1.1.2.
Hint.
In part (d), complete a square.
Appendix B Hints for Exercises
1 Vectors and Geometry in Two and Three Dimensions
1.1 Points
Exercises
1.1.4.
Hint.
In part (d) you are being asked to find the value of \(x\) that minimizes the distance from \((x,0,0)\) to \(A\text{.}\) You found a formula for that distance in part (c).
1.1.8.
Hint.
It is not necessary to solve the equation \(x^2-2px+y^2=3p^2\) for \(p(x,y)\text{.}\) For example, a point \((x,y)\) is on the isobar \(p(x,y)=1\) if and only if \(x^2-2x+y^2=3\text{.}\) This curve can be easily identified if one first completes a square.
1.2 Vectors
1.2.9 Exercises
1.2.9.2.
Hint.
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).
1.2.9.3.
Hint.
Review Theorem 1.2.11.
1.2.9.4.
1.2.9.5.
1.2.9.15.
Hint.
(a) The three line segments 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.
(b) Let \((\bar x,\bar y)\) be the coordinates of \(C\text{.}\) Write down the equations that say that \((\bar x,\bar y)\) is equidistant from the three vertices \(O\text{,}\) \(A\) and \(B\text{.}\)
1.2.9.16.
Hint.
The centre of the sphere is the midpoint of the diameter.
1.2.9.17.
Hint.
Draw a sketch. Call the vertices of the triangle \(A\text{,}\) \(B\) and \(C\) with \(C\) being the vertex that joins the two sides. Let \(\va\) be the vector from \(C\) to \(A\) and \(\vb\) be the vector from \(C\) to \(B\text{.}\) Determine, in terms of \(\va\) and \(\vb\text{,}\)
- the vector from \(A\) to \(B\text{,}\)
- the two vectors from \(C\) to the two midpoints and finally
- the vector joining the two midpoints.
1.2.9.18.
Hint.
Review § 1.2.4.
1.2.9.19. (✳).
Hint.
Determine the four corners of the parallelogram.
1.2.9.20.
Hint.
Review §1.2.4.
1.2.9.30. (✳).
Hint.
Evaluate \(\diff{L}{t}\) by differentiating \(\vr(t)\times\vr'(t)\text{.}\)
1.2.9.40.
Hint.
Choose 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. Express \(A\text{,}\) \(B\text{,}\) \(C\) and \(D\text{,}\) which are areas of triangles, as one half times cross products of vectors built from \(\va\text{,}\) \(\vb\) and \(\vc\text{.}\)
1.2.9.41.
Hint.
Do problem 1.2.9.40 first.
1.3 Equations of Lines in 2d
Exercises
1.3.1.
Hint.
What, exactly, is \(t\text{?}\)
1.3.2.
Hint.
What, exactly, is \(\mathbf c\text{?}\)
1.3.3.
Hint.
Set \(t=0\) in both equation to get two different points with integer coordinates; show that these two points are on both lines.
1.3.4.
Hint.
A line is specified by two things: one point on the line, and a vector parallel to the direction of the line.
1.3.5.
Hint.
Remember that the parametric equation of a line with direction \(\mathbf d\text{,}\) passing through point \(\mathbf c\text{,}\) is \(\llt x,y\rgt =\mathbf c + t\mathbf d \text{.}\)
1.3.6.
Hint.
Review Equation 1.3.3.
1.3.7.
Hint.
Review Example 1.3.5.
1.3.9.
Hint.
The radius of the circle will serve as a normal vector to the line.
1.4 Equations of Planes in 3d
Exercises
1.4.1.
Hint.
You are looking for a vector that is perpendicular to \(z=0\) and hence is parallel to \(\hk\text{.}\)
1.4.2.
Hint.
A point \((x,y,z)\) is on the \(y\)-axis if and only if \(x=z=0\text{.}\) Similarly, a point \((x,y,z)\) is on the \(z\)-axis if and only if \(x=y=0\text{.}\)
1.4.4. (✳).
Hint.
Guess.
1.4.5.
Hint.
a. See Question 1.4.4 — or just have a guess!
1.4.6.
Hint.
Three points don’t always determine a plane — why?
