%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Scientific Word Wrap/Unwrap Version 2.5 % % % % If you are separating the files in this message by hand, you will % % need to identify the file type and place it in the appropriate % % directory. The possible types are: Document, DocAssoc, Other, % % Macro, Style, Graphic, PastedPict, and PlotPict. Extract files % % tagged as Document, DocAssoc, or Other into your TeX source file % % directory. Macro files go into your TeX macros directory. Style % % files are used by Scientific Word and do not need to be extracted. % % Graphic, PastedPict, and PlotPict files should be placed in a % % graphics directory. % % % % Graphic files need to be converted from the text format (this is % % done for e-mail compatability) to the original 8-bit binary format. % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Files included: % % % % "/document/exam3Sol_99f.tex", Document, 9603, 11/7/1999, 17:31:02, ""% % "/document/shadow1.jpg", ImportPict, 9921, 11/3/1999, 21:20:54, "" % % "/document/FKDE9W00.wmf", ImportPict, 108470, 11/3/1999, 20:17:34, ""% % "/document/shpere-cyl.wmf", ImportPict, 135054, 11/3/1999, 22:54:46, ""% % "/document/FKNBGP04.wmf", ImportPict, 7654, 11/3/1999, 23:48:30, "" % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% Start /document/exam3Sol_99f.tex %%%%%%%%%%%%%%%%%% %% This document created by Scientific Word (R) Version 3.0 \documentclass[12pt,thmsa]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{sw20jart} \usepackage{amsmath} \usepackage{graphicx} %TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart} %TCIDATA{Created=Mon Aug 19 14:52:24 1996} %TCIDATA{LastRevised=Sun Nov 07 12:30:59 1999} %TCIDATA{Language=American English} %TCIDATA{CSTFile=webmath.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{MapleDefs= %$F \left( x\right) =2\ln x-3\cos x-5x^{-1}+6x+\allowbreak C $ %$V \left( y\right) =\pi \left( R^{2}-y^{2}\right) 2y $ %$f \left( x\right) =\ln \left( x^{2}+1\right) $ %} %TCIDATA{AllPages= %F=36,\PARA{038

\hfill \thepage} %} \input{tcilatex} \begin{document} \section{\protect\vspace{1pt} Ma 115 \qquad \qquad \qquad Exam III\qquad \qquad \qquad \qquad 11/4/99} \subsection{Name:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \ \qquad \qquad \qquad\ IDN:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \subsection{E-Mail:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\qquad \qquad \qquad \qquad \qquad Lecturer:\_\_\_\_\_\_\_\_\_\_\_} \subsection{\protect\small I pledge my honor that I have abided by the Stevens Honor System. \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_% \_} \subsection{SHOW\ ALL WORK! You may not use a calculator on this exam.\qquad \qquad} \begin{description} \item[I a. $\left( 10\text{ pts}\right) $] Find $\dfrac{dy}{dx}$ if $\ \ y=\arcsin (e^{5x}).$ \end{description} \qquad Solution: \qquad $y^{\prime }=\dfrac{1}{\sqrt{1-(e^{5x})^{2}}}\cdot 5e^{5x}$ \vspace{1pt} \begin{description} \item[ b. $\left( 14\text{ pts}\right) $] Find the derivative of $y=(\sec 5x)^{x}.$ \end{description} \qquad Solution: \vspace{1pt}$\ln y=x\ln (\sec 5x)$ $\dfrac{1}{y}\dfrac{dy}{dx}=\ln (\sec 5x)+x\left( \dfrac{1}{\sec 5x}\right) \cdot 5\sec 5x\tan 5x=\ln (\sec 5x)+5x\tan 5x$ $\vspace{1pt}$ $\dfrac{dy}{dx}=\left[ \ln (\sec 5x)+5x\tan 5x\right] (\sec 5x)^{x}$ \vspace{1pt} also solved as: $\frac{d}{dx}(\sec 5x)^{x}=\allowbreak \dfrac{\ln \frac{1}{\cos 5x}\cos 5x+5x\sin 5x}{\cos 5x}\left( \dfrac{1}{\cos 5x}\right) ^{x}$ \vspace{1pt} \begin{description} \item[ c. $\left( 10\text{ pts}\right) $] Find the most general antiderivative of \ $f(x)=\dfrac{2}{x}+3\sin x+\dfrac{5}{x^{2}}+6.$ \end{description} \qquad \vspace{1pt}Solution: $F\left( x\right) =2\ln x-3\cos x-5x^{-1}+6x+C$ $\vspace{1pt}$ \vspace{1pt}\newpage \begin{description} \item[ II $\left( 22\text{ pts}\right) $] \vspace{1pt}A man, $6$ feet tall, walks at a speed of $4$ ft./sec. toward a street lamp which is $14$ feet above the ground. \ Let $\theta $ be the angle subtended by the man's shadow at his head (This is also the angle between the lamp post and a light ray from the lamp to his head. See the diagram below.) \ How fast is $\theta $ changing with respect to time when the man is $12$ feet from the post? \end{description} \vspace{1pt}\FRAME{dtbpFX}{1.7123in}{1.548in}{0pt}{}{}{shadow1.