\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}{Definition} \newtheorem{example}{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.4in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Related Rates} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright \; Hidegkuti, Powell, 2009} \rfoot{\small Last revised: December 6, 2013} \textwidth 7.6in \textheight 9.5in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\LARGE Sample Problems\bigskip } \end{center} \begin{enumerate} \item A city is of a circular shape. \ The area of the city is growing at a constant rate of $2\dfrac{\unit{mi}^{2}}{\unit{y}}$ \ (square miles per year). \ How fast is the radius growing when it is exactly $15\unit{mi}$? \item A sphere is growing in such a manner that its radius increases at $0.2% \dfrac{\unit{m}}{\unit{s}}$ (meters per second). \ How fast is its volume increasing when its radius is $4\unit{m}$ long? \item A sphere is growing in such a manner that its volume increases at $0.2% \dfrac{\unit{m}^{3}}{\unit{s}}$ (cubic meters per second). \ How fast is its radius increasing when it is $7\unit{m}$ long? \item The area of a rectangle is kept fixed at $100$ square meters while the legths of the sides vary. \ Express the rate of change of the length of the vertical side in terms of the rate of change in the length of the other side when a) \ the horizontal side is $18$ meters long \ \ \ b) \ the rectangle is a square \item A ladder $20\unit{ft}$ long leans a gainst a vertical building. If the top of the ladder slides down at a rate of $\sqrt{3}\dfrac{\unit{ft}}{\unit{s% }}$, how fast is the bottom of the ladder sliding away from the building when the top of the ladder is $10\unit{ft}$ above the ground? \item A tank, shaped like a cone shown on the picture below, is being filled up with water. \ The top of the tank is a circle with radius $5~\unit{ft}$, its height is $15~\unit{ft}$. \ Water is added to the tank at the rate of $% V^{\prime }\left( t\right) =2\pi \dfrac{\unit{ft}^{3}}{\unit{min}}$. How fast is the water level rising when the water level is $6$ $\unit{ft}$ high? \ (The volume of a cone with height $h$ and base radius $r$ is $V=\dfrac{\pi r^{2}h}{3}$.)\FRAME{dtbpF}{1.8005in}{2.066in}{0pt}{}{}{insert21.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8005in;height 2.066in;depth 0pt;original-width 1.8663in;original-height 2.1465in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert21.bmp';file-properties "XNPEU";}} \item Two quantities $p$ and $q$ depending on $t$ are subject to the relation $\dfrac{1}{p}+\dfrac{1}{q}=1.$ a) \ Express $p^{\prime }\left( t\right) $ in terms of $q^{\prime }\left( t\right) $. \ \newline b) At a certain moment, $p\left( t_{0}\right) =\dfrac{4}{3}$ and $p^{\prime }\left( t_{0}\right) =2.$ \ Find $q\left( t_{0}\right) $ and $q^{\prime }\left( t_{0}\right) .$ \item The base radius and height of a cylinder are constantly changing but the volume of the cylinder is kept at a constant $600\pi $ $\unit{in}^{3}$. \ a) \ At a time $t_{1}$ the base radius is $r\left( t_{1}\right) =10\unit{in}$ and its rate of change is $r^{\prime }\left( t_{1}\right) =0.2\dfrac{\unit{in% }}{\unit{s}}.