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    Hey, I have a few questions..

    1. (1 + x) (2 + y) = x² + y²

    (a) Work ouy dy/dx. I got dy/dx =
    2x + y - 2
    1 + x - 2y

    And then.. (the bits I'm stuck on)
    (b) Find the gradient of the curve (1 + x) (2 + y) = x² + y² at each of the 2 points where the curve meets the axis

    I did:
    Meets y axis at x=0, so dy/dx = y-2/1+x
    Meets x axis at y=0, so dy/dx = 2x-2/1+x

    But I dunno what to do next

    (c) Show also that there are 2 points at which the tangents to this curve are parallel to the y axis.

    Parallel to the y axis.. Does that mean dy/dx = 0? I'm not sure to do with it though..


    2. A curve has equation 7x² + 48xy - 7y² + 75 = 0.
    A and B are 2 distinct points on the curve and at each of the points the gradient = 2/11. Use implicit differentiation to show that x + 2y = 0 at the points A and B.

    By implicit differentiation i got dy/dx =
    - 14x - 48y
    48x - 14y

    And i substitued dy/dx = 2/11 into i dont think its right cos i started getting really big numbers... What do I do after differentiating... Or did i differentiate wrong? :confused:

    Someone help me please.. Thanks
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    2. You've done the differentiation correctly. Just note that the division can be simplified, since each term contains a factor of 2, so you get:

    dy/dx = (-7x-24y)/(24x-7y)

    Now you'll avoid the 'big numbers' since everything will be a factor of 2 less. But you've done the right thing. Just need to keep plugging away.
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    (Original post by mockel)
    2. You've done the differentiation correctly. Just note that the division can be simplified, since each term contains a factor of 2, so you get:

    dy/dx = (-7x-24y)/(24x-7y)

    Now you'll avoid the 'big numbers' since everything will be a factor of 2 less. But you've done the right thing. Just need to keep plugging away.
    Oh yeh..! Wow, that was easy, I thought it was lots of working out!
    Thanks
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    1.
    a)
    2 + 2x + y +xy = x² + y²
    2 + dy/dx + y + x(dy/dx) = 2x + 2y(dy/dx)

    dy/dx[1+x-2y] = 2x-y-2

    dy/dx = (2x-y-2)/(1+x-2y)

    [Slightly different to yours]

    b) At x-axis, y=0
    => (1+x)(2) = x²
    x²-2x-2 = 0
    x = (2 ± √12)/2 = 1 ± √3

    At y-axis, x=0
    => 2+y = y²
    (y-2)(y+1) = 0
    y = 2 or y = -1

    Hmm. This suggests that there are 4 points at which the curve meets the axis. Sure it doesn't say just 1 of the axis?
    Anyway, now sub in these x and y values into dy/dx (which you've calculated in b) )


    c) Parallel to y-axis means dy/dx = ∞, since it's just a vertical line.

    dy/dx = (2x-y-2)/(1+x-2y)

    dy/dx = ∞, when the denominator = 0

    1+x-2y = 0
    y = (x+1)/2

    Erm...not too sure why that means "2 points at which...."
    Meh, someone else will figure it out.
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    Ooh i see..
    So does that mean gradients parallel to the x axis = 0 ?
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    (Original post by violet)
    Ooh i see..
    So does that mean gradients parallel to the x axis = 0 ?
    Yes, common sense really, since x-axis is 0 gradient, any lines parallel will also be 0, according to the definition of a gradient! :rolleyes:
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    Yep, i totally knew that.. I just wanted to, erm, check if you did! :rolleyes:
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    Sorry, i've just downloaded mathtype and wanted something to test it out on
 
 
 
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