Question

PARAMETRIC EQNS...
http://gyazo.com/d62915dbbd5acee002a95340a49951e8

Answer 1

i have tried simplifying my answer but i am not getting cot theta....

Answer 2

Easy,\[x=2 \theta - \sin2 \theta,~~~y=2-\cos2 \theta\]
\[\frac{dx}{d \theta}=4 \left( \sin ^2 \theta \right),~~~ \frac{dy}{d \theta}= 2 ( \sin (2 \theta) )\]

Answer 3

\[\frac{dy}{dx}=\frac{2\sin(2 \theta)}{4\sin^2 \theta} \\ \\ \frac{dy}{dx}=\frac{1}{2}\frac{2\sin \theta \cos \theta}{\sin^2 \theta} \\ \\ \frac{dy}{dx}=\frac{\cos \theta}{\sin \theta} \\ \\ \frac{dy}{dx}= \cot \theta\]
Got it?

Answer 4

\[\frac{ dx }{ d \theta } = \frac{ 1 }{ 2-2\cos \theta }\]
this is what i got for dx/d theta

Answer 5

\[x=2 \theta -\sin2 \theta\]
\[\frac{dx}{d \theta}=2-\cos(2 \theta)(2) \\ \\ \frac{dx}{d \theta}=2(1-\cos(2\theta))\]
Using identity, \(cos(2 \theta)=1-2sin^2 \theta\)
\[\frac{dx}{d \theta}= 2(1-(1-2\sin^2 \theta)) \\ \\ \frac{dx}{d \theta}= 4\sin^2 \theta\]

Answer 6

thank you