$ \ Compute the rate of change of the height of the cylinder $% h\left( t\right) $ at time $t_{1}$. b) \ At a time $t_{2}$ the height is $h\left( t_{2}\right) =12\unit{in}$ and its rate of change is $r^{\prime }\left( t_{2}\right) =-0.5\dfrac{\unit{in}}{% \unit{s}}.$ \ Compute the rate of change of the radius of the cylinder $% r\left( t\right) $ at time $t_{2}$. \item An object, dropped from a height of $h$ has a location of $y\left( t\right) =-16t^{2}+h$ feet after $t$ seconds. \ We dropped a small object from a height of $60$ feet.\FRAME{dtbpF}{2.4275in}{2.527in}{0pt}{}{}{% related1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.4275in;height 2.527in;depth 0pt;original-width 2.5728in;original-height 2.6801in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../Math207/math207_fa13/quizzes/sample_quiz14/related1.bmp';file-properties "XNPEU";}% } a) \ Where is the object and what is its velocity after $1.5$ seconds? b) \ Suppose there is a $30$ feet tall street light $10$ feet away from the point where the object will land. \ How far is the shadow of the object from the base of the street light at $t=1.5$? c) \ How fast is the obejct's shadow moving at $t=1.5$? \ {\LARGE \pagebreak } \end{enumerate} \begin{center} {\LARGE Sample Problems - Answers\bigskip \bigskip } \end{center} 1.) $\ \dfrac{1}{15\pi }\dfrac{\unit{mi}}{\unit{y}}$ \ \ \ \ \ \ \ \ \ 2.) $% \ 12.\,\allowbreak 8\pi \dfrac{\unit{m}^{3}}{\unit{s}}\approx 40.\,\allowbreak 212\,385\,965\,\allowbreak 949\,4\dfrac{\unit{m}^{3}}{\unit{% s}}$ \ \ \ \ \ \ \ \ 3.) \ $\dfrac{1}{980\pi }\dfrac{\unit{m}}{\unit{s}}% \approx \allowbreak 3.\,\allowbreak 248\,06\times 10^{-4}\dfrac{\unit{m}}{% \unit{s}}$\bigskip 4.) \ a) \ $v^{\prime }\left( t\right) =-\dfrac{25}{81}h^{\prime }\left( t\right) $\ \ \ \ \ \ \ \ \ \ \ \ b) \ $v^{\prime }\left( t\right) =-h^{\prime }\left( t\right) $ \ \ \ \ \ \ \ 5.) \ $1\dfrac{\unit{ft}}{\unit{% s}}$ \ \ \ \ \ \ \ 6.) \ $\dfrac{1}{2}\dfrac{\unit{ft}}{\unit{min}}$ \ \ \ \ \ \ \bigskip 7.) \ a) \ $p^{\prime }=-\dfrac{p^{2}}{q^{2}}q^{\prime }$ \ \ \ \ \ \ b) \ $% -18$ \ \ \ \ \ \ \ \ \ 8.) \ a) \ $-0.24\dfrac{\unit{in}}{\unit{s}}$ \ \ \ \ b) \ $\dfrac{5}{48}\sqrt{2}\dfrac{\unit{in}}{\unit{s}}\approx \allowbreak 0.147\,314\dfrac{\unit{in}}{\unit{s}}$ \ \ \ \ \bigskip 9.) \ a) \ \ $y\left( 1.5\right) =24\unit{ft}$ \ and $y^{\prime }\left( 1.5\right) =-48\dfrac{\unit{ft}}{\unit{s}}$ \ \ \ \ b) $50\unit{ft}$\ \ \ \ c) \ $-400\dfrac{\unit{ft}}{\unit{s}}$\ \ \ \ {\LARGE \bigskip \bigskip \bigskip \bigskip } \begin{center} {\LARGE Sample Problems - Solutions\bigskip }\bigskip \end{center} \begin{enumerate} \item A city is of a circular shape. \ The area of the city is growing at a constant rate of $2\dfrac{\unit{mi}^{2}}{\unit{y}}$ \ (square miles per year). \ How fast is the radius growing when it is exactly $15\unit{mi}$? \ \ \ \ \newline Solution: The area of a circle with radius $r$ is $A=\pi r^{2}.$ \ Only this time, both $A$ and $r$ are functions of time: \ $A\left( t\right) =\pi r^{2}\left( t\right) .$ \ It is also given that $A^{\prime }\left( t\right) =2\dfrac{\unit{mi}^{2}}{\unit{y}}.$ \ We differentiate both sides of $% A\left( t\right) =\pi r^{2}\left( t\right) $ with respect to $t$.