqrt{a^2-s^2} \qquad\text{ where } s=r-b\\ &= \frac{2M}{\pi a^2b} \int_{-a}^{a}\dee{s}\ (s^3+3s^2b+3sb+b^3)\sqrt{a^2-s^2} \end{align*}

Again, by oddness, the $s^3$ and $3sb$ integrals are zero. For the others, substitute in $s=a\sin t\text{,}$ $\dee{s}=a\cos t\text{.}$

\begin{align*} \text{moment} &= \frac{2M}{\pi a^2b} \int_{-{\pi\over 2}}^{{\pi\over 2}}(a\cos t\, \dee{t}) \ (3a^2b\sin^2 t+b^3)a\cos t\\ &= \frac{2M}{\pi } \int_{-{\pi\over 2}}^{{\pi\over 2}}\dee{t}\ (3a^2\sin^2 t+b^2)\cos^2 t\\ &= \frac{4M}{\pi} \int_0^{{\pi\over 2}}\dee{t}\ (3a^2\cos^2 t-3a^2\cos^4 t+b^2\cos^2 t)\\ &\hskip1in \text{since } \sin^2t=1-\cos^2t \end{align*}

To integrate

⁹

For a general discussion of trigonometric integrals see §1.8 in the CLP-2 text. In particular the integral $\int \cos^4 x\ \dee{x}$ is evaluated in Example 1.8.8 in the CLP-2 text. For an efficient, sneaky, way to evaluate $\int_0^{{\pi\over 2}} \cos^2 t\ \dee{t}$ see Remark 3.3.5.

$\cos^2t$ and $\cos^4t\text{,}$ we use the double angle formulae $\sin^2 x= \frac{1-\cos(2x)}{2}$ and $\cos^2 x= \frac{1+\cos(2x)}{2}$ to write

\begin{equation*} \cos^2 t= \frac{1+\cos(2t)}{2} \end{equation*}

and

\begin{align*} \cos^4 t &= \left[ \frac{1+\cos(2t)}{2} \right]^2\\ &= \frac{1}{4} + \frac{1}{2} \cos(2t) + \frac{1}{4}\cos^2(2t)\\ &= \frac{1}{4} + \frac{1}{2} \cos(2t) + \frac{1}{8}\left(1 + \cos(4t)\right)\\ &= \frac{3}{8} + \frac{1}{2} \cos(2t) + \frac{1}{8}\cos(4t) \end{align*}

\begin{align*} \text{moment} &= \frac{4M}{\pi} \bigg[3a^2\left(\frac{t}{2}\!+\!\frac{\sin(2t)}{4}\right) \!-\!3a^2\left(\frac{3t}{8}\!+\! \frac{1}{4} \sin(2t)\!+\! \frac{1}{32}\sin(4t)\right)\\ &\hskip3.0in +b^2\left(\frac{t}{2}+\frac{\sin(2t)}{4}\right)\!\bigg]_0^{{\pi\over 2}}\\ &= \frac{4M}{\pi} \Big[3a^2\frac{\pi}{4}-3a^2\frac{3\pi}{16}+b^2\frac{\pi}{4}\Big] =M\left(\frac{3}{4}a^2+b^2\right) \end{align*}

3.7 Triple Integrals in Spherical Coordinates
3.7.5 Exercises

3.7.5.1.

Solution.

Since the spherical coordinate $\varphi(x,y,z)$ of a point $(x,y,z)$ is the angle between the positive $z$-axis and the radius vector from $(0,0,0)$ to $(x,y,z)\text{,}$ the sets

\begin{align*} \Set{(x,y,z)}{\varphi(x,y,z)=0} &=\text{the positive $z$-axis}\\ \Set{(x,y,z)}{\varphi(x,y,z)=\frac{\pi}{2}} &=\text{the $xy$-plane}\\ \Set{(x,y,z)}{\varphi(x,y,z)=\pi} &=\text{the negative $z$-axis} \end{align*}

Alternatively, $\tan\varphi(x,y,z)=\frac{z}{\sqrt{x^2+y^2}}\text{,}$ so that, for any $0 \lt \Phi \lt \pi\text{,}$

\begin{align*} \Set{(x,y,z)}{\varphi(x,y,z)=\Phi} &=\Set{(x,y,z)}{z=\tan\Phi\sqrt{x^2+y^2}}\\ &\hskip-0.5in=\text{the cone that makes the angle $\Phi$ with the positive $z$-axis} \end{align*}

3.7.5.2.

Solution.

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

The spherical coordinate $\rho$ is the distance of the point from the origin $(0,0,0)\text{.}$ In particular if $\rho=0\text{,}$ then the point is the origin (regardless of the values of $\theta$ and $\varphi$). If $\rho=1$ then the point lies on the sphere of radius $1$ centred on the origin.
The spherical coordinate $\varphi$ is the angle between the positive $z$-axis and the radial line segment from the origin to $(x,y,z)\text{.}$ In particular, all points with $\varphi=0$ lie on the positive $z$-axis (regardless of the value of $\theta$). All points with $\varphi=\frac{\pi}{2}$ lie in the $xy$-plane.

3.7.5.3.

Solution.

(a) The point $(-2,0,0)$

lies in the $xy$-plane (i.e. has $z=\rho\cos\varphi=0$) and so has $\varphi=\frac{\pi}{2}$ and
lies on the negative $x$-axis and so has $\theta=\pi$ and
is a distance $2$ from the origin and so has $\rho=2\text{.}$

(b) The point $(0,3,0)$

lies in the $xy$-plane (i.e. has $z=\rho\cos\varphi=0$) and so has $\varphi=\frac{\pi}{2}$ and
lies on the positive $y$-axis and so has $\theta=\frac{\pi}{2}$ and
is a distance $3$ from the origin and so has $\rho=3\text{.}$

lies on the negative $z$-axis and so has $\varphi=\pi$ and $\theta$ arbitrary and
is a distance $4$ from the origin and so has $\rho=4\text{.}$

(d) The point $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{3}\right)$

has $\rho=\sqrt{x^2+y^2+z^2} =\sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 +\left(\frac{1}{\sqrt{2}}\right)^2 +\left(\sqrt{3}\right)^2}=\sqrt{4}=2$ and
has $\sqrt{3}=z=\rho\cos\varphi = 2\cos\varphi$ so that $\cos\varphi = \frac{\sqrt{3}}{2}$ and $\varphi=\frac{\pi}{6}$ and
has $-\frac{1}{\sqrt{2}}=x=\rho\sin\varphi\cos\theta = 2\big(\frac{1}{2}\big)\cos\theta$ so that $\cos\theta = -\frac{1}{\sqrt{2}}\text{.}$ As $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ is in the second quadrant, we have $\frac{\pi}{2}\le\theta\le\pi$ and so $\theta=\frac{3\pi}{4}\text{.}$

3.7.5.4.

Solution.

(a) The Cartesian coordinates corresponding to $\rho=1\text{,}$ $\theta=\frac{\pi}{3}\text{,}$ $\varphi=\frac{\pi}{6}$ are

\begin{align*} x&=\rho\sin\varphi\cos\theta =\sin\frac{\pi}{6}\cos\frac{\pi}{3} =\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) =\frac{1}{4}\\ y&=\rho\sin\varphi\sin\theta =\sin\frac{\pi}{6}\sin\frac{\pi}{3} =\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) =\frac{\sqrt{3}}{4}\\ z&=\rho\cos\varphi =\cos\frac{\pi}{6} =\frac{\sqrt{3}}{2} \end{align*}

(b) The Cartesian coordinates corresponding to $\rho=2\text{,}$ $\theta=\frac{\pi}{2}\text{,}$ $\varphi=\frac{\pi}{2}$ are

\begin{align*} x&=\rho\sin\varphi\cos\theta =2\sin\frac{\pi}{2}\cos\frac{\pi}{2} =0\\ y&=\rho\sin\varphi\sin\theta =2\sin\frac{\pi}{2}\sin\frac{\pi}{2} =2\\ z&=\rho\cos\varphi =2\cos\frac{\pi}{2} =0 \end{align*}

Alternatively, we could just observe that

as $\varphi=\frac{\pi}{2}$ the point lies in the $xy$-plane and so has $z=0$ and
as $\rho=2\text{,}$ $\theta=\frac{\pi}{2}$ the point lies on the positive $y$-axis and is a distance $2$ from the origin and so is $(0,2,0)\text{.}$

3.7.5.5.

Solution.

(a) In spherical coordinates

\begin{align*} z^2=3x^2+3y^2 &\iff \rho^2\cos^2\varphi = 3\rho^2\sin^2\varphi\cos^2\theta +3\rho^2\sin^2\varphi\sin^2\theta\\ &\phantom{\iff \rho^2\cos^2\varphi\ } =3\rho^2\sin^2\varphi\\ &\iff \tan^2\varphi=\frac{1}{3}\\ &\iff \tan\varphi=\pm\frac{1}{\sqrt{3}}\\ &\iff \varphi=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{align*}

The surface $z^2=3x^2+3y^2$ is a cone. The upper half of the cone, i.e. the part with $z\ge 0\text{,}$ is $\varphi=\frac{\pi}{6}\text{.}$ The lower half of the cone, i.e. the part with $z\le 0\text{,}$ is $\varphi=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\text{.}$

(b) In spherical coordinates

\begin{align*} &x^2+y^2+(z-1)^2=1\\ &\hskip0.5in\iff \rho^2\sin^2\varphi\cos^2\theta +\rho^2\sin^2\varphi\sin^2\theta +\big(\rho\cos\varphi-1\big)^2 =1\\ &\hskip0.5in\iff \rho^2\sin^2\varphi +\rho^2\cos^2\varphi-2\rho\cos\varphi =0\\ &\hskip0.5in\iff \rho^2-2\rho\cos\varphi =0\\ &\hskip0.5in\iff \rho=2\cos\varphi \end{align*}

\begin{align*} x^2+y^2=4 &\iff \rho^2\sin^2\varphi\cos^2\theta +\rho^2\sin^2\varphi\sin^2\theta =4\\ &\iff \rho^2\sin^2\varphi =4\\ &\iff \rho\sin\varphi=2 \end{align*}

since $\rho\ge 0$ and $0\le\varphi\le\pi$ so that $\sin\varphi\ge 0\text{.}$

3.7.5.6. (✳).

Solution.

In spherical coordinates, the sphere in question is

\begin{align*} B =\big\{\ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 1\,,\,&0\le\varphi\le\pi\,,\,\\ &\hskip0.25in 0\le\theta\le 2\pi\ \big\} \end{align*}

As $\dee{V} =\rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta}\text{,}$

\begin{align*} \text{Volume}(S) =\tripInt_B\dee{V} &=\int_0^{2\pi}\dee{\theta}\int_0^{\pi}\dee{\varphi}\int_0^1\dee{\rho}\ \rho^2\sin\varphi\\ &=\left[\int_0^{2\pi}\dee{\theta}\right] \left[\int_0^{\pi}\dee{\varphi}\ \sin\varphi\right] \left[\int_0^1\dee{\rho}\ \rho^2\right]\\ &= 2\pi\Big[-\cos\varphi\Big]_0^\pi \left[\frac{\rho^3}{3}\right]_0^1 =(2\pi)(2)\left(\frac{1}{3}\right)\\ &=\frac{4\pi}{3} \end{align*}

3.7.5.7. (✳).

