1.1.1.
Answer.
- The sphere of radius 3 centered on \((1,-2,0)\text{.}\)
- The interior of the sphere of radius 3 centered on \((1,-2,0)\text{.}\)
Appendix C Answers to Exercises
1 Vectors and Geometry in Two and Three Dimensions
1.1 Points
Exercises
1.1.2.
Answer.
-
\(x=y\) 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.
-
\(x^2+y^2=4\) is the circle with centre \((0,0)\) and radius 2. It is sketched in the figure on the left below.
- \(x^2+y^2=2y\) is the circle with centre \((0,1)\) and radius 1. It is sketched in the figure on the right above.
-
\(x^2+y^2 \lt 2y\) is the set of points that are strictly inside the circle with centre \((0,1)\) and radius 1. It is the shaded region (not including the dashed circle) in the sketch below.
1.1.3.
Answer.
-
The set \(z=x\) is 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.
-
\(x^2+y^2+z^2=4\) 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.
-
(d) \(x^2+y^2+z^2=4\text{,}\) \(z=1\) is the 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
-
\(x^2+y^2=4\) 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.
-
\(z=x^2+y^2\) is a paraboloid consisting of 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.
Answer.
- \(\displaystyle 3\)
- \(\displaystyle 1\)
- \(\displaystyle \sqrt{(x-2)^2 + 10}\)
- \(\displaystyle (2,0,0)\)
- \(\displaystyle \sqrt{10}\)
1.1.5.
Answer.
The circumscribing circle has centre \((\bar x,\bar y)\) and radius \(r\) with \(\bar x=\frac{a}{2}\text{,}\) \(\bar y=\frac{b^2+c^2-ab}{2c}\) and \(r=\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}\text{.}\)
1.1.6. (✳).
Answer.
\(x^2+y^2=4z\) The surface is a paraboloid consisting of a stack of horizontal circles, starting with a point at the origin and with radius increasing vertically. The circle in the plane \(z=z_0\) has radius \(2\sqrt{z_0}\text{.}\)
1.1.7.
Answer.
The sphere has radius 3 and is centered on \((1,2,-1)\text{.}\)
1.1.8.
Answer.
1.2 Vectors
1.2.9 Exercises
1.2.9.1.
Answer.
\(\va+\vb=\llt 3,1\rgt \text{,}\) \(\va+2\vb=\llt 4,2\rgt \) \(2\va-\vb=\llt 3,-1\rgt \)
1.2.9.2.
Answer.
- not collinear
- collinear
1.2.9.3.
Answer.
- perpendicular
- perpendicular
- not perpendicular
1.2.9.4.
Answer.
- \(\displaystyle \frac{1}{5}\llt 3,4 \rgt\)
- \(\displaystyle \pm\frac{1}{5}\llt 3,4 \rgt\)
- \(\displaystyle \pm \llt 6,8 \rgt\)
- \(\displaystyle \pm\frac{1}{5}\llt 4,-3 \rgt\)
1.2.9.5.
Answer.
- \(\displaystyle \frac{1}{5}\llt 3,4,0 \rgt\)
- \(\displaystyle \pm\frac{1}{5}\llt 3,4,0 \rgt\)
- A vector is of length one and perpendicular to \(\vb\) if and only if it is of the form \(\llt x,-\frac{3}{4}x,z \rgt\) with \(\sqrt{\frac{25}{16}x^2+z^2}=1\text{.}\) There are infinitely many such vectors. Four of them are\begin{equation*} \pm\llt 0,0,1\rgt\qquad \pm\frac{1}{5}\llt 4,-3,0\rgt \end{equation*}
1.2.9.6.
Answer.
\(\text{proj}_{\hi}\va=a_1\hi\) \(\text{proj}_{\hj}\va=a_2\hj\text{.}\)
1.2.9.7.
Answer.
Yes.
1.2.9.8.
Answer.
See the solution.
1.2.9.9.
Answer.
See the solution
1.2.9.10.
Answer.
See the solution.
1.2.9.11.
Answer.
This statement is false. One counterexample is \(\va=\llt 1,0,0\rgt \text{,}\) \(\vb=\llt 0,1,0\rgt ,\ \vc=\llt 0,0,1\rgt \text{.}\) Then \(\va\cdot\vb=\va\cdot\vc=0\text{,}\) but \(\vb\ne\vc\text{.}\) There are many other counterexamples.
1.2.9.12.
Answer.
True.
1.2.9.13.
Answer.
None. The given equation is nonsense.
1.2.9.14.
Answer.
If \(\vb\) and \(\vc\) are parallel, then \(\va\cdot(\vb\times\vc)=0\) for all \(\va\text{.}\) If \(\vb\) and \(\vc\) are not parallel, then \(\va\) must be of the form \(\al\vb+\be\vc\) with \(\al\) and \(\be\) real numbers.
1.2.9.15.
Answer.
(a), (c)
\(\text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC}
=\overrightarrow{OP_A}=\llt a/2,0\rgt\) \(\text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC}
=\overrightarrow{OP_A}=\llt b/2,c/2\rgt\)
(b) The centre of the circumscribing circle is \((\bar x,\bar y)\) with \(\bar x=\frac{a}{2}\) and \(\bar y =\frac{b^2+c^2-ab}{2c}\text{.}\)
1.2.9.16.
Answer.
\((x-3)^2+(y-2)^2+(z-7)^2=11\)
1.2.9.17.
Answer.
See the solution.
1.2.9.18.
Answer.
- \(\displaystyle 13\)
- \(\displaystyle 20\)
1.2.9.19. (✳).
Answer.
\(2\sqrt{19}\)
1.2.9.20.
Answer.
- \(\displaystyle 126\)
- \(\displaystyle 5\)
1.2.9.21.
Answer.
\begin{alignat*}{2}
\va\cdot\vb&=4\qquad &
\theta &= 60.25^\circ
\tag{a}\\
\va\cdot\vb&=0 &
\theta &= 90^\circ
\tag{b}\\
\va\cdot\vb&=4 &
\theta &= 0^\circ
\tag{c}\\
\va\cdot\vb&=2 &
\theta &= 61.87^\circ
\tag{d}\\
\va\cdot\vb&=0 &
\theta &= 90^\circ
\tag{e}
\end{alignat*}
1.2.9.22.
Answer.
- \(\displaystyle 10.3^\circ\)
- \(\displaystyle 61.6^\circ\)
- \(\displaystyle 82.6^\circ\)
1.2.9.23.
Answer.
- \(\displaystyle -1\)
- \(\displaystyle 0,4\)
- \(\displaystyle -2,-3\)
1.2.9.24.
Answer.
- \(\displaystyle -5\)
- \(\displaystyle 0.8\)
- none
1.2.9.25.
Answer.
- \(\displaystyle \frac{42}{\sqrt{14}}\)
- \(\displaystyle \llt 3,6,9\rgt\)
- \(\displaystyle \llt 1,4,-3\rgt\)
1.2.9.26.
Answer.
\(-3\hi+6\hj-3\hk\)
1.2.9.27.
Answer.
- \(\displaystyle \llt -27,-9,-9\rgt\)
- \(\displaystyle \llt -31,-34,8\rgt\)
- \(\displaystyle \llt -4,5,-4\rgt\)
1.2.9.28.
Answer.
(a) See the solution.
(b) \(\vp\times\vq = -\vq\times\vp = \llt -6,5,-13\rgt\)
(c) \(\vp \times (3\vr) = 3(\vp\times\vr) = \llt 6,9,-15\rgt\)
(d) \(\vp\times(\vq+\vr) = \vp\times\vq+\vp\times\vr = \llt -4,8,-18\rgt\)
(e) \(\vp\times(\vq\times\vr) = \llt -46,-19,15\rgt\text{,}\) \((\vp\times\vq)\times\vr = \llt -44,-32,8\rgt\)
1.2.9.29.
Answer.
\(2\sqrt{6}\)
1.2.9.30. (✳).
Answer.
See the solution.
1.2.9.31.
Answer.
See the solution.
1.2.9.32.
Answer.
- All \(6\) edges have length \(\sqrt{2}s\text{.}\)
- \(\displaystyle 109.5^\circ\)
1.2.9.33.
Answer.
\(35.26^\circ\) or \(90^\circ\) or \(144.74^\circ\)
1.2.9.34.
Answer.
\(\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)}\)
1.2.9.35.
Answer.
The marble rolls in the directionn\(\llt ac,bc,-a^2-b^2\rgt\text{.}\) 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.
Answer.
See the solution.
1.2.9.37.
Answer.
See the solution.
1.2.9.38.
Answer.
\((\va\times\vb)\cdot(\vc\times\vd)=
(\va\cdot\vc)(\vb\cdot\vd)
-(\va\cdot\vd)(\vb\cdot\vc)\)
1.2.9.39.
Answer.
- \(AA'B'B\) is a parallelogram, but not a rectangle.
- \(AA'C'C\) is a rectangle.
- \(BB'C'C\) is a parallelogram, but not a rectangle.
- \(\displaystyle \sqrt{17}\)
- \(\displaystyle \frac{13}{2}\)
- \(\displaystyle \frac{51}{2}\)
1.2.9.40.
Answer.
See the solution.
1.2.9.41.
Answer.
See the solution.
1.3 Equations of Lines in 2d
Exercises
1.3.1.
Answer.
Both!
1.3.2.
Answer.
Generally, only the first.
1.3.3.
Answer.
Since both lines pass through \((1,9)\) and \((9,13)\text{,}\) the lines are identical.
1.3.4.
Answer.
\(\llt d_x,d_y\rgt\) can be any nonzero scalar multiple of \(\llt 9,7\rgt\text{,}\) and \(\llt x_0,y_0\rgt\) can be any point on the line, i.e. any pair that satisfies \(7x_0+24=9y_0\text{.}\)
1.3.5.
Answer.
(a) \(\llt x,y\rgt=\llt 1,2\rgt+t\llt 3,2\rgt \text{,}\) \(x=1+3t,\ y=2+2t\text{,}\) \(\frac{x-1}{3}=\frac{y-2}{2}\)
(b) \(\llt x,y\rgt=\llt 5,4\rgt+t\llt 2,-1\rgt \text{,}\) \(x=5+2t,\ y=4-t\text{,}\) \(\frac{x-5}{2}=\frac{y-4}{-1}\)
(c) \(\llt x,y\rgt=\llt -1,3\rgt+t\llt -1,2\rgt \text{,}\) \(x=-1-t,\ y=3+2t\text{,}\) \(\frac{x+1}{-1}=\frac{y-3}{2}\)
1.3.6.
Answer.
(a) \(\llt x,y\rgt=\llt 1,2\rgt+t\llt -2,3\rgt \text{,}\) \(x=1-2t,\ y=2+3t\text{,}\) \(\frac{x-1}{-2}=\frac{y-2}{3}\)
(b) \(\llt x,y\rgt=\llt 5,4\rgt+t\llt 1,2\rgt \text{,}\) \(x=5+t,\ y=4+2t\text{,}\) \(x-5=\frac{y-4}{2}\)
(c) \(\llt x,y\rgt=\llt -1,3\rgt+t\llt 2,1\rgt \text{,}\) \(x=-1+2t,\ y=3+t\text{,}\) \(\frac{x+1}{2}=y-3\)
1.3.7.
Answer.
\(14/5\)
1.3.8.
Answer.
(a)
\begin{align*}
\vx(t)&=\va+t\big(\half\vb+\half\vc-\va\big)\\
\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) \(\frac{1}{3}(\va+\vb+\vc)\)
1.3.9.
Answer.
One way of writing the equation is \(x+{\sqrt3}y=4+{\sqrt3}\text{.}\)
1.4 Equations of Planes in 3d
Exercises
1.4.1.
Answer.
Any vector of the form \(c\,\hk\) with \(c\ne 0\) and \(c\ne 1\) works. Three possible choices are \(-\hk\text{,}\) \(2\,\hk\text{,}\) \(7.12345\,\hk\text{.}\)
1.4.2.
Answer.
- \(\displaystyle (0,8,0)\)
- \(\displaystyle (0,0,4)\)
1.4.3.
Answer.