1.5 Equations of Lines in 3d
Exercises
1.5.1.
Hint.
What’s fishy about those normal vectors?
1.5.2.
Hint.
The two lines \(\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =t\vd\) and \(\llt x-x'_0\,,\,y-y'_0\,,\,z-z'_0\rgt =t\vd'\) are not parallel if \(\vd\times\vd'\ne\vZero\text{.}\)
1.5.15. (✳).
Hint.
All three of the points \(A\text{,}\) \(B\text{,}\) \(C\) lie in the plane \(y=2\text{.}\)
1.5.17.
Hint.
Review Example 1.5.7.
1.6 Curves and their Tangent Vectors
1.6.2 Exercises
1.6.2.1.
Hint.
Find the value of \(t\) at which the three points occur on the curve.
1.6.2.2.
Hint.
The curve “crosses itself” when \((\sin t,t^2)\) gives the same coordinate for different values of \(t\text{.}\) When these crossings occur will depend on which crossing you’re referring to, so your answers should all depend on \(t\text{.}\)
1.6.2.3.
Hint.
Draw sketches. Don’t forget the range that the parameter runs over.
1.6.2.4.
Hint.
For part (b), find the position of \(P\) relative to the centre of the circle. Then combine your answer with part (a).
1.6.2.5.
Hint.
We aren’t concerned with \(x\text{,}\) so we can eliminate it by solving one equation for \(x\) as a function of \(y\) and \(z\) and plugging the result into the other equation.
1.6.2.6.
Hint.
To determine whether the particle is rising or falling, we only need to consider its \(z\)-coordinate.
1.6.2.7.
Hint.
This is the setup from Lemma 1.6.12. The two quantities you’re labelling are related, but different.
1.6.2.8.
Hint.
See the note just before Example 1.6.14.
1.6.2.9. (✳).
Hint.
To simplify your answer, remember: the cross product of \(\va\) and \(\vb\) is a vector orthogonal to both \(\va\) and \(\vb\text{;}\) the cross product of a vector with itself is zero; and two orthogonal vectors have dot product 0.
1.6.2.10. (✳).
Hint.
Just compute \(|\vv(t)|\text{.}\) Note that \(\big(e^{at}+e^{-at}\big)^2
=e^{2at} + 2 + e^{-2at}\text{.}\)
1.6.2.11.
Hint.
To figure out what the curves look like, first detemine what curve \(\big(x(t),y(t)\big)\) traces out. For part (b) this will be easier if trig identities are first used to express \(x(t)\) and \(y(t)\) in terms of \(\sin(2t)\) and \(\cos(2t)\text{.}\)
1.6.2.12. (✳).
Hint.
Review Lemma 1.6.12. The arc length should be positive.
1.6.2.13.
Hint.
\begin{equation*}
\int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t
\end{equation*}
The notation looks a little confusing at first, but we can break it down piece by piece: \(\diff{\vr}{t}(t)\) is a vector, whose components are functions of \(t\text{.}\) If we take its magnitude, we’ll get one big function of \(t\text{.}\) That function is what we integrate. Before integrating it, however, we should simplify as much as possible.
1.6.2.14.
Hint.
\(\vr(t)\) is the position of the particle, so its acceleration is \(\vr''(t)\text{.}\)
1.6.2.15. (✳).
1.6.2.16. (✳).
Hint.
Review Lemma 1.6.12.
1.6.2.17. (✳).
Hint.
If you got the answer \(0\) in part (b), you dropped some absolute value signs.
1.6.2.19. (✳).
Hint.
The integral you get can be evaluated with a simple substitution. You may want to factor the integrand first.
1.6.2.20.
Hint.
Given the position of a particle, you can find its velocity.
1.6.2.23. (✳).
Hint.
If \(\vr(u)\) is the parametrization of \(\cC\) by \(u\text{,}\) then the position of the particle at time \(t\) is \(\vR(t) = \vr\big(u(t)\big)\text{.}\)
1.6.2.24. (✳).
Hint.
By Newton’s law, \(\vF=m\va\text{.}\)
1.6.2.25. (✳).
Hint.
