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Q22E

Expert-verifiedFound in: Page 139

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**The radius of a circular disk is given as \({\rm{24\;cm}}\) with a maximum error in measurement of \({\rm{0}}{\rm{.2\;cm}}\).**

**\(({\rm{a}})\) Use differentials to estimate the maximum error in the calculated area of the disk.**

**\(({\rm{b}})\) What is the relative error? What is the percentage error?**

Part \((a)\) The maximum error in the calculated area of the disk is \(9.6\pi c{m^2}\).

Part \((b)\) The relative error is \(0.0167\) and the percentage error is \(1.67\% \).

** **The radius of a circular disk is given as \(24\;cm\) with a maximum error in measurement of \(0.2\;cm\).

Calculus relies heavily on derivatives. The derivative of a function of a real variable measures a quantity's sensitivity to change as defined by another quantity.

Apply the formula of the surface area of the disk

\(\begin{array}{c}S = \pi {r^2}{\rm{ }}\\{S^\prime }(r) = 2\pi r\end{array}\)

Calculate the maximum error in \(S\)

\(\begin{array}{c}\Delta S \approx {S^\prime }(r)\Delta r\\ = 2\pi r\Delta r\\ = 2\pi (24)0.2\\ = 9.6\pi \end{array}\)

Apply the formula of the area of the circle

\(\begin{array}{c}A = \pi {r^2}\\ = \pi {(24\;cm)^2}\\ = 576\pi c{m^2}\end{array}\)

By definition, calculate the relative error

\(\begin{array}{c}\frac{{\Delta A}}{A} = \frac{{9.6\pi c{m^2}}}{{576\pi c{m^2}}}\\ = \frac{1}{{60}}\\ \approx 0.0167\end{array}\)

The percentage error is simply the relative error, which is rewritten in the percentage form. To rewrite a number in the percentage form, multiply that number by \(100\) Therefore

\(\begin{array}{c}0.0167 = 0.0167 \cdot 100\% \\ = 1.67\% \end{array}\)

Part \((a)\) The maximum error in the calculated area of the disk is \(9.6\pi c{m^2}\).

Part \((b)\) The relative error is \(0.0167\) and the percentage error is \(1.67\% \).

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