% \documentclass[12pt]{article} \usepackage{amssymb,amsmath} \newcommand{\Title}{Math 234, Final Exam, Tuesday December 18, 2001} \newif\ifanswer \answertrue %\answerfalse \def\Points{ Points} \newcount\total \def\Qi{40} \advance\total by \Qi % \def\Qii{50} \advance\total by \Qii % \def\Qiii{50} \advance\total by \Qiii % \def\Qiv{60} \advance\total by \Qiv % \def\Qv{50} \advance\total by \Qv % \def\Qvi{50} \advance\total by \Qvi % \def\Qvii{50} \advance\total by \Qvii % \def\Qviii{50} \advance\total by \Qviii % %\def\Qvii{30} \advance\total by \Qvii % %\def\Qviii{30} \advance\total by \Qviii % %\def\Qix{15} \advance\total by \Qx % \newcommand{\0}{\mathbf{0}} \newcommand{\bfa}{\mathbf{a}} \newcommand{\bfc}{\mathbf{c}} \newcommand{\bfh}{\mathbf{h}} \newcommand{\bfi}{\mathbf{i}} \newcommand{\bfj}{\mathbf{j}} \newcommand{\bfk}{\mathbf{k}} \newcommand{\bfr}{\mathbf{r}} \newcommand{\bfn}{\mathbf{n}} \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfF}{\mathbf{F}} \newcommand{\bfN}{\mathbf{N}} \newcommand{\bfR}{\mathbf{R}} \newcommand{\bfT}{\mathbf{T}} \newcommand{\p}{\partial} \newcommand{\prob}[2]{\pagebreak[3]\noindent{\bf #1.} (#2 points.) } \newcommand{\medskipnoindent}{\par\medskip\par\noindent} \newcommand{\ds}{\displaystyle} \ifanswer\else \setlength{\topmargin}{0in} \setlength{\textwidth}{6.8in} \setlength{\textheight}{9in} \setlength{\oddsidemargin}{-.2in} \setlength{\evensidemargin}{-.2in} \pagestyle{empty} \fi \newcommand{\Answer}[1]{\par\medskip\par\noindent{\bf Answer: }{\small #1}\par\medskip\par} \newcommand{\PageBreak}[1]{\par\bigskip\par\pagebreak[#1]\par} \begin{document} \vspace*{-0.5in} \ifanswer \else Name \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \fi \medskip \centerline{\large \Title} \medskip \ifanswer \centerline{\large Answers} \bigskip (The page references below are to Thomas and Finney fifth edition.) \bigskip \else \medskip \fi \ifanswer\else \begin{center} {\em Circle your section.} \begin{tabular}{|llr|llr|} \hline 321 & Cho & 7:45 T & 322 & Cho & 7:45 R \\ 323 & Cho & 8:50 T & 324 & Cho & 8:50 R \\ 325 & Raichev & 9:55 T & 326 & Raichev & 9:55 R \\ 329 & Russell & 12:05 T & 330 & Russell & 12:05 R \\ 331 & Russell & 1:20 T & 332 & Russell & 1:20 R \\ 333 & Raichev & 2:25 T & 334 & Raichev & 2:25 R \\ \hline \end{tabular} \large \renewcommand{\arraystretch}{1.8} \bigskip \begin{tabular}{|l|l|@{\qquad\qquad}c|} \hline\hline I & \Qi \Points & \\ \hline II & \Qii \Points & \\ \hline III & \Qiii \Points & \\ \hline IV & \Qiv \Points & \\ \hline V & \Qv \Points & \\ \hline VI & \Qvi \Points & \\ \hline VII & \Qvii \Points & \\ \hline VIII & \Qviii \Points & \\ \hline % IX & \Qix \Points & \\ \hline % X & \Qx \Points & \\ \hline \hline \hline Total & \the\total \Points & \\ \hline \hline % Total = 250 \end{tabular} \end{center} \bigskip {\sc You may leave iterated integrals unevaluated so long as you correctly indicate the limits of integration. Show your reasoning. Your grade will be based in part on the clarity of your writing. YOU MUST EXPLAIN HOW YOU GOT YOUR ANSWER.