Question

A spherical balloon is being inflated at the rate of 12 cubic feet per second. What is the radius of the balloon when its surface area is increasing at a rate of 8 feet square feet per second? Volume= (4/3)(pi)r^3 Area= 4(pi)r^2?

Answer 1

Let
`S` be the surface area (in square feet);
`V` be the volume of the sphere (in cubic feet);
`t` be the time (in seconds); and
`r` be the radius (in feet).

`(d S)/(dt) = 8` (given)
`rarr (dt)/(d S) = 1/8`
`(dV)/(dt) = 12` (given)
Therefore
`(dV)/(d S) = (d V)/(dt) * (dt)/(dS) = (12)/(8) = 3/2`

Using the Volume and Surface Area formulae (as given)
`(dV)/(dr) = 4 pi r^2`
and
`(dS)/(dr) = 8 pi r`
`rarr (dr)/(dS) = 1/(8 pi r)`
Therefore
`(dV)/(dS) = (dV)/(dr) * (dr)/(dS) = (4 pi r^2)/(8 pi r) = r/2`

Equating our two solutions for `(dV)/(dS)`
`r/2 = 3/2`

and the radius (`r`) is 3 feet.