Solution.

(a) First observe that both boundaries of $E\text{,}$ namely $\rho=1$ and $\rho= 1 + \cos\varphi\text{,}$ are independent of the spherical coordinate $\theta\text{.}$ So $E$ is invariant under rotations about the $z$-axis. To sketch $E$ we

first sketch the part of the boundary of $E$ with $\theta=0$ (i.e. in the half of the $xz$-plane with $x \gt 0$), and then
rotate about the $z$-axis.

The part of the boundary of $E$ with $\theta=0$ (i.e. in the half-plane $y=0\text{,}$ $x\ge 0$), consists of two curves.

$\rho=1+\cos\varphi\text{,}$ $\theta=0\text{:}$
- When $\varphi=0$ (i.e. on the positive $z$-axis), We have $\cos\varphi =1$ and hence $\rho=2\text{.}$ So this curve starts at $(0,0,2)\text{.}$
- As $\varphi$ increases $\cos\varphi\text{,}$ and hence $\rho\text{,}$ decreases.
- When $\varphi$ is $\frac{\pi}{2}$ (i.e. in the $xy$-plane), we have $\cos\varphi =0$ and hence $\rho=1\text{.}$
- When $\frac{\pi}{2} \lt \varphi\le \pi\text{,}$ we have $\cos\varphi \lt 0$ and hence $\rho \lt 1\text{.}$ All points in $E$ are required to obey $\rho\ge 1\text{.}$ So this part of the boundary stops at the point $(1,0,0)$ in the $xy$-plane.
- The curve $\rho=1+\cos\varphi\text{,}$ $\theta=0\text{,}$ $0\le\varphi\le\frac{\pi}{2}$ is sketched in the figure on the left below. It is the outer curve from $(0,0,2)$ to $(1,0,0)\text{.}$
$\rho=1\text{,}$ $\theta=0\text{:}$
- The surface $\rho=1$ is the sphere of radius $1$ centred on the origin.
- As we observed above, the conditions $1\le\rho\le 1+\cos\varphi$ force $0\le\varphi\le\frac{\pi}{2}\text{,}$ i.e. $z\ge 0\text{.}$
- The sphere $\rho=1$ intersects the quarter plane $y=0\text{,}$ $x\ge 0\text{,}$ $z\ge 0\text{,}$ in the quarter circle centred on the origin that starts at $(0,0,1)$ on the $z$-axis and ends at $(1,0,0)$ in the $xy$-plane.
- The curve $\rho=1\text{,}$ $\theta=0\text{,}$ $0\le\varphi\le\frac{\pi}{2}$ is sketched in the figure on the left below. It is the inner curve from $(0,0,1)$ to $(1,0,0)\text{.}$
To get $E\text{,}$ rotate the shaded region in the figure on the left below about the $z$-axis. The part of $E$ in the first octant is sketched in the figure on the right below. The part of $E$ in the $xz$-plane (with $x\ge 0$) is lightly shaded and the part of $E$ in the $yz$-plane (with $y\ge 0$) is shaded a little more darkly.

(b) In $E$

$\varphi$ runs from $0$ (i.e. the positive $z$-axis) to $\frac{\pi}{2}$ (i.e. the $xy$-plane).
For each $\varphi$ in that range $\rho$ runs from $1$ to $1+\cos\varphi$ and $\theta$ runs from $0$ to $2\pi\text{.}$
In spherical coordinates $\dee{V} = \rho^2\,\sin\varphi\,\dee{\rho}\, \dee{\theta}\,\dee{\varphi}\text{.}$

\begin{align*} \text{Volume}(E) &=\int_0^{\pi/2}\dee{\varphi} \int_1^{1+\cos\varphi}\dee{\rho} \int_0^{2\pi}\dee{\theta}\ \rho^2\sin\varphi\\ &=2\pi \int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\ \frac{(1+\cos\varphi)^3-1^3}{3}\\ &=-\frac{2\pi}{3} \int_2^1 \big(u^3-1\big)\ \dee{u} \quad\text{with }u=1+\cos\varphi,\ \dee{u}=-\sin\varphi\,\dee{\varphi}\\ &= -\frac{2\pi}{3}\left[\frac{u^4}{4}-u\right]_2^1\\ &=-\frac{2\pi}{3} \left[\frac{1}{4}-1-4+2\right]\\ &=\frac{11 \pi}{6} \end{align*}

3.7.5.8. (✳).

Solution.

Recall that in spherical coordinates,

\begin{align*} x&=\rho\sin\varphi\cos\theta\\ y&=\rho\sin\varphi\sin\theta\\ z&=\rho\cos\varphi\\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that $x^2+y^2+z^2= 4$ becomes $\rho = 2\text{,}$ and $\sqrt{x^2+y^2}= z$ becomes

\begin{gather*} \rho\sin\varphi = \rho\cos\varphi \iff \tan\varphi= 1 \iff \varphi=\frac{\pi}{4} \end{gather*}

Here is a sketch of the $y=0$ cross-section of $D\text{.}$

Looking at the figure above, we see that, on $D$

$\varphi$ runs from $0$ (the positive $z$-axis) to $\frac{\pi}{4}$ (on the cone), and
for each $\varphi$ is that range, $\rho$ runs from $0$ to $2$ and $\theta$ runs from $0$ to $2\pi\text{.}$

\begin{align*} D&=\Set{(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)} {0\le\varphi\le\frac{\pi}{4},\ 0\le \theta\le 2\pi,\ \rho\le 2} \end{align*}

and, as $\dee{V} =\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}\text{,}$

\begin{align*} I&= \int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2}\dee{\rho}\ \rho^2\sin\varphi\ \overbrace{\rho\cos\varphi}^{z}\\ &= \int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2}\dee{\rho}\ \rho^3\sin\varphi\,\cos\varphi\\ &= 2\pi\ \frac{2^4}{4} \int_0^{\pi/4}\dee{\varphi}\ \sin\varphi\,\cos\varphi\\ &= 2\pi\ \frac{2^4}{4}\left[\frac{\sin^2\varphi}{2}\right]_0^{\pi/4}\\ &= 2\pi \end{align*}

3.7.5.9.

Solution.

(a) Recall that in spherical coordinates,

\begin{align*} x&=\rho\sin\varphi\cos\theta\\ y&=\rho\sin\varphi\sin\theta\\ z&=\rho\cos\varphi\\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that $x^2+y^2+z^2= a^2$ becomes $\rho = a\text{,}$ and $\sqrt{x^2+y^2}= z$ becomes

\begin{gather*} \rho\sin\varphi = \rho\cos\varphi \iff \tan\varphi= 1 \iff \varphi=\frac{\pi}{4} \end{gather*}

Here is a sketch of the $y=0$ cross-section of the specified region.

Looking at the figure above, we see that, on that region,

$\varphi$ runs from $0$ (the positive $z$-axis) to $\frac{\pi}{4}$ (on the cone), and
for each $\varphi$ is that range, $\rho$ runs from $0$ to $a$ and $\theta$ runs from $0$ to $2\pi\text{.}$

so that

\begin{align*} \text{Volume} &= \int_{0}^{a}\dee{\rho} \int_{0}^{2\pi} \dee{\theta} \int_0^{{\pi\over 4}}\dee{\varphi}\ \rho^2\sin\varphi\\ &= \Big\{\int_{0}^{a}\dee{\rho}\ \rho^2 \Big\}\Big\{\int_{0}^{2\pi} \dee{\theta}\Big\}\Big\{ \int_0^{{\pi\over 4}}\dee{\varphi}\ \sin\varphi\Big\}\\ &= \frac{a^3}{3}\ 2\pi\ \Big[-\cos\varphi\Big]_0^{{\pi\over 4}} =2\pi\frac{a^3}{3}\left(1-\frac{1}{\sqrt{2}}\right) \end{align*}

(b) The part of the sphere in question is

\begin{align*} R&=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ x\ge0,\ y\ge 0,\ z\ge 0}\\ &=\Set{(\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{\rho\le a,\ 0\le\varphi\le\tfrac{\pi}{2}, \ 0\le\theta\le\tfrac{\pi}{2}} \end{align*}

By symmetry, the two specified integrals are equal, and are

\begin{align*} \int_{0}^{a}\dee{\rho}\ \rho^2 \int_0^{{\pi\over 2}}\dee{\varphi}\ \sin\varphi \int_{0}^{{\pi\over 2}} \dee{\theta}\ \overbrace{\rho\cos\varphi}^{z} &=\frac{a^4}{4}\frac{\pi}{2} \int_0^{{\pi\over 2}}\dee{\varphi}\ \sin\varphi\cos\varphi\\ &=\frac{\pi a^4}{8} \int_0^{1}\dee{t}\ t\\ &\hskip0.5in\text{ where } t=\sin\varphi, \dee{t}=\cos\varphi\,\dee{\varphi}\\ &=\frac{\pi a^4}{16} \end{align*}

\begin{align*} P&=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2}\\ &=\Set{(\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)}{\rho\le a,\ 0\le\varphi\le\pi, \ 0\le\theta\le2\pi} \end{align*}

So the

\begin{align*} \text{mass} &= \int_{0}^{a}\dee{\rho}\ \rho^2 \int_0^{\pi}\dee{\varphi}\ \sin\varphi \int_{0}^{2\pi} \dee{\theta}\ \overbrace{\frac{A}{B+\rho^2}}^{\rm density}\\ &=2\pi A\Big\{ \int_0^{\pi}\dee{\varphi}\ \sin\varphi\Big\} \Big\{\int_{0}^{a}\dee{\rho}\ \frac{\rho^2}{B+\rho^2} \Big\}\\ &=4\pi A\int_{0}^{a}\dee{\rho}\ \left(1-\frac{B}{B+\rho^2} \right)\\ &=4\pi Aa-4\pi A\sqrt{B}\int_{0}^{a/\sqrt{B}}\dee{s}\ \frac{1}{1+s^2} \quad\hbox{ where } \rho=\sqrt{B}\,s, \dee{\rho}=\sqrt{B}\,\dee{s}\\ &=4\pi A\left(a-\sqrt{B}\tan^{-1}\frac{a}{\sqrt{B}}\right) \end{align*}

(d) Observe that

when $\varphi=0$ (i.e. on the positive $z$-axis), $\cos\varphi=1$ so that $\rho=a(1-\cos\varphi)=0$ and
as $\varphi$ increases from $0$ to $\frac{\pi}{2}\text{,}$ $\cos\varphi$ decreases so that $\rho=a(1-\cos\varphi)$ increases and
when $\varphi=\frac{\pi}{2}$ (i.e. on the $xy$-plane), $\cos\varphi=0$ so that $\rho=a(1-\cos\varphi)=a$ and
as $\varphi$ increases from $\frac{\pi}{2}$ to $\pi\text{,}$ $\cos\varphi$ continues to decrease so that $\rho=a(1-\cos\varphi)$ increases still more and
when $\varphi=\pi$ (i.e. on the negative $z$-axis), $\cos\varphi=-1$ so that $\rho=a(1-\cos\varphi)=2a$

So we have the following sketch of the intersection of the specified volume with the right half of the $yz$-plane.

