Question

Water is flowing into a conical tank at a rate of 3 ft3/min. The height of the tank is 10 feet and its diameter is 6 feet. Find the rate at which the height of the water level is changing at the instant its height is 5 feet.

Answer 1

Dh/dt = 1.273 ft/min

Explanation:

Volume of cone V = 1/3 * π *r² * h

Then

DV/dt = 1/3* π *r² * Dh/dt (1)

We have to find out values of r when h = 5

By symmetry in a cone (we have proportion between h and r

when h = 10 ft r = 3 ft from problem statement

Then h = 5 ft r = 1.5 ft

That is from proportion 3/10 = X/5

Then by subtitution in (1)

DV/dt = 3 ft³/min r = 1,5 ft

3 = 1/3*π*(1.5)²* Dh/dt

3 = 2.355 Dh/dt

Dh/dt = 1.273 ft/min