Gas is being pumped into a a spherical balloon at the rate of `30 ft^(3)//min`. Then the rate at which the radius increases when it reaches the value 15 ft, is

Options

A. `(1)/(30pi)ft`/min
B. `(1)/(15pi)ft`/min
C. `(1)/(20)ft`/min
D. `(1)/(25)ft`/min

Answer

A

Explanation

`because(dV)/(dt)=30ft^(3)`/min, r=15 ft
volume of spherical balloon,
`V=(4)/(3)pir^(3)`
On differentiatinig w.r.t. t, we get
`(dV)/(dt)=4pir^(2)(dr)/(dt)`
`implies30=4pixx(15)^(2)(dr)/(dt)`
`implies(dr)/(dt)=(30)/(4pixx15xx15)`
`=(1)/(30pi)ft`/min