NEED HELP WITH QUESTION!

Question

A water tank at a filtration plant is built in the shape of an inverted cone with height 5.2 m and diameter 5 m at the top. Water is being pumped into the tank at a rate of 1.2 m^3 /min. Find the rate at which the water level is rising when there is 8π m^3 of water

Answer 1

Formula for Volume of a cone is:

V = (1/3)π•r²h

Given:

d = 5 m

∴ r = 2.5 m

h = 5.2 m

Therefore:

r = (25/52) h

When V = 8π m³/min., the height is:

V = (1/3) π((25/52) h)² h

V = (1/3)π(625/2704) h³

V = π(625/8112) h³

8π = π(625/8112) h³

h³ = 103.8336

h ≈ 4.7

Using derivative in V = (1/3)π(625/2704) h³

dV/dt = π (625/2704) h² (dh/dt)

When dV/dt = 1.2 m³ /min, dh/dt = the rate of water level rising.

1.2 = π (625/2704) (4.7)² (dh/dt)

dh/dt ≈ 0.07 m /min.