I need the solutions to two questions from the GRE Subject Test:
Mathematics practice booklet.  One relates to differential equations
and the other analysis.  They should be simple for anyone versed in
math (and I know Google's got alot of Math Ph.Ds, something I'm hoping
to start this coming year).

This is word for word from the Math Subject Test Practice booklet
available from the ETS website

54.  The inside of a certain water tank is a cube measuring 10 feet on
each edge and having verticle sides and no top.  Let h(t) denote the
water level in feet above the floor of the tank at time t seconds. 
Starting at time t=0, water pours into the tank at a constant rate of
1 cubic foot per second, and simultaneously, water is removed from the
tank at a rate of 0.25*h(t) cubic feet per second. As t->infinity,
what is the limit of the volume of the water in the tank?

A. 400 cubic feet
B. 600 cubic feet 
C. 1000 cubic feet
D. The limit does not exist
E. The limit exists but cannot be determined without knowing h(0).

63.  At how many points in the xy plane do the graphs of y = x^(12) and y = 2^x
intersect?

A. None
B. One
C. Two
D. Three
E. Four

Again...I have the answers, I need to know how they were determined. 
Any help would be very much appreciated.

I'm not a mathematician, so there's probably an elegant way to do
this. My method is partly visual and partly analytical.

For #63, there should be two intersections.

Here is how I solved #63:

If you think of graphing the functions, the two curves meet when they
have the same y value for a given x.

So, it they meet, x^(12)=2^x at that point(s)

It helped me to graph these roughly, but I can only describe that here.

y=x^(12) is like y=x^2 but rises much more rapidly (parabola-like). Y
values are positive for all x except x=0 where y=0 (the shape should
be somewhat like a squished parabola).

2^x is always positive. For x=0, y=2^x=1; for x>0 y is greater than 1
and curves upward; for x<0 y is between 1 and 0, steadily decreasing
in value. The curve is shaped somewhat like a hyperbola intersecting
the y axis at y=1; it decreases steadily in value in going from x=0 to
large negative x values and increases in value in going from x to
positive infinity.

Intersection point 1: Since y=x^(12) is shaped like a parabola with a
y value of 0 at x=0 and steadily increases to positive infinity as x
goes from 0 (y=0) to minus infinity and since y=2^x decreases from 1
to zero steadily in the same negative x range), the two curves must
cross once.

Intersection point 2: For positive values of x, y=x^(12) increases
more rapidly than y=2^x and the curves cross once. You can see that
this point is between x=1 and x=2 if you roughly plot the values of
the two functions.

x      y=x^(12)      y=2^x
0         0                  1
1         1                  2
2       4096              4

No lemmas, no theorems, but this reasoning should give the answer.

I initially shied away from looking at #54, but I think I have the
solution to that one too.

Suppose the tank is initially empty; then no water is draining out at
that instant but it is coming in a 1 ft^3/s. At this point, the water
will start rising in the tank.

Suppose the tank is full initially; water is leaving at 10 (0.25)
ft^3/s and water is coming in at 1 ft^3/s. Hence, the water level will
fall.

Equilibrium will be reached somewhere in between -- when the rate of
water removal equals the rate of water addition, that is, when 1
ft^3/s = 0.25 h. Solving for h gives a level of 4 feet, or 400 cubic
feet in the tank.

There is probably a way to solve this with differential equations, but
I'll quit with the above.

For water tank.....
The question says at any time t seconds, the change in volume of the
water in the tank is 1 CF minus 0.25h, so,
dV/dt = 1 -0.25h  -----***
If a maximum V can be attained, then dV/dt at that time is zero, hence,
0 = 1 -0.25h
h = 1/0.25 = 4 ft.
That means when h=4ft, there is no more change in volume of the water
in the tank, so, at h=4ft, V is max.
Now, V = 10*10*h, so, max V = 100*4 = 400 CF  ------answer.

That means also that Vo is immaterial. If there were a Vo, and it were
less than 400CF, then the h will rise and settle at 4ft. If Vo were
more than 400CF, then the h will fall down to the 4-ft level and
settle there.

--------------------
For the two graphs.......

There will be two intersection points, one at the 1st quadrant, and
another at the 2nd quadrant, as seen from their graphs on the same x,y
plane.

If you want to solve for the intersection points, you can use the
Newton's Method of iteration.

samirhc-ga,

I need to correct an English mistake. I should not have used the term,
"equilibrium," as it implies too much. I should have used
"steady-state" or "approaching a limit."

ticbol-ga,

Thanks for the differential equation solution. I am enlightened.

You guys are correct about problem 54, but not about problem 63.

I just did #63 again, this time using Excel to plot a number of points
rather than the few points hand-one points and the general reasoning
that I used before. As far as I can tell, there can be no more than 1
crossover point in the negative x range (x between -1 and -.09) and it
sure looks like there can't be more than one crossover point in the
positive x range (x between 1.05 and 1.1).

