Question

A swimming pool is nearly​ empty, holding only 5400 gallons of water. A system is set up so that the water in the pool starts to increase by 13​% per hour. After how many hours is the pool filling at a rate of 2848 gallons per​ hour?

Answer 1

11.964 hours

Explanation:

The number of gallons p in the pool after x hours can be written as ...

p(x) = 5400·1.13^x

The derivative of this is ...

p'(x) = 5400·1.13^x·ln(1.13)

We want that equal to 2848, so we want to solve ...

2848 = 5400·1.13^x·ln(1.13)

Dividing by the coefficient of the exponential term gives ...

2848/(5400·ln(1.13)) = 1.13^x

Taking the natural log, we get ...

ln(2848/(5400·ln(1.13))) = x·ln(1.13)

x = ln(2848/(5400·ln(1.13))) / ln(1.13) ≈ 11.9637 . . . hours

After about 11.964 hours, the pool is filling at the rate 2848 gph.