Problem 14 - 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. $$y=2 x-x^{3}, \quad \text { at }(1,1)$$

Step-by-Step Solution

Answer

The equation of the tangent line at \((1, 1)\) is \( y = -x + 2 \).

Step 1: Write the function

We are given the function \( y = 2x - x^3 \). We need to find the equation of the tangent line to this curve at the point \((1, 1)\).

Step 2: Find the derivative of the function

To find the equation of the tangent line, we need the slope at the given point. The derivative of a function at a point gives us the slope of the tangent line at that point. Let's differentiate \( y = 2x - x^3 \) with respect to \( x \): \( \frac{dy}{dx} = 2 - 3x^2 \).

Step 3: Evaluate the derivative at the given point

Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \((1, 1)\): \( \frac{dy}{dx} \bigg|_{x=1} = 2 - 3(1)^2 = 2 - 3 = -1 \). Thus, the slope of the tangent line is \(-1\).

Step 4: Use the point-slope form of a line

The point-slope form of the equation of a line is given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. Substituting \( m = -1 \), \( x_1 = 1 \), and \( y_1 = 1 \) gives us: \( y - 1 = -1(x - 1) \).

Step 5: Simplify the equation

Simplify the equation from Step 4: Begin by distributing \(-1\): \( y - 1 = -x + 1 \). Then, solving for \( y \) results in \( y = -x + 2 \). This is the equation of the tangent line.

Key Concepts

Derivative

When working with curves and trying to find tangent lines, the concept of the derivative is essential. A derivative is a mathematical tool used to measure how a function changes as its input changes. Specifically, it tells us the slope of the tangent line to the curve at any point. For the function given in the exercise, which is \( y = 2x - x^3 \), we need to determine the slope of this function at the point \((1, 1)\).

By differentiating the function with respect to \( x \), we calculate \( \frac{dy}{dx} = 2 - 3x^2 \). This derivative represents the rate of change of \( y \) with respect to \( x \). Evaluating this derivative at \( x = 1 \), we find \( \frac{dy}{dx} \bigg|_{x=1} = -1 \), meaning the slope of the tangent line at that point is \(-1\).

• Derivatives help identify the rates of change.
• They provide the slope of the tangent line to the curve at a specific point.
• In our example, the derivative \( 2 - 3x^2 \) tells us how the curve's slope changes with \( x \).

Point-Slope Form

Once we know the slope of the tangent line, the next step is to use the point-slope form to write the tangent line's equation. This form is helpful because it allows us to take the slope from the derivative and the given point to find the line equation. The point-slope form is expressed as \( y - y_1 = m(x - x_1) \).

In this expression:

• \( m \) represents the slope of the tangent line, found using the derivative.
• \( (x_1, y_1) \) is the specific point on the curve where the tangent is drawn. For our problem, this point is \((1, 1)\).

Substituting the slope \(-1\) and point \((1, 1)\) into the form, we get \( y - 1 = -1(x - 1) \). Simplifying this equation gives us the final equation of the tangent line \( y = -x + 2 \). This concise formula helps create a linear representation of the curve's local behavior at our specified point.

Slope of a Line

The slope of a line is a fundamental concept when discussing tangent lines and derivatives. It directly measures the steepness or incline of a line. In simpler terms, it tells us how much \( y \) changes for a unit change in \( x \). In the context of tangent lines, this slope is that of the line which just "touches" the curve at, and only at, a single point without crossing over the curve.

Calculating the slope for the given problem:

• The derivative \( \frac{dy}{dx} = 2 - 3x^2 \) is used to find the slope at any given \( x \).
• Plugging \( x = 1 \) into this equation, we find the slope at the point \( (1, 1) \) is \(-1\).

This means the tangent line falls one unit vertically for each unit it runs horizontally to the right when moving in the positive direction along the x-axis. The negative sign indicates the line is sloping downwards. Knowing the slope is important, as it helps us understand and predict the behavior not only of the tangent line but potentially of the curve itself at that particular point.