The volume in question is invariant under rotations about the $z$-axis so that

\begin{align*} \text{Volume} &= \int_{0}^{2\pi} \dee{\theta}\int_0^{\pi}\dee{\varphi}\ \sin\varphi\int_{0}^{a(1-\cos\varphi)}\dee{\rho}\ \rho^2\\ &=2\pi\frac{a^3}{3} \int_0^{\pi}\dee{\varphi}\ \sin\varphi(1-\cos\varphi)^3\\ &=2\pi\frac{a^3}{3} \int_0^2\dee{t}\ t^3 \qquad\hbox{ where } t=1-\cos\varphi, \dee{t}=\sin\varphi\,\dee{\varphi}\\ &=2\pi\ \frac{a^3}{3}\ \frac{2^4}{4} =\frac{8}{3}\pi a^3 \end{align*}

3.7.5.10. (✳).

Solution.

Let’s use $H$ to denote the hemispherical shell. On that shell, the spherical coordinate $\varphi$ runs from $0$ (on the $z$-axis) to $\pi/2$ (on the $xy$-plane, $z=0$) and the spherical coordinate $\rho$ runs from $2\text{,}$ on $x^2+y^2+z^2=4\text{,}$ to $3\text{,}$ on $x^2+y^2+z^2=9\text{.}$ So, in spherical coordinates,

\begin{align*} H=\big\{\ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\,\rho\cos\varphi)\ \big|\ 2\le \rho\le 3,\ &0\le\varphi\le\pi/2,\\ & \hskip0.25in 0\le\theta\le 2\pi\ \big\} \end{align*}

(a) In spherical coordinates $\dee{V}=\rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta}\text{,}$ so that, as the density is the constant $D\text{,}$

\begin{align*} \text{Mass}(H) &=\int_2^3\dee{\rho} \int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi}\ D\ \rho^2\sin\varphi\\ &=D\ \left[\int_2^3\dee{\rho}\ \rho^2\right]\ \left[\int_0^{2\pi}\dee{\theta}\right]\ \left[\int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\right]\\ &= D\left[\frac{3^3}{3}-\frac{2^3}{3}\right]\ \big[2\pi\big]\ \big[\cos 0 -\cos(\pi/2)\big]\\ &=\frac{38}{3}\pi D \end{align*}

We could have gotten the same result by expressing the mass as

one half, times
the density $D\text{,}$ times
the difference between the volume of a sphere of radius $3$ and a sphere of radius $2\text{.}$

That is

\begin{gather*} \text{Mass}(H) =\frac{1}{2}D\left[\frac{4}{3}\pi 3^3-\frac{4}{3}\pi 2^3\right] =\frac{38}{3}\pi D \end{gather*}

(b) By definition, the centre of mass is $(\bar x,\bar y,\bar z)$ where $\bar x\text{,}$ $\bar y$ and $\bar z$ are the weighted averages of $x\text{,}$ $y$ and $z\text{,}$ respectively, over $H\text{.}$ That is

\begin{gather*} \bar x =\frac{\tripInt_H x\,D\,\dee{V}}{\tripInt_H D\,\dee{V}}\qquad \bar y =\frac{\tripInt_H y\,D\,\dee{V}}{\tripInt_H D\,\dee{V}}\qquad \bar z =\frac{\tripInt_H z\,D\,\dee{V}}{\tripInt_H D\,\dee{V}} \end{gather*}

As $H$ is invariant under reflection in the $yz$-plane (i.e. under $x\rightarrow-x$) we have $\bar x=0\text{.}$ As $H$ is also invariant under reflection in the $xz$-plane (i.e. under $y\rightarrow-y$) we have $\bar y=0\text{.}$ So we just have to find $\bar z\text{.}$ We have already found the denominator in part (a), so we just have evaluate the numerator

\begin{align*} \tripInt_H z\,D\,\dee{V} &=\int_2^3\dee{\rho} \int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi}\ D\ \rho^2\sin\varphi\ \overbrace{\rho\cos\varphi}^{z}\\ &=D\ \left[\int_2^3\dee{\rho}\ \rho^3\right]\ \left[\int_0^{2\pi}\dee{\theta}\right]\ \left[\int_0^{\pi/2}\dee{\varphi}\ \sin\varphi\ \cos\varphi\right]\\ &= D\left[\frac{3^4}{4}-\frac{2^4}{4}\right]\ \big[2\pi\big]\ \left[\frac{1}{2}\sin^2\frac{\pi}{2} -\frac{1}{2}\sin^2 0\right]\\ &=\frac{81-16}{4}\pi D =\frac{65}{4}\pi D \end{align*}

All together

\begin{gather*} \bar x = \bar y=0\qquad \bar z =\frac{\frac{65}{4}\pi D}{\frac{38}{3}\pi D} =\frac{195}{152} \approx 1.28 \end{gather*}

3.7.5.11. (✳).

Solution.

(a) Here is a sketch

(b) On $T\text{,}$

the spherical coordinate $\varphi$ runs from $0$ (the positive $z$-xis) to $\frac{\pi}{2}$ (the $xy$-plane), and
for each fixed $\varphi$ in that range, $\theta$ runs from $0$ to $\frac{\pi}{2}\text{,}$ and
for each fixed $\varphi$ and $\theta\text{,}$ the spherical coordinate $\rho$ runs from $0$ to $1\text{.}$
In spherical coordinates $\dee{V}=\rho^2\,\sin\varphi\,\dee{\rho}\,\dee{\theta}\, \dee{\varphi}$ and
\begin{equation*} xz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta \end{equation*}

\begin{gather*} I=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{gather*}

\begin{align*} I&= \left[\int_0^{\pi/2}\dee{\varphi}\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\dee{\theta}\ \cos\theta\right] \left[\int_0^1\dee{\rho}\ \rho^4\right]\\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1\\ &=\frac{1}{15} \end{align*}

3.7.5.12. (✳).

Solution.

We’ll use spherical coordinates. On $Q\text{,}$

the spherical coordinate $\varphi$ runs from $0$ (the positive $z$-axis) to $\frac{\pi}{2}$ (the $xy$-plane),
the spherical coordinate $\theta$ runs from $0$ (the half of the $xz$-plane with $x\ge 0$) to $\frac{\pi}{2}$ (the half of the $yz$-plane with $y\ge 0$) and
the spherical coordinate $\rho$ runs from $0$ to $3\text{.}$

As $\dee{V}=\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}\text{,}$

\begin{align*} W&= \tripInt_Q xz\ \dee{V} =\int_0^3 \dee{\rho} \int_0^{\pi/2}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi\cos\theta}^{x} \overbrace{\rho\cos\varphi}^{z}\\ &=\int_0^3 \dee{\rho} \int_0^{\pi/2}\dee{\theta} \ \rho^4\cos\theta\left[\frac{\sin^3\varphi}{3}\right] _{\varphi=0}^{\varphi=\pi/2}\\ &=\frac{1}{3}\int_0^3 \dee{\rho}\ \rho^4\Big[\sin\theta\Big]_0^{\pi/2}\\ &=\frac{3^5}{15}=\frac{81}{5} \end{align*}

3.7.5.13. (✳).

Solution.

Let’s use spherical coordinates. This is an improper integral. So, to be picky, we’ll take the limit as $R\rightarrow\infty$ of the integral over $0\le\rho\le R\text{.}$

\begin{align*} &\tripInt_{\bbbr^3} {\big[1+{(x^2+y^2+z^2)}^3\big]}^{-1}\ \dee{V}\\ &\hskip0.5in=\lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \int_0^{2\pi}\dee{\theta} \int_0^\pi\dee{\varphi}\ \rho^2\sin\varphi\ \frac{1}{1+\rho^6}\\ &\hskip0.5in=\lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \int_0^{2\pi}\dee{\theta} \ \frac{\rho^2}{1+\rho^6}\Big[-\cos\varphi\Big]_0^\pi\\ &\hskip0.5in=4\pi \lim_{R\rightarrow\infty} \int_0^R \dee{\rho} \ \frac{\rho^2}{1+\rho^6}\\ &\hskip0.5in=\frac{4\pi}{3} \lim_{R\rightarrow\infty} \int_0^{R^3} \dee{u} \ \frac{1}{1+u^2}\qquad \text{with }u=\rho^3,\ \dee{u}=3\rho^2\,\dee{\rho}\\ &\hskip0.5in=\frac{4\pi}{3}\lim_{R\rightarrow\infty}\Big[\arctan u\Big]_0^{R^3}\\ &\hskip0.5in=\frac{2\pi^2}{3}\qquad\text{since } \lim_{R\rightarrow\infty}\arctan R^3=\frac{\pi}{2} \end{align*}

3.7.5.14. (✳).

Solution.

On the domain of integration

$x$ runs from $-1$ to $1\text{.}$
For each fixed $x$ in that range, $y$ runs from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}\text{.}$ In inequalities, that is $-\sqrt{1-x^2}\le y\le \sqrt{1-x^2}\text{,}$ which is equivalent to $x^2+y^2\le 1\text{.}$
For each fixed $(x,y)$ obeying $x^2+y^2\le 1\text{,}$ $z$ runs from $1-\sqrt{1-x^2-y^2}$ to $1+\sqrt{1-x^2-y^2}\text{.}$ In inequalities, that is $1-\sqrt{1-x^2-y^2}\le z\le 1+\sqrt{1-x^2-y^2}\text{,}$ which is equivalent to $x^2+y^2+(z-1)^2\le 1\text{.}$

So the domain of integration is

\begin{equation*} V = \Set{(x,y,z)}{x^2+y^2+(z-1)^2\le 1} \end{equation*}

In spherical coordinates, the condition $x^2+y^2+(z-1)^2\le 1$ is

\begin{align*} &(\rho\sin\varphi\cos\theta)^2 +(\rho\sin\varphi\sin\theta)^2 +(\rho\cos\varphi-1)^2\le 1\\ &\iff \rho^2\sin^2\varphi + (\rho\cos\varphi-1)^2\le 1\\ &\iff \rho^2\sin^2\varphi + \rho^2\cos^2\varphi -2 \rho\cos\varphi +1 \le 1\\ &\iff \rho^2\le 2\rho\cos\varphi\\ &\iff \rho\le 2\cos\varphi \end{align*}

Note that $V$ is contained in the upper half, $z\ge 0\text{,}$ of $\bbbr^3$ and that the $xy$-plane in tangent to $V\text{.}$ So as $(x,y,z)$ runs over $V\text{,}$ the spherical coordinate $\varphi$ runs from $0$ (the positive $z$-axis) to $\frac{\pi}{2}$ (the $xy$-plane). Here is a sketch of the side view of $V\text{.}$

As $\dee{V} = \rho^2\sin\varphi\ \dee{\rho}\,\dee{\varphi}\,\dee{\theta}$ and $\big(x^2+y^2+z^2\big)^{5/2}=\rho^5\text{,}$ the integral is

\begin{align*} &\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{1-\sqrt{1-x^2-y^2}}^{1+\sqrt{1-x^2-y^2}} (x^2+y^2+z^2)^{5/2} \ \dee{z}\,\dee{y}\,\dee{x}\\ &\hskip0.5in=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi \ \rho^5\\ &\hskip0.5in=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/2}\dee{\varphi} \ \frac{2^8\cos^8\varphi}{8}\sin\varphi\\ &\hskip0.5in=32\int_0^{2\pi}\dee{\theta}\ \left[-\frac{\cos^9\varphi}{9}\right]_0^{\pi/2}\\ &\hskip0.5in=\frac{32}{9}(2\pi) =\frac{64\pi}{9} \end{align*}

3.7.5.15.

