# Calculus

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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)

• Calculus -

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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