Question

the radius ???? of a sphere is expanding at a rate of 70 cm/min. The volume of a sphere is ????=43????????3 and its surface area is 4????????2. Determine the rate at which the volume is changing with respect to time at ????=2 min, assuming that ????=0 at ????=0.

Answer 1

Answer:

$\frac{dV}{dt}=1120 \pi cm^3/min$

Step-by-step explanation:

We are given that

Radius of sphere expanding at the rate=$\frac{dr}{dt}=70 cm/min$

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

Surface area of sphere=$S=4\pi r^2$

We have to determine the rate  at which the volume is changing with respect to time at r= 2 cm.

$V=\frac{4}{3}\pi r^3$

Differentiate w.r.t time

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

Substitute the values then we get

$\frac{dV}{dt}=4\pi (2)^2(70)=1120 \pi cm^3/min$

Hence, the rate at which the volume of sphere is changing is given by

$\frac{dV}{dt}=1120 \pi cm^3/min$