Answer:
[tex]\frac{dH}{dt}=0.044\;\;feet\;per\;min[/tex]
Explanation:
Given,
Radius R = 6 feet
Height H = 9 feet
Depth of the water = 5 feet
Volume of right circular cylinder,
[tex]V=\pi R^2 H\\[/tex]
Differentiate with respect to the H
[tex]\frac{dV}{dH}=\pi R^2\\\frac{dV}{dH}=36\pi[/tex]
Now,
[tex]\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[/tex]
Hence, the rate of chage of height of water is 0.044 feet per min.