If #63 is supposed to be purely analytical, I could see how you might
get only one crossover point. Analytically, I think you could go
through logarithms, and since logarithms of negative x are undefined
(I think), you will get only one crossover point that way (the one in
the positive x region). If the instructions restrict you to using such
an analytical method only, then I could see how they got an answer of
one crossover point.

Can you give me a hint about the answer, that is, is it supposed to be
more or fewer than the number of crossover points that we got (two)?

Thanks!

There are three.

Oops! I just noticed that today's first comment was not from the
original questioner.

rracecarr-ga,

How do you know we are wrong about #63? Do you have the book or do you
have a different solution? If you have a different solution, please
share it with us.

Thanks!

Well, I'm sure there is an elegant analytical way to show that there
are 3 points of intersection.  I haven't thought of one though.  The
intersection you are missing is near x=75.

On (-oo,0), x^12 is monotonically decreasing and 2^x is monotonically
increasing.  Both are continuous, so since they must cross somewhere
on that interval, there is exactly one intersection for negative x.

For the positive solutions, take the log of both functions.  x^12
becomes 12 ln x, and 2^x becomes x ln 2.  Plugging in a few points
near x=1 shows that there is an intersection in that region.  The end
behavior of the linear function (x ln 2) is to grow faster than the
logarithmic function (12 ln x), so there must be at least 1 additional
intersection.  As both functions are continuous and differentiable on
(0,oo), one is linear and the other has a second derivative that is
strictly negative, there must be exactly one additional point.  So
three total.  As I said earlier, this is messy.

OK, I see a possibility between 76 and 77. To do that, though, I had
to compare logarithms since the numbers were so large. Before, I was
estimating that the y=2^x curve wouldn't bend upwards more than the
y=x^(12) curve.

To get the third point, I had to switch to comparing (12 log x) and (x
log 2) because the numbers are so large.

The original request was for how we got the answers. How did you go
about solving the problem?

Thanks!

Our comments keep crossing. 

Thanks!

rracecarr-ga,

After a break, I took a look at your solution - very nice! Especially
for showing that there is exactly one more solution.

I'll probably never use this line of reasoning again, but this problem
taught me something about making estimations too hastily and inferring
too much from previous experience.

I hope that samirhc-ga enjoyed this as much as I have.

Re #63: rracecarr's analysis is correct and elegant.  I will only add
some somewhat more precise decimal approximations for x to the ones so
far put forth:

x = -0.94678
x = 1.06335
x = 74.6693256

I got the first two from graphing x^12 = 2^x on the graphing
calculator on my Mac.  The last I got from running a short iteration
program, using x = 12logx/log(2), on an old programmable computer.

Problem 54.  The correct answer is A (400 ft^3)

The differential equation expressing the time rate of change in the
height of the liquid is:

dh(t)/dt  + h(t) * (0.25 *ft^3)/(100 ft^2) - (1ft^3/s)/(100 ft^2) = 0

This equation simply says that the rate of change in the height is
equal to the volumetric rate at which water is entering the tank 
minus the volumetric rate at which water is leaving the tank, all
divided by the cross sectional area of the tank (to convert volumes to
heights).

This is a linear first-order inhomogeneous differential equation.  See
<http://hyperphysics.phy-astr.gsu.edu/hbase/math/deinhom.html#c1> for
the straightforward method of solution.

The homogeneous equation:

dh(t)/dt  + 0.0025 * h(t) = 0

has solution:

h(t) = h_0 * exp(-t/400)

where h_0 is the height of the water at t = 0

The particular (aka nonhomogeneous) solution is given by:

h_p(t) = 0.01/0.0025 = 4.

The complete solution is given by the sum of the homogeneous and
particular solutions:

h(t) = h_0*exp(-t/400) + 4

As t -> infinity, the exponential term goes to zero, and we are left
with h(t-> infinity)-> 4.  The volume is simply the height times the
area:  4*ft* (100 ft^2) = 400 ft^3

As ticbol and brix indicated in their comments, another way to solve
this problem without having to find the complete solution to the
differential equation is to recognize that *if* the the height ever
reaches a steady-state value, then dh(t)/dt must be equal to zero. 
(If it never reaches steady state, then one will encounter an
expression that can never be true, when dh(t)/dt is set to zero,
indicating that the assumption of the existence of a steady state was
incorrect.)  This converts the differential equation into an algebraic
equation that can be solved with some simply algebra:

dh(t-> infinity)/dt = 0 = -h(t) * (0.25 *ft^3)/(100 ft^2) + (1ft^3/s)/(100 ft^2)
h(t-> infinity) = [(1ft^3/s)/(100 ft^2)]/[(0.25 *ft^3)/(100 ft^2)] = 4 ft
V(t-> infinity = h(t-> infinity) * 100ft^2 = 400 ft^3



Problem 63.  The correct answer is D, (three)

The problem can be solved using the "Lambert function", (see
<http://mathworld.wolfram.com/LambertW-Function.html>).  The Lambert
function is the inverse function of f(w) = w*exp(w), and is usually
written as W(z).  For any complex number z, z = W(z)*exp(W(z)).  For
real arguments, W is two-valued in the interval [-1/e,0).  One branch
has W>= -1, and is called the principal branch denoted W_)(z), and the
other branch has W(z)< -1, and is denoted W_-1(z).

