Posted by **Shania** on Monday, May 6, 2013 at 8:08pm.

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.

- Related Rates -
**Damon**, Monday, May 6, 2013 at 8:22pm
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

- Related Rates -
**Shania**, Monday, May 6, 2013 at 8:32pm
Thank you very much. I understand it now. :-)

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