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Slope of tangent line (DE)

  1. Jun 27, 2005 #1
    Here's the problem:

    detrmine DE giving the slope of the tangent line at point (x,y) for the given family of curves:

    [tex] 2cy = x^2 - c^2 [/tex]

    Here's what I have:

    slope of this curve:

    [tex] y' = \frac{x}{c} [/tex]

    So, then the slope of tangent line is negative inverse:

    [tex] y' = \m \frac{-c}{x} [/tex]

    so now I need c, but I don't see how to pull it out from the original equation!

    Thanks for any help.
     
    Last edited: Jun 27, 2005
  2. jcsd
  3. Jun 27, 2005 #2

    AKG

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    The negative reciprocal would be the slope of the perpendicular line. The slope of the tangent line would just be x/c. What you want to do now is express c in terms of x and y (use the quadratic formula).
     
  4. Jun 27, 2005 #3
    That is precisely the problem.
    and I'm not seeing how I can do that, even with quadratic formula.
     
  5. Jun 27, 2005 #4

    AKG

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    [tex] 2cy = x^2 - c^2 [/tex]

    This is a quadratic equation in c.
     
  6. Jun 27, 2005 #5
    Are you talking about [tex] (a + b)^2[/tex]? but there's y.
    RHS has [tex] (a + b) (a - b)[/tex] but I don't know how that's helping me.
     
  7. Jun 27, 2005 #6

    AKG

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    A quadratic equation in x takes the form ax² + bx + c = 0, and the quadratic formula allows you to find the two complex solutions to this equation:

    [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
     
  8. Jun 27, 2005 #7
    Oh, yeah, complex... I didn't think of them being here.
    Thanks a lot.
     
  9. Jun 27, 2005 #8

    AKG

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    Well your solutions for this quadratic equation will be real, but real numbers are complex numbers.
     
  10. Jun 27, 2005 #9
    English terminology was the hardest part here: quadratic equation. It's called somth. else in my language and does't translate word for word, so I made a wrong association.
    Thanks for your help, that is really like hischool algebra, I feel dumb.... :redface:
     
  11. Jun 27, 2005 #10

    HallsofIvy

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    You are completely off base here. You need to eliminate the constant c, getting a differential equation that does not involve c. Yes, 2cy'= 2x. Now, [tex]c^2+ 2cy= x^2[/tex] so [tex]c^2+ 2cy+ y^2= (c+y)^2= x^2+ y^2[/tex]. That is: [tex]c+ y= \pm\sqrt{x^2+ y^2}[/tex]
    and [tex]c= -y+ \sqrt{x^2+ y^2}[/tex] or [tex] c= -y- \sqrt{x^2+ y^2}[/tex]. The slope satisfies [tex](-y+ \sqrt{x^2+ y^2})y'= x[/tex] or [tex](-y- \sqrt{x^2+ y^2}= x[/tex].

    No, the slope of the tangent line is the derivative. You are thinking of the slope of the normal line.

     
    Last edited: Jun 27, 2005
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