Denote by \(\vr(x)\) the parametrization of \(C\) by \(x\text{.}\) If the \(x\)--coordinate of the particle at time \(t\) is \(x(t)\text{,}\) then the position of the particle at time \(t\) is \(\vR(t)=\vr\big(x(t)\big)\text{.}\) Also, though the particle is moving at a constant speed, it doesn’t necessarily have a constant value of \(\diff{\vx}{t}\text{.}\)
1.6.2.26.
Hint.
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 tangent function gives a clean-looking relation between \(\theta(t)\text{,}\) \(x(t)\text{,}\) and \(y(t)\text{.}\)
1.7 Sketching Surfaces in 3d
1.7.2 Exercises
1.7.2.2.
Hint.
For each \(C\text{,}\) redraw the level curve \(f(x,y)=C\) in the plane \(z=C\text{.}\)
1.7.2.3.
Hint.
Draw in the plane \(z=C\) for several values of \(C\text{.}\)
1.7.2.12.
Hint.
Since the level curves are circles centred at the origin (in the \(xy\)-plane), the equation will have the form \(x^2+y^2=g(z)\text{,}\) where \(g(z)\) is a function depending only on \(z\text{.}\)
2 Partial Derivatives
2.1 Limits
2.1.2 Exercises
2.1.2.1.
Hint.
How does the behaviour of a function far away from \((0,0)\) affect its limit at \((0,0)\text{?}\)
2.1.2.2.
Hint.
In this analogy, \(f(x,y)\) is the diameter of a particle taken from the position \((x,y)\) in the basin.
2.1.2.3.
Hint.
You can probably solve (a) and (b) by just staring at \(f(x,y)\text{.}\)
2.1.2.4.
Hint.
Recall \(\cos^2\theta-\sin^2\theta=\cos(2\theta)\)
2.1.2.5.
Hint.
Theorem 2.1.6
2.1.2.6.
Hint.
For parts (b), (c), (d), (e), switch to polar coordinates. For part (f),
\begin{equation*}
\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{equation*}
2.1.2.7. (✳).
Hint.
Switch to polar coordinates.
2.1.2.8. (✳).
Hint.
(a) Switch to polar coordinates.
(b) What are the limits when (i) \(x=0\) and \(y\rightarrow 0\) and when (ii) \(y=0\) and \(x\rightarrow 0\text{?}\)
2.1.2.9. (✳).
Hint.
For part (a) switch to polar coordinates. For part (b), switch to polar coordinates centred on \((0,1)\text{.}\) That is, make the change of variables \(x=r\cos\theta\text{,}\) \(y=1+r\sin\theta\text{.}\)
2.1.2.10.
Hint.
For part (c), does there exist a 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 \((0,0)\text{?}\)
2.1.2.11. (✳).
Hint.
For part (b), consider the ratio of \(\frac{\sin(xy)}{x^2+y^2}\) (from part (b)) and \(\frac{xy}{x^2+y^2}\) (from part (a)), and recall that \(\ds\lim_{t\rightarrow 0}\tfrac{\sin t}{t}=1\text{.}\)
For part (d) consider the limits along the positive \(x\)- and \(y\)-axes.
2.1.2.12.
Hint.
For part (a), determine what happens as \((x,y)\) tends to \((0,0)\) along the curve \(y=x+\frac{x^2}{a}\text{,}\) where \(a\) is any nonzero constant.
2.2 Partial Derivatives
2.2.2 Exercises
2.2.2.1.
Hint.
Review Definition 2.2.1.
2.2.2.2.
Hint.
What happens if you move “backwards,” in the negative \(y\) direction?
2.2.2.3. (✳).
Hint.
For (a) and (b), remember \(\pdiff{f}{x}(x,y)=\lim\limits_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}\) and \(\pdiff{f}{y}(x,y)=\lim\limits_{h\to0}\frac{f(x,y+h)-f(x,y)}{h}\text{.}\) For (c), you’re finding the derivative of a function of one variable, say \(g(t)\text{,}\) where
\begin{equation*}
g(t)=f(t,t)=\begin{cases}
\frac{t^2t}{t^2+t^2} & \text{if } t\ne 0 \\
0 & \text{if } t= 0
\end{cases}
\end{equation*}
2.2.2.5.
