Related Rates Inverted Cone problem

Question

Water is leaking out of an inverted conical tank at a rate of 7200 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 13 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 25 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

In order to find the rate at which water is being pumped into the tank, we need to find the net rate that the volume of water in the tank is increasing based the given information that the water level is rising at a rate of 25 cm/min when the height of the water is 1.0 m (100 cm).

We need to derive an equation that relates

1. the rate that the water level is rising, dh/dt
2. the rate that the volume of water is increasing, dV/dt

Step 1: express the volume of water (V) as a function of water level (h)

V = (1/3)πr²h

Since the tank dimensions are h = 13 cm and r = d/2 = 6.5/2 = 3.25 cm this means r = h/4

So V = (1/3)πr²h = (1/3)π(h/4)²h = (1/48)πh³

Step 2: Use the expression from step 1 to find an expression that relates dV/dt to dh/dt

dV/dt = (1/48)π(3h²)(dh/dt) = (1/16)πh²(dh/dt)

When h = 1m = 100 cm, dh/dt = 25 cm/min

This means dV/dt = (1/16)π(100)²(25) = 15,625π ≈ 49,087 cm³/min

dV/dt = (dV/dt)_in – (dV/dt)_out where

1. (dV/dt)_in is the rate that water is being pumped into the tank, which we are trying to find
2. (dV/dt)_out is the rate that water is leaking out of the tank, which is given to be 7200 cm³/min
3. dV/dt is the net rate that the volume of water in the tank is increasing

So the rate that water is being pumped into the tank is given by

(dV/dt)_in = dV/dt + (dV/dt)_out = 49,087 + 7200 = 56,287 cm³/min

Answer (rounded to 2 significant figures)

The rate at which water is being pumped into the tank is 56,000 cm³/min