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}$