% \begin{eqnarray*} A\left( t\right) &=&\pi r^{2}\left( t\right) \\ A^{\prime }\left( t\right) &=&2\pi r\left( t\right) r^{\prime }\left( t\right) \end{eqnarray*}% Let $t_{1}$ be the time when $r\left( t_{1}\right) =15\unit{mi}.$ Then% \begin{eqnarray*} A^{\prime }\left( t_{1}\right) &=&2\pi r\left( t_{1}\right) r^{\prime }\left( t_{1}\right) \\ \dfrac{A^{\prime }\left( t_{1}\right) }{2\pi r\left( t_{1}\right) } &=&r^{\prime }\left( t_{1}\right) ~~~~~~\Longrightarrow ~~r^{\prime }\left( t_{1}\right) =\dfrac{A^{\prime }\left( t_{1}\right) }{2\pi r\left( t_{1}\right) }=\dfrac{2\dfrac{\unit{mi}^{2}}{\unit{y}}}{2\pi 15\unit{mi}}=% \dfrac{1}{15\pi }\dfrac{\unit{mi}}{\unit{y}} \end{eqnarray*} \item A sphere is growing in such a manner that its radius increases at $0.2% \dfrac{\unit{m}}{\unit{s}}$ (meters per second). \ How fast is its volume increasing when its radius is $4\unit{m}$ long?% \begin{eqnarray*} V\left( t\right) &=&\dfrac{4}{3}\pi r^{3}\left( t\right) \\ V^{\prime }\left( t\right) &=&\dfrac{4}{3}\pi \left( 3r^{2}\left( t\right) r^{\prime }\left( t\right) \right) =4\pi r^{2}\left( t\right) r^{\prime }\left( t\right) \end{eqnarray*}% Let $t_{1}$ be the time when $r\left( t_{1}\right) =4\unit{m}.$ Then\ \begin{equation*} V^{\prime }\left( t_{1}\right) =4\pi r^{2}\left( t_{1}\right) r^{\prime }\left( t_{1}\right) =4\pi \left( 4\unit{m}\right) ^{2}\left( 0.2\dfrac{% \unit{m}}{\unit{s}}\right) =12.8\pi \dfrac{\unit{m}^{3}}{\unit{s}}\cong \allowbreak 40.\,\allowbreak 212\,386\dfrac{\unit{m}^{3}}{\unit{s}} \end{equation*} \item A sphere is growing in such a manner that its volume increases at $0.2% \dfrac{\unit{m}^{3}}{\unit{s}}$ (cubic meters per second). \ How fast is its radius increasing when it is $7\unit{m}$ long?% \begin{eqnarray*} V\left( t\right) &=&\dfrac{4}{3}\pi r^{3}\left( t\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ \ \ \ \ \ }V=\dfrac{4}{3}\pi r^{3} \\ V^{\prime }\left( t\right) &=&\dfrac{4}{3}\pi \left( 3r^{2}\left( t\right) r^{\prime }\left( t\right) \right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }V^{\prime }=\dfrac{4}{3}\pi \left( 3r^{2}r^{\prime }\right) \\ V^{\prime }\left( t\right) &=&4\pi r^{2}\left( t\right) r^{\prime }\left( t\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }% V^{\prime }=4\pi r^{2}r^{\prime } \\ r^{\prime }\left( t\right) &=&\dfrac{V^{\prime }\left( t\right) }{4\pi r^{2}\left( t\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }r^{\prime }=\dfrac{V^{\prime }}{4\pi r^{2}} \end{eqnarray*}% Let $t_{1}$ be the time when $r\left( t_{1}\right) =7\unit{m}.$ Then% \begin{equation*} r^{\prime }\left( t_{1}\right) =\dfrac{V^{\prime }\left( t_{1}\right) }{4\pi r^{2}\left( t_{1}\right) }=\dfrac{0.2\dfrac{\unit{m}^{3}}{\unit{s}}}{4\pi \left( 7\unit{m}\right) ^{2}}=0.00032481\dfrac{\unit{m}}{\unit{s}} \end{equation*} \item The area of a rectangle is kept fixed at $100$ square meters while the legths of the sides vary. \ Express the rate of change of the length of the vertical side in terms of the rate of change in the length of the other side when a) \ the horizontal side is $18$ meters long \ \ \ b) \ the rectangle is a square\newline Solution: \ Let $v$ and $h$ denote the length of the horizontal and