\hfill \thepage} %} \input{tcilatex} \begin{document} \subsection*{Ma116 \hspace{2.00in} Exam I -- \textsl{Solutions}\hfill 2/8/00} \subsection*{{\protect\normalsize Name: \protect\underline{\hspace{2.5in}} \hfill ID: \protect\underline{\hspace{2.0in}}}} \subsection*{{\protect\normalsize E-Mail: \protect\underline{\hspace{2.5in}} \hfill Lecturer: \protect\underline{\hspace{2.0in}}}} \hfill {\small \textit{I pledge my honor that I have abided by the Stevens Honor System.}\hspace{2pt}\hfill\underline{\hspace{2.0in}}} \hfill {\small \textbf{SHOW ALL WORK! You may not use a calculator on this exam.}} \hfill \begin{description} \item[1. ] Consider a particle moving along a circle with position vector $% \,\vec{r}(t)=(R\sin (\pi t/2),R\cos (\pi t/2))$\/. \end{description} \begin{description} \item[a) \lbrack 7 pts.\rbrack ] Calculate the unit tangent vector for all $% t$ and using the figure provided, sketch the tangent vectors at $t=0$ and $% t=1$. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \begin{array}[b]{lcl} \vec{r}\,^{\prime }(t) & = & \left( \dfrac{\pi R}{2}\cos (\pi t/2),\,-\dfrac{% \pi R}{2}\sin (\pi t/2)\right) \\ |\,\vec{r}\,^{\prime }(t)\,| & = & \dfrac{\pi R}{2} \\ \vec{T}(t) & = & (\cos (\pi t/2),-\sin (\pi t/2)) \\ & & \\ & & \\ & & \end{array} \hspace{1in}\FRAME{itbpF}{1.62in}{1.62in}{0in}{}{}{circle.gif}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.62in;height 1.62in;depth 0in;original-width 203.25pt;original-height 203.25pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'q1/circle.gif';file-properties "XNPEU";}} \] \end{description} \begin{description} \item[b) \lbrack 7 pts.\rbrack ] Determine the angle between the velocity vector and the acceleration vector. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \vec{r}\,^{\prime \prime }(t)\;=\;\left( -\dfrac{\pi ^{2}R}{4}\sin (\pi t/2),\,-\dfrac{\pi ^{2}R}{4}\cos (\pi t/2)\right) \] \[ \vec{r}\,^{\prime }(t)\cdot \vec{r}\,^{\prime \prime }(t)\;=\;\dfrac{% R^{2}\pi ^{3}}{8}\left( -\sin (\pi t/2)\cos (\pi t/2)+\cos (\pi t/2)\sin (\pi t/2)\right) \;=\;0 \] \[ \Rightarrow \vec{r}\,^{\prime }(t)\perp \vec{r}\,^{\prime \prime }(t) \] \end{description} \hfill \begin{description} \item[c) \lbrack 6 pts.\rbrack ] Suppose that a particle's position vector in 3-space satisfies $|\,\vec{r}(t)\,|=c$ ($c$ is a constant). Show that $% \vec{r}(t)$ and $\vec{r}\,^{\prime }(t)$ are orthogonal to one another.\\[5pt% ] \underline{\textit{Hint}}\/: Recall that $|\,\vec{r}(t)\,|^{2}=\vec{r}% (t)\cdot \vec{r}(t)$. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \begin{array}{lcl} |\,\vec{r}\,|^{2} & = & \vec{r}(t)\cdot \vec{r}(t)=c^{2} \\ \dfrac{d}{dt} & \longrightarrow & 2\vec{r}(t)\cdot \vec{r}\,^{\prime }(t)=0\;\;\Longrightarrow \;\vec{r}(t)\perp \vec{r}\,^{\prime }(t) \end{array} \] \end{description} \newpage \begin{description} \item[2. \lbrack 20 pts.