Answer:
[tex]744876\pi in^3/min[/tex].
Step-by-step explanation:
We are given that
[tex]\frac{dr}{dt}=19 in/min[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
[tex]r=4 at t=0[/tex]
[tex]dr=19 dt[/tex]
Integrating on both sides then we get
[tex]r=19 t+C[/tex]
Substitute r=4 and t=0
[tex]4=C[/tex]
[tex]r=19t+4[/tex]
[tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex]
[tex]\frac{dV}{dt}=4\pi (19t+4)^2(19)[/tex]
Substitute t=5
[tex]\frac{dV}{dt}=4\pi (19(5)+4)^2(19)[/tex]
[tex]\frac{dV}{dt}=4 \pi (99)^2 (19)=744876\pi in^3/min [/tex]
Hence, the volume is changing at the rate of [tex]744876\pi in^3/min[/tex].
