The point P(1, 0) lies on the curve y = sin(10π/x) . (a) If Q is the point (x, sin(10π/x)) , fin

The point P(1, 0) lies on the curve y = sin(10π/x) . (a) If...

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 $.

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 $.

Transcript

00:01 Given f of x which is equal to sign of tent by over x which passes through the point p which is coordinates 10.

00:08 We need to find the following.

00:09 1st the slope of the second lines through the x values.

00:14 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.

00:23 And lastly we need to choose appropriate sequence lines to estimate the slope at p.

00:30 For the first one to find the slope of the 2nd line.

00:34 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.

00:49 That will be x minus one.

00:52 Since f of one is zero.

00:54 Then we have f of x.

00:56 All over x -1 is the slope of the second line.

01:00 So we will use this formula to find me slope of the second lines.

01:05 Through that these given x values.

01:09 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.

01:31 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.

01:43 We get -2.1651 at x equals 0.7.

01:50 We have signed up 10 pi over 0.7 over 0.7 -1.

01:57 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.

02:14 This gives us -5.

02:17 And if it's your .9 we have sign of 10 by over 0.9 over 0.9 -1.

02:27 This gives us 3.4-0 20.

02:35 If access 1.1 we have signs of 10 pi over 1.1 divided by 1.1 -1, we have negative two point 8173.

02:51 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.

03:05 Yes access 1.3 we have signs of 10 pi over 1.3 over 1.3 -1.

03:13 This gives us -2.7433...

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