- \(\displaystyle x+2y+3z=0\)
- \(\displaystyle x+y+3z=3\)
- There is no plane that passes through both \((1,2,3)\) and \((1,0,0)\) and has normal vector \(\llt 4,5,6\rgt\text{.}\)
- \(\displaystyle 2x+y+z=7\)
1.4.4. (✳).
Answer.
\(x+y+z=1\)
1.4.5.
Answer.
- \(\displaystyle x+y+z=2\)
- No.
- No.
- Yes.
1.4.6.
Answer.
All three points \((1,2,3),\ (2,3,4)\) and \((3,4,5)\) are on the line \(\vx(t)=(1,2,3)+t(1,1,1)\text{.}\) There are many planes through that line.
1.4.7.
Answer.
- \(\displaystyle 9x-y-z=8\)
- \(\displaystyle 14x-7y-8z=52\)
(c) For any real numbers \(a\) and \(b\text{,}\) the plane \(ax+by-(a+b)z=4a+b\) contains the three given points.
1.4.8.
Answer.
- \(\displaystyle \sqrt{3}\)
- \(\displaystyle 7/\sqrt{6}\)
1.4.9. (✳).
Answer.
- \(\displaystyle 3x+2y+z = 8\)
- \(\displaystyle \big(3\,,\,-1\,,\,1\big)\)
1.4.10. (✳).
Answer.
- \(\displaystyle x-y+z=3\)
- \(\displaystyle 5x+y-4z=-3\)
1.4.11. (✳).
Answer.
\(4x + 2y - 4z=15\) and \(4x + 2y - 4z=-9\)
1.4.12. (✳).
Answer.
\(2\)
1.4.13. (✳).
Answer.
- \(\displaystyle (x,y,z) = (-3,1,0) +t \llt 0,-1,1\rgt\)
- \(\displaystyle \sqrt{17}\)
1.4.14.
Answer.
\((x-1)^2+(y-2)^2+(z-3)^2=3\)
1.4.15.
Answer.
\(3x-y+z=-5\)
1.4.16.
Answer.
\(|c-\vn\cdot\vp|/|\vn|\)
1.4.17.
Answer.
It is the plane \(x+z=8\text{,}\) which is the plane through \((3,2,5)=\half(1,2,3)+\half(5,2,7)\) with normal \(\llt 1,0,1\rgt=\frac{1}{4}\big(\llt 5,2,7\rgt-\llt 1,2,3\rgt \big)\text{.}\)
1.4.18.
Answer.
It is the plane \(2(\vb-\va)\cdot\vx=|\vb|^2-|\va|^2\text{,}\) which is the plane through \(\half\va+\half\vb\) with normal vector \(\vb-\va\text{.}\)
1.4.19. (✳).
Answer.
- \(\displaystyle \half\sqrt{11}\approx 1.658\)
- \(\displaystyle \frac{3}{\sqrt{11}}\approx 0.9045\)
1.4.20. (✳).
Answer.
Any positive constant times \(\llt 1,1,-1\rgt\times\llt 1,0,1\rgt
=\llt 1, -2, -1\rgt\)
1.5 Equations of Lines in 3d
Exercises
1.5.1.
Answer.
There are infinitely many planes satisfying the condition described, but we’re asked for “the” line.
We need two normal directions to figure out the direction of a line in \(\mathbb R^3\text{.}\) Since the given normal vectors are parallel to each other, they really only specify one normal direction.
1.5.2.
Answer.
There are infinitely many correct answers. One is
\begin{align*}
\llt x\,,\,y\,,\,z-1\rgt &= t\llt 1,0,0\rgt &
\llt x\,,\,y\,,\,z-2\rgt &= t\llt 0,1,0\rgt\\
\llt x\,,\,y\,,\,z-3\rgt &= t\llt 1,1,0\rgt &
\llt x\,,\,y\,,\,z-4\rgt &= t\llt 1,-1,0\rgt
\end{align*}
1.5.3.
Answer.
(a) \(\llt x,y,z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt t\)
(b) \(\llt x,y,z\rgt = \llt -2,-1,0\rgt +\llt \frac{3}{2},1,1\rgt t\)
1.5.4.
Answer.
(a) \(\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt \)
(b) The two planes are parallel and do not intersect.
1.5.5.
Answer.
(a) \((1,0,3)\) lies on both lines. \(x+y+2z=7\) is the only plane containing both lines.
(b) The two lines do not intersect. No plane contains the two lines.
(c) The two lines do not intersect. \(x+z=1\) is the only plane containing both lines.
(d) The two lines are identical. For arbitrary \(a\) and \(b\) (not both zero) the plane \(ax+by+(a+b)z=a\) contains both lines.
1.5.6.
Answer.
vector parametric equation: \(\llt x-2,y+1,z+1\rgt= t\llt 1,-1,-1\rgt\)
scalar parametric equation: \(x=2+t\text{,}\) \(y=-1-t\text{,}\) \(z=-1-t\)
symmetric equation: \(x-2=-y-1=-z-1\)
1.5.7. (✳).
Answer.
\(\llt x,y,z\rgt = \llt 1,0,2\rgt +t\llt-1,1,-1 \rgt\)
1.5.8.
Answer.
(a) \(\llt x,y,z\rgt=\llt 5,3,0\rgt+t\llt 4,1,-2\rgt =\llt 5+4t,3+t,-2t\rgt\text{.}\)
(b) \(\sqrt{5}\)
1.5.9.
Answer.
(a) \(8x+2y-11z=59\)
(b) \(\frac{64}{\sqrt{189}} \approx 4.655\)
1.5.10. (✳).
Answer.
(a) Any nonzero constant times \(\llt 1 \,,\, -5 \,,\, -3 \rgt\text{.}\)
(b) \(x = -4 + t\text{,}\) \(y = 3 - 5t\text{,}\) \(z = 2 - 3t\)
1.5.11. (✳).
Answer.
(a) \((3,3,0)\text{,}\) \((12,0,-6)\text{,}\) \((0,4,2)\)
(b) \(x=10+t\text{,}\) \(y=11+t\text{,}\) \(z=13+t\text{.}\)
1.5.12. (✳).
Answer.
(a) \(\frac{x-2}{3}=\frac{y}{4}\qquad z=-1\)
(b) \(\al =\frac{\pi}{2}-\arccos\frac{1}{5\sqrt{6}}
\approx 0.08\,\text{radians}\)
1.5.13. (✳).
Answer.
\((x,y,z) = \big(12\,,\,-1-t\,,\,t\big)\)
1.5.14. (✳).
Answer.
(a) \((0,4,0)\)
(b) \(2x-y-z=-2\)
1.5.15. (✳).
Answer.
(a) \(x=0,\ y=2+t,\ z=2\)
(b) The sphere \((x-1)^2 +(y-2)^2+(z-2)^2 = 1\)
(c) \((0,4,4)\)
1.5.16.
Answer.
See the solution.
1.5.17.
Answer.
\(3\)
1.6 Curves and their Tangent Vectors
1.6.2 Exercises
1.6.2.1.
Answer.
\((1,25)\text{,}\) \((-1/\sqrt2,0)\text{,}\) \((0,25)\text{.}\)
1.6.2.2.
Answer.
The curve crosses itself at all points \((0,(\pi n)^2)\) where \(n\) is an integer. It passes such a point twice, \(2\pi n\) time units apart.
1.6.2.3.
Answer.
(a) \(\vr(y)=\sqrt{a^2-y^2}\,\hi+ y\,\hj\text{,}\) \(0\le y\le a\)
(b) \(\big(x(\phi),y(\phi)\big)
=\big(a\sin \phi ,-a\cos \phi \big)\text{,}\) \(\frac{\pi}{2}\le\phi\le\pi\)
(c) \(\big(x(s),y(s)\big)
=\big(a\cos(\tfrac{\pi}{2}-\frac{s}{a}),
a\sin(\tfrac{\pi}{2}-\tfrac{s}{a})\big)\text{,}\) \(0\le s\le\tfrac{\pi}{2}a\)
1.6.2.4.
Answer.
(a) \((a+a\theta,a)\)
(b)\((a+a\theta+a\sin\theta,a+a\cos\theta)\)
1.6.2.5.
Answer.
\(z=-\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}\)
1.6.2.6.
Answer.
The particle is moving upwards from \(t=1\) to \(t=2\text{,}\) and from \(t=3\) onwards. The particle is moving downwards from \(t=0\) to \(t=1\text{,}\) and from \(t=2\) to \(t=3\text{.}\)
The particle is moving faster when \(t=1\) than when \(t=3\text{.}\)
1.6.2.7.
Answer.
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.
Answer.
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. (✳).
Answer.
(c)
1.6.2.10. (✳).
Answer.
(d)
1.6.2.11.
Answer.
(a)
\begin{align*}
\vv(t)&= -a \sin t\,\hi+a\cos t\,\hj+c\,\hk\\
\diff{s}{t}(t)&= \sqrt{a^2+c^2}\\
\va(t)&= -a \cos t\,\hi-a\sin t\,\hj
\end{align*}
The path is a helix with radius \(a\) and with each turn having height \(2\pi c\text{.}\)
(b)
\begin{align*}
\vv(t)&= a \cos 2t\,\hi +a\sin 2t\,\hj-a\sin t\,\hk\\
\diff{s}{t}(t)&= a\sqrt{1+\sin^2t}\\
\va(t)&= -2a \sin 2t\,\hi +2a\cos 2t\,\hj-a\cos t\,\hk
\end{align*}
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 addition, the curve lies on the intersection of the cylinder \(x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4}\) and the sphere \(x^2+y^2+z^2=a^2\text{.}\)
1.6.2.12. (✳).
Answer.
(a) \(\hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}}\)
(b) \(\frac{1}{3}\big[5^{3/2}-8\big]\)
1.6.2.13.
Answer.
2
1.6.2.14.
Answer.
1
1.6.2.15. (✳).
Answer.
(a) \(\frac{20}{3}\)
(b) \(x(t) = -2\pi -2t,\
y(t) = -2\pi t,\
z(t) = \frac{\pi^3}{3} + \pi^2 t\)
1.6.2.16. (✳).
Answer.
(a) \(\vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big)\)
(b) 5
1.6.2.17. (✳).
Answer.
(a) \(\frac{1}{27}\big(10\sqrt{10}-1\big)\)
(b) \(\frac{2}{27}\big(10\sqrt{10}-1\big)\)
1.6.2.18. (✳).
Answer.
\(s(t)=\frac{t^3}{3} +\frac{t}{2}\)
1.6.2.19. (✳).
Answer.
\(\frac{8}{27}\Big[\Big(2 + \frac{9}{4}b^m\Big)^{3/2}
-\Big(2 + \frac{9}{4}a^m\Big)^{3/2}\Big]\)
1.6.2.20.
Answer.
\(|t|\)
1.6.2.21. (✳).
Answer.
(a) \(\vr(2)=2\hj+4\hk\)
(b) any nonzero multiple of \(\vr'(2)=2\pi\,\hi+\hj+4\,\hk\)
(c) \(\Gamma\) and \(E\) do not intersect at right angles.
1.6.2.22. (✳).
Answer.
\(\diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big]=0\)
1.6.2.23. (✳).
Answer.
- \(\displaystyle \vr(u)=u^3\,\hi+3u^2\,\hj+6u\,\hk\)
- \(\displaystyle 7\)
- \(\displaystyle 2\)
- \(\displaystyle 1\)
1.6.2.24. (✳).
Answer.
(a) \(\vr(t) = \big(\frac{\pi^2 t}{2}-\frac{t^3}{2}\big)\,\hi +
(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk\)
(b) \(t=\pi\)
(c) \(-\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk\)
1.6.2.25. (✳).
Answer.
- \(\displaystyle 21\)
- \(\displaystyle 6\)
- \(\displaystyle 2\hi+4\,\hj+4\,\hk\)
- \(\displaystyle -\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big)\)
1.6.2.26.
Answer.
\(\frac{x(t)y'(t)-y(t)x'(t)}{x^2+y^2}\)
1.6.2.27.
Answer.
\(\vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0
+g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk\)
1.6.2.28. (✳).
Answer.