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.7123in;height 1.548in;depth 0pt;original-width 240.875pt;original-height 217.5625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/shadow1.jpg';file-properties "XNPEU";}} \vspace{1pt}Solution: Want to find $\dfrac{d\theta }{dt}$ The solution can be found using any af the three triangles; the small lower triangle, the small upper triangle, or the large triangle. Let \ $s$ be the length of the shadow and $m$ be the distance from the lamp to the man as labeled above. \vspace{1pt}Then \ $\dfrac{dm}{dt}=-4$ ft./sec. \ is given Observe that $\ $\ for the large triangle $\ \tan \theta =\dfrac{s+m}{14}$ , for the lower triangle$\ \tan \theta =\dfrac{s}{6}$ ,\ and for the upper triangle $\tan \theta =\frac{m}{8}$ If you use the large or lower triangles to solve the problem note the relationship \ $\dfrac{s}{6}=\dfrac{s+m}{14}$ \ from the above observation or from similar triangles. Thus \ $14s=6\left( s+m\right) $ \ or \ $4s=3m$ \vspace{1pt}It is now possible to rewrite \ $\tan \theta $ in terms of \ $m$ $\tan \theta =\dfrac{\left( \frac{3}{4}m\right) +m}{14}=\allowbreak \dfrac{1% }{8}m$ \ \ \ \ \ (Note that this was obtained directly above from the definition of tangent.) $\theta =\arctan \dfrac{m}{8}$ \ Differentiate using the chain rule: $\dfrac{d\theta }{dt}=\dfrac{1}{1+\left( \frac{m}{8}\right) ^{2}}\left( \dfrac{1}{8}\right) \dfrac{dm}{dt}$ When \ $m=12$ \ \ and \ $\dfrac{dm}{dt}=-4$ \ we have: $\dfrac{d\theta }{dt}=\dfrac{1}{1+\left( \frac{12}{8}\right) ^{2}}\left( \dfrac{1}{8}\right) \left( -4\right) =\allowbreak -\dfrac{2}{13}$ radians/sec. \newpage \begin{description} \item[ III $\left( 22\text{ pts}\right) $] Find the height of the right circular cylinder of largest volume that can be inscribed in a sphere of radius $R$. \ Justify that your value of $h$ indeed gives the cylinder of largest volume. \ (The volume of a cylinder with height $h$ and radius $r$ is $\pi r^{2}h$).\vspace{1pt} \end{description} \vspace{1pt} \ \ \FRAME{itbpFX}{3.0441in}{2.0384in}{0in}{}{}{fkde9w00.wmf}{% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.0441in;height 2.0384in;depth 0in;original-width 216.8125pt;original-height 144.5625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/FKDE9W00.wmf';file-properties "XNPEU";}}% \ \ \ \ $\FRAME{itbpFX}{2.0297in}{2.0358in}{0in}{}{}{shpere-cyl.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.0297in;height 2.0358in;depth 0in;original-width 116.9375pt;original-height 117.25pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/shpere-cyl.wmf';file-properties "XNPEU";}% }$ \ \vspace{1pt} Solution: \vspace{1pt} Draw the cross-section of the sphere as above. Note that the height of the cylinder is \ $h=2y$ \ \ so the volume is \ $% V=\pi r^{2}\left( 2y\right) $ Note also that \ $r^{2}+y^{2}=R^{2}$ \ so \ $r^{2}=R^{2}-y^{2}$ \vspace{1pt}Volume of the cylinder can now be expressed as a function of \ $% y $: \ \ $V\left( y\right) =\pi \left( R^{2}-y^{2}\right) \left( 2y\right) $ $\vspace{1pt}$ $V^{\prime }\left( y\right) =\allowbreak -6\pi y^{2}+2\pi R^{2}=0$ \ \ will locate local extremum $-6\pi y^{2}+2\pi R^{2}=0$ \ Solution is : $y=\frac{1}{3}\sqrt{3}R$ $\ $or $% \ y=-\frac{1}{3}\sqrt{3}R$ \vspace{1pt}Note that \ $0\leq y\leq R$ Critical points are: \vspace{1pt}$y=\frac{1}{3}\sqrt{3}R$ \ \ $y=0$ \ \ $y=R$ $V^{\prime \prime }\left( y\right) =\allowbreak -12\pi y$ when $\ y=\frac{1}{3}\sqrt{3}R$ \ \ \ $V^{\prime \prime }\left( y\right) =-12\pi \left( \frac{1}{3}\sqrt{3}R\right) $ \ is negative so the function is concave down \ thus $y=\frac{1}{3}\sqrt{3}R$ \ is a local maximum \vspace{1pt} \ $0$ \ and \ $R$ \ result in cylinders with no volume. \vspace{1pt} Thus the height of the cylinder of maximum volume is: $h=2y=\frac{2}{3}\sqrt{3}R$ $\ \ \ \ \ \ $(and its volume is:$\ \ \ \ V\left( \frac{1}{3}\sqrt{3}R\right) =\allowbreak \frac{4}{9}\pi R^{3}\sqrt{3} $ ) \newpage \begin{description} \item[ IV $\left( 22\text{ pts}\right) $] \vspace{1pt}Find all maxima, minima, and inflection points of $f(x)=\ln (x^{2}+1)$. \ On what interval or intervals is the graph concave up? \ On what interval or intervals is the graph concave down? \end{description} \vspace{1pt}Solution: The domain of the natural log function is \ $0