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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 126 C - Spring 2009 \hfil \pp
\hfil Mid-Term Exam Number Two\hfil  \pp
\hfil May 14, 2009 \hfil \pp
\hfil Solutions \hfil \pp
\normalsize

\begin{enumerate}

\item {\textit{The curve $y=x^x$ has one local extremum.  Find the curvature at that point.}}

We have
\[
y = x^x = e^{x \ln x}
\]
so
\[
y' = e^{x \ln x}( \ln x +1)
\]
and
\[
y'' = e^{x \ln x} ((\ln x + 1)^2 + \frac{1}{x}).
\]

At a local extremum, $y'=0$.  Solving $y'=0$, we find
\[
x = \frac{1}{e}
\]
At this point we have curvature
\[
\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} = |y''| = 
\left( \frac{1}{e} \right)^{\frac{1}{e}} \left( \frac{1}{\frac{1}{e}} \right) = 
e\left( \frac{1}{e} \right)^{1/e} =
e^{1-\frac{1}{e}}.
\]


\item {\textit{An object moves so that its position at time $t$ is given by}}
\[
\vec{r}(t) = \langle t^2, t^3-4t, 0 \rangle.
\]
{\textit{
A portion of its path is shown below.
}}
\epsfig{file=fig1.ps, 
 width=8cm,
 angle=270 }

{\textit{
Find all times when the object's velocity vector is orthogonal to its acceleration vector. 
}}

We find
\[
\vec{r}'(t) = \langle 2t, 3t^2-4,0 \rangle
\]
and
\[
\vec{r}''(t) = \langle 2, 6t, 0 \rangle
\]
so that
\[
\vec{r}' \vec{r}'' = 4t + 6t(3t^2-4) = t (-20 +18t^2) = 0
\]
when $t=0$ or 
\[
t = \pm \frac{\sqrt{10}}{3}.
\]

\item {\textit{Let }}
\[
z = xe^y + y \sin x + \frac{x}{y}.
\]

\begin{enumerate}

\item {\textit{Find}} $\dsps \frac{\partial z}{\partial x}$.

\[
\frac{\partial z}{\partial x} = e^y + y \cos x + \frac{1}{y}.
\]


\item {\textit{Find}} $\dsps \frac{\partial z}{\partial y}$.

\[
\frac{\partial z}{\partial y} = xe^y + \sin x - \frac{x}{y^2}.
\]

\end{enumerate}



\item {\textit{You wish to build a four-sided box like the one shown in the figure,
with three rectangular sides perpendicular to a rectangular base.}}

{\textit{You want the box to have a volume of 100 cubic centimeters.  }}

{\textit{If all sides are to be made from the same thin material, what dimensions
will minimize the amount of material used?}}

{\textit{Be sure to justify your answer using the Hessian (i.e., the second derivatives test).
}}

\epsfig{file=box01.eps, 
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If we label the height of the box $z$, and label the sides of the base $y$ and $x$,
then the volume of the box is 
\[
xyz = 100
\]
and the amount of material needed is 
\[
M = xy + 2yz + xz
\]
Since $xyz=100$, we have
\[
z = \frac{100}{xy}
\]
so we may write
\[
M = M(x,y) = xy + \frac{200}{x} + \frac{100}{y}.
\]
Taking partial derivatives, we have
\[
\frac{\partial M}{\partial x} = y - \frac{200}{x^2}
\]
and 
\[
\frac{\partial M}{\partial y} = x - \frac{100}{y^2}.
\]
Setting these equal to zero, we can conclude first that
\[
y = \frac{200}{x^2}
\]
and then that
\[
x^3 = 400
\]
or $x \approx 7.36806...$, from which we find
\[
y=z= 50^{1/3} \approx 3.684... .
\]
Thus, we have one critical point.  Taking second partials we find that
\[
\frac{\partial^2 M}{\partial x^2} = \frac{400}{x^3}
\]
and
\[
\frac{\partial^2 M}{\partial y^2} = \frac{200}{y^3}
\]
and so
\[
D = \frac{400 \cdot 200}{x^3y^3} - 1 = 3 > 0
\]
at the critical point, so we have a local min, and since it is the only
critical point, we may conclude that it yields the global min.


\item {\textit{Find the volume of the solid under the surface $z=xy^2$ and above
the triangle with vertices $(0,0)$, $(0,5)$ and $(2,3)$.}}

The required volume can be expressed as the following double integral
\[
V = \int_0^2 \int_{\frac{3}{2}x}^{5-x} xy^2 \, dy \, dx
\]
When evaluated, we find 
\[
V = \frac{82}{3}.
\]

\end{enumerate}

\end{document}


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