\rbrack ] Set up, \emph{but do not evaluate}\/, a definite integral which determines the area in the \emph{first quadrant} that lies between $0\leq x\leq 2$\/ and between the two curves, \[ y=\frac{3x^{2}}{4}-1\hspace{0.25in}\text{and}\hspace{0.25in}y=\sqrt{2x}\,. \] \underline{\textit{Note}}\/:\hspace{2pt} You must include a sketch of the region using the grid provided. \end{description} \hfill \hspace{3.40in} \FRAME{itbpF}{2.50in}{2.53in}{0in}{}{}{prob2_sol.gif} {% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 2.50in;height 2.53in;depth 0in;original-width 2.50in;original-height 2.53in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'q1/prob2_sol.gif';file-properties "XNPEU";}} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \begin{array}[b]{lll} \multicolumn{3}{l}{\text{\textsl{Using horizontal strips:}}} \\ A & = & \dint_{0}^{2}\left( \sqrt{4(y+1)/3}-y^{2}/2\right) \,dy\medskip \\ & = & \left. \left( \dfrac{4\left( y+1\right) ^{3/2}}{3\sqrt{3}}-\dfrac{y^{3}% }{6}\right) \right| _{0}^{2}\medskip \\ & = & \dfrac{8\sqrt{3}-4}{3\sqrt{3}} \\ \rule{0pt}{10pt} & & \\ \multicolumn{3}{l}{\text{\textsl{Using vertical strips:}}} \\ A & = & \dint_{0}^{2}\sqrt{2x}\,dx\,-\,\dint_{2/\sqrt{3}}^{2}\left( \dfrac{% 3x^{2}}{4}\,-\,1\right) \,dx\medskip \\ & = & \left. \dfrac{2\sqrt{2}x^{3/2}}{3}\right| _{0}^{2}-\left. \left( \dfrac{x^{3}}{4}-x\right) \right| _{2/\sqrt{3}}^{2}\medskip \\ & = & \dfrac{8\sqrt{3}-4}{3\sqrt{3}} \end{array} \] \end{description} \newpage \begin{description} \item[{3. [20 pts.]}] If the area between $y=\sqrt{x}$ and $y = x^2$ for $0 \leq x \leq 1$ is rotated around the $y$-axis, what is the volume of the resulting solid?\\[5pt] \underline{\textit{Note}}\/:\hspace{2pt} You must \emph{evaluate} the integral and include a sketch of the region to be rotated. \end{description} \hfill \hspace{3.40in} \FRAME{itbpF}{2.48in}{2.54in}{0in}{}{}{prob3_sol.gif} {% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 2.48in;height 2.54in;depth 0in;original-width 2.48in;original-height 2.54in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'q1/prob3_sol.gif';file-properties "XNPEU";}} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \begin{array}[b]{lcl} \multicolumn{3}{l}{\text\QTR{sl}{Using cross-sections:}} \\ V & = & \pi\dint_0^1 \left((\sqrt{y})^2 - (y^2)^2\right)\,dy \smallskip \\ & = & \pi\dint_0^1 \left( y - y^4\right)\,dy \smallskip \\ & = & \pi \left.\left( \dfrac{y^2}{2} - \dfrac{y^5}{5}\right) \right|_0^1\;=\; \pi\left(\dfrac{1}{2} - \dfrac{1}{5}\right) \;=\; \dfrac{% 3\pi}{10} \\ & & \\ \multicolumn{3}{l}{\text\QTR{sl}{Using cylindrical shells:}} \\ V & = & \dint_0^1 2\pi x\left(\sqrt{x} - x^2\right)\,dx \smallskip \\ & = & 2\pi\dint_0^1 \left(x^{3/2} - x^3\right)\,dx \smallskip \\ & = & 2\pi\left.\left(\dfrac{2x^{5/2}}{5} - \dfrac{x^4}{4}\right)\right|_0^1 \;=\; 2\pi\left(\dfrac{2}{5} - \dfrac{1}{4}\right) \;=\; \dfrac{3\pi}{10} \end{array} \] \end{description} \newpage \begin{description} \item[4. ] The figure below shows a portion of the curve described by the parametric equations, \[ \begin{array}[b]{lcl} x & = & \alpha\,t^{2} \hspace{0.65in} (\alpha > 0) \\ y & = & 3t-\dfrac{t^{3}}{3} \\ \rule{0in}{1.25in} & & \end{array} \hspace{0.5in} \FRAME{itbpF}{2.51in}{1.67in}{0in}{}{}{cubic.gif} {\special {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 2.51in;height 1.67in;depth 0in;original-width 2.51in;original-height 1.67in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'q1/cubic.gif';file-properties "XNPEU";}} \] \end{description} \begin{description} \item[{a) [12 pts.]