Related Rates
posted by Shania .
A particle moves along a path described by y=x^2. At what point along the curve are x and y changing at the same rate? Find this rate if at time t we have x=sin t and y= sin^2t.
I solved the first part and got (1/2, 1/4), but I have no idea how to tackle the second part. Help is much appreciated.

dy/dt = dy/dx * dx/dt
but
dy/dx = 2x
so
dy/dt = 2 x dx/dt
so when 2 x = 1 (when the slope = 1 of course)
x = 1/2 then y = 1/4
x = sin t
when is sin t = 1/2
when t = 30 degrees or pi/6 radians
dx/dt = cos t = cos 30 = (sqrt 3 )/ 2
dy/dt = 2 sin t cos t = 2(1/2)(sqrt 3/2) sure enough 
Thank you very much. I understand it now. :)
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