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\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}
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\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}
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\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