Problem 30 - Calculus Single Variable

Question

A function \(f\) and a point \(P\) are given. Find the slope-intercept form of the equation of the tangent line to the graph of \(f\) at \(P\). $$ f(x)=x^{3} / 3 \quad P=(-3,-9) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 9x + 18 \).

Step 1: Differentiate the Function

To find the equation of the tangent line, we first need the derivative of the function, which gives us the slope of the tangent line at a given point. The function is \( f(x) = \frac{x^3}{3} \). Differentiating this with respect to \( x \) gives the derivative: \( f'(x) = \frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2 \).

Step 2: Evaluate the Derivative at Point P

Next, we calculate the slope of the tangent line at the given point \( P = (-3, -9) \). Substitute \( x = -3 \) into the derivative: \( f'(-3) = (-3)^2 = 9 \). This means the slope of the tangent line at \( P \) is 9.

Step 3: Write the Point-Slope Form

Use the point-slope form of the equation of a line: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point \( P \). Substitute \( m = 9 \), \( x_1 = -3 \), and \( y_1 = -9 \) into the equation: \( y + 9 = 9(x + 3) \).

Step 4: Convert to Slope-Intercept Form

To express the equation in slope-intercept form \( y = mx + b \), simplify the equation from Step 3: \( y + 9 = 9x + 27 \). Solve for \( y \) to get the slope-intercept form: \( y = 9x + 18 \).

Key Concepts

Differentiation

Differentiation is a powerful concept in calculus, primarily used to find the rate at which a function is changing at any given point. This idea is essential when finding the slope of a tangent line to a curve at a particular point. When you differentiate a function, you're essentially figuring out how the function behaves as the input changes by a small amount.

To understand differentiation, consider a function like \( f(x) = \frac{x^3}{3} \). By differentiating this function, we calculate its derivative, \( f'(x) \). The derivative here is \( x^2 \), which tells us the slope of the tangent line at any point \( x \).

For example:

• Take the derivative: \( f'(x) = x^2 \).
• Evaluate it at the point of interest, say \( x = -3 \), so \( f'(-3) = 9 \).

This means the slope of the tangent line at \( x = -3 \) is 9. Differentiation gives us the slope, which is a crucial step in determining the equation of the tangent line.

Slope-Intercept Form

The slope-intercept form of a linear equation is a very convenient way to express the equation of a line. It's written as \( y = mx + b \), where \( m \) is the slope of the line, and \( b \) is the y-intercept, or the point where the line crosses the y-axis.

Once we have the slope from differentiation and a point on the line, we can rewrite the equation in slope-intercept form. For the tangent line in our example:

• The slope \( m \) is 9, as calculated from differentiation.
• Starting from the point-slope form \( y + 9 = 9x + 27 \), simplify to get \( y = 9x + 18 \).

Here, the slope \( m = 9 \) confirms the line rises steeply, and the y-intercept \( b = 18 \) shows where the line crosses the y-axis. Using slope-intercept form makes graphing lines easy and helps quickly identify the line's steepness and vertical position.

Point-Slope Form

The point-slope form is a great tool for deriving the equation of a line when you know the slope and a specific point on the line. It's expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the known point.

This form is especially helpful for calculating the equation of a tangent line to a curve at a particular point. For example, in finding the tangent line to the graph of \( f(x) = \frac{x^3}{3} \) at point \( P = (-3, -9) \):

• The slope \( m \) is 9, determined through differentiation.
• Using point \((x_1, y_1) = (-3, -9)\), the equation becomes \( y + 9 = 9(x + 3) \).

This form is straightforward and sets up perfectly for converting into the slope-intercept form \( y = mx + b \). It bridges the step from knowing the slope and point to writing a concise equation that clearly represents the line. The point-slope form is an intermediate step that links the mathematical approach of differentiation with expressing the line in standard form.