Solution.

The top of the cylinder has equation $z=h\text{,}$ i.e. $\rho\cos\varphi=h\text{.}$ The side of the cylinder has equation $x^2+y^2=a^2\text{,}$ i.e. $\rho\sin\varphi=a\text{.}$ The bottom of the cylinder has equation $z=0\text{,}$ i.e. $\varphi=\frac{\pi}{2}\text{.}$

For each fixed $\varphi\text{,}$ $\theta$ runs from $0$ to $2\pi$ and $\rho$ runs from $0$ to either $\frac{h}{\cos\varphi}$ (at the top of the can, when $\varphi \lt \tan^{-1}\frac{a}{h}$) or $\frac{a}{\sin\varphi}$ (at the side of the can, when $\varphi \gt \tan^{-1}\frac{a}{h}$). So the

\begin{align*} \text{Volume} &= \int_{0}^{\tan^{-1}{a\over h}} \dee{\varphi} \int_{0}^{2\pi} \dee{\theta} \int_0^{h/\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi\\ &\hskip1in +\int_{\tan^{-1}{a\over h}}^{{\pi\over 2}} \dee{\varphi} \int_{0}^{2\pi} \dee{\theta} \int_0^{a/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi\\ &= 2\pi\int_{0}^{\tan^{-1}{a\over h}} \dee{\varphi} \ \frac{h^3\sin\varphi}{3\cos^3\varphi} +2\pi\int_{\tan^{-1}{a\over h}}^{{\pi\over 2}} \dee{\varphi} \ \frac{a^3\sin\varphi}{3\sin^3\varphi}\\ &=2\pi\Big\{\int_0^{{a\over h}}\dee{t}\ \frac{h^3}{3}t -\int^0_{{h\over a}}\dee{s}\ \frac{a^3}{3}\Big\}\\ &\hskip0.5in\hbox{ where } t=\tan\varphi, \dee{t}=\sec^2\varphi\,\dee{\varphi}, s=\cot\varphi, \dee{s}=-\csc^2\varphi\,\dee{\varphi}\\ &=2\pi\left\{\frac{h^3}{3}\,\frac{1}{2}\left(\frac{a}{h}\right)^2 +\frac{a^3}{3}\,\frac{h}{a}\right\} =2\pi\left\{\frac{ah^2}{6}+\frac{a^2h}{3}\right\} =\pi a^2h \end{align*}

3.7.5.16. (✳).

Solution.

In spherical coordinates,

\begin{equation*} x=\rho\sin\varphi\cos\theta\qquad y=\rho\sin\varphi\sin\theta\qquad z=\rho\cos\varphi \end{equation*}

so that the sphere $x^2+y^2+z^2=4$ is $\rho^2=4$ or $\rho=2$ and the cone $x^2+y^2=z^2$ is $\rho^2\sin^2\varphi=\rho^2\cos^2\varphi$ or $\tan\varphi=\pm1$ or $\varphi=\frac{\pi}{4},\ \frac{3\pi}{4}\text{.}$ So

\begin{align*} \text{moment} &=\int_0^2\dee{\rho}\int_0^{\pi/4}\dee{\varphi}\int_0^{2\pi}\dee{\theta}\ \rho^2\sin\varphi\ (\rho\cos\varphi)^2\\ &=2\pi\int_0^2 \dee{\rho}\ \rho^4\int_0^{\pi/4}\dee{\varphi}\ \sin\varphi\cos^2\varphi\\ &=2\pi\left[\frac{\rho^5}{5}\right]_0^2 \left[-\frac{1}{3}\cos^3\varphi\right]_0^{\pi/4} =\frac{64}{15}\pi\left(1-\frac{1}{2\sqrt{2}}\right)\\ &\approx 8.665 \end{align*}

3.7.5.17. (✳).

Solution.

(a) In spherical coordinates,

\begin{gather*} x=\rho\sin\phi\cos\theta\qquad y=\rho\sin\phi\sin\theta\qquad z=\rho\cos\phi \end{gather*}

so that

the sphere $x^2+y^2+z^2=1$ is $\rho=1\text{,}$
the $xy$-plane, $z=0\text{,}$ is $\phi=\frac{\pi}{2}\text{,}$
the positive half of the $xz$-plane, $y=0\text{,}$ $x \gt 0\text{,}$ is $\theta=0$ and
the positive half of the $yz$-plane, $x=0\text{,}$ $y \gt 0\text{,}$ is $\theta=\frac{\pi}{2}\text{.}$

\begin{align*} \tripInt_\Om z\,\dee{V}&= \int_0^1 \dee{\rho}\int_0^{\pi/2}d\phi\int _0^{\pi/2}\dee{\theta}\ \rho^2\sin\phi\overbrace{(\rho\cos\phi)}^{z}\\ &=\frac{\pi}{2}\int_0^1 \dee{\rho}\int_0^{\pi/2}d\phi\ \rho^3\sin\phi \cos\phi\\ &=\frac{\pi}{2}\int_0^1 \dee{\rho}\ \rho^3\ \frac{1}{2}\sin^2\phi \Big|_0^{\pi/2} =\frac{\pi}{4}\int_0^1 \dee{\rho}\ \rho^3 =\frac{\pi}{16} \end{align*}

(b) The hemispherical ball given by $z\ge 0\text{,}$ $x^2+y^2+z^2\le 1$ (call it $H$) has centroid $(\bar x,\bar y,\bar z)$ with $\bar x=\bar y=0$ (by symmetry) and

\begin{gather*} \bar z=\frac{\tripInt_H z\,\dee{V}}{\tripInt_H\,\dee{V}} =\frac{4\tripInt_\Om z\,\dee{V}}{\half\times\frac{4}{3}\pi} =\frac{\frac{\pi}{4}}{\frac{2\pi}{3}} =\frac{3}{8} \end{gather*}

3.7.5.18. (✳).

Solution.

(a) In spherical coordinates,

\begin{equation*} x=\rho\sin\phi\cos\theta\qquad y=\rho\sin\phi\sin\theta\qquad z=\rho\cos\phi \end{equation*}

the sphere $x^2+y^2+z^2=4$ is $\rho^2=4$ or $\rho=2$ and the $xy$-plane is $\phi=\frac{\pi}{2}\text{.}$ So

\begin{equation*} \text{mass} =\int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\ \rho^2\sin\phi \ \overbrace{(9 \rho\cos\phi)}^{\rm density} \end{equation*}

(b) The mass of the half ball is

\begin{align*} &9 \int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\ \rho^3\sin\phi \cos\phi\\ &\hskip1in=9 \bigg[\int_0^2d\rho\ \rho^3\bigg] \bigg[\int_0^{\pi/2}d\phi\ \sin\phi \cos\phi\bigg] \bigg[\int_0^{2\pi}d\theta\bigg] \end{align*}

In spherical coordinates, the cone $x^2+y^2=z^2$ is $\rho^2\sin^2\phi=\rho^2\cos^2\phi$ or $\tan\phi=\pm1$ or $\phi=\frac{\pi}{4},\ \frac{3\pi}{4}\text{.}$ So the mass of the part that is inside the cone is

\begin{align*} &9 \int_0^2d\rho\int_0^{\pi/4}d\phi\int_0^{2\pi}d\theta\ \rho^3\sin\phi \cos\phi\\ &\hskip1in=9 \bigg[\int_0^2d\rho\ \rho^3\bigg] \bigg[\int_0^{\pi/4}d\phi\ \sin\phi \cos\phi\bigg] \bigg[\int_0^{2\pi}d\theta\bigg] \end{align*}

The fraction inside the cone is

\begin{equation*} \frac{\int_0^{\pi/4}d\phi\ \sin\phi \cos\phi}{\int_0^{\pi/2}d\phi\ \sin\phi \cos\phi} =\frac{\half\sin^2\phi\big|_0^{\pi/4}}{\half\sin^2\phi\big|_0^{\pi/2}} =\frac{1}{2} \end{equation*}

3.7.5.19. (✳).

Solution.

In spherical coordinates, $x=\rho\sin\varphi\cos\theta\text{,}$ $y=\rho\sin\varphi\sin\theta\text{,}$ $z=\rho\cos\varphi$ so that

\begin{align*} &\frac{\rho^2\sin^2\varphi\cos\theta\sin\theta +\rho^3\sin\varphi\sin\theta\,\cos^2\varphi +\rho^3\sin\varphi\,\cos^2\varphi} {\rho^2\sin^2\varphi+\rho^4\cos^4\varphi}\\ &\hskip0.5in =\frac{\sin^2\varphi\cos\theta\sin\theta +\rho\sin\varphi\sin\theta\,\cos^2\varphi +\rho\sin\varphi\cos\theta\,\cos^2\varphi} {\sin^2\varphi+\rho^2\cos^4\varphi} \end{align*}

As $(x,y,z)\to (0,0,0)\text{,}$ the radius $\rho\to 0$ and the second and third terms in the numerator and the second term in the denominator converge to $0\text{.}$ But that leaves

\begin{equation*} \frac{\sin^2\varphi\cos\theta\sin\theta}{\sin^2\varphi} =\cos\theta\sin\theta \end{equation*}

which takes many different values. In particular, if we send $(x,y,z)\to (0,0,0)$ along either the $x$- or $y$-axis, that is with $z=0$ and either $x=0$ or $y=0\text{,}$ then

\begin{equation*} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\Atop{x=0\text{ or }y=0}{z=0}}=0 \end{equation*}

converges to $0\text{.}$ But, if we send $(x,y,z)\to (0,0,0)$ along the line $y=x\text{,}$ $z=0$

\begin{equation*} \frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\Atop{y=x}{z=0}}=\frac{x^2}{2x^2} =\frac{1}{2} \end{equation*}

converges to $1/2\text{.}$ So $\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$ does not approach a single value as $(x,y,z)\to(0,0,0)$ and the limit does not exist.

3.7.5.20. (✳).

Solution.

The disk of radius $2$ centred at the origin in the $xy$-plane is $x^2+y^2\le 4\text{.}$ So

\begin{align*} V&=\Set{(x,y,z)}{x^2+y^2\le 4,\ 0\le z\le 2} \end{align*}

The cone with vertex at the origin that contains the top edge, $x+y^2=4\text{,}$ $z=2\text{,}$ of $U$ is $x^2+y^2=z^2\text{.}$ So

\begin{align*} U&=\Set{(x,y,z)}{x^2+y^2\le 4,\ 0\le z\le 2,\ x^2+y^2\ge z^2} \end{align*}

Here are sketches of the $y=0$ cross-section of $V\text{,}$ on top, and $U\text{,}$ on the bottom.

