Chapter 2.1, Problem 9E

The point P(1, 0) lies on the curve y = sin(l0π/x).

(a) If Q is the point (x, sin(10π/x)), find the slope of the secant line PQ (correct to four decimal places) for x = 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5. 0.6, 0.7, 0 .8, and 0.9.

Do the slopes appear to be approaching a limit?

(b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) arc not close to the slope of the tangent line at P.

(c) By choosing appropriate secant lines, estimate the slope of the tangent line at P.

To determine

To find: The slope of the secant line PQ for the given values of x.

Answer to Problem 9E

The slope of the secant line PQ for the following values of x is given below:

x	$\text{Slope }{m}_{PQ}$
2	0
1.5	1.7321
1.4	−1.0847
1.3	−2.7433
1.2	−4.3301
1.1	−2.8173
0.5	0
0.6	−2.1651
0.7	−2.6061
0.8	−5
0.9	3.4202

Explanation of Solution

Given:

The equation of the curve is $y=\sin \left(\frac{10\pi }{x}\right)$.

The point P(1, 0) lies on the curve y.

The Q is the point $\left(x,\sin \left(\frac{10\pi }{x}\right)\right)$.

Calculation:

The slope of the secant lines between the points, P(1, 0) and Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)$ is

$m_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}$ (1)

Obtain the slope of the secant line PQ for the value of $x=2$.

Substitute 2 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{2}\right)\\ =\sin \left(5\pi \right)\\ =0\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(2,0\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0-0}{2-1}\\ =\frac{0}{1}\\ =0\end{array}$

Thus, the slope of the secant line PQ for the value of $x=2$ is $0$.

Obtain the slope of the secant line PQ for the value of $x=1.5$.

Substitute 1.5 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.5}\right)\\ =\sin \left(\frac{20}{3}\pi \right)\\ =0.866025\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.5,0.866025\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0.866025-0}{1.5-1}\\ =\frac{0.866025}{0.5}\\ \approx 1.7321\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.5$ is $1.7321$.

Obtain the slope of the secant line PQ for the value of $x=1.4$.

Substitute 1.4 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.4}\right)\\ =-0.43388\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.4,-0.43388\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{-0.43388-0}{1.4-1}\\ =\frac{-0.43388}{0.4}\\ \approx -1.0847\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.4$ is $-1.0847$.

Obtain the slope of the secant line PQ for the value of $x=1.3$.

Substitute 1.3 for x in $\sin \left(\frac{10\pi }{x}\right)$.

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.3}\right)\\ =-0.82298\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.3,-0.82298\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{-0.82298-0}{1.3-1}\\ =\frac{-0.82298}{0.3}\\ \approx -2.7433\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.3$ is $-2.7433$.

Obtain the slope of the secant line PQ for the value of $x=1.2$.

Substitute 1.2 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.2}\right)\\ =0.86602\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.2,0.86602\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0.86602-0}{1.2-1}\\ =\frac{0.86602}{0.2}\\ \approx 4.3301\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.2$ is $4.3301$.

Obtain the slope of the secant line PQ for the value of $x=1.1$.

Substitute 1.1 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.1}\right)\\ =-0.28173\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.1,-0.28173\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{-0.28173-0}{1.1-1}\\ =\frac{-0.28173}{0.1}\\ \approx -2.8173\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.1$ is $-2.8173$.

Obtain the slope of the secant line PQ for the value of $x=0.5$.

Substitute 0.5 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.5}\right)\\ =\sin \left(20\pi \right)\\ =0\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.5,0\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0-0}{0.5-1}\\ =\frac{0}{-0.5}\\ \approx 0\end{array}$

Thus, the slope of the secant line PQ for the value of $x=0.5$ is $0$.

Obtain the slope of the secant line PQ for the value of $x=0.6$.

Substitute 0.5 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.6}\right)\\ =0.86603\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.6,0.86603\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0.86603-0}{0.6-1}\\ =\frac{0.86603}{-0.4}\\ \approx -2.1651\end{array}$

Thus, the slope of the secant line PQ for the value of $x=0.6$ is $-2.1651$.

Obtain the slope of the secant line PQ for the value of $x=0.7$.

Substitute 0.7 for x in $\sin \left(\frac{10\pi }{x}\right)$.

