Answer:
Volume will decrease at a rate of [tex]12\pi r^2m^3/sec[/tex]
Step-by-step explanation:
We have given rate of change of radius is 3 cm per second
So [tex]\frac{dr}{dt}=3cm/sec[/tex]
Volume of the ball is given as [tex]V=\frac{4}{3}\pi r^3[/tex]
So [tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex]
As [tex]\frac{dr}{dt}=3cm/sec[/tex]
So [tex]\frac{dv}{dt}=4\pi r^2\times 3=12\pi r^2m^3/sec[/tex]
So volume will decrease at a rate of [tex]12\pi r^2m^3/sec[/tex]
