% \documentclass[12pt]{article} \usepackage{amssymb,amsmath} \newcommand{\Title}{Math 234, Midterm Exam, October 23, 2001} \newif\ifanswer \answertrue %\answerfalse \def\Points{ Points} \newcount\total \def\Qi{30} \advance\total by \Qi % \def\Qii{30} \advance\total by \Qii % \def\Qiii{70} \advance\total by \Qiii % \def\Qiv{30} \advance\total by \Qiv % \def\Qv{30} \advance\total by \Qv % \def\Qvi{30} \advance\total by \Qvi % \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{\abf}{\mathbf{a}} \newcommand{\cbf}{\mathbf{c}} \newcommand{\hbf}{\mathbf{h}} \newcommand{\ibf}{\mathbf{i}} \newcommand{\jbf}{\mathbf{j}} \newcommand{\kbf}{\mathbf{k}} \newcommand{\rbf}{\mathbf{r}} \newcommand{\ubf}{\mathbf{u}} \newcommand{\vbf}{\mathbf{v}} \newcommand{\Fbf}{\mathbf{F}} \newcommand{\Nbf}{\mathbf{N}} \newcommand{\Rbf}{\mathbf{R}} \newcommand{\Tbf}{\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}{10in} \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} \medskip \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 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} (1) Find the gradient $\nabla U=\mathrm{grad}\, U$ of the function $$ U=(x^2+y^2+z^2)^{-1/2}. $$ \ifanswer \Answer{ $U_x=-(x^2+y^2+z^2)^{-3/2}x$, $U_y=-(x^2+y^2+z^2)^{-3/2}y$, and $U_z=-(x^2+y^2+z^2)^{-3/2}z$, so $$ \nabla U=U_x\ibf+U_y\jbf+U_z\kbf =-\frac{x\ibf }{\rho^3}-\frac{y\jbf }{\rho^3}-\frac{z\kbf}{\rho^3} $$ where $\rho=(x^2+y^2+z^2)^{1/2}$. } \else \vfill \fi \noindent(2) Find the total differential $dU$ of this function. \ifanswer \Answer{ $\ds d U=U_x\,dx+U_y\,dy+U_z\,dz =-\frac{x\,dx }{\rho^3}-\frac{y\,dy }{\rho^3}-\frac{z\,dz}{\rho^3} $ } \else \vfill \newpage \fi \prob{II}{\Qii} A planet moves in space according to Newton's law $ m\, d^2\rbf/dt^2= - mGM|\rbf|^{-3}\rbf $ where $\rbf=x\ibf+y\jbf+z\kbf$ is the position vector of the planet. Let $\vbf$ be the velocity vector, $K=m|\vbf|^2/2$, and $W=-mGM/|\rbf|$. Show that the quantity $E=K+W$ is constant. \ifanswer \Answer{ By the previous problem $$ \nabla W=-\nabla\left(\frac{mGM}{\rho}\right) = \frac{mGM}{\rho^3}(x\ibf+ y\jbf + z\kbf) =\frac{mGM}{|\rbf|^3}\rbf=-\Fbf=-m\abf. $$ Now by the product rule for differentiation $$ \frac{dK}{dt} = \frac{m}{2}\frac{d}{dt}(\vbf\cdot\vbf) = \frac{m}{2}\left(\frac{d\vbf}{dt}\cdot\vbf+\vbf\cdot\frac{d\vbf}{dt}\right) =m\abf\cdot\vbf $$ and by the chain rule $$ \frac{dW}{dt} = (\nabla W)\cdot\vbf= -m\abf\cdot\vbf $$ so $$ \frac{dE}{dt}=\frac{dK}{dt}+\frac{dW}{dt}= m\abf\cdot\vbf-m\abf\cdot\vbf=0. $$ } \else \newpage \fi \prob{III}{\Qiii} The position vector $\Rbf$ of a moving point is given by $$ \Rbf= (2t+3)\ibf+(t^2-1)\jbf. $$ \ifanswer(This is example~3 on page 554 and problem~6 on page 564 of the text.)