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.
$y-y_1=m(x-x_1)$y−y1=m(x−x1)
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.
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 $x-axis$x−axis. 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:
$m=\tan(\theta)$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=tan−1(x)
Worked examples
example 3
Find the gradient of a line with an angle of inclination of $45°$45° with the positive $x-axis$x−axis
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.
Find $f'\left(x\right)$f′(x).
Find the gradient of the tangent to the curve at the point $\left(4,63\right)$(4,63).
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.
Find the gradient of the normal to the curve at the point $\left(4,63\right)$(4,63).
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.