(a) In cylindrical coordinates, $x^2+y^2\le 4$ becomes $r\le 2$ and $x^2+y^2\ge z^2$ is $r\ge |z|\text{,}$ and the density is $\sqrt{x^2+y^2}=r\text{.}$ So

\begin{align*} U&=\Set{(r\cos\theta,r\sin\theta,z)}{r\le 2,\ 0\le z\le 2,\ r\ge z} \end{align*}

Looking at the figure below, we see that, on $U$

$z$ runs from $0$ to $2\text{,}$ and
for each $z$ is that range, $r$ runs from $z$ to $2$ and $\theta$ runs from $0$ to $2\pi\text{.}$
$\displaystyle \dee{V} =r\,\dee{r}\,\dee{\theta}\,\dee{z}$

\begin{align*} \text{Mass}&= \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r\ \overbrace{r}^{\text{density}} = \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r^2 \end{align*}

(b) Recall that in spherical coordinates,

\begin{align*} x&=\rho\sin\varphi\cos\theta\\ y&=\rho\sin\varphi\sin\theta\\ z&=\rho\cos\varphi\\ x^2+y^2 &=\rho^2\sin^2\varphi \end{align*}

so that $x^2+y^2\le 4$ becomes $\rho\sin\varphi\le 2\text{,}$ and $x^2+y^2\ge z^2$ becomes

\begin{gather*} \rho\sin\varphi \ge \rho\cos\varphi \iff \tan\varphi\ge 1 \iff \varphi\ge\frac{\pi}{4} \end{gather*}

and the density $\sqrt{x^2+y^2}=\rho\sin\varphi\text{.}$ So

\begin{align*} U=\big\{\ (\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi) \ \big|\ \frac{\pi}{4}\le\varphi\le\frac{\pi}{2},\ &0\le \theta\le 2\pi,\ \\ &\hskip0.25in\rho\sin\varphi\le 2\ \big\} \end{align*}

Looking at the figure below, we see that, on $U$

$\varphi$ runs from $\frac{\pi}{4}$ (on the cone) to $\frac{\pi}{2}$ (on the $xy$-plane), and
for each $\varphi$ is that range, $\rho$ runs from $0$ to $\frac{2}{\sin\varphi}$ and $\theta$ runs from $0$ to $2\pi\text{.}$
$\displaystyle \dee{V} =\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}$

\begin{align*} \text{Mass}&= \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi}^{\text{density}}\\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{2/\sin\varphi}\dee{\rho}\ \rho^3\sin^2\varphi \end{align*}

\begin{align*} \text{Mass} &= \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\ r^2\\ &=2\pi \int_0^2\dee{z}\ \frac{8-z^3}{3}\\ &=\frac{2\pi}{3}\ \left[16-\frac{2^4}{4}\right]\\ &=8\pi \end{align*}

3.7.5.21. (✳).

Solution.

(a) Call the solid $V\text{.}$ In cylindrical coordinates

$x^2+y^2+z^2\le 2$ is $r^2+z^2\le 2$ and
$\sqrt{x^2+y^2}\le z$ is $r\le z$ and
the density $\de=r^2\text{,}$ and
$\dee{V}$ is $r\,\dee{r}\,\dee{\theta}\,\dee{z}$

Observe that $r^2+z^2= 2$ and $r= z$ intersect when $2r^2=2$ so that $r=z=1\text{.}$ Here is a sketch of the $y=0$ cross-section of $E\text{.}$

\begin{equation*} V =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le 1,\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{2-r^2}} \end{equation*}

and

\begin{align*} M &=\tripInt_V \rho(x,y,z)\ \dee{V} =\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{2-r^2}}\dee{z}\ r\overbrace{(r^2)}^{\de}\\ &=\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{2-r^2}}\dee{z}\ r^3 \end{align*}

(b)

In spherical coordinates

$x^2+y^2+z^2\le 2$ is $\rho\le\sqrt{2}\text{,}$ and
$\sqrt{x^2+y^2}\le z$ is $\rho\sin\varphi\le \rho\cos\varphi\text{,}$ or $\tan\varphi\le 1$ or $\varphi\le\frac{\pi}{4}\text{,}$ and
the density $x^2+y^2=\rho^2\sin^2\varphi\text{,}$ and
$\dee{V}$ is $\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}$

\begin{align*} V =\big\{\ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le \sqrt{2},\ &0\le\theta\le2\pi,\\ &\hskip 0.25in 0\le \varphi\le \frac{\pi}{4}\ \big\} \end{align*}

and, since the integrand $x^2+y^2=\rho^2\sin^2\varphi\text{,}$

\begin{align*} M &=\tripInt_V \big(x^2+y^2\big)\ \dee{V} =\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^2\sin\varphi\ \rho^2\sin^2\varphi\\ &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi \end{align*}

\begin{align*} M &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi\\ &=\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big]\\ &=2\pi \int_0^{\sqrt{2}}\dee{\rho}\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/4} =2\pi \left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] \int_0^{\sqrt{2}}\dee{\rho}\ \rho^4\\ &=2\pi \frac{4\sqrt{2}}{5}\left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] =\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503 \end{align*}

3.7.5.22. (✳).

Solution.

(a) On $E\text{,}$

the spherical coordinate $\varphi$ runs from $0$ (the positive $z$-xis) to $\frac{\pi}{2}$ (the $xy$-plane), and
for each fixed $\varphi$ in that range, $\theta$ runs from $0$ to $\frac{\pi}{2}\text{,}$ and
for each fixed $\varphi$ and $\theta\text{,}$ the spherical coordinate $\rho$ runs from $0$ to $1\text{.}$
In spherical coordinates $\dee{V}=\rho^2\,\sin\varphi\,\dee{\rho}\,\dee{\theta}\, \dee{\varphi}$ and
\begin{equation*} xz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta \end{equation*}

\begin{gather*} I=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{gather*}

(b) In cylindrical coordinates, the condition $x^2+y^2+z^2\le 1$ becomes $r^2+z^2\le 1\text{.}$ So, on $E$

the cylindrical coordinate $z$ runs from $0$ (in the $xy$-plane) to $1$ (at $(0,0,1)$) and
for each fixed $z$ in that range, $\theta$ runs from $0$ to $\pi/2$ and
for each such fixed $z$ and $\theta\text{,}$ the cylindrical coordinate $r$ runs from $0$ to $\sqrt{1-z^2}$ (recall that $r^2+z^2\le 1$).
In cylindrical coordinates $\dee{V}=r\,\dee{r}\,\dee{\theta}\, \dee{z}$ and
\begin{equation*} xz=\big(r\cos\theta\big)\big(z\big) =r\,z\, \cos\theta \end{equation*}
So
\begin{gather*} I=\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \ r^2\,z\,\cos\theta \end{gather*}

(c) Both spherical and cylindrical integrals are straight forward to evaluate. Here are both. First, in spherical coordinates,

Now in cylindrical coordinates

\begin{align*} I&=\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \ r^2\,z\,\cos\theta\\ &=\frac{1}{3}\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\ z{(1-z^2\big)}^{3/2}\, \cos\theta\\ &=\frac{1}{3}\int_0^1\dee{z}\ z{(1-z^2\big)}^{3/2}\\ &=\frac{1}{3}\left[-\frac{1}{2}\ \frac{(1-z^2)^{5/2}}{5/2}\right]_0^1\\ &=\frac{1}{15} \end{align*}

3.7.5.23. (✳).

Solution.

(a) Recall that in spherical coordinates

\begin{gather*} x=\rho\,\sin\varphi\,\cos\theta\qquad y=\rho\,\sin\varphi\,\sin\theta\qquad z=\rho\,\cos\varphi \end{gather*}

so that

$x^2+y^2+z^2\le 9$ is $\rho\le 3\text{,}$ and
$\sqrt{3x^2+3y^2}\le z$ is $\sqrt{3}\rho\sin\varphi\le \rho\cos\varphi\text{,}$ or $\tan\varphi\le \frac{1}{\sqrt{3}}$ or $\varphi\le\frac{\pi}{6}\text{,}$ and
the integrand $x^2+y^2=\rho^2\sin^2\varphi\text{,}$ and
$\dee{V}$ is $\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}$

\begin{align*} T =\big\{\ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 3,\ &0\le\theta\le2\pi,\\ &\hskip0.25in 0\le \varphi\le \frac{\pi}{6}\ \big\} \end{align*}

and,

\begin{align*} I &=\tripInt_T \big(x^2+y^2\big)\ \dee{V} =\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^2\sin\varphi\ \overbrace{\rho^2\sin^2\varphi}^{x^2+y^2}\\ &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi \end{align*}

(b) In cylindrical coordinates

$x^2+y^2+z^2\le 9$ is $r^2+z^2\le 9$ and
$\sqrt{3 x^2+3 y^2}\le z$ is $\sqrt{3}\,r\le z$ and
the integand $x^2+y^2=r^2\text{,}$ and
$\dee{V}$ is $r\,\dee{r}\,\dee{\theta}\,\dee{z}$

Observe that $r^2+z^2= 9$ and $\sqrt{3}\,r= z$ intersect when $r^2+3r^2=9$ so that $r=\frac{3}{2}$ and $z=\frac{3\sqrt{3}}{2}\text{.}$ Here is a sketch of the $y=0$ cross-section of $T\text{.}$

\begin{equation*} T =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le \tfrac{3}{2},\ 0\le\theta\le2\pi,\ \sqrt{3}\,r\le z\le \sqrt{9-r^2}} \end{equation*}

and

\begin{align*} I &=\tripInt_V (x^2+y^2)\ \dee{V} =\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ r\overbrace{(r^2)}^{x^2+y^2}\\ &=\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ \ r^3 \end{align*}

\begin{align*} I &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi\\ &=\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big]\\ &=2\pi \int_0^3\dee{\rho}\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/6}\\ &=2\pi \left[-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8}+1-\frac{1}{3}\right] \int_0^3\dee{\rho}\ \rho^4\\ &=2\pi \frac{3^5}{5}\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] =\overbrace{81}^{3^4}\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24 \end{align*}

3.7.5.24. (✳).

Solution.

(a) In cylindrical coordinates

$x^2+y^2+z^2\le 1$ is $r^2+z^2\le 1$ and
$x^2+y^2\le z^2$ is $r^2\le z^2$ and
$\dee{V}$ is $r\,\dee{r}\,\dee{\theta}\,\dee{z}$

Observe that $r^2+z^2= 1$ and $r^2= z^2$ intersect when $r^2=z^2=\frac{1}{2}\text{.}$ Here is a sketch of the $y=0$ cross-section of $E\text{.}$

\begin{equation*} E =\Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le\frac{1}{\sqrt{2}},\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{1-r^2}} \end{equation*}

and

\begin{align*} J &=\tripInt_E \sqrt{x^2+y^2+z^2}\ \dee{V} =\int_0^{1/\sqrt{2}}\dee{r}\int_0^{2\pi}\dee{\theta} \int_r^{\sqrt{1-r^2}}\dee{z}\ r\sqrt{r^2+z^2} \end{align*}

(b) In spherical coordinates

$x^2+y^2+z^2\le 1$ is $\rho\le 1$ and
$x^2+y^2\le z^2$ is $\rho^2\sin^2\varphi\le \rho^2\cos^2\varphi\text{,}$ or $\tan\varphi\le 1$ or $\varphi\le\frac{\pi}{4}\text{,}$ and
$\dee{V}$ is $\rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}$

\begin{align*} E =\big\{\ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 1,\ &0\le\theta\le2\pi,\\ &\hskip0.25in 0\le \varphi\le \frac{\pi}{4}\ \big\} \end{align*}

and, since the integrand $\sqrt{x^2+y^2+z^2}=\rho\text{,}$

\begin{align*} J &=\tripInt_E \sqrt{x^2+y^2+z^2}\ \dee{V} =\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^2\sin\varphi\ \rho\\ &=\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi \end{align*}

\begin{align*} J &=\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta} \int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi\\ &=2\pi \int_0^{1}\dee{\rho}\ \rho^3 \Big[-\cos\varphi\Big]_0^{\pi/4} =2\pi\ \frac{1}{4}\ \left[1-\frac{1}{\sqrt{2}}\right]\\ &=\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right] \end{align*}

3.7.5.25. (✳).