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.7}\right)\\ =0.78183\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.7,0.78183\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0.78183-0}{0.7-1}\\ =\frac{0.78183}{-0.3}\\ \approx -2.6061\end{array}$

Thus, the slope of the secant line PQ  for the value of $x=0.7$ is $-2.6061$.

Obtain the slope of the secant line PQ  for the value of $x=0.8$.

Substitute 0.8 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.8}\right)\\ =1\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.8,1\right)$ in equation (1).,

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{1-0}{0.8-1}\\ =\frac{1}{-0.2}\\ \approx -5\end{array}$

Thus, the slope of the secant line PQ for the value of $x=0.8$ is $-5$.

Obtain the slope of the secant line PQ for the value of $x=0.9$.

Substitute 0.9 for x in $\sin \left(\frac{10\pi }{x}\right)$,

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.9}\right)\\ =-0.34202\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.9,-0.34202\right)$ in equation (1),

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{-0.34202-0}{0.9-1}\\ =\frac{-0.34202}{-0.1}\\ \approx 3.4202\end{array}$

Thus, the slope of the secant line PQ for the value of $x=0.9$ is $3.4202$.

Conclusion:

The slope does not appear to be approaching a limit. Suppose x approaches 1, then the slope $m_{PQ}$ do not approach  any specific value.

To determine

To explain: The slopes of the secant lines in part (a) are not close to the slope of the tangent line at P by using a graph.

Explanation of Solution

The graph of the curve $y=\sin \left(\frac{10\pi }{x}\right)$ is shown below in Figure 1.

From Figure 1, it is observed that there seems to be the frequent oscillations of the graph. Moreover, the tangent line is so steep at the point P(1, 0). Thus, the slopes of the secant lines are not closer to the slope of the tangent line at P. Therefore, it is necessary to consider the values of x much closer to 1 for better accurate estimates of the slope.

To determine

To estimate: The slope of the tangent line to the curve at P(1, 0).

Answer to Problem 9E

The estimated value of the slope of the tangent line to the curve at P (1, 0) is −31.4.

Explanation of Solution

The graph of the curve $y=\sin \left(\frac{10\pi }{x}\right)$ between the points 0.5 and 2 is shown below in Figure 2.

The secant line is close to the tangent line at P(1, 0) when $x=0.999$ and $x=1.001$.

The value of the slope of the tangent line to the curve at $P\left(1,0\right)$ is close to the average value of the slope of the secant lines closest to P.

Obtain the slope of the secant line PQ for the value of $x=1.001$.

Substitute 1.001 for x in $\sin \left(\frac{10\pi }{x}\right)$.

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{1.001}\right)\\ =-0.031379\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(1.001,-0.031379\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{-0.0313794-0}{1.001-1}\\ =\frac{-0.0313794}{0.001}\\ \approx -31.3794\end{array}$

Thus, the slope of the secant line PQ for the value of $x=1.001$ is $-31.3794$.

Obtain the slope of the secant line PQ for the value of $x=0.999$.

Substitute 0.999 for x in $\sin \left(\frac{10\pi }{x}\right)$.

$\begin{array}{c}\sin \left(\frac{10\pi }{x}\right)=\sin \left(\frac{10\pi }{0.999}\right)\\ =0.03144219\end{array}$

Substitute Q$\left(x,\sin \left(\frac{10\pi }{x}\right)\right)=\left(0.999,0.03144219\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\sin \left(\frac{10\pi }{x}\right)-0}{x-1}\\ =\frac{0.03144219-0}{0.999-1}\\ =\frac{0.03144219}{-0.001}\\ \approx -31.4422\end{array}$

Thus, the slope of the secant line PQ for the value of $x=0.999$ is $-31.4422$.

The slope of the secant line PQ for the value of $x=1.001$ is $-31.3794$ and the slope of the secant line PQ for the value of $x=0.999$ is $-31.4422$.

Take the average of two slopes of the secant lines,

$\begin{array}{c}m\approx \frac{-31.3794+\left(-31.4422\right)}{2}\\ \approx \frac{-62.8216}{2}\\ \approx -31.4108\\ \approx -31.4\end{array}$

Thus, the estimated value of the slope of the tangent line to the curve at P (1, 0) is −31.4.