Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 6m 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?

Hmm… since we are dealing with a conical shape the Volume (V) should = 1/3*pi*r^2*h
And the Area should = pi*r*sqr(r^2+h^2)

We are given dh/dt = 20 dV/dt = C – 10,000 where V = 1/3*pi*r^2*h at time t
A picture will help us figure this out, review my picture below:



r = 2 m
let h = 2 m
height = 6 m
The Diameter is 4 m
1) We have V = 1/3*pi*r^2*h
2) By similar triangles, h/6 = r/2 so
h = r/2 * 6 = 6r/2 = 3r and
r = 2h/6 = 1/3h
4) So we have V = 1/3*pi*(1/3*h)^(2)*h = 1/3*pi*(1/3*h)^3 = pi/27*h^3
5) So dV/dt = pi/9*h^2 *dh/dt
We were given that
6) dV/dt = pi/9*h^2 *dh/dt
We are given that dh/dt = 20
7) So we have dV/dt = C – 10,000 = pi/9*h^2*dh/dt
8) All of these figures are in cm so where as our h value = 2 is in meters so we must convert 2 meters to centimeters = 200 cm
9) Now we can plug in our values h = 200,and dh/dt = 20 to solve for C
C-10,000 = pi/9*(200)^2*(20)
C =10,000 + pi/9*(200)^2*(20) = 289252.68 cm^3/min
Therefore the constant rate at which the water is pored in to the tank is 289,252.68cm^3/min