\def\frac#1#2{{{#1}\over{#2}}}
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{\sl These are the problems of the 1997 Texas A\& M University
Freshman-Sophomore Contest. Your mission: to solve boldly where no solver has
solved before. 
Please record both your answers, and the logical basis for your conclusions, on
the data-storage sheets provided. Problems designated FS are for both freshmen
and sophomores. Problems denoted F are for freshmen only; those denoted S are
for sophomores only. All contestants compete in a single pool for prizes,
though the problems set contestants at different stages are in part different.
Calculators, pagers, etc. are not permitted. You have two hours.}

\item{1FS.} A conical pile of sand grows beneath a sand-dumping hose. The hose
drops a cubic foot of sand per second onto the pile. As the pile grows, sand
shifts and slides down the sides so that it maintains its shape, with a slope
of 45 degrees from horizontal. How many seconds must go by, from when the first
sand spilled onto the flat dirt, before the sandpile height is growing at the
rate of 1 inch per second?

If the angle is 45 degrees, then the cone has height equal to its radius. This
means that the volume formula $V=(1/3)\pi r^2h$ simplifies to $V=\pi h^3/3$
here. Now the time derivative of the volume is $\pi h^2 (dh/dt)$ on the one
hand, and on the other hand from the story, it's one. (So $V=t$). Also, at the
time required, $dh/dt=1/12$. So at the time which represents the answer, $\pi
h^2/12=1$ and so $h=\sqrt{12/\pi}$. Now solving $V=t=\pi h^3$ with this value
of $h$ gives an elapsed time of $4\sqrt{12/\pi}$ which is the answer. 

\item{2FS.} Let $f(x):=\int_0^x\sin^4t\, dt$. 
\itemitem{(a)} Find $f(2\pi)$. 
\itemitem{(b)} Sketch $f(x)$ over the interval $0\le x\le 2\pi$. 

Probably the best way to do this integral is with the trig identity
$\sin^2u=(1-\cos u)/2$. Then $f(x)=(1/4)\int_0^{2\pi}(1-2\cos(2t)+\cos^2t)\,
dt$. The first term evaluates to $\pi/2$, the second to zero, and with a second
application of the same identity to $\cos^2t=1-\sin^2t$ the last term evaluates
to $\pi/4$. Adding gives $3\pi/4$. 

If those identities don't come to mind, another approach is to integrate by
parts. Taking $U=\sin^3t$ and $V=\sin t\, dt$ gives $-\sin^3x\cos
x+3\int_0^x\sin^2t\cos^2t\, dt$. Now replacing $\cos^2t$ with $1-\sin^2t$ in
the integral gives a formula with $f(x)$ on one side, and $-3f(x)$ together
with other, easily evaluated terms, on the other side. Collect all $f(x)$ terms
to get an identity for $4f(x)$. You get the same answer either way.

For the graph, first graph $\sin^4 t$. Now plot a curve whose slope is that
other graph. Since $\sin^4t\ge 0$ for all $t$, the graph of $f(x)$ will never
have a negative slope. It will rise most rapidly when $t=\pi/2$ or $3\pi/2$,
and it will be essentially flat when $t$ is near $0$, $\pi$, and $2\pi$. 
It's no trouble to plot this with a computer algebra system, if you'd like to
see the exact shape. Just key in 

f:=(sin(t))\^\ 4;g:=int(f,t=0..x);plot(g,x=0..2$*$Pi);

and sit back and look. 


\item{3FS.} Prove 
 $$\eqalign{\sum_{n=1}^{\infty}&\frac{1}{n(n+1)}=1\cr
\sum_{n=1}^{\infty}&\frac{1}{n(n+1)\dots (n+k)}=\frac{1}{k!k}\cr
\sum_{n=1}^{\infty}&\frac{k!}{n^2(n+1)\dots
(n+k)}=\frac{\pi^2}{6}-\sum_{j=1}^k\frac{1}{j^2}\cr}$$ For the third identity,
take as given that
$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.[Note...the $k!$ was missing
in the original. This last part required the previous part which went unsolved,
so fortunately did not affect the outcome of the contest.]

