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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions for Midterm Exam 3 -- December 2, 2005}
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\noindent
I.  $f(x) = x e^{-x^2}$.
\item{A)}  By the product and chain rules:
$$f'(x) = e^{-x^2} + xe^{-x^2} (-2x) = (1 - 2x^2) e^{-x^2}.$$
\item{B)}  The critical points of $f$ occur when $1 - 2x^2 = 0$, so
$x = {\pm 1\over \sqrt{2}} = {\pm \sqrt{2} \over 2} \doteq \pm .707$.
Of these, only $-\sqrt{2}/2 \doteq -.707$ is in our interval.  
Evaluating $f$ at the critical
point and at the endpoints:
$$\eqalign{f(-2) &= -2 e^{-4} \doteq -.0366\cr
           f(-\sqrt{2}/2) &= -\sqrt{2}/2\cdot e^{-1/2} \doteq -.4289\cr
           f(0) = 0\cr}$$
So the maximum value is $f(0) = 0$ and the minimum
is $f(-\sqrt{2}/2) = -\sqrt{2}/2\cdot e^{-1/2} \doteq -.4289$.
\item{C)}  Yes it does.  From the form of $f'$ we can
see that $f'$ changes from negative to positive at
$-\sqrt{2}/2$, so by the First Derivative Test, this
is a local minimum.
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\noindent
II.
\item{A)} By the quotient rule:
$$\eqalign{f'(x) &= {(\cos(x) + \sin(x))\cos(x) - \sin(x)(-\sin(x) + \cos(x))\over 
(\sin(x) + \cos(x))^2}\cr
&= {\cos^2(x) + \sin(x)\cos(x) + \sin^2(x) - \sin(x)\cos(x)\over
(\sin(x) + \cos(x))^2}\cr
&={1\over (\sin(x) + \cos(x))^2}\cr}$$
\item{B)}  By the rule for natural log plus the chain rule:
$$g'(x) = {1\over 3x + {1\over x}} (3 - {1\over x^2}) = {3x^2 - 1\over 3x^3 + x}$$
\item{C)} By the rule for arcsin and the chain rule:
$$h'(x) = {1\over \sqrt{1 - (4x)^2}}\cdot 4 = {4\over \sqrt{1 - 16x^2}}.$$
\item{D)}  By implicit differentiation (using product rule):
$$ y^2 + 2xy{dy\over dx} + 2xy + x^2 {dy\over dx} = 0$$
Solving for ${dy\over dx}$:
$$ (2xy + x^2){dy\over dx} = -2xy - y^2,$$
so 
$${dy\over dx} = {-2xy - y^2\over 2xy + x^2}.$$
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\noindent
III.  Call the dimensions of the rectangle $x,y$ (in yards).  Say $x$
is the side parallel to the existing fence.  Then we have
$xy = 500$ and we want to minimize $L = x + 2y = x + {1000\over x}$.
Then $L' = 1 - {1000\over x^2} = 0$ when $x = \sqrt{1000} = 10\sqrt{10}$.
This is a minimum, since it is the only critical point with positive $x$,
and $L'' = {1000\over x^3}$, so $L'' > 0$ for all $x > 0$.  The
dimensions of the lot with minimum fence are $x = 10\sqrt{10} \doteq 31.6$ 
and $y = 5\sqrt{10}\doteq 15.8$ (from $xy = 5000$).
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\noindent
IV.  For the sphere, $V = {4\pi\over 3} r^3$.  So taking time
derivatives:
$${dV\over dt} = 4\pi r^2 {dr\over dt}$$
We are given ${dr\over dt} = 3$, so when $r = 6$,
$${dV\over dt} = 4\pi\cdot (6)^2 \cdot 3 = 432\pi$$
(units are mm${}^3$/min). 

\bye