iGCSE (2021 Edition)

# 11.04 Tangents and normals

Lesson

So far this chapter you have learnt that, given a function $f(x)$f(x), it is possible to find a derivative function $f'(x)$f(x) which we can use to find the gradient of the tangent at any point on the original function. Remember, the tangent is defined as a straight line or plane that touches a curve or curved surface at a single point.

In this lesson we will look at the definition of the angle of inclination of a line and how the derivative allows us to determine the equations of tangents and normals to a function given a single point.

A normal at a point is a line that is perpendicular to the tangent at the same point on the curve.

Since the normal and tangent at a point are perpendicular to each other their gradients are related by:

$m_1\ m_2=-1$m1 m2=1

Or:

$m_2=-\frac{1}{m_1}$m2=1m1

Where $m_1$m1 and $m_2$m2 are the gradients of the tangent and normal. That is, the gradient of the normal is the negative reciprocal of the gradient of the tangent at the same point on a function.

### Finding equations of tangents and normals

Finding equations of tangents and normals requires knowing how to find the equation of a straight line.  Remember that we need a point $(x_1,y_1)$(x1,y1)  and the gradient $m$m.

Knowing these two things we can use the point gradient formula to determine the equation of a line.

$y-y_1=m(x-x_1)$yy1=m(xx1)

#### Worked examples

##### Example 1

Find the point on the curve $y=3-4x^2$y=34x2  where the gradient is $16.$16.

Think: We will take the derivative of $y$y and equate it to the gradient given.

Do:

 $\frac{dy}{dx}$dydx​ $=$= $-8x$−8x

Substituting $\frac{dy}{dx}=16$dydx=16 gives:

 $-8x$−8x $=$= $16$16

Solving for $x$x:

 Solving for $x$x: $x$x $=$= $-2$−2

To find the $y$y coordinate of the point at $x=-2$x=2 we substitute the $x$x value back into the equation for $y$y:

 $y$y $=$= $3-4x^2$3−4x2 $y$y $=$= $3-4(-2)^2$3−4(−2)2 $y$y $=$= $-13$−13

Therefore, the point on the curve $y=3-4x^2$y=34x2  where the gradient is $16$16 is $(-2,\ -13).$(2, 13).

##### Example 2

Find the equation of the tangent and normal to the curve with equation $y=x^3-3x^2+2$y=x33x2+2, at the point $(1,0)$(1,0)

Think: To find the equations of the tangent and the normal we need to know the gradient of the two lines and a single point on each line. The gradient will be given by the derivative of the curve evaluated at that point.

Do: Find the gradient function by differentiating $y=x^3-3x^2+2$y=x33x2+2:

$y'=3x^2-6x$y=3x26x

Evaluate the gradient at the point $(1,0)$(1,0):

$y'(1)=3(1)^2-6\times1=-3$y(1)=3(1)26×1=3

Identify the gradient of the tangent and the gradient of the normal.

The value of the gradient at the point is $m_1=-3$m1=3 as found above. The gradient of the normal will be:

$m_2=-\frac{1}{m_1}=-\frac{1}{-3}=\frac{1}{3}$m2=1m1=13=13

Find the equation of the tangent using the point gradient formula with point $(1,0)$(1,0) and gradient $-3$3

 $y-y_1$y−y1​ $=$= $m(x-x_1)$m(x−x1​) $y-0$y−0 $=$= $-3(x-1)$−3(x−1) $y$y $=$= $-3x+3$−3x+3

This is the equation of the tangent.

Find the equation of the normal using the point gradient formula with point $(1,0)$(1,0) and gradient $\frac{1}{3}$13:

 $y-y_1$y−y1​ $=$= $m(x-x_1)$m(x−x1​) $y-0$y−0 $=$= $\frac{1}{3}(x-1)$13​(x−1) $y$y $=$= $\frac{x}{3}-\frac{1}{3}$x3​−13​

This is the equation of the normal.

Let's just confirm that these all look correct on a graph.

### Angle of inclination

Another way to think about the gradient of a line is to look at the angle of inclination, which is the angle that a line makes with the positive $x-axis$xaxis. The following diagram illustrates the angle of inclination$(\theta)$(θ) in a right-angled triangle $ABC.$ABC.

Using right-angled trigonometry and equating to our existing definition of gradient, with the rise $=\ AB$= AB and the run $=\ BC$= BC we can say that:

$\tan(\theta)=\frac{opposite}{adjacent}=\frac{AB}{BC}=\frac{rise}{run}=m$tan(θ)=oppositeadjacent=ABBC=riserun=m

This leads us to the result that:

Formula for finding the gradient using the angle of inclination

$m=\tan(\theta)$m=tan(θ)

To find an angle using $m=\tan(\theta)$m=tan(θ), rearrange the formula such that it becomes

That is:

$\theta=\tan^{-1}(x)$θ=tan1(x)

#### Worked examples

##### example 3

Find the gradient of a line with an angle of inclination of $45°$45° with the positive $x-axis$xaxis

Think: The formula $m=\tan(\theta)$m=tan(θ) relates the gradient and angle of inclination.

Do:

$m=\tan45°$m=tan45°

$m=1$m=1

Therefore, the gradient of the line is $1$1

#### Practice questions

##### Question 1

By considering the graph of $f\left(x\right)=2x$f(x)=2x, find $f'$f$\left(-5\right)$(5).

##### Question 2

Consider the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15.

1. Find $f'\left(x\right)$f(x).

2. Find the gradient of the tangent to the curve at the point $\left(4,63\right)$(4,63).

3. Determine the equation of the tangent to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).

Express the equation of the tangent line in the form $y=mx+c$y=mx+c.

4. Find the gradient of the normal to the curve at the point $\left(4,63\right)$(4,63).

5. Determine the equation of the normal to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).

##### Question 3

A line passing through the points $\left(2,-3\right)$(2,3) and $\left(x,9\right)$(x,9) makes an angle of $120^\circ$120° with the positive $x$x-axis. Solve for the value of $x$x, expressing your answer in simplest rationalised form.

### Outcomes

#### 0606C14.5A

Apply differentiation to gradients, tangents and normals.