Calculate the equation of the tangent and normal lines for the curve f(x) = tan 2x, at the point where the x-coordinate is: x = π/8.

asked by Bonaventutre
  1. dy/dx = 2sec^2 (2x)
    when x = π/8
    f(π/8) = tan π/4 = 1
    so our point of contact is (π/8, 1)

    slope of tangent = 2 sec^2 (π/4)
    = 2(√2/1)^2 = 4

    equation of tangent:
    y - 1 = 4(x - π/8)
    y - 1 = 4x - π/2
    y = 4x + 1-π/2

    normal has slope -1/4
    y = (-1/4)x + b
    but the point (π/8,1) lies on it, so
    1 = (-1/4)(π/8) + b
    b = 1 + π/32
    y = (-1/4)x + 1+π/32

    posted by Reiny

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