Question

A swimming pool is 25 ft wide, 40 ft long, 3 ft deep at one end and 9 ft deep at the other end. If water is pumped into the pool at the rate of 10 cubic feet/min, how fast is the water level rising when it is 4 ft deep at the deep end?

Answer 1

From the given information we can calculate the ratio `(dh)/(dV)` where `h` is the height of water in the deep end and `V` is the volume of water.
We are given `(dV)/(dt)` and multiplying the two ratios together we can calculate `(dh)/(dt)`

The Volume of water in the pool is given by the formula
`V= (wxxlxxh)/2` where `h` is the height of water at its deepest point, `l` is the length of the surface area of the water, and `w` is the (constant) width of the water surface.

`w=25`
`l/h = 40/6` provided `h<=6` (by similar triangles)
`rarr l = 20/3 h`

and the formula for the Volume of water can be rewritten as
`V(h) = (25xx 20/3 hxx h)/2`
or
`V(h) = (250 h^2)/3`

`(dV)/(dh) = (500 h)/3`

`rarr (dh)/(dV) = 3/(500 h)`

The rate of change in depth of the water per unit of time is
`(dh)/(dt) = (dh)/(dV) * (dV)/(dt)`

We are told `(dV)/(dt) = 10` cu.ft./min.

So when the water is 4 feet deep (unfortunately I labelled this `h` for height), the rate of change in depth (aka height) of the water is
`(dh)/(dt) = 3/(500*4)*10`

`= 3/200 = 0.015` ft/min.