Question

A spherical balloon is inflated so that it’s radius (r) increases at a rate of 2/r cm/sec. How fast is the volume of the balloon increasing when the radius is 4 cm?

Answer 1

32π cm³/sec

Explanation:

List given information

`\displaystyle V=(4)/(3)\pi r^3\\\\r=4\\\\(dr)/(dt)=(2)/(r)\\ \\(dV)/(dt)=\,\,?`

Solve for dV/dt

`\displaystyle V=(4)/(3)\pi r^3\\\\(dV)/(dt)=4\pi r^2\biggr((dr)/(dt)\biggr)\\\\(dV)/(dt)=4\pi r^2\biggr((2)/(r) \biggr)\\\\(dV)/(dt)=4\pi (4)^2\biggr((2)/(4) \biggr)\\\\(dV)/(dt)=4\pi(16)\biggr((1)/(2) \biggr)\\\\(dV)/(dt)=4\pi(8)\\\\(dV)/(dt)=32\pi`

Hence, the volume of the balloon is increasing at a rate of 32π cm³/sec when the radius is 4 cm.