Equation for the Normal Line to a Curve – Step-by-step

The equation of the normal line to a curve can be found by finding the slope of the tangent line at the given point using the derivative of the function. Then, the slope of the normal line is equal to -1/m. Finally, we use the form y=mx+b with the coordinates of the point to find the value of b.

Here, we will look at the process we can use to find the equation of the normal line to a curve. In addition, we will use this process to solve some practice problems.

CALCULUS
Diagram of the equation of the normal to a curve in a point P

Relevant for

Learning to find the equation of the normal line to a curve.

See process

CALCULUS
Diagram of the equation of the normal to a curve in a point P

Relevant for

Learning to find the equation of the normal line to a curve.

See process

Process for finding the equation of the normal line to a curve

The normal line to a curve at a point P is the straight line through point P that is perpendicular to the tangent line to the curve at point P.

Diagram of the equation of the normal to a curve in a point P

Since the tangent and normal are perpendicular to each other, if the slope of the tangent is $latex m$, then the slope of the normal is equal to $latex -\frac{1}{m}$ .

Therefore, if we want to find the equation of the normal line to a curve at the point $latex (x_{1},~y_{1})$, we can follow these steps:


Step 1: Find the derivative of the function that represents the curve.

Step 2: Find the slope of the tangent line to the curve at the point $latex (x_{1},~y_{1})$. For this, we use the x-coordinate of the point in the derivative of the function. That is, we have $latex m_{1}=f'(x_{1})$.

Step 3: Use the slope from Step 2 to find the slope of the normal line to the curve. The slope of the normal line is equal to $latex m=-\frac{1}{m_{1}}$.

Step 4: Substitute the slope from step 3 into the form $latex y=mx+b$ and use the x and y coordinates of the given point to find the value of b. That is, we have $latex y_{1}=mx_{1}=b$.

Step 5: Use the values of m and b in the form $latex y=mx+b$ to obtain the equation of the line.


Examples of the equation of the normal line to a curve

The following examples are solved using the process seen above to find the equation of the normal line to a curve. Each example has a detailed solution.

EXAMPLE 1

What is the equation of the normal line to the curve $latex f(x)=x^2$ at the point P=(3, 2)?

Step 1: We have to start by finding the derivative of the function:

$latex f(x)=x^2$

$latex f'(x)=2x$

Step 2: The slope of the tangent line at the point (3, 2) is equal to $latex f'(3)$. Thus, we have:

$latex m_{1}=f'(3)=2(3)$

$latex m_{1}=6$

Step 3: The slope of the normal line is equal to $latex m=-\frac{1}{6}$.

Step 4: Using the slope from step 3, we have the equation $latex y=-\frac{1}{6}x+b$. To find the value of b, we use the point (3, 2) in the equation:

$latex y=-\frac{1}{6}x+b$

$latex 2=-\frac{1}{6}(3)+b$

$latex b=\frac{5}{2}$

Step 5: The equation of the normal line at the point (3, 2) is $latex y=-\frac{1}{6}x+\frac{5}{2}$.

EXAMPLE 2

Find the equation of the normal line to $latex f(x)=2x^3-3x$ at the point (1, 3).

Step 1: We have to use the derivative of the function. Therefore, we have:

$latex f(x)=2x^3-3x$

$latex f'(x)=6x-3$

Step 2: To find the slope of the tangent line at (1, 3) we evaluate $latex f'(1)$. Then, we have:

$latex m_{1}=f'(1)=6(1)-3$

$latex m_{1}=3$

Step 3: We use the slope from step 2 to find the slope of the normal line. Thus, the slope of the normal line is $latex m=-\frac{1}{3}$.

Step 4: Now, we can form the equation $latex y=-\frac{1}{3}x+b$. Now, let’s find the value of b, using the point (1, 3) in the equation:

$latex y=3x+b$

$latex 3=-\frac{1}{3}(1)+b$

$latex b=\frac{10}{3}$

Step 5: The equation of the normal line at the point (1, 3) is $latex y=-\frac{1}{3}x+\frac{10}{3}$.

EXAMPLE 3

What is the equation of the normal line to $latex f(x)=x^3+\frac{8}{x}$ at the point (2, -2)?

Step 1: To get the derivative of the function, we write the radical as a numerical exponent:

$latex f(x)=x^3+8x^{-1}$

$latex f'(x)=x^2-8x^{-2}$

$latex f'(x)=x^2-\frac{8}{x^2}$

Step 2: Let’s find the slope of the tangent at (2, -2) using $latex f'(2)$:

$latex m_{1}=f'(2)=(2)^2-\frac{8}{2^2}$

$latex =4-2$

$latex m_{1}=2$

Step 3: Using the slope from step 2, we find the slope of the normal line. Therefore, we have $latex m=-\frac{1}{2}$.

