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Math Help - Differential Equations.

  1. #1
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    Differential Equations.

    Example 4: Find the curves that are perpendicular to the parabolas in Example 3.

    It is said:

    \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y}

    My question is about \frac{-x}{2y} and why it is so.

    The equations for the parabolas in Example 3 are:

    x^2
    2x^2
    -x^2
    0x^2
    y=-2y^2
    100x^2

    I read through Example 3 and I understand to the extent that I follow the arithmetic and I see how the equations for the parabolas were derived.

    The \frac{-x}{2y} in \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y} I am having some trouble understanding why that is or what is really being said. I don't think I've really been through this topic of a curve being perpendicular to another curve before.

    Thank You...
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  2. #2
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    Re: Differential Equations.

    Hey sepoto.

    I think you should keep in mind that two lines are perpendicular in two dimensions if m1*m2 = -1 where m1 is the gradient of one line and m2 is the gradient of the other. They must satisfy this relationship for the perpendicular attribute to hold.
    Thanks from sepoto
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  3. #3
    MHF Contributor

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    Re: Differential Equations.

    The family of parabolas you given are of the form y= ax^2, varying as a varies. y'= 2ax so that at any point where a curve Y(x) is perpendicular to that, we must have Y'= -1/(2ax). Since y= ax^2, x/y= x/ax^2= 1/ax and so Y'= -1/(2ax)= -x/2y
    Thanks from sepoto
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