The trick solving the type of problem at hand in terms of the Lambert
Function involves manupulating the equation (in this case, x^12 = 2^x)
into a form that involves something that looks like w*exp(w).

Starting with the equation:

x^12 = 2^x, take the twelfth root of both sides yields two equations:

x = 2^(x/12)  and x = -2^(x/12)

Now divide both sides by 2^(x/12), noting that this expression can be
written as exp(x*ln(2)/12).  This yields:

x*exp(-x*ln(2)/12) = 1 and x*exp(-x*ln(2)/12) = -1

Now multiply both sides by -ln(2)/12 to get:

-[x*ln(2)/12]*exp(-x*ln(2)/12) = -ln(2)/12  and
-[x*ln(2)/12]*exp(-x*ln(2)/12) = +ln(2)/12

The left hand side of these equations are of the form z*exp(z), where
z = -x*ln(2)/12, so the solutions are given by

-x*ln(2)/12 = W(ln(2)/12) and -x*ln(2)/12 = W(-ln(2)/12)

x = -[12/ln(2)]*W(ln(2)/12) and x = -[12/ln(2)]*W(-ln(2)/12)

Now, remember that on the interval [-1/e,0), W(z) is multivalued. 
-ln(2)/12 ~= 0.058, and -1/e ~= -0.368, so -1/e < -ln(2)/12 < 0, and
the second equation above actually has two solutions, corresponding to
the two branches of the W function.  Writing these in terms of the
principal and second branch terminology, the real three solutions are:

x =  -[12/ln(2)]*W_0(ln(2)/12)
x =  -[12/ln(2)]*W_0(-ln(2)/12)
x =  -[12/ln(2)]*W_-1(-ln(2)/12)

Oh, yeah, numerically, the coordinates of the intersections turn out
to be at approximately:

(1.06334683, 2.08977386)  {x =  -[12/ln(2)]*W_0(-ln(2)/12)}
(-0.94678033, 0.51878895) {x =  -[12/ln(2)]*W_-1(-ln(2)/12)}
(74.669325535, 3.00404713*10^22) {x =  -[12/ln(2)]*W_0(ln(2)/12)}

If one considers complex values for x, then there are actually 10
additional solutions.

First, thanks to all of you for all your help.  

These problems were getting really frustrating.  I wish I'd posted on
Google Answers a little before 2 days prior to the Math GRE, but oh
well.  The exam itself went decent, but there's still some ground I
need to cover before I'm entirely comfortable with all the fundemental
material.  Similarly, I need to be much more intuitive when approached
with a problem.

On second thought I wonder if I should've posted the answers with the
questions I asked, but then, if anyone went to the ETS website to find
a print copy of this test, the answers are also available there. 
rracecar-ga had the correct answers for both questions first.  However
everyone here was able to answer the Diff. Eq. question.  One that,
now, seems so simple I can't believe I overlooked the fact that the
rate should be zero at equilibrium.

hfshaw provided a great answer for the intersection problem (#63). 
The Lambert equation seems to fit well, and I'm sure there are other
methods.  I think the exam, however, is looking for the application of
rracecar's intuitive concepts regarding the monotonicity of the
functions and the functional behavior past x=50.

I'll post some more, and hopefully get this tips thing figured out in
a short while.  Thanks alot for all your help, I very much appreciate
it.

Sincerely,
Samir

After all comments except Samir's latest one, I decided to see if I
could google these types of problems for methods of solution. I was
not very successful; I was trying something general like "fill empty
'steady state level'."  I may have gotten a generic answer, but it was
fairly abstract for me and difficult to relate to the problem at hand.
I got plenty of tank discussions, but, to me, at least, nothing with
the specific issue at hand.

After Samir's comment "if anyone went to the ETS website to find a
print copy of this test, the answers are also available there," I
googled the following exact phrases from the problems, "certain water
tank is a cube" and "At how many points in the xy plane do the graphs
of y."  However, I didn't pick up the ETS material (even though it's
on the web). I did, though, get links to sites where both of these
specific problems were discussed. Overall, I prefer the discussion
here.

Google wants to make the world's information available, but sometimes
it's much more fun to think about something than just to look up how
someone else solved the problem. Of course, if my thinking gets stuck,
then seeing someone else's answer is helpful - which, I hope, is part
of Google's goal.