Solution.

(a) As a check, the body of the snow man has radius $\sqrt{12} = 2\sqrt{3} \approx 3.46\text{,}$ which is between $2$ (the low point of the head) and $4$ (the center of the head). Here is a sketch of a side view of the snowman.

We want to determine the volume of the intersection of the body and the head, whose side view is the darker shaded region in the sketch.

The outer boundary of the body and the outer boundary of the head intersect when both $x^2 + y^2 + z^2 = 12$ and $x^2 + y^2 + (z - 4)^2 = 4\text{.}$ Subtracting the second equation from the first gives
\begin{equation*} z^2-(z-4)^2 =12 -4 \iff 8z-16 =8 \iff z=3 \end{equation*}
Then substituting $z=3$ into either equation gives $x^2+y^2=3\text{.}$ So the intersection of the outer boundaries of the head and body (i.e. the neck) is the circle $x^2+y^2=3\text{,}$ $z=3\text{.}$
The top boundary of the intersection is part of the top half of the snowman’s body and so has equation $z=+\sqrt{12-x^2-y^2}\text{.}$
The bottom boundary of the intersection is part of the bottom half of the snowman’s head, and so has equation $z=4-\sqrt{4-x^2-y^2}$

The intersection of the head and body is thus

\begin{gather*} \cV = \Set{(x,y,z}{x^2+y^2\le 3,\ 4-\sqrt{4-x^2-y^2}\le z\le \sqrt{12-x^2-y^2}} \end{gather*}

We’ll compute the volume of $\cV$ using cylindrical coordinates

\begin{align*} &\text{Volume}(\cV) = \int_0^{\sqrt{3}}\dee{r}\int_0^{2\pi}\dee{\theta} \int_{4-\sqrt{4-r^2}}^{\sqrt{12-r^2}}\dee{z}\ r\\ &\hskip0.5in =\int_0^{\sqrt{3}}\dee{r}\ 2\pi\ r\big[\sqrt{12-r^2}-4+\sqrt{4-r^2}\big]\\ &\hskip0.5in=2\pi\left[-\frac{1}{3}{\big(12-r^2\big)}^{3/2} -2r^2 -\frac{1}{3}{\big(4-r^2\big)}^{3/2}\right]_0^{\sqrt{3}}\\ &\hskip0.5in= 2\pi\left[-\frac{1}{3}{\big(9\big)}^{3/2} -2(3) -\frac{1}{3}{\big(1\big)}^{3/2} +\frac{1}{3}{\big(12\big)}^{3/2} +2(0)^2 +\frac{1}{3}{\big(4\big)}^{3/2} \right]\\ &\hskip0.5in= 2\pi\left[-9 -6 -\frac{1}{3} +\frac{1}{3}{\big(12\big)}^{3/2} +\frac{8}{3} \right]\\ &\hskip0.5in=2\pi\left[\frac{1}{3}{\big(12\big)}^{3/2} -\frac{38}{3} \right] \end{align*}

So the volume of the snowman is

\begin{align*} &\frac{4\pi}{3}\big(12\big)^{3/2} +\frac{4\pi}{3} 2^3 -2\pi\left[\frac{1}{3}{\big(12\big)}^{3/2} -\frac{38}{3} \right]\\ &\hskip0.5in=\frac{2\pi}{3}\left[\big(12\big)^{3/2}+54\right] \end{align*}

(b) The top figure below is another side view of the snowman. This time it is divided into a lighter gray top part, a darker gray middle part and a lighter gray bottom part. The bottom figure below is an enlarged view of the central part of the figure on the left.

i. The top part is the Pac-Man

part of the snowman’s head. It is the part of the sphere

\begin{equation*} x^2+y^2+(z-4)^2\le 4 \end{equation*}

that is above the cone

\begin{equation*} z-4 = - \sqrt{\frac{x^2+y^2}{3}} \end{equation*}

(which contains the points $(0,0,4)$ and $(\sqrt{3},0,3)$).

ii. The middle part is the diamond shaped

part of the snowman’s head and body. It is bounded on the top by the cone

\begin{equation*} z-4 = - \sqrt{\frac{x^2+y^2}{3}} \end{equation*}

(which contains the points $(0,0,4)$ and $(\sqrt{3},0,3)$) and is bounded on the bottom by the cone

\begin{equation*} z = \sqrt{3(x^2+y^2)} \end{equation*}

(which contains the points $(0,0,0)$ and $(\sqrt{3},0,3)$).

iii. The bottom part is the Pac-Man

part of the snowman’s body. It is the part of the sphere

\begin{equation*} x^2+y^2+z^2\le 12 \end{equation*}

that is below the cone

\begin{equation*} z = \sqrt{3(x^2+y^2)} \end{equation*}

(which contains the points $(0,0,0)$ and $(\sqrt{3},0,3)$).

3.7.5.26. (✳).

Solution.

(a) Recall that, in spherical coordinates, $\varphi$ runs from $0$ (that’s the positive $z$-axis) to $\pi$ (that’s the negative $z$-axis), $\theta$ runs from $0$ to $2\pi$ ($\theta$ is the regular polar or cylindrical coordinate) and $\dee{V} = \rho^2\ \sin\varphi\ \dee{\rho}\,\dee{\theta}\,\dee{\varphi}\text{.}$ So

\begin{align*} \text{Volume} & = \int_0^\pi \dee{\varphi}\int_0^{2\pi}\dee{\theta} \int_0^{8\sin\varphi}\dee{\rho}\ \rho^2\ \sin\varphi\\ & = \int_0^\pi \dee{\varphi}\int_0^{2\pi}\dee{\theta}\ \frac{(8\sin\varphi)^3}{3} \sin\varphi\\ & = \frac{2(8^3)\,\pi}{3} \int_0^\pi \dee{\varphi}\ \sin^4\varphi\\ &= \frac{2(8^3)\,\pi}{3} \left[\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big)\right]_0^\pi\\ &=\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32} =128\pi^2 \end{align*}

(b) Fix any $\varphi$ between $0$ and $\pi\text{.}$ If $\rho=8\sin\varphi\text{,}$ then as $\theta$ runs from $0$ to $2\pi\text{,}$

\begin{align*} (x,y,z) &= \big(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi\big)\\ &= \big(8\sin^2\varphi\cos\theta\,,\,8\sin^2\varphi\sin\theta\,,\, 8\sin\varphi\cos\varphi\big)\\ &= \big(R\cos\theta\,,\,R\sin\theta\,,\,Z\big)\qquad \text{with }R=8\sin^2\varphi,\ Z= 8\sin\varphi\cos\varphi \end{align*}

sweeps out a circle of radius $R=8\sin^2\varphi$ contained in the plane $z=Z=8\sin\varphi\cos\varphi$ and centred on $\big(0,0,Z=8\sin\varphi\cos\varphi\big)\text{.}$ So the surface is a bunch of circles stacked one on top of the other. It is a surface of revolution. We can sketch it by

first sketching the $\theta=0$ section of the surface (that’s the part of the surface in the right half of the $xz$-plane)
and then rotate the result about the $z$-axis.

The $\theta=0$ part of the surface is

\begin{align*} &\Set{(x,y,z)}{x=8\sin^2\varphi,\ y=0,\ z=8\sin\varphi\cos\varphi,\ 0\le\varphi\le \pi}\\ &=\Set{(x,y,z)}{x=4-4\cos(2\varphi),\ y=0,\ z=4\sin(2\varphi),\ 0\le\varphi\le \pi} \end{align*}

It’s a circle of radius $4\text{,}$ contained in the $xz$-plane (i.e. $y=0$) and centred on $(4,0,0)\text{!}$ The figure on the left below is a sketch of the top half of the circle. When we rotate the circle about the $z$-axis we get a torus (a donut) but with the hole in the centre shrunk to a point. The figure on the right below is a sketch of the part of the torus in the first octant.

3.7.5.27. (✳).

Solution.

(a) In cylindrical coordinates $0\le z\le \sqrt{x^2+y^2}$ becomes $0\le z\le r\text{,}$ and $x^2+y^2\le 1$ becomes $0\le r\le 1\text{.}$ So

\begin{equation*} E = \Set{(r\cos\theta\,,\,r\sin\theta\,,\,z)}{0\le r\le 1,\ 0\le\theta\le 2\pi, \ 0\le z\le r} \end{equation*}

and, since $\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}\text{,}$

\begin{align*} I &= \tripInt_E z \sqrt{x^2 + y^2 + z^2}\ \dee{V} = \int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_0^r\dee{z}\ r\ z\ \sqrt{\underbrace{r^2+z^2}_{x^2+y^2+z^2}} \end{align*}

(b) Here is a sketch of a constant $\theta$ section of $E\text{.}$

Recall that the spherical coordinate $\varphi$ is the angle between the $z$-axis and the radius vector. So, in spherical coordinates $z=r$ (which makes an angle $\frac{\pi}{4}$ with the $z$ axis) becomes $\varphi=\frac{\pi}{4}\text{,}$ and the plane $z=0\text{,}$ i.e. the $xy$-plane, becomes $\varphi=\frac{\pi}{2}\text{,}$ and $r=1$ becomes $\rho\sin\varphi = 1\text{.}$ So

\begin{align*} E = \Big\{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \Big|\ \frac{\pi}{4}\le \varphi\le \frac{\pi}{2}, \ &0\le\theta\le 2\pi,\\ &\hskip0.1in 0\le \rho\le \frac{1}{\sin\varphi}\ \Big\} \end{align*}

and, since $\dee{V} = \rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}\text{,}$

\begin{align*} I &= \tripInt_E z \sqrt{x^2 + y^2 + z^2}\ \dee{V}\\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z}\ \rho\\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^4\sin\varphi \ \cos\varphi \end{align*}

\begin{align*} I &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{1/\sin\varphi}\dee{\rho}\ \rho^4\sin\varphi \ \cos\varphi\\ &= \int_{\pi/4}^{\pi/2}\dee{\varphi} \int_0^{2\pi}\dee{\theta}\ \frac{1}{5\sin^5\varphi}\sin\varphi \ \cos\varphi\\ &= \frac{2\pi}{5}\int_{\pi/4}^{\pi/2}\dee{\varphi}\ \frac{\cos\varphi}{\sin^4\varphi}\\ &= \frac{2\pi}{5}\int_{1/\sqrt{2}}^{1} \frac{\dee{u}}{u^4}\qquad\text{with } u=\sin\varphi,\ \dee{u}=\cos\varphi\,\dee{\varphi}\\ &= \frac{2\pi}{5} \left[\frac{u^{-3}}{-3}\right]_{1/\sqrt{2}}^{1}\\ &= \frac{2(2\sqrt{2}-1)\pi}{15} \end{align*}

3.7.5.28. (✳).

Solution.

The main step is to figure out what the domain of integration looks like.

