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Find y' if (x-y)/(x+y)=(x/y)+1 and show that there are no points on that curve

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Use implicit differentiation to find y' if (x-y)/(x+y)=(x/y)+1. Now show that there are, in fact, no points on that curve, so the derivative you calculated is meaningless.

    2. Relevant equations



    3. The attempt at a solution

    I managed to get it into the form:
    dy/dx = [(x+y)2-2y3]/[(x)(-2y2+(x+y)2]
    This doesn't seem like it proves anything though. Someone help please!
     
  2. jcsd
  3. Oct 28, 2012 #2

    SammyS

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    Find an equation equivalent to [itex]\displaystyle \frac{x-y}{x+y}=\frac{x}{y}+1[/itex] which shows that this equation cannot be satisfied by any combination of x & y.

    Alternatively, solve this equation for x or y.
     
  4. Oct 28, 2012 #3
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  5. Oct 28, 2012 #4
    I understand this, but how would I go about doing so? I already derived it and what I got wasn't all that nice.



    Also the picture is very helpful! How did you go about simplifying it down so far?
     
  6. Oct 28, 2012 #5
    Nvm, I simplified it down lol. My brain froze for a minute there.
     
  7. Oct 28, 2012 #6
    So where I substitute in (2) do I have to get (2) in terms of one of the variables and then substitute that in for the variable?
     
  8. Oct 28, 2012 #7
    Substitute -1/4 for y/x and see that (2) is not satisfied.
     
  9. Oct 28, 2012 #8
    I just realized that right before I seen your post (I'm not having a good night lol) So last question, I promise, how did you come up with the -1/4 ?
     
  10. Oct 28, 2012 #9
    the denominator of y' is 0 where x+4y=0,that is,y/x=-1/4
     
  11. Oct 28, 2012 #10
    Ok, I see that. So when you plug that into the initial equation (2), it shows that both sides are not equal. Wouldn't that show that there are in fact points on that curve? I'm just totally missing this whole question in general. I really appreciate all you help for this by the way!
     
  12. Oct 28, 2012 #11

    SammyS

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    The question as to whether or not there are points which fit [itex]\displaystyle \frac{x-y}{x+y}=\frac{x}{y}+1[/itex] has nothing to do with finding y'.

    The thing is that there is no pair or real numbers, (x, y), which can make the left side of the equation equal to the right side. The problem asks you to show this.
     
  13. Oct 28, 2012 #12
    Ok, thanks Sammy, making some sense now (the question asks to find y' I guess just to show you can and whatnot) So now, how would I begin to show this?
     
  14. Oct 28, 2012 #13

    Dick

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    Start by trying to simplify it. Put everything to one side of the equation and put everything over a common denominator.
     
  15. Oct 28, 2012 #14
    I think the point in this exercise is to practice implicit differentiation.It seems that the choice of the equation is not so good.
     
  16. Oct 28, 2012 #15
    Ok Dick, I have that so far, everything to one side simplified. I'm so fed up with this question.
     
    Last edited: Oct 28, 2012
  17. Oct 28, 2012 #16
    I've spent hours on this, many, many hours, can someone please help me. I've been looking at it so long my brain is mush.
     
  18. Oct 28, 2012 #17
    I know that no matter which number I select for x or y, I cant solve for the other one, but I need a way to prove this for ALL values. I'm really stumped.
     
  19. Oct 28, 2012 #18

    Dick

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    What did you get? The idea will be to write the numerator as a sum of squares.
     
  20. Oct 28, 2012 #19
    See answer 3.The simplification is by common denominator and canceling.Then perform implicit differentiation and find y'.y' is valid when its denominator is not zero,i.e when x is not equal -4y.Finally substitute x=4y in the given equation and show that it is not satisfied.Conclude that the derivative exists everywhere on the given curve.
     
  21. Oct 28, 2012 #20
    I got : x2+xy+2y2=0
     
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