← Back to Calculus
Grade 12
CAPS Syllabus
Calculus
Derivative Applications

Tangents & Normals to Curves

Learn how to find the equations of tangent and normal lines to curves using derivatives. Master this essential calculus skill with interactive visualizations and worked examples.

What You'll Learn
  • The gradient of a curve equals the gradient of its tangent at any point
  • How to find the equation of a tangent using derivatives
  • How to find the equation of a normal (perpendicular to tangent)
  • The relationship: $m_{\text{tangent}} \times m_{\text{normal}} = -1$

What is a Tangent?

A tangent to a curve at a given point is a straight line that "just touches" the curve at that point. It has the same gradient as the curve at that exact point.

Key Concept

At a given point on a curve, the gradient of the curve equals the gradient of the tangent to the curve.

$$\text{Gradient of curve} = \text{Gradient of tangent} = \frac{dy}{dx}$$

Think About It:

Imagine a car driving along a curved road. At any moment, if the car drove straight, it would follow the tangent line to the curve at that point!

The Derivative = Gradient of the Tangent

This is a fundamental connection in calculus: the derivative of a function tells us the gradient of the tangent line at any point.

The Connection

If we have a function $y = f(x)$:

Function:$y = f(x)$
Derivative:$\frac{dy}{dx} = f'(x)$
Gradient at point:$m = f'(x_1)$

The Power of This Idea

Once you can find the derivative, you can find the gradient of the tangent at any point. And once you have the gradient and a point, you can find the equation of the tangent line!

Finding the Equation of a Tangent

Here's the step-by-step method for finding the equation of a tangent to a curve at a given point:

4-Step Method
1Find the derivative

Use the rules of differentiation to find $\frac{dy}{dx}$ or $f'(x)$.

2Calculate the gradient

Substitute the x-coordinate of the given point into the derivative to find the gradient $m$ of the tangent.

3Use the straight line equation

Substitute the gradient and the coordinates of the point into the gradient-point form:

$$y - y_1 = m(x - x_1)$$
4Simplify

Make $y$ the subject of the formula and simplify to get the final equation in the form $y = mx + c$.

Normals to Curves

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

Perpendicular Lines

For two lines to be perpendicular, the product of their gradients must equal −1:

$$m_{\text{tangent}} \times m_{\text{normal}} = -1$$

Therefore, if the tangent has gradient $m$, the normal has gradient:

$$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$$

Important:

If the tangent has gradient 3, the normal has gradient $-\frac{1}{3}$.
If the tangent has gradient $-\frac{1}{2}$, the normal has gradient 2.

Finding the Equation of a Normal

Follow the same steps as for a tangent, but with one key difference:

  1. Find the derivative and calculate the gradient of the tangent
  2. Calculate the gradient of the normal: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$
  3. Use $y - y_1 = m_{\text{normal}}(x - x_1)$
  4. Simplify to standard form

Worked Examples

Example 1: Finding the Equation of a Tangent
Find the equation of the tangent to the curve $y = 3x^2$ at the point $(1; 3)$.

Step 1: Find the derivative

$$\begin{aligned} y &= 3x^2 \\ \frac{dy}{dx} &= 3(2x) \\ &= 6x \end{aligned}$$

Step 2: Calculate the gradient

At $x = 1$:

$$\begin{aligned} m &= 6x \\ &= 6(1) \\ &= 6 \end{aligned}$$

Step 3: Use the straight line equation

Using $y - y_1 = m(x - x_1)$ with point $(1; 3)$:

$$\begin{aligned} y - 3 &= 6(x - 1) \\ y - 3 &= 6x - 6 \\ y &= 6x - 6 + 3 \\ y &= 6x - 3 \end{aligned}$$

Answer:

$y = 6x - 3$

Example 2: Finding a Tangent at a Given x-value
Given $g(x) = (x + 2)(2x + 1)^2$, determine the equation of the tangent at $x = -1$.