\fi \medskipnoindent{\bf(1)} Find the velocity vector $\vbf$. \ifanswer \Answer{ $\ds\vbf=\frac{d\Rbf}{dt}=2\ibf+2t\jbf$. } \else \vfill \fi \medskipnoindent{\bf(2)} Find the acceleration vector $\abf$. \ifanswer \Answer{ $\ds\abf=\frac{d\vbf}{dt}=2\jbf$. } \else \vfill \fi \medskipnoindent{\bf(3)} Find the speed. \ifanswer \Answer{ $\ds\frac{ds}{dt}=|\vbf|=2\sqrt{1+t^2}$. } \else \vfill \fi \medskipnoindent{\bf(4)} Find the unit tangent vector $\Tbf$. \ifanswer \Answer{ $\ds\Tbf=\vbf/|\vbf|=\frac{\ibf}{\sqrt{1+t^2}}+\frac{t\jbf}{\sqrt{1+t^2}}$. } \else \vfill \newpage \centerline{\small (Problem III continued)} \medskip \fi \medskipnoindent{\bf(5)} Find the unit normal vector $\Nbf$. \ifanswer \Answer{ The unit normal vector is perpendicular to the unit tangent vector so $$ \Nbf= \pm\left(\frac{-t\ibf}{\sqrt{1+t^2}}+\frac{\jbf}{\sqrt{1+t^2}}\right) $$ This vector points to the concave side of the curve so that $\ds\frac{d\Tbf}{ds}=\kappa\Nbf$ where $\kappa>0$. The curve is a parabola which opens up so the plus sign is correct. } \else \vfill \fi \medskipnoindent{\bf(6)} Find the normal and tangential components of the acceleration. \ifanswer \Answer{ The tangential component is $\ds\abf\cdot\Tbf=2\jbf\cdot\Tbf=\frac{2t}{\sqrt{1+t^2}}$. The normal component is $\ds\abf\cdot\Nbf=2\jbf\cdot\Nbf=\frac{2}{\sqrt{1+t^2}}$. Hence $\ds\abf=\frac{2t}{\sqrt{1+t^2}}\Tbf+\frac{2}{\sqrt{1+t^2}}\Nbf$. } \else \vfill \fi \medskipnoindent{\bf(7)} Find the curvature. \ifanswer \Answer{ From the formula $\ds\abf=\frac{d^2s}{dt^2}\Tbf+\kappa\left(\frac{ds}{dt}\right)^2\Nbf$ we get that $$ \kappa =\frac{\abf\cdot\Nbf}{(ds/dt)^2} = \frac{2}{\sqrt{1+t^2}}\cdot\frac{1}{4(1+t^2)}=\frac{1}{2(1+t^2)^{3/2}}. $$ } \else \vfill \newpage \fi \prob{IV}{\Qiv} Find the local maxima and local minima of the function $$ f(x,y)= x^2-xy+y^2+2x+2y-4. $$ \ifanswer (This is the example on page 611 of the text.)\fi \ifanswer \Answer{$f_x=2x-y+2$, $f_y=-y+2y+2$, so $f_x=f_y=0$ if and only of $x=-2$, $y=-2$. For $(x,y)$ near the point $(a,b)=(-2,-2)$ we have the quadratic approximation $$ Q(x,y)=-8+(x+2)^2-(x+2)(y+2)+(y+2)^2. $$ (In this example $f(x,y)=Q(x,y)$.) The discriminant $b^2-4ac=-3$ is negative so we have with a local maximum or a local minimum. When $y=-2$ we have $Q(x,y)=-8+(x+2)^2\ge-8=Q(-2,-2)$ so $(-2,-2)$ is a local minimum. (In fact it is a global minimum.) As $(-2,-2)$ is the only critical point there is no other local extremum. } \else \newpage \fi \prob{V}{\Qv} The temperature $T$ at any point $(x,y,z)$ in space is $$ T=400xyz^2. $$ Find the highest temperature on the unit sphere $$ x^2+y^2+z^2=1. $$ \ifanswer (This is problem~57 on page 643 of the text.)