Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 9

Given the curve $\displaystyle y= \sin \left(\frac{10 \pi}{x}\right)$ and the point $A(1,0)$ that lies on the curve.

(a) If $B$ is the point $\displaystyle \left( x, \sin \left(\frac{10 \pi}{x}\right) \right)$, find the slope of the secant line $AB$ (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?

$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|}
\hline\\
x & B_1 (x) & B_2 \left(\sin \left(\frac{10 \pi}{x}\right)\right) & A_1 & A_2 & \text{slope} = \frac{B_2 - A_2}{B_1 - A_1} \\
\hline\\
2 & 2 & 0 & 1 & 0 & 0 \\
\hline\\
1.5 & 1.5 & \frac{\sqrt{3}}{2} & 1 & 0 & \sqrt{3} \\
\hline\\
1.4 & 1.4 & -0.4339 & 1 & 0 & -1.0848 \\
\hline\\
1.3 & 1.3 & -0.8230 & 1 & 0 & -2.7433 \\
\hline\\
1.2 & 1.2 & \frac{\sqrt{3}}{2} & 1 & 0 & 4.3301 \\
\hline\\
1.1 & 1.1 & -0.2817 & 1 & 0 & -2.817 \\
\hline\\
0.5 & 0.5 & 0 & 1 & 0 & 0 \\
\hline\\
0.6 & 0.6 & \frac{\sqrt{3}}{2} & 1 & 0 & -2.1651 \\
\hline\\
0.7 & 0.7 & 0.7818 & 1 & 0 & -2.606 \\
\hline\\
0.8 & 0.8 & 1 & 1 & 0 & -5 \\
\hline\\
0.9 & 0.9 & -0.3420 & 1 & 0 & 3.42\\
\hline
\end{array}

\end{aligned}
\end{equation}
$


Based from the values we obtain in the table, the slopes of the secant lines do not approach any limit since their values
are not close to each other. Also, some lines have positive slope while most of the lines have negative slope.



(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 $A$.








Based from the graph of the function, the slopes of the secant line are not close to the slope of the tangent line at $A$
it's because of the infinite number of cycles/oscillations; thus, unlike a more linear or hyperbolic relationship,
slopes of the secant lines of the chosen points do not approach the slope of the tangent line at $A$ because they lie
on the part of the graph that is already going to a new cycle/oscillation than the one $A$ is in.

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







We let the secant line connects the point M(0.95,1) and A(1,0) as shown from the graph. Therefore, slope can be computed as..

$\displaystyle m = \frac{1-0}{0.95-1}$

$m = -20$