Question

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 3 centimeters. Assuming the balloon is filled with helium at a rate of 10 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops. (The volume of a sphere of radius r is V = 4/3 πr3. Round your answer to two decimal places.)

Answer 1

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.

Explanation:

We are given the following information in the question:

A spherical balloon is designed to pop at the instant its radius has reached 3 centimeters.

The balloon is filled with helium at a rate of 10 cubic centimeters per second .

`\displaystyle(dV)/(dt) = 10~cm^3/sec.`

We have to find fast the radius is growing at the instant it pops.

Volume of spherical balloons =

`\displaystyle(4)/(3)\pi r^3`

Differentiating, we get,

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

Putting the values, we get,

`10 = 4* 3.14* (3)^2* \displaystyle(dr)/(dt)\\\\(dr)/(dt) = (10)/(4* 3.14* (3)^2) = 0.0884 \approx 0.09\\\\(dr)/(dt) = 0.09\text{ centimeter per second}`

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.