### 2 Tangents and normals to a curve

As we know, the relationship between an independent variable $x$ and a dependent variable $y$ is denoted by

$\phantom{\rule{2em}{0ex}}y=f\left(x\right)$

As we also know, the geometrical interpretation of this relation takes the form of a curve in an $xy$ plane as illustrated in Figure 4.

Figure 4

We know how to calculate a value of $y$ given a value of $x$ . We can either do this graphically (which is inaccurate) or else use the function itself. So, at an $x$ value of ${x}_{0}$ the corresponding $y$ value is ${y}_{0}$ where

$\phantom{\rule{2em}{0ex}}{y}_{0}=f\left({x}_{0}\right)$

Let us examine the curve in the neighbourhood of the point $\left({x}_{0},{y}_{0}\right)$ . There are two important constructions of interest

• the tangent line at $\left({x}_{0},{y}_{0}\right)$
• the normal line at $\left({x}_{0},{y}_{0}\right)$

These are shown in Figure 5.

Figure 5

We note the geometrically obvious fact: the tangent and normal lines at any given point on a curve are perpendicular to each other.

The curve $y={x}^{2}$ is drawn below. On this graph draw the tangent line and the normal line at the point $\left({x}_{0}=1,{y}_{0}=1\right)$ :

From your graph, estimate the values of $\theta$ and $\psi$ in degrees. (You will need a protractor.)

$\phantom{\rule{2em}{0ex}}\theta \approx 63.{4}^{o}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\psi \approx 153.{4}^{o}$

Returning to the curve $y=f\left(x\right):$ we know, from the geometrical interpretation of the derivative that

$\phantom{\rule{2em}{0ex}}\frac{df}{dx}{\left|\right}_{{x}_{0}}=tan\theta$

(the notation $\frac{df}{dx}{\left|\right}_{{x}_{0}}$ means evaluate $\frac{df}{dx}$ at the value $x={x}_{0}$ )

Here $\theta$ is the angle the tangent line to the curve $y=f\left(x\right)$ makes with the positive $x$ -axis. This is highlighted in Figure 6:

Figure 6