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April 19, 2014

April 19, 2014

Posted by **Laura** on Tuesday, April 16, 2013 at 10:42pm.

x=sin(t); y=(cos(t))^2; t=(pi/3)

- Calculus -
**Reiny**, Tuesday, April 16, 2013 at 11:12pmx = sin t

dx/dt = cost

y = cos^2 t

dy/dt = 2cost(-sint) = -2sint cost

slope = dy/dx = (dy/dt) / (dx/dt

= cost/(-2sintcost) = -1/(2sint)

when t = π/3

dy/dt = -1/2sin(π/3) = -1/(2√3/2) = -1/√3

when t = π/3 , x = sin π/3 = √3/2

and y = cos^2 (π/3) = 1/4

so we have a point (√3/2 , 1/4) with slope √3/2

y - 1/4 = (√3/2)(x-√3/2)

2y - 1/2 = √3x - 3/4

√3x - 2y = 1/4

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