Problem 39 - Calculus Single Variable

Question

A function \(f\) and a point \(P\) are given. Find the slope-intercept form of the equation of the normal line to the graph of \(f\) at \(P\). $$ f(x)=3 x^{3}-2 x^{2}-10 \quad P=(2,6) $$

Step-by-Step Solution

Answer

The normal line's equation is \( y = -\frac{1}{28}x + \frac{85}{14} \).

Step 1: Differentiate the Function

To find the equation of the normal line, we first need the derivative of the function, which gives us the slope of the tangent line at any point. Let's differentiate the given function. Given \( f(x) = 3x^3 - 2x^2 - 10 \), its derivative is \( f'(x) = 9x^2 - 4x \).

Step 2: Evaluate the Derivative at P

Substitute \( x = 2 \) into \( f'(x) \) to find the slope of the tangent line at point \( P = (2, 6) \). This gives \( f'(2) = 9(2)^2 - 4(2) = 36 - 8 = 28 \). So, the slope of the tangent line at \( P \) is 28.

Step 3: Determine the Slope of the Normal Line

The normal line is perpendicular to the tangent line. If the slope of the tangent line is \( m \), then the slope of the normal line is \( -\frac{1}{m} \). Thus, the slope of the normal line is \( -\frac{1}{28} \).

Step 4: Use the Point-Slope Form

We use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (2, 6) \) and \( m = -\frac{1}{28} \). Substituting these values, we have \( y - 6 = -\frac{1}{28}(x - 2) \).

Step 5: Convert to Slope-Intercept Form

Solve the equation from the previous step to put it into slope-intercept form \( y = mx + b \). Distribute \( -\frac{1}{28} \) through the parentheses: \( y - 6 = -\frac{1}{28}x + \frac{2}{28} \). Simplify to get \( y = -\frac{1}{28}x + \frac{2}{28} + 6 \). Simplifying further gives \( y = -\frac{1}{28}x + \frac{85}{14} \), which is in the desired form.

Key Concepts

Normal Line Equation

The normal line is a straight line that is perpendicular to the tangent line at a given point on a curve. To find the equation of the normal line, we first need to know the slope of the tangent line at the point of interest. Once we know this, the slope of the normal line is found using the concept of perpendicular slopes. The basic approach involves:

• Finding the slope of the tangent line through differentiation.
• Calculating the slope of the normal line as the negative reciprocal of the tangent slope.
• Using this slope along with the given point to write the equation of the normal line in point-slope form.

The final step often involves converting the point-slope form equation into the more familiar slope-intercept form for ease of use.

Slope-Intercept Form

The slope-intercept form of a line's equation is one of the most widely used forms and offers a lot of conveniences. It is written as \( y = mx + b \), where:

• \( m \) represents the slope of the line.
• \( b \) is the y-intercept, the point where the line crosses the y-axis.

This form allows us to quickly identify both the steepness of the line (slope) and its position relative to the y-axis (intercept). When we have a line equation in a different form, converting it to the slope-intercept form often helps in visualizing the line and understanding its properties more intuitively.
Applying this format facilitates graphing the line and comparing it with other linear relationships.

Differentiation

Differentiation is a fundamental concept in calculus, used to find the rate at which a function is changing at any point on its curve. When we differentiate a function, we obtain its derivative, which represents the slope of the tangent line at each point. In this problem, the function \( f(x) = 3x^3 - 2x^2 - 10 \) was differentiated to get \( f'(x) = 9x^2 - 4x \).
Differentiation unfolds in several key steps:

• Apply power rules to each term: for instance, the derivative of \( x^n \) is \( nx^{n-1} \).
• Combine derivatives linearly: the derivative of a sum is the sum of the derivatives.

In specific exercises like this one, differentiation is essential for finding the slope of the tangent line, and thus, leads us toward determining the parameters needed for the equation of the normal to the curve.

Perpendicular Slopes

When two lines are perpendicular, their slopes have a very special relationship described as negative reciprocals of each other. This simply means that if the slope of one line is \( m \), then the slope of the line perpendicular to it is \( -\frac{1}{m} \). For example, if a tangent line to a curve at some point has a slope of 28, the slope of the line normal to it at that same point will be \(-\frac{1}{28}\).

• Perpendicular slopes are crucial in geometry and calculus when we need to deal with orthogonal trajectories or find pathways that travel directly away from tangent directions.
• This concept helps us determine the orientation and equations of lines that maintain right angles to specified linear features.

Understanding perpendicular slopes is foundational in tasks requiring precision in directionality and construction of perpendicular intersecting lines.