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The Swimming Pool
We wish to find the change in volume of a 20footwide pool as it fills up with water. A crosssection of the pool is shown below.
1) How would you express dV/dt in terms of h, V, and dh/dt ?
2) What addtional information would you need to find dh/dt at t=10 minutes ?

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)(h16)] (ii)

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)(h16)] (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)

" 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).