} \newpage \fi \prob{I}{\Qi} Find an equation for the plane that is tangent to the surface $z=x^2-xy-y^2$ at the point $P_0(1,1,-1)$. \ifanswer \Answer{The equation for the surface has form $z=f(x,y)$ so the answer is given by the linearization $$ z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0), $$ i.e. $$ z= -1+ (x-1)-3(y-1). $$ Alternatively the surface has equation $F(x,y,z)=0$ where $F(x,y,z)=z-x^2+xy+y^2$ so the equation for the tangent plane is $$ 0=\nabla F(x_0,y_0,z_0)\cdot\overrightarrow{P_0P} =-(x-1)+3(y-1)+(z+1). $$ (This is problem~3 on page~583.) \PageBreak{1} }\else \vfill\par\newpage \fi \prob{II}{\Qii} {\bf(1)} Find the linear function $L(x,y)$ which best approximates the function $$ f(x,y)=\frac{1}{1+x-y} $$ near the point $(2,1)$. \ifanswer \Answer{This is the linearization $$ L(x,y)=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1), $$ i.e. $$ L(x,y)=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}. $$ (This is Example~3 on page~589.) }\else \vfill\par \fi \medskipnoindent {\bf(2)} Find the quadratic function $Q(x,y)$ which best approximates the function $f(x,y)$ of part~(1) near the point $(2,1)$. \ifanswer \Answer{This is the quadratic approximation $$ Q(x,y)=L(x,y)+\frac{f_{xx}(2,1)(x-2)^2+2f_{xy}(2,1)(x-2)(y-1)+f_{yy}(2,1)(y-1)^2}{2}, $$ i.e. $$ Q(x,y)= \frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}+\frac{(x-2)^2}{8}-\frac{(x-2)(y-1)}{4}+\frac{(y-1)^2}{8}. $$ (This was explained in the lecture and in section 16-11.) \PageBreak{1} }\else \vfill\par\newpage \fi \prob{III}{\Qiii} {\bf(1)} Find the total differential $dw$ of the function $$ w=e^{2x+3y}\cos4z. $$ \ifanswer \Answer{$\ds dw=\frac{\p w}{\p x}\,dx+\frac{\p w}{\p y}\,dy+\frac{\p w}{\p z}\,dz$ so $$ dw =2e^{2x+3y}\cos4z\,dx+3e^{2x+3y}\cos4z\,dy-4e^{2x+3y}\sin4z\,dz. $$ }\else \vfill \fi \medskipnoindent {\bf(2)} Find the derivative $\ds\frac{dw}{dt}$ of the function $w$ of part~(1) along the curve given by the parametric equations $$ x=\ln t, \qquad y=\ln(t^2+1),\qquad z=t. $$ \ifanswer \Answer{$\ds \frac{dw}{dt}= \frac{\p w}{\p x}\,\frac{dx}{dt}+ \frac{\p w}{\p y}\,\frac{dy}{dt}+ \frac{\p w}{\p z}\,\frac{dz}{dt}$ so $$ \frac{dw}{dt} =2e^{2\ln t+3\ln(t^2+1)}\cos4t\,\frac{1}{t}\,+ 3e^{2\ln t+3\ln(t^2+1)}\cos4t\,\frac{2t}{t^2+1}\,- 4e^{2\ln t+3\ln(t^2+1)}\sin4t. $$ Alternatively, along the curve we have $$ w=e^{2\ln t+3\ln(t^2+1)}\cos4t=t^2(t^2+1)^3\cos4t $$ so, by the product rule, $$ \frac{dw}{dt}= 2t(t^2+1)^3\cos4t+t^2(t^2+1)^26t\cos4t-4t^2(t^2+1)^3\sin4t. $$ (This is exercise~3 on page~604.) \PageBreak{1} }\else \vfill\par\newpage \fi \prob{IV}{\Qiv} The unit tangent vector and curvature vector of a curve whose position vector is $\bfR$ are defined by $$ \bfT=\frac{d\bfR}{ds},\qquad \kappa\bfN=\frac{d\bfT}{ds}=\frac{d^2\bfR}{ds^2} $$ where $s$ is the arclength. The osculating circle to a curve at a point $P$ on the curve is that circle through $P$ which has the same unit tangent vector and curvature vector at $P$ as the curve. \medskipnoindent{\bf(1)} Find $\bfT$ and $\kappa\bfN$ for the curve $y=e^x$. Express your answers as functions of the $x$ coordinate. If either (or both) of your answers depend on the direction of parameterization so indicate by placing a $\pm$ in front of your answer. \ifanswer \Answer{ $\bfR=x\bfi+e^x\bfj$ so $$ \bfT=\frac{d\bfR}{dx}\frac{dx}{ds}= \pm(\bfi+e^x\bfj)(1+e^{2x})^{-1/2} $$ and $$ \kappa\bfN=\frac{d\bfT}{dx}\frac{dx}{ds}= \left(e^x\bfj(1+e^{2x})^{-1/2}-(\bfi+e^x\bfj)(1+e^{2x})^{-3/2}e^{2x}\right)(1+e^{2x})^{-1/2}. $$ } \else\par\vfill\par \centerline{(Problem IV continues on next