Question

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 5 centimeters. Assuming the balloon is filled with helium at a rate of 12 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops.

Answer 1

`V=\frac43\pi r^3`

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

You're given that the volume of the balloon is increasing at a rate of 12 cm^3/sec, which means
`(\mathrm dV)/(\mathrm dt)=12`. The balloon pops when
`r=5`, so you have enough information to find the rate of change of the radius at this time:


`12=4\pi(5)^2(\mathrm dr)/(\mathrm dt)\implies (\mathrm dr)/(\mathrm dt)=\frac3{25\pi}\approx0.0382` cm/sec