Question

Blowing Soap Bubbles Carlos is blowing air into a spherical soap bubble at the rate of 10 cm3/sec. How fast is the radius of the bubble changing when the radius is 11 cm? (Round your answer to four decimal places.)

Answer 1

Answer:

The rate of change of the radius is 0.0066 cm/s when the radius is 11 cm.

Step-by-step explanation:

The change in volume is according to the inflow of air, that has a rate of 10 cm3/s.

The volume of the sphere can be defined as:

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

To calculate the rate of change of the radius (dr/dt), we have to derive the last expression.

$\frac{dV}{dt}=(\frac{4\pi}{3})\cdot (3r^2)\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$

Then we have an expression for dr/dt:

$\frac{dr}{dt}=\frac{1}{4\pi r^2} \frac{dV}{dt}=\frac{10}{4\pi r^2}$

For r = 11 cm, we have:

$\frac{dr}{dt}=\frac{10}{4\pi (11)^2}=\frac{10}{4*3.14*121} =\frac{10}{1520}=0.0066$

The rate of change of the radius is 0.0066 cm/s when the radius is 11 cm.