### 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