Differentiation - Parametric Equations?

[TwinkleTits]

I musta been out when they did this cuz Ive never even heard of a parametric equation. Heres the question

Find dy/dx where the parametric equations of a curve are given by
x = t - 2cost
y = 2sint - cos2t

Give your answer in the form p cost, p e R
That e in p e R is actually a greek letter. Lowercase epsilon.

[Mantikore]

ahhh, this is an easy one

dy/dt = (dy/dx)*(dx/dt)................ this is the chain rule. notice how on the right hand side, the dx cancels out. now just rearrange it to isolate dy/dx
dy/dx = (dy/dt)/(dx/dt)............. just differentiate both equations with respect to t, then substitute in

dy/dx = (2cost + 2sin2t)/(1 + 2sint)...........heres where knowing your double angle results come in handy.
sin2t = 2sintcost
dy/dx = (2cost + 2*2sintcost)/(1+2sint)..... factorise the top taking out 2cost has the highest common factor
dy/dx = (2cost (1+2sint))/(1+2sint)...cancel out (1+2sint)
dy/dx = 2cost

wow the magic of math!

edit: by the way, you can find help on google. i just did a quick search for 'parametric differentiation' and it came up with this
http://www.maths.abdn.ac.uk/~igc/tch...es/node46.html which i think explains it quite well