A conical water tank with vertex down has a radius of 12 feet at the top and is 27 feet high. If water flows into the tank at a rate of 10 ft3/min

Question

A conical water tank with vertex down has a radius of 12 feet at the top and is 27 feet high. If water flows into the tank at a rate of 10 ft3/min, how fast is the depth of the water increasing when the water is 16 feet deep?

Answer 1

A picture might look like this:

We are give the rate at which the water is filing the tank. That is dV/dt. We are being asked to find dh/dt at a particular height. So if we develop a function for the volume of the water at any particular time "t", in terms of the height of the water "h", and differentiate it with respect to "t", then the result will have dV/dt and dh/dt in it and will be able to find dh/dt.

The volume of the water at any time "t" is V = 1/3πr²h but there are two variables, "r" and "h"

Using the cross section, we can say that (because there are two similar triangles) 12/27 = r/h and cross multiplying gives us 12h = 27r or r = (12/27)h = (4/9)h. So now we can substitute into the volume equation above:

V = 1/3πr²h but r = (4/9)h so:

V = 1/3π(4/9)h)²h

V = 1/3π•16/81h³

V = (16π/243)h³

Differentiating with respect to time (using the chain rule) gives us:

dV/dt = 3(16π/243)h²•dh/dt

dV/dt = (16π/81)h²•dh/dt

and solving for dh/dt we get:

dh/dt = (dV/dt)/((16π/81)h²)

dh/dt = 81(dV/dt)/(16πh²)

Plugging in the numbers we get:

dh/dt = 81(10)/(16π(16²))

dh/dt = 405/(2048π) ft/min which is approx 0.0629 ft/min