A swimming pool is 20ft wide, 40ft long, 3ft deep at the shallow end and 9ft deep at its deepest point.?

Question

A swimming pool is 20ft wide, 40ft long, 3ft deep at the shallow end and 9ft deep at its deepest point. A cross-section is shown in the figure. If the pool is being filled at a section is shown in the figure. If the pool is being filled at a rate of .8 ft ^3/min, how fast is the water level rising whne the depth at the deepest point is 5 ft.

i know that you can not see the picture but is there any way that you can help me start the problem... i don't know how to put the picture up...

Answer 1

I'll assume the pool slopes smoothly from 3ft to 9ft deep and is otherwise uniform in cross-section.

The floor of the pool goes up 6ft in a 40ft distance, so when the water is h ft deep at the deepest point it will reach (40 h / 6) ft along the pool (for h between 0 and 6). The cross-section will be a right-angled triangle with the two smaller sides h and 40h/6 = 20h/3 feet in length. So the area is 20h^2/3; the total volume is this multiplied by the width of the pool, to give
V = 400h^2 / 3
dV/dt = (400/3) (2h) dh/dt = (800/3) h dh/dt.
So dh/dt = 3/800 (dV/dt) / h
= 3/800 (0.8) / 5
= 0.0006 ft/min.