Tangents and Normals

Equation of a line is given by: $ y = mx + c $

where m is the gradient and c is the y intercept

The gradient at a particular point on a curve can be found using $ \frac{dy}{dx} $

To find the equation of a line, you will need a point on the line and the gradient of the line

Example 1

The equation of a curve is $ y = \frac{5}{1 + x^2} $ Find the equation of the tangent to the curve at the point where x = 2 .

First we differentiate y wrt x

To make our job easier, we rewrite y as $ y = 5( 1 + x^2)^{-1} $

$ \begin{aligned}

\frac{dy}{dx} &= 5(-1)(1+x^2)^{-1-1}(2x) \qquad \text{applying chain rule} \\

&= (-10x)(1+x^2)^{-2} \\

&= \frac{-10x}{(1+x^2)^{2}}\\

\end{aligned} $

In order to find the equation of the tangent, we need to determine the gradient of the tangent.

When x = 2 , $ \frac{dy}{dx} = \frac{-10(2)}{1+2^2}} = -4 $

Now that we have determined the gradient, we need to find a point on the tangent

The point on the tangent is also the point at which the tangent interse…