(a) \(\vr(t)=\llt -\cos t, -\sin t, t\rgt\)
(b) \(\vv(t)\cdot\va(t)=0\)
(c) \(\vr(u)=\llt 0,1,-\frac{\pi}{2}\rgt+u\llt -1,0,1\rgt\)
(d) True
1.6.2.29. (✳).
Answer.
(a) \(x(t)^2+y(t)^2=z(t)^2\) for all \(t\)
(b)
\begin{equation*}
\text{velocity}= \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
\end{equation*}
\begin{equation*}
\text{speed}=\sqrt{2+\frac{\pi^2 t^2}{4}}
\end{equation*}
(c) \(\llt x,y,z\rgt = \llt 0,1,1\rgt +(t-1)\llt -\frac{\pi}{2},1,1\rgt\)
(d) \(\frac{2}{\pi}\) seconds
1.6.2.30. (✳).
Answer.
(a) \(90^\circ\)
(b) \(2\sqrt{3}\)
1.7 Sketching Surfaces in 3d
1.7.2 Exercises
1.7.2.1. (✳).
Answer.
(a) \(\leftrightarrow\) (C)
(b) \(\leftrightarrow\) (F)
(c) \(\leftrightarrow\) (D)
(d) \(\leftrightarrow\) (B)
(e) \(\leftrightarrow\) (A)
(f) \(\leftrightarrow\) (E)
1.7.2.2.
Answer.
(a)
(b)
1.7.2.3.
Answer.
1.7.2.4.
Answer.
(a)
(b)
(c)
1.7.2.5. (✳).
Answer.
1.7.2.6. (✳).
Answer.
1.7.2.7. (✳).
Answer.
(a)
(b)
1.7.2.8. (✳).
Answer.
1.7.2.9.
Answer.
(a) If \(c \gt 0\text{,}\) \(f(x,y,z)=c\) 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) \(f(x,y,z)=c\) is the plane normal to \((1,2,3)\) passing through \((c,0,0)\text{.}\)
(c) If \(c \gt 0\text{,}\) \(f(x,y,z)=c\) 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.
Answer.
(a)
(b)
(c)
1.7.2.11.
Answer.
(a) 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) This is a hyperboloid of one sheet with axis the \(x\)-axis.
(d) 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) 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) This is an elliptic paraboloid with axis the \(x\)-axis.
(h) This is an upward openning parabolic cylinder.
1.7.2.12.
Answer.
\(\displaystyle x^2+y^2=\left( \frac{|z|}{3}+1\right)^2\)
2 Partial Derivatives
2.1 Limits
2.1.2 Exercises
2.1.2.1.
Answer.
in general, false.
2.1.2.2.
Answer.
(a) the position of the particle in the basin
(b) the position in the basin that the millstone hits
(c) 50 \(\mu\)m
2.1.2.3.
Answer.
(a) along the \(x\)-axis
(b) along the \(y\)-axis
(c) \(\lim\limits_{(x,y)\to(0,0)}f(x,y)\) does not exist
2.1.2.4.
Answer.
(a) \(r^2\cos(2\theta)\)
(b) \(\text{min}=-1,\ \text{max}=1\)
(c) \(\text{min}=-r^2,\ \text{max}=r^2\)
(d) \(r \lt \sqrt\epsilon\)
(e) \(\lim\limits_{(x,y)\to(0,0)}f(x,y)=0\)
2.1.2.5.
Answer.
\(f(a,b)\)
2.1.2.6.
Answer.
(a) \(2\)
(b) undefined
(c) undefined
(d) \(0\)
(e) \(0\)
(f) \(1\)
2.1.2.7. (✳).
Answer.
(a) \(0\)
(b) See the solution.
2.1.2.8. (✳).
Answer.
(a) \(0\)
(b) The limit does not exist since the limits (i) \(x=0\text{,}\) \(y\rightarrow 0\) and (ii) \(y=0\text{,}\) \(x\rightarrow 0\) are different.
2.1.2.9. (✳).
Answer.
(a) \(2\)
(b) The limit does not exist. See the solution.
2.1.2.10.
Answer.
(a) \(0\)
(b) \(\frac{1}{2}\)
(c) No.
2.1.2.11. (✳).
Answer.
(a), (b), (d) Do not exist. See the solutions.
(c) \(0\)
2.1.2.12.
Answer.
(a), (b). The limit does not exist. See the solution.
2.2 Partial Derivatives
2.2.2 Exercises
2.2.2.1.
Answer.
- \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*}and\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*}
- \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*}and\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*}
- \(\pdiff{f}{x}(0,0) =1 \) and \(\pdiff{f}{y}(0,0) = 0\)
2.2.2.2.
Answer.
No: you can go higher by moving in the negative \(y\) direction.
2.2.2.3. (✳).
Answer.
(a) \(0\)
(b) \(0\)
(c) \(\frac{1}{2}\)
2.2.2.4.
Answer.
(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.
Answer.
See the solution.
2.2.2.6. (✳).
Answer.
(a) \(\pdiff{z}{x} = \frac{z(1-x)}{x(yz-1)}\text{,}\)
\(\pdiff{z}{y} = \frac{z(1+y-yz)}{y(yz-1)}\)
(b) \(\pdiff{z}{x}(-1,-2) =\frac{1}{2}\text{,}\)
\(\pdiff{z}{y}(-1,-2) =0\text{.}\)
2.2.2.7. (✳).
Answer.
\(\pdiff{U}{T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)}\)
\(\pdiff{T}{V}(1,2,4) = 1 -\frac{1}{4\ln(2)}\)
2.2.2.8. (✳).
Answer.
\(24\)
2.2.2.9.
Answer.
\(f_x(0,0)=1\text{,}\)
\(f_y(0,0)=2\)
2.2.2.10.
Answer.
Yes.
2.2.2.11.
Answer.
(a) \(\pdiff{f}{x}(0,0)=1\text{,}\) \(\pdiff{f}{y}(0,0)=4\)
(b) Nope.
2.2.2.12.
Answer.
1 resp. 0
2.3 Higher Order Derivatives
2.3.3 Exercises
2.3.3.1.
Answer.
See the solution.
2.3.3.2.
Answer.
No such \(f(x,y)\) exists.
2.3.3.3.
Answer.
(a) \(f_{xx}(x,y) = 2y^3\)
\(f_{yxy}(x,y) = f_{xyy}(x,y) = 12xy\)
(b) \(f_{xx}(x,y)= y^4e^{xy^2}\)
\(f_{xy}(x,y)= \big(2y+2xy^3\big)e^{xy^2}\)
\(f_{xxy}(x,y)= \big(4y^3 + 2xy^5\big)e^{xy^2}\)
\(f_{xyy}(x,y) = \big(2+10xy^2+4x^2y^4\big)e^{xy^2}\)
(c) \(\displaystyle\frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w)
= -\frac{36}{(u+2v+3w)^4}\)
\(\displaystyle\frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1)
= -0.0036
= -\frac{9}{2500}\)
2.3.3.4.
Answer.
\(f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}}\)
\(f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}\)
\(f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}}\)
2.3.3.5.
Answer.
(a) \(f_{xyz}(x,y,z)=0\)
(b) \(f_{xyz}(x,y,z)=0\)
(c) \(f_{xx}(1,0,0)=0\)
2.3.3.6. (✳).
Answer.
(a) \(f_{rr}(1,0)=m(m-1),\
f_{r\theta}(1,0)=0,\
f_{\theta\theta}(1,0)=-m^2\)
(b) \(\la=1\)
2.3.3.7.
Answer.
See the solution.
2.4 The Chain Rule
2.4.5 Exercises
2.4.5.1.
Answer.
(a) \(\pdiff{h}{x}(x,y)
=\pdiff{f}{x}\big(x,u(x,y)\big)
+\pdiff{f}{u}\big(x,u(x,y)\big)
\pdiff{u}{x}(x,y)\)
(b)
\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*}
(c)
\begin{align*}
\pdiff{h}{x}(x,y,z)
\amp =\pdiff{f}{u}\big(u(x,y,z),v(x,y),w(x)\big)
\pdiff{u}{x}(x,y,z)\\
\amp\ \ +\pdiff{f}{v}\big(u(x,y,z),v(x,y),w(x)\big)
\pdiff{v}{x}(x,y)\\
\amp\ \ +\pdiff{f}{w}\big(u(x,y,z),v(x,y),w(x)\big)
\diff{w}{x}(x)
\end{align*}
2.4.5.2.
Answer.
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 approximation \(f_x(x,y)\approx \frac{\Delta f}{\Delta x}\text{,}\) starting at the point \(P_0\text{,}\) \(\Delta x=x_2-x_1\) and \(\Delta f=z_2-z_1\text{.}\)
In the approximation \(\diff{w}{x}\approx \frac{\Delta w}{\Delta x}\text{,}\) starting at the point \(P_0\text{,}\) \(\Delta x=x_2-x_1\) again, and \(\Delta w=z_1-z_1=0\text{.}\)
2.4.5.3. (✳).
Answer.
\(\diff{w}{t}=1\) and \(f_t=5\text{.}\) \(f_t\) gives the rate of change of \(f(x,y,t)\) as \(t\) varies while \(x\) and \(y\) are held fixed. \(\diff{w}{t}\) 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.
Answer.
See the solution.
2.4.5.5.
Answer.
The problem is that \(\pdiff{w}{x}\) is used to represent two completely different functions in the same equation. See the solution for more details.
2.4.5.6.
Answer.
\(w_s(s,t)=2s(t^2+1)\)
\(w_t(s,t)=s^2(2t)\)
2.4.5.7.
Answer.
We have
\begin{align*}
\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy)
&=6f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112}\\
&\hskip0.25in+(6xy+2x^2)\,f_{122}+x^2y\,f_{222}
\end{align*}
All functions on the right hand side have arguments \((2x+3y,xy)\text{.}\) The notation \(f_{21}\text{,}\) for example, means first differentiate with respect to the second argument and then differentiate with respect to the first argument.
2.4.5.8.
Answer.
\begin{align*}
g_{ss}(s,t)&=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)\\
&\hskip1in+9f_{22}(2s+3t,3s-2t)\\
g_{st}(s,t)&=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)\\
&\hskip1in-6f_{22}(2s+3t,3s-2t)\\
g_{tt}(s,t)&=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. (✳).
Answer.
See the solutions.
2.4.5.10. (✳).
Answer.
\(a=5\) and \(b=c=2\text{.}\)
2.4.5.11. (✳).
Answer.
\begin{align*}
\frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy)
&= 2\, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy)\\
&\hskip0.5in
+4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy)\\
&\hskip0.5in
+4xy\, F_{vv}(x^2 - y^2 , 2xy)
\end{align*}
2.4.5.12. (✳).
Answer.
(a) \(\pdiff{u}{x}(x,y) = \frac{e^y}{x}\text{,}\) \(\pdiff{u}{y}(x,y) = e^y\,\ln (x) -y^2\,e^y -2y e^y\)
(b) See the solution.
2.4.5.13. (✳).
Answer.
\(-54\)
2.4.5.14. (✳).
Answer.
See the solution.
2.4.5.15. (✳).
Answer.
(a)
\begin{align*}
\pdiff{z}{t}(r,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)
\begin{align*}
\frac{\partial^2 z}{\partial t^2}(r,t)
&= -r\cos t \ \pdiff{f}{x}
-r\sin t \ \pdiff{f}{y}\\
&\hskip0.5in
+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. (✳).
Answer.
\(28\)
2.4.5.17. (✳).
Answer.
\(A=2\text{.}\)
2.4.5.18. (✳).
Answer.
\(10a\)
2.4.5.19. (✳).
Answer.
See the solution.
2.4.5.20. (✳).
Answer.
(a)
\begin{align*}
w_{ss}(s,t) &=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) \(A=-1\)
2.4.5.21. (✳).
Answer.
(a)
\begin{align*}
\pdiff{}{\theta}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big]
&=-r\sin\theta\, f_x
+r\cos\theta\, f_y\\
\pdiff{}{r}\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big]
&= \cos\theta\, f_x
+\sin\theta\, f_y\\
\frac{\partial^2}{\partial \theta\,\partial r}
\big[f\big(r\cos\theta\,,\,r\sin\theta\big)\big]
&=-\sin\theta\, f_x
+\cos\theta\, f_y\\
&\hskip0.5in
-r\sin\theta\cos\theta\, f_{xx}
+r[\cos^2\theta-\sin^2\theta]\, f_{xy}\\
&\hskip0.5in
+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)
\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. (✳).
