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Equation of normal line.

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the equation of each of the normal lines to the curve y = x^3 - 4x that is parallel to the line x + 8y - 8 = 0.


    2. Relevant equations
    Differentiation, y - y1 = m(x - x1)


    3. The attempt at a solution
    Well, clearly I start by differentiating y = x^3 - 4x to gey dy/dx = 3x^2 - 4. Then, this is the slope of the tangent to the graph y. This slope is equal to the slope of the line x + 8y - 8 = 0. Which is -8. So, I then evaluate for x and sub back into dy/dx, giving me the slope of the tangent. The slope of the normal line is the negative reciprocal of this. If you solve for x using
    3x^2 - 4 = -8, you get an imaginary value though. Where have I gone wrong?
     
  2. jcsd
  3. May 22, 2013 #2

    CAF123

    User Avatar
    Gold Member

    You have the gradient of the given line wrong.
     
  4. May 22, 2013 #3
    Cool. See where I went wrong now. Stupid slip.
     
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