How do you use related rates to find rate of water level in a spherical tank?

Question

The tank of radius A contains water that has a max depth H, volume V of water in the tank is V=(1/3)pi(H^2)(3A-H). Now the tank has radius 16 feet and is being filled at rate of 13.37 ft^3/min. What is the rate when the water is rising at H= 4ft

I know how to do related rates and got almost everything down, but I just don't know how to start this problem

Answer 1

you already have the equation...

just get the derivative...
dV/dt = f'(H) dH/dt... get the derivative, A is constant (it has a value there, read the question).

then plug in H = 4, dV/dt =13.37, now you can get dH/dt.

You need to practice to be able to solve questions by yourself confidently... you can now do this.


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Answer 2

Volume is a function of H and H is a function of t, since you want the rate (I assume) at which H is increasing, dH/dt. Let's multiply out the volume equation:

V = πH²(3A-H)/3, with A = 16,
V = 16πH² - πH³/3
dV/dH = 32πH - πH²
we know dV/dt = 13.37 ft³/min, and H = 4, so
dV/dt = dV/dH • dH/dt
13.37 = (32πH - πH²)dH/dt
13.37 = (128π - 16π)dH/dt
13.37 = (112π)dH/dt
dH/dt = 13.37/(112π) = 0.038 ft/min

I don't really like that answer, but 112π is the area of the circular slice of the sphere 4 feet from the bottom, and if a cylinder with that base had a volume of 13.37 ft³, it's height would be, yes, exactly 13.37/(112π) = 0.038 ft, so I guess the answer does make sense.

Answer 3

So substituting A=16 into your equation gives:
V = 1/3 π H²(48-H)
Additionally dV/dt = 13.37 which was given.
Now dH/dt = (dV/dt) / (dV/dH) so you need to find dV/dH
Differentiate and you will get:
dV/dH = πH(32-H)
So dH/dt = (dV/dt) / (dV/dH)
= 3/(πH(32-H))
When H=4 we get:
dH/dt (at H=4) = 13.37/(π4(32-4))
=13.37/(112π)
0.038 ft/min