vertical sides, respectively. \ Then clearly $vh=100$. \ We differentiate both sides of this with respect of time, and then solve for $v^{\prime }$ in terms of $% h^{\prime }$. \begin{eqnarray*} h\left( t\right) v\left( t\right) &=&100\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }vh=100 \\ h^{\prime }\left( t\right) v\left( t\right) +h\left( t\right) v^{\prime }\left( t\right) &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }v^{\prime }h+vh^{\prime }=0 \\ v^{\prime }\left( t\right) &=&-\dfrac{h^{\prime }\left( t\right) v\left( t\right) }{h\left( t\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }v^{\prime }=-\dfrac{h^{\prime }v}{h} \end{eqnarray*}% a) \ When $h=18$, then $v=\dfrac{100}{18}=\dfrac{50}{9}$ and so% \begin{equation*} v^{\prime }\left( t\right) =-\dfrac{h^{\prime }\left( t\right) v\left( t\right) }{h\left( t\right) }=-\dfrac{h^{\prime }\left( t\right) \left( \dfrac{50}{9}\right) }{18}=-\dfrac{25}{81}h^{\prime }\left( t\right) \end{equation*}% b) \ When the rectangle is a square, then $v=h=10$. \begin{equation*} v^{\prime }\left( t\right) =-\dfrac{h^{\prime }\left( t\right) v\left( t\right) }{h\left( t\right) }=-\dfrac{h^{\prime }\left( t\right) \left( 10\right) }{10}=-h^{\prime }\left( t\right) \end{equation*} \item A ladder $20\unit{ft}$ long leans a gainst a vertical building. If the top of the ladder slides down at a rate of $\sqrt{3}\dfrac{\unit{ft}}{\unit{s% }}$, how fast is the bottom of the ladder sliding away from the building when the top of the ladder is $10\unit{ft}$ above the ground? Solution: \ Let $x\left( t\right) $ and \ $y\left( t\right) $denote the horizontal and vertical position of the endpoints of the ladder. \ With this notation , \ $y^{\prime }\left( t\right) =-\sqrt{3}\dfrac{\unit{ft}}{\unit{s}% }.$ \ By the Pythagorean theorem, $x\left( t\right) ^{2}+y\left( t\right) ^{2}=400$. \ We differentiate both sides and solve for $x^{\prime }\left( t\right) .$ \begin{eqnarray*} x\left( t\right) ^{2}+y\left( t\right) ^{2} &=&400\text{ \ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x^{2}+y^{2}=400 \\ 2x\left( t\right) x^{\prime }\left( t\right) +2y\left( t\right) y^{\prime }\left( t\right) &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }2xx^{\prime }+2yy^{\prime }=0 \\ x^{\prime }\left( t\right) &=&-\dfrac{y\left( t\right) y^{\prime }\left( t\right) }{x\left( t\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x^{\prime }=-\dfrac{yy^{\prime }}{x} \end{eqnarray*}% Let $t_{1}$ denote the time when $x\left( t_{1}\right) =10\unit{ft}.$% \begin{equation*} x^{\prime }\left( t_{1}\right) =-\dfrac{y\left( t_{1}\right) y^{\prime }\left( t_{1}\right) }{x\left( t_{1}\right) }=-\dfrac{\left( 10\unit{ft}% \right) \left( -\sqrt{3}\dfrac{\unit{ft}}{\unit{s}}\right) }{\sqrt{\left( 20% \unit{ft}\right) ^{2}-\left( 10\unit{ft}\right) ^{2}}}=-\dfrac{-10\sqrt{3}% \dfrac{\unit{ft}^{2}}{\unit{s}}}{\sqrt{300\unit{ft}^{2}}}=1\dfrac{\unit{ft}}{% \unit{s}} \end{equation*} \item A tank, shaped like a cone shown on the picture below, is being filled up with water. \ The top of the tank is a circle with radius $5\unit{ft}$, its height is $15\unit{ft}$. \ Water is added to the tank at the rate of $% V^{\prime }\left( t\right) =2\pi \dfrac{\unit{ft}^{3}}{\unit{min}}$. How fast is the water level rising when the water level is $6$ $\unit{ft}$ high? \ (The volume of a cone with height $h$ and base radius $r$ is $V=\dfrac{\pi r^{2}h}{3}$.) \ \ \ \ \FRAME{dtbpF}{1.9043in}{2.1854in}{0pt}{}{}{insert21.bmp% }{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.9043in;height 2.1854in;depth 0pt;original-width 1.8663in;original-height 2.1465in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert21.bmp';file-properties "XNPEU";}}Solution: \ Let $h\left( t\right) $ and $r\left( t\right) $ denote the height and radius of the surface. \ There is a simple connection between these two:\FRAME{dtbpF}{1.1338in}{1.9043in}{0pt}{}{}{insert2.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.1338in;height 1.9043in;depth 0pt;original-width 1.1in;original-height 1.8663in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert2.bmp';file-properties "XNPEU";}}The two triangles shown on the picture above are similar. \ $% \dfrac{r}{5}=\dfrac{h}{15}$ \ \ and so \ $r=\dfrac{h}{3}$. \begin{eqnarray*} V\left( t\right) &=&\dfrac{1}{3}\pi r^{2}\left( t\right) h\left( t\right) =% \dfrac{1}{3}\pi \left( \dfrac{h\left( t\right) }{3}\right) ^{2}h\left( t\right) =\dfrac{1}{27}\pi h^{3}\left( t\right) \\ V^{\prime }\left( t\right) &=&\dfrac{1}{27}\pi \left( 3h^{2}\left( t\right) h^{\prime }\left( t\right) \right) =\dfrac{1}{9}\pi h^{2}\left( t\right) h^{\prime }\left( t\right) \\ V^{\prime }\left( t\right) &=&\dfrac{1}{9}\pi h^{2}\left( t\right) h^{\prime }\left( t\right) \\ \dfrac{9V^{\prime }\left( t\right) }{\pi h^{2}\left( t\right) } &=&h^{\prime }\left( t\right) \end{eqnarray*}% Let $t_{1}$ denote the time when $h\left( t_{1}\right) =6\unit{ft}.$% \begin{equation*} h^{\prime }\left( t_{1}\right) =\dfrac{9V^{\prime }\left( t_{1}\right) }{\pi h^{2}\left( t_{1}\right) }=\dfrac{9\left( 2\pi \dfrac{\unit{ft}^{3}}{\unit{% min}}\right) }{\pi \left( 6\unit{ft}\right) ^{2}}=\dfrac{1}{2}\dfrac{\unit{ft% }}{\unit{min}} \end{equation*} \item Two quantities $p$ and $q$ depending on $t$ are subject to the relation $\dfrac{1}{p}+\dfrac{1}{q}=1.$ a) \ Express $p^{\prime }$ in terms of $q^{\prime }$. \ \begin{equation*} \dfrac{1}{p}+\dfrac{1}{q}=0~~~\Longrightarrow ~~~-\dfrac{p^{\prime }}{p^{2}}-% \dfrac{q^{\prime }}{q^{2}}=0~~~\Longrightarrow ~~~p^{\prime }=-\dfrac{p^{2}}{% q^{2}}q^{\prime } \end{equation*}% b) At a certain moment, $p\left( t_{0}\right) =\dfrac{4}{3}$ and $p^{\prime }\left( t_{0}\right) =2.$ \ Find $q\left( t_{0}\right) $ and $q^{\prime }\left( t_{0}\right) .$% \begin{equation*} \dfrac{1}{p\left( t_{0}\right) }+\dfrac{1}{q\left( t_{0}\right) }% =1~~~\Longrightarrow ~~~\dfrac{1}{\dfrac{4}{3}}+\dfrac{1}{q\left( t_{0}\right) }=1~~~\Longrightarrow ~~~q\left( t_{0}\right) =4 \end{equation*}% \begin{equation*} q^{\prime }=-\dfrac{q^{2}}{p^{2}}p^{\prime }=-\dfrac{4^{2}}{\left( \dfrac{4}{% 3}\right) ^{2}}\left( 2\right) =-18 \end{equation*} \item The base radius and height of a cylinder are constantly changing but the volume of the cylinder is kept at a constant $600\pi $ $\unit{in}^{3}$. \ a) \ At a time $t_{1}$ the base radius is $r\left( t_{1}\right) =10\unit{in}$ and its rate of change is $r^{\prime }\left( t_{1}\right) =0.2\dfrac{\unit{in% }}{\unit{s}}.