### 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: $y = -\frac{1}{3} x + 5\frac{2}{3}$