Question

A spherical balloon is inflated so that its volume is increasing at the rate of 12 cubicfeet per minute. How fast is the radius of the balloon increasing when the radius is 6feet?

Answer 1

Given:

The rate of change in volume = 12 cubic feet per minute.

We need to find the rate of change of radius at radius = 6 feet.

Consider the formula to find the volume of the sphere.

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

Differentiate with respect to t.

`(dV)/(dt)=(4)/(3)\pi*3r^2*(dr)/(dt)`
`\text{ Substitute }(dV)/(dt)=12\text{ and r=6 in the formula.}`

`12=(4)/(3)\pi*3(6)^2*(dr)/(dt)`

`12=144\pi*(dr)/(dt)`

Dividing both sides by 144pi, we get

`(12)/(144\pi)=(dr)/(dt)`

`(dr)/(dt)=(1)/(12\pi)`

`(dr)/(dt)=(1)/(37.68)=0.0265\text{ feet per minute}`

Hence the radius of the balloon increases by 0.03 feet per minute.