Answer.
\(\vnabla z(1,2)=\llt -47,108\rgt\)
2.4.5.23. (✳).
Answer.
(a) See the solution.
(b) \(\frac{4}{15}\)
2.4.5.24. (✳).
Answer.
(a) \(\pdiff{x}{u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi}\text{,}\) \(\pdiff{y}{u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}\)
(b) \(\frac{8}{\pi}\)
2.4.5.25. (✳).
Answer.
\(\al=2\)
2.4.5.26.
Answer.
See the solutions.
2.4.5.27.
Answer.
(a) \(\displaystyle\pdiff{y}{z}(x,z)
=\frac{x^2 \ln y(x,z)-y(x,z)e^{y(x,z)\,z}}
{ze^{y(x,z)\,z}-\frac{x^2 z}{y(x,z)}}\)
(b) \(\displaystyle\diff{y}{x}(x)
=-\frac{F_1\big(x,y(x),x^2-y(x)^2\big)+2x\,F_3\big(x,y(x),x^2-y(x)^2\big)}
{F_2\big(x,y(x),x^2-y(x)^2\big)-2y(x)\,F_3\big(x,y(x),x^2-y(x)^2\big)}\)
(c) \(\displaystyle\left(\pdiff{y}{x}\right)_u\!\!(x,u)
=\frac{y(x,u)\,v(x,u)-x\,y(x,u)}{x\,y(x,u)-x\,v(x,u)}\)
2.5 Tangent Planes and Normal Lines
2.5.3 Exercises
2.5.3.1.
Answer.
Yes. The plane \(z=0\) is the tangent plane to both surfaces at \((0,0,0)\text{.}\)
2.5.3.2.
Answer.
See the solution.
2.5.3.3.
Answer.
The normal plane is \(\vn\cdot\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =0\text{,}\) where the normal vector \(\vn = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0)\text{.}\)
2.5.3.4.
Answer.
Tangent line is
\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. (✳).
Answer.
\(2x+y+9z=2\)
2.5.3.6. (✳).
Answer.
\(2x+y+z = 6\)
2.5.3.7.
Answer.
(a) The tangent plane is \(4x+2y+z=-3\) and the normal line is \(\llt x,y,z\rgt=\llt -2,1,3\rgt+t\llt 4,2,1\rgt\text{.}\)
(b) The tangent plane is \(2y-z=-1\) and the normal line is \(\llt x,y,z\rgt=\llt 2,0,1\rgt+t\llt 0,2,-1\rgt\text{.}\)
2.5.3.8. (✳).
Answer.
\(z = -\frac{3}{4} x- \frac{3}{2} y + \frac{11}{4}\)
2.5.3.9. (✳).
Answer.
(a) \(2ax -2ay +z = -a^2\)
(b) \(a=\frac{1}{2}\text{.}\)
2.5.3.10. (✳).
Answer.
The tangent plane is \(\frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5}\text{.}\)
The normal line is \(\llt x,y,z\rgt = \llt -1,2,\frac{4}{5}\rgt
+t \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt\text{.}\)
2.5.3.11. (✳).
Answer.
\(\pm(1,0,-2)\)
2.5.3.12. (✳).
Answer.
\(\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)\) and \(\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)\)
2.5.3.13. (✳).
Answer.
\(\pm \big(\half,-1,-1\big)\)
2.5.3.14.
Answer.
\(\pm\sqrt{3}\frac{\llt 3,14,-30\rgt}{|\llt 3,14,-30\rgt|}
=\pm\sqrt{\frac{3}{1105}}\llt 3,14,-30\rgt\)
2.5.3.15.
Answer.
The horizontal tangent planes are \(z=0\text{,}\) \(z=e^{-1}\) and \(z=-e^{-1}\text{.}\) The largest and smallest values of \(z\) are \(e^{-1}\) and \(-e^{-1}\text{,}\) respectively.
2.5.3.16. (✳).
Answer.
(a) \(y+z=3\)
\(\vr(t)=\llt 0,2,1\rgt+t\llt 0,5,5\rgt\)
(b) \(z_x(x,y)=\frac{2-2x-y}{y+3z(x,y)^2}\)
\(z_y(x,y)=-\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\)
\(z_y(0,2)=-1\)
(c) \(z_{xy}(x,y)=\frac{1}{y+3z(x,y)^2}
-\frac{2-2x-y}{{[y+3z(x,y)^2]}^2}
\left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right)\)
2.5.3.17. (✳).
Answer.
(a) \(\llt 1,0,3\rgt\)
(b) \(\llt 3,3,-1\rgt\)
(c) \(\vr(t)=\llt 1,1,3\rgt+t\llt 3,3,-1\rgt\)
2.5.3.18. (✳).
Answer.
\(49.11^\circ\) (to two decimal places)
2.5.3.19.
Answer.
\(\frac{\sqrt{3}}{2}\)
2.6 Linear Approximations and Error
2.6.3 Exercises
2.6.3.1.
Answer.
(a) \(P(x_0+\De x,y_0+\De y)
\approx P(x_0,y_0) + mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y\)
(b) \(P_\% \le |m|\,x_\% + |n|\,y_\%\)
2.6.3.2.
Answer.
We used that \(\diff{}{\theta}\sin\theta = \cos\theta\text{.}\) That is true only if \(\theta\) is given in radians, not degrees.
2.6.3.3.
Answer.
\(0.01\,\pi + 0.05 \approx 0.0814\)
2.6.3.4. (✳).
Answer.
\(0.3\)
2.6.3.5.
Answer.
\(25000\)
2.6.3.6. (✳).
Answer.
\(0.22\)
2.6.3.7. (✳).
Answer.
\(-0.06\)
2.6.3.8.
Answer.
\(\frac{1}{17}\approx 0.059\)
2.6.3.9.
Answer.
\(\frac{13}{40}\%=0.325\%\)
2.6.3.10. (✳).
Answer.
Method 1 is better.
2.6.3.11.
Answer.
The sag will be more sensitive to changes in height.
2.6.3.12. (✳).
Answer.
\(0.84\)
2.6.3.13. (✳).
Answer.
(a) \(f_x(x,y) = -\frac{y\,f(x,y) +1}{3f(x,y)^2 +xy}\)
(b) \(f(x,y) \approx -1 -\frac{3}{2} (y-1)\)
(c) \(-0.955\)
2.6.3.14. (✳).
Answer.
(a) The differential at \(x=a\text{,}\) \(y=b\) is \(\frac{\dee{x}}{e^{f(a,b)}+b} + \frac{1-f(a,b)}{e^{f(a,b)}+b} \,\dee{y}\)
(b) \(f\big(0.99\,,\,0.01\big) \approx 0\)
2.6.3.15. (✳).
Answer.
\(\frac{\pi}{75}+0.14\approx 0.182\)
2.6.3.16. (✳).
Answer.
\(59.560\)
2.6.3.17. (✳).
Answer.
\(\pi\times 128\times 0.04=
5.12\pi\approx 16.1\)cc
2.6.3.18. (✳).
Answer.
(a) \(z(x,y)\approx 1-4x+2y\)
(b) \(0.84\)
2.6.3.19. (✳).
Answer.
(a) \(\pdiff{z}{x} = \frac{4+yz^2}{3z^2-2xyz}\text{,}\) \(\pdiff{z}{x} = \frac{xz^2}{3z^2-2xyz}\)
(b) \(\pdiff{z}{x}(1,1) = 1\text{,}\) \(\pdiff{z}{y}(1,1) = \frac{1}{2}\)
(c) \(\pm 0.04\)
(d) At \(A\text{,}\) \(\diff{z}{\theta} =\frac{2}{3}\text{.}\)
At \(B\text{,}\) \(\diff{z}{\theta} = -1\text{.}\)
2.6.3.20. (✳).
Answer.
(a) \(\frac{1+e^{-1}}{10}\)
(b) any nonzero constant times \(\llt -1\,,\,e^{-1}\,,\,-1\rgt\)
2.6.3.21. (✳).
Answer.
(a) \(\pdiff{z}{x} = \frac{y^2 z^2}{4z^3-2xy^2z}\text{,}\) \(\pdiff{z}{y} = \frac{2xy\, z^2-1}{4z^3-2xy^2 z}\)
(b) \(\pdiff{z}{x}(2,-1/2) =\frac{1}{12}
\text{,}\) \(\pdiff{z}{y}(2,-1/2) =-1\)
(c) \(f(1.94,-0.4) - 1 \approx -0.105\)
(d) \(\frac{x}{12} -y -z = -\frac{1}{3}\)
2.6.3.22. (✳).
Answer.
(a) \(1.4\)
(b) \(3x-2y-z = -4\)
2.6.3.23. (✳).
Answer.
\(0.1\)
2.6.3.24. (✳).
Answer.
(a) \(z=1-4x+8y\)
(b) \(1.12\)
2.6.3.25. (✳).
Answer.
(a) \(f(0.1,1.2) \approx 0.4\)
(b) \(\left(\frac{1}{5}\,,\,\frac{4}{5}\,,\,\ln\frac{4}{5}\right)\)
2.6.3.26. (✳).
Answer.
(a) \(-(2+\pi)x +2y +3z = -\pi-3\)
(b) \(\pdiff{z}{x}(1,1) = \frac{\pi+2}{3}\)
(c) \(z(0.97,1) \approx -\frac{\pi+102}{100}\)
2.6.3.27. (✳).
Answer.
(a) \(F_y(1,2)=-2\text{,}\)\ \(F_z(1,2)=-16\)
(b) \(0.3\)
2.7 Directional Derivatives and the Gradient
2.7.2 Exercises
2.7.2.1. (✳).
Answer.
\(0\)
2.7.2.2. (✳).
Answer.
\(y\cos(xy)\,\hi + [2y + x\cos(xy)]\,\hj\)
2.7.2.3.
Answer.
(a) \(-3\)
(b) \(-\frac{61}{144\sqrt{3}}\approx-0.2446\)
2.7.2.4.
Answer.
(a) \(\llt\pm\frac{\sqrt{3}}{2} ,-\frac{1}{2} \rgt\)
(b) \(\llt 0 ,-1 \rgt\)
(c) No direction works!
2.7.2.5.
Answer.
\(\vnabla f(a,b)=\llt 7,-1\rgt\)
2.7.2.6. (✳).
Answer.
\(-2\)
2.7.2.7. (✳).
Answer.
(a) \(6\hi - 3\hj + 3\hk\)
(b) \(\frac{3}{\sqrt{2}}\)
2.7.2.8. (✳).
Answer.
(a) 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 \(|\vnabla f(2,1)| = |\llt -1\,,\, -4\rgt| = \sqrt{17}
\text{.}\)
(b) So the hiker descends with slope \(4\text{.}\)
(c) \(\pm\frac{1}{\sqrt{17}}\llt 4, -1\rgt\)
2.7.2.9.
Answer.
\(a=1\text{,}\) \(b=\frac{3}{2}\)
2.7.2.10. (✳).
Answer.
(a) 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) \(-\frac{\llt 6\,,\,2\,,\,-1\rgt}{\sqrt{41}}\)
2.7.2.11. (✳).
Answer.
(a) \(10\)
(b) \(\pm \frac{1}{5}\llt 3\,,-4\,,0\rgt\)
2.7.2.12. (✳).
Answer.
(a) \(\frac{1}{3}\)
(b) \(\llt \pm\frac{4}{5}\,,\,\frac{3}{5}\rgt\)
2.7.2.13. (✳).
Answer.
\(\frac{0.04 e}{\sqrt{5}} \)
2.7.2.14. (✳).
Answer.
(a) \(0\)
(b) \(\frac{1}{\sqrt{6}}\llt -1\,,\,1\,,\,-2 \rgt \)
(c) \(4\)
(d) \(a=2\)
2.7.2.15. (✳).
Answer.
(a) \(\vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt\text{,}\) \(\vnabla g(0,1,-1) =\llt -1,1,-1\rgt\)
(b) \(10\)
(c) Any vector which is a (strictly) positive constant times \(\llt -1 \,,\, 0 \,,\, 1 \rgt\) is fine.