page)} \newpage \centerline{(Problem IV continued from previous page)}\par\medskip \fi \medskipnoindent{\bf(2)} Find an equation for the osculating circle to the curve of part~(1) at the point $P(0,1)$. \ifanswer \Answer{At the point $P(0,1)$ we have $$ \bfT=\pm\frac{\bfi+\bfj}{\sqrt{2}}, \qquad \kappa\bfN=\frac{\bfj}{2}-\frac{\bfi+\bfj}{4}=-\frac{\bfi}{4}+\frac{\bfj}{4}. $$ The osculating circle has curvature $\kappa=\frac{\sqrt{2}}{4}$ and hence radius $\frac{4}{\sqrt{2}}$ and thus has an equation of form $(x-a)^2+(y-b)^2=8$. The center $Q(a,b)$ of this circle lies on the concave side of the curve $y=e^x$ and the line $\overline{PQ}$ from the center $Q$ to the point $P$ on the curve is perpendicular to the tangent line $y=x+1$ to the curve at $P$. Hence the line $\overline{PQ}$ has slope $\ds\frac{b-1}{a-0}=-1$ so $b=1-a$. Since the distance from $Q$ to $P$ is $\frac{4}{\sqrt{2}}=\sqrt{(a-0)^2+(b-1)^2}$ we have $8=a^2+(b-1)^2=2a^2$ so $a=-2$ and $b=3$. Thus an equation for the circle is $$ (x+2)^2+(y-3)^2=8. $$ (This is problem~13 on page~557.) \PageBreak{1} }\else \vfill\par\newpage \fi \prob{V}{\Qv} {\bf(1)} If the vector field $\bfv=M\bfi+N\bfj$ represents the velocity vector field of a fluid in the plane then the flux across a curve $C$ measures the rate at which the fluid flows across the curve. Find the flux of the field $$ \bfv=2x\bfi-3y\bfj $$ outward across the ellipse $$ 16x^2+y^2=16. $$ \ifanswer \Answer{Let $R$ denote the interior of the ellipse, $\bfn$ denote the unit outward normal to the ellipse, and $ds$ denote the arclength element of the ellipse. Orient the ellipse counter clockwise. The unit tangent and the outward normal are $$ \bfT=\frac{dx}{ds}\bfi+\frac{dy}{ds}\bfj,\qquad \bfn=\frac{dy}{ds}\bfi-\frac{dx}{ds}\bfj. $$ By the Divergence Theorem the flux of $\bfv=M\bfi+N\bfj$ across the ellipse is $$ \oint_{\p R}\bfv\cdot\bfn\, ds = \int\!\!\!\int_R\left(\frac{\p M}{\p x}+\frac{\p N}{\p y}\right)\, dA =\int_{x=-1}^1\int_{y=-\sqrt{16-16x^2}}^{\sqrt{16-16x^2}} (2-3)\,dy\,dx. $$ The value of the integral is $-4\pi$ since the area of the ellipse $\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi ab$. (This is example~3 on page~708.) \PageBreak{1} } \else\par\vfill\par \centerline{(Problem V continues on next page)} \newpage \centerline{(Problem V continued from previous page)}\par\medskip \fi \medskipnoindent{\bf(2)} Find the work done by the force field $$ \bfF= 3y\bfi+2x\bfj, $$ in moving a particle once\ifanswer\footnote{The wording of the question in Thomas Finney fifth would be ``$\ldots$ when the point of application moves once $\ldots$''.} \fi counter clockwise around the ellipse in part~(1). Hint: The vector field $\bfv$ arises from $\bfF$ by rotation through ninety degrees. \ifanswer \Answer{The force field is $\bfF=-N\bfi+M\bfj$ where $\bfv=M\bfi+N\bfj$ is the field in part~(1). The work is $$ \oint_{\p R}\bfF\cdot\bfT\, ds= \oint_{\p R}\bfv\cdot\bfn\, ds $$ so the answer is the same as for part~(1). (This is example~4 on page~709.) \PageBreak{1} }\else \vfill\par\newpage \fi \prob{VI}{\Qvi} Find the area cut off the surface $y^2+z^2=2x$ by the plane $x=1$. \ifanswer \Answer{This is the same as the area of the surface $x^2+y^2=2z$ cut off by the plane $z=1$ and the area is $$ \int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+ \left(\frac{\p z}{\p x}\right)^2+\left(\frac{\p z}{\p y}\right)^2 }\, dx\,dy =\int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+x^2+y^2}\,dx\,dy. $$ In cylindrical coordinates the surface is $r^2=2z$ and $z\le 1$ when $r\le\sqrt{2}$ the area is $$ \int\!