Question

Water is being pumped into an inverted conical tank at some constant rate. The tank has a height of 600 cm and the diameter of the tank across the top is 400 cm. If the water level is rising at a rate of 20 cm/min when the height of the water is 200 cm, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)
cm3/min

Answer 1

Answer:

The water is being pumped at a speed of

$\boxed{\bf \frac{10592000*\pi}{27}\;cm^3/min}$

Step-by-step explanation:

By congruence of triangles, the radius r of the cone base when its height is 200 cm satisfies the relation

r/200 = 400/600

(See picture attached)

So, r = 400/3 cm when the water is 200 cm high.

The volume of a cone with radius of the base = R is given by

$\bf V=\frac{\pi R^2h}{3}$

So, the volume of water when it is 200 cm high is

$\bf V_1=\frac{\pi* (400/3)^2*200}{3}=\frac{32000000*\pi}{27}\;cm^3$

One minute later, the height of the water is 200 cm + 20 cm = 220 cm

The radius now satisfies

r/220 = 400/600

and now the radius of the base is

r = 440/3

and the new volume of water is  

$\bf V_2=\frac{\pi* (440/3)^2*220}{3}=\frac{42592000*\pi}{27}\;cm^3$

So, the water is raising (being pumped) at a rate (speed) of

$\bf V_2-V_1=(\frac{42592000*\pi}{27}-\frac{32000000*\pi}{27})=\frac{10592000*\pi}{27}\;cm^3/min$