houdini...Due tonight

[ilovemath142]

Houdini wanted to finally escape from his multiple shackles as the water reached the top of his head in exactly 15 minutes.
Your group has been contracted by Houdini to complete the following tasks.
a) Compute how high the block should be given that:
- The water is poured at a rate of 20? cubic feet per minute
- The cross-sectional radius of the flask (r), measured in feet, as a function of
Height (h) from the ground, measured in feet, is given by 15/(h^(1/2))
- The bottom of the flask is at h = 2 feet
- Houdini is 66 inches tall.
Hint: Express the volume of the water in the flask as a function of the height of the water above ground level. What is the volume of the water when it reaches the top of Houdini’s head? State any assumptions you make (e.g. accounting for the volume of Houdini or the block)

>>>>for part A i figured out it the block would be around 2.08 feet.

b) Let h(t) be the height of the water above ground level at time t. Derive the differential equation for the rate of change of the height of the water with respect to time. Express this as a function of height above the ground. How fast is the water level changing when it first starts to fill? How fast is it changing when it reaches the top of his head?

c) Houdini wants to have an option in case he can’t get the chains and locks off in time (a good contingency plan like any Air Force Officer would have). Towards the bottom of the flask, he has designed a plug to be pulled if the water gets too far above his head (he has specified to pull the plug when the water is 11 feet above ground level). Water flows out of this hole at a rate such that the rate of change of the height of the water is proportional to the cube root of the height of the water. How long will it take for the water to be 5 inches below the top of his head, allowing him to breathe again?* How long will he have held his breath at this point? Do you think he could hold his breath that long?


*Houdini and his assistant found the following in testing: 30 seconds after pulling the plug, the water level had dropped 1 foot.

I need help finding the related rates for part b and a differential equation for part C. or at least that is what i think I need to do for it.

Thank you!

 

[Deleted member 4993]

Do a little search - and you would have found -

viewtopic.php?f=3&t=34595

 

[galactus]

This problem has become quite cliche. If you google 'Houdini Calculus problem' or 'Houdini Escape Problem' , many things will pop up. There is even a Power Point presentation out there.

Here is a nice little tidbit to know.

The rate of change of the volume, dV/dt, is equal to the cross-sectional area(A(t)) at that instant times the rate of change of the height, dh/dt.

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

We are given \(\displaystyle \frac{dV}{dt}=20\pi\)

The cross-sectional area of the water at some instant is given by \(\displaystyle A={\pi}\cdot (\frac{15}{\sqrt{h}})^{2}\)

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

\(\displaystyle \frac{dh}{dt}=\frac{20\pi}{{\pi}\cdot (\frac{15}{\sqrt{h}})^{2}}=\frac{4h{\pi}}{45}\)