Question

A large container has the shape of a frustum of a cone with top radius 5 m, bottom radius 3m, and height 12m.
The container is being filled with water at the constant rate 4.9 m^3/min. At what rate is the level of water rising at the instant the water is 2 deep?

Answer 1

When the water has depth y, the radius of the surface is

3 + (y/12)(5-3) = 3 + y/6

Its volume at that point is
v = 1/3 (R^2-r^2) h
= (1/3)((3+y/6)^2-3^2)y = y^3/108 + y^2/3

Now we are ready to begin. When y=2,
v = 8/108 + 4/3 = 38/27

dv/dt = (y^2/36 + 2y/3) dy/dt
4.9 = (4/36 + 8/3) dy/dt
dy/dt = 1.76 m/min

Answer 2

Sorry Mr Steve
I submitted your answer and It was wrong answer
the right answer = 14.0 cm/min

Could You please tell me how to do this question because I have to do it again with different values.

Answer 3

I guess you should have checked my math. The actual formula for the volume is

(π/3)((18+y)(3+y/6)^2-18*3^2)
= π(y^3/108 + y^2/2 + 9y)

because the water is in the bottom of the cone

dv/dt = π/36(y+18)^2 dy/dt
4.9 = π/36*20^2 dy/dt
dy/dt = .14 m/min or 14.0 cm/min