Question

A spherical balloon is inflating with helium at a rate of 8080piπ StartFraction ft cubed Over min EndFraction ft3 min. How fast is the​ balloon's radius increasing at the instant the radius is 22 ​ft? Write an equation relating the volume of a​ sphere, V, and the radius of the​ sphere, r.

Answer 1

`(dr)/(dt)`=0.04132 ft/min

Explanation:

V = volume of the sphere

r = radius of the sphere

`(dV)/(dt)` = 80π ft³/min

now,

For sphere, V =
`(4)/(3)\pi r^3`

for rate of change of volume with respect to time (t)

`(dV)/(dt)=(d((4)/(3)\pi r^3))/(dt)`

or

`(dV)/(dt)=3*(4)/(3)\pi r^2*(dr)/(dt)`

or

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

at r = 22ft,
`(dV)/(dt)` = 80π ft³/min

80π =
`4\pi(22)^2*(dr)/(dt)`

or

`(dr)/(dt)`=0.04132 ft/min