# Math Help - Applications of Differentiation

1. ## Applications of Differentiation

The parametric equations of a curve are x = cos 4A + 2cos 2A, y = sin 4A - 2sin2A where A is the parameter. Find (a ) dy/dx and (b) equation of the normal to the curve at the point where A = pi/4

2. Originally Posted by puggie
The parametric equations of a curve are x = cos 4A + 2cos 2A, y = sin 4A - 2sin2A where A is the parameter. Find (a ) dy/dx and (b) equation of the normal to the curve at the point where A = pi/4
For part a) you need to use the Chain Rule.

Recall that $\frac{dy}{dx} = \frac{dy}{dA}\times\frac{dA}{dx}$.

$\frac{dy}{dA} = 4\cos{(4A)} - 4\cos{(2A)}$

$\frac{dA}{dx} = \frac{1}{\frac{dx}{dA}}$

$= \frac{1}{-4\sin{(4A)} - 4\sin{(2A)}$.

So $\frac{dy}{dx} = \frac{4\cos{(4A)} - 4\cos{(2A)}}{-4\sin{(4A)} - 4\sin{(2A)}}$

$= -\frac{\cos{(4A)} - \cos{(2A)}}{\sin{(4A)} + \sin{(2A)}}$.

For part b) , you need to remember that a normal is a LINE.

So it is of the form $y - y_1 = m(x - x_1)$.

At the point $A = \frac{\pi}{4}$...

$x_1 = \cos{\pi} + 2\cos{\frac{\pi}{2}} = -1$

$y_1 = \sin{\pi} - 2\sin{\frac{\pi}{2}} = -2$.

The gradient of the curve at point A is $\frac{dy}{dx} = -\frac{\cos{\pi} - \cos{\frac{\pi}{2}}}{\sin{\pi} + \sin{\frac{\pi}{2}}}$

$= -\frac{-1}{1}$

$= 1$.

The gradient of the normal is the negative reciprocal. So $m = -1$.

Thus the normal's equation is given by

$y - (-2) = -1[x - (-1)]$

$y + 2 = -x - 1$

$y = - x - 3$.

Hope that helped.