Hint.
Just evaluate \(x\pdiff{z}{x}(x,y)+y\pdiff{z}{y}(x,y)\text{.}\)
2.2.2.10.
Hint.
Just evaluate \(y\pdiff{z}{x}(x,y)\) and \(x\pdiff{z}{y}(x,y)\text{.}\)
2.2.2.12.
Hint.
You can find an equation for the surface, or just look at the diagram.
2.3 Higher Order Derivatives
2.3.3 Exercises
2.3.3.1.
Hint.
Repeatedly use (Clairaut’s) Theorem 2.3.4.
2.3.3.2.
Hint.
If \(f(x,y)\) obeying the specified conditions exists, then it is necessary that \(f_{xy}(x,y)=f_{yx}(x,y)\text{.}\)
2.3.3.5.
Hint.
(a) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing the order of evaluation of the derivatives.
(b) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing a different order of evaluation of the derivatives for each of the three terms.
(c) Set \(g(x) = f(x,0,0)\text{.}\) Then \(f_{xx}(1,0,0)=g''(1)\text{.}\)
2.4 The Chain Rule
2.4.5 Exercises
2.4.5.1.
Hint.
Review §2.4.1.
2.4.5.2.
Hint.
This is a visualization, in a simplified setting, of Example 2.4.10.
2.4.5.3. (✳).
Hint.
Pay attention to which variables change, and which are held fixed, in each context.
2.4.5.4.
Hint.
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 \(z=h(x,y)\)).
For example, 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 \(z\text{.}\) The equation
\begin{equation*}
\left(\pdiff{y}{x}\right)
\left(\pdiff{z}{y}\right)
\left(\pdiff{x}{z}\right)=-1
\end{equation*}
really means
\begin{equation*}
\left(\pdiff{g}{x}\right)
\left(\pdiff{h}{y}\right)
\left(\pdiff{f}{z}\right)=-1
\end{equation*}
So use \((*)\) to compute \(\pdiff{f}{z}\text{.}\) Use other equations similar to \((*)\) to compute \(\pdiff{g}{x}\) and \(\pdiff{h}{y}\text{.}\)
2.4.5.5.
Hint.
Is the \(\pdiff{w}{x}\) on the left hand side really the same as the \(\pdiff{w}{x}\) on the right hand side?
2.4.5.6.
Hint.
To avoid the chain rule, write \(w\) explicitly as a function of \(s\) and \(t\text{.}\)
2.4.5.7.
Hint.
Start by setting \(F(x,y)=f(2x+3y,xy)\text{.}\) It might also help to define \(g(x,y)=2x+3y\) and \(h(x,y)=xy\text{.}\)
2.4.5.9. (✳).
Hint.
Start by showing that, because \(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0\text{,}\) the second derivative \(\frac{\partial^2g}{\partial s^2} = 2\frac{\partial^2f}{\partial x \partial y}\text{.}\)
2.4.5.10. (✳).
Hint.
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. Until you get used to it, undo this notation conflict by renaming the function of \(s\) and \(t\) to \(F(s,t)\text{.}\) That is, \(F(s,t) = f\big(2s+t\,,\,s-t\big)
\text{.}\)
Then, evaluate each term on the right-hand side of the equation.
2.4.5.11. (✳).
Hint.
Let \(u(x,y) = x^2 - y^2\) , and \(v(x,y) = 2xy\text{.}\) Then \(F(x^2 - y^2 , 2xy)
=F\big(u(x,y),v(x,y)\big)\text{.}\)
2.4.5.12. (✳).
Hint.
(b) Since \(F\) is a function of only one variable, the chain rule for (say) \(\frac{\partial}{\partial x} F\big(xe^{-y^2}\big)\) has only one term.
2.4.5.13. (✳).
Hint.
At some point, you’ll be using the chain rule you learned in first-semester calculus.
2.4.5.14. (✳).
Hint.
Just compute the first order partial derivatives of \(w(x,y,z)\text{.}\)
2.4.5.15. (✳).
Hint.
The function \(\pdiff{f}{x}\) depends on both \(x\) and \(y\text{,}\) so don’t forget to account for both of these when you take its partial derivative.
