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

Answer:

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

Explanation:

Given,

The volume of the balloon ( spherical ),

$V=\frac{4}{3}\pi r^3$

Where,

r = radius of the balloon,

Differentiating with respect to t ( time ),

$\frac{dV}{dt}=\frac{4}{3}\times 3\pi r^2 \frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Here, $\frac{dV}{dt}=15\text{ cubic cm per sec}$ and r = 4 cm,

$15=4\pi (4)^2 \frac{dr}{dt}$

$\implies \frac{dr}{dt}=\frac{15}{64\pi}=0.0746\approx 0.07\text{ cm per sec}$