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\begin{document}
\begin{center}{\large\bf Solutions to final review problems}\end{center}
% \author{The Author}
{\large\bf Solutions to problems from the book}
\\[2ex]
{\large\bf Chapter 11.}
\tf 2. False. 4. False (true in quadrants 1 and 4, false in quadrants 2 and 3). 6. True. 8. True. 10. True (sorry; we didn't do this stuff.)\\[2ex]
\begin{tabular}{p{2.5in}@{\hspace{.4in}}l}
{\bf Problem 10}& {\bf Problem 14}\\
\includegraphics[scale=.5]{f1110}&
\includegraphics[scale=.5]{f1114}\\
\end{tabular}
\prob16. $r=2$.
\prob20. $\dydx = \frac{dy/dt}{dx/dt} = \frac{2-2t}{3t^2 +6}$. So when $t=-1$, the slope is $4/9$.
\prob24. $\dydx = \frac{dy/dt}{dx/dt} = \frac{1-3t^2}{2t} = \frac1{2t} -\frac32 t$.
$$
\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\dydx)}{dx/dt} = \frac{-\frac12t^{-2}-\frac32}{2t} = -\frac{3t^2+1}{4t^3}
$$
\prob38. $ L = \int_0^\pi \sqrt{\sin^6(\frac{\theta}{3}) + \sin^4(\frac{\theta}{3})\cos^2(\frac{\theta}{3})}\; d\theta
= \int_0^\pi \sin^2 (\frac{\theta}{3})\; d\theta = \left[ \frac12(\theta -\frac32\sin(\frac{2\theta}{3}))\right]_0^\pi =\frac{\pi}{2} -\frac38\sqrt3$.\\[2ex]
{\bf Problems Plus}
\prob1. By the fundamental theorem of calculus, $dx/dt = \cos(t)/t$ and $dy/dt = \sin(t)/t$. The tangent is vertical when $dx/dt = 0$, which first happens at $t=\pi/2$. The length of the curve is
$$
L = \int_1^{\pi/2}\sqrt{\frac{\cos^2t}{t^2} + \frac{\sin^2t}{t^2}}\; dt = \int_1^{\pi/2}\frac{1}{t}\; dt = \ln(\pi/2)
$$
\prob5. (a) If you replace the parameter value $t$ by $1/t$, then the values of $x$ and $y$ are interchanged. So if $(a,b) = (x(t),y(t))$, then $(b,a) = (x(1/t),y(1/t))$ is also on the curve. This doesn't work for $t=0$, but $t=0$ gives the point $(0,0)$ on the curve. (The symmetry is also easy to see if you do part (e).\\
(b) $dy/dt = \frac{6t-3t^4}{(1+t^3)^2}$ is $0$ when $t=0$ or $t=\sqrt[3]{2}$, so those values of $t$ give horizontal tangents at $(0,0)$ and $(\sqrt[3]{4},\sqrt[3]{2})$. From the symmetry of the curve (part (a)) we have vertical tangents at $(0,0)$ and $(\sqrt[3]{2},\sqrt[3]{4})$.\\
(d)\vspace*{-.4in}
\begin{center}
\includegraphics[scale=.5]{pp115b}
\end{center}
(e) $x^3+y^3$ and $3xy$ both come out to $\frac{2t^3}{(1+t^3)^2}$.\\
(f) Use the equation from part (e) and substitute to get $r^3\cos^3\theta +r^3\sin^3\theta =3r^2\cos\theta\sin\theta$, or $r= \frac{3\cos\theta\sin\theta}{\cos^3\theta+\sin^3\theta} = \frac{3\sec\theta\tan\theta}{1+\tan^3\theta}$.\\
(g) The loop in question corresponds to $\theta\in(0,\pi/2)$, so the area is
$$
\int_0^{\pi/2} r^2/2\; d\theta = \frac12\int_0^{\pi/2}\frac{3\sec^2\theta\tan^2\theta}{(1+\tan^3\theta)^2}\; d\theta = \frac92\int_0^\infty \frac{u^2\; du}{(1+u^3)^2} = \frac32
$$
using the substitution $u=\tan\theta$.\\
{\large\bf Chapter 13.}
\tf 2 False. 4 True. 6 True. 8 False; for example $\ii \times (\ii \times \jj) \neq (\ii\times \ii)\times \jj$. 10 True. 12 False. 14 False.
