Chapter 3.9, Problem 32E

(a)

To determine

To estimate: The error in computing the length of the hypotenuse by using differential.

Answer to Problem 32E

The maximum error to computing the length of the hypotenuse is $dx\approx \pm 1.21\text{ cm}$.

Explanation of Solution

Given:

One side of the triangle is 20 cm and opposite angle is $30°$ with error is $\pm 1°$.

Calculation:

Obtain the hypotenuse of the triangle.

Form Figure 1, it is observed that

$\begin{array}{l}\sin \theta =\frac{20}{x}\\ x=\frac{20}{\sin \theta }\\ x=20\csc \theta \end{array}$

The differential is $dx=f^{\prime }\left(\theta \right)d\theta$.

The derivative of the function $f\left(\theta \right)=20\csc \theta$ is computed as follows,

$\begin{array}{c}{f}^{\prime }\left(\theta \right)=\frac{d}{d\theta }\left(\csc \theta \right)\\ =20\frac{d}{d\theta }\left(\csc \theta \right)\\ =20\left(-\cot \theta \csc \theta \right)\\ =-20\cot \theta \csc \theta \end{array}$

Substitute the $f^{\prime }\left(\theta \right)=-20\cot \theta \csc \theta$ in $dx=f^{\prime }\left(\theta \right)d\theta$,

$dx=\left(-20\cot \theta \csc \theta \right)d\theta$

Substitute the value $\theta =30°\text{ and }d\theta =\pm 1°$,

$\begin{array}{c}dx=\left(-20\cot 30°\csc 30°\right)\left(\pm 1°\right)\\ =-20\left(\sqrt{3}\right)\left(2\right)\left(\pm \frac{\pi }{180}\right)\left(Q1°=\frac{\pi }{180}\right)\\ =\pm \frac{2\sqrt{3}\pi }{9}\\ \approx \pm 1.21\text{ cm}\end{array}$

Therefore, the maximum error to computing the length of the hypotenuse is $dx\approx \pm 1.21\text{ cm}$.

(b)

To determine

To find: The percentage error.

Answer to Problem 32E

The percentage error is $\pm 3\%$.

Explanation of Solution

Calculation:

Obtain the percentage error.

The relative error $\frac{\Delta x}{x}$ is computed as follows,

$\frac{\Delta x}{x}\approx \frac{dx}{x}$

Substitute the value $dx\approx \pm 1.21\text{ cm}$ (from part (a)) and $x=20\csc 30°$,

$\begin{array}{c}\frac{\Delta x}{x}\approx \frac{\pm 1.21}{20\csc 30°}\\ =\frac{\pm 1.21}{20\left(2\right)}\\ =\pm 0.03\end{array}$

Thus, the relative error is $\pm 0.03$.

Since the percentage error is product of relative error and 100%, $\pm 0.03\times 100\%=3\%$.

Therefore, the percentage error is $\pm 3\%$.