Question

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 4 centimeters. Assuming the balloon is filled with helium at a rate of 17 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops.

Answer 1

Use V = (4/3)(pi)(r^3) for the volume of a sphere whose radius is r.

You want dr/dt when r = 4 when dV/dt = 17.

Answer 2

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

Answer 3

17 = 4pi(16) dr/dt

Answer 4

dr/dt = 17/(64pi) cm per second

Answer 5

Done.

Answer 6

Wow! thank you so much! I had no clue how to before but the work helps a ton!

Answer 7

I took derivative with respect to time on both sides of the equation and plugged in the given. You're welcome.

Answer 8

wait was it a decimal?

Answer 9

fraction of 17/(64pi) you can convert that to a decimal if you want. I dont have a calculator handat the moment. But that answer id fine.

Answer 10

okay thank you!

Answer 11

welcome.