\prob2. \begin{center}
\includegraphics[scale=.8]{f132}
\end{center}
\prob4. (a) $11\ii-4\jj-\kk$. (b) $\sqrt{14}$. (c) $-1$. (d) $-3\ii-7\jj-5\kk$. (e) $4\sqrt{35}$. (f) $18$. (g) ${\bf 0}$. (h) $33\ii -21\jj+6\kk$. (i) $-1/\sqrt6$. (j) $-\frac16(\ii+\jj -2\kk)$. (k) $\cos\theta =\frac{-1}{2\sqrt{21}}$ so $\theta \approx 96^\circ$.
\prob10. $\vec{AB} = \la1,3,-1\ra$, $\vec{AC} = \la-2,1,3\ra$ and $\vec{AD} = \la -1,3,1\ra$. The volume is the (absolute value of) $\vec{AB}\cdot(\vec{AC} \times\vec{AD}) = |-6| = 6$.
\prob24. (a) The normal vectors are not parallel (by direct examination), so the planes are not parallel. They are not perpendicular (their dot product is $-5$) so the planes are not perpendicular. (b) The angle between the planes is the angle between the normal vectors, or $\cos^{-1}(\frac{-5}{\sqrt{87}}) \approx 122^\circ$. Actually, we usually use the angle less than $90^\circ$, so we should say $58^\circ$.
\prob34. Complete the square to get $x=(y-1)^2 +(z-2)^2$, which is a circular paraboloid opening up in the positive $x$ direction.
\prob46. Cylindrical: $r^2 + z^2 = 2r\cos\theta$. Spherical : $\rho^2 =2\rho\sin\phi\cos\theta$ or $\rho =2\sin\phi\cos\theta$\\[2ex]
{\bf Problems Plus}
\prob2. First do it in 2 dimensions. We have\\[-2ex]
\begin{tabular}{p{4in}@{\hspace{.4in}}c}
\vspace*{-.8in}
a rectangle of size $L \times W$, two rectangles of size $L \times 1$, two of size $1 \times W$, plus four quarter circles of radius $1$. In three dimensions, the central part is a rectangular solid of dimensions $L \times W \times H$. In addition we have rectangular solids on the sides of dimensions $L\times W \times 1$, $L \times 1 \times H$, and $1\times W \times H$. &
\includegraphics[scale=.75]{pp132.eps}\\
\end{tabular}\\
The corners are made of four quarter cylinders of radius $1$ and height $H$, four of height $L$, and four of height $W$. Finally, the corners get $8$ eighths of a sphere of radius $1$. All this adds up to $LWH+2LW+2WH+2LH +\pi 1^2W+\pi 1^2L+\pi 1^2H + \frac43\pi1^3 = LWH+2(LW +WH+LH) + \pi(L+W+H) + \frac43\pi$.
\prob4. (a) Let $\vv_c = 180\jj$ be the velocity shown by the compass, let $\ww$ be the velocity of the wind, and let $\vv_g=\vv_c+\ww$ be the actual velocity relative to the ground. Since the plane flew 80km in a half-hour, $|\vv_g| = 160$. So $\vv_g = (160\cos85^\circ)\ii+ (160\sin85^\circ)\jj \approx 13.9\ii + 159.4\jj$. So $\ww=\vv_g-\vv_c \approx 13.9\ii -20.6\jj$. \\
(b) The pilot should correct course by traveling with velocity $\vv_c -\ww \approx -13.9\ii +200.6\jj$.\\
{\large\bf Chapter 14.}
\tf 2 True. 4 True. 6 False.
\prob4. The point corresponds to parameter value $t=1$. The velocity vector is $\la2,4,3\ra$, so the equation of the line is $\xx(t) = (1+2t,1+4t,1+3t)$.
\prob6. (a) We want $y=0$, ie $t=1/2$, so the point is $(15/8,0,-\ln2)$. (b) The tangent vector is $\rr'(1) = \la -3,2,1\ra$. So the tangent line (in vector form) is $\xx(t) =(1-3t,1+2t,t)$.
