Standard level

# 10.08 Equations of 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.