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Fairly simple calculus question

  1. Jun 2, 2005 #1
    I'm given a curve defined by [tex]2y^3 + 6x^2y - 12x^2 + 6y = 1[/tex]

    Its derivative is [tex]\frac{4x - 2xy}{x^2 + y^2 + 1}[/tex]

    One of the problems is as follows: "Write an equation of each horizontal tangent line to the curve."

    I reasoned that if the top of the derivative is equal to zero, the tangent line will be horizontal at that point. Therefore, I solved for [tex]4x - 2xy = 0[/tex]

    [tex]4x = 2xy[/tex]

    [tex]y = \frac{4x}{2x} = 2[/tex]

    This is the only solution I can find, but the wording of the question, "EACH horizontal line" makes me feel as though I'm missing something and there are actually more. Is y = 2 the only one?

    The next problem is as follows: "The line through the origin with slope -1 is tangent to the curve at point P. Find the x- and y-coordinates of point P."

    Correct me if I'm wrong, but doesn't this mean that we have to solve for x AND y?

    [tex]-1 = \frac{4x - 2xy}{x^2 + y^2 + 1}[/tex]

    Am I going the wrong way about this, or is there a way to simplify this equation that I'm not seeing?

  2. jcsd
  3. Jun 2, 2005 #2


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    what about the subtle solution, x = 0? 2y^3 + 6y = 1
  4. Jun 3, 2005 #3
    Is that saying that the CURVE has a slope of -1 as it passes through the origin? I interpreted it as saying that a line drawn through the origin with a slope of -1 would be parallel to the curve itself at point P...
  5. Jun 3, 2005 #4


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    It's not just parallel, it's tangent...
  6. Jun 3, 2005 #5
    Oh, I see. Wouldn't x and y be zero then, though?
  7. Jun 3, 2005 #6


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    Why do you say this?

    That it's a tangent simplifies things greatly. You knew that at P the slope had to be -1, but you also know that P lies on the line through the origin with slope -1 as well, so you can use this to simplify [tex]-1 = (4x - 2xy)(x^2 + y^2 + 1)^{-1}[/tex] by removing either x or y.
  8. Jun 3, 2005 #7


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    The line through the origin with slope -1 is simply y= -x. You need to find a point P= (x,y) on the curve so that the derivative is -1 and y= -x.
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