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Math Help - Another Differentiation

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    Another Differentiation

    Find the gradient of the curve (1+x) (2+y) = x^2 + y^2 at each of the two points where the curve meets the y-axis.

    Show also that there are two points at which the tangents to this curve are parallel to the y-axis.


    Iíve done 1st part of this question. But how can I solve 2nd part?
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  2. #2
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    Quote Originally Posted by geton View Post
    Find the gradient of the curve (1+x) (2+y) = x^2 + y^2 at each of the two points where the curve meets the y-axis.

    Show also that there are two points at which the tangents to this curve are parallel to the y-axis.


    Iíve done 1st part of this question. But how can I solve 2nd part?
    Find the derivative:
    (1 + x)(2 + y) = x^2 + y^2

    (1)(2 + y) + (1 + x) \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}

    2 + y - 2x = [-(1 + x) + 2y] \frac{dy}{dx}

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

    Now, you are looking for tangents where the slope is undefined. The derivative is undefined for
    -x + 2y - 1 = 0

    y = \frac{1}{2}x + \frac{1}{2}

    So any points where the tangent line is parallel to the y-axis is going to be on this line. Thus find the intersections of this line with the given curve.

    -Dan
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  3. #3
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    Thank you so much 'topsquark' for help.
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  4. #4
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    Quote Originally Posted by topsquark View Post
    [snip]

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

    Now, you are looking for tangents where the slope is undefined. The derivative is undefined for
    -x + 2y - 1 = 0

    y = \frac{1}{2}x + \frac{1}{2}

    So any points where the tangent line is parallel to the y-axis is going to be on this line. Thus find the intersections of this line with the given curve.

    -Dan
    I'd just like to add a little something here:

    It's trivially obvious in this instance that when the denominator is zero, the numerator is NON-zero.

    However, there can be times when some of the solutions that make the denominator equal to zero also make the numerator equal to zero. At such points the derivative becomes indeterminant: \frac{dy}{dx} \rightarrow \frac{0}{0}. The tangent at these points is NOT necessarily vertical (and there may even be more than one tangent). Such singular points are called multiple points - there are three types:

    1. Cusp.

    2. Crunode.

    3. Acnode.

    This existence of a multiple point should always be considered.

    A simple example: y^2 = x^3 + ax^2.

    a = 0: The origin is a cusp.

    a > 0: The origin is a crunode.

    a < 0: The origin is an acnode.

    By the way, a crunode is a point where the curve intersects itself.
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