Standard level

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:

$m_1\ m_2=-1$`m`1 `m`2=−1

Or:

$m_2=-\frac{1}{m_1}$`m`2=−1`m`1

Where $m_1$`m`1 and $m_2$`m`2 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 requires knowing how to find the equation of a straight line. Remember that we need a point $(x_1,y_1)$(`x`1,`y`1) 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

$y-y_1=m(x-x_1)$`y`−`y`1=`m`(`x`−`x`1)

Find the point on the curve $y=3-4x^2$`y`=3−4`x`2 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 $\frac{dy}{dx}=16$`d``y``d``x`=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`=3−4`x`2 where the gradient is $16$16 is $(-2,\ -13).$(−2, −13).

Find the equation of the tangent and normal to the curve with equation $y=x^3-3x^2+2$`y`=`x`3−3`x`2+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`=`x`3−3`x`2+2:

$y'=3x^2-6x$`y`′=3`x`2−6`x`

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 $m_1=-3$`m`1=−3 as found above. The gradient of the normal will be:

$m_2=-\frac{1}{m_1}=-\frac{1}{-3}=\frac{1}{3}$`m`2=−1`m`1=−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.