Problem 40 - 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 \(P\). $$ f(x)=x^{2}-3 / x \quad P=(-3,10) $$

Step-by-Step Solution

Answer

The equation of the normal line is \( y = \frac{3}{17}x + \frac{179}{17} \).

Step 1: Find the derivative of the function

First, we need to differentiate the function to find the slope of the tangent line at point \( P \). The function given is \( f(x) = x^2 - \frac{3}{x} \). To find the derivative, apply the power rule for the terms:\[ f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}\left(\frac{3}{x}\right) = 2x + \frac{3}{x^2} \]

Step 2: Calculate the slope of the tangent line at P

Evaluate the derivative at \( x = -3 \) to determine the slope of the tangent line at the given point. Substitute and calculate:\[ f'(-3) = 2(-3) + \frac{3}{(-3)^2} = -6 + \frac{3}{9} = -6 + \frac{1}{3} = -\frac{18}{3} + \frac{1}{3} = -\frac{17}{3} \] Therefore, the slope of the tangent line at \( P(-3, 10) \) is \( -\frac{17}{3} \).

Step 3: Determine the slope of the normal line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope. Thus:\[ m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{-\frac{17}{3}} = \frac{3}{17} \] So, the slope of the normal line is \( \frac{3}{17} \).

Step 4: Use point-slope form to find the equation of the normal line

Use the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found in the previous step and \( (x_1, y_1) \) is the point \( P(-3, 10) \).Substitute to get:\[ y - 10 = \frac{3}{17}(x + 3) \]

Step 5: Convert to slope-intercept form

Simplify the equation to get it in \( y = mx + b \) form. Distribute the slope and simplify:\[ y - 10 = \frac{3}{17}x + \frac{3}{17} \times 3 \]\[ y - 10 = \frac{3}{17}x + \frac{9}{17} \]Add 10 to both sides:\[ y = \frac{3}{17}x + \frac{9}{17} + \frac{170}{17} \]\[ y = \frac{3}{17}x + \frac{179}{17} \] This is the slope-intercept form of the normal line.

Key Concepts

Derivative Calculation

To find the equation of the normal line to the graph of a function at a specific point, we first need to understand the concept of a derivative. The derivative of a function at any given point provides the slope of the tangent line at that point.

For the given function, \( f(x) = x^2 - \frac{3}{x} \), we differentiate it with respect to \( x \). Differentiating involves applying the power rule: this rule states that the derivative of \( x^n \) is \( nx^{n-1} \). We apply the power rule to each term in the function.

• For \( x^2 \), the derivative is \( 2x \).
• For \(-\frac{3}{x} \), rewrite it as \(-3x^{-1} \) and differentiate to get \(3x^{-2} \).

Putting it together, the derivative of the function is \( f'(x) = 2x + \frac{3}{x^2} \). This expression will help us find the tangent line slope.

Tangent Line Slope

Once we have the derivative, we can determine the slope of the tangent line at point \( P \) by evaluating the derivative at the x-coordinate of the point, \( x = -3 \).

Calculating \( f'(-3) \), we substitute \( -3 \) into our derivative \( f'(x) = 2x + \frac{3}{x^2} \). This calculation gives:

• \( 2(-3) = -6 \)
• \( \frac{3}{(-3)^2} = \frac{3}{9} = \frac{1}{3} \)

So, \( f'(-3) = -6 + \frac{1}{3} = -\frac{18}{3} + \frac{1}{3} = -\frac{17}{3} \). Thus, the tangent line's slope at \( P(-3, 10) \) is \(-\frac{17}{3}\).

Negative Reciprocal

For lines that are perpendicular, such as the tangent and normal lines, the slopes have a special relationship: they are negative reciprocals of each other. In simple terms, if you have the slope \( m \) for one line, the slope for a line perpendicular to it is \(-\frac{1}{m} \).

In this exercise, we already found the slope of the tangent line as \(-\frac{17}{3} \). To find the normal line's slope, we calculate the negative reciprocal:

• The negative reciprocal of \(-\frac{17}{3}\) is \(\frac{3}{17} \).

This slope \( \frac{3}{17} \) will help us write the equation of the normal line.

Slope-Intercept Form

To express the equation of the normal line in slope-intercept form, recall the equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

We first use the point-slope form, \( y - y_1 = m(x - x_1) \), with point \( P(-3, 10) \) and slope \( \frac{3}{17} \). This gives:

• \( y - 10 = \frac{3}{17}(x + 3) \)

Next, we simplify to get it into \( y = mx + b \) form:

• Distribute \( \frac{3}{17} \): \( y - 10 = \frac{3}{17}x + \frac{9}{17} \).
• Add 10 to get \( y \) alone on the left: \( y = \frac{3}{17}x + \frac{9}{17} + \frac{170}{17} \)
• Simplify to \( y = \frac{3}{17}x + \frac{179}{17} \).

And that is how you get the slope-intercept form of the normal line equation for the given point and function.