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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

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given F of X which is equal to sign of tent by over X which passes through the point P Which is coordinates 10. We need to find the following. 1st the slope of the second lines through the X values. And then we need to explain why these slopes of the second lines in part A are not close to the slope of the tangent line at P. And lastly we need to choose appropriate sequence lines to estimate the slope at P. For the first one to find the slope of the 2nd line. We used the formula for the slope of the line which is just the change and why over the change in X, which in this case will be F of x minus ever won over. That will be x minus one. Since f of one is zero. Then we have F of X. All over X -1 is the slope of the second line. So we will use this formula to find me slope of the second lines. Through that these given X values. Now for us to see a pattern, we will do a table and in the stable we will write our X values in order So for extra 0.5 we have signs of 10 pie Over 0.5 that's divided by 0.5 -1. We have Value equal to zero And if it's 0.6 we have Sign of 10 by over 0.6 Divided by 0.6 -1. We get -2.1651 At X equals 0.7. We have signed up 10 pi over 0.7 over 0.7 -1. That gives us -2.6061 at X 0.8 We have signs of 0.8 Rather 10 pi over 0.8 Over 0.8 -1. This gives us -5. And if it's your .9 we have sign of 10 by Over 0.9 over 0.9 -1. This gives us 3.4-0 20. If access 1.1 We have signs of 10 pi over 1.1 divided by 1.1 -1, we have negative two point 8173. and if X is 1.2 we have signs of 10 pi over 1.2, All over 1.2 -1, that gives us 4.3301. Yes Access 1.3 We have signs of 10 pi over 1.3 Over 1.3 -1. This gives us -2.7433. and if it's 1.4 we have a sign of Then by over 1.4 divided by 1.4 -1 we have -1.0847. And if it's 1.5 we have sign of 10 by Over 1.5 over 1.5 -1. This is just one point 7321. And lastly if it's too we have signs of 10 by Over to over 2 -1, we have Slope equal to zero Know that as we move closer to x equals one. The value of the slope is not really approaching a value close to each other. So we're not really seeing a pattern here. So to estimate the slope of the tangent line at P. Let's look at the graph and from there, choose the correct points to form the second lines, the lowest a graph of the function. And here we see that the function is changing rapidly from increasing to decreasing. And so this is the reason why we're not seeing a pattern for the slope of the tangent line. And so for part C to choose the correct points, we will use the X values just right here. That will be X values in between 0.95 and 1.05. Let's say X is 0.9 brian 0.999. And then for X values right of one, let's say we have one point 0001 And then 1.001. And then from here we find these slopes of the second lines. We do the same process as what they did in party. We make a table and then find the slope there below is the table of values now, yes, X 0.99 We have signs of 10 by over 0.99 over 0.99 -1. We get negative 31 point 2033. And if it's 0.999, we have Sign of 10 by over 3.999 Over 0.999 -1. This gives us -31.4422. Now, if X is 1.0001, we have sign of 10 by over 1.0001 Over 1.00 1 -1. That gives us -31.41 27 At X equals 1.001. We have sign of 10 pies over 1.001 Over 1.001 -1. We get -31.3794. Therefore the slope of the tangent line at p is approximately -31.4. This is because as we Move closer to one, The value of the slope is approaching negative 31.4