\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][P roof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{ Ma 115 \qquad \qquad Exam II Solutions\qquad \qquad 10/14/99} \subsection{Name:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \ \ IDN:\_\_\_\_\_\_\_\_\_\_\_\_\_ E-Mail:\_\_\_\_\_\_\_\_\_\_\_\_} \subsection{\protect\small I pledge my honor that I have abided by the Stevens Honor System. \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_% \_} {\Large You may not use a calculator on this exam. You must show all of your work.} \begin{description} \item[1 a (25 points)] Find the equation of the tangent line to the curve \end{description} \vspace{1pt} \begin{equation*} x^{3}+y^{3}=\dfrac{9xy}{2} \end{equation*} at the point $(2,1)$. \vspace{1pt} First, use implicit differentiation to find the derivative: \begin{equation*} 3x^{2}+3y^{2}y^{\prime }=\frac{9}{2}\left( y+xy^{\prime }\right) \end{equation*} \begin{equation*} y^{\prime }\left( 3y^{2}-\frac{9x}{2}\right) =\frac{9y}{2}-3x^{2} \end{equation*} \ \ \begin{equation*} \text{or \ \ }y^{\prime }=\frac{9y-6x^{2}}{6y^{2}-9x} \end{equation*} \vspace{1pt} at the point $\left( 2,1\right) $ \ $\ \ \ y^{\prime }=\frac{5}{4}$ \ which is the slope of the tangent line at that point. Thus the equation of the tangent line is: \begin{equation*} \text{\ }\ \ y-1=\frac{5}{4}\vspace{1pt}\left( x-2\right) \end{equation*} \vspace{1pt} \begin{description} \item[2 a (10 points)] Let $y=\sec t$ and $x=\tan t.$ Find $\dfrac{dy}{dx}$ at $x=1.$ \end{description} \vspace{1pt}Solution $1:$ \vspace{1pt}Use parametric differentiation: \ \ \ \begin{equation*} \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \end{equation*} \begin{equation*} \frac{dy}{dt}=\sec t\tan t \end{equation*} \begin{equation*} \frac{dx}{dt}=\sec ^{2}t \end{equation*} thus: \ \begin{equation*} \frac{dy}{dx}=\frac{\sec t\tan t}{\sec ^{2}t}=\frac{\tan t}{\sec t}=\sin t \end{equation*} when \ \ $x=1$ \ then we know that $\tan t=1$ \ \ (which means that \ $\sin t=\cos t$ ) \ so \ \ $t=\dfrac{\pi }{4}$ \qquad Thus \begin{equation*} \left. \frac{dy}{dx}\right| _{x=1}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \end{equation*} Solution $2:$ Rather that use parametric differentiation one write $y$ in terms of $x$ and then differentiate. Since $x=\tan t,$ we have the diagram below: \begin{center} \FRAME{dtbpF}{3.3797in}{3.0545in}{0pt}{}{}{two.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.3797in;height 3.0545in;depth 0pt;original-width 3.333in;original-height 3.0104in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'two.jpg';file-properties "XNPEU";}% } \vspace{1pt} \end{center} Thus $y=\sec t=\sqrt{1+x^{2}},$ so $y^{\prime }=\left( \dfrac{1}{2}\right) \left( 1+x^{2}\right) ^{-\frac{1}{2}}\left( 2x\right) .$ Setting $x=1,$ we again get $y^{\prime }=\dfrac{1}{\sqrt{2}}$ \begin{description} \item[2 b $\left( 10\text{ points}\right) $] \vspace{1pt}Find the derivative of$\ \ y=\sqrt[4]{x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}}.