2.4.5.17. (✳).
Hint.
Just compute \(\frac{\partial G}{\partial t}\text{,}\) \(\frac{\partial^2 G}{\partial \gamma^2}\) and \(\frac{\partial^2 G}{\partial s^2}\text{.}\)
2.4.5.19. (✳).
Hint.
Use implicit differentiation to find \(\pdiff{z}{x}(x,y)\) and \(\pdiff{z}{y}(x,y)\text{.}\)
2.4.5.21. (✳).
Hint.
This question uses bad (but standard) notation, in that the one symbol \(f\) is used for two different functions, namely \(f(x,y)\) and \(f(r,\theta)=f(x,y)\big|_{x=r\cos\theta,\,y=r\sin\theta}\text{.}\) Until you get used to it, undo this notation conflict by renaming the function of \(r\) and \(\theta\) to \(F(r,\theta)\text{.}\) That is, \(F(r,\theta) = f\big(r\cos\theta\,,\,r\sin\theta\big)
\text{.}\) Similarly, rename \(g\text{,}\) viewed as a function of \(r\) and \(\theta\text{,}\) to \(G(r,\theta)\text{.}\) That is, \(G(r,\theta) = g\big(r\cos\theta\,,\,r\sin\theta\big)
\text{.}\)
2.4.5.23. (✳).
Hint.
(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 \(x=u(x,y)^3-3u(x,y)\,v(x,y)^2\text{,}\) \(y=3u(x,y)^2v(x,y)-v(x,y)^3\text{.}\) Then \(z(x,y)\) is determined by \(z(x,y)=u(x,y)^2-v(x,y)^2\text{.}\)
2.4.5.24. (✳).
Hint.
The question tells us that \(x(u,v)\) and \(y=y(u,v)\) ar eimplicitly determined by
\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*}
at least near \((x,y)=(1,1)\text{,}\) \((u,v)=\big(\frac{\pi}{2},0\big)\text{.}\) Then, in part (b), \(z=x^4+y^4\) really means \(z(u,v)=x(u,v)^4+y(u,v)^4\text{.}\)
2.4.5.25. (✳).
Hint.
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*}
2.4.5.26.
Hint.
Use the chain rule to show that \(\frac{\partial^2 u}{\partial x^2}(x,t)
-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)
=4\frac{\partial^2 v\,}{\partial\xi\partial\eta}
\big(\xi(x,t),\eta(x,t)\big)\text{.}\)
2.4.5.27.
Hint.
For each part, first determine which variables \(y\) is a function of.
2.5 Tangent Planes and Normal Lines
2.5.3 Exercises
2.5.3.1.
Hint.
What are the tangent planes to the two surfaces at \((0,0,0)\text{?}\)
2.5.3.2.
Hint.
Apply the chain rule to \(G\big(\vr(t)\big)=0\text{.}\)
2.5.3.3.
Hint.
To find a tangent vector to the curve of intersection of the surfaces \(F(x,y,z)=0\) and \(G(x,y,z)=0\) at \((x_0,y_0,z_0)\text{,}\) use Q[2.5.3.2] twice, once for the surface \(F(x,y,z)=0\) and once for the surface \(G(x,y,z)=0\text{.}\)
2.5.3.4.
Hint.
To find a tangent vector to the curve of intersection of the surfaces \(z=f(x,y)\) and \(z=g(x,y)\) at \((x_0,y_0,z_0)\text{,}\) use Q[2.5.3.2] twice, once for the surface \(z=f(x,y)\) and once for the surface \(z=g(x,y)\text{.}\)
2.5.3.11. (✳).
Hint.
Let \((x,y,z)\) be a desired point. Then
- \((x,y,z)\) must be on the surface and
- the normal vector to the surface at \((x,y,z)\) must be parallel to the plane’s normal vector.
2.5.3.12. (✳).
Hint.
First find a parametric equation for the normal line to \(S\) at \((x_0,y_0,z_0)\text{.}\) Then the requirement that \((0,0,0)\) lies on that normal line gives three equations in the four unknowns \(x_0,y_0,z_0\) and \(t\text{.}\) The requirement that \((x_0,y_0,z_0)\) lies on \(S\) gives a fourth equation. Solve this system of four equations.
