An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 19 ft3/min. At what rate is the water rising in the tank when the depth of the water is:

Question

An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 19 ft³/min. At what rate is the water rising in the tank when the depth of the water is:

1 foot? Answer	=
10 feet? Answer	=
51 feet? Answer	=

Answer 1

We need to find dh/dt. We know dV/dt = 19 ft³/min.

Volume of a right circular cone = (1/3)(π)r²h

We need to get the volume in terms of one variable, h, since we're looking for dh/dt

The diameter at the top of the cone is 26' so the radius is 13'

We can find h by similar ratios using the height and radius of the water in the tank.

13/52 = r/h ⇒ 1/4 = r/h ⇒ r= h/4

Substituting into the volume we get V = (1/3)(π)(h/4)²h = (1/3)(π)h³ / (16)

dV/dt = [(1/3)(3)(π)(h)² / 16](dh/dt) = [(π)(h)² / 16](dh/dt)

Solving for (dh/dt) we get

dh/dt = (dV/dt)(16) / [((π)(h)²]

We know that the change in volume is dV/dt = 19 ft³/min

dh/dt = (19 ft³/min)(16) / [((π)(h)²]

When h = 1 ft, dh/dt = (19 ft³/min)(16) / [((π)(1 ft)²] ≈ 112.05 ft/min

When h = 10 ft, dh/dt = (19 ft³/min)(16) / [((π)(10 ft)²] = (304 ft³)/(100 ft² π) ≈ 0.97 ft/min

When h = 51 ft, dh/dt = (19 ft³/min)(16) / [((π)(51 ft)²] = (304 ft³)/(2601 ft² π) ≈ 0.04 ft/min