# Thread: differentiation, parametric equation help

1. ## differentiation, parametric equation help

Find $\frac{dy}{dx}$ for the following, leaving your answer in terms of the parameter t.

$x = 2t-sin2t$

$y = 1-cos2t$

$\frac{dx}{dt} = 2-2cos2t$

$\frac{dy}{dt} = 2sin2t$

$\frac{dy}{dx} = \frac{2sin2t}{2-2cos2t}$

I can factor and cancel the two's but that still does give me the correct answer. Can someone please show me what I am doing wrong?

Thank you.

2. Hello, Tweety!

It would help if you told us the correct answer.
I see nothing wrong with your work.

Find $\frac{dy}{dx}$ for the following, leaving your answer in terms of the parameter $t.$

. . $\begin{array}{ccc}x &=& 2t-\sin2t \\
y & =& 1-\cos2t \end{array}$

My work:

. . $\begin{array}{ccc}\dfrac{dx}{dt} &=& 2-2\cos2t \\ \\[-3mm]
\dfrac{dy}{dt} &=& 2\sin2t \end{array}$

. . $\dfrac{dy}{dx} \:=\: \frac{2\sin2t}{2-2\cos2t}$

I can factor and cancel the two's but that still does give me the correct answer.

We have: . $\frac{dy}{dx} \:=\:\frac{\sin2t}{1 - \cos2t} \;=\;\frac{2\sin t\cos t}{2\sin^2\!t} \;=\;\frac{\cos t}{\sin t} \;=\;\cot t$

Is that it?

3. Originally Posted by Soroban
Hello, Tweety!

It would help if you told us the correct answer.
I see nothing wrong with your work.

We have: . $\frac{dy}{dx} \:=\:\frac{\sin2t}{1 - \cos2t} \;=\;\frac{2\sin t\cos t}{2\sin^2\!t} \;=\;\frac{\cos t}{\sin t} \;=\;\cot t$

Is that it?

yes that is the correct answer cott, but how did you manage to get that?

I see you have replace sin2t with 2sin2tcost, but what happened ti the other number two? And the bottom bit, how did you get it?

thanks