}] Set up, \emph{but do not evaluate}\/, a definite integral with respect to the variable $t$ which expresses the exact arc length of the \emph{closed loop}. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \begin{array}[b]{lcl} \text{Setting }t=0 & \longrightarrow & (x,y) = (0,0) \\ \text{Setting }y=0 & \longrightarrow & 0 = \dfrac{t}{3}(9 - t^2) \\ & \Longrightarrow & t = \pm3 \\ \text{Closed loop:} & \longleftrightarrow & -3 \leq t \leq 3 \\ & & \end{array} \] \vspace{10pt} \[ \begin{array}{rcl} L & = & \dint_{-3}^{3} \left(\left(x\,^\prime(t)\right)^2 \,+\, \left(y\,^\prime(t)\right)^2\right)^{1/2}\,dt \medskip \\ & = & \dint_{-3}^{3} \left(\left(2\alpha t\right)^2 \,+\, \left(3 - t^2\right)^2\right)^{1/2}\,dt \medskip \\ & = & \dint_{-3}^{3}\left(t^4 +(4\alpha^2 - 6)t^2 + 9\right)^{1/2}\,dt \end{array} \] \end{description} \hfill \begin{description} \item[{b) [8 pts.]}] Determine the value of $\alpha$ which makes the curvature at the origin equal to 1. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ \vec{r}(t) = (\alpha t^2, 3t - t^3/3, 0) \hspace{.50in} \vec{r}\,^\prime(t) = (2\alpha t, 3 - t^2, 0) \hspace{.50in} \vec{r}\,^{\prime\prime}(t) = (2\alpha, -2t, 0) \] Evaluating at $t=0$, corresponding to $(x,y)=(0,0)$, \[ \vec{r}\,^\prime(0) = (0,3,0), \hspace{.50in} |\,\vec{r}\,^\prime(0)\,| = 3, \hspace{.50in} \vec{r}\,^{\prime\prime}(0) = (2\alpha, 0, 0). \] Calculating the curvature $\kappa$, \[ \kappa = \dfrac {\left|\,\vec{r}\,^\prime(0) \times \vec{r}% \,^{\prime\prime}(0)\,\right|} {\left|\,\vec{r}\,^\prime(0)\,\right|^3} = \dfrac{\left|\,(0,0,-6\alpha)\right|}{3^3}=\dfrac{2\alpha}{9} \Longrightarrow \alpha = 9/2\,. \] \end{description} \pagebreak \begin{description} \item[5. ] \hspace{1pt} \end{description} \begin{description} \item[{a) [7 pts.]}] Evaluate the improper integral $\,\dint_e^\infty \dfrac{1}{x(\ln x)^2}\,dx$\/. \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} Apply the substitution $u = \ln x$, \[ \dint_e^\infty \dfrac{1}{x(\ln x)^2}\,dx\, = \lim_{t\rightarrow\infty}\dint_1^t \dfrac{du}{u^2} = \lim_{t\rightarrow\infty}\left.\dfrac{-1}{u}\right|_1^t = 1 - \lim_{t\rightarrow\infty}\dfrac{1}{t} = 1 \] \end{description} \hfill \hfill \hfill \begin{description} \item[{b) [7 pts.]}] If the curve $\,y = 1/x$, for $1 \leq x < \infty$, is rotated around the $x$-axis, what is the resulting volume? \end{description} \begin{description} \item \textbf{\textsl{Solution}}\/:\hspace{5pt} \[ V = \int_1^\infty \pi\frac{1}{x^2}\,dx = \pi \hspace{.25in} \text{see result from part (a)} \] \end{description} \hfill \hfill \hfill \begin{description} \item[{c) [6 pts.]}] Given that the surface area of the solid in part (b) is bigger than $\int_1^\infty 2\pi y\,dx$\/, determine which is larger, the volume of the solid in (b) or the surface area of this solid? \end{description} \begin{description} \item \hspace{2pt} \textbf{\textsl{Solution}}\/:\hspace{5pt} Letting $A$ denote the surface area of the infinite hyperbolic cylinder, \[ A > \int_1^\infty 2\pi y\,dx = 2\pi\int_1^\infty\dfrac{dx}{x} = 2\pi\lim_{t\rightarrow\infty}\,\left.\ln x\right|_1^t = 2\pi\lim_{t\rightarrow\infty}\,\ln t \] The limit on the right-hand side \textit{diverges}. 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