Answer
(a)
(i) $172\pi~\mu m^2$
(ii) $121.3\pi~\mu m^2$
(iii) $102\pi~\mu m^2$
(b) When $r = 5~\mu m$, the instantaneous rate of change of the volume with respect to the radius is $~~100\pi~\mu m^2$
(c) The rate of change of the volume with respect to the radius is equal to the surface area.
Work Step by Step
(a) $V(r) = \frac{4}{3}\pi~r^3$
(i) $V(5) = \frac{4}{3}\pi~(5)^3 = \frac{500}{3}\pi~\mu m^3$
$V(8) = \frac{4}{3}\pi~(8)^3 = \frac{2048}{3}\pi~\mu m^3$
We can find the average rate of change:
$\frac{\Delta V}{\Delta r} = \frac{1548\pi/3}{8-5} = 172\pi~\mu m^2$
(ii) $V(5) = \frac{4}{3}\pi~(5)^3 = \frac{500}{3}\pi~\mu m^3$
$V(6) = \frac{4}{3}\pi~(6)^3 = \frac{864}{3}\pi~\mu m^3$
We can find the average rate of change:
$\frac{\Delta V}{\Delta r} = \frac{364\pi/3}{6-5} = 121.3\pi~\mu m^2$
(iii) $V(5) = \frac{4}{3}\pi~(5)^3 = \frac{500}{3}\pi~\mu m^3$
$V(5.1) = \frac{4}{3}\pi~(5.1)^3 = \frac{530.6}{3}\pi~\mu m^3$
We can find the average rate of change:
$\frac{\Delta V}{\Delta r} = \frac{30.6\pi/3}{5.1-5} = 102\pi~\mu m^2$
(b) $V(r) = \frac{4}{3}\pi ~r^3$
$\frac{dV}{dr} = 4\pi~r^2$
We can find the instantaneous rate of change when $r = 5~\mu m$:
$\frac{dV}{dr} = 4\pi~r^2 = 4\pi(5)^2 = 100\pi~\mu m^2$
When $r = 5~\mu m$, the instantaneous rate of change of the volume with respect to the radius is $~~100\pi~\mu m^2$
(c) The surface area of a circle is $4\pi~r^2$
Since $\frac{dV}{dr} = 4\pi~r^2$, the rate of change of the volume with respect to the radius is equal to the surface area.
Suppose the original volume of a circle is $\frac{4}{3}\pi~r^3$
After we increase the radius by $\Delta r$, the new volume is $\frac{4}{3}\pi~(r+\Delta r)^3$
We can find the change in volume:
$\Delta V = \frac{4}{3}\pi~(r+\Delta r)^3-\frac{4}{3}\pi~r^3$
$\Delta V = \frac{4}{3}\pi~r^3+4\pi~r^2~(\Delta r)+4\pi~r~(\Delta r)^2+\frac{4}{3}\pi(\Delta r)^3-\frac{4}{3}\pi~r^2$
$\Delta V = 4\pi~r^2~(\Delta r)+4\pi~r~(\Delta r)^2+\frac{4}{3}\pi(\Delta r)^3$
$\Delta V \approx 4\pi~r^2~(\Delta r)$
Note that we can say this because $4\pi~r~(\Delta r)^2+\frac{4}{3}\pi(\Delta r)^3 \approx 0$
Then:
$\Delta V \approx 4\pi~r^2~(\Delta r)$
$\frac{\Delta V}{\Delta r} \approx 4\pi~r^2$