\!\!\int_{x^2+y^2\le 2}\sqrt{1+x^2+y^2}\,dx\,dy = \int_0^{2\pi}\int_0^{\sqrt2} \sqrt{1+r^2}\, r\, dr\,d\theta. $$ The integral can be evaluated with the substitutions $u=1+r^2$, $du=2r\,dr$ so $u=1$ when $r=0$ and $u=3$ when $r=\sqrt2$ and $$ \int_0^{\sqrt2}\sqrt{1+r^2}\, r\, dr=\frac12\int_1^3\sqrt{u}\,du= \left.\frac{1}{2}\cdot\frac{2}{3}u^{3/2}\right|_1^3=\frac{\sqrt{27}-1}{3}. $$ (This is problem~48 on page 678.) }\else \vfill\par\newpage \fi \prob{VII}{\Qvii} Let $A$, $B$, $C$ be twice continuously differentiable functions of $(x,y,z)$ and let $\bfF$ be the vector field defined by $$ \bfF=A\bfi+B\bfj+C\bfk. $$ The curl of $\bfF$ is denoted by $\nabla\times\bfF$ and the divergence of $\bfF$ is denoted by $\nabla\cdot\bfF$. \medskipnoindent{\bf(1)} Write formulas for the divergence and curl of the vector field $\bfF$ and show that the divergence of the curl is zero. \ifanswer \Answer{ $$ \nabla\cdot\bfF=\frac{\p A}{\p x}+\frac{\p B}{\p y}+\frac{\p C}{\p z}, $$ $$ \nabla\times\bfF= \left(\frac{\p C}{\p y}-\frac{\p B}{\p z}\right)\bfi+ \left(\frac{\p A}{\p z}-\frac{\p C}{\p x}\right)\bfj+ \left(\frac{\p B}{\p x}-\frac{\p A}{\p y}\right)\bfk, $$ so $$ \nabla\cdot\left(\nabla\times\bfF\right) =(C_y-B_z)_x+(A_z-C_x)_y+(B_x-A_y)_z=0. $$ (This is problem~24 on page~702.) \PageBreak{1} } \else\par\vfill\par \centerline{(Problem VII continues on next page)} \newpage \centerline{(Problem VII continued from previous page)}\par\medskip \fi \medskipnoindent{\bf(2)} State Stokes' Theorem and the Divergence Theorem for the vector field $\bfF$. Be sure to define any notation you use. \ifanswer \Answer{ For any three dimensional region $R$ with boundary $\p R$ the Divergence Theorem (see page~722) says that $$ \int\!\!\!\int\!\!\!\int_R \nabla\cdot\bfF\,dV = \int\!\!\!\int_{\p R} (\bfF\cdot\bfn)\,d\sigma $$ where $dV$ is the volume element, $\bfn$ is the outward unit normal, and $d\sigma$ is the area element. For any surface $S$ with boundary $\p S$ Stokes' Theorem (see page~729) says that $$ \int\!\!\!\int_S (\nabla\times\bfF)\cdot\bfn\,d\sigma = \int_{\p S} \bfF\cdot d\bfR $$ where $d\bfR=dx\,\bfi+dy\,\bfj+dz\,\bfk=\bfT\,ds$; here $\bfT$ is the unit tangent vector to the boundary $\bfR$. The relation between the direction of the unit normal $\bfn$ and the direction of the unit tangent $\bfT$ is such that if the normal is the thumb of your right hand, the forefinger of your right hand indicates the direction around the boundary. \PageBreak{1} }\else \vfill\par\newpage \fi \prob{VIII}{\Qviii} Let $\bfF$ denote the force field in the Kepler problem, i.e. $$ \bfF= -\frac{x\bfi}{\rho^3}- \frac{y\bfj}{\rho^3}-\frac{z\bfk}{\rho^3}, \qquad \rho=\sqrt{x^2+y^2+z^2}. $$ \medskipnoindent{\bf(1)} Is there a function $W$ such $\nabla W=\bfF$? If so, find it; if not, say why not. \ifanswer \Answer{As we learned in our study of the Kepler problem, $\bfF=\nabla W$ where $W=\rho^{-1}$. }\else \vfill \fi \medskipnoindent{\bf(2)} Write an iterated integral (you need not evaluate it) for the outward flux $$ \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma $$ over the ellipsoid $S$ defined by $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1. $$ Here $d\sigma$ denotes the area element on the ellipsoid and $\bfn$ denotes the outward unit normal to the ellipsoid. If you don't see how to do this, do the special case of the sphere ($a=b=c$) for partial credit.