Problem 3 - Calculus: Early Transcendentals

Question

The point \( P(2, -1) \) lies on the curve \( y = 1/(1-x) \). (a) If \( Q \) is the point \( (x, 1/(1-x)) \), 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 \( P(2, -1) \). (c) Using the slope from part (b), find an equation of the tangent line to the curve at \( P(2, -1) \).

Step-by-Step Solution

Answer

(a) Slopes calculated for each x; (b) Approaching negative infinity; (c) Vertical line, x = 2.

Step 1: Calculate the y-coordinate of Q

For each given value of \( x \), calculate the y-coordinate of point \( Q(x, 1/(1-x)) \) using the equation \( y = \frac{1}{1-x} \).

Step 2: Find the slope of the secant line PQ

The slope \( m \) of the secant line \( PQ \) is given by the formula \( m = \frac{y_Q - y_P}{x_Q - x_P} \), where \( (x_P, y_P) = (2, -1) \) and \( (x_Q, y_Q) = (x, \frac{1}{1-x}) \). Calculate \( m \) for each \( x \) value.

Step 3: Calculate slopes for each x value

Use the results from Step 1 and Step 2 to find: (i) For \( x = 1.5 \), \( m = \frac{2 - (-1)}{1.5 - 2} = -6.000000 \).(ii) For \( x = 1.9 \), \( m = \frac{10 - (-1)}{1.9 - 2} = -11.000000 \).(iii) For \( x = 1.99 \), \( m = \frac{100 - (-1)}{1.99 - 2} = -101.000000 \).(iv) For \( x = 1.999 \), \( m = \frac{1000 - (-1)}{1.999 - 2} = -1001.000000 \).(v) For \( x = 2.5 \), \( m = \frac{-2 - (-1)}{2.5 - 2} = -2.000000 \).(vi) For \( x = 2.1 \), \( m = \frac{-10 - (-1)}{2.1 - 2} = -9.000000 \).(vii) For \( x = 2.01 \), \( m = \frac{-100 - (-1)}{2.01 - 2} = -99.000000 \).(viii) For \( x = 2.001 \), \( m = \frac{-1000 - (-1)}{2.001 - 2} = -999.000000 \).

Step 4: Guess the slope of the tangent line

Based on the values of the slopes of the secant lines as \( x \) approaches 2, observe that the slopes are increasing in magnitude and negative. They suggest that the slope of the tangent line approaches \( -rac{1}{0}(or very large) \) as \( x \) gets closer to 2 from either side.

Step 5: Confirm and find the equation of the tangent line

Use the point-slope form of a line equation \( y - y_1 = m (x - x_1) \) with the point \((2, -1)\) and an estimated slope of \(-rac{1}{0} \). However, due to the undefined nature of the slope (as \( x = 2 \) results in vertical slope), the tangent is a vertical line passing through \( x = 2 \).Hence, the equation is \( x = 2 \).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point. In calculus, the concept of a tangent line is crucial because it represents the idea of the rate of change or the derivative at a specific point.
For instance, in the original exercise, the point of tangency is at point \((2, -1)\) on the curve \(y = \frac{1}{1-x}\). At this point, the tangent line would provide the instantaneous rate of change of the curve.
To find this tangent line, you calculate what happens as another nearby point (not on the tangent line) gets closer and closer to our point of interest. This essentially gives us a glimpse of the behavior of the curve around that one point.

Secant Line

A secant line intersects a curve at two or more points. It's different from a tangent line which only touches the curve at one point.
In problems like the one in the exercise, we use secant lines to approximate the slope of the tangent line by seeing what happens as the two points on the secant line get closer together.
Think of a secant line as a tool that helps us understand how the curve behaves between two points. In the exercise, by calculating the slope of the secant line for different values approaching \(x = 2\), you approximate the slope of the tangent line.

Slope Calculation

Calculating the slope is a key step in understanding the relationship between points on a curve. For any two points \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) is given by the formula:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]This formula tells us how much \(y\) changes per unit change in \(x\).
In the case of a secant line, this slope gives us an average rate of change between the two points. By observing the slopes as they approach the tangent point, you can infer the slope of the tangent line, which represents an instantaneous rate of change.
In the exercise, this process is used to estimate the behavior at the point \(P(2, -1)\).

Vertical Line

A vertical line is a line where all points have the same x-coordinate. This means it goes straight up and down without tilting. A vertical line cannot have a slope in the traditional sense; we often describe the slope of a vertical line as undefined or infinitely large.
In the given exercise, as the value of \(x\) tends towards 2, the slope values of the secant line become extremely large and negative, indicating that the tangent line at \((2, -1)\) is vertical.
This leads to the conclusion that the equation of the tangent line is simply \(x = 2\), as there is no change in the x-coordinate along this line, serving as an important characteristic of vertical lines in calculus.