Step 1: Find the y-coordinate

$$\begin{aligned} g(-1) &= (-1 + 2)[2(-1) + 1]^2 \\ &= (1)(-2 + 1)^2 \\ &= (1)(-1)^2 \\ &= 1 \end{aligned}$$

So the point is $(-1; 1)$.

Step 2: Expand the function

$$\begin{aligned} g(x) &= (x + 2)(2x + 1)^2 \\ &= (x + 2)(4x^2 + 4x + 1) \\ &= 4x^3 + 4x^2 + x + 8x^2 + 8x + 2 \\ &= 4x^3 + 12x^2 + 9x + 2 \end{aligned}$$

Step 3: Find the derivative

$$\begin{aligned} g'(x) &= 4(3x^2) + 12(2x) + 9 \\ &= 12x^2 + 24x + 9 \end{aligned}$$

Step 4: Calculate the gradient

$$\begin{aligned} g'(-1) &= 12(-1)^2 + 24(-1) + 9 \\ &= 12 - 24 + 9 \\ &= -3 \end{aligned}$$

Step 5: Find the equation

$$\begin{aligned} y - 1 &= -3(x - (-1)) \\ y - 1 &= -3(x + 1) \\ y - 1 &= -3x - 3 \\ y &= -3x - 2 \end{aligned}$$

Answer:

$y = -3x - 2$

Example 3: Finding the Equation of a Normal
Determine the equation of the normal to the curve $xy = -4$ at $(-1; 4)$.

Step 1: Make y the subject and differentiate

$$\begin{aligned} xy &= -4 \\ y &= -\frac{4}{x} = -4x^{-1} \\ \frac{dy}{dx} &= -4(-1x^{-2}) \\ &= 4x^{-2} = \frac{4}{x^2} \end{aligned}$$

Step 2: Calculate the gradient of the tangent

$$\begin{aligned} \frac{dy}{dx} &= \frac{4}{(-1)^2} \\ &= \frac{4}{1} \\ &= 4 \end{aligned}$$

Step 3: Calculate the gradient of the normal

$$\begin{aligned} m_{\text{tangent}} \times m_{\text{normal}} &= -1 \\ 4 \times m_{\text{normal}} &= -1 \\ m_{\text{normal}} &= -\frac{1}{4} \end{aligned}$$

Step 4: Find the equation of the normal

$$\begin{aligned} y - 4 &= -\frac{1}{4}(x - (-1)) \\ y - 4 &= -\frac{1}{4}(x + 1) \\ y - 4 &= -\frac{1}{4}x - \frac{1}{4} \\ y &= -\frac{1}{4}x - \frac{1}{4} + 4 \\ y &= -\frac{1}{4}x + \frac{15}{4} \end{aligned}$$

Answer:

$y = -\frac{1}{4}x + \frac{15}{4}$

Interactive Visualization

Interactive Tangent Visualization
Move the slider to see how the tangent changes at different points

At x = 1.0:

y = 1.00

Gradient = 2.00

Tangent equation:

y = 2.00x + -1.00

Practice Questions

Additional Practice Questions
Access more practice problems with PDF export functionality. Download questions-only or with complete solutions for offline study.
View Practice Page

Apply what you've learned with these practice problems.

Question 1
Question 2
Question 3

Quick Self-Check Quiz

Test your understanding with these questions.

Question 1
What does the derivative of a function represent?
Question 2
If the tangent to a curve has gradient 5, what is the gradient of the normal?
Question 3
What is the first step in finding the equation of a tangent to a curve?
Question 4
The normal to a curve at a point is:
Question 5
For $f(x) = x^2$, what is $f'(3)$?
Question 6
If two lines are perpendicular, the product of their gradients is:
Question 7
Which form of the straight line equation is most useful for finding tangent equations?

Summary

"The tangent is the curve's best straight-line approximation at a point."

Key Formula:

Gradient of tangent = f'(x)

For Perpendicular Lines:

$m_{\text{tangent}} \times m_{\text{normal}} = -1$

Steps:

  1. Find derivative
  2. Calculate gradient at point
  3. Use $y - y_1 = m(x - x_1)$
  4. Simplify
More Calculus LessonsReview Differentiation Rules