A conical tank (with vertex down) is 12 feet across the top and 18 feet deep. If water is flowing into the tank at a rate of 18 cubic feet per minute, find the rate of change of the depth of the water when the water is 10 feet deep.

Step-by-step explanation:

Volume of a cone V = 1/3πR²h

r is the radius

h is the height

Given

dV/dt = 18ft³/min

h = 10ft

dimeter = 12ft

radius = 6ft

according to similar triangles

R = 6/10 h

R = 3/5 h

The volume becomes

V = 1/3πR²h

V = π/3(3h/5)²h

V = π/3 9h²h/25

V = 9h³π/75

dV/dt =  dV/dh * dh/dt

dV/dt = 27h²π/75dh/dt

Substitute the given values

18 = 27(10)²π/75dh/dt

18 = 2700π/75dh/dt

18 = 36π dh/dt

dh/dt = 18/36π

dh/dt = 1/2π

dh/dt = 1/2(3.14)

dh/dt = 1/6.28

dh/dt = 0.1592ft/min