$ \ Compute the rate of change of the height of the cylinder $% h\left( t\right) $ at time $t_{1}$. Solution: \ \ $V=\pi r^{2}h$ \ \ \ \ \ \ \ \ \ $h=\dfrac{V}{\pi r^{2}}$% \begin{eqnarray*} V &=&\pi r^{2}h \\ 0 &=&\pi \left( 2rr^{\prime }h+r^{2}h^{\prime }\right) \\ 0 &=&2rr^{\prime }h+r^{2}h^{\prime } \\ \dfrac{-2rr^{\prime }h}{r^{2}} &=&h^{\prime } \\ h^{\prime } &=&\dfrac{-2r^{\prime }h}{r}=\dfrac{-2r^{\prime }\left( \dfrac{V% }{\pi r^{2}}\right) }{r}=\dfrac{-2r^{\prime }V}{\pi r^{3}}=\dfrac{-2\left( 0.2\dfrac{\unit{in}}{\unit{s}}\right) \left( 600\pi \unit{in}^{3}\right) }{% \pi \left( 10\unit{in}\right) ^{3}}=-0.24\dfrac{\unit{in}}{\unit{s}} \end{eqnarray*}% b) \ At a time $t_{2}$ the height is $h\left( t_{2}\right) =12\unit{in}$ and its rate of change is $r^{\prime }\left( t_{2}\right) =-0.5\dfrac{\unit{in}}{% \unit{s}}.$ \ Compute the rate of change of the radius of the cylinder $% r\left( t\right) $ at time $t_{2}$. Solution: \ $V=\pi r^{2}h$ \ \ \ \ \ \ \ \ \ $r=\sqrt{\dfrac{V}{\pi h}}$% \begin{eqnarray*} V &=&\pi r^{2}h \\ 0 &=&\pi \left( 2rr^{\prime }h+r^{2}h^{\prime }\right) \\ 0 &=&2rr^{\prime }h+r^{2}h^{\prime } \\ r^{\prime } &=&\dfrac{-r^{2}h^{\prime }}{2rh}=\dfrac{-rh^{\prime }}{2h}=% \dfrac{-\left( \sqrt{\dfrac{V}{\pi h}}\right) h^{\prime }}{2h}=\dfrac{% -\left( \sqrt{\dfrac{600\pi \unit{in}^{3}}{\pi \left( 12\unit{in}\right) }}% \right) \left( -0.5\dfrac{\unit{in}}{\unit{s}}\right) }{2\left( 12\unit{in}% \right) }=\dfrac{-\left( \sqrt{50\unit{in}^{2}}\right) \left( -0.5\dfrac{% \unit{in}}{\unit{s}}\right) }{24\unit{in}} \\ &=&\dfrac{\left( 5\sqrt{2}\unit{in}\right) \left( 0.5\dfrac{\unit{in}}{\unit{% s}}\right) }{24\unit{in}}=\dfrac{5}{48}\sqrt{2}\dfrac{\unit{in}}{\unit{s}}% \approx \allowbreak 0.147\,314\dfrac{\unit{in}}{\unit{s}} \end{eqnarray*} \item An object, dropped from a height of $h$ has a location of $y\left( t\right) =-16t^{2}+h$ feet after $t$ seconds. \ We dropped a small object from a height of $60$ feet.\FRAME{dtbpF}{2.0245in}{2.1075in}{0pt}{}{}{% related1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.0245in;height 2.1075in;depth 0pt;original-width 2.5728in;original-height 2.6801in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../Math207/math207_fa13/quizzes/sample_quiz14/related1.bmp';file-properties "XNPEU";}% } a) \ Where is the object and what is its velocity after $1.5$ seconds? Solution: we substitute $t=1.5$ into the formula $y\left( t\right) =-16t^{2}+60$ and get \ $y\left( 1.5\unit{s}\right) =24\unit{ft}$ For the velocity, we differentiate $y\left( t\right) $ with resepct to $t$ and evaluate the derivative at $t=1.5\unit{s}$. \ \ \begin{equation*} y^{\prime }\left( t\right) =-32t\text{ \ \ \ \ \ \ \ }y^{\prime }\left( 1.5\right) =-32\left( 1.5\right) =-48 \end{equation*}% So $y^{\prime }\left( 1.5\unit{s}\right) =-48\dfrac{\unit{ft}}{\unit{s}}$% \pagebreak b) \ Suppose there is a $30$ feet tall street light $10$ feet away from the point where the object will land. \ How far is the shadow of the object from the base of the street light at $t=1.5$? Solution: \ Let us denote by $y\left( t\right) $ the vertical position of the object and by $x\left( t\right) $ the horizontal position of its shadow. \ \FRAME{dtbpF}{5.521in}{2.2883in}{0pt}{}{}{related2.