Solve Weird Depth Problem: Conical Tank Water Flow Rate

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I've looked everywhere to try to solve this problem and I can't find anything. It is:

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rateof 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Thanks for the help.
 
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this is a pretty easy related rates problem
 
Why would you consider that weird? In general, if you can find a formula for volume as a function of height, V(h), then, by the chain rule, dV/dt= (dV/dh) dh/dt. You are given dV/dt and asked to find dh/dt. You need to be able to calculate dV/dh. Do you know (or can you look up) the formula for volume of a cone in terms of radius and height? Here you are told that the tank itself has height 12 feet and radius 5 feet. Do you see that any "cone of water" contained by that tank will have h/r= 12/5?
 
sorry, i just started rates on friday
 
Good! Then you should know exactly how to do this!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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