SOLUTION 4: Draw a circle of radius $r$, and assume that $r$ is a function of time $t$.

$\ \ \ \$ a.) The circumference of a circle is $$C=2 \pi r$$ GIVEN: $\ \ \ \displaystyle{ dr \over dt } = 10 \ meters/min.$

FIND: $\ \ \ \displaystyle{ dC \over dt }$ when $r =20 \ m.$

Now differentiate the circumference equation with respect to time $t$ getting

$$D \{ C \} = D \{ 2 \pi r \} \ \ \ \longrightarrow$$ $$\displaystyle{ dC \over dt } = 2 \pi \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dr \over dt } = 10. \Big)$

$$\displaystyle{ dC \over dt } = 2 \pi (10) \ \ \ \longrightarrow$$ $$\displaystyle{ dC \over dt } = 20 \pi \ \ m/min. \approx 62.8 \ \ meters/min.$$

$\ \ \ \$ b.) The area of a circle is $$A = \pi r^2$$ GIVEN: $\ \ \ \displaystyle{ dr \over dt } = 10 \ meters/min.$

FIND: $\ \ \ \displaystyle{ dA \over dt }$ when $r =20 \ m.$

Now differentiate the area equation with respect to time $t$ getting

$$D \{ A \} = D \{ \pi r^2 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dA \over dt } = \pi \cdot 2 r \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dr \over dt } = 10$ and $r=20. \Big)$

$$\displaystyle{ dA \over dt } = 2 \pi (20)(10) = 400 \pi \ \ meters^2/min. \approx 1256.6 \ \ meters^2/min.$$