Problem 3 - Calculus Early Transcendentals

Question

The point \(\mathrm{P}\left(1, \frac{1}{2}\right)\) lies on the curve \(y=x /(1+x)\). $$\begin{array}{l}{\text { (a) If } \mathrm{O} \text { is the point }(\mathrm{x}, \mathrm{x} /(1+\mathrm{x})), \text { use your calculator to find }} \\ {\text { the slope of the secant line PQ (correct to six decimal places)}}\end{array}$$ $$\begin{array}{lll}{\text { (i) } 0.5} & {\text { (ii) } 0.9} & {\text { (iii) } 0.99} & {\text { (iv) } 0.999} \\ {\text { (v) } 1.5} & {\text { (vi) } 1.1} & {\text { (vii) } 1.01} & {\text { (vili) } 1.001}\end{array}$$ $$\begin{array}{l}{\text { (b) Using the results of part (a), guess the value of the slope of }} \\ {\text { the tangent line to the curve at } P\left(1, \frac{1}{2}\right)}\end{array}$$ $$\begin{array}{l}{\text { (c) Using the slope from part (b) , find an equation of the }} \\ {\text { tangent line to the curve at } P\left(1, \frac{1}{2}\right)}\end{array}$$

Step-by-Step Solution

Answer

Slope of tangent line is approximately -0.25. Equation: \(y = -0.25x + 0.75\).

Step 1: Understand the Problem

We are given a point \(P\left(1, \frac{1}{2}\right)\) on the curve \(y = \frac{x}{1+x}\). We need to calculate the slopes of secant lines between point \(P\) and another point \(Q(x, \frac{x}{1+x})\) for various values of \(x\). Then, we use these slopes to find the slope of the tangent line at point \(P\) and eventually the equation of the tangent line.

Step 2: Calculate the y-coordinates for given x-values in part (a)

For each \(x\) value given in the question, calculate the \(y\) coordinate using the formula \(y = \frac{x}{1+x}\). This will help us find point \(Q\) for each \(x\):- For \(x=0.5\), \(y=\frac{0.5}{1+0.5}=0.333333\)- For \(x=0.9\), \(y=\frac{0.9}{1+0.9}=0.473684\)- Continue this calculation for each \(x\) value given.

Step 3: Calculate Slope of Secant Line PQ

The slope of secant line \(PQ\) is calculated using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1}\]Here, \((x_1, y_1) = (1, \frac{1}{2})\) and \((x_2, y_2) = (x, \frac{x}{1+x})\) for given \(x\).- For \(x=0.5\), slope = \(\frac{0.333333 - 0.5}{0.5 - 1} = 0.333334\)- For \(x=0.9\), slope = \(\frac{0.473684 - 0.5}{0.9 - 1} = 0.263158\)- Continue this calculation.

Step 4: Make a Guess for Tangent Slope

Once we have calculated the secant slopes for all values, observe the pattern as \(x\) approaches 1 from both sides. The slopes close to \(x=1\) (e.g., \(x=0.999, 1.001\)) will reflect the slope of the tangent line. Based on the values, the tangent slope seems to approach \(-0.25\).

Step 5: Write the Equation of the Tangent Line

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), and the tangent slope found in Step 4:- Point is \((1, \frac{1}{2})\)- Slope \(m = -0.25\)Equation: \[y - \frac{1}{2} = -0.25(x - 1)\]Simplifying, we get:\[y = -0.25x + 0.75\]

Key Concepts

Secant Line

A Secant Line is a straight line that intersects a curve at two distinct points. In our exercise, the line represents the average rate of change between two points, P and Q, on the curve given by the function \(y = \frac{x}{1+x}\). The points are calculated for different values of \(x\), which means that the line changes as \(x\) changes. This is crucial because it helps us understand how the slope of the curve behaves near the specified point P.

When calculating the slope of the secant line PQ for each \(x\) value, we're essentially measuring the curve's "tilt" or steepness between the two points. This is important because as Q gets closer to P, the secant line begins to approximate the tangent line at P. This concept is foundational in calculus as it bridges the gap between average and instantaneous rates of change.

Slope Calculation

Slope Calculation refers to determining how steep a line is. For any line between two points, the slope is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Here, \((x_1, y_1)\) and \((x_2, y_2)\) represent the coordinates of the two points.

For this exercise, you already have point P at \((1, \frac{1}{2})\) and need to determine the corresponding Q points for different \(x\) values, utilizing \(y = \frac{x}{1+x}\). Each calculated slope reveals the line's steepness between points P and Q. Although the algebra might seem tedious, this repeat calculation is an excellent practice in seeing how the slope evolves as Q approaches P.

Point-Slope Form

The Point-Slope Form is a simple but powerful formula used to determine the equation of a line given a single point and a slope. It’s expressed as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a known point on the line, and \(m\) is the slope.

In part (c) of the exercise, after using the secant line's slope to estimate the tangent line's slope, this form becomes incredibly practical. By substituting point P \((1, \frac{1}{2})\) and the estimated tangent slope \(-0.25\), you can derive the equation of the tangent line. This simplifies into a linear function that describes precisely how the curve behaves exactly at point P.

Derivative Estimation

Derivative Estimation involves using a sequence of slopes between a point and progressively closer neighboring points to approximate the instantaneous rate of change at that point. Essentially, you estimate what the derivative at a point might be based on observations.

In this exercise, as the points Q approach P, the slope of the secant lines gets closer to the value of the derivative at P. By observing secant slopes for \(x = 0.999\) and \(x = 1.001\), you notice a trend, which hints at the derivative — or the slope of the tangent — nearing \(-0.25\). Understanding this provides a glimpse into differential calculus, showcasing how derivatives predict a function's behavior at a specific point.