Question

A machinist is required to manufacture a circular metal disk with area 1000 cm².

a. What radius produces such a disk?

b. If the machinist is allowed an error tolerance of \( \pm {\bf{5}}\,{\bf{c}}{{\bf{m}}^{\bf{2}}}\) 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 \(\varepsilon \), \(\delta \) definition of \(\mathop {{\bf{lim}}}\limits_{x \to a} f\left( x \right) = L\), 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 17.8412 cm.

b. The error tolerance in the disk area is 0.0445 cm.

c. If x represents the radius, then \(f\left( x \right)\) is the area. arepresents radius for which the disk area will be \(1000\;{\rm{c}}{{\rm{m}}^2}\). \(\varepsilon \) is the error tolerance in the area of the disk and \(\delta \) is the tolerance in the radius as in part (b).

Step-by-step solution

Step 1:Find the value of the radius

The equation gives the area of the disk:

\(\begin{aligned}A &= \pi {r^2}\\1000 &= \pi {r^2}\\r &= \sqrt {\frac{{1000}}{\pi }} \\ \approx 17.8412\end{aligned}\)

So, the radius of the disk is approximate\(17.8412\;{\rm{cm}}\).

Find the answer for part (b)

As the machinist is allowed an error of tolerance \( \pm 5\;{\rm{c}}{{\rm{m}}^2}\)in the area of the disk, so the variation in the radius is computed as shown below:

\(\begin{aligned}\left| {A - 1000} \right| \le 5\\ - 5 \le \pi {r^2} - 1000 \le 5\\995 \le \pi {r^2} \le 1005\\\sqrt {\frac{{995}}{\pi }} \le r \le \sqrt {\frac{{1005}}{\pi }} \\17.8010 \le r \le 17.8458\end{aligned}\)

So, the variation in the radius of the disk is \(\left( {17.8458 - 17.8010} \right)\), i.e. \(0.0445\;{\rm{cm}}\).

Therefore, if the machinist gets a radius within 0.0445 cm of 17.8412, the area will be within \(5\;{\rm{c}}{{\rm{m}}^2}\) of 1000.

Find the answer for part (c)

If x represents the radius, then \(f\left( x \right)\) is the area. arepresents radius for which the area of the disk will be \(1000\;{\rm{c}}{{\rm{m}}^2}\). \(\varepsilon \) is the error tolerance in the area of the disk and \(\delta \) is the tolerance in radius