Question

A machinist is required to manufacture a circular metal disk with area\(1000\;{\rm{c}}{{\rm{m}}^2}\).

(a) What radius produces such a disk?

(b) If the machinist is allowed an error tolerance of \( \pm 5\;{\rm{c}}{{\rm{m}}^2}\)in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius?

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

Short answer

(a) The radius of the disc is\(\sqrt {\frac{{1000}}{\pi }} \;{\rm{cm}}\).

(b) The mechanist must control the radius by approximately\(0.0445\;{\rm{cm}}\).

(c) The values are\(x = r\), \(f\left( x \right) = \pi {x^2}\), \(a = \sqrt {\frac{{1000}}{\pi }} \), \(L = 1000\), \(\varepsilon = 5\), and \(\delta = 0.0445\).

Step-by-step solution

Area of the circular disc(a)

The area of the circular disc is given by,

\(A = \pi {r^2}\)

The given area is\(1000\;{\rm{c}}{{\rm{m}}^2}\).

Find the radius of the disc

Substitute all the values in the area formula to calculate the area of the disc.

\(\begin{array}{c}1000 = \pi {r^2}\\{r^2} = \frac{{1000}}{\pi }\\r = \sqrt {\frac{{1000}}{\pi }} \;{\rm{cm}}\end{array}\)

Therefore, the radius of the disc is\(\sqrt {\frac{{1000}}{\pi }} \;{\rm{cm}}\).

(b) Step 3: Find the variation in radius

For the upper end of the range, the area is,

\(\begin{array}{c}A = 1000 + 5\\ = 1005\;{\rm{c}}{{\rm{m}}^2}\end{array}\)

Find the radius as follows.

\(\begin{array}{c}{r^2} = \frac{{1005}}{\pi }\\r = \sqrt {\frac{{1005}}{\pi }} \;{\rm{cm}}\end{array}\)

Calculate the difference between the ideal radius and this radius.

\(\sqrt {\frac{{1005}}{\pi }} - \sqrt {\frac{{1000}}{\pi }} \approx 0.04454748797\)

For the lower end of the range, the area is,

\(\begin{array}{c}A = 1000 - 5\\ = 995\;{\rm{c}}{{\rm{m}}^2}\end{array}\)

Find the radius as follows.

\(\begin{array}{c}{r^2} = \frac{{995}}{\pi }\\r = \sqrt {\frac{{995}}{\pi }} \;{\rm{cm}}\end{array}\)

Calculate the difference between the ideal radius and this radius.

\(\sqrt {\frac{{995}}{\pi }} - \sqrt {\frac{{1000}}{\pi }} \approx 0.0446589966\)

Therefore, the mechanism must control the radius by approximately\(0.0445\;{\rm{cm}}\).

(c) Step 3: Find the values of x,\(f\left( x \right)\), a, L, \(\varepsilon \), and the corresponding value of \(\delta \)

From the above result, all the characteristics of\(\mathop {\lim }\limits_{r \to \sqrt {\frac{{1000}}{\pi }} } A = 1000\). That is to say, for\(\varepsilon = 5\), there exists a \(\delta = 0.0445\) such that if \(\left| {r - \sqrt {\frac{{1000}}{\pi }} } \right| < 0.0445\) then\(\left| {A - 1000} \right| < 5\).

The \(x\) will be equal to the radius of the disc. So,

\(x = r\)

The function will be equal to the area of the disc. So,

\(f\left( x \right) = \pi {x^2}\)

The value of \(a\) will be equal to\(a = \sqrt {\frac{{1000}}{\pi }} \).

The value of \(L\) will be equal to the target area. So,

\(L = 1000\)

The value of \(\varepsilon \) will be equal to the tolerance. So,

\(\varepsilon = 5\)

The value of \(\delta \) corresponding to \(\varepsilon \) is equal to\(0.0445\).

Therefore, the values are\(x = r\),\(f\left( x \right) = \pi {x^2}\),\(a = \sqrt {\frac{{1000}}{\pi }} \),\(L = 1000\),\(\varepsilon = 5\), and\(\delta = 0.0445\).