\\[2ex]
{\bf Problems Plus}\\
\prob1. (a, b) $\vv(t) = \rr'(t) = R\omega\cos\omega t \ii - R\omega\sin\omega t\jj$ and direct computation shows $\rr\cdot \rr'=0$ and $|\vv| = R\omega$. The time of revolution is the circumference of the circle divided by the speed, ie $T=\frac{2\pi R}{|\vv|} = \frac{2\pi}{\omega}$.\\
(c) $\aa(t) = -R\omega^2\sin\omega t \ii - R\omega^2\cos\omega t\jj = -\omega^2\rr(t)$ which clearly points inward. \\
(d) Use Newton's law $\mathbf{F}=m\aa$; then $|\mathbf{F}| = m|\aa| = m R\omega^2 = \frac{mR^2\omega^2}{R} = \frac{m|\vv|^2}{R}$.
\prob3. (a) Taking the given formulas for granted, the maximum height is achieved when the $\jj$-component of the velocity is $0$. This occurs when $v_0\sin\alpha = gt$, ie at $t= \frac{v_0\sin\alpha}{g}$. The height at this time is given by $y =\frac{v_0^2\sin^2\alpha}{2g}$. As a function of $\alpha$, this is maximized when $\alpha = \pi/2$, so that the maximum height is $\frac{v_0^2}{2g}$.
\prob5(a). Divide by m, and take the integral of both sides to give
\begin{align*}
\int \frac{d^2 \R}{dt^2}\, dt + \frac k m \int \frac{d\R}{dt}\, dt &= -g\int \jj\, dt + \mathbf{C} \quad \Rightarrow \\
\frac{d\R}{dt} +\frac k m \R &= -gt\jj + \mathbf{C}
\end{align*}
Plugging at $t=0$ gives $\mathbf{C} = \vv(0)$. (The problem assumes that $\mathbf{R}(0) = \vec{0}$.)
\\
{\large\bf Chapter 15.}
\tf 2 False (Clairaut). 4 True. 6 False. 8 False. 10 True. 12 False (see exercise 15.7.35).
\prob6. Parabolas opening down.
\prob12. We estimate $T_x(6,4) $ by averaging the rates of change for $\Delta x = \pm 2$, which are $\frac{T(8,4)-T(6,4)}{2} = 3$ and $\frac{T(4,4)-T(6,4)}{-2} = 4$ respectively, so $T_x(6,4)\approx3.5 $. Similarly, $T_y(6,4) \approx -3$. So a linear approximation is $L(x,y) = 80 + 3.5(x-6)-3(y-4)$. This gives an approximation (to the actual value of $T(5,3.8)$) of $L(5,3.8) = 80 -3.5 -3(-0.2) = 77.1^\circ C$.
\prob14. $u_r = -e^{-r}\sin2\theta$. $u_\theta =2e^{-r}\cos2\theta$.
\prob20. $z_{xx} = 0$, $z_{yy} = 4xe^{-2y}$, and $z_{xy}=z_{yx} = -2e^{-2y}$.
\prob26.$z_x=1$ and $z_y=0$, so an equation of the tangent plane is $z-1 = x$.\\
(b) A normal vector to the tangent plane is $\la 1,0,-1\ra$, so parametric equations for the normal line are $x=t,\ y=0,\ z=1-t$.
\prob34. (a) $dA = \frac{\partial A}{\partial x}dx + \frac{\partial A}{\partial y}dy = \frac12ydx+ \frac12xdy. $Since $|\Delta x| \leq .002m$ and $|\Delta y| \leq .002m$ (mind your units!) the maximum error would be about $6(.002) + \frac52(.002) = .017m^2$.\\
(b) For the hypotenuse, $h$, we have $dh = \frac{x}{\sqrt{x^2+y^2}}dx+\frac{y}{\sqrt{x^2+y^2}}dy$. So the maximum error is about $dh = \frac{5}{13}(.002) + \frac{12}{13}(.002)\approx .0026m$.
\prob40. $A=\frac12\sin\theta$, so by the chain rule, $\frac{dA}{dt} = \frac12\left[(y\sin\theta)\frac{dx}{dt} +(x\sin\theta)\frac{dy}{dt} + (xy\cos\theta)\frac{d\theta}{dt}\right]$. For the given values, this works out to $\approx 60.8 in^2/s$.
