Answer:
So rate of pump after t =4 sec will be [tex]1607.68cm^3/sec[/tex]
Explanation:
We have given that volume of the sphere [tex]V=\frac{4}{3}\pi R^3[/tex]
It is given that at after t sec radius of the sphere is 2t cm
So volume of the sphere [tex]V=\frac{4}{3}\pi (2t)^3=\frac{32}{3}\pi t^3[/tex]
After differentiating [tex]\frac{dV}{dt}=32\pi t^2[/tex]
We have to find the rate of pump at t = 4 sec
So [tex]\frac{dV}{dt}[/tex] at t = 4 sec
[tex]\frac{dV}{dt}=32\times 3.14\times 4^2=1607.68cm^3/sec[/tex]
So rate of pump after t =4 sec will be [tex]1607.68cm^3/sec[/tex]
