Water runs into a conical tank at the rate of `9 ft^(3)//"min"`. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep ?

Explanation

`V=(1)/(3) pi x^(2) y`
` (dV)/(dt)=(pi y)/(3) 2x (dx)/(dt) +(pi)/(3) x^(2) (dy)/(dt)`
`"but" (x)/(y)=(1)/(2) rArr 2x=y rArr (2dx)/(dt) =(dy)/(dt)`
` rArr " " 9= (pi)/(3) [xy xx ((dy)/(dt)) + x^(2) (dy)/(dt) ]`
` rArr " " 9= (pi)/(3) (dy)/(dt) [xy + x^(2)]`
`rArr " " (dy)/(dt)=(9xx3)/(pi xx (18+9))=(9xx3)/(pi xx 27)`
`=(1)/(pi)= 0.318 ft // "min" `