Question

Given r=9cos(2theta) find dy/dx in polar form

Answer 1

Polar differentiation... Annoying as hell. let\[f(\theta) = r = 9 \cos (2 \theta)\]\[f'(\theta) = -18\sin(2\theta)\]Then just plug in f(theta) and f'(theta) into the polar derivative formula.\[\frac {dy}{dx} = \frac {f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)}{f'(\theta) \cos (\theta) - f(\theta) \sin (\theta)}\]

Answer 2

I had everything right but the f'\[\theta\]. Thanks so much I got it now!

Answer 3

Alright, no problem, be careful with these since the formula is so complicated, its easy to make little mistakes :)

Answer 4

Ok I'm stuck again. I'm trying to simplify, do I use trig identities? I've been trying to use double-angle formulas but cant seem to get anything to cancel out.

Answer 5

If possible, you can use trig identities to simplify. Its fine it you can't get things to cancel out. Some of the derivatives are just very messy.

Answer 6

According to wolfram, this simplifies out to\[\frac {dy}{dx} = \frac { \cos (\theta) - 3 \cos(3 \theta)}{\sin(\theta) + 3 \sin (3\theta)}\]

Answer 7

Here are the identities that would be used...\[\sin 3x = 3\sin x - 4\sin^3 x\]\[\cos 3x = 4\cos^3 x - 3\cos x. \]