An inflatable backyard swimming pool is being filled from a garden hose with a flowrate of 0.12 gal/s. (a) If the pool is 8 ft in diameter, determine the time rate of change of the depth of the water in the pool.

0.00318 ft/s

Explanation:

Volume of the inflatables swimming pool, V = πr²h

V = πr²h

a) (dh/dt) = (dh/dV) × (dV/dt)

V = πr²h

(dV/dh) = πr²

(dh/dV) = (1/πr²)

So, we require the rate of change of depth when diameter = 8 ft, radius = (diameter)/2 = 4 ft

(dh/dt) = (dh/dV) × (dV/dt)

(dh/dV) = (1/πr²) = (1/π(4²)) = 0.0199 ft⁻²

(dV/dt) = 0.12 gal/s = 0.016 ft³/s

(dh/dt) = (0.0199 × 0.016) = 0.00318 ft/s

Explanation:

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