Question

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Answer 1

`(dh)/(dt)`≅
`0.286(ft^(3) )/(min)`

Explanation:

`V=(\pi )/(3)r^(2)h`; rate of change
`(dV)/(dt)=10(ft^(3) )/(min)`, we must find the rate of change of the depth
`(dh)/(dt) =?;h=8ft`

5h=12r;
`V=(\pi )/(3)\({\((5h}/12} )} ^(2)h=(\pi )/(3)((25h^(2) )/(144))h; V=(25\pi h^(3))/(432)`; deriving
`(dV)/(dt) = (25\pi )/(432)(3h^(2))(dh)/(dt)` →
`10=(25\pi h^(2))/(144) (dh)/(dt)` → h=8 then
`(dh)/(dt)=(1440)/(25\pi 64)=(9)/(10\pi)`≅ 0.286
`(ft^(3) )/(min)`