Parametric Equation?

Question

Hey, I'll just jump right in
x=t-sin(t) y=1-cos(t)

dy/dx=sin(t)/1-cos(t)
=?/y
Can't find an expression for sin(t) so i am stuck can anyone help?

Answer 1

dy/dx=sin(t)/1-cos(t)
now, y= 1-cost
on the other hand
x= t - sint
so sint = t-x
and we can get an expression of t using y:
cost = 1-y
t=arccos(1-y)
so sint = arccos(1-y) -x
and
dy/dx=sin(t)/1-cos(t)
=( arccos(1-y) -x ) /y .

Answer 2

1+ cos t= 2 - y so sin^2 t= 2y -y^2 and sin t is + or - sqrt {2y -y^2}.