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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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