2.7.2.16. (✳).
Answer.
(a) \(\vv = \llt -2 \,,\, -4 \,,\, 4 \rgt\)
(b) \(-10\)
2.7.2.17. (✳).
Answer.
(a) \(25\,\sqrt{2}\,e^{-9}\)
(b) \(\frac{\llt 2,2,-3\rgt}{\sqrt{17}}\)
(c) \(-10\sqrt{17} e^{-9}\)
2.7.2.18. (✳).
Answer.
The unit vector in the direction of maximum rate of change is \(\frac{\llt 2\,,\,-4\,,\,-3\rgt}{\sqrt{29}}\text{.}\) The maximum rate of change is \(\sqrt{29}\text{.}\)
2.7.2.19. (✳).
Answer.
\(\pm\frac{1}{\sqrt{5}}\llt 2, -1\rgt\)
2.7.2.20. (✳).
Answer.
(a) South
(b) \(2\)
(c) \(-2\sqrt{5}\)
2.7.2.21. (✳).
Answer.
(a) \(-16\)
(b) \(C \llt 4\,,\, 1 \,,\, -2\rgt\) for any nonzero constant \(C\)
(c) Any positive non zero multiple of \(-\llt 1\,,\, 2\,,\,1\rgt\) will do.
2.7.2.22. (✳).
Answer.
\(\vnabla f(a,b,c) =\sqrt{3} \llt 2,6,-4\rgt\)
2.7.2.23. (✳).
Answer.
(a) \(\frac{1}{\sqrt{2}}\llt 1,1\rgt\)
(b) \(\vv=c\llt 1,-1,0\rgt\) for any nonzero constant \(c\)
(c) \(\vv=\frac{1}{\sqrt{2}}\llt 1,1,2e^{-2}\rgt\text{.}\) Any positive multiple of this vector is also a correct answer.
2.7.2.24. (✳).
Answer.
(a) \(\llt 1,2,2\rgt\)
(b) \(14^\circ/{\rm s}\)
(c) any positive constant times \(-\llt 2,2,4\rgt\)
(d) any positive constant times \(\pm\llt 2,0,-1\rgt\)
2.7.2.25. (✳).
Answer.
(a) \(\vv=-\llt \frac{3}{5},\frac{4}{5},0\rgt\)
(b) \(\frac{500}{169}\approx 2.96^\circ/{\rm s}\)
2.7.2.26. (✳).
Answer.
(a) \(360\)
(b) \(-\frac{40}{3\sqrt{3}}\approx-7.70\)
(c) \(\vnabla T(x,y,z)=-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}}\)
2.7.2.27. (✳).
Answer.
(a)
(b) \(\pm\frac{1}{\sqrt{17}}\llt 4,-1\rgt\)
(c) \(-\frac{1}{\sqrt{17}}\llt 1,4\rgt\)
2.7.2.28.
Answer.
(a) 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 \((2,-1)\text{.}\)
(b) \(\llt -1,-1\rgt/\sqrt{2}\)
(c) \(4\sqrt{2}\,v\)
(d) \(\frac{12}{\sqrt{5}}\,v\)
(e) \(y=-\frac{4}{x^2}\)
2.7.2.29. (✳).
Answer.
(a) \(\hi\)
(b) \(x=1\)
(c) \(1\)
(d) \(\frac{\pi}{4}\)
2.7.2.30. (✳).
Answer.
(a) \(2\sqrt{3}\)
(b) \(5.5\)
(c) \(\pm\frac{\llt 1,1,-1\rgt}{\sqrt{3}}\)
2.7.2.31. (✳).
Answer.
(a) \(-2.1\)
(b) \(F\) increases.
(c) Any nonzero constant times \(\llt 4 \,,\, 8 \,,\, -1 \rgt\text{.}\)
2.7.2.32. (✳).
Answer.
(a)
(b) \(-\llt 1,4\rgt\)
2.7.2.33. (✳).
Answer.
(a) any positive multiple of \(\llt 3,4\rgt\)
(b) \(-800 e^{-17}\)
(c) \(\pm\big(\frac{4}{5},-\frac{3}{5}\big)\)
(d) \(y=\frac{2}{9}x^2\)
2.7.2.34. (✳).
Answer.
(a) See the solution.
(b) \(4\sqrt{5}\approx 8.944\)
(c) \(7.35\)
2.8 A First Look at Partial Differential Equations
2.8.3 Exercises
2.8.3.1.
Answer.
\(g(x) = 1-4x^2\)
2.8.3.2.
Answer.
(a) \(u(x,y)=C(y)\) with \(C(y)\) being any function of the single variable \(y\text{.}\)
(b) \(u(x,y)=F(x)+C(y)\) where \(F(x)\) is any function obeying \(F'(x)=f(x)\) (i.e. any antiderivative of \(f(x)\)) and \(C(y)\) is any function of the single variable \(y\text{.}\)
2.8.3.3.
Answer.
(a), (d) and (e) are harmonic. (b) and (c) are not harmonic.
2.8.3.4. (✳).
Answer.
\(a=\pm 2\)
2.8.3.5.
Answer.
\(a=\pm 5\)
2.8.3.6.
Answer.
\(a=\pm b\text{,}\) for any real number \(b\text{.}\)
2.8.3.7.
Answer.
Yes it is. For the justification, see the solution.
2.8.3.8.
Answer.
\(f(t) = Ce^{-4t}\) with \(C\) being an arbitrary constant.
2.8.3.9.
Answer.
See the solution.
2.8.3.10.
Answer.
perpendicular
2.8.3.11.
Answer.
\(n=0,-1\)
2.8.3.12.
Answer.
(a), (b) See the solutions.
(c) \(T(t) = Ce^{\la t}\text{,}\) \(X(x) = K x^\la\text{,}\) \(u(x,t)= D\,e^{\la t}\,x^\la\) with \(C\text{,}\) \(D\) and \(K\) being arbitrary positive constants.
2.8.3.13.
Answer.
See the solution.
2.8.3.14.
Answer.
(a) \(v_X(X,Y)=\frac{1}{3} v(X,Y)\)
(b) \(v_X(X,Y)=\frac{1}{X} v(X,Y)\)
2.9 Maximum and Minimum Values
2.9.4 Exercises
2.9.4.1. (✳).
Answer.
(a) (i) \(T\text{,}\) \(U\)
(a) (ii) \(U\)
(a) (iii) \(S\)
(a) (iv) \(S\)
(b) (i) \(F_x(1,2) \gt 0\)
(b) (ii) \(F\) does not have a critical point at \((2,2)\text{.}\)
(b) (iii) \(F_{xy}(1,2) \lt 0\)
2.9.4.2.
Answer.
The minimum height is zero at \((0,0,0)\text{.}\) The derivatives \(z_x\) and \(z_y\) do not exist there. The maximum height is \(\sqrt{2}\) at \((\pm 1,\pm 1,\sqrt{2})\text{.}\) There \(z_x\) and \(z_y\) exist but are not zero — those points would not be the highest points if it were not for the restriction \(|x|,|y|\le 1\text{.}\)
2.9.4.3.
Answer.
\(\vnabla g(\vx_0)\cdot\vd=0\text{,}\) 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. (✳).
Answer.
(a)
(b) \((0,0)\) is a local (and also absolute) minimum.
(c) No. See the solutions.
2.9.4.5. (✳).
Answer.
\(|c| \gt 2\)
2.9.4.6. (✳).
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | saddle point |
| \(\left(-\frac{2}{3},\frac{2}{3}\right)\) | local max |
2.9.4.7. (✳).
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,3)\) | saddle point |
| \((0,-3)\) | saddle point |
| \((-2,1)\) | local max |
| \((2,-1)\) | local min |
2.9.4.8. (✳).
Answer.
The minimum value is \(0\) on
\begin{equation*}
\Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5,\
\text{at least one of $x,y,z$ zero}}
\end{equation*}
The maximum value is \(4\) at \((1,2,1)\text{.}\)
2.9.4.9.
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | local min |
| \((\sqrt{2},-1)\) | saddle point |
| \((-\sqrt{2},-1)\) | saddle point |
2.9.4.10. (✳).
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \(\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\) | local min |
| \(-\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)\) | saddle point |
2.9.4.11. (✳).
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | local max |
| \((2,0)\) | saddle point |
2.9.4.12.
Answer.
\(\text{min}=0\qquad
\text{max}=\frac{2}{3\sqrt{3}}\approx0.385\)
2.9.4.13.
Answer.
\(\text{min}=-\frac{1}{\sqrt{e}}\qquad
\text{max}=\frac{1}{\sqrt{e}}\)
2.9.4.14. (✳).
Answer.
(i)
(ii)
(iii)
(iv)
2.9.4.15. (✳).
Answer.
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \(\big(\frac{3}{2},-\frac{1}{4}\big)\) | local min |
| \((-1,1)\) | saddle point |
2.9.4.16. (✳).
Answer.
(a)
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \(\left(0,\frac{2}{\sqrt{3}}\right)\) | local max |
| \(\left(0,-\frac{2}{\sqrt{3}}\right)\) | local min |
| \((2,0)\) | saddle point |
| \((-2,0)\) | saddle point |
(b) 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. (✳).
Answer.
The minimum is \(-2\) and the maximum is \(6\text{.}\)
2.9.4.18. (✳).
Answer.
\(6-2\sqrt{5}\)
2.9.4.19. (✳).
Answer.
(a) (0,0) and (3,0) and (0,3) are saddle points
(1,1) is a local min
(b) The minimum is \(-1\) at \((1,1)\) and the maximum is \(80\) at \((4,4)\text{.}\)
2.9.4.20. (✳).
Answer.
(0,0) is a local max
(0,2) is a local min
(1,1) and (-1,1) are saddle points
2.9.4.21. (✳).
Answer.
(a) \((1,1)\) is a saddle point and \((2,4)\) is a local min
(b) The min and max are \(\frac{19}{27}\) and \(5\text{,}\) respectively.
2.9.4.22. (✳).
Answer.
\((0,0)\) is a saddle point and \(\pm(1,1)\) are local mins
2.9.4.23. (✳).
Answer.
(a) \((0,0)\text{,}\) \((6,0)\text{,}\) \((0,3)\) are saddle points and \((2,1)\) is a local min
(b) The maximum value is \(0\) and the minimum value is \(4(4\sqrt{2}-6)
\approx -1.37\text{.}\)
2.9.4.24. (✳).
Answer.
\((0,0)\) is a saddle point and \(\pm(1,1)\) are local mins
2.9.4.25. (✳).
Answer.
The coldest temperture is \(-0.391\) and the coldest point is \((0,2)\text{.}\)
2.9.4.26. (✳).
Answer.
\((0,\pm 1)\) are saddle points, \(\big(\frac{1}{\sqrt{3}},0\big)\) is a local min and \(\big(-\frac{1}{\sqrt{3}},0\big)\) is a local max
2.9.4.27. (✳).
Answer.
(a) \((0,-5)\) is a saddle point
(b) 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. (✳).
Answer.
\((-1,\pm\sqrt{3})\) and \((2,0)\) are saddle points and \((0,0)\) is a local max.
2.9.4.29. (✳).
Answer.
\(e^{-1}\approx0.368\)
2.9.4.30.
Answer.
\(\frac{2500}{\sqrt{3}}\)
2.9.4.31.
Answer.
The box has dimensions \((2V)^{1/3}\times(2V)^{1/3}\times 2^{-2/3}V^{1/3}\text{.}\)
2.9.4.32. (✳).
Answer.
(a) 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) \(\frac{1}{\sqrt{17}}\llt -4,-1\rgt\)
(c) \(18\)
(d) \((0,2)\)
2.9.4.33. (✳).
Answer.
Case \(k \lt \frac{1}{2}\text{:}\)
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | local max |
| \((0,2)\) | saddle point |
Case \(k=\frac{1}{2}\text{:}\)
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | local max |
| \((0,2)\) | unknown |
Case \(k \gt \frac{1}{2}\text{:}\)
| \(\Atop{\text{critical}}{\text{point}}\) | type |
| \((0,0)\) | local max |
| \((0,2)\) | local min |
| \(\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)\) | saddle point |
| \(\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)\) | saddle point |
2.9.4.34. (✳).
