Applications of Differentiation Part I

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 intersect the curve

When x =2 , $ y = \frac{5}{1+2^2} = 1 $

Using $\ y = mx + c $,
Sub x =2, y= 1 and m= -4 to find c
$ 1 = -4(2) + c $ , c= 9

Therefore equation of tangent is $ y = -4x + 9 $

Example 2
The equation of a curve is $ y = \frac{5}{1 + x^2} $ Find the equation of the normal to the curve at the point where x = 3 .

The normal is perpendicular to the tangent. Hence the gradient of the normal is negative reciprocal of the gradient of the tangent.

Gradient of tangent can be found by substituting x = 3 into $\frac{dy}{dx} $

$\begin{aligned}
\frac{dy}{dx} &= \frac{-10x}{(1+x^2)^2} \\
&= \frac{-10(3)}{(1 + 3^2)^2} \\
&= -\frac{3}{10} \\
\end{aligned} $

Gradient of normal = $ \frac{10}{3} $

The point at which the normal intersects the curve can be obtained by substituting x =3 into $ y = \frac{5}{1 + x^2} =\frac{5}{10} = \frac{1}{2} $

Using $\ y = mx + c $,
Sub $ x = 3 \qquad y= \frac{1}{2} \qquad m= \frac{10}{3} $ to find c

$ \frac{1}{2} = \frac{10}{3} (3) + c \qquad c= -9\frac{1}{2} $

Therefore equation of tangent is $y = \frac{10}{3}x -9\frac{1}{2} $

What's Next ?
For more Applications of Differentiation
click here for Applications of Differentiation Part III: Stationary Values
click here for Applications of Differentiation Part IV: Connected Rates of Change