2.5.3.13. (✳).
Hint.
Two (nonzero) vectors \(\vv\) and \(\vw\) are parallel if and only if there is a \(t\) such that \(\vv=t\,\vw\text{.}\) Don’t forget that the point has to be on the hyperboloid.
2.5.3.14.
Hint.
The curve lies in the surface \(z^2=4x^2+9y^2\text{.}\) So the tangent vector to the curve is perpendicular to the normal vector to \(z^2=4x^2+9y^2\) at \((2,1,-5)\text{.}\)
The curve also lies in the surface \(6x+3y+2z=5\text{.}\) So the tangent vector to the curve is also perpendicular to the normal vector to \(6x+3y+2z=5\) at \((2,1,-5)\text{.}\)
2.5.3.15.
Hint.
At the highest and lowest points of the surface, the tangent plane is horizontal.
2.5.3.17. (✳).
Hint.
(b) If \(\vv\) is tangent, at a point \(P\text{,}\) to the curve of intersection of the surfaces \(S_1\) and \(S_2\text{,}\) then \(\vv\)
- has to be tangent to \(S_1\) at \(P\text{,}\) and so must be perpendicular to the normal vector to \(S_1\) at \(P\) and
- has to be tangent to \(S_2\) at \(P\text{,}\) and so must be perpendicular to the normal vector to \(S_2\) at \(P\text{.}\)
2.5.3.18. (✳).
Hint.
The angle between the curve and the surface at \(P\) is \(90^\circ\) minus the angle between the curve and the normal vector to the surface at \(P\text{.}\)
2.5.3.19.
Hint.
Let \(D(x,y)\) be the distance (or the square of the distance) from \((1,1,0)\) to the point \(\big(x,y, x^2+y^2)\) on the paraboloid. 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.
2.6 Linear Approximations and Error
2.6.3 Exercises
2.6.3.1.
Hint.
Review Example 2.6.11. Be careful when taking absolute values.
2.6.3.2.
Hint.
Units!
2.6.3.5.
Hint.
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 \(x_4+\veps_4\text{.}\) 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 \(\big|P(x_1+\veps_1,x_2+\veps_2,x_3+\veps_3,x_4+\veps_4)
-P(x_1,x_2,x_3,x_4)\big|\text{.}\)
2.6.3.6. (✳).
Hint.
Use Pythagoras to express the length of the hypotenuse in terms of the lengths of the other two sides.
2.6.3.8.
Hint.
Review the relationship between absolute error and percentage error given in Definition 2.6.6.
2.6.3.9.
Hint.
Be very careful about signs. There is a trap hidden in this question. As an example of the trap, suppose you know that \(|\veps_1|\le 0.2\) and \(|\veps_2|\le 0.1\text{.}\) It does not follow from this that \(\big|\veps_1-\veps_2|\le 0.2 -0.1 =0.1\text{.}\) The reason is that it is possible to have \(\veps_1=0.2\) and \(\veps_2=-0.1\) and then \(\veps_1-\veps_2=0.3\text{.}\) The correct way to bound \(\big|\veps_1-\veps_2|\) is
\begin{equation*}
\big|\veps_1-\veps_2| \le |\veps_1|+ |\veps_2|\le 0.2+0.1\le 0.3
\end{equation*}
2.6.3.11.
Hint.
Determine, approximately, the change in sag when the height changes by a small amount \(\veps\) and also when the width changes by a small amount \(\veps\text{.}\) Which is bigger?
2.6.3.15. (✳).
Hint.
\(1^\circ = \frac{\pi}{180}\) radians
2.7 Directional Derivatives and the Gradient
2.7.2 Exercises
2.7.2.4.
Hint.
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 \(\llt \cos\theta,\sin\theta\rgt \cdot\vnabla f(2,0)
\text{.}\)
2.7.2.5.
Hint.
Denote \(\vnabla f(a,b)=\llt \al,\be\rgt\text{.}\)
2.7.2.6. (✳).
Hint.
