If the volume of spherical ball is increasing at the rate of `4pi" cc/s"`, then the rate of change of its surface area when the volume is 288 `pi` cc is

Options

A. `(4)/(3)pi" "cm^(2)//s`
B. `(2)/(3)pi" "cm^(2)//s`
C. `4pi" "cm^(2)//s`
D. `2pi" "cm^(2)//s`

Answer

A

Explanation

Let V and r be the volume and radius of spherical ball, respectively,
volume of spherical ball `=(4)/(3)pir^(3)`
`impliesV=(4)/(3)pir^(3)` . . . (i)
`implies288pi=(4)/(3)pir^(3)` . . . [given, `V=288cm^(3)`]
`implies288=(4)/(3)r^(3)`
`impliesr^(3)=72xx3=8xx27`
`impliesr^(3)=72xx3=8xx27`
`impliesr^(3)=2xx3` [taking cube root both sides]
On differentiating Eq. (i) w.r.t. 't', we get
`(dV)/(dt)=4pir^(2)(dr)/(dt)`
`therefore4pi=4pir^(2)(dr)/(dt)` [given `(dV)/(dt)=4pi` cubic cm/s]
`implies1=(6)^(2)=(dr)/(dt)` . . [`becauser=6`]
`implies(dr)/(dt)=(1)/(36)`
now, surface area of spherical ball, (s)`=4pir^(2)`
`implies s=4pir^(2)`
On differentiating both sides, w.r.t. 't', we get
`(ds)/(dt)=4xx2pir(dr)/(dt)`
`=8xxpixx6xx(1)/(36)" "[becauser=6 and (dr)/(dt)=(1)/(36)]`
`implies(ds)/(dt)=(4pi)/(3)cm^(2)//s`