Question

Water is leaking out of an inverted conical tank at a rate of 9,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)

Answer 1

Answer:

Step-by-step explanation:

rate of leaking water = 9500 cm³/min

height of tank, H = 6 m = 600 cm

diameter of tank, = 4 m

radius of tank, R = 2 m = 200 cm

dh/dt = 20 cm/min

at any time, h = 2 m Let at that time radius of the tank is r.

According to the diagram

R / H = r / h

200 / 600 = r / h

r = h/3

The volume of tank is given by

$V = \frac{1}{3}\pi\times r^{2}h$

$V = \frac{1}{27}\pi\times h^{3}$

Differentiate with respect to t

$\frac{dV}{dt} = \frac{1}{9}\pi\times h^{2} \frac{dh}{dt}$

$\frac{dV}{dt} = \frac{\pi}{9}\times 200^{2}\times 20$

dV/dt = 279111.11 cm³/min

This is the volume of water increasing per minute in the tank.

Let C be the rate of volume of water pumped into the tank.

So, C = 279111.11 + 9500

C = 288611.11 cm³/min