Question

The point \(P\left( {{\bf{1}},{\bf{0}}} \right)\) lies on the curve \(y = {\bf{sin}}\left( {\frac{{{\bf{10}}\pi }}{x}} \right)\).

a. If Qis the point \(\left( {x,{\bf{sin}}\left( {\frac{{{\bf{10}}\pi }}{x}} \right)} \right)\), find the slope of the secant line PQ (correct to four decimal places) for \(x = {\bf{2}}\), 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) are 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.

Short answer

(a) The table is shown below:

x	\(Q \equiv \left( {x,\sin \left( {\frac{{10\pi }}{x}} \right)} \right)\)	\({m_{PQ}}\)(slope)
2	\(\left( {2,0} \right)\)	0
1.5	\(\left( {1.5,0.8660} \right)\)	1.7321
1.4	\(\left( {1.4, - 0.4339} \right)\)	-1.0847
1.3	\(\left( {1.5, - 0.8230} \right)\)	-2.7433
1.2	\(\left( {1.2,0.8660} \right)\)	4.3301
1.1	\(\left( {1.1, - 0.2817} \right)\)	-2.8173
0.5	\(\left( {0.5,0} \right)\)	0
0.6	\(\left( {0.6,0.8660} \right)\)	-2.1651
0.7	\(\left( {0.7,0.7818} \right)\)	-2.6061
0.8	\(\left( {0.8,0.1} \right)\)	-5
0.9	\(\left( {0.9, - 0.3420} \right)\)	3.4202

The value of x is approaching one, whereas the slope is not approaching any particular value.

(b) The tangent is steep at P. We need to take x-values much closer to 1 in order to get accurate estimates.

(c) Slope of the tangent line is \( - 31.4\).

Step-by-step solution

Step 1:Find the slope of the secant line

For the curve \(y = \sin \left( {\frac{{10\pi }}{x}} \right)\), the slope of the secant is obtained as shown below:

For \(x = 2\),

\(\begin{aligned}{c}{m_{PQ}} &= \frac{{0 - 0}}{{2 - 1}}\\ &= 0\end{aligned}\)

For \(x = 1.5\),

\(\begin{aligned}{c}{m_{PQ}} &= \frac{{0 - 0.8660}}{{2 - 1.5}}\\ &= - 1.732\end{aligned}\)

The table below represents the slope of secant line PQ as shown:

x	\(Q \equiv \left( {x,\sin \left( {\frac{{10\pi }}{x}} \right)} \right)\)	\({m_{PQ}}\)(slope)
2	\(\left( {2,0} \right)\)	0
1.5	\(\left( {1.5,0.8660} \right)\)	-1.7321
1.4	\(\left( {1.4, - 0.4339} \right)\)	-1.0847
1.3	\(\left( {1.5, - 0.8230} \right)\)	-2.7433
1.2	\(\left( {1.2,0.8660} \right)\)	4.3301
1.1	\(\left( {1.1, - 0.2817} \right)\)	-2.8173
0.5	\(\left( {0.5,0} \right)\)	0
0.6	\(\left( {0.6,0.8660} \right)\)	-2.1651
0.7	\(\left( {0.7,0.7818} \right)\)	-2.6061
0.8	\(\left( {0.8,0.1} \right)\)	-5
0.9	\(\left( {0.9, - 0.3420} \right)\)	3.4202

The value of x is approaching one, but the slope is not approaching any particular value.

Explain the slopes of the secant line

The curve below represents the curve for \(y = \sin \left( {\frac{{10\pi }}{x}} \right)\) and the secant line from point P.

Use the following steps to plot the graph of given functions:

1.In the graphing calculator, select “STAT PLOT” and enter the equation \(\sin \left( {\frac{{10\pi }}{X}} \right)\).

2.Enter the graph button in the graphing calculator.

The tangent is steep at P. We need to take x-values much closer to 1 in order to get accurate estimates.

Estimate the slope at point P

For \(x = 1.001\), the point Q is \(\left( {1.001, - 0.0314} \right)\) and the slope is \({m_{PQ}} \approx - 31.3794\). Similarly, for \(x = 0.999\), the point Q is \({m_{PQ}} \approx - 31.4422\).

The average slope is \( - 31.4108\). Therefore, the slope of the tangent line at P is \( - 31.4\).