Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 11

Suppose that an engineer is required to manufacture a circular metal disk with area $1000 cm^2$

(a) What radius produces such disk?
Recall that,

$
\begin{equation}
\begin{aligned}

A =& \pi r^2 \qquad ; \text{ where } && A = \text{ area of the circle } \\

& && r = \text{ radius }\\

1000 =& \pi r^2 && \\

r =& \sqrt{\frac{1000}{\pi}} &&= 17.8412 cm


\end{aligned}
\end{equation}
$



(b) If the allowed tolerance in the area of the disk is $\pm 5 cm^2$, how close to the ideal radius in part (a) must the engineer control the radius?


$
\begin{equation}
\begin{aligned}

A_1 =& \pi r^2
&& A = \pi r^2\\
1000 + 5 =& \pi r_1^2
&& 1000-5 = \pi r_2^2\\
r_1 =& \sqrt{\frac{1005}{\pi}} = 17.8858 cm
&& r_2 = \sqrt{\frac{995}{\pi}} = 17.7966 cm

\end{aligned}
\end{equation}
$


The allowed radius should be the closer to the ideal value which is $17.8858 cm$ and the tolerance can be computed as 17.8858-17.8412 = 0.0446 $cm$.
Therefore, in order to fit in the error tolerance of $\pm 5 cm^2$ in the area of the disk, the engineer should only have $a \pm 0.04 cm$ tolerance for the radius

(c) In terms if the $\varepsilon, \delta$ definition of $\lim \limits_{x \to a} f(x) = L$, what is $x$? What is $f(x)$?
What is $a$? What is $L$? What value of $\varepsilon$ is given? What is the corresponding value of $\delta$?

In terms of the precise definition of a limit


$
\begin{equation}
\begin{aligned}
& x && \text{ corresponds to the radius}\\
& f(x) && \text{ for Area} \\
& a && \text{ is the ideal radius}\\
& L && \text{ is the 1000 $cm^2$}\\
& \varepsilon && \text{ is for the tolerance $\pm 5cm^2$ in the area }\\
& \delta && \text{ is for the tolerance $\pm 0.04 cm$ in the radius}

\end{aligned}
\end{equation}
$