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The point $ P(1, 0) $ lies on the curve $ y = \sin (10\pi /x) $.

(a) If $ Q $ is the point $ (x, \sin (10\pi /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) 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 $.

A. (a) For the curve $y=\sin (10 \pi / x)$ and the point $P(1,0)$$$\begin{array}{|c|l|c|}\hline x & Q & m_{P Q} \\\hline 2 & (2,0) & 0 \\1.5 & (1.5,0.8660) & 1.7321 \\1.4 & (1.4,-0.4339) & -1.0847 \\1.3 & (1.3,-0.8230) & -2.7433 \\1.2 & (1.2,0.8660) & 4.3301 \\1.1 & (1.1,-0.2817) & -2.8173 \\\hline\end{array}$$$$\begin{array}{|c|l|c|}\hline x & Q & m_{P Q} \\\hline 0.5 & (0.5,0) & 0 \\0.6 & (0.6,0.8660) & -2.1651 \\0.7 & (0.7,0.7818) & -2.6061 \\0.8 & (0.8,1) & -5 \\0.9 & (0.9,-0.3420) & 3.4202 \\\hline\end{array}$$B. We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at $P$ that we need to take $x$ -values much closer to 1 in order to get accurate estimates of its slope.C.. If we choose $x=1.001,$ then the point $Q$ is (1.001,-0.0314) and $m p Q \approx-31.3794 .$ If $x=0.993,$ then $Q$ is (0.999,0.0314) and $m p Q=-31.4422 .$ The average of these slopes is -31.4108 .So we estimate that the slope of the tangent line at $P$ is about -31.4

08:11

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 1

The Tangent and Velocity Problems

Limits

Derivatives

Campbell University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

03:09

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Mhm. For this problem we have a .10 lying on the curve of y equals sign of 10 pie over X. Let's given right here, we know that Q is a point. We want to find the slope of the secret line When x equals to 1.5. So we can plug in these values. If we let this be ffx, We can plug in f of two 1.5, 1.4, 1.3 1 .21 .1. Um And we see that as we get closer. Okay? And we're approaching this value about one buddy keeps oscillating so there's not really a clear value that we're actually approaching.

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