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 \((x,y)\text{.}\) Express the various slopes in terms of \(\vnabla z(0,0)\text{.}\)
2.7.2.9.
Hint.
In order for \(y=ax^b\) to give the \((x,y)\) coordinates of the path of steepest ascent, the tangent vector to \(y=ax^b\) must be parallel to the height gradient \(\vnabla h(x,y)\) at all points on \(y=ax^b\text{.}\) Also, don’t forget that \((1,1)\) must be on \(y=ax^b\text{.}\)
2.7.2.28.
Hint.
Review §2.7.
(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. This gives a separable differential equation for the function \(y(x)\text{.}\) Also, don’t forget that \((2,-1)\) must be on the curve.
2.8 A First Look at Partial Differential Equations
2.8.3 Exercises
2.8.3.1.
Hint.
Just evaluate \(u_t(x,t)\) and \(u_{xx}(x,t)+u(x,t)\) and stare at them for a while.
2.8.3.2.
Hint.
(a), (b) Fix any \(y_0\) and set \(v(x)=u(x,y_0)\text{.}\) What is \(\diff{v}{x}(x)\text{?}\)
2.8.3.4. (✳).
Hint.
Just substitute the given \(u(x,t)\) into the given PDE.
2.8.3.5.
Hint.
Just substitute the given \(u(x,y,z)\) into the given PDE.
2.8.3.6.
Hint.
Just substitute the given \(u(x,t)\) into the given PDE.
2.8.3.7.
Hint.
Just substitute the given \(z(x,y)\) into the given PDE.
2.8.3.8.
Hint.
Substitute the given \(u(x,t)\) into the given PDE. Review Theorem 3.3.2 in the CLP-1 text.
2.8.3.10.
Hint.
Evaluate \(u_{xx}+u_{yy}\) for the given \(u(x,y)\text{.}\)
2.8.3.12.
Hint.
(b) The left hand side is independent of \(x\) and the right hand side is independent of \(t\text{.}\)
(c) Review Section 3.3 in the CLP-1 text and Section 2.4 in the CLP-2 text.
2.8.3.13.
Hint.
Evaluate \(\diff{}{t}u\big(X(t),Y(t)\big)\text{.}\)
2.8.3.14.
Hint.
Evaluate \(v_X\text{.}\)
2.9 Maximum and Minimum Values
2.9.4 Exercises
2.9.4.2.
Hint.
Interpret the height \(\sqrt{x^2+y^2}\) geometrically.
2.9.4.3.
Hint.
Define \(f(t)=g(\va+t\vd)\text{.}\)
2.9.4.4. (✳).
Hint.
Write down the equations of specified level curves.
2.9.4.13.
Hint.
One way to deal with the boundary \(x^2+y^2=1\) is to parametrize it by \(x=\cos\theta\text{,}\) \(y=\sin\theta\text{,}\) \(0\le\theta \lt 2\pi\text{.}\)
2.9.4.30.
Hint.
Suppose that the bends are made a distance \(x\) from the ends of the fence and that the bends are through an angle \(\theta\text{.}\) Draw a sketch of the enclosure and figure out its area, as a function of \(x\) and \(\theta\text{.}\)
2.9.4.31.
Hint.
Suppose that the box has side lengths \(x\text{,}\) \(y\) and \(z\text{.}\)
2.10 Lagrange Multipliers
2.10.2 Exercises
2.10.2.2.
Hint.
(a) The function \(f\) decreases, or at least does not increase, as you leave \((x_0,y_0,z_0)\) in the direction \(\vd\text{.}\) The function \(f\) also decreases, or at least does not increase, as you leave \((x_0,y_0,z_0)\) in the direction \(-\vd\text{.}\)
2.10.2.4.
Hint.
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 \(\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=1\text{.}\)
2.10.2.19.
Hint.
The ends of the major axes are the points on the ellipse which are farthest from the origin. The ends of the minor axes are the points on the ellipse which are closest to the origin.
3 Multiple Integrals
3.1 Double Integrals
3.1.7 Exercises
3.1.7.2.
Hint.
Be careful to match each integration variable with its own limits. Remember that the integral with respect to \(x\) treats \(y\) as a constant and the integral with respect to \(y\) treats \(x\) as a constant.