\prob50. From problem N, we want the normal to both surfaces. One is given by $\NN = \nabla(z-2x^2+y^2) =\la8,4,1\ra$ and the other by $\nn = \la 0,0,1\ra$. So a tangent vector is given by $\NN\times \nn = 4\ii -8\jj$. (You can figure this out without problem N, by saying that the tangent vector is perpendicular to the normal vector to the curve $2x^2 +y^2 =4$ in the plane $z=4$.) Hence parametric equations are given by $x=-2+4t,\ y=2-8t,\ z=4$.
\prob52. $f_x = 3x^2-6y$, $f_y = -6x+24y^2$. Solving $f_x=0$ gives $y=\frac12x^2$, and then substituting into $f_y=0$ gives $6x(x^3-1)=0$. The critical points are then $(0,0)$ and $(1, \frac12)$. The second derivative test shows the first is a saddle, and the second is a local minimum.
\prob54. $f_x=2xe^{y/2}$, $f_y=e^{y/2}(2+x^2+y)/2$. So the only critical point is $(0,-2)$, which is a local minimum by the second derivative test.\\[2ex]
{\bf Problems Plus}
\prob4. Let's show something less than was asked for, namely that the function is not continuous if $r\leq 2$. To do this, we need to show that the limit as $(x,y,z) \to (0,0,0)$ either does not exist, or is not $0$ if it does exist. Consider the function along the line $(x,0,0)$, ie $g(x) = \frac{x^r}{x^2}$. If $r=2$, this has limit $1$, but that is not the value of the function at $(0,0,0)$, so the function is not continuous. If $r<2$, then the limit certainly doesn't exist, so again the function is not continuous. (It turns out to be continuous if $r>2$, but that takes considerable work to prove.)
\prob7. There's nothing else to do but plug in the expressions for $x,\, y$, and $z$ in cylindrical (part a) or spherical (b) coordinates and grind away with the chain rule. I'll check your work if you like.
\\
{\large\bf Chapter 16.}
\tf 2 False. If you change the order of integration, the limits will be $\int_0^1\int_y^1$. 4. True (use the fact that $e^{y^2}\sin y$ is an odd function). 6. True: The integrand is $\leq 3$, and the area of the base is 3. (We didn't actually discuss this in class, but it should make intuitive sense.)
\prob 4. $\int_0^1 \int_0^1 ye^{xy}dxdy = \int_0^1(e^y-1)dy = e-2$.
\prob14. The region is to the right of the curve $x=\sqrt{y}$ with $0\leq y\leq 1$. This can also be described as $0\leq x \leq 1$, and below the curve $y=x^2$. So the integral is $\int_0^1\int_0^{x^2}\frac {ye^{x^2}}{x^3} dy\;dx = \int_0^1\frac12xe^{x^2}dx = \frac14\left[ e^{x^2}\right]_0^1 = \frac14(e-1)$.
\prob20. A picture would help.\\[1ex]
\begin{tabular}{p{3in}@{\hspace{.4in}}c}
$\ds \int_1^2 \int_{1/y}^y y\;dx\;dy = \int_1^2y\left(y-\frac{1}{y}\right)dy
= \int_1^ (y^2-1) dy =\frac43$.&\\
&
\includegraphics[scale=.75]{f1620.eps}\\
\end{tabular}
\prob22. Use polar coordinates: $\int_0^{\pi/2}\int_1^{\sqrt{2}} (r\cos\theta)rdr d\theta = \frac13(2^{3/2}-1)$.
