The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Explanation

Let the rate of change of the balloon be constant `k`.
Then, `(d)/(dt)` (volume)`=` constant
`implies (d)/(dt)((4)/(3)pir^(3))=k` (`:'` Volume of balloon `=(4)/(3)pir^(3)`)
`implies ((4)/(3)pi)(3r^(2)(dr)/(dt))=k`
`implies 4pir^(2)dr=k dt`………`(1)`
`implies 4piintr^(2)dr=kintdt`
`implies 4pi(r^(3))/(3)=kt+Cimplies4pir^(3)=3(kt+C)`...........`(2)`
Initially when `t=0` then `r=3`
`:. 4pi(3)^(3)=3(kxx0+C)`
`implies 108pi=3CimpliesC=36pi`
Again, when `t=3`, then `r=6` , now from equation `(2)`,
`4pi(6)^(3)=3(kxx3+C)`
`implies 864pi=3(3k+36pi)`
`3k=288pi-36pi=252pi`
`impliesk=84pi`
put the value of `k` and `C` in equation `(2)`,
`4pir^(3)=3(84pit+36pi)`
`implies 4pir^(3)=4pi(63t+27)`
`implies r^(3)=63t+27`
`impliesr=(63t+27)^(1//3)`
which is the required radius of the balloon at time`t`.