SOLUTION 23: Consider a circle of radius $r$. Then its circumference is $$C = 2 \pi r \ \ \ \ \longrightarrow \ \ \ \ C'= 2 \pi$$ and its area is $$A= (length)(width) = x \cdot x = x^2 \ \ \ \ \longrightarrow \ \ \ \ A'=2x$$

GIVEN: The absolute percentage error of $x$ is $$\displaystyle{ |\Delta x| \over x } \le \ 3\%$$
$\ \ \ \$ a.) Use a Differential to estimate the absolute percentage error in perimter, i.e., estimate $\ \displaystyle{ |\Delta P| \over P }.$ Then $$\displaystyle{ |\Delta P| \over P } \approx { | dP| \over P } = { | P' \cdot \Delta x| \over P } = { | 4 \cdot \Delta x| \over 4x } = { | \cdot \Delta x| \over x } \le \ 3\%$$ i.e., $$\displaystyle{ |\Delta P| \over P } \le \ 3\%$$
$\ \ \ \$ b.) Use a Differential to estimate the absolute percentage error in area, i.e., estimate $\ \displaystyle{ |\Delta A| \over A }.$ Then $$\displaystyle{ |\Delta A| \over A } \approx { | dA| \over A } = { | A' \cdot \Delta x| \over A } = { | 2x \cdot \Delta x| \over x^2 } = 2{ |\Delta x| \over x } \le \ 2 (3\%) = \ 6\%$$ i.e., $$\displaystyle{ |\Delta A| \over A } \le \ 6\%$$