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 5.521in;height 2.2883in;depth 0pt;original-width 6.5069in;original-height 2.6801in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../Math207/math207_fa13/quizzes/sample_quiz14/related2.bmp';file-properties "XNPEU";}% }By similar triangles, we have that% \begin{eqnarray*} \dfrac{y\left( t\right) }{x\left( t\right) -10} &=&\dfrac{30}{x\left( t\right) }\text{ \ clear denominators} \\ x\left( t\right) y\left( t\right) &=&30\left( x\left( t\right) -10\right) \end{eqnarray*}% We use this equation first to solve for $x\left( t\right) $% \begin{eqnarray*} x\left( t\right) y\left( t\right) &=&30x\left( t\right) -300 \\ 300 &=&30x\left( t\right) -x\left( t\right) y\left( t\right) \\ 300 &=&x\left( t\right) \left( 30-y\left( t\right) \right) \\ \dfrac{300}{30-y\left( t\right) } &=&x\left( t\right) \end{eqnarray*}% In particular, $x\left( 1.5\right) =\dfrac{300}{30-y\left( 1.5\right) }=% \dfrac{300}{30-24}=50$. \ c) \ How fast is the obejct's shadow moving at $t=1.5$? Solution: \ To compute $x^{\prime }\left( 1.5\right) $, we differentiate $% x\left( t\right) y\left( t\right) =30\left( x\left( t\right) -10\right) $% \begin{eqnarray*} y\left( t\right) x\left( t\right) &=&30\left( x\left( t\right) -10\right) \\ x^{\prime }\left( t\right) y\left( t\right) +x\left( t\right) y^{\prime }\left( t\right) &=&30x^{\prime }\left( t\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solve for }x^{\prime }\left( t\right) \\ x\left( t\right) y^{\prime }\left( t\right) &=&30x^{\prime }\left( t\right) -x^{\prime }\left( t\right) y\left( t\right) \\ x\left( t\right) y^{\prime }\left( t\right) &=&x^{\prime }\left( t\right) \left( 30-y\left( t\right) \right) \\ \dfrac{x\left( t\right) y^{\prime }\left( t\right) }{30-y\left( t\right) } &=&x^{\prime }\left( t\right) \end{eqnarray*}% So $x^{\prime }\left( 1.5\right) $ is \ \begin{equation*} x^{\prime }\left( 1.5\right) =\dfrac{\left( 50\unit{ft}\right) \left( -48% \dfrac{\unit{ft}}{\unit{s}}\right) }{30\unit{ft}-24\unit{ft}}=-400\dfrac{% \unit{ft}}{\unit{s}} \end{equation*}% The negative sign indicates that the shadow is traveling toward the left. \item Two boats leave a port at the same time, one traveling west at 20 mi/hr and the other traveling southwest at 15 mi/hr. At what rate is the distance between them changing 30 min after they leave the port? Solution: \ Let $\left( x_{1},y_{1}\right) $ denote the location of the car moving to the west. \ Let $\left( x_{2},y_{2}\right) $ denote the location of the car moving southwest. \ \ \ \ \ \ \ \FRAME{itbpF}{2.2598in}{1.8645in}{0in}{}{}{pic10.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.2598in;height 1.8645in;depth 0in;original-width 2.22in;original-height 1.8265in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '//psf/Home/Desktop/pic10.bmp';file-properties "XNPEU";}} \ \ \ \ \ \ \FRAME{% itbpF}{2.1387in}{1.8645in}{0in}{}{}{pic10b.