For the first part, you have a telescoping sum. That is,
$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. With this observation, (which is
part of the integration technique known as partial fractions), the original sum
can be rewritten as
$1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+\frac{1}{3}-\frac{1}{4}\dots$ and
everything except the initial 1 cancels out, leaving a sum of 1. 

For the second part, you use the same idea, only with the identity
$\frac{1}{n(n+k)}=\frac{1}{k}(\frac{1}{n}-\frac{1}{n+k})$. Applying this to the
sum given gives another telescoping sum. This time, it's
$$\frac{1}{k}(\frac{1}{1\cdot 2\cdot 3\dots k}-\frac{1}{2\cdot 3\cdot 4\dots
(k+1)}+\frac{1}{2\cdot 3\cdot 4\dots(k+1)}-\frac{1}{3\cdot 4\cdot 5\dots
(k+2)}+\dots)$$which telescopes to $\frac{1}{k\cdot k!}$. 


\item{4F.} A graduated cylinder in the chemistry storeroom has a leaky bottom.
The cylinder was full of water yesterday evening at 5PM, but the next morning
at 8AM, it was only half full. Water seeps out of the crack at the bottom at a
rate proportional to the square root of the pressure. How many more hours, from
8 AM today, will it be until the cylinder is completely empty? Neglect
evaporation, surface tension and so on. 

The pressure at the bottom of the cylinder is clearly (it should be clear)
directly proportional to the depth of water in the cylinder. Thus the leakage
rate is proportional to the square root of the depth. This gives rise to a
differential equation $dh/dt=-K\sqrt{h}$, since the rate at which the height
$h$ in the cylinder drops is also directly proportional to the leakage rate. 

Solving this differential equation by separation of variables (you don't need
the techniques of 308 here, you just convert it to $h^{-1/2}\, dh=-K\, dt$ and
integrate) gives $h=K(C-t)^2$. Assuming $h=1$ when $t=0$ and $h=1/2$ when
$t=15$ gives $K=C^{-2}$ and then $C=15/(1-1/\sqrt{2})$. Thus the total elapsed
time until the cylinder leaks dry is this $C$, and the EXTRA time after the
first 15 hours is $C-15$. 

\item{5F.} Assume that $h(t)$ is positive for all $t\ge 0$, continuous, and has
the property that for all $x>0$, $\int_x^{\infty}h(t)\, dt=2 h(x)$. Find
$h(4)$. Typo...meant to ask for $h(4)/h(0)$. As it is, the problem is
indeterminate, since any value for $h(0)$ leads to a different value for
$h(4)$. The real question, though, remains: what is $h(t)$ like? Invoking the
fundamental theorem of calculus on an integral where the bottom end of
integration is moving gives $-h(x)=2h'(x)$. Thus $h'(x)/h(x)=-1/2$ and so $\log
h(x)=-(x/2)+C$ and $h(x)=Ke^{-x/2}$. (Thus $h(4)=e^{-2}h(0)$.)

\item{6F.} A new delivery van costs $\$\,\!20000$. Once bought, there is a
steady stream of maintenance costs.  The company has a formula for the costs to
be expected for a van aged $t$ years, over a short interval of further time
$[t,t+\Delta t]$: $\$\,\!20t^2\Delta t$. Buying a new van every year, and
scrapping the old one, is ridiculous. Holding on to a van that costs more in
yearly maintenance than a new van is still sillier. The company wants to
minimize the quantity: $${{\hbox{total cost of owning the van over\ } T \hbox{\
years}}\over{\hbox{number of years\ } T{\hbox{\  van is held}}}}$$At what age
should an old van be scrapped and a new one bought?

