Question

A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 feet deep at its deepest point. If the pool is being filled at a rate of 0.8ft3/min, how fast is the water level rising when the depth at the deepest point is 5 ft

Answer 1

Using implicit differentiation, it is found that the water level is rising a rate of 0.001 ft/sec.

The volume of a pool of length l, width w and height h is given by:

$V = lwh$

Applying implicit differentiation, the rate of change of the volume is given by:

$\frac{dV}{dt} = wh\frac{dl}{dt} + lh\frac{dw}{dt} + lw\frac{dh}{dt}$

Neither the width nor the length changes, hence $\frac{dl}{dt} = \frac{dw}{dt} = 0$, and:

$\frac{dV}{dt} = lw\frac{dh}{dt}$

We are given that:

$\frac{dV}{dt} = 0.3, l = 20, w = 40$, hence:

$\frac{dV}{dt} = lw\frac{dh}{dt}$

$0.8 = 20(40)\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{0.8}{800}$

$\frac{dh}{dt} = 0.001$

Water level is rising at rate of 0.001 ft/sec.

A similar problem is given at brainly.com/question/9543179