So a right circular cone is 25m deep with a radius of 9m. It's being filled with water at a rate of $3m^3$ per minute. a) find dh/dt when its height is 10m.
What I did first was use a proportion to find the new radius: $9/25 = r/10$ therefore $r = 3.6$. Next, since I don't have dr/dt, I put $r^2$ in terms of $h$ in the volume equation $V = (π/3) * r^2 * h$. The new equation, in context should be $V = (π/3)*1.296h^2$.
Deriving yields $dv/dt = 1.296(π/3) * 2h *dh/dt$. Plugging $h$ in and solving should give $dh/dt = .111$
This answer seems incorrect. Is it? If so, how?
now b): when the radius of the water surface is 4m, what is $dr/dt$? Can I use a similar process to what I did with the height?