A gas barrel has the shape of a right cylinder with the radius of 5 ft. and the height of 15 ft. If gas is being pumped into the barrel at the rate of 3 ft.3/min, find the rate at which the gas is rising when the gas is 8 ft. deep.

Question

any help would be great!!

Answer 1

Since the tank is a right circular cylinder, the cross-section does not change.

The volume of a 1-ft-high section is 25pi ft^3, so the liquid rises at a rate of 3/25pi ft/min.

Now, if you want to use calculus, you can say that

v = pi r^2 h
v' = pi(2rr'h + r^2h')
But, r is constant, so r' = 0
v' = pi r^2 h'
3 = pi * 25 * h'

h' = 3/25pi

Answer 2

okay, thank you!!!