Problem 47 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the equation of the line through the points \((2, \ln 2)\) and \((2+\epsilon, \ln (2+\epsilon))\).

Step-by-Step Solution

Answer

The equation of the line through the points \((2, \ln 2)\) and \((2 + \epsilon, \ln (2 + \epsilon))\) is \(y = \ln[(2 + \epsilon) / 2] / \epsilon (x - 2) + \ln 2 \)

Step 1: Calculate the slope

The slope (m) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) can be calculated using the formula \(m = (y_2 - y_1) / (x_2 - x_1)\). Here, \(x_1 = 2\), \(y_1 = \ln 2\), \(x_2 = 2 + \epsilon\), \(y_2 = \ln (2 + \epsilon)\). Substituting these values in the slope formula gives: \(m = (\ln (2+\epsilon) - \ln 2 ) / (\epsilon)\). Using the property of logarithms that \(\ln a - \ln b = \ln(a/b)\), this simplifies to: \(m = \ln [(2 + \epsilon) / 2] / \epsilon\).

Step 2: Find the line equation using point-slope form

The point-slope form of the equation of a line is \(y - y_1 = m(x - x_1)\). Substituting the known values \(x_1 = 2\), \(y_1 = \ln 2\), and \(m = \ln[(2 + \epsilon) / 2] / \epsilon\) gives the equation: \( y - \ln 2 = \ln[(2 + \epsilon) / 2] / \epsilon (x - 2)\).

Step 3: Simplify the equation

Solve the equation for y to get it in slope-intercept form \(y = mx + b\). So we have: \(y = \ln[(2 + \epsilon) / 2] / \epsilon (x - 2) + \ln 2 \)

Key Concepts

Slope Calculation

To find the equation of a line, calculating the slope is an essential first step. The slope (often represented by the letter \(m\)) tells us how steep the line is. It is defined as the change in \(y\)-values divided by the change in \(x\)-values between two points on the line. In mathematical terms, the slope is given by the formula:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]Where:

• \(x_1, y_1\) are the coordinates of the first point.
• \(x_2, y_2\) are the coordinates of the second point.

For the points \((2, \ln 2)\) and \((2 + \epsilon, \ln (2 + \epsilon))\), we substitute these into the formula to find the slope:\[m = \frac{\ln(2 + \epsilon) - \ln 2}{(2 + \epsilon) - 2}\]This simplifies to:\[m = \frac{\ln\left(\frac{2 + \epsilon}{2}\right)}{\epsilon}\]This expression uses a property of logarithms: \(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\), allowing us to understand how the slope of the line is composed using logarithmic differences.

Point-Slope Form

Once the slope is calculated, finding the line equation using the point-slope form follows. This form is especially useful because it directly uses the slope and a single point on the line.The point-slope form of a line’s equation is:\[y - y_1 = m(x - x_1)\]Where:

• \((x_1, y_1)\) is a known point on the line.
• \(m\) is the slope we calculated.

Applying this form to our specific points \((2, \ln 2)\) and slope \(m = \frac{\ln[(2 + \epsilon) / 2]}{\epsilon}\), the equation becomes:\[y - \ln 2 = \frac{\ln[(2 + \epsilon) / 2]}{\epsilon}(x - 2)\]This representation gives us a way to describe a line’s behavior around any given point along its length, based on its slope.

Logarithmic Functions

To fully understand the process, it’s important to delve into logarithmic functions and their properties. Logarithms are the inverse of exponential functions, which means they answer the question: "to what exponent must a base number be raised, to yield a particular value?"Here are some key properties of logarithms that prove essential in equation manipulations:

• Difference of Logs: \(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\). This turns subtraction of logs into division inside a single log expression.
• Base 'e' Log: When dealing with natural logs (\(\ln\)), the base is \(e\), which is an irrational constant approximately equal to 2.71828.

In our problem, by employing the difference of logarithms, we simplified the slope’s calculation. Recognizing these properties allows us to manage complex expressions and convert them into more workable forms. Understanding logarithms also aids in interpreting the relationship between variables, especially when they experience growth or decay patterns, making them a powerful tool in mathematical equations.