Need Help w/ Part C of Houdini's Escape (due tonight)

[4yi]

Here is a link to the problem:
http://barzilai.org/cr/houdini.html

My Teacher gave me this problem and cut out the formulas meant to be used for it so I had to search for it online, but even after finding the problem part C has me stumped. The project is due tomorrow but I must have the work done by tonight (before midnight), I have part a and b completed and here is my work:

Part A

V=??[R(x)]2dx
V=??(10/?z)2(h)dy, z=1
V=?[(10/?z)3/(3)](h2/2)dy
=?[(1/3)(10/?z)3](1/2h2)dy
220?=333.3(1/2h2)dy
.66?=(1/2)h2dy
?1.32?=?h2
1.15?=h
h=3.61 ft.

To determine the height of the actual block Houdini’s height must be subtracted from the height found in the last slide.
h=3.61ft
6ft-3.61ft=
The height of the block Houdini was standing on was 2.39 ft.
To find the volume need to reach the top of Houdini’s head, simply evaluate the volume function, after the intergration, in the last slide with z=1 & h=3.61 therefore,
V= ?[(1/3)(10/?z)3](1/2h2)dy
?[(1/3)(10/?1)3](1/2(3.612)dy
=68.24 cubic ft.

Part B

In order to derive Houdini’s function we must integrate the volume in water, (dv/dt), evaluate r(z)=(10/?z) in for x, and then differentiate to get Houdini’s function.
dv/dt =?22? = ?(22x/1) = 22?x
r(x)=22?(10/?z)
r?(z)=22?(10z-1/2) = 22?(-5z-3/2) = -110?z-3/2

Now to find how fast the water is rising when the flask starts to fill and when it reaches the top of Houdini’s head…use Houdini’s equation derived in the previous slide and substitute 22? in for dv/dt and 10 min in for dz/dt.
dv/dt= -110?/z-3/2dz/dt
22? = -110?/z-3/2(10)
(z3/2)(22?) = (-1100?/z3/2)(z3/2)
(22z3/2?)/(22?) = (-1100?/22?)
(2/3)z3/2 = -50(2/3)
z= 33.3 ft.3 per min.

To see how fast the water rises when it reaches the top of Houdini’s head subtract his height and the height og the block he stands on.
z= (33.3)-(6)-(2.39)=24.91ft.3 per min.

So there is part A and part B, but I am at a loss for part C.
Please help

 

[galactus]

Are they still assigning this problem?. Apparently so. Anyway, here is a link that may help a little.

http://www.mathnerds.com/mathnerds/archive/archive.aspx

Here's a power point regarding this problem:

http://www.chipola.edu/instruct/math/cr ... Escape.ppt

Google it and you'll probably find more. As a matter of fact, it was asked on this forum before. Search around for it.


But part c is just a generalization of part b. The volume of the flask would be \(\displaystyle {\pi}\int_{1}^{h}[r(z)]^{2}dz\)

 

[4yi]

So, if I'm generalizing part B, can my work for C start by going through all the steps of that part substituting a for 22??
If I do that, this is what I get for the first part:

dv/dt =?a = (ax/1) = ax
r(x)=a(10/?z)
r?(z)=a(10z^-1/2) = a(-5z^-3/2) = -5az^-3/2

 

[galactus]

No, not quite. You used the values form the first part, \(\displaystyle \frac{10}{\sqrt{z}}\).

One thing to note is that the rate of change of volume is equal to the cross sectional area at that instant times the rate of change of the height.

\(\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}\)

 

[4yi]

so then without using anything from the first part of the problem and substituting I can come up with
dV/dt = ?a = ax
dV/dt=A(t)(dh/dt)
(dV/dt)/A(t)=dh/dt
which leaves me at dh/dt = ax/A(t) but I don' think I'm getting anywhere or moving anywhere near dh/dt as a function of h(t)

Maybe if you help me by figuring out the answer, I can figure out the work once I've seen what I'm working toward. I'm not sure where I'm going wrong...