Question

A swimming pool is 60 ft long and 25 ft wide. Its depth varies uniformly from 3 ft at the shallow
end to 15 ft at the deep end. It is being filled with water at the rate of 800 ft3
/min. At what rate is
the depth of water increasing at the deep end when that end is 5 ft deep?

Answer 1

The pool bottom rises uniformly 12 ft over 60 ft, so the slope is 1/5. When the water depth at the deep end is h = 5 ft, the water extends 5h = 25 ft toward the shallow end, forming a triangular prism of water.

1.) ANSWER: 1.28 ft/min
Volume as a function of h: V =25 x 5(5h)(h)=62.5h2

2.)
Differentiate: dV dt = 125h dh dt.

3.)Given dy = 800 ft 3/min and h = 5: 800=125(5) - 1.28 ft/min

Answer 2

Let s(t) be the depth of the water at the deep end. For s ≤ 12 ft the water reaches only part of the pool: the water extends from the deep end a distance x1 where the bottom slope reaches the water surface. The bottom rises 12 ft over 60 ft, so b(x)=x/5 (with x measured from the deep end). The water extends to x1 where b(x1)=s, so x1=5s.

The longitudinal cross-section of the water is a right triangle with base 5s and height s, so its area is (1/2)(5s)(s)=2.5 s^2. Multiplying by the width 25 ft gives the volume
V = 25·2.5 s^2 = 62.5 s^2.

Differentiate: dV/dt = 125 s (ds/dt). Given dV/dt = 800 ft^3/min and s = 5 ft,
ds/dt = 800/(125·5) = 800/625 = 1.28 ft/min.

Thus the depth at the deep end is increasing at 1.28 ft/min (while the water level is still below the shallow end).