If water flows into the tank at a rate of 20 ft3/min, how fast is the depth of the water increasing when the water is 18 feet deep?

Question

A conical water tank with vertex down has a radius of 15 feet at the top and is 24 feet high. If water flows into the tank at a rate of 20 ft3/min, how fast is the depth of the water increasing when the water is 18 feet deep?

Answer 1

The formula for the volume of a cone is:

V = (1/3) π r² h

Given:

h = 24 ft.

r = 15 ft.

dV/dt = 20 cu.ft./ min.

Find dh/dt at 18 ft. depth

First find the function r in terms of h'

24•r = 15•h

r = (15/24)•h

r = (5/8) h

Substitute the value of r in the formula:

V = (1/3) π (5/8)²h² h

V = (25/192)•π•h³

Get the derivative of V with respect to h, we'll have:

dV/dt = 3•(25/192)•π•h² (dh/dt)

at 18 feet deep:

20 = 3•(25/192)•π•(18)² (dh/dt)

20 = (24300π / 192)(dh/dt)

dh/dt ≈ 0.05 ft./ min.