The cost to buy and then maintain the van for $T$ years is
$20000+\int_0^T20t^2\, dt=20000+(20/3)T^3$. Dividing by $T$, the question is
asks for the minimum of $20000/T+20T^2/3$. The derivative is $-20000/T^2+40T/3$
which is zero when $T=1500^{1/3}$. That's between 11 and 12, incidentally.
\vfill\eject

\item{4S.} On the Mir space station, a pingpong ball is currently at $(0,0,0)$
and is moving forward along the $x$ axis. The paddle wielder wishes the ball to
bounce off a paddle which is to be held motionless in some plane, so that the
ball will bounce off and continue along the line $(x-2)/2=y/2=z$. Give the
equation of the plane in which the paddle should be held.

The ping-pong ball will bounce off the plane at the same angle it hit it, since
the plane is not moving. Thus the vector from which the ball came, and the
vector along which it proceeds after it bounces, should make the same angle
with the normal to the plane. The line along which the ball is supposed to
proceed goes through $(2,0,0)$, and so does the $x$ axis. Thus this is a point
in the plane of the paddle. Now the unit vector from which the ball comes is
$-i$, while the unit vector parallel to the direction it bounces is
$(2/3)i+(2/3)j+(1/3)k$. The sum of these unit vectors makes the same angle with
each of them, so the direction of the normal to the paddle plane is parallel to
$(-1/3)i+(2/3)j+(1/3)k$. The plane is thus of the form $-x+2y+z=C$. Choosing
$C=-2$ means that the point $(2,0,0)$ is in the plane as required, so the plane
is $-x+2y+z=-2$ or equivalently $x-2y-z=2$. 

\item{5S.} Consider the set $S$ of all points in a cube which are closer to the
center than to any of the corners. 
\itemitem{(a)} What fraction of the volume of the cube is in $S$?
\itemitem{(b)} If now $T$ denotes the points of the cube which are closer to
the center than to any of the {\it faces}, sketch the region $T$ and describe
its boundary with an equation or set of equations. 

Consider another cube, oriented the same and of the same size, but with its
center at one of the corners of the original cube. (Thus the two cubes
overlap). Their intersection is a cube with opposite corners at their
respective centers. Clearly the portion of this cube nearest the one center is
equal to the portion nearest the center of the other. For each of the other
corners the situation is the same, so the original cube loses half of its
volume in each octant. The answer is $1/2$. 

A more pedestrian solution is to analyze the region slice
by slice. Set up a coordinate system so the cube has corners $(\pm 1,\pm 1\pm
1)$ and center $(0,0,0)$> In the plane $z=h$, when $h\ge 1/2$ the cross section
of the points nearest the origin is a square with its edges parallel to the
diagonal of the
square with corners $(\pm 1,\pm 1,h)$. The corners of the little square are at
$(0,\pm j,h)$ and at $(\pm j,0,h)$, where $j^2+h^2$ is the square of the
distance to the center, and $(1-j)^2+(1-h)^2+1$ is the square of the distance
to the nearest corner. Setting the two expressions equal gives $j=(3-2h)/2$.
The area of the square with these corners is $\frac{1}{2}(3-2h)^2$. That
describes the situation for large values of $h$. When $0<h<1/2$, the part of
the cube nearest the center, and in the plane $z=h$, is an octagon got by
cutting four corners off the original square cross section. The corner at
$(1,1,h)$ loses the triangle $[(1,1,h),(1/2-h,1,h),(1,1/2-h,h)]$ and so it has
area $(1+2h)^2/8$ and the four lost corners together have area $(1/2)(1+2h)^2$.
Thus the area of the octagon cross section is $4-(1/2)(1+2h)^2$. Now
$\int_0^{1/2}(4-(1/2)(1+2h)^2)+\int_{1/2}^1(1/2)(3-2h)^2$ works out to 2. This
is half the volume of the upper half of the cube, and the situation for
negative values of $h$ is the same. The volume of the central region is half
that of the entire region. 

You have two choices for problem 6. Work one or the other, but not both. 