The outside integral says that $x$ runs from $-a$ to $0\text{.}$
The middle integrals says that, for each $x$ in that range, $y$ runs from $-\sqrt{a^2-x^2}$ to $0\text{.}$ We can rewrite $y=-\sqrt{a^2-x^2}$ in the more familiar form $x^2+y^2=a^2\text{,}$ $y\le 0\text{.}$ So $(x,y)$ runs over the third quadrant part of the disk of radius $a\text{,}$ centred on the origin.
Finally, the inside integral says that, for each $(x,y)$ in the quarter disk, $z$ runs from $0$ to $\sqrt{a^2-x^2-y^2} \text{.}$ We can also rewrite $z=\sqrt{a^2-x^2-y^2}$ in the more familiar form $x^2+y^2+z^2=a^2\text{,}$ $z\ge 0\text{.}$

So the domain of integration is the part of the interior of the sphere of radius $a\text{,}$ centred on the origin, that lies in the octant $x\le 0\text{,}$ $y\le 0\text{,}$ $z\ge 0\text{.}$

\begin{align*} V&=\Big\{\ (x,y,z)\ \Big|\ -a\le x\le 0,\ -\sqrt{a^2-x^2}\le y\le 0,\\ &\hskip0.1in 0\le z\le\sqrt{a^2-x^2-y^2} \Big\}\\ &=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ x\le 0,\ y\le 0,\ z\ge 0 } \end{align*}

(a) Note that, in $V\text{,}$ $(x,y)$ is restricted to the third quadrant, which in cylindrical coordinates is $\pi\le\theta\le\frac{3\pi}{2}\text{.}$ So, in cylindrical coordinates,

\begin{align*} V&=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ r^2+z^2\le a^2,\ \pi\le\theta\le\frac{3\pi}{2},\ z\ge 0 \Big\}\\ &=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ 0\le z\le a,\ \pi\le\theta\le\frac{3\pi}{2},\ 0\le r\le\sqrt{a^2-z^2} \Big\} \end{align*}

and

\begin{align*} I&= \tripInt_V \big(x^2+y^2+z^2\big)^{2014}\ \dee{V} = \tripInt_V \big(r^2+z^2\big)^{2014}\ r\,\dee{r}\,\dee{\theta}\,\dee{z}\\ &=\int_0^a \dee{z}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^{\sqrt{a^2-z^2}}\dee{r}\ r\big(r^2+z^2\big)^{2014} \end{align*}

(b) The spherical coordinate $\varphi$ runs from $0$ (when the radius vector is along the positive $z$-axis) to $\frac{\pi}{2}$ (when the radius vector lies in the $xy$-plane) so that

\begin{align*} I&= \tripInt_V \big(x^2+y^2+z^2\big)^{2014}\ \dee{V} = \tripInt_V \rho^{2\times 2014}\ \rho^2\sin\varphi\ \dee{\rho}\,\dee{\theta}\,\dee{\varphi}\\ &=\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi \end{align*}

\begin{align*} I&=\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi\\ &=\frac{a^{4031}}{4031} \int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta} \ \sin\varphi\\ &=\frac{a^{4031}\pi}{8062} \int_0^{\pi/2} \dee{\varphi}\ \sin\varphi\\ &=\frac{a^{4031}\pi}{8062} \end{align*}

3.7.5.29. (✳).

Solution.

(a) In cylindrical coordinates, the paraboloid is $z=r^2$ and the cone is $z=r\text{.}$ The two meet when $r^2=r\text{.}$ That is, when $r=0$ and when $r=1\text{.}$ So, in cylindrical coordinates

\begin{gather*} I = \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\ z(r^2+z^2) \end{gather*}

(b) In spherical coordinates, the paraboloid is

\begin{equation*} \rho\cos\varphi=\rho^2\sin^2\varphi\qquad\text{or}\qquad \rho=\frac{\cos\varphi}{\sin^2\varphi} \end{equation*}

and the cone is

\begin{equation*} \rho\cos\varphi=\rho\sin\varphi\quad\text{or}\quad \tan\varphi=1\quad\text{or}\quad \varphi=\frac{\pi}{4} \end{equation*}

The figure below shows a constant $\theta$ cross-section of $E\text{.}$ Looking at that figure, we see that $\varphi$ runs from $\frac{\pi}{4}$ (i.e. the cone) to $\frac{\pi}{2}$ (i.e. the $xy$-plane).

So, is spherical coordinates,

\begin{align*} I &= \int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z} \overbrace{\rho^2}^{x^2+y^2+z^2}\\ &= \int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta} \int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^5\sin\varphi \cos\varphi \end{align*}

\begin{align*} I &= \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\ z(r^2+z^2)\\ &= \int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta}\ \left[r^2\frac{z^2}{2}+\frac{z^4}{4}\right]_{r^2}^r\\ &= 2\pi \int_0^1 \dee{r}\ r \left[r^2\frac{r^2}{2}+\frac{r^4}{4} -r^2\frac{r^4}{2}-\frac{r^8}{4}\right]\\ &=2\pi \left[\frac{1}{12}+\frac{1}{24} -\frac{1}{16}-\frac{1}{40}\right] =\frac{\pi}{2} \left[\frac{1}{3}+\frac{1}{6} -\frac{1}{4}-\frac{1}{10}\right]\\ &=\frac{3\pi}{40} \end{align*}

3.7.5.30. (✳).

Solution.

Note that both the sphere $x^2+y^2+(z-1)^2=1$ and the cone $z=\sqrt{x^2+y^2}$ are invariant under rotations around the $z$-axis. The sphere $x^2 + y^2 + (z-1)^2 = 1$ and the cone $z = \sqrt{x^2 + y^2}$ intersect when $z=\sqrt{x^2+y^2}\text{,}$ so that $x^2+y^2=z^2\text{,}$ and

\begin{align*} x^2 + y^2 + (z-1)^2 = z^2+(z-1)^2=1 &\iff 2z^2-2z=0\\ &\iff 2z(z-1)=0\\ &\iff z=0,1 \end{align*}

So the surfaces intersect on the circle $z=1\text{,}$ $x^2+y^2=1$ and

\begin{equation*} S = \Set{(x,y,z)}{x,y\ge 0,\ x^2+y^2\le 1,\ \sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2}} \end{equation*}

Here is a sketch of the $y=0$ cross section of S.

(a) In cylindrical coordinates

the condition $x,y\ge 0$ is $0\le\theta\le\frac{\pi}{2}\text{,}$
the condition $x^2+y^2\le 1$ is $r\le 1\text{,}$ and
the conditions $\sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2}$ are $r\le z\le 1+\sqrt{1-r^2}\text{,}$ and
$\dee{V} = r\,\dee{r}\,\dee{\theta}\,\dee{z}\text{.}$

\begin{gather*} V = \tripInt_S\dee{V} = \int_0^1\dee{r} \int_0^{\pi/2}\dee{\theta}\int_r^{1+\sqrt{1-r^2}}\dee{z} \ r \end{gather*}

(b) In spherical coordinates,

the cone $z=\sqrt{x^2+y^2}$ becomes
\begin{align*} &\rho\,\cos\varphi =\sqrt{\rho^2\sin^2\varphi\,\cos^2\theta +\rho^2\sin^2\varphi\,\sin^2\theta} =\rho\sin\varphi\\ &\hskip0.5in\iff \tan\varphi=1\\ &\hskip0.5in\iff \varphi=\frac{\pi}{4} \end{align*}
so that, on $S\text{,}$ the spherical coordinate $\varphi$ runs from $\varphi=0$ (the positive $z$ -axis) to $\varphi=\frac{\pi}{4}$ (the cone $z=\sqrt{x^2+y^2}$), which keeps us above the cone,
the condition $x,y\ge 0$ is $0\le\theta\le\frac{\pi}{2}\text{,}$
the condition $x^2+y^2+(z-1)^2\le 1\text{,}$ (which keeps us inside the sphere), becomes
\begin{align*} &\rho^2\sin^2\varphi\,\cos^2\theta + \rho^2\sin^2\varphi\,\sin^2\theta +\big(\rho\cos\varphi-1\big)^2\le 1\\ &\iff \rho^2\,\sin^2\varphi +\rho^2\cos^2\varphi -2\rho\cos\varphi +1 \le 1\\ &\iff \rho^2-2\rho\cos\varphi\le 0\\ &\iff \rho\le 2 \cos\varphi \end{align*}
and $\dee{V} = \rho^2\sin\varphi\,\dee{\rho}\,\dee{\theta}\,\dee{\varphi}\text{.}$

\begin{gather*} V = \tripInt_S\dee{V} = \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi \end{gather*}

\begin{align*} V &= \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta} \int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi\\ &=\frac{8}{3} \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta}\ \cos^3\varphi\ \sin\varphi\\ &=\frac{8}{3}\ \frac{\pi}{2}\ \left[-\frac{\cos^4\varphi}{4}\right]_0^{\pi/4}\\ &=\frac{\pi}{3} \left[1-\frac{1}{(\sqrt{2})^4}\right]\\ &=\frac{\pi}{4} \end{align*}

3.7.5.31. (✳).

Solution.

(a) In cylindrical coordinates, the density of is $\de=x^2+y^2=r^2\text{,}$ the bottom of the solid is at $z=\sqrt{3x^2+3y^2}=\sqrt{3}\,r$ and the top of the solid is at $z=\sqrt{9-x^2-y^2}=\sqrt{9-r^2}\text{.}$ The top and bottom meet when

\begin{equation*} \sqrt{3}\,r=\sqrt{9-r^2}\iff 3r^2=9-r^2\iff 4r^2=9\iff r=\frac{3}{2} \end{equation*}

The mass is

\begin{gather*} m=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ \overbrace{r^2}^{\de} \end{gather*}

(b) In spherical coordinates, the density of is $\de=x^2+y^2=\rho^2\sin^2\varphi\text{,}$ the bottom of the solid is at

\begin{equation*} z=\sqrt{3}\,r\iff \rho\cos\varphi=\sqrt{3}\,\rho\sin\varphi\iff \tan\varphi=\frac{1}{\sqrt{3}}\iff\varphi=\frac{\pi}{6} \end{equation*}

and the top of the solid is at $x^2+y^2+z^2=\rho^2=9\text{.}$ The mass is

\begin{equation*} m=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\int_0^3\dee{\rho}\ \big(\rho^2\sin\varphi\big) \overbrace{\big(\rho^2\sin^2\varphi\big)}^{\de} \end{equation*}

(c) Solution 1: Making the change of variables $s=\cos\varphi\text{,}$ $\dee{s}=-\sin\varphi\ \dee{\varphi}\text{,}$ in the integral of part (b),

\begin{align*} m&=\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi} \int_0^3\dee{\rho}\ \rho^4\sin\varphi \big(1-\cos^2\varphi\big)\\ &=\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\ \sin\varphi \big(1-\cos^2\varphi\big)\\ &=-\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\int_1^{\sqrt{3}/2}\dee{s}\ (1-s^2)\\ &=-\frac{3^5}{5}\int_0^{2\pi}\dee{\theta}\ \left[s-\frac{s^3}{3}\right]_1^{\sqrt{3}/2}\\ &=2\pi \frac{3^5}{5} \left[1-\frac{1}{3}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{8}\right]\\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{align*}

\begin{align*} m&=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2\cr &=\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big)\cr &=2\pi\int_0^{3/2}\dee{r}\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big) \end{align*}

