Find the equation of the normal line to the curve y^6 + secx - 2siny = 2y + x + π at the origin.?

y^6 means "y to the 6th power"

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  • 1 decade ago
    Favorite Answer

    First, find the slope of the tangent line. Use implicit differentiation.

    6y^5y' + tanx secx -2cosyy' = 2y' + 1 + 0

    Solving for y',

    y' = (1 - tanx secx)/(6y^5 - 2cosy - 2)

    Plug in the point (0,0) to get

    y' = 1/(-4)

    The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal.

    m=4

    Since the normal line passes through the origin, the normal line is y=4x.

    On further inspection, the origin does not even lie on the curve. Are you sure the problem is as stated? It is a much more difficult question if they ask for normal lines containing the origin.

  • Anonymous
    1 decade ago

    Your equation.

    y^6 - 2y - 2 sin y = x - sec x + pi

    Implicit differentiation.

    6 y^5 y' - 2 y' - 2 y' cos y = 1 - sec x tan x

    y' (6 y^5 - 2 cos y - 2) = 1 - sec x tan x

    dy/dx = (1 - sec x tan x) / (6 y^5 - 2 cos y - 2)

    When x=0,

    y^6 - 2y - 2 sin y + 1 - pi = 0

    Use Newton's method to find y.

    y = -0.5421355

    y is the y-intercept of the normal line.

    Use x=0, y=-0.5421355 to find dy/dx @ x=0.

    dy/dx = (1 - sec x tan x) / (6 y^5 - 2 cos y - 2)

    dy/dx = -0.2503626

    m = slope of the normal line = -(dy/dx)^(-1)

    m = 3.994207

    Equation of the normal line:

    y = 3.994207 x - 0.5421355

  • 1 decade ago

    ask ur parents

  • Anonymous
    1 decade ago

    huh?

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