Question

Air is pumped into a spherical balloon, so the balloon expands. The volume of a sphere of radius R is 4πR3/3. If the radius of the sphere after t seconds is 2t centimeters, at what rate is air being pumped in when t=4? (Hint: the rate air is pumped in equals the rate that the volume of the sphere increases).

Answer 1

So rate of pump after t =4 sec will be
`1607.68cm^3/sec`

Step-by-step explanation:

We have given that volume of the sphere
`V=(4)/(3)\pi R^3`

It is given that at after t sec radius of the sphere is 2t cm

So volume of the sphere
`V=(4)/(3)\pi (2t)^3=(32)/(3)\pi t^3`

After differentiating
`(dV)/(dt)=32\pi t^2`

We have to find the rate of pump at t = 4 sec

So
`(dV)/(dt)` at t = 4 sec

`(dV)/(dt)=32* 3.14* 4^2=1607.68cm^3/sec`

So rate of pump after t =4 sec will be
`1607.68cm^3/sec`