\magnification=\magstep1
\noindent
I.
\item{A)}  Cross section diameter is $y = x/2$ (similar triangles), 
so cross section area is
$$A(x) = (1/2)\pi (x/4)^2 = \pi x^2/32.$$
\item{B)}  The volume is 
$$\int_0^2 \pi x^2/32\, dx = \pi/12$$
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\noindent
II.
\item{A)}  Subdividing the wire into $n$ equal pieces of length $\Delta x = 3/n$,
the mass of the segment of the wire between $x_{i-1}$ and $x_i$
is approximately 
$$({\rm density})\cdot({\rm length}) \doteq (4 - x_i)\sqrt{1 + (2x)^2}\Delta x$$
\item{B)}
$$\int_{-1}^2 (4-x)\sqrt{1+4x^2}\, dx.$$
\item{C)} 
$$ {32\over 12}\sqrt{17} - \ln(\sqrt{17}-4) + {29\over 12}\sqrt{5} - 
\ln(-2 + \sqrt{5}) \doteq 19.6.$$
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\noindent
III.
\item{A)}  Letting $P$ be the population as a function of time, 
${dP\over dt} = -k \sqrt{P}$.
\item{B)}  All slopes on a horizontal line are equal.  For 
$P = 0$, slope is 0, $P = 1$, slope is $-1$, $P = 2$, slope is $-\sqrt{2}$, etc.
\item{C)}  Solving by separation of variables, 
$$P(t) = (-{k\over 2}t + c)^2$$
for some constant $c$.  Letting $t = 0$, using $P(0) = 900$, $c = 30$.
Letting $t = 6$, using $P(6) = 441$ solving for $k$, $k = 3$.  So
$P$ reaches $0$ when $t = 20$.
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\noindent
IV.  
\item{A)}  
$$\eqalign{ {dx(t)\over dt} &= 1 + 3\cos(t)/2 = y(t)\cr
            {dy(t)\over dt} &= -3\sin(t)/2 = -x(t) + t\cr}$$
\item{B)}  $t$ solving $\cos(t) = -2/3$, so $t \doteq 2.3$ radians.
\item{C)}  Three times; once before $t = \pi$, once before and once after.
(This is a prolate cycloic)
\bye