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:
[tex]V=\frac{4}{3} \pi r^3[/tex]
To calculate the rate of change of the radius (dr/dt), we have to derive the last expression.
[tex]\frac{dV}{dt}=(\frac{4\pi}{3})\cdot (3r^2)\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}[/tex]
Then we have an expression for dr/dt:
[tex]\frac{dr}{dt}=\frac{1}{4\pi r^2} \frac{dV}{dt}=\frac{10}{4\pi r^2}[/tex]
For r = 11 cm, we have:
[tex]\frac{dr}{dt}=\frac{10}{4\pi (11)^2}=\frac{10}{4*3.14*121} =\frac{10}{1520}=0.0066[/tex]
The rate of change of the radius is 0.0066 cm/s when the radius is 11 cm.
