Applications of Differentiation Part II

Tangents and Normal

Example 4
The point A lies on the curve $ y = x^2 -x + c $ where c is a constant. The tangent to the curve at A is $ y = 3x - 1 $. Find the equation of the normal to the curve at A.

Recall equation of straight line: $ y = mx +c \qquad \text{m is gradient} $
Since tangent to the the curve at A is $ y = 3x + 5 $ , gradient of tangent is 3

We can calculate the gradient using $\frac{dy}{dx} $
$\frac{dy}{dx} = 2x -1 $
To obtain x coordinates of A, we set $ \frac{dy}{dx} = 3 $
$2x -1 = 3 \qquad x = 2 $

To determine the equation of normal we need to know the gradient of the normal and a point on the normal

Gradient of the normal is the negative reciprocal of gradient of tangent
Gradient of normal = $-\frac{1}{3} $

A is a point on the tangent as well as the normal
When $ x = 2 \qquad y = 3(2)-1 = 5 $

Sub $ x= 2 \quad y = 5 \quad m=-\frac{1}{3} $ into $ y = mx +c $

$\begin{aligned}
5 &= -\frac{1}{3}(2) + c \\
c &= 5\frac{2}{3} \\
\end{aligned}$

Equation of normal: $ y = -\frac{1}{3} x + 5\frac{2}{3} $