Gradients, Tangents and Normals

Finding equations of tangent and normal lines to curves using differentiation

SPEC 7.3 • DIFFERENTIATION

Gradient of a Curve

The gradient of a curve at any point is given by the derivative

\frac{d y}{d x}

evaluated at that point.

Key Concept

To find the gradient at a specific point

(a, b)

:

Gradient = \frac{d y}{d x}_{x = a}

Method

Differentiate the function

Substitute the x-value

Calculate the gradient

Remember

The gradient changes at every point on a curve - unlike a straight line which has constant gradient.

Worked Example01

Find the gradient of the curve $y = x^{3} - 4 x + 2$ at the point where $x = 2$ .

1

Differentiate:

\frac{d y}{d x} = 3 x^{2} - 4

2

Substitute

x = 2

:

\frac{d y}{d x} = 3 (2)^{2} - 4 = 12 - 4 = 8

AnswerGradient = 8

Tangent Lines

A tangent is a straight line that touches a curve at exactly one point and has the same gradient as the curve at that point.

Tangent Formula

y - y_{1} = m (x - x_{1})

where

m = \frac{d y}{d x}

at the point

(x_{1}, y_{1})

Method for Finding Tangent

Find

\frac{d y}{d x}

Substitute the x-value to find gradient

m

Find the y-coordinate (if not given)

Use

y - y_{1} = m (x - x_{1})

Rearrange to required form

Worked Example02

Find the equation of the tangent to the curve $y = x^{2} - 3 x + 5$ at the point $(2, 3)$ .

1

Differentiate:

\frac{d y}{d x} = 2 x - 3

2

At

x = 2

:

m = 2 (2) - 3 = 1

3

Using

y - y_{1} = m (x - x_{1})

with point

(2, 3)

and

m = 1

:

4

y - 3 = 1 (x - 2)

5

y = x - 2 + 3 = x + 1

Answer

y = x + 1

Normal Lines

A normal is a straight line that is perpendicular (at right angles) to the tangent at the point of contact.

Key Relationship

If the tangent has gradient

m

, the normal has gradient:

m_{normal} = - \frac{1}{m}

Because perpendicular gradients multiply to give

- 1

Common Error

Don't forget the negative sign! If tangent gradient is

3

, normal gradient is

- \frac{1}{3}

, NOT

\frac{1}{3}

.

Worked Example03

Find the equation of the normal to the curve $y = x^{2} - 3 x + 5$ at the point $(2, 3)$ .

1

From Example 2, tangent gradient

m = 1

2

Normal gradient:

m_{n} = - \frac{1}{1} = - 1

3

Using

y - y_{1} = m (x - x_{1})

with

(2, 3)

and

m = - 1

:

4

y - 3 = - 1 (x - 2)

5

y = - x + 2 + 3 = - x + 5

Answer

y = - x + 5

or

x + y = 5

Worked Example04

Find the equations of the tangent and normal to $y = 2 x^{3} - x$ at the point where $x = 1$ .

1

Find y when

x = 1

:

y = 2 (1)^{3} - 1 = 1

, so point is

(1, 1)

2

Differentiate:

\frac{d y}{d x} = 6 x^{2} - 1

3

At

x = 1

:

m = 6 (1)^{2} - 1 = 5

4

Tangent:

y - 1 = 5 (x - 1)

→

y = 5 x - 4

5

Normal:

m_{n} = - \frac{1}{5}

, so

y - 1 = - \frac{1}{5} (x - 1)

→

y = - \frac{1}{5} x + \frac{6}{5}

Tangent:

y = 5 x - 4

Normal:

5 y + x = 6

Worked Example05

Find the equation of the tangent to $y = e^{2 x}$ at the point where $x = 0$ .

1

Find y when

x = 0

:

y = e^{0} = 1

, so point is

(0, 1)

2

Differentiate:

\frac{d y}{d x} = 2 e^{2 x}

3

At

x = 0

:

m = 2 e^{0} = 2

4

Tangent:

y - 1 = 2 (x - 0)

Answer

y = 2 x + 1

Worked Example06

Find the equation of the normal to $y = sin x$ at the point where $x = \frac{π}{3}$ .

1

Find y:

y = sin \frac{π}{3} = \frac{3}{2}

, so point is

(\frac{π}{3}, \frac{3}{2})

2

Differentiate:

\frac{d y}{d x} = cos x

3

At

x = \frac{π}{3}

:

m = cos \frac{π}{3} = \frac{1}{2}

4

Normal gradient:

m_{n} = - \frac{1}{\frac{1}{2}} = - 2

5

Normal:

y - \frac{3}{2} = - 2 (x - \frac{π}{3})

Answer

y = - 2 x + \frac{2 π}{3} + \frac{3}{2}

Worked Example07

The tangent to $y = x^{2} + a x + b$ at $x = 1$ has equation $y = 5 x - 3$ . Find the values of $a$ and $b$ .

1

Tangent

y = 5 x - 3

has gradient 5 and passes through

x = 1

2

\frac{d y}{d x} = 2 x + a

. At

x = 1

:

2 (1) + a = 5

, so

a = 3

3

Point on tangent at

x = 1

:

y = 5 (1) - 3 = 2

, so point is

(1, 2)

4

Point must be on curve:

2 = (1)^{2} + 3 (1) + b

5

2 = 1 + 3 + b = 4 + b

, so

b = - 2

Answer

a = 3

,

b = - 2

Practice

Question 1: Find the equation of the tangent to

y = x^{3} - 2 x + 1

at the point

(1, 0)

.

\frac{d y}{d x} = 3 x^{2} - 2

. At

x = 1

:

m = 3 (1)^{2} - 2 = 1

Tangent:

y - 0 = 1 (x - 1)

Answer

y = x - 1

Practice

Question 2: Find the equation of the normal to

y = 4 x - x^{2}

at the point where

x = 3

.

When

x = 3

:

y = 12 - 9 = 3

, so point is

(3, 3)

\frac{d y}{d x} = 4 - 2 x

. At

x = 3

:

m = 4 - 6 = - 2

Normal gradient:

m_{n} = - \frac{1}{- 2} = \frac{1}{2}

Normal:

y - 3 = \frac{1}{2} (x - 3)

Answer

y = \frac{1}{2} x + \frac{3}{2}

or

2 y = x + 3

Practice

Question 3: Find the equations of the tangent and normal to

y = ln x

at the point where

x = e

.

When

x = e

:

y = ln e = 1

, so point is

(e, 1)

\frac{d y}{d x} = \frac{1}{x}

. At

x = e

:

m = \frac{1}{e}

Tangent:

y - 1 = \frac{1}{e} (x - e)

→

y = \frac{x}{e}

Normal:

m_{n} = - e

, so

y - 1 = - e (x - e)

→

y = - e x + e^{2} + 1

Summary

Key Formulas

ConceptFormula

Gradient at a point

\frac{d y}{d x}_{x = a}

Tangent equation

y - y_{1} = m (x - x_{1})

where

m = \frac{d y}{d x}

Normal gradient

m_{n} = - \frac{1}{m}

Exam Skills

Always find BOTH coordinates before forming line equation

Remember the negative sign for normal gradient

Check your answer by substituting the point into your equation