Since h is height of the water from the bottom of the pool, there are two separate cross sections to consider.

1) The section from h=0 to h=16.

V = 20[25*h +(1/2)(h)*(25h/16)] ----(i)

This 25h/16 is the length of the horizonral water line above the slanting bottom. You can get it by proportion. Say this horizontal line is x,

x/h = 25/16

Multiply both sides by h,

x = 25h/16.

2.) the section from h=0 to h=20, but the "effective h" is from h=16 to h=20 only.

V = 20[25*16 +(1/2)(25)(16) +(75)(h-16)] ---(ii)

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V = 20[25*h +(1/2)(h)*(25h/16)] ----(i)

V = 20[25h +(25h^2)/32]

V = 500h +(500h^2)/32

V = 500h +(125/8)(h^2) -----(ia)

Differentiate both sides with respect to time t,

dV/dt = 500(dh/dt) +(125/8)[2h*(dh/dt)]

dV/dt = [500 +(125h/4)](dh/dt) -----------------------(1)

V = 20[25*16 +(1/2)(25)(16) +(75)(h-16)] ---(ii)

V = 20[400 +200 +75h -1200]

V = 20[75h -600]

V = 1500h -12,000 -----(iia)

Differentiate both sides with respect to time t,

dV/dt = 1500(dh/dt) ----------------------------------(2)

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" 2) What addtional information would you need to find dh/dt at t=10 minutes ? "

We need the dV/dt, which is the rate of change of the volume.

We can derive that if we can only find the rate of flow of the water filling the pool, like 40 cubic feet per minute, for example.

Here, at t=10min, we can get the V. Then, using Eq.(ia) or (iia), we can get the h.

So if we know already the dV/dt and h, then we can find dh/dt in Eq.(1) or (2).