\item{6Sa.} Let $LB$ denote the both-directional Laplace transform: $LB$
applied to $y(t)$ gives the function
$Y(s):=\int_{-\infty}^{\infty}\exp(-st)y(t)\, dt$. Take as given that
$\int_{-\infty}^{\infty}\exp(-\pi t^2)\, dt=1$ (you may have seen this in
connection with polar coordinates.)
\itemitem{(a)} Find $LB(e^{-\pi t^2})$. 
\itemitem{(b)} Find $LB(f(t))$ if $f(t)=1-|t|$ for $|t|<1$ and $f(t)=0$
otherwise. 

The integral to be evaluated is $\int_{-\infty}^{\infty}\exp(-\pi t^2-st)\,
dt$. With the change of variables $u=t-s/2\pi$, this integral becomes
$\int_{-\infty}^{\infty}\exp(-\pi u^2+s^2/4\pi)\,
du=\exp(s^2/4\pi)\int_{-\infty}^{\infty}\exp(-\pi u^2)\, du$. Since the last
integral here is 1, the answer to the original question is that $LB(e^{-\pi
t^2})=\exp(s^2/4\pi)$. For the other integral, direct evaluation is doable, but
the fun way is to treat this as the transform of the convolution of the
function $u(t+1)-u(t-1)$ with itself. Thus the transform is
$((1/s)(e^{-s}-e^s))^2$. 



\item{6Sb.} Consider the system of differential equations:
$$\eqalign{y_1'&=y_2-y_1\cr y_2'&=y_3-y_2\cr y_3'&=y_1-y_3\cr}$$Your faithful
assistant has turned the crank and got some responses from the machine; these
are supplied in an abbreviated appendix (see next page).  Assuming your
assistant chose $m$ correctly, what was $m$? What were the eigenvalues of $m$?
Discuss the behavior of the solutions as time passes: Is there a limit? Will
the initially largest of the $y_k$'s always be the largest? 

The eigenvalues were $-\frac{3}{2}\pm\frac{\sqrt{3}}{2}i$ and $0$. The first two
eigenvalues have negative real part, so the terms of the solution corresponding
to those eigenvalues tend to zero. The eigenvector for $0$ is $[1,1,1]$, so the
solutions tend toward a state in which all three variables are equal.
Furthermore, the sum $y_1+y_2+y_3$ is constant, since its derivative is
$y_2-y_1+y_3-y_2+y_1-y_3$. The initially largest of the $y_k$ will not remain
largest: This is an autonomous system, so for any solution $Y(t)$, the vector
functions $Y(t+c)$ are also solutions. We can ignore the constant solution, and
just look at the path which the nonconstant part of the solution takes to zero.
Equivalently, we can assume that initially, and thus always, $y_1+y_2+y_3=0$. 
The claim is that no such solution involves one of the three quantities always
being largest. Now the first component of the given nonconstant solution is
$\exp(-3t/2)\cos(wt-\theta)$. All these solutions have first entry or component
%$y_1(t)$ of the form $c_1\exp(-3t/2)\cos(wt-\theta)+c_2\exp(-3t/2)\cos(wt)$,
on taking $c$ so that $wc=\theta$. But the sum of two sine waves of the same
frequency is again a sine wave of that frequency, so that $y_1$ moves in a
damped sine wave. Likewise with $y_2$ and $y_3$. Since the total is identically
zero,  whenever one of these damped sine waves is on a downswing, it cannot be
largest of the three.  \vfill\eject





 

 \bye


Let $g(x):=\sum_{n=1}^{\infty}\frac{x^{n!}}{n!}$ whenever the series is
convergent. 
\itemitem{(a)} Find integers $a$ and $b$ (explicitly) so that
$$\left|\frac{a}{b}-f(1/2)\right|<10^{-10}$$
\itemitem{(b)} Find $f(1)$ exactly. 
\itemitem{(c)} Find the radius of convergence of the series.