10.08 Equations of tangents and normals

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.

Gradient of a normal

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.

 

Worked examples

Example 1

Find the point on the curve $y=3-4x^2$y=3−4x2  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=3−4x2  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=x3−3x2+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=x3−3x2+2:

$y'=3x^2-6x$y′=3x2−6x

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

$y'(1)=3(1)^2-6\times1=-3$y′(1)=3(1)2−6×1=−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​​=−1−3​=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.