how fast is surface area rising when there is 5 inches deep worth of water in the funnel?

Question

Funnel that is 10 inches across 8 inches deep; has a flow of water in =12 in^3/sec and flow of water going out = 4 in^3/sec; how fast is surface area rising when there is 5 inches deep worth of water in the funnel.

I solved for dh/dt by using the volume formula, because my teacher said you could solve it without the surface area of a cone equation, I got 1/125pi as my answer.

I can explain my work further, but I'm more concerned about whether I got the right answer.

Any help would be much appreciated.

Answer 1

I am assuming that the funnel has zero radius at the bottom and that "10 inches across" means that the diameter of the funnel at the top is D = 10 in - so the radius is R = 5 in. We also know that the height is H = 8cm. Note that at any level of the funnel with height h and radius r, the following relationship will be true:

r/R = h/H ⇒ r/5 = h/8 ⇒ r = 5/8*h

Now the volume of the funnel is given by

V = π r^2 h/3 ⇒ V = π * 25/64 * h^2 * h/3 ⇒ V = 25/192 * π * h^3

If we differentiate this equation w.r.t. time t, we get the following (differential) equation

dV/dt = 25/192 * π * 3 * h^2 dh/dt ⇒ dV/dt = 25/64 * π * h^2 dh/dt

Of course we also know that the derivative dV/dt is constant and equal with dV/dt = 12 - 4 = 8 in^3/sec.

As a result, at h = 5 in, we get

8 = 25/64 * π * 25 * dh/dt ⇒ 1 = 125/8 * π * dh/dt which leads to

dh/dt = 8/(125*π) in/sec.

It seems that the solution is similar but slightly different than yours. Let me know if you need any additional help or feel free to post details about your solution. I'd be glad to help more :)