Question

In physics textbooks, the period \(T\) of a pendulum of length \(L\) is often given as \(T \approx 2\pi \sqrt {L/g} \), provided that the pendulum swings through a relatively small arc. In the course of deriving this formula, the equation \({a_T} = - g\sin \theta \) for the tangential acceleration of the bob of the pendulum is obtained, and then \(\sin \theta \) is replaced by \(\theta \) with the remark that for small angles, \(\theta \) (in radians) is very close to \(\sin \theta \).

(a) Verify the linear approximation at 0 for the sine function:

\(\sin \theta \approx \theta \)

(b) If \(\theta = \pi /18\) (equivalent to 10°) and we approximate \(\sin \theta \)by \(\theta \), what is the percentage error?

(c) Use a graph to determine the values of \(\theta \) for which \(\sin \theta \) and \(\theta \) differ by less than 2%. What are the values in degrees?

Short answer

(a) The function \(f\left( \theta \right) = \sin \theta \) is approximated as \(f\left( \theta \right) \approx \theta \).

(b) The percentage error is \(0.51\% \).

(c) The two functions have a relative error less than 2% for \( - 19.7^\circ < \theta < 19.7^\circ \)

Step-by-step solution

Find the linearization of function \(f\left( \theta  \right) = \sin \theta \)

It is given that \(f\left( \theta \right) = \sin \theta \) and \(f'\left( \theta \right) = \cos \theta \). Thus,\(f\left( 0 \right) = 0\) and \(f'\left( 0 \right) = 1\). The linearization of the function \(f\left( \theta \right)\)at \(\theta = 0\) is written as \(L\left( \theta \right) = f\left( 0 \right) + f'\left( 0 \right)\left( {\theta - 0} \right)\).On plugging the values of \(f'\left( 0 \right)\), \(f\left( 0 \right)\) in \(L\left( \theta \right)\) and simplifying, we obtain :

^.\(\begin{aligned}{c}L\left( \theta \right) &= f\left( 0 \right) + f'\left( 0 \right)\left( {\theta - 0} \right)\\ &= 0 + 1\left( {\theta - 0} \right)\\ &= \theta \end{aligned}\)

Thus around \(\theta = 0\), the function \(f\left( \theta \right) = \sin \theta \) is approximated as \(f\left( \theta \right) \approx \theta \)

Find the relative error

The relative error in the above approximation for any value around\(\theta = 0\)is given as \(\frac{{\theta - \sin \theta }}{{\sin \theta }}\)on plugging \(\theta = \pi /18\) in \(\frac{{\theta - \sin \theta }}{{\sin \theta }}\)and simplifying, we obtain as follows:

\(\frac{{\pi /18 - \sin \left( {\pi /18} \right)}}{{\sin \left( {\pi /18} \right)}} \approx 0.0051\)

Multiplying relative error with 100 to obtain percentage error as:

\(0.0051\left( {100} \right) \approx 0.51\% \).

Find the inequality for at most 2% relative error

For the relative error, in the approximation of\(f\left( \theta \right) = \sin \theta \), to be less than 2%, \(\left| {\frac{{\theta - \sin \theta }}{{\sin \theta }}} \right| < 0.02\) condition must be satisfied. On simplifying it, we obtain following condition:

\(\begin{aligned}{c}\left| {\frac{{\theta - \sin \theta }}{{\sin \theta }}} \right| < 0.02 &= - 0.02 < \frac{{\theta - \sin \theta }}{{\sin \theta }} < 0.02\\ &= \left\{ \begin{aligned}{l} - 0.02\sin \theta < \theta - \sin \theta < 0.02\sin \theta ,\,\,\,\,\sin \theta > 0\\ - 0.02\sin \theta > \theta - \sin \theta > 0.02\sin \theta ,\,\,\,\,\sin \theta < 0\end{aligned} \right\}\\ &= \left\{ \begin{aligned}{l}0.98\sin \theta < \theta < 1.02\sin \theta ,\,\,\,\,\sin \theta > 0\\1.02\sin \theta < \theta < 0.98\sin \theta ,\,\,\,\,\sin \theta < 0\end{aligned} \right\}\end{aligned}\)

Draw the graph of \(f\left( \theta  \right) = \theta \)and \(f\left( \theta  \right) = \sin \theta \)

Draw the graph of the functions\(f\left( \theta \right) = \sin \theta \)and\(f\left( \theta \right) = \theta \)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\sin X\)in the\({Y_1}\)tab,\(X\)in the\({Y_2}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.
3. Change the “Window” to obtain three different graphs in three viewing windows.

Visualization of graphs of the functions \(f\left( \theta \right) = \sin \theta \)and \(f\left( \theta \right) = \theta \) is shown below:

Analyze the graphs

In the first viewing window, the graphs of two functions are approximated to be equal near \(\theta = 0\). In the second viewing window, the graph of functions \(f\left( \theta \right) = 1.02\sin \theta \)and \(f\left( \theta \right) = \theta \)intersects at around \(\theta \approx 0.344\) and \(\theta \approx - 0.344\)functions due to symmetry, which can be seen in the third viewing window.

Draw the conclusion

As\(f\left( \theta \right) = 1.02\sin \theta \)and\(f\left( \theta \right) = \theta \)intersects at around\(\theta \approx 0.344\)and\(\theta \approx - 0.344\), so, the two functions have a relative error less than 2% for\( - 0.344 < \theta < 0.344\). After converting this range of\(\theta \approx 0.344\)in degrees, we have \( - 0.344\left( {180^\circ /\pi } \right) < \theta < 0.344\left( {180^\circ /\pi } \right)\), which is equal to \( - 19.7^\circ < \theta < 19.7^\circ \).