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 15 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops. HINT [See Example 1.] (The volume of a sphere of radius r is V = 4 3 πr3. Round your answer to two decimal places.)

Answer 1

The rate at which the radius of the balloon is growing at the instant it pops is approximately 0.024 cm/s.

Step-by-step explanation:

To calculate how fast the radius of the balloon is growing at the instant it pops, we need to find the rate at which the volume of the balloon is increasing. The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius of the balloon. Differentiating this equation with respect to time, we get dV/dt = 4πr²(dr/dt). Since we know that dV/dt = 15 cm³/s, we can substitute this value into the equation and solve for dr/dt.

dV/dt = 4π(4²)(dr/dt) = 15

192π(dr/dt) = 15

dr/dt = 15 / (192π) ≈ 0.024 cm/s

Answer 2

The radius is growing with the rate of 0.07 cm per sec

Step-by-step explanation:

Given,

The volume of the balloon ( spherical ),

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

Where,

r = radius of the balloon,

Differentiating with respect to t ( time ),

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

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

Here,
`(dV)/(dt)=15\text{ cubic cm per sec}` and r = 4 cm,

`15=4\pi (4)^2 (dr)/(dt)`

`\implies (dr)/(dt)=(15)/(64\pi)=0.0746\approx 0.07\text{ cm per sec}`