Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the equation of the line tangent to \(y=\frac{2}{x+1}\) at the point \((1,1)\).

Step-by-Step Solution

Answer

The equation of the line tangent to \(y=\frac{2}{x+1}\) at the point (1,1) is \(y = -0.5x + 1.5\).

Step 1: Compute the derivative

First, calculate the derivative of the function \(y=\frac{2}{x+1}\). The derivative is computed as follows: \(y' = \frac{-2}{{(x+1)}^2}\).

Step 2: Evaluate the derivative at the given point

Then, the slope of the tangent line at the point (1,1) is found by evaluating the derivative at x=1. This yields \(y'(1) = \frac{-2}{{(1+1)}^2} = -0.5\). So, the slope of the tangent line is -0.5.

Step 3: Use slope-point equation to get the line equation

To find the tangent line's equation, the slope-point form of the line equation, \(y - y1 = m(x - x1)\), is used. In this case, \(m=-0.5\), \(x1=1\), and \(y1=1\). Substituting these values into the formula yields the equation as \(y - 1 = -0.5*(x - 1)\). Upon simplifying this equation, we get \(y = -0.5x + 1.5\) as the equation of the tangent line.

Key Concepts

Tangent Line Equation

In calculus, a tangent line to a curve at a specific point is a straight line that touches the curve precisely at that point without crossing it nearby. Finding the equation of a tangent line is a fundamental application of derivatives.
To discover the tangent line equation, one must understand the function of your curve and its derivative. Here, we consider the function \( y = \frac{2}{x+1} \). To find the tangent line equation at a certain point, say \((1,1)\), we need:

• To evaluate the derivative, which gives the slope of the tangent line at that point.
• Utilize this slope in conjunction with the coordinates of the given point in a linear equation form.

This process ensures the tangent line mirrors the direction and steepness of the curve exactly at the contact point, offering a precise local linear approximation.

Derivative Evaluation

The derivative of a function provides crucial information about its slope at any point along its curve. For our given function, \( y = \frac{2}{x+1} \), computing the derivative helps us find the slope of the tangent line.
The function derivative is calculated using rules of differentiation applicable to rational functions, leading to: \[ y' = \frac{-2}{{(x+1)}^2} \] This expression represents the rate of change of \( y \) with respect to \( x \). Evaluating this derivative at a specific \( x \)-value yields the slope of the tangent line at that point. When \( x = 1 \), the value of the derivative is \( -0.5 \).

• This signifies that at \((1,1)\), the curve is declining with a slope of \(-0.5\).
• This slope information is essential for identifying the exact tangent line equation at the particular point.

Slope-Point Form

The slope-point form of a line's equation is a practical tool for finding the equation of a line when you know one point the line passes through and the slope of the line. The form is expressed as: \[ y - y_1 = m(x - x_1) \] Here, \( m \) is the slope, and \((x_1, y_1)\) is a point on the line. This format directly integrates the calculated slope and known point from the curve. To formulate the tangent line equation for our earlier problem, use the slope \( m = -0.5 \) found from the derivative evaluation at point \((1,1)\). Substitute these into the slope-point form: \[ y - 1 = -0.5(x - 1) \] Upon simplifying, you obtain the tangent line equation: \[ y = -0.5x + 1.5 \] This equation reflects the tangent line's path that precisely kisses the curve at \((1,1)\). Using this format allows for straightforward calculations, making it a staple in solving calculus problems related to tangents.