Step 4: We have formed the equation $latex y=-\frac{1}{2}x+b$. To find the value of b, we use the point (2, -2) in the equation:

$latex y=-\frac{1}{2}x+b$

$latex -2=-\frac{1}{2}(2)+b$

$latex b=-1$

Step 5: The equation of the normal line at the point (2, -2) is $latex y=-\frac{1}{2}x-1$.

EXAMPLE 4

What is the normal line to $latex f(x) = -x^{-2}+\sqrt{x}$ at the point (1, 3)?

Step 1: We get the derivative of the function by writing the square root as a numerical exponent:

$latex f(x)=-x^{-2}+x^{\frac{1}{2}}$

$$f'(x)=2x^{-3}+\frac{1}{2}x^{-\frac{1}{2}}$$

$$f'(x)=\frac{2}{x^3}+\frac{1}{2\sqrt{x}}$$

Step 2: The slope of the tangent line at (1, 3) is equal to $latex f'(1)$:

$$m_{1}=f'(1)=\frac{2}{(1)^3}+\frac{1}{2\sqrt{1}}$$

$latex =2+\frac{1}{2}$

$latex m_{1}=\frac{5}{2}$

Step 3: The slope of the normal line is equal to $latex m=-\frac{2}{5}$.

Step 4: Using the slope from step 3 we form the equation $latex y=-\frac{2}{5}x+b$. To find the value of b, we use the point (1, 3) in the equation:

$$y=-\frac{2}{5}x+b$$

$$3=-\frac{2}{5}(1)+b$$

$latex b=\frac{17}{2}$

Step 5: The equation of the normal line at the point (1, 3) is $latex y=-\frac{2}{5}x+\frac{17}{2}$.

EXAMPLE 5

If we have the function $latex f(x)=\sin(x)-\cos(x)$, find the normal line at the point (0, 1).

Step 1: We need the derivative of the function to find the slope of the line. Therefore, we have:

$latex f(x)=\sin(x)-\cos(x)$

$latex f'(x)=\cos(x)+\sin(x)$

Step 2: The slope of the tangent line at the point (0, 1) is found by evaluating $latex f'(0)$. Then, we have:

$latex m=f'(0)=\cos(0)+\sin(0)$

$latex m_{1}=1+0$

$latex m_{1}=1$

Step 3: The slope of the normal line is equal to $latex m=-1$.

Step 4: We have the equation $latex y=-x+b$. Now, we find the value of b, using the point (0, 1) in the equation:

$latex y=-x+b$

$latex 1=-0+b$

$latex b=1$

Step 5: The equation of the normal line at the point (0, 1) is $latex y=-x+1$.

EXAMPLE 6

If we have the function $latex f(x)=x^2-3x+1$, find the equation of the normal line at the point where the curve intersects the y-axis.

Step 1: The derivative of the function is:

$latex f(x)=x^2-3x+1$

$latex f'(x)=2x-3$

Step 2: In this case, we don’t have the coordinates of the point directly. However, we know that the curve intersects the y-axis at that point. Then, let’s find the coordinates of that point.

When the curve intersects the y-axis, we know that the x-coordinates must be 0. Therefore, we have:

$latex y=x^2-3x+1$

$latex y=0^2-3(0)+1$

$latex y=1$

Thus, the point is (0, 1). The slope of the tangent line at the point (0, 1) is equal to $latex f'(0)$:

$latex m_{1}=f'(0)=2(0)-3$

$latex m_{1}=-3$

Step 3: The slope of the normal line is $latex m=\frac{1}{3}$.

Step 4: We have found the equation $latex y=\frac{1}{3}x+b$. To find the value of b, we use the point (0, 1) in the equation:

$latex y=\frac{1}{3}x+b$

$latex 1=\frac{1}{3}(0)+b$

$latex b=1$

Step 5: The equation of the normal line at the point (0, 1) is $latex y=\frac{1}{3}x+1$.


Equation of the normal line to a curve – Practice problems

Use everything learned in this article to solve the following practice problems and find the normal line to the given functions.

Find the equation of the normal line to $latex f(x)=x^2-3x$ where $latex x=2$.

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What is the equation of the normal line to $latex f(x)=x^3+4$ where $latex x=-1$?

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Find the equation of the normal line to $latex f(x)=\frac{6}{x}$ where $latex x=3$.

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Find the equation of the normal line to the function $latex f(x)=2\sqrt{x}$ where $latex x=9$.

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Find the equation of the normal line to $latex f(x)=6-\frac{1}{x^2}$ where $latex x=1$.

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See also

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