A LevelAQAEdexcelOCR

## Gradients, Tangents and Normals Revision

As mentioned in the Differentiation section, we can find a derivative to give the gradient of a graph at any given point.

From that information, we can create a tangent and a normal.

A LevelAQAEdexcelOCR

## How to Compare the Tangent and Normal

Let’s denote the gradient of the tangent $m_T$ and gradient of the normal $m_N$.

For a pair of tangent and normal lines at one point, we have one rule:

The two must be perpendicular.

This means that we must have

$m_T\cdot m_N = -1$

A LevelAQAEdexcelOCR

@mmerevise

A LevelAQAEdexcelOCR

## Example 1: Finding the Tangent

tangent is a straight line which touches our graph, but doesn’t pass through it at the meeting point. By definition, a straight line graph (i.e. $y = x$) cannot have a tangent – only a curved graph can have a tangent.

So, as an example, here’s the graph of $\textcolor{blue}{y = (x - 1)^2 - 2}$. Find the equation of the tangent line at $x = \dfrac{3}{2}$.

[4 marks]

To find the tangent, we first find the gradient at our point of interest.

Since we have

$\textcolor{blue}{y = (x - 1)^2 - 2}$

$= \textcolor{blue}{x^2 - 2x - 1}$

$\dfrac{dy}{dx} = 2x - 2$

at any point, and $\dfrac{dy}{dx} = 1$ when $x = \dfrac{3}{2}$.

Since our gradient is a straight line, it must have the general equation

$y = mx + c$

When $x = \dfrac{3}{2}$, $y = -\dfrac{7}{4}$ and $\dfrac{dy}{dx} = m_T = 1$.

Then we can plug in these values to find $c$:

$-\dfrac{7}{4} = \dfrac{3}{2} + c$

$c = -\dfrac{7}{4} - \dfrac{3}{2} = -\dfrac{13}{4}$

Therefore, our tangent to $y = (x - 1)^2 - 2$ at $x = \dfrac{3}{2}$ is given by the equation $\textcolor{red}{y = x - \dfrac{13}{4}}$.

A LevelAQAEdexcelOCR

## Example 2: Finding the Normal

The normal is a straight line which is exactly perpendicular to the tangent.

Say we have the same graph from Example 1, but now we wish to find the normal at the point $x = \dfrac{3}{2}$.

[3 marks]

Since we already know that the gradient of the tangent, $m_T = 1$, we can conclude that the gradient of the normal, $m_N = -1$.

Given, also, that the point of interest is at $\left( \dfrac{3}{2}, -\dfrac{7}{4} \right)$, we can form a straight line equation for the normal:

$-\dfrac{7}{4} = \left(-1 \times \dfrac{3}{2} \right) + c$

$c = -\dfrac{7}{4} + \dfrac{3}{2} = -\dfrac{1}{4}$

The equation of the normal is given by $\textcolor{limegreen}{y = -x - \dfrac{1}{4}}$.

A LevelAQAEdexcelOCR

## Gradients, Tangents and Normals Example Questions

$f(x) = -x^2$ gives $f'(x) = -2x$

When $x = 2$,

$m_T = -4$ and $m_N = \dfrac{-1}{-4} = \dfrac{1}{4}$

Gold Standard Education

$y = x^3$ gives a gradient of

$\dfrac{dy}{dx} = 3x^2$

When $x = 1$,

$\dfrac{dy}{dx} = 3$ and $y = 1$

Then

$1 = (3 \times 1) + c$

$c = 1 - 3 = -2$

The equation of the tangent at $x = 1$ is $y = 3x - 2$.

At $x = -1$, $\dfrac{dy}{dx} = 3 \times (-1)^2 = 3$.

Therefore, we can see that the gradients of the tangents at $x = 1$ and $x = -1$ are the same, so the two must run parallel.

Gold Standard Education

When $x = \dfrac{-1}{2}$, $y = \sqrt{\dfrac{7}{2}}$.

We have $m_T = \dfrac{1}{2\sqrt{\dfrac{7}{2}}} = \dfrac{1}{\sqrt{14}}$. Then $m_N = -2\sqrt{\dfrac{7}{2}} = -\sqrt{14}$.

$y = mx + c$ gives

$\sqrt{\dfrac{7}{2}} = \left( -\sqrt{14} \times \dfrac{-1}{2} \right) + c$

$\sqrt{\dfrac{7}{2}} = \sqrt{\dfrac{14}{4}} + c$

$\sqrt{\dfrac{7}{2}} = \sqrt{\dfrac{7}{2}} + c$

$c = 0$

So, the normal intercepts the $y$-axis at $y = 0$, or, more appropriately, the origin.

Gold Standard Education