3.1.7.9. (✳).
Hint.
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 \(y\text{.}\)
3.1.7.10. (✳).
Hint.
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 try exchanging the order of integration.
3.1.7.14. (✳).
Hint.
The inside integral, \(\int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}\text{,}\) of the given integral looks pretty nasty. Try reversing the order of integration.
3.1.7.15. (✳).
Hint.
(b) The inside integral, \(\int_{\frac{1}{2}\sqrt{x}}^1 e^{y^3}\ \dee{y}\text{,}\) looks pretty nasty because \(e^{y^3}\) does not have an obvious antiderivative. Try reversing the order of integration.
3.1.7.16. (✳).
Hint.
(b) The inside integral, \(\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}\) looks nasty. Try reversing the order of integration.
3.1.7.18. (✳).
Hint.
\(\frac{1}{9-x^2} =\frac{1}{6}\left(\frac{1}{x+3}-\frac{1}{x-3}\right)\text{.}\)
3.1.7.19. (✳).
Hint.
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. Try reversing the order of integration.
3.1.7.22. (✳).
Hint.
The inside integral, over \(y\text{,}\) looks pretty nasty because \(\sin(y^3)\) does not have an obvious antiderivative. So try reversing the order of integration.
3.1.7.24. (✳).
Hint.
The inside integral, \(\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}\text{,}\) in the given integral looks really nasty. So try exchanging the order of integration.
3.1.7.26. (✳).
Hint.
The inside integral, \(\ds\int_{\sqrt{y}}^3\sin\big(\pi x^3\big)\ \dee{x}\text{,}\) in the given integral looks really nasty. So try exchanging the order of integration.
3.2 Double Integrals in Polar Coordinates
3.2.5 Exercises
3.2.5.2.
Hint.
\(r\) is allowed to be negative.
3.2.5.3.
Hint.
Compute, for each angle \(\theta\text{,}\) the dot product \(\he_r(\theta)\cdot\he_\theta(\theta)\text{.}\)
3.2.5.4.
Hint.
Sketch \(\llt a, b\rgt\) and \(\llt A, B\rgt\text{.}\) The trigonometric addition formulas
\begin{align*}
\sin(\theta+\varphi)
&=\sin\theta\cos\varphi+\cos\theta\sin\varphi\\
\cos(\theta+\varphi)
&=\cos\theta\cos\varphi-\sin\theta\sin\varphi
\end{align*}
will help.
3.2.5.11. (✳).
Hint.
\begin{equation*}
\int \sin^n u\ \dee{u} =-\frac{1}{n}\sin^{n-1}u\ \cos u
+\frac{n-1}{n}\int\sin^{n-2}u\ \dee{u}
\end{equation*}
3.3 Applications of Double Integrals
3.3.4 Exercises
3.3.4.9. (✳).
Hint.
Try using polar coordinates.
3.4 Surface Area
Exercises
3.4.1.
Hint.
\(S\) is a very simple geometric object.
3.4.3.
Hint.
The triangle is part of the plane \(\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1\text{.}\)
3.4.8. (✳).
Hint.
The total surface area of (b) (ii) can be determined without evaluating any integrals.
3.4.9.
Hint.
Rewrite the equation of the cone in the form \(x=g(y,z)\text{.}\)
3.5 Triple Integrals
Exercises
3.5.20. (✳).
Hint.
Sketch \(E\text{.}\) You can picture \(E\) by thinking of the region bounded by the planes \(x=0\text{,}\) \(y=0\text{,}\) \(z=0\) and \(x+y=2\) as a large wedge of cheese and thinking of the cylinder \(y^2+z^2=1\) as a drill hole in the wedge. Then to set up the limits of integration, first sketch a top view of \(E\text{.}\)
3.6 Triple Integrals in Cylindrical Coordinates
3.6.4 Exercises
3.6.4.12.
Hint.
Use cylindrical coordinates.
3.7 Triple Integrals in Spherical Coordinates
3.7.5 Exercises
3.7.5.19. (✳).
Hint.
Switch to spherical coordinates.
3.7.5.26. (✳).
Hint.
(b) it is a solid of revolution.
