The point P(2, ln 2) lies on the curve y = ln x.
(a) If Q is the point (x, in x), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x:
(i) 1.5
(ii) 1.9
(iii) 1.99
(iv) 1.999
(v) 2.5
(vi) 2.1
(vii) 2.01
(viii) 2.001
(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(2, ln 2).
(c) Using the slope from part (b), find an equation of the tangent line to the curve at P(2, ln 2).
(d) Sketch the curve, two of the secant lines, and the tangent line.
The Correct Answer and Explanation is :
(a) Finding the slope of the secant line PQ for the given values of ( x ):
The slope of the secant line joining two points ( P(x_1, y_1) ) and ( Q(x_2, y_2) ) on the curve is given by the formula:
[
m_{\text{secant}} = \frac{y_2 – y_1}{x_2 – x_1}
]
where ( P(2, \ln 2) ) and ( Q(x, \ln x) ). Here, ( y_1 = \ln 2 ) and ( x_1 = 2 ).
So the slope of the secant line becomes:
[
m_{\text{secant}} = \frac{\ln x – \ln 2}{x – 2}
]
Now we calculate for each value of ( x ):
(i) For ( x = 1.5 ):
[
m_{\text{secant}} = \frac{\ln 1.5 – \ln 2}{1.5 – 2} = \frac{\ln \frac{1.5}{2}}{-0.5} = \frac{\ln 0.75}{-0.5} \approx 0.287682
]
(ii) For ( x = 1.9 ):
[
m_{\text{secant}} = \frac{\ln 1.9 – \ln 2}{1.9 – 2} = \frac{\ln \frac{1.9}{2}}{-0.1} \approx 1.156820
]
(iii) For ( x = 1.99 ):
[
m_{\text{secant}} = \frac{\ln 1.99 – \ln 2}{1.99 – 2} = \frac{\ln \frac{1.99}{2}}{-0.01} \approx 6.273197
]
(iv) For ( x = 1.999 ):
[
m_{\text{secant}} = \frac{\ln 1.999 – \ln 2}{1.999 – 2} = \frac{\ln \frac{1.999}{2}}{-0.001} \approx 69.314718
]
(v) For ( x = 2.5 ):
[
m_{\text{secant}} = \frac{\ln 2.5 – \ln 2}{2.5 – 2} = \frac{\ln \frac{2.5}{2}}{0.5} \approx 0.405465
]
(vi) For ( x = 2.1 ):
[
m_{\text{secant}} = \frac{\ln 2.1 – \ln 2}{2.1 – 2} = \frac{\ln \frac{2.1}{2}}{0.1} \approx 0.487227
]
(vii) For ( x = 2.01 ):
[
m_{\text{secant}} = \frac{\ln 2.01 – \ln 2}{2.01 – 2} = \frac{\ln \frac{2.01}{2}}{0.01} \approx 0.497588
]
(viii) For ( x = 2.001 ):
[
m_{\text{secant}} = \frac{\ln 2.001 – \ln 2}{2.001 – 2} = \frac{\ln \frac{2.001}{2}}{0.001} \approx 0.499751
]
(b) Guessing the slope of the tangent line at P(2, ln 2):
By examining the slopes of the secant lines as ( x ) approaches 2, we see that the slopes increase rapidly as ( x ) approaches 2 from the left and stabilize around 0.5 as ( x ) approaches 2 from the right. The slopes near ( x = 2 ) are:
- For ( x = 2.01 ): 0.497588
- For ( x = 2.001 ): 0.499751
Therefore, we can reasonably guess that the slope of the tangent line at ( P(2, \ln 2) ) is approximately ( 0.5 ).
(c) Finding the equation of the tangent line at P(2, ln 2):
The general equation of a line is:
[
y – y_1 = m(x – x_1)
]
where ( m ) is the slope, and ( (x_1, y_1) ) is a point on the line. Using ( P(2, \ln 2) ) and the slope ( m = 0.5 ), we get:
[
y – \ln 2 = 0.5(x – 2)
]
Solving for ( y ):
[
y = 0.5(x – 2) + \ln 2
]
[
y = 0.5x – 1 + \ln 2
]
Thus, the equation of the tangent line is:
[
y = 0.5x – 1 + \ln 2
]
(d) Sketching the curve, two secant lines, and the tangent line:
For this step, we would plot the curve ( y = \ln x ), the tangent line ( y = 0.5x – 1 + \ln 2 ), and two secant lines for chosen values of ( x ) such as ( x = 2.01 ) and ( x = 1.99 ). The secant lines should approach the tangent line as ( x ) gets closer to 2.
The tangent line approximates the curve closely at ( x = 2 ), and as the points ( Q(x, \ln x) ) approach ( P(2, \ln 2) ), the secant lines become increasingly similar to the tangent line, indicating the concept of the derivative at a point.
Explanation:
The secant lines show how the slope between two points on the curve changes as the second point gets closer to the point of tangency. As we move closer to ( x = 2 ), the slope of the secant lines approaches the slope of the tangent line. The derivative of ( \ln x ) at ( x = 2 ) provides the slope of the tangent line at that point, and we approximate this value by computing the secant slopes for values of ( x ) close to 2. The value ( 0.5 ) is a reasonable estimate for the derivative of ( \ln x ) at ( x = 2 ), since the actual derivative of ( \ln x ) is ( \frac{1}{x} ), and at ( x = 2 ), this is exactly ( \frac{1}{2} ).