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. `80 pi cu m//s`
B. `144 pi 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 respectively.
Given that `(dr)/(dt) = 3m//s`
and `(dh)/(dt) = - 4m//s`
Also, `" "V = pi r^(2)h`
On differentiating w.r.t , we get
`(dV)/(dt) = pi [r^(2) (dh)/(dt )+ h2r (dr)/(dt)]`
`(dV)/(dt) = pi [r^(2) (dh)/(dt )+ h2r (dr)/(dt)]`
At r = 4 m and h = 6, m , them
`(dV)/(dt) = pi [ - 64 + 144]`
`80 pi cu m//s`