The radius of a cylinder is increasing at the rate of `3ms^(-1)` and its altitude is decreasing at the rate of `4ms^(-1)`. The rate of change of volume when radius is `4m` and altitude is 6m is

Options

A. `80pi` cu m/s
B. `144pi` cu m/s
C. 80 cu m/s
D. 64 cu m/s

Answer

A

Explanation

Let, h and r be the height and radius of cylinder.
Given that, `(dr)/(dt)=3m//s`
`(dh)/(dt)=-4m//s`
Let volume of cylinder, `V=pir^(2)h`
`implies(dV)/(dt)=pi[r^(2)(dh)/(dt)+h*2r(dr)/(dt)]`
At `r=4m and h=6m `
`therefore (dV)/(dt)=pi[-64+144]=80picum//s`