Answer.
(a) \(x=1\text{,}\) \(y=\half\text{,}\) \(f\big(1,\half\big)=6\)
(b) 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.
Answer.
\(m=\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2}\) and \(b=\frac{S_yS_{x^2}-S_xS_{xy}}{nS_{x^2} -S_x^2}\) where \(S_y=\smsum\limits_{i=1}^n y_i\text{,}\) \(S_{x^2}=\smsum\limits_{i=1}^n x^2_i\) and \(S_{xy}=\smsum\limits_{i=1}^n x_iy_i\text{.}\)
2.10 Lagrange Multipliers
2.10.2 Exercises
2.10.2.1. (✳).
Answer.
(a) \(f\) does not have a maximum. It does have a minimum.
(b) The minima are at \(\pm (1,1)\text{,}\) where \(f\) takes the value \(2\text{.}\)
2.10.2.2.
Answer.
(a), (b) \(0\)
2.10.2.3.
Answer.
The max is \(f=\sqrt{3}\) and the min is \(f=-\sqrt{3}\text{.}\)
2.10.2.4.
Answer.
\(a=c=\sqrt{3},\ b=2\sqrt{3}\text{.}\)
2.10.2.5. (✳).
Answer.
The minimum value is \(2^{\frac{1}{3}} + 2^{-\frac{2}{3}}
=\frac{3}{2}\root{3}\of{2}
=\frac{3}{\sqrt[3]{4}}\) at \(\big(\pm 2^{\frac{1}{6}}\,,\, 2^{-\frac{1}{3}}\big)\text{.}\)
2.10.2.6. (✳).
Answer.
\(\text{radius}=\sqrt{\frac{2}{3}}\) and \(\text{height}=\frac{2}{\sqrt{3}}\text{.}\)
2.10.2.7. (✳).
Answer.
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. (✳).
Answer.
min\(=1\text{,}\) max\(=\sqrt{2}\text{.}\)
2.10.2.9. (✳).
Answer.
\((1,-2,1)\) is the closest point. \((-1,4,-1)\) is the farthest point.
2.10.2.10. (✳).
Answer.
The maximum is \(5\) and the minimum is \(0\text{.}\)
2.10.2.11. (✳).
Answer.
\(\big(\sqrt{3}\,,\,\sqrt{2}\,,\,1\big)\)
2.10.2.12. (✳).
Answer.
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. (✳).
Answer.
\(\sqrt{6-4\sqrt{2}}\approx 0.59\)
2.10.2.14. (✳).
Answer.
(a) \(\frac{4}{3}\)
(b) \((1,1)\)
2.10.2.15. (✳).
Answer.
(a) The min is \(6\) and the max is \(54\text{.}\)
(b) \((-1,1,-2)\)
2.10.2.16. (✳).
Answer.
(a) \(\big(\sqrt{5}-1\big)^2=6-2\sqrt{5}\)
(b) 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. (✳).
Answer.
The maximum value is \(2e^2\) and the minimum value is \(-2 e^2\text{.}\)
2.10.2.18. (✳).
Answer.
The farthest points are \(\pm\sqrt{6}(-2,1)\text{.}\) The nearest points are \(\pm(1,2)\text{.}\)
2.10.2.19.
Answer.
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. (✳).
Answer.
\(x=y=4,\ z=6\ \text{meters}\)
2.10.2.21. (✳).
Answer.
The hottest temperature is \(+5\) and the coldest temperature is \(-5\text{.}\)
2.10.2.22. (✳).
Answer.
\(2\sqrt{3} \times 4\times 24\)
2.10.2.23. (✳).
Answer.
\(2{\rm m}\times 2{\rm m}\times 3{\rm m}\)
2.10.2.24. (✳).
Answer.
\(\frac{2}{\root 3\of 3}{\rm m}\times\frac{2}{\root 3\of 3}{\rm m}
\times 3^{2/3}{\rm m}\)
2.10.2.25. (✳).
Answer.
\(a=\frac{1}{3}\text{,}\) \(b=\frac{1}{6}\text{,}\) \(c=\frac{1}{9}\text{,}\) max volume\(= 27\)
2.10.2.26. (✳).
Answer.
\(\sqrt{11}\)
2.10.2.27. (✳).
Answer.
The min is \(-27\sqrt{3}\) at \((-3\sqrt{3},0,3)\) and the max is \(48\) at \((4,\pm 4,2)\text{.}\)
2.10.2.28. (✳).
Answer.
(a - i)
\begin{align*}
2x\,e^y &=\la (2x)\\
e^y\big(x^2+y^2+2y\big) &=\la (2y)\\
x^2+y^2&=100
\end{align*}
(a-ii) The warmest point is \((0,10)\) and the coolest point is \((0,-10)\text{.}\)
(b-i)
\begin{align*}
2x\,e^y &=0\\
e^y\big(x^2+y^2+2y\big) &=0
\end{align*}
(b-ii) \((0,0)\) and \((0,-2)\)
(c) \((0,0)\)
2.10.2.29. (✳).
Answer.
(a) \(c=\pm 12\)
(b) \(\pm 12\)
(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.
Answer.
See the solution.
3 Multiple Integrals
3.1 Double Integrals
3.1.7 Exercises
3.1.7.1.
Answer.
(a) \(20\)
(b) \(\pi\)
(c) \(9\pi\)
3.1.7.2.
Answer.
(a) \(108 y^3\)
(b) \(48x^2\)
(c), (d) \(432\)
(e) \(648\)
3.1.7.3.
Answer.
(a) \(\frac{1}{3}\big(a^3b+ab^3\big)\)
(b) \(\frac{a^2b}{6}-\frac{ab^2}{2}\)
(c) \(\frac{3}{56}\)
(d) \(\frac{1}{2}(1-\cos 1)\)
(e) \(\frac{1}{2}(e-2)\)
(f) \(\frac{1}{4}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right)\)
3.1.7.4.
Answer.
(a) \(e^2-3\)
(b) \(\frac{8}{3}\)
(c) \(\frac{9}{2}\)
3.1.7.5. (✳).
Answer.
\(\int_{x=0}^{x=1}\int_{y=x}^{y=2-x} f(x,y)\ \dee{y}\,\dee{x}\)
3.1.7.6. (✳).
Answer.
(a)
(b) \(\half(e-1)\)
3.1.7.7. (✳).
Answer.
(a) The region \(R\) is the shaded region in the figure
(b) \(I= \int_{1/\sqrt{2}}^1 \int_{1/x}^{2x} f(x,y)\,\dee{y}\,\dee{x}
+\int_1^2 \int_{x^2}^{2x} f(x,y)\,\dee{y}\,\dee{x}\)
(c) \(\frac{1}{2}\)
3.1.7.8. (✳).
Answer.
(a) \(\frac{32}{3}\)
(b)
(c) \(I = \int_0^1\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ x
+\int_1^9\dee{y} \int_{(y-3)/2}^{\sqrt{y}}\dee{x}\ x\)
3.1.7.9. (✳).
Answer.
\(\frac{3}{4}\big[1-\cos(16)\big]\)
3.1.7.10. (✳).
Answer.
(a)
(b) \(\frac{1}{\pi}\)
3.1.7.11. (✳).
Answer.
(a) \(I=\int_0^1\dee{x}\int_x^1\dee{y}\ y^2\sin xy
=\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy\)
(b) \(\frac{1-\sin 1}{2}\)
3.1.7.12. (✳).
Answer.
\(\frac{1-e^{-1}}{2}\)
3.1.7.13. (✳).
Answer.
(a)
(b) \(I=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x}
+\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x}\)
(c) \(2\ln 2 -1\)
3.1.7.14. (✳).
Answer.
(a)
(b) \(\frac{2\big(2\sqrt{2}-1\big)}{9}\)
3.1.7.15. (✳).
Answer.
(a) \(J=\frac{27}{4}\)
(b) \(I=\frac{4}{3}\big[e-1\big]\)
3.1.7.16. (✳).
Answer.
(a)
(b) \(\frac{\sin(8)}{3}\)
3.1.7.17. (✳).
Answer.
(a) \(I = \int_{-1}^2\int_{x^2}^{x+2} f(x,y)\ \dee{y}\,\dee{x}\)
(b) \(2e^2 + \frac{1}{e}\)
3.1.7.18. (✳).
Answer.
(a)
(b) \(\int_0^2\int_0^{x^2} f(x,y)\,\dee{y}\,\dee{x}
+\int_2^{\sqrt{8}}\int_0^{8-x^2} f(x,y)\,\dee{y}\,\dee{x}\)
(c) \(\sqrt{8}-\arctan 2 -\frac{1}{6}\left[\ln\frac{3+\sqrt{8}}{3-\sqrt{8}}
-\ln 5\right]\)
3.1.7.19. (✳).
Answer.
\(\frac{1}{4}\big[e^4-1\big]\)
3.1.7.20. (✳).
Answer.
\(I = \int_0^2 \int_y^{6-y^2} f(x,y)\ \dee{x}\,\dee{y}\)
3.1.7.21. (✳).
Answer.
(a)
(b) \(\int_0^1 \int_{-\sqrt{1-y}}^{\sqrt{1-y}} f(x,y)\ \dee{x}\,\dee{y}\text{,}\) \(\int_{-1}^1 \int_0^{1-x^2} f(x,y)\ \dee{y}\,\dee{x}\)
(c) \(e^{2/3}-e^{-2/3}\)
3.1.7.22. (✳).
Answer.
(a)
(b) \(\frac{1-\cos(1)}{12}\)
3.1.7.23. (✳).
Answer.
(a)
(b) \(27+18\ln\frac{3}{2}\approx 34.30\)
3.1.7.24. (✳).
Answer.
\(\frac{4}{3}\sin 8\approx 1.319\)
3.1.7.25. (✳).
Answer.
(a) \(\ds I=\dblInt_D (8+2xy)\ \dee{x}\dee{y}\) where \(D=\Set{(x,y)}{x^2+(y-1)^2\le 1}\)
(b) \(\ds I
=\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ (8+2xy)
=\int_{-1}^1 \dee{x}\int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\dee{y}\ (8+2xy)\)
(c) \(8\pi\)
3.1.7.26. (✳).
Answer.
\(\frac{2}{3\pi}\approx 0.212\)
3.1.7.27. (✳).
Answer.
(a)
(b) \(\ds\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y}\)
(c) \(0\)
3.1.7.28. (✳).
Answer.
\(36\)
3.1.7.29.
Answer.
\(48\sqrt{2}\,\pi\)
3.2 Double Integrals in Polar Coordinates
3.2.5 Exercises
3.2.5.1.
Answer.
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.
\(r_1 = 3\text{,}\) \(\theta_1=0\)
\(r_2 = \sqrt{2}\text{,}\) \(\theta_2=\frac{\pi}{4}\)
\(r_3 = 1\text{,}\) \(\theta_3=\frac{\pi}{2}\)
\(r_4 = \sqrt{2}\text{,}\) \(\theta_4=\frac{3\pi}{4}\)
\(r_5 = 2\text{,}\) \(\theta_5=\pi\)
3.2.5.2.
Answer.
