Water is leaking out of an inverted conical tank at a rate of 1.5 cm3 /min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 cm and the diameter at the top is 6 cm. If the water level is rising at a rate of 1.0 cm/min when the height of the water is 2 cm, find the rate at which water is being pumped into the tank. (a) Draw a picture of the situation for any time t. (b) What quantities are given in the problem

a) Check Explanation

b) Check Explanation

c) The rate at which water is being pumped into the tank = 2.631 cm³/min

Explanation:

Let the rate of flow of water into the tank be k cm³/min

a) The image of the conical tank is presented in the attached image

Note, the radius and height of a cone are related through the similar triangles principle.

As shown in the attached image, it is evident that

r/h = 3/10

r = 3h/10 = 0.3 h

b) The quantities given in the problem.

- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3

- total height of the tank, H = 10 cm

- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm

- rate at which water is leaking from the tank = 1.5 cm³/min

- water is being pumped into the tank at constant rate of k cm³/min

- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min

c) volume of the tank at any time = πr²h/3

Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)

dV/dt = k - 1.5

V = πr²h/3 and r = 0.3 h, r² = 0.09 h²

V = 0.03πh³

dV/dt = (dV/dh) × (dh/dt)

dV/dh = 0.09π h²

dV/dt = 0.09π h² (dh/dt)

dV/dt = k - 1.5

0.09π h² (dh/dt) = k - 1.5

But at h = 2 cm, (dh/dt) = 1.0 cm/min

0.09π h² (dh/dt) = k - 1.5

0.09π 2² (1) = k - 1.5

k - 1.5 = 1.131

k = 1.5 + 1.131 = 2.631 cm³/min