Question

Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43????????3V=43πr3 and its surface area is 4????????2.4πr2. Determine the rate at which the volume is changing with respect to time at ????=2 min,t=2 min, assuming that ????=0r=0 at ????=0.t=0. (Use symbolic notation and fractions where needed.)

Answer 1

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

$\frac{d\textup{r}}{\textup{dt}}$  = 70 cm/min

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

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, $\frac{d\textup{V}}{\textup{dt}}$

 = $\frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70$

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min