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Please help me to find an equation for the tangent to the curve xsin2y=ycos2x at the point (phi/4, phi/2)

  • math -

    first, it's PI, not PHI
    pi(π) = 3.14, the ratio of circumference to diameter
    phi(φ) = 1.62 (1+√5)/2, the golden ratio

    So, we use implicit differentiation to get

    xsin2y = ycos2x
    sin2y + 2xcos2y y' = cos2x y' - 2ysin2x

    y' = -(sin2y + 2y sin2x)/(2x cos2y - cos2x)

    at (π/4,π/2), that gives a slope of

    y' = -(0 + π*(1))/(π/2*(-1) - 0)
    = -π/(-π/2)
    = 2


    So, bow you want the line with slope=2 through (π/4,π/2):

    y - π/2 = 2(x - π/4)
    y = 2x

  • math -

    xsin2y=ycos2x

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