Question

Assume that the radius rr of a sphere is expanding at a rate of 19 in./min.19 in./min. The volume of a sphere is V=43πr3V=43πr3. Determine the rate at which the volume is changing with respect to time when t=5 min.t=5 min. assuming that r=4r=4 at t=0t=0. The volume is changing at a rate of

Answer 1

Answer:

$744876\pi in^3/min$.

Step-by-step explanation:

We are given that

$\frac{dr}{dt}=19 in/min$

Volume of sphere=$\frac{4}{3}\pi r^3$

$r=4 at t=0$

$dr=19 dt$

Integrating on both sides then  we get

$r=19 t+C$

Substitute r=4 and t=0

$4=C$

$r=19t+4$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi (19t+4)^2(19)$

Substitute t=5

$\frac{dV}{dt}=4\pi (19(5)+4)^2(19)$

$\frac{dV}{dt}=4 \pi (99)^2 (19)=744876\pi in^3/min$

Hence, the volume is changing at the rate of $744876\pi in^3/min$.