Problem 37 - 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 \(P\). $$ f(x)=3 x^{2} \quad P=(-1,3) $$

Step-by-Step Solution

Answer

The normal line equation is \(y = \frac{1}{6}x + \frac{19}{6}\).

Step 1: Find the Derivative of the Function

The first step in finding the slope of the normal line is to compute the derivative of the function. Given \( f(x) = 3x^2 \), we find the derivative with respect to \( x \). Using the power rule, the derivative is \( f'(x) = 6x \).

Step 2: Evaluate the Derivative at Point P

Substitute the \( x \)-coordinate of point \( P \) into the derivative to find the slope of the tangent line at \( P \). Since \( P = (-1, 3) \), substitute \( -1 \) into \( f'(x) = 6x \) to get the slope \( f'(-1) = 6(-1) = -6 \).

Step 3: Determine the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, if the slope of the tangent line is \(-6\), the slope of the normal line is \( \frac{1}{6} \).

Step 4: Use Point-Slope Form to Find the Equation

Now use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is point \(P\) and \(m\) is the slope of the normal line. Substitute \((x_1, y_1) = (-1, 3)\) and \(m = \frac{1}{6}\) to get \( y - 3 = \frac{1}{6}(x + 1) \).

Step 5: Convert to Slope-Intercept Form

Simplify the equation from the point-slope form to the slope-intercept form \(y = mx + b\). Start by distributing \(\frac{1}{6}\): \(y - 3 = \frac{1}{6}x + \frac{1}{6}\). Add 3 to both sides to solve for \(y\): \(y = \frac{1}{6}x + \frac{1}{6} + 3\). Simplify to get \(y = \frac{1}{6}x + \frac{19}{6}\).

Key Concepts

Derivative of a Function

The derivative of a function measures how the function value changes as its input changes. In simple terms, it tells us the rate at which a function is changing at any given point. When given a function like \( f(x) = 3x^2 \), the derivative helps us find the slope of the tangent line at any point on its graph. To find the derivative of this function, we use the power rule. The power rule states that for any function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \). For our function, applying the power rule gives us \( f'(x) = 6x \). This derivative provides the slope of the tangent line at any point \( x \) on the curve. So essentially, by calculating this, we can find how steep the curve is at a particular point.

Slope of Tangent Line

The slope of the tangent line at a given point on a function's graph reflects the rate of change of the function at that point. It's a fundamental concept in calculus that helps understand the behavior of functions. To find this slope for the function \( f(x) = 3x^2 \) at point \( P = (-1, 3) \), we evaluate the derivative at \( x = -1 \).
- Start with the derivative \( f'(x) = 6x \). - Substitute \( x = -1 \) to find \( f'(-1) = 6(-1) = -6 \).
This calculation reveals that the slope of the tangent line at the point \( P \) is \(-6\). This means that at the point \((-1, 3)\), the curve is decreasing steeply.

Negative Reciprocal

The concept of the negative reciprocal is essential when determining the slope of the normal line to a curve at a given point. The normal line is perpendicular to the tangent line and shares a unique relationship with its slope. When you have the slope of a tangent line, the slope of the normal line is the negative reciprocal of that slope.

• If the tangent line's slope is \(-6\), then the normal line's slope will be \( \frac{1}{6} \).

The negative reciprocal simply means turning the number upside down (reciprocal) and then changing the sign (negative). For example, if the tangent slope is \( m = -6 \), the negative reciprocal is found by inverting \( 6 \) to \( \frac{1}{6} \) and changing the sign, resulting in a normal line slope of \( \frac{1}{6} \).
This relationship ensures that the two lines are perfectly perpendicular to each other.

Point-Slope Form

The point-slope form of a line's equation is an efficient way to write the equation 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 point.
In our scenario, we use the slope of the normal line, \( \frac{1}{6} \), and the point \( P = (-1, 3) \). Plugging these values into the point-slope form, the equation becomes:- \( y - 3 = \frac{1}{6}(x + 1) \).
This form makes it easier to transition to the slope-intercept form \( y = mx + b \), which is more commonly used. By distributing \( \frac{1}{6} \) and solving for \( y \), we convert the equation to \( y = \frac{1}{6}x + \frac{19}{6} \). This step reveals how the normal line behaves on a coordinate plane.