(a) \(\big(r_1=2\,,\,\theta_1= n\pi,\ n\text{ odd integer }\big)\) or \(\big(r_1=-2\,,\,\theta_1= n\pi,\ n\text{ even integer }\big)\text{.}\) In particular, \(\big(r_1=-2\,,\,\theta_1= 0\big)\) has \(r_1\lt 0\) and \(0\le\theta_1\lt 2\pi\text{.}\)
(b) \(\big(r_2=\sqrt{2}\,,\,\theta_2= \frac{\pi}{4} + 2n\pi\big)\) or \(\big(r_2=-\sqrt{2}\,,\,\theta_2= \frac{5\pi}{4} + 2n\pi\big)\text{,}\) with \(n\) integer. In particular, \(\big(r_2=-\sqrt{2}\,,\,
\theta_2= \frac{5\pi}{4}\big)\) has \(r_2\lt 0\) and \(0\le\theta_2\lt 2\pi\text{.}\)
(c) \(\big(r_3=\sqrt{2}\,,\,\theta_3= \frac{5\pi}{4} + 2n\pi\big)\) or \(\big(r_3=-\sqrt{2}\,,\,\theta_3= \frac{\pi}{4} + 2n\pi\big)\text{,}\) with \(n\) integer. In particular, \(\big(r_3=-\sqrt{2}\,,\,
\theta_3= \frac{\pi}{4}\big)\) has \(r_3\lt 0\) and \(0\le\theta_3\lt 2\pi\text{.}\)
(d) \(\big(r_4=3\,,\,\theta_4= 0 + 2n\pi\big)\) or \(\big(r_4=-3\,,\,\theta_4= \pi + 2n\pi\big)\text{,}\) with \(n\) integer. In particular, \(\big(r_4=-3\,,\,
\theta_4= \pi\big)\) has \(r_4\lt 0\) and \(0\le\theta_4\lt 2\pi\text{.}\)
(e) \(\big(r_5=1\,,\,\theta_5= \frac{\pi}{2} + 2n\pi\big)\) or \(\big(r_5=-1\,,\,\theta_5= \frac{3\pi}{2} + 2n\pi\big)\text{,}\) with \(n\) integer. In particular, \(\big(r_5=-1\,,\,
\theta_5= \frac{3\pi}{2}\big)\) has \(r_5\lt 0\) and \(0\le\theta_5\lt 2\pi\text{.}\)
3.2.5.3.
Answer.
(a) Both \(\he_r(\theta)\) and \(\he_\theta(\theta)\) have length 1. The angle between them is \(\frac{\pi}{2}\text{.}\) The cross product is \(\he_r(\theta) \times \he_\theta(\theta)=\hk\text{.}\)
(b) 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.
Answer.
(a) \(a=r\cos\theta\text{,}\) \(b=r\sin\theta\)
(b) \(A=a\cos\varphi-b\sin\varphi\text{,}\) \(B=b\cos\varphi+a\sin\varphi\)
3.2.5.5.
Answer.
(a)
\begin{align*}
\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)\\
&=\int_0^2\dee{r}
\int_0^{\frac{\pi}{4}}\dee{\theta}
\ r\ f(r\cos\theta,r\sin\theta)
\end{align*}
(b)
\begin{align*}
\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)\\
&=\int_1^2\dee{r}
\int_0^{\frac{\pi}{2}}\dee{\theta}
\ r\ f(r\cos\theta,r\sin\theta)
\end{align*}
(c)
\begin{align*}
\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)\\
&=\int_0^2\dee{r}
\int_0^{\arccos\frac{r}{2}}\dee{\theta}
\ r\ f(r\cos\theta,r\sin\theta)
\end{align*}
(d)
\begin{align*}
\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)\\
&=\int_0^2\dee{r}
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dee{\theta}
\ r\ f(r\cos\theta,r\sin\theta)\\
&\hskip0.5in+\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.
Answer.
(a)
(b)
(c)
3.2.5.7.
Answer.
(a) \(\frac{a^3}{6}\big[\sqrt{3}+1\big]\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{3}\)
(d) \(-\pi\)
3.2.5.8.
Answer.
\(\pi \left[\frac{4}{3}\sqrt{2}-\frac{7}{6}\right]
\approx 2.26\)
3.2.5.9.
Answer.
\(\frac{64}{9}a^3\)
3.2.5.10.
Answer.
\(\frac{128}{15}a^3\)
3.2.5.11. (✳).
Answer.
(a)
(b) \(\frac{32}{9}\)
3.2.5.12. (✳).
Answer.
\(\frac{16\pi}{5}\)
3.2.5.13. (✳).
Answer.
(a)
(b) \(M = \int_0^{\pi/4}\dee{\theta}\int_{\sqrt{2}}^2\dee{r}\ r\,
\rho(r\cos\theta\,,\,r\sin\theta)\)
(c) \(\frac{1}{2}\)
3.2.5.14. (✳).
Answer.
\(\pi\)
3.2.5.15. (✳).
Answer.
\(1-\frac{1}{\sqrt{2}}\)
3.2.5.16. (✳).
Answer.
(a)
(b) \(\frac{\pi}{12}\big[5\ln(5)-4\big]\)
3.2.5.17. (✳).
Answer.
\(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\)
3.2.5.18. (✳).
Answer.
(a) \(D = \Set{(r\cos\theta\,,\,r\sin\theta)}
{-\frac{\pi}{2}\le\theta\le\frac{\pi}{4},\ 0\le r\le 2\cos\theta}\)
(b) \(\text{Volume}=\frac{40}{18\sqrt{2}} +\frac{16}{9}\)
3.2.5.19. (✳).
Answer.
\(2\frac{\pi+44/9}{\pi+8}\approx 1.442\)
3.2.5.20. (✳).
Answer.
(a)
(b)
\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*}
(c)
\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. (✳).
Answer.
(a)
(b) \(J = \int_0^{\sqrt{2}}\int_0^x \frac{y}{x} e^{x^2+y^2}\ \dee{y}\,\dee{x}
+ \int_{\sqrt{2}}^2\int_0^{\sqrt{4-x^2}} \frac{y}{x} e^{x^2+y^2}\
\dee{y}\,\dee{x}\)
(c) \(\frac{1}{4}\left[e^4-1\right]\ln 2\)
3.2.5.22.
Answer.
\(\frac{\pi}{8}ab\)
3.2.5.23.
Answer.
About 3.5’’ above the bottom
3.2.5.24. (✳).
Answer.
(a) \(\pi\big(e^9-1\big)\approx 25,453\)
(b) \(\int_0^1 \dee{x}\int_x^{2-x}\dee{y}\ e^{x^2+y^2}\)
3.3 Applications of Double Integrals
3.3.4 Exercises
3.3.4.1.
Answer.
(a) \(6\pi\)
(b) \(\half ab (a+b)\)
3.3.4.2. (✳).
Answer.
\(\bar x=0\) and \(\bar y = \frac{5}{7}\text{.}\)
3.3.4.3. (✳).
Answer.
(a)
(b) \(\text{mass} = \frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big)\)
(c) \(\bar x = \frac{2\sqrt{3}- \ln(2+\sqrt{3})}
{\frac{4\pi}{3} - 2\ln(2+\sqrt{3})}
\approx 1.38\text{,}\) \(\bar y=0\text{.}\)
3.3.4.4. (✳).
Answer.
\(\bar x = \frac{10}{3\pi+8} \approx 0.57\)
3.3.4.5. (✳).
Answer.
\(\bar x = \bar y =\frac{4}{3\pi}\)
3.3.4.6. (✳).
Answer.
\(\frac{6}{5}\)
3.3.4.7. (✳).
Answer.
\(\frac{2}{3}\)
3.3.4.8. (✳).
Answer.
(a) \(\frac{3\pi}{4}\)
(b) \(-\frac{1}{6}\)
3.3.4.9. (✳).
Answer.
\(4\sqrt{2} -\sqrt{2}\pi\approx 1.214\)
3.3.4.10. (✳).
Answer.
(a) \(\frac{1}{3}(b-a\,,\,c)\)
(b) See the solution.
3.4 Surface Area
Exercises
3.4.1.
Answer.
\(ab\sqrt{1+\tan^2\theta}=ab\sec\theta\)
3.4.2.
Answer.
\(\frac{\sqrt{a^2+b^2+c^2}}{c} A(D)\)
3.4.3.
Answer.
(a) \(\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2}\)
(b) See the solution.
3.4.4. (✳).
Answer.
\(\frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right]\)
3.4.5. (✳).
Answer.
\(\frac{\pi}{6}\big[{(1+4a^2)}^{3/2}-1\big]\)
3.4.6. (✳).
Answer.
\(5\sqrt{2}\pi\)
3.4.7. (✳).
Answer.
\(\frac{4}{15}\big[9\sqrt{3}-8\sqrt{2}+1\big]\)
3.4.8. (✳).
Answer.
(a) \(F(x,y) = \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\)
(b)-(i) \(\int_0^{2\pi}\dee{\theta}\int_0^1\dee{r}\ \frac{2r}{\sqrt{4-r^2}}\)
(b)-(ii) \(16\pi\)
3.4.9.
Answer.
\(6\pi\approx 18.85\)
3.4.10. (✳).
Answer.
\(a^2[\pi-2]\)
3.5 Triple Integrals
Exercises
3.5.1.
Answer.
\(\frac{1}{4}\pi ab^2\)
3.5.2. (✳).
Answer.
\(3\)
3.5.3.
Answer.
\(\frac{a^2bc}{24}\)
3.5.4.
Answer.
\(\frac{5}{24}\)
3.5.5.
Answer.
(a) \(\ds\int_0^1\dee{x}\int_0^{x}\dee{y}\int_0^{1-x}\hskip-10pt \dee{z}\hskip3pt f(x,y,z)
+\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{1-y}\hskip-10pt \dee{z}\hskip3pt f(x,y,z)\)
(b) \(\ds\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{y^2}\!\! \dee{z}\ f(x,y,z)\)
3.5.6. (✳).
Answer.
(a)
(b)
\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. (✳).
Answer.
(a)
(b)
\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. (✳).
Answer.
\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. (✳).
Answer.
(a) \(I = \int_{x=-1}^{x=0}\int_{y=-2(1+x)}^{y=0}\int_{z=0}^{z=3(1+x+y/2)}
f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x}\)
(b) \(I = \int_{z=0}^{z=3}\int_{x=-(1-z/3)}^{x=0}\int_{y=-2(1+x-z/3)}^{y=0}
f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}\)
3.5.10. (✳).
Answer.
\(\frac{1}{48}\)
3.5.11. (✳).
Answer.
\(2\)
3.5.12. (✳).
Answer.
\(\frac{1}{12}\)
3.5.13. (✳).
Answer.
\(\frac{1}{60}\)
3.5.14. (✳).
Answer.
\(\frac{13}{2}-\frac{e^{-6}}{2}\)
3.5.15. (✳).
Answer.
(a)
(b) \(\int_{-1}^1\dee{x}\int_0^{1-x^2}\dee{y}\int_y^{1-x^2}\dee{z}\ f(x,y,z)\)
3.5.16. (✳).
Answer.
\(J = \int_0^1 \int_z^1\int_y^1 f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}\)
3.5.17. (✳).
Answer.
\(\int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{x=y^2/2}^{x=3}\int_{z=y^2}^{z=2x}
f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y}\)
\(\int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{z=y^2}^{z=6}\int_{x=z/2}^{x=3}
f(x,y,z)\ \dee{x}\,\dee{z}\,\dee{y}\)
\(\int_{z=0}^{z=6}\int_{x=z/2}^{x=3}\int_{y=-\sqrt{z}}^{y=\sqrt{z}}
f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}\)
3.5.18. (✳).
Answer.
(a) \(\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}
\int_{-1}^y f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \)
(b) \(\int_{-1}^1\int_z^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}
f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}\)
(c)
\begin{align*}
&\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. (✳).
Answer.
(a) \(\int_0^2 \int_{-\sqrt{3-y}}^{\sqrt{3-y}} \int_{x^2}^{3-y}
f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y}\)
(b) \(\int_0^1 \int_0^2 \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\
\dee{x}\,\dee{y}\,\dee{z}
+\int_1^3 \int_0^{3-z} \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\
\dee{x}\,\dee{y}\,\dee{z}\)
(c) \(\int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}}\int_0^2 f(x,y,z)\
\dee{y}\,\dee{x}\,\dee{z}
+\int_1^3 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^{3-z} f(x,y,z)\
\dee{y}\,\dee{x}\,\dee{z}\)
3.5.20. (✳).
Answer.
\(\frac{13}{24}\approx 0.5417\)
3.5.21. (✳).
Answer.
\(\frac{1}{8}=0.125\)
3.5.22. (✳).
Answer.
(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) \(\frac{17}{60}\)
3.6 Triple Integrals in Cylindrical Coordinates
3.6.4 Exercises
3.6.4.1.
Answer.
(a), (b)
(c), (d)
3.6.4.2.
Answer.
3.6.4.3.
Answer.
(a) \((1,0,0)\)
(b) \(\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)\)
(c) \((0,1,0)\)
(d) \((0,0,1)\)
(e) \(\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right)\)
3.6.4.4.
Answer.
