Question

Assume that the radius rr of a sphere is expanding at a rate of 19 in./min.19 in./min. The volume of a sphere is V=43πr3V=43πr3. Determine the rate at which the volume is changing with respect to time when t=5 min.t=5 min. assuming that r=4r=4 at t=0t=0. The volume is changing at a rate of

Answer 1

`744876\pi in^3/min`.

Explanation:

We are given that

`(dr)/(dt)=19 in/min`

Volume of sphere=
`(4)/(3)\pi r^3`

`r=4 at t=0`

`dr=19 dt`

Integrating on both sides then we get

`r=19 t+C`

Substitute r=4 and t=0

`4=C`

`r=19t+4`

`(dV)/(dt)=4\pi r^2(dr)/(dt)`

`(dV)/(dt)=4\pi (19t+4)^2(19)`

Substitute t=5

`(dV)/(dt)=4\pi (19(5)+4)^2(19)`

`(dV)/(dt)=4 \pi (99)^2 (19)=744876\pi in^3/min`

Hence, the volume is changing at the rate of
`744876\pi in^3/min`.