\prob32. $z$ ranges from $0$ to $3-y = r-r\sin\theta$. So the volume is
$$
\int_0^{2\pi}\int_0^2\int_0^{2-r\sin\theta} rdzdrd\theta = \int_0^{2\pi}\int_0^2 (3r-r^2\sin\theta)drd\theta = 12\pi.$$
{\bf Problems Plus}
\prob2. Divide the rectangle into two pieces, $R_1$ and $R_2$, according to whether $x$ or $y$ is larger. (This is the same as whether $x^2$ or $y^2$ is larger.) The integral splits accordingly, with the function being $e^{x^2}$ on $R_1$ and $e^{y^2}$ on $R_2$ . The integral over $R_1$ is $\int_0^1\int_0^x e^{x^2} dydx$, while the integral over $R_2$ is $\int_0^1\int_0^y e^{y^2} dxdy$. The first one gives $ \int_0^1 xe^{x^2}dx = \frac12(e-1)$; the second gives $ \int_0^1 ye^{y^2}dy = \frac12(e-1)$ so the total is $(e-1)$.
\begin{center}
\includegraphics[scale=1]{pp162.eps}
\end{center}
{\large\bf Chapter 17.}
\prob2. Use the parameterization $x=t, \ y=t^2$ for $t\in [0,1]$. We get $ds = \sqrt{1+4t^2}dt$ so the integral becomes
$$
\int_0^1 t \sqrt{1+4t^2}dt = \frac23 \frac18 \left[(1+4t^2)^{3/2}\right]_0^1 = \frac1{12}(5^{3/2} -1)
$$
\prob4. Set $x=t$ and $y= \sin t$. Then (use integration by parts for $\int t \sin t \; dt$)
$$
\int_C xy\; dx + y\; dy = \int_0^{\pi/2} t\sin t \; dt + \sin t \cos t \; dt
= [-t\cos t + \sin t + \frac12\sin^2(t)]_0^{\pi/2} = \frac32
$$
\prob6.$\ds \int_C \sqrt{xy}\; dx + e^y\; dy + xz\; dz = \int_0^1 \sqrt{t^6}\,4t^3\,dt + e^{t^2}\,2t\,dt + t^7\,3t^2\, dt = \left[\frac47t^7 + e^{t^2} + \frac3{10} t^{10}\right]_0^1 = \frac47 + e-1 + \frac3{10}$.
\prob8. First compute $\rr'(t) = \cos t\ii +\jj$. Then (again with integrating by parts)
\begin{align*}\ds \int_C ((1+t)\sin t \ii + \sin ^2 t \jj) \cdot (\cos t\ii +\jj)\;dt &= \int_0^\pi (1+t)\sin t \cos t\; dt + \sin^2 t \; dt = \\
\frac12\int_0^\pi(1+t)\sin 2t ; dt + (1-\cos 2t) \; dt &= \frac12\left[-\frac12(1+t)\cos 2t + \frac14\sin 2t + t -\frac12\sin 2t\right]_0^\pi = \frac\pi4.
\end{align*}
\prob10. The problem asks for the line integral $\int_C \mathbf{F} \cdot d\rr$ for two paths $C = C_1$ or $C_2$. Let $C_1$ be the straight line, where $\rr_1(t) = (1-t)(3,0,0) + t(0,\pi/2,3)$ and let $C_2$ be the spiral where $\rr_2(t)$ is given in the book. For the first integral, $x = 3(1-t)$, $y= \pi/2t$ and $z= 3t$, so the integral
$$
\int_{C_1} \mathbf{F} \cdot d\rr_1 = \int_0^1 (3t)(-3)dt + (3-3t)(\pi/2)dt + (\pi/2t)(3)dt = \left[-\frac92 t^2 +\frac{3\pi}{2}t -\frac{3\pi}{4}t^2 +\frac{3\pi}{4}t^2 \right]_0^1 = -\frac92 + \frac{3\pi}{2}.
$$
For the second integral, use the given parameterization (and note that $t \in [0, \pi/2]$):
$$
\int_0^{\pi/2} (-9\sin^2 t + 3\cos t + 3t\cos t) dt = \left[-\frac92 +\frac94 \sin 2t + 3\sin t + 3(t\sin t -\sin t)\right]_0^{\pi/2} = -\frac{3\pi}{4}.
