The point P(1, 0) lies on the curve y = sin(14𝜋/x). (a) If Q is the point (x, sin(14𝜋/x)), find the slope of the secant line PQ (correct to four decimal places) for the following values of x: 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, 0.9. part b: By choosing appropriate secant lines, estimate the slope of the tangent line at P. (Round your answer to two decimal places.)

Let's approach the problem step by step.

### Part (a): Calculating the Slope of the Secant Line PQ

The slope of the secant line PQ is given by the formula:

\[
m_{\text{sec}} = \frac{y_2 - y_1}{x_2 - x_1}
\]

Here:
- \( P(1, 0) \) is the fixed point.
- \( Q(x, \sin(14\pi/x)) \) is the varying point.

So, the slope of the secant line can be expressed as:

\[
m_{\text{sec}} = \frac{\sin\left(\frac{14\pi}{x}\right) - \sin\left(14\pi\right)}{x - 1} = \frac{\sin\left(\frac{14\pi}{x}\right)}{x - 1}
\]

This is because \(\sin(14\pi)\) equals zero.

We will calculate the slope for each given value of \(x\):

1. \(x = 1.5\)
2. \(x = 1.4\)
3. \(x = 1.3\)
4. \(x = 1.2\)
5. \(x = 1.1\)
6. \(x = 0.5\)
7. \(x = 0.6\)
8. \(x = 0.7\)
9. \(x = 0.8\)
10. \(x = 0.9\)

Let's compute these slopes.Here are the slopes of the secant line \(PQ\) for each value of \(x\), correct to four decimal places:

1. \(x = 1.5\): Slope = \(-1.7321\)
2. \(x = 1.4\): Slope = \(0.0000\)
3. \(x = 1.3\): Slope = \(2.2104\)
4. \(x = 1.2\): Slope = \(-4.3301\)
5. \(x = 1.1\): Slope = \(7.5575\)
6. \(x = 0.5\): Slope = \(0.0000\)
7. \(x = 0.6\): Slope = \(2.1651\)
8. \(x = 0.7\): Slope = \(0.0000\)
9. \(x = 0.8\): Slope = \(5.0000\)
10. \(x = 0.9\): Slope = \(9.8481\)

### Part (b): Estimating the Slope of the Tangent Line at \(P(1, 0)\)

To estimate the slope of the tangent line at \(P(1, 0)\), we observe how the slopes of the secant lines change as \(x\) approaches 1 from both sides.

Notice the trend in the slopes as \(x\) gets closer to 1. The sign changes and the magnitude increases significantly.

By choosing appropriate secant lines (particularly those closer to \(x = 1\)), we estimate the slope of the tangent line as:

\[
m_{\text{tan}} \approx \text{average of slopes as } x \text{ approaches } 1
\]

Given the results, a reasonable estimate for the slope of the tangent line at \(P(1, 0)\) would be approximately \(\boxed{0.00}\).

This indicates that the tangent at \(P(1, 0)\) might be horizontal.

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Would you like more details on any of the steps? Here are some related questions:

1. What happens to the secant slope if \(x\) is very close to 1 from the left or right?
2. How would the slopes change if the function were modified slightly, say \(y = \sin(10\pi/x)\)?
3. Can the sine function influence the behavior of the tangent differently at different points on the curve?
4. What is the relationship between the magnitude of the slope and the periodic nature of the sine function?
5. Could there be a scenario where the slope of the secant line tends to infinity or negative infinity?

**Tip:** When estimating tangent slopes using secant lines, the closer the secant points are to the point of tangency, the more accurate the estimate.

Explore how to calculate secant and tangent lines for the curve y = sin(14𝜋/x) given a point on the curve and varying x-values. Understand the slope formula for secant lines and how to estimate the slope of the tangent line at a specific point. Suitable for advanced high school students.

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