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.1387in;height 1.8645in;depth 0in;original-width 2.0998in;original-height 1.8265in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '//psf/Home/Desktop/pic10b.bmp';file-properties "XNPEU";}} Then $x_{1}\left( t\right) =-20t$, $\ y_{1}\left( t\right) =0$ and $\ x_{2}\left( t\right) =-\dfrac{15}{\sqrt{2}}t$ and $y_{2}\left( t\right) =-% \dfrac{15}{\sqrt{2}}t$. \ \ And then $\dfrac{dx_{1}}{dt}=-20$, $\dfrac{dy_{1}% }{dt}=0$, \ $\dfrac{dx_{2}}{dt}=-\dfrac{15}{\sqrt{2}}=-\dfrac{15\sqrt{2}}{2}$ \ and \ $\dfrac{dy_{2}}{dt}=-\dfrac{15}{\sqrt{2}}=-\dfrac{15\sqrt{2}}{2}$. First we will express the distance between the cars. \ For that, we will use the Pythagorean theorem.% \begin{equation*} s^{2}=\left( x_{2}-x_{1}\right) ^{2}+\left( -y_{2}\right) ^{2} \end{equation*} \end{enumerate} We differentiate both sides with respect to $t$ and solve for $\dfrac{ds}{dt} $% \begin{eqnarray*} 2s\dfrac{ds}{dt} &=&2\left( x_{2}-x_{1}\right) \left( \dfrac{dx_{2}}{dt}-% \dfrac{dx_{1}}{dt}\right) +2\left( -y_{2}\right) \left( -\dfrac{dy_{2}}{dt}% \right) \\ \dfrac{ds}{dt} &=&\dfrac{2\left( x_{2}-x_{1}\right) \left( \dfrac{dx_{2}}{dt}% -\dfrac{dx_{1}}{dt}\right) +2y_{2}\left( \dfrac{dy_{2}}{dt}\right) }{2s} \\ \dfrac{ds}{dt} &=&\dfrac{\left( x_{2}-x_{1}\right) \left( \dfrac{dx_{2}}{dt}-% \dfrac{dx_{1}}{dt}\right) +y_{2}\left( \dfrac{dy_{2}}{dt}\right) }{\sqrt{% \left( x_{2}-x_{1}\right) ^{2}+\left( -y_{2}\right) ^{2}}} \end{eqnarray*}% After $30$ minutes, $x_{1}=-10$ \ \ $y_{1}=0$ \ \ and \ $x_{2}=-\dfrac{15}{% \sqrt{2}}\left( \dfrac{1}{2}\right) =-\dfrac{15}{4}\sqrt{2}$ and $% y_{2}\left( t\right) =-\dfrac{15}{\sqrt{2}}\left( \dfrac{1}{2}\right) =-% \dfrac{15\sqrt{2}}{4}$. \ At that time, \ \begin{eqnarray*} \dfrac{ds}{dt} &=&\dfrac{\left( x_{2}-x_{1}\right) \left( \dfrac{dx_{2}}{dt}-% \dfrac{dx_{1}}{dt}\right) +y_{2}\left( \dfrac{dy_{2}}{dt}\right) }{\sqrt{% \left( x_{2}-x_{1}\right) ^{2}+\left( -y_{2}\right) ^{2}}} \\ \dfrac{ds}{dt} &=&\dfrac{\left( -\dfrac{15}{4}\sqrt{2}-\left( -10\right) \right) \left( -\dfrac{15\sqrt{2}}{2}-\left( -20\right) \right) +\left( -% \dfrac{15\sqrt{2}}{4}\right) \left( -\dfrac{15}{\sqrt{2}}\right) }{\sqrt{% \left( -\dfrac{15}{4}\sqrt{2}-\left( -10\right) \right) ^{2}+\left( \dfrac{15% \sqrt{2}}{4}\right) ^{2}}} \\ &=&\dfrac{\left( -3.75\sqrt{2}+10\right) \left( -7.5\sqrt{2}+20\right) +% \dfrac{225}{4}}{\sqrt{\left( -3.75\sqrt{2}+10\right) ^{2}+2\left( \dfrac{15}{% 4}\right) ^{2}}}\approx 14.\,\allowbreak 168\,131 \end{eqnarray*}% \bigskip \bigskip Or: Let $a$ be the disance from the origin of one car and $b$ the distance of the origin from the other car. \ Then \begin{equation*} s^{2}=a^{2}+b^{2}-2ab\cos 45^{\circ } \end{equation*}% we differentiate both sides:% \begin{eqnarray*} 2ss^{\prime } &=&2aa^{\prime }+2bb^{\prime }-2\left( \dfrac{\sqrt{2}}{2}% \right) \left( a^{\prime }b+ab^{\prime }\right) \\ s^{\prime } &=&\dfrac{aa^{\prime }+bb^{\prime }-\left( \dfrac{\sqrt{2}}{2}% \right) \left( a^{\prime }b+ab^{\prime }\right) }{s}= \\ &=&\dfrac{10\left( 20\right) +7.5\left( 15\right) -\left( \dfrac{\sqrt{2}}{2}% \right) \left( 20\left( 7.5\right) +10\left( 15\right) \right) }{\sqrt{% 10^{2}+7.5^{2}-2\left( 10\right) \left( 7.5\right) \cos 45^{\circ }}}\approx 14.\,\allowbreak 168\,131 \end{eqnarray*} \bigskip \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}