Answer
(a)
(i) $5~\pi$
(ii) $4.5~\pi$
(iii) $4.1~\pi$
(b) When $r = 2$, the instantaneous rate of change of the area with respect to the radius is $~~4\pi$
(c) The rate of change of the area with respect to the radius is equal to the circumference.
Work Step by Step
(a) $A(r) = \pi ~r^2$
$\frac{dA}{dr} = 2\pi~r$
(i) $A(2) = \pi(2)^2 = 4\pi$
$A(3) = \pi(3)^2 = 9\pi$
We can find the average rate of change:
$\frac{\Delta A}{\Delta r} = \frac{9\pi-4\pi}{3-2} = 5\pi$
(ii) $A(2) = \pi(2)^2 = 4\pi$
$A(2.5) = \pi(2.5)^2 = 6.25\pi$
We can find the average rate of change:
$\frac{\Delta A}{\Delta r} = \frac{6.25\pi-4\pi}{2.5-2} = 4.5\pi$
(iii) $A(2) = \pi(2)^2 = 4\pi$
$A(2.1) = \pi(2.1)^2 = 4.41\pi$
We can find the average rate of change:
$\frac{\Delta A}{\Delta r} = \frac{4.41\pi-4\pi}{2.1-2} = 4.1\pi$
(b) We can find the instantaneous rate of change when $r = 2$:
$\frac{dA}{dr} = 2\pi~r = 2\pi(2) = 4\pi$
When $r = 2$, the instantaneous rate of change of the area with respect to the radius is $~~4\pi$
(c) The circumference of a circle is $2\pi~r$
Since $\frac{dA}{dr} = 2\pi~r$, the rate of change of the area with respect to the radius is equal to the circumference.
Suppose the original area of a circle is $\pi~r^2$
After we increase the radius by $\Delta r$, the new area is $\pi~(r+\Delta r)^2$
We can find the change in area:
$\Delta A = \pi~(r+\Delta r)^2-\pi~r^2$
$\Delta A = \pi~r^2+2\pi~r~(\Delta r)+\pi(\Delta r)^2-\pi~r^2$
$\Delta A = 2\pi~r~(\Delta r)+\pi(\Delta r)^2$
$\Delta A \approx 2\pi~r~(\Delta r)$
Note that we can say this because $\pi(\Delta r)^2 \approx 0$
Then:
$\Delta A \approx 2\pi~r~(\Delta r)$
$\frac{\Delta A}{\Delta r} \approx 2\pi~r$
