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Math Help - isolating a variable...difficulties

  1. #1
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    isolating a variable...difficulties

    Once again, you guys are helping me out heaps. Thanks! My maths teacher I believe has put many tricks into these questinos. I eventually need to find the gradient of this curve. I would normally isolate y, and then take the derivative. But I'm having issues isolating y. Any help is appreciated...thanks

    xy(x+y)+16=0
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  2. #2
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    Quote Originally Posted by solarscott View Post
    Once again, you guys are helping me out heaps. Thanks! My maths teacher I believe has put many tricks into these questinos. I eventually need to find the gradient of this curve. I would normally isolate y, and then take the derivative. But I'm having issues isolating y. Any help is appreciated...thanks

    xy(x+y)+16=0
    that is the main reason for using implicit differentiation.

    but, if you insist on isolating y ...

    xy^2 +  x^2 y + 16 = 0

    y = \frac{-x^2 \pm \sqrt{x^4 - 64x}}{2x}
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  3. #3
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    ahhh, I knew there was another way to do it...just forgot the name..implicit differentiation! Thanks. I will look that up...Im believe that is the way he is expecting it to be done...
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  4. #4
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    It looks more like a FOIL problem to me...

    xy(x+y)+16=0

    x^2+y^2+2y+2x+16=0
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  5. #5
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    Quote Originally Posted by Caturdayz View Post
    It looks more like a FOIL problem to me...

    xy(x+y)+16=0

    x^2+y^2+2y+2x+16=0
    it's not a FOIL problem.
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  6. #6
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    Yep, I just worked it through to find what I was supposed to (the equation of the tangent where the gradient = -1, and the answer I think makes sense...

    I found:
    y'=(-y^2-2xy)/(x^2+2xy)

    subbing in y'=-1, I solve for:

    -1=(-y^2-2xy)/(x^2+2xy)
    -x^2-2xy=-2xy-y^2
    x^2=y^2
    y=-+x


    So the equation of the tangent where the gradient =-1 would be y=+-x? Does this make sense? Thanks
    Last edited by solarscott; September 13th 2009 at 06:28 PM. Reason: corrected mistake
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