Thread: parametric dy/dx

Hi
Can somebody tell me if i am correct?

find dy/dx in terms of t for the curve with the parametric representation:

x=(1-t)(1+3t) y=(1+t)(1-3t)

My solution;

x=1 + 2t - 3t^2 y= 1 - 2t - 3t^2

dx/dt= 2-6t dy/dt= -2-6t


dy/dx=(dy/dt)/ (dx/dt)

dy/dx=(-2-6t)/(2-6t)

Can somebody tell me if this is correct-and can somebody recomend a good website with more parametric dy/dx examples-my maths book is not very good.
thanks
John

I think it's correct!
Sorry I don't know any site with more exercices.

thanks Siron-yet again?

The problem with my maths book is he examples are really basic and the self tests at the end of the chapters are really difficult. Not all the answers are available.

thanks for your help
John

You're welcome! You can create exercices by yourself if you want and make them more difficult .

You could, of course, cancel. dy/dx= (-1-3t)/(1-3t) might be considered a better answer.