The second term

\begin{equation*} -2\pi\int_0^{3/2}\dee{r}\ \sqrt{3}\,r^4=-2\pi\sqrt{3}\,\frac{r^5}{5}\bigg|_0^{3/2} =-2\pi\sqrt{3}\,\frac{3^5}{5\times 2^5} \end{equation*}

For the first term, we substitute $s=9-r^2\text{,}$ $\dee{s}=-2r\,\dee{r}\text{.}$

\begin{align*} 2\pi\int_0^{3/2}\dee{r}\ r^3\sqrt{9-r^2} &=2\pi\int_9^{27/4}\overbrace{\frac{\dee{s}}{-2}}^{r\,\dee{r}} \overbrace{(9-s)}^{r^2} \sqrt{s} =-\pi \left[6s^{3/2}-\frac{2}{5}s^{5/2}\right]_9^{27/4}\cr &=-\pi \left[\frac{3^5\sqrt{3}}{4}-2\times 3^4 -\frac{3^7}{2^4 5}\sqrt{3}+2\frac{3^5}{5}\right] \end{align*}

Adding the two terms together,

\begin{align*} m&=-2\pi\frac{3^5}{5} \frac{\sqrt{3}}{32} -2\pi\frac{3^5}{5} \frac{5\sqrt{3}}{8} +2\pi\frac{3^5}{5} \frac{5}{3} +2\pi\frac{3^5}{5} \frac{9\sqrt{3}}{32} -2\pi\frac{3^5}{5}\\ &=2\pi\frac{3^5}{5}\left[\left(\frac{5}{3}-1\right) -\sqrt{3}\left(\frac{1}{32}+\frac{5}{8}-\frac{9}{32}\right)\right]\\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{align*}

Prev TopNext

\(\Atop{\text{critical}}{\text{point}}\)	\(f_{xx}f_{yy}-f_{xy}^2\)	\(f_{xx}\)	type
\((0,0)\)	\(0\times 0 -(-2)^2 \lt 0\)		saddle point
\(\left(-\frac{2}{3},\frac{2}{3}\right)\)	\((-4)\times (-4)-(-2)^2 \gt 0\)	\(-4\)	local max

\(\Atop{\text{critical}}{\text{point}}\)	\(f_{xx}f_{yy}-f_{xy}^2\)	\(f_{xx}\)	type
\((0,3)\)	\((6)\times (0)-(6)^2 \lt 0\)		saddle point
\((0,-3)\)	\((-6)\times (0)-(-6)^2 \lt 0\)		saddle point
\((-2,1)\)	\((-10)\times (-4)-(-2)^2 \gt 0\)	\(-10\)	local max
\((2,-1)\)	\((10)\times (4)-(2)^2 \gt 0\)	\(10\)	local min

\(\Atop{\text{critical}}{\text{point}}\)	\(f_{xx}f_{yy}-f_{xy}^2\)	\(f_{xx}\)	type
\((0,0)\)	\(2\times 2-0^2 \gt 0\)	\(2 \gt 0\)	local min
\((\sqrt{2},-1)\)	\(0\times 2-(2\sqrt{2})^2 \lt 0\)		saddle point
\((-\sqrt{2},-1)\)	\(0\times 2-(-2\sqrt{2})^2 \lt 0\)		saddle point

\(\Atop{\text{critical}}{\text{point}}\)	\(f_{xx}f_{yy}-f_{xy}^2\)	\(f_{xx}\)	type
\(\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\)	\((2\sqrt{3}+2)\times (2)-(-2)^2 \gt 0\)	\(2\sqrt{3}+2 \gt 0\)	local min
\(-\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\)	\((-2\sqrt{3}+2)\times (2)-(-2)^2 \lt 0\)		saddle point

\(\Atop{\text{critical}}{\text{point}}\)	\(f_{xx}f_{yy}-f_{xy}^2\)	\(f_{xx}\)	type
\((0,0)\)	\((-6)\times(-8)-(0)^2 \gt 0\)	\(-6\)	local max
\((2,0)\)	\(6\times(-4)-(0)^2 \lt 0\)		saddle point

point	\((0,0)\)	\(x=0\)	\(y=0\)	\((1,0)\)	\((1,\half)\)	\((1,1)\)	\((0,1)\)	\((\frac{1}{\sqrt{3}},1)\)	\((1,1)\)
value of \(f\)	\(0\)	\(0\)	\(0\)	\(0\)	\(\frac{1}{4}\)	\(0\)	\(0\)	\(\frac{2}{3\sqrt{3}}\)	\(0\)
	min	min	min	min		min	min	max	min

point	\((\half,\half)\)	\((-\half,-\half)\)	\((\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\)	\((-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\)
value of \(T\)	\(\frac{1}{\sqrt{e}}\approx0.61\)	\(-\frac{1}{\sqrt{e}}\)	\(\frac{\sqrt{2}}{e}\approx0.52\)	\(-\frac{\sqrt{2}}{e}\)
	max	min

\(\Atop{\text{critical}}{\text{point}}\)	\(h_{xx}h_{yy}-h_{xy}^2\)	\(h_{xx}\)	type
\(\left(0,\frac{2}{\sqrt{3}}\right)\)	\(\left(\frac{-4}{\sqrt{3}}\right)\times \left(-\frac{12}{\sqrt{3}}\right)-(0)^2 \gt 0\)	\(\frac{-4}{\sqrt{3}}\)	local max
\(\left(0,-\frac{2}{\sqrt{3}}\right)\)	\(\left(\frac{4}{\sqrt{3}}\right)\times \left(\frac{12}{\sqrt{3}}\right)-(0)^2 \gt 0\)	\(\frac{4}{\sqrt{3}}\)	local min
\((2,0)\)	\(0\times 0-(-4)^2 \lt 0\)		saddle point
\((-2,0)\)	\(0\times 0-(4)^2 \lt 0\)		saddle point

point	\((1,0)\)	\((-1,0)\)	\((1,\pm 1)\)	\((-1,\pm 1)\)	\((3,0)\)	\((3,\pm 1)\)
value of \(f\)	\(6\)	\(2\)	\(2\)	\(-2\)	\(2\)	\(-2\)
	max			min		min

point(s)	\((1,1)\)	\(L_1\)	\(L_2\)	\((0,8)\)	\((8,0)\)	\((4,4)\)
value of \(f\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(80\)
	min					max

\((x,y)\)	\(f(x,y)\)
\((2,2)\)	\(2\times 2(2+2\times2-6)=0\)
\((4,1)\)	\(4\times 1(4+2\times 1-6)=0\)
\((2\sqrt{2},\sqrt{2})\)	\(4(2\sqrt{2}+2\sqrt{2}-6) \lt 0\)

\((x,y)\)	\(g(x,y)\)
\((0,-2)\)	\(16\)
\((\pm 4,0)\)	\(16\)
\((\pm \sqrt{12},-1)\)	\(21\)
\((0,0)\)	\(0\)

\((x,y)\)	\(g(x,y)\)
\((\sqrt{2},\sqrt{2})\)	\(2e^{-2}\approx0.271\)
\((x,0)\)	\(0\)
\((0,y)\)	\(0\)
\((1,1)\)	\(e^{-1}\approx0.368\)

point	\((-2\sqrt{2},\sqrt{2}, 10)\)	\((2\sqrt{2},-\sqrt{2}, 10)\)	\((-2,-4,20) \)	\((1,2,5) \)
value of \(f\)	\(8+2-5\)	\(8+2-5\)	\(4+16-20\)	\(1+4-\frac{25}{20}\)
	max	max	min

point	\(\big(-1 + \sqrt{2}, 0, 1-\sqrt{2}\big)\)	\(\big(-1 - \sqrt{2}, 0, 1+\sqrt{2}\big)\)	\((-2,2,2)\)	\((-2,-2,2) \)
value of \(f\)	\(2\big(3-2\sqrt{2}\big)\)	\(2\big(3+2\sqrt{2}\big)\)	\(12\)	\(12\)
	min		max	max

point	\((0,0, -6)\)	\((3\sqrt{3},0,3)\)	\((-3\sqrt{3},0,3)\)	\((4,4,2) \)	\((4,-4,2) \)
value of \(f\)	\(0\)	\(27\sqrt{3}\)	\(-27\sqrt{3}\)	\(48\)	\(48\)
			min	max	max

point	\((0,0, \pm 6)\)	\((3\sqrt{3},0,3)\)	\((-3\sqrt{3},0,3)\)	\((4,4,2) \)	\((4,-4,2) \)
value of \(f\)	\(0\)	\(27\sqrt{3}\)	\(-27\sqrt{3}\)	\(48\)	\(48\)
			min	max	max

point	\((0,10)\)	\((0,-10)\)
value of \(T\)	\(100 e^{10}\)	\(100 e^{-10}\)
	max	min

point	\((1,-1,2)\)	\(-(1,-1,2)\)
value of \(f\)	\(6\)	\(54\)
	min	max

point	\(\frac{1}{\sqrt{5}}(2,1,0)\)	\(-\frac{1}{\sqrt{5}}(2,1,0)\)
value of \(f\)	\(\big(1-\sqrt{5}\big)^2\)	\(\big(1+\sqrt{5}\big)^2\)
	min	max

point	\((1,2,1)\)	\((-1,2,-1)\)
value of \(f\)	\(2e^2\)	\(-2e^2\)
	max	min

Appendix D Solutions to Exercises

1 Vectors and Geometry in Two and Three Dimensions1.1 Points

Exercises

1.1.1.

1.1.2.

1.1.3.

1.1.4.

1.1.5.

1.1.6. (✳).

1.1.7.

1.1.8.

1.2 Vectors1.2.9 Exercises

1.2.9.1.

1.2.9.2.

1.2.9.3.

1.2.9.4.

1.2.9.5.

1.2.9.6.

1.2.9.7.

1.2.9.8.

1.2.9.9.

1.2.9.10.

1.2.9.11.

1.2.9.12.

1.2.9.13.

1.2.9.14.

1.2.9.15.

1.2.9.16.

1.2.9.17.

1.2.9.18.

1.2.9.19. (✳).

1.2.9.20.

1.2.9.21.

1.2.9.22.

1.2.9.23.

1.2.9.24.

1.2.9.25.

1.2.9.26.

1.2.9.27.

1.2.9.28.

1.2.9.29.

1.2.9.30. (✳).

1.2.9.31.

1.2.9.32.

1.2.9.33.

1.2.9.34.

1.2.9.35.

1.2.9.36.

1.2.9.37.

1.2.9.38.

1.2.9.39.

1.2.9.40.

1.2.9.41.

1.3 Equations of Lines in 2d

Exercises

1.3.1.

1.3.2.

1.3.3.

1.3.4.

1.3.5.

1.3.6.

1.3.7.

1.3.8.

1.3.9.

1.4 Equations of Planes in 3d

Exercises

1.4.1.

1.4.2.

1.4.3.

1.4.4. (✳).

1.4.5.

1.4.6.

1.4.7.

1.4.8.

1.4.9. (✳).

1.4.10. (✳).

1.4.11. (✳).

1.4.12. (✳).

1.4.13. (✳).

1.4.14.

1 Vectors and Geometry in Two and Three Dimensions
1.1 Points

1.2 Vectors
1.2.9 Exercises

1.6 Curves and their Tangent Vectors
1.6.2 Exercises

1.7 Sketching Surfaces in 3d
1.7.2 Exercises