## Differentiation – Normal of a Tangent

January 26, 2012 in Additional Mathematics, Differentiation, Normal of a Tangent by Shivakkumar Vadiveyl

## Section I Differentiation – Normal of a Tangent

In this lesson, we learn how to derive the gradient for a normal of a tangent of a curve and subsequently the equation for the normal.

The normal of a tangent is perpendicular to the tangent. That is, the normal is at 90 degrees to the tangent.

The gradient of a normal to the tangent is -1 divided by the gradient of the tangent.

The formula is $m_{n}=-dfrac{1}{m_{t}}$

where $m_{n}$= gradient of normal,

and $m_{t}$= gradient of the tangent.

This means that if we are given the gradient of the tangent, we can find the gradient of the normal and with that, we can derive the equation for the normal. As the normal is also a straight line, we can make use of this formula for the deriving the equation of the normal.

$begin{array}{ccccc}m_{n}= & -dfrac{1}{m_{t}} & = & dfrac{y-y_{1}}{x-x_{1}}& y-y_{1} & = & -dfrac{1}{m_{t}}(x-x_{1})end{array}$

Note that both the x and y coordinates must be part of the normal.

## Normal of a Tangent Example A

Given that a function of x is expressed with this equation,$begin{array}{ccccc}& y & = & 2x^{2}+x+1end{array}$, find the gradient of the normal at (2,11) and with that, define the expression for the normal line in x and y.

$begin{array}{ccccc}& y & = & 2x^{2}+x+1m_{t}= & dfrac{dy}{dx} & = & 4x+1at,(2,11) & & = & 4(2)+1& m_{t} & = & 9end{array}$

$begin{array}{ccccc}& y-y_{1} & = & -dfrac{1}{m_{t}}(x-x_{1})vspace{2mm}using,(2,11) & y-11 & = & -dfrac{1}{9}(x-2)vspace{2mm}& y & = & -dfrac{1}{9}x+dfrac{2}{9}+11vspace{2mm}& y & = & -dfrac{1}{9}x+11dfrac{2}{9}end{array}$