Question

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 5 cubic feet per minute. If the pool has radius 6 feet and height 9 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet

Answer 1

Answer:

$\frac{dH}{dt}=0.044\;\;feet\;per\;min$

Explanation:

Given,

Radius R = 6 feet

Height H = 9 feet

Depth of the water = 5 feet

Volume of right circular cylinder,

$V=\pi R^2 H$

Differentiate with respect to the H

$\frac{dV}{dH}=\pi R^2\\\frac{dV}{dH}=36\pi$

Now,

$\frac{dV}{dH}.\frac{dt}{dt} =36\pi\\\frac{dV}{dt}.\frac{dt}{dH}=36\pi\\ 5.\frac{dt}{dH}=36\pi\\\frac{dH}{dt}=\frac{5}{36\pi}\\\frac{dH}{dt}=0.044\;\;feet\;per\;min$

Hence, the rate of chage of height of water is 0.044 feet per min.