$$
\vfil\eject
{\large\bf Additional Problems.}\\
\prob J. The first thing is to get a picture of the graph\\[-10ex]
\begin{tabular}{p{4in}@{\hspace{.4in}}c}
\vspace*{-1.1in}
of the polar curve $r^2=9\cos5\theta$.This one is similar to the graph of $r=\cos2\theta$, done in section 11.3; the result is pictured at right. The area is 5 times the area of one loop, which is described by $-\frac{\pi}{10} \leq \theta \leq \frac{\pi}{10}$. Thus the area is &
\includegraphics[scale=.75]{fj.eps}\\
\end{tabular}
$$
45 \int_{-\frac{\pi}{10}}^{\frac{\pi}{10}} \int_0^{\cos^{1/2}(5\theta)} rdrd\theta = \frac{45}{2} \int_{-\frac{\pi}{10}}^{\frac{\pi}{10}}\cos(5\theta)d\theta = \frac{9}{2} \left[\sin(5\theta)\right]_{-\frac{\pi}{10}}^{\frac{\pi}{10}} = 9.
$$
(The answer in the book is 18; I don't see where the factor of $2$ went.)\\
For problem 34, we need a picture of the graphs of the polar curves \\[-4ex]
\begin{tabular}{p{4in}@{\hspace{.4in}}c}
\vspace*{-1.1in}
$r=2+\cos2\theta$ and $r=2+\sin\theta$. They are pictured together at right; the first one is the peanut shaped region, and the second is the more circular region. The area is twice the area of the crescent-shaped region at lower right, which ranges from $\theta = -\pi/2$ to the $\theta$ that we get from solving $2+\cos2\theta=2+\sin\theta$. This is given by $\sin\theta = \frac12$, i.e., by $\theta = \frac{\pi}{6}$. So&
\includegraphics[scale=.6]{fj34.eps}\\
\end{tabular}
$$
A =2 \int_{-\pi/2}^{\pi/6}\int_{2+\sin\theta}^{2+\cos2\theta} rdr d\theta % = \int_{-\pi/2}^{\pi/6} d\theta
$$
which is really too messy to work out.
\prob K. I claim that $f$ must be a constant. This seems pretty obvious, but requires a bit of proof. First, note that $f_x =0$, so by 1-variable calculus, $f(x,y,z)$ is constant in $x$, ie can be written as a function $g(y,z)$. Now $g_y=f_y=0$, so by the same argument $g$ is constant in $y$, ie can be written as a function $h(z)$. Finally, $h_z=f_z=0$, so again $h$ is a constant. It follows that $f$ is this same constant.
\prob L. The hypothesis is just a way of writing that $g=f_x$ and $h=f_y$. Now $g_y=f_{xy} = f_{yx}$ (by Clairaut's theorem) which is in turn just $h_x$. Thus $g_y= h_x$ as required.
\prob M. This is just a particularly succinct form of the chain rule. The less succinct version is
$ g'(t) = \frac{\del F}{\delx} x'(t) + \frac{\del F}{\dely} y'(t) + \frac{\del F}{\delz} z'(t) = (\nabla F)\cdot \rr'(t)$.
\prob N. We showed earlier in the course that the tangent vector to a curve in a level surface is orthogonal to the normal vector to that surface. (This followed from the chain rule: write the curve as $\rr(t) = (x(t),y(t),z(t))$, and take $\frac{d}{dt}$ of both sides of the equation $F(x(t),y(t),z(t)) = 0$. By the chain rule, the left-hand side is $\nabla F \cdot \rr'(t)$. But $\nabla F $ is the normal vector to the surface, and $\rr'$ is the tangent vector to the curve.) Since the curve lies in both surfaces, it is orthogonal to $\nabla F$ and also to $\nabla G$, ie orthogonal to both normals.
\prob O. Following the hint, let's look at the product rule(s):
$$
\frac{d}{dt} (\uu(t) \cdot \vv(t)) = \uu'(t) \cdot \vv(t) + \uu(t) \cdot \vv'(t), \quad \frac{d}{dt} (\uu(t) \times \vv(t)) = \uu'(t) \times \vv(t) + \uu(t) \times \vv'(t).
$$
From the first one, we have $\int \uu(t) \cdot \vv'(t) \; dt = \int \frac{d}{dt} (\uu(t) \cdot \vv(t)) \; dt - \int \uu'(t) \cdot \vv(t)\; dt$, and using the fundamental theorem of calculus, $ \int \frac{d}{dt} (\uu(t) \cdot \vv(t)) \; dt = \uu(t) \cdot \vv(t)$. The same argument works for the cross product product rule.
\end{document}