Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 13

a.) Determine the average rate of change of the area of a circle with respect to its radius $r$ as $r$ changes from
$(i) 2 \text{ to } 3 \qquad (ii) 2 \text{ to } 2.5 \qquad (iii) 2 \text{ to } 2.1$

b.) Find the instantaneous rate of change when $r=2$
c.) Show that the rate of change of the area of a circle with respect to its radius (at any $r$) is equal to the circimference of the circle.

Recall that the Area of the circle is $A = \pi r^2$ where $r$ is radius. where $r$ is radius
$(i)$ $r$ from 2 to 3
$\displaystyle \text{Average rate} = \frac{A(3) - A(2)}{3-2} = 5 \pi = 15.7080 \frac{\text{unit area}}{\text{unit radius}}$

$(ii)$ $r$ from 2 to 2.5
$\displaystyle \text{Average rate} = \frac{A(2.5) - A(2)}{2.5-2} = \frac{9}{2} \pi = 14.1372 \frac{\text{unit area}}{\text{unit radius}}$

$(iii)$ $r$ from 2 to 2.1
$\displaystyle \text{Average rate} = \frac{A(2.1) - A(2)}{2.1-2} = \frac{41}{10} \pi = 12.8805 \frac{\text{unit area}}{\text{unit radius}}$


$
\begin{equation}
\begin{aligned}
\text{b.) } \frac{dA}{dr} &= \pi \frac{d}{dr} (r^2)\\
\\
\frac{dA}{dr} &= \pi(2r)\\
\\
\frac{dA}{dr} &= 2\pi r
\end{aligned}
\end{equation}
$


when $r = 2$
$\displaystyle \frac{dA}{dr} = 2\pi(2)$
$\displaystyle \frac{dA}{dr} = 4\pi \frac{\text{unit area}}{\text{unit radius}}$

c.) Recall that the circumference of the circle is equal to $c = 2 \pi r$ and the area of the circle $A=\pi r^2$ has rate of change of $A'(r) = \pi \frac{d}{dr}(r^2) = \pi (2r)$. We have,
$A'(r) = 2 \pi r$ which equals the circumference of the circle.