A fish bowl is constructed out of a spherical bowl with diameter 20cm.

Question

The bowl was initially empty and water flows into the bowl at a rate of 10cm^3/s. Let V and h be the volume and depth of water in the bowl at time t seconds respectively. Find dh/dt when h is increasing at the slowest rate.

Answer 1

The volume of a spherical cap of radius r and height h (the water depth) is

v = π/3 h^2(3r-h)
= π/3 h^2(30-h)
= π/3 (30h^2 - h^3)

dv/dt = π(20h - h^2) dh/dt

Now, when is h increasing most slowly? When the water surface area is greatest (since dv/dt is constant, and dv = a*dh)

That is when the bowl is half full: h=10.

10 = π/3 (20*10 - 100) dh/dt
dh/dt = 3/(10π) cm/s

Answer 2

Last few steps should be:
10 = π(20*10 - 100) dh/dt
dh/dt = 1/(10π) cm/s

Answer 3

thanks. leftover from a previous attempt.