Question

Water is leaking out of an inverted conical tank at a rate of 6800.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 30.0 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

Answer:

6.82 $mt^3/min$

Step-by-step explanation:

By congruence of triangles, the radius r of the cone base when its height is 1 meter satisfies the relation

r/1 = 2.5/6

(See picture attached)

So, r = 0.4166 meters when the water is 1 meter high.

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

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

So, the volume of water when it is 1 meter high is

$V=\frac{\pi* 0.(4166)^2}{3}=0.1817\;mt^3$

One minute later, the height of the water is 1 meter + 30 centimeters = 1.30 mt

The radius now satisfies

r/1.3 = 2.5/6

and now the radius of the base is

r = 0.5416 mt

and the new volume of water is $0.3071\;mt^3$

So, the water is raising at a speed of

0.3071-0.1871 = 0.12 $mt^3/min$

This speed equals the difference between the water being pumped and the water leaking,

So,  

0.12 = Speed of water being pumped -6.8  

and

Speed of water being pumped = 6.8+0.12 = 6.82 $mt^3/min$