Calculus
posted by Laura .
Find the equation of the tangent line to the curve at the given value of t:
x=sin(t); y=(cos(t))^2; t=(pi/3)

x = 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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