Problem 11 - Precalculus Mathematics for Calculus

Question

Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=\frac{x}{x-1} \quad \text { at }(2,2)$$

Step-by-Step Solution

Answer

The tangent line is \( y = -x + 4 \).

Step 1: Differentiate the Function

Find the derivative of the function to determine the slope of the tangent line. Given the function \( y = \frac{x}{x-1} \), use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = x \) and \( v = x-1 \). So, the derivative is \( y' = \frac{(x-1) \cdot 1 - x \cdot 1}{(x-1)^2} = \frac{x-1-x}{(x-1)^2} = \frac{-1}{(x-1)^2} \).

Step 2: Substitute the Point into the Derivative

Substitute \( x = 2 \) into the derivative to find the slope (m) of the tangent line at the point (2,2). So we have: \( y' = \frac{-1}{(2-1)^2} = \frac{-1}{1} = -1 \).

Step 3: Use the Point-Slope Form of a Line

Using the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \), substitute the point \((x_1, y_1) = (2,2)\) and \(m = -1\) to get the equation of the tangent line: \( y - 2 = -1(x - 2) \). Simplify to get \( y = -x + 4 \).

Step 4: Graph the Curve and Tangent Line

To visually represent the solution, plot the curve \( y = \frac{x}{x-1} \) and the tangent line \( y = -x + 4 \) on the same axes. The point of tangency should be visible at (2,2) on both the curve and the line.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, crucial for finding the slope of tangent lines. It involves calculating the derivative of a function to ascertain how the function changes at any given point. In our case, the function is given as \( y = \frac{x}{x-1} \). The differentiation process here uses the quotient rule.
The quotient rule applies when you need to differentiate a function that can be expressed as a ratio of two functions. If \( u \) and \( v \) are functions of \( x \), the derivative of \( \frac{u}{v} \) is given by:

• \( \frac{v \cdot u' - u \cdot v'}{v^2} \)

Applying this rule to our function, \( u = x \) and \( v = x-1 \), gives us the derivative \( y' = \frac{-1}{(x-1)^2} \). This expression will help us find the slope at any point on the curve.

Slope of Tangent Line

The slope of a tangent line is essentially the value of the derivative at a particular point on a curve. It represents the instantaneous rate of change of the function at that point. For our problem, after finding the derivative as \( y' = \frac{-1}{(x-1)^2} \), we substitute \( x = 2 \) to find the slope at the point (2,2).
Substituting gives us \( y' = \frac{-1}{1} = -1 \). Thus, the slope of the tangent line at (2,2) is \(-1\). This slope tells us that the tangent line is decreasing; for every unit increase in \( x \), \( y \) decreases by one unit.

Point-Slope Form of a Line

To find the equation of the tangent line, we use the point-slope form of a line, given by:

• \( y - y_1 = m(x - x_1) \)

In this formula, \( (x_1, y_1) \) are the coordinates of a point on the line, and \( m \) is the slope. In our problem, \( (x_1, y_1) = (2,2) \), and the slope \( m = -1 \).
Substituting these into the point-slope formula gives:
\( y - 2 = -1(x - 2) \).
Simplifying, we arrive at the equation of the tangent line: \( y = -x + 4 \). This is a straight line that touches the curve at exactly one point, (2,2), both having the same slope there, ensuring it's a tangent.