$ Do \textsl{not% } simplify your result. \end{description} \vspace{1pt}Think of this as \ $\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{\frac{1}{4}}$ and use the chain rule where \ $u=x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}$ \vspace{1pt}then $\ \ y=\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{\frac{1}{4}}=u^{\frac{1}{4}}$ \ and \ $y^{\prime }=\frac{1}{4}% u^{-\frac{3}{4}}\cdot u^{\prime }.$ But \ \ $u^{\prime }$ \ can be found by using the sum rule and another application of the power and chain rule and use of the quotient rule: \vspace{1pt}$u^{\prime }=3x^{2}+6\left( \dfrac{x+1}{x-1}\right) ^{5}\cdot \left( \dfrac{1\cdot \left( x-1\right) -\left( x+1\right) \cdot 1}{\left( x-1\right) ^{2}}\right) $ Thus \begin{equation*} y^{\prime }=\frac{1}{4}\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{-\frac{3}{4}}\left[ 3x^{2}+6\left( \dfrac{x+1}{x-1}\right) ^{5}\cdot \left( \dfrac{1\cdot \left( x-1\right) -\left( x+1\right) \cdot 1}{% \left( x-1\right) ^{2}}\right) \right] \end{equation*} \begin{description} \item[3] Find the derivatives of the following functions. You need not simplify your results. \item[3 a (7 points)] $\dfrac{e}{\sqrt{x^{3}+2x}}$ \end{description} \vspace{1pt} Think \ \begin{equation*} \frac{d}{dx}\left( e\left( x^{3}+2x\right) ^{-\frac{1}{2}}\right) =e\left[ \frac{-1}{2}\left( x^{3}+2x\right) ^{-\frac{3}{2}}\left( 3x^{2}+2\right) % \right] \end{equation*} (Note: \ $e$ is a constant not a variable and not a function in the same way as $\pi $ is a constant: \ \ $e\approx \allowbreak 2.\,\allowbreak 718\,281\,828.$\ ) \vspace{1pt} \begin{description} \item[3 b (7 points)] $\dfrac{\sin ^{2}\left( t^{3}+t\right) }{t^{3}+t}$ \end{description} \vspace{1pt} This is a quotient rule with a chain rule with a power rule with a sum rule ... \vspace{1pt} \begin{equation*} \frac{d}{dx}\left( \dfrac{\sin ^{2}\left( t^{3}+t\right) }{t^{3}+t}\right) =% \frac{2\sin \left( t^{3}+t\right) \cdot \cos \left( t^{3}+t\right) \cdot \left( 3t^{2}+1\right) \left( t^{3}+t\right) -\sin ^{2}\left( t^{3}+t\right) \left( 3t^{2}+1\right) }{\left( t^{3}+t\right) ^{2}} \end{equation*} \begin{description} \item[3 c (11 points)] $\left( \tan \left( x^{2}+2\right) \right) \cos \left( \dfrac{e}{\sqrt{x^{3}+2x}}\right) $ \end{description} \vspace{1pt}Product rule, chain rule ... \vspace{1pt} \begin{equation*} \frac{d}{dx}\left[ \left( \tan \left( x^{2}+2\right) \right) \cos \left( \dfrac{e}{\sqrt{x^{3}+2x}}\right) \right] =\sec ^{2}\left( x^{2}+2\right) \cdot 2x\left( \cos \left( \dfrac{e}{\sqrt{x^{3}+2x}}\right) \right) +\left( \tan \left( x^{2}+2\right) \right) \left[ -\sin \left( \dfrac{e}{\sqrt{% x^{3}+2x}}\right) \cdot \left\{ \frac{-e}{2}\left( x^{3}+2x\right) ^{-\frac{3% }{2}}\left( 3x^{2}+2\right) \right\} \right] \end{equation*} \begin{description} \item[4 a (10 points)] Show that $(fgh)^{\prime }=f^{\prime }gh+fg^{\prime }h+fgh^{\prime }$. \end{description} \vspace{1pt} \vspace{1pt}Start by applying the product rule to \ $\left[ \left( fg\right) h\right] ^{\prime }=\left( fg\right) ^{\prime }h+\left( fg\right) h^{\prime } $ But then since $\left( fg\right) ^{\prime }=f^{\prime }g+fg^{\prime },$ we have \begin{equation*} \left[ \left( fg\right) h\right] ^{\prime }=\left( f^{\prime }g+fg^{\prime }\right) h+\left( fg\right) h^{\prime }=f^{\prime }gh+fg^{\prime }h+fgh^{\prime } \end{equation*} \ as desired. \begin{description} \item[4 b (10 points)] Find the equation of the line tangent to $y=\sqrt{% x^{2}+1}\left( \dfrac{1}{2x-1}\right) ^{2}\left( x+4\right) ^{3}$ at $% (0,64). $ \end{description} \vspace{1pt} \vspace{1pt}First apply the above rule to find \ \begin{equation*} y^{\prime }=\left[ \frac{1}{2}\left( x^{2}+1\right) ^{-\frac{1}{2}}\cdot 2x% \right] \left( \dfrac{1}{2x-1}\right) ^{2}\left( x+4\right) ^{3}+\sqrt{% x^{2}+1}\left[ 2\left( \dfrac{1}{2x-1}\right) \cdot \left( -1\right) \left( 2x-1\right) ^{-2}\cdot 2\right] \left( x+4\right) ^{3}+\sqrt{x^{2}+1}\left( \dfrac{1}{2x-1}\right) ^{2}\left[ 3\left( x+4\right) ^{2}\cdot 1\right] \end{equation*} When \ \ $x=0$ ,\ \ \ \ $y^{\prime }=\left[ \frac{1}{2}\left( 1\right) \cdot 2\left( 0\right) \right] \left( -1\right) \left( 4\right) ^{3}+\left( 1\right) \left[ 2\left( -1\right) \cdot \left( -1\right) \left( 1\right) \cdot 2\right] \left( 4\right) ^{3}+\left( 1\right) \left( 1\right) \left[ 3\left( 4\right) ^{2}\right] =\allowbreak 304$ \ is the slope of the tangent line. Thus at the point \ $\left( 0,64\right) $ \ the equation of the tangent line is \ \begin{equation*} \ y-64=304\left( x-0\right) \end{equation*} \begin{description} \item[5 $\left( 10\text{ points}\right) $] Find the derivative of $y=$ $% e^{\cos \left( \arctan 2x^{2}\right) }$. \end{description} Hint: You may want to simplify the exponent before differentiating. \vspace{1pt} If $\theta =\arctan 2x^{2},$ then $\tan \theta =2x^{2}.$ Thus if you draw a right triangle and call one acute angle $\theta ,$ then the side opposite $% \theta $ has length $2x^{2},$ and the side adjacent to $\theta $ has length $% 1.$ Thus the hypotenuse has length $\sqrt{1+4x^{4}}.$ \begin{center} $\vspace{1pt}\FRAME{itbpF}{3.3797in}{3.0545in}{0in}{}{}{five.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.3797in;height 3.0545in;depth 0in;original-width 3.333in;original-height 3.0104in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/five.jpg';file-properties "XNPEU";}}$ \end{center} \QTP{Body Math} $\vspace{1pt}$ Therefore \QTP{Body Math} $\vspace{1pt}$ \QTP{Body Math} \begin{equation*} \cos \left( \arctan 2x^{2}\right) =\dfrac{1}{\sqrt{1+4x^{4}}} \end{equation*} \QTP{Body Math} $\vspace{1pt}$ and \begin{equation*} y=e^{\cos \left( \arctan 2x^{2}\right) }=e^{\left( 1+4x^{4}\right) ^{-\frac{1% }{2}}} \end{equation*} \QTP{Body Math} $\vspace{1pt}$ Thus \QTP{Body Math} \begin{equation*} y^{\prime }=e^{\left( 1+4x^{4}\right) ^{-\frac{1}{2}}}\left( -\dfrac{1}{2}% \right) \left( 1+4x^{4}\right) ^{-\frac{3}{2}}\left( 16x^{3}\right) \end{equation*} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/exam2Sol_99f.tex %%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% Start /document/two.jpg %%%%%%%%%%%%%%%%%%%%%%% c}`C@DJYTRFAP@ADP@lD@K@@p[CpP@``AF\`AE`pAG\PBI``BLPQCLlpBLdaDS|@E]hqG^ta F\pAHdxrI`HBKcpAGh\SJl@SLtPCM_\RN}`cL|xrLtHs[CpPAdPBIppBL`QCM`aLapQHrHcLrH 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