\footnote{ In grading this question, it was sometimes difficult to decide if the student was doing the special case or the general case.} \ifanswer \Answer{The problem is easy when $a=b=c$; then the outward unit normal is $\bfn=a^{-1}\bfR$ and $\bfF=-a^{-3}\bfR=-a^{-2}\bfn$ so $\bfF\cdot\bfn=-a^{-2}$. As this is constant, the integral is this constant times the area of the sphere, i.e. $$ \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma = -a^{-2}(4\pi a^2)=-4\pi. $$ For the general case, parameterize the ellipsoid by the equations $$ x=a\cos\theta\sin\phi,\qquad y=b\sin\theta\sin\phi,\qquad z=c\cos\phi. $$ Let $\bfR=x\bfi+y\bfj+z\bfk$ be the position vector. Then $$ \frac{\p\bfR}{\p\phi}=a\cos\theta\cos\phi\bfi+b\sin\theta\cos\phi\bfj-c\sin\phi\bfk $$ and $$ \frac{\p\bfR}{\p\theta}=-a\sin\theta\sin\phi\bfi+b\cos\theta\sin\phi\bfj $$ so $$ \frac{\p\bfR}{\p\phi}\times \frac{\p\bfR}{\p\theta} = bc\sin\theta\sin^2\phi\bfi+ac\cos\theta\sin^2\phi\bfj+ab\cos\phi\sin\phi\bfk. $$ Now $\ds\bfn \,d\sigma=\frac{\p\bfR}{\p\phi}\times \frac{\p\bfR}{\p\theta}d\phi\,d\theta$ and $$ \bfF= -\frac{a\cos\theta\sin\phi\bfi}{\rho^3}- \frac{b\sin\theta\sin\phi\bfj}{\rho^3}-\frac{c\cos\phi\bfk}{\rho^3}, $$ where $$ \rho=\sqrt{x^2+y^2+z^2}=(a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi+c^2\cos^2\phi)^{1/2} $$ so $$ \int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma = -abc\int_0^{2\pi}\int_0^\pi \frac{2\cos\theta\sin\theta\sin^3\phi+\cos^2\phi}{ (a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi+c^2\cos^2\phi)^{3/2} } \,d\phi\,d\theta $$ {\bf Remark.} Some of you may have learned in a course in electrostatics that the integral is $4\pi$ times the total charge inside. This is actually a consequence of the Divergence Theorem as follows. A direct calculation (see problem~11 on page~632) shows that for $W=\rho^{-1}$ we have $$ \nabla\cdot\bfF=\nabla\cdot\nabla W= \frac{\p^2 W}{\p x^2}+\frac{\p^2 W}{\p y^2}+\frac{\p^2 W}{\p z^2}=0. $$ Let $R$ be the region outside the tiny sphere $\rho=h$ and inside the ellipsoid $S$. Then $$ 0=\int\!\!\!\!\int\!\!\!\!\int_R\nabla\cdot\bfF\,dV =\int\!\!\!\!\int_{\p R} \bfF\cdot\bfn \,d\sigma =\int\!\!\!\!\int_S \bfF\cdot\bfn \,d\sigma-\int\!\!\!\!\int_{\rho=h} \bfF\cdot\bfn \,d\sigma. $$ The integral over the sphere $\rho=h$ is $-4\pi$. If you used this method to answer the question you will get full credit provided you indicated that the divergence of $\bfF$ vanishes and that you are using the Divergence Theorem. }\else \vfill % last page \centerline{(Continue on reverse side if you need more space.)} \fi \ifanswer \begin{verbatim} ____ Thu Dec 20 17:43:32 2001 -- There are 253 scores grade range count percent A 320...400 31 12.3% AB 300...319 15 5.9% B 230...299 65 25.7% BC 180...229 60 23.7% C 150...179 30 11.9% D 100...149 32 12.6% F 0... 99 20 7.9% Mean score = 217.1. Mean grade = 2.42. -- \end{verbatim} The cutoffs for the toal score (out of 1000) were A 865, AB 785, B 705, BC 603, C 500, D 385. \fi \end{document}