(a) \(r=\sqrt{2}\text{,}\) \(z=2\text{,}\) \(\theta=\frac{\pi}{4}\) (plus possibly any integer multiple of \(2\pi\))
(b) \(r=\sqrt{2}\text{,}\) \(z=2\text{,}\) \(\theta=\frac{5\pi}{4}\) (plus possibly any integer multiple of \(2\pi\))
(c) \(r=2\text{,}\) \(z=0\text{,}\) \(\theta=\frac{2\pi}{3}\) (plus possibly any integer multiple of \(2\pi\))
(d) \(r=0\text{,}\) \(z=1\text{,}\) \(\theta=\text{arbitrary}\)
3.6.4.5.
Answer.
(a) \(z=r^2\sin(2\theta)\)
(b) \(r^2+z^2=1\)
(c) \(r=2\cos\theta\)
3.6.4.6.
Answer.
(a) \(a^3\big(2\pi-\frac{32}{9}\big)\)
(b) \(\frac{7}{48}\pi\)
(c) \(\frac{\pi}{2}\)
3.6.4.7. (✳).
Answer.
\(\frac{8\pi}{35}\)
3.6.4.8. (✳).
Answer.
\(\big(0,0,\frac{7}{16\sqrt{2}-14}
\approx 0.811\big)\)
3.6.4.9. (✳).
Answer.
\(\pi\left[\frac{48}{5}\sqrt{6}-\frac{328}{15}\right]
\approx 1.65\pi\)
3.6.4.10. (✳).
Answer.
\(\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right]\)
3.6.4.11. (✳).
Answer.
\(\frac{16\pi}{3}\)
3.6.4.12.
Answer.
\(\bar x=\bar y=\bar z=\frac{3}{8}a\)
3.6.4.13. (✳).
Answer.
(a) \(\ds\text{mass} =
\int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\
\frac{5}{\sqrt{3}}(z^2+1)r\)
(b) \(\frac{525}{24}\sqrt{5}\pi\approx 153.7\text{kg}\)
3.6.4.14. (✳).
Answer.
(a) \(\ds\int_{-2}^2 \dee{x}
\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\dee{y}\int_0^{e^{-x^2-y^2}}\dee{z}\)
(b) \(\pi\big[1-e^{-4}\big]\approx 3.084\)
3.6.4.15. (✳).
Answer.
\(\frac{8}{3}\)
3.6.4.16. (✳).
Answer.
(a) The unit vector in the direction of maximum rate of increse is \(\frac{1}{\sqrt{10}}(3,0,1)\text{.}\)
(b) \(2\pi\)
3.6.4.17.
Answer.
\(M\big(\frac{3}{4}a^2+b^2\big)\)
3.7 Triple Integrals in Spherical Coordinates
3.7.5 Exercises
3.7.5.1.
Answer.
3.7.5.2.
Answer.
3.7.5.3.
Answer.
(a) \(\rho=2\text{,}\) \(\theta=\pi\text{,}\) \(\varphi=\frac{\pi}{2}\)
(b) \(\rho=3\text{,}\) \(\theta=\frac{\pi}{2}\text{,}\) \(\varphi=\frac{\pi}{2}\)
(c) \(\rho=4\text{,}\) \(\theta=\text{arbitrary}\text{,}\) \(\varphi=\pi\)
(d) \(\rho=2\text{,}\) \(\theta=\frac{3\pi}{4}\text{,}\) \(\varphi=\frac{\pi}{6}\)
3.7.5.4.
Answer.
(a) \(\left(\frac{1}{4}\,,\,\frac{\sqrt{3}}{4}\,,\,\frac{\sqrt{3}}{2}\right)\)
(b) \((0,2,0)\)
3.7.5.5.
Answer.
(a) \(\varphi=\frac{\pi}{6}\text{ or }\frac{5\pi}{6}\)
(b) \(\rho=2\cos\varphi\)
(c) \(\rho\sin\varphi=2\)
3.7.5.6. (✳).
Answer.
See the solution.
3.7.5.7. (✳).
Answer.
(a)
(b) \(\frac{11\pi}{6}\)
3.7.5.8. (✳).
Answer.
\(2\pi\)
3.7.5.9.
Answer.
(a) \(2\pi\frac{a^3}{3}\big(1-\frac{1}{\sqrt{2}}\big)\)
(b) \(\frac{\pi a^4}{16}\)
(c) \(4\pi A\big(a-\sqrt{B}\tan^{-1}\frac{a}{\sqrt{B}}\big)\)
(d) \(\frac{8}{3}\pi a^3\)
3.7.5.10. (✳).
Answer.
(a) \(\frac{38}{3}\pi D\)
(b) \(\bar x = \bar y=0\)
\(\bar z =\frac{195}{152} \approx 1.28\)
3.7.5.11. (✳).
Answer.
(a)
(b) \(I=\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\
\rho^4\sin^2\varphi\cos\varphi\ \cos\theta\)
(c) \(\frac{1}{15}\)
3.7.5.12. (✳).
Answer.
\(\frac{81}{5}\)
3.7.5.13. (✳).
Answer.
\(\frac{2\pi^2}{3}\)
3.7.5.14. (✳).
Answer.
\(\frac{64\pi}{9}\)
3.7.5.15.
Answer.
\(\pi a^2h\)
3.7.5.16. (✳).
Answer.
\(\frac{64}{15}\pi\big(1-\frac{1}{2\sqrt{2}}\big)\approx 8.665 \)
3.7.5.17. (✳).
Answer.
(a) \(\frac{\pi}{16}\)
(b) The centroid is \((\bar x,\bar y,\bar z)\) with \(\bar x=\bar y=0\) and \(\bar z=\frac{3}{8}\text{.}\)
3.7.5.18. (✳).
Answer.
(a) \(\ds 9\int_0^2d\rho\int_0^{\pi/2}d\phi\int_0^{2\pi}d\theta\
\rho^3\,\sin\phi\,\cos\phi\)
(b) \(\frac{1}{2}\)
3.7.5.19. (✳).
Answer.
It does not exist.
3.7.5.20. (✳).
Answer.
(a) \(\text{Mass} = \int_0^2\dee{z}\int_0^{2\pi}\dee{\theta}\int_z^2\dee{r}\
r^2\)
(b) \(\text{Mass} = \int_{\pi/4}^{\pi/2}\dee{\varphi}\int_0^{2\pi}\dee{\theta}
\int_0^{2/\sin\varphi}\dee{\rho}\
\rho^3\sin^2\varphi\)
(c) \(8\pi\)
3.7.5.21. (✳).
Answer.
(a) \(\int_0^1\dee{r}\int_0^{2\pi}\dee{\theta}
\int_r^{\sqrt{2-r^2}}\dee{z}\ r^3\)
(b) \(\int_0^{\sqrt{2}}\dee{\rho}\int_0^{2\pi}\dee{\theta}
\int_0^{\pi/4}\dee{\varphi}\ \rho^4\sin^3\varphi\)
(c) \(\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right]
\approx 0.5503\)
3.7.5.22. (✳).
Answer.
(a) \(\int_0^{\pi/2}\dee{\varphi}\int_0^{\pi/2}\dee{\theta}\int_0^1\dee{\rho}\
\rho^4\sin^2\varphi\cos\varphi\ \cos\theta\)
(b) \(\int_0^1\dee{z}\int_0^{\pi/2}\dee{\theta}\int_0^{\sqrt{1-z^2}}\dee{r} \
r^2\,z\,\cos\theta\)
(c) \(\frac{1}{15}\)
3.7.5.23. (✳).
Answer.
(a) \(\int_0^{3}\dee{\rho}\int_0^{2\pi}\dee{\theta}
\int_0^{\pi/6}\dee{\varphi}\ \rho^4\sin^3\varphi\)
(b) \(\int_0^{3/2}\dee{r}\int_0^{2\pi}\dee{\theta}
\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\dee{z}\ \ r^3\)
(c) \(81\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right]
\approx 5.24\)
3.7.5.24. (✳).
Answer.
(a) \(\int_0^{1/\sqrt{2}}\dee{r}\int_0^{2\pi}\dee{\theta}
\int_r^{\sqrt{1-r^2}}\dee{z}\ r\sqrt{r^2+z^2}\)
(b) \(\int_0^{1}\dee{\rho}\int_0^{2\pi}\dee{\theta}
\int_0^{\pi/4}\dee{\varphi}\ \rho^3\sin\varphi\)
(c) \(\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right]\)
3.7.5.25. (✳).
Answer.
(a) \(\frac{2\pi}{3}\left[\big(12\big)^{3/2}+54\right]\)
(b) i. The top part is the part of the snowman’s head that is inside the sphere
\begin{equation*}
x^2+y^2+(z-4)^2 = 4
\end{equation*}
and above the cone
\begin{equation*}
z-4 = - \sqrt{\frac{x^2+y^2}{3}}.
\end{equation*}
ii. The middle part is the part of the snowman’s head and body that is bounded on the top by the cone \(z-4 = - \sqrt{\frac{x^2+y^2}{3}}\) and is bounded on the bottom by the cone \(z = \sqrt{3(x^2+y^2)}
\text{.}\)
iii. The bottom part is the part of the snowman’s body that is inside the sphere \(x^2+y^2+z^2 = 12\) and is below the cone \(z = \sqrt{3(x^2+y^2)}
\text{.}\)
3.7.5.26. (✳).
Answer.
(a) \(\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32}=128\pi^2\)
(b) The surface is a torus (a donut) but with the hole in the centre shrunk to a point. The figure below is a sketch of the part of the surface in the first octant.
3.7.5.27. (✳).
Answer.
(a) \(I = \int_0^1\dee{r} \int_0^{2\pi}\dee{\theta} \int_0^r\dee{z}\
r\ z\ \sqrt{r^2+z^2}\)
(b) \(I = \int_{\pi/4}^{\pi/2}\dee{\varphi}
\int_0^{2\pi}\dee{\theta}
\int_0^{1/\sin\varphi}\dee{\rho}\
\rho^4\sin\varphi \ \cos\varphi\)
(c) \(\frac{2(2\sqrt{2}-1)\pi}{15}\)
3.7.5.28. (✳).
Answer.
(a) \(\int_0^a \dee{z}\int_{\pi}^{3\pi/2}\dee{\theta}
\int_0^{\sqrt{a^2-z^2}}\dee{r}\ r\big(r^2+z^2\big)^{2014}\)
(b) \(\int_0^{\pi/2} \dee{\varphi}\int_{\pi}^{3\pi/2}\dee{\theta}
\int_0^a\dee{\rho}\ \rho^{4030}\sin\varphi\)
(c) \(\frac{a^{4031}\pi}{8062}\)
3.7.5.29. (✳).
Answer.
(a) \(\int_0^1 \dee{r}\ r \int_0^{2\pi}\dee{\theta} \int_{r^2}^r\dee{z}\
z(r^2+z^2)\)
(b) \(\int_{\pi/4}^{\pi/2} \dee{\varphi} \int_0^{2\pi}\dee{\theta}
\int_0^{\cos\varphi/\sin^2\varphi} \dee{\rho}\ \rho^5\sin\varphi
\cos\varphi\)
(c) \(\frac{3\pi}{40}\)
3.7.5.30. (✳).
Answer.
(a) \(V = \int_0^1\dee{r} \int_0^{\pi/2}\dee{\theta}\int_r^{1+\sqrt{1-r^2}}\dee{z}
\ r\)
(b) \(V = \int_0^{\pi/4}\dee{\varphi} \int_0^{\pi/2}\dee{\theta}
\int_0^{2\cos\varphi}\dee{\rho}\ \rho^2\sin\varphi\)
(c) \(\frac{\pi}{4}\)
3.7.5.31. (✳).
Answer.
(a) \(\int_0^{2\pi}\dee{\theta}\int_0^{3/2}\dee{r}\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2\)
(b) \(\int_0^{2\pi}\dee{\theta}\int_0^{\pi/6}\dee{\varphi}\int_0^3\dee{\rho}\ \big(\rho^2\sin\varphi\big)
\big(\rho^2\sin^2\varphi\big)\)
(c) \(2\pi \frac{3^5}{5} \big[\frac{2}{3}-\frac{3\sqrt{3}}{8}\big]\)
