# Thread: Differentiation of parametric equation

1. ## Differentiation of parametric equation

2day i doing this question...den i stuck.

x=t^2 , y = 1/(1+t)
1-2t

My solution,
dx/dt =2t(1-t)/(1-2t)^2 , dy/dt= -1/(1+t)^2

Then,
dy/dx = dy/dt * dt/dx

=-1/(1+t)^2 * (1-2t)^2/2t(1-t)
=?????
at there i stuck.

You've got an answer why do you say you're stuck.
Maybe you could simplify it a little but
dx/dt = {2t*(1-2t)+2t*t^2}/(1-2t)^2
dy/dt = -1/(1+t)^2 as you said
You did the good calculation dy/dx = (dy/dt)/(dx/dt)
Simplify as much as you can and that is your answer : (1/2)*{(2t-1)^2}/{t*(t^2-1)*(t+1)}

3. Originally Posted by vincisonfire
You've got an answer why do you say you're stuck.
Maybe you could simplify it a little but
dx/dt = {2t*(1-2t)+2t*t^2}/(1-2t)^2
dy/dt = -1/(1+t)^2 as you said
You did the good calculation dy/dx = (dy/dt)/(dx/dt)
Simplify as much as you can and that is your answer : (1/2)*{(2t-1)^2}/{t*(t^2-1)*(t+1)}
Ok, oic...like tat ar, becouse i expand all and wan to simplify but canot.
Haha...didnt see tat. Thx a alot...my problem solved. I tot the answer should be very simple.