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8.06 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.

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.

Point gradient formula

$y-y_1=m(x-x_1)$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$34x2
$y$y $=$= $3-4(-2)^2$34(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

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-0$y0 $=$= $-3(x-1)$3(x1)
$y$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:

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-0$y0 $=$= $\frac{1}{3}(x-1)$13(x1)
$y$y $=$= $\frac{x}{3}-\frac{1}{3}$x313

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

MA11-1

uses algebraic and graphical techniques to solve, and where appropriate, compare alternative solutions to problems

MA11-5

interprets the meaning of the derivative, determines the derivative of functions and applies these to solve simple practical problems

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