Answer:
[tex]\frac{dV}{dt}=525 cm^{3}/min[/tex]
Step-by-step explanation:
The volume of a sphere is:
[tex]V=\frac{4}{3}\pi r^{3}[/tex] (1)
We know that:
If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.
[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex] (2)
We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.
Solving (1) for r, we have:
[tex]r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm[/tex]
Finally dV/dt will be:
[tex]\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min[/tex]
I hope it helps you!