\fi \ifanswer \Answer{At the maximum we have $\nabla T= \lambda\nabla (x^2+y^2+z^2-1)$ or \begin{eqnarray*} 400 yz^2 &=& \lambda 2x \\ 400 xz^2 &=& \lambda 2y \\ 400 2xyz &=& \lambda 2z \\ x^2+y^2+z^2 &=& 1. \end{eqnarray*} If $x=0$ or $y=0$ or $z=0$, then $T=0$. But the function $T$ takes positive values so we may assume that $x\ne 0$ and $y\ne 0$ and $z\ne 0$. Dividing the first two equations by the third gives $$ \frac{z}{2x}=\frac{x}{z},\qquad \frac{z}{2y}=\frac{y}{z} $$ so $2x^2=z^2$ and $2y^2=z^2$. From the last equation we get $1=x^2+y^2+z^2= 2z^2$ so $z=\pm1/\sqrt2$. The sign of $x$ and $y$ must be the same (otherwise $T<0$) so the maximum occurs at one of the four points $$ (x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt2}\right),\; \left(-\frac{1}{2},-\frac{1}{2},\frac{1}{\sqrt2}\right),\; \left(\frac{1}{2},\frac{1}{2},-\frac{1}{\sqrt2}\right),\; \left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{\sqrt2}\right). $$ The value of $T$ at each of these four points is $400/8=50$. } \else \newpage \fi \prob{VI}{\Qvi} {\bf(a)} Find an equation of the plane tangent to the surface $$ x^3+xy^2+y^3+z^3+1=0 $$ at the point $P_0(-2,1,2)$. \ifanswer{(This is problem~10 on page~641 of the text.)}\fi \ifanswer \Answer{The point $P(x,y,z)$ lies on this plane if and only if the vector from $P_0$ to $P$ is perpendicular to the gradient at $P_0$ of the defining function $F(x,y,z)=x^3+xy^2+y^3+z^3+1$. But $$ \nabla F=(3x^2+y^2)\ibf+(2xy+3y^2)\jbf+3z^2\kbf $$ so $$ \left.\nabla F\right|_{(-2,1,2)}=13\ibf-\jbf+12\kbf $$ so the tangent plane is $13(x+2)-(y-1)+12(z-2)=0$. } \else \vfill \fi \medskipnoindent{\bf(b)} Find equations of the straight line through $P_0(-2,1,2)$ perpendicular to the tangent plane found in part~(a). \ifanswer \Answer{A point $P$ lies on the line if and only if the vector from $P_0$ to $P$ is a multiple of $t\nabla F$ of the gradient vector $\nabla F$ at $P_0$; hence the line has parametric equations $x=-2+13t$, $y=1-t$, $z=2+12t$. Equations for the line are $$ \frac{x+2}{13}=1-y=\frac{z-2}{12}. $$ } \else \vfill \newpage \fi \prob{VII}{\Qvii}{\bf(a)} Find the linear function which best approximates $$ f(x,y)= \frac{1}{1+x-y} $$ near the point $(x,y)=(2,1)$. \ifanswer(This is example~3 on page 589 of the text.)\fi \ifanswer \Answer{ $\ds L(x,y)=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}. $ } \else \vfill \fi \medskipnoindent{\bf(b)} Find the tangent plane at the point $(x,y)=(2,1)$ to the surface $$ z=\frac{1}{1+x-y} $$ \ifanswer \Answer{$z=L(x,y)$, i.e. $\ds z=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}$. } \else \vfill \fi \ifanswer \bigskip \hrule \begin{verbatim} _________ Mon Oct 29 15:09:28 2001 -- There are 261 scores grade range count percent A 225...250 59 22.6% AB 215...224 16 6.1% B 200...214 30 11.5% BC 160...199 77 29.5% C 125...159 48 18.4% D 100...124 11 4.2% F 0... 99 20 7.7% Mean score = 181.5. Mean grade = 2.61. -- \end{verbatim} \fi \end{document}