Question

Assume the radius r of a sphere is expanding at a rate of 30 cm/min. The volume of a sphere is V = 4/3(pi)(r^3) and it's surface area is 4(pi)(r^2).

Determine the rate of change of the volume with respect to time at t = 2 min, assuming that r = 0 at t = 0.

Answer 1

let's keep in mind that r = 0 at t = 0, namely the radius is 0 at 0 seconds, and we know the radius is expanding at 30 cm/min, so in t = 2, in 2 minutes the radius is then 30 + 30 = 60 cm wide.

$\bf \stackrel{\textit{volume of a sphere}}{V=\cfrac{4\pi r^3}{3}\implies V=\cfrac{4\pi }{3}r^3}~\hspace{7em}\cfrac{dV}{dt}=\cfrac{4\pi }{3}\left( \stackrel{chain-rule}{3r^2\cdot \cfrac{dr}{dt}} \right) \\\\\\ \cfrac{dV}{dt}=4\pi r^2\cdot \cfrac{dr}{dt}\implies \cfrac{dV}{dt}=4\pi (60)^2(30)\implies \cfrac{dV}{dt}\approx 1357168.02635$

Answer 2

Using implicit differentiation, it is found that the rate of change of the volume at time t = 2 is of 1,357,168 cm³/min.

The volume of the sphere is given by:

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

The rate of change is found applying implicit differentiation, so:

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

Radius expanding at a rate of 30 cm/min, thus $\frac{dr}{dt} = 30$.

Radius starts at 0 cm, thus, at 2 min, it will be of 60 cm, that is, $r = 60$.

Then:

$\frac{dV}{dt} = 4\pi(60)^2(30)$

$\frac{dV}{dt} = 1357168$

The rate of change in the volume is of 1,357,168 cm³/min.

A similar problem is given at https://brainly.com/question/9543179