A Machinist's Imperfect Disk Exercise

A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.

*

*What radius produces such a disk?


*If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?


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


Solution
Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).

1. What radius produces such a disk?

$\pi r_0^2 = 1000 \implies r_0 = \sqrt{\frac{1000}{\pi}} \approx 17.8412$

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

$a = r_0 \approx 17.8412$
$L = \pi r_0^2 = 1000$
$x = r$
$f(x) = \pi r^2$
$|r - r_0| < \delta$
$|\pi r^2 - \pi r_0^2| < \epsilon$

2. If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?

$|\pi r^2 - \pi r_0^2| < \epsilon$
$\implies |\pi r^2 - 1000| < 5$
$\implies -5 < \pi r^2 - 1000 < 5$
$\implies 995 < \pi r^2 < 1005$
$\implies \frac{995}{\pi} < r^2 < \frac{1005}{\pi}$
$\implies \sqrt{\frac{995}{\pi}} < r < \sqrt{\frac{1005}{\pi}}$
$\implies 17.7966 < r < 17.8858$
$|r - r_0| < \delta \implies |r - 17.8412| < \delta$
$|17.7966 - 17.8412| < \delta \implies 0.0446 < \delta$
$|17.8858 - 17.8412| < \delta \implies 0.0446 < \delta$
$\delta = \min(0.0446, 0.0446) = 0.0446$

Answer

1. What radius produces such a disk?

$$r_0 = 17.8412 \text{ cm}$$

2. If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?

$$\delta = 0.0446 \text{ cm}$$

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

$$|r - r_0| < \delta$$
$$|\pi r^2 - \pi r_0^2| < \epsilon$$

Request
Is my answer correct? If not, in what part of my solution did I make a mistake?
 A: In part 3 you are asked about four items: $x$, $f(x)$, $\epsilon$, and $\delta$. Your answer omits these. To make it clear:


*

*The independent variable $x$ here is the radius $r$ of the disk

*The function $f$ maps radius to area, i.e., $r\mapsto \pi r^2$. In other words, $f(x)=\pi x^2$.

*The given $\epsilon$ is $5\,\text{cm}^2$

*The corresponding (or rather: a suitable) value of $\delta$ is the $\approx 0.0446\,\text{cm}$ you found


On a sidenote, we may observe that picking $\delta$ such that $\frac\epsilon\delta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2\pi x$, so we quickly obtain $\delta=\frac{\epsilon}{2\pi r_0}\approx 0.0446$. 
Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $\frac 5{1000}=\frac1{200}$, the relative input error should be bounded by $\frac{1}{400}$, and indeed $\frac1{400}\cdot 17.8412\approx 0.0446$.
