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Math Help - Diffrential equation question on tangents.

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    Diffrential equation question on tangents.

    Ok,
    \frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}
    1. Show that curve has no tangent parallel to y axis.
    Well, when thats the case, dy/dx = 0.
    Giving
    2 - y^2 = 0
    And therefore y = +-/sqrt{2}
    Well you were meant to do
    2xy - 3 = 0 and sub into original equation to show there is no solution.

    And my question is Why? Do you always do 1/(dy/dx) and equate that to 0 to find a line parallel to y axis? Why do you do that?
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Ok,
    \frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}
    1. Show that curve has no tangent parallel to y axis.
    Well, when thats the case, dy/dx = 0.
    Giving
    2 - y^2 = 0
    And therefore y = +-/sqrt{2}
    Well you were meant to do
    2xy - 3 = 0 and sub into original equation to show there is no solution.

    And my question is Why? Do you always do 1/(dy/dx) and equate that to 0 to find a line parallel to y axis? Why do you do that?
    If the tangent is parallel to the vertical axis it is a vertical line. What's the gradient of a vertical line ....?

    You're getting mixed up with parallel to the x-axis, in which case the tangent is a horizontal line with gradient zero ....
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    Quote Originally Posted by mr fantastic View Post
    If the tangent is parallel to the vertical axis it is a vertical line. What's the gradient of a vertical line ....?

    You're getting mixed up with parallel to the x-axis, in which case the tangent is a horizontal line with gradient zero ....
    Oh...I see...Thank-you!
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