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Solving an equation with respect to y, where y is twice [hard]

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Given the following equation

    X(y) = y/(t*(sqrt(b^2+y^2)) - 1/p

    How would I go solving that equation X(y) = 0 with respect to y?


    3. The attempt at a solution

    I can choose a commen dominator called p*t*(b^2+y^2)

    But I when end up with

    ((y*p - t*p*(sqrt(y^2+b^2))/(p*t(b^2+y^2)) = 0

    How would you guys surgest I proceed from where in order to isolate y?
     
  2. jcsd
  3. Jan 26, 2014 #2

    Dick

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    Now that you have it over a common denominator, if f(y)/g(y)=0, then f(y)=0.
     
  4. Jan 26, 2014 #3
    Thank you for your answer.

    That implies that I need to solve the equation

    (y*p - t*p*(sqrt(y^2+b^2)) = 0

    Which allows me to arrive at the solution (using my graphical calculator)

    y = -b * p *(sqrt(-1/(p^2-1)

    But solution is suppose to be:

    y = b/(sqrt(p/t +1) * sqrt(p/t-1))

    I can't quite comprehend which step I need to use arrive at that solution. But cause to the best of my knowledge there aren't any variables which I substitute in order to arrive at that solution.

    Any idears?
     
  5. Jan 26, 2014 #4

    Dick

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    You aren't going to get there by being sloppy and relying on a calculator.

    There's an extra 'p' in that equation.

    Extra 'p' here also, and what happened to the 't'? I think you should fix those up and figure out how your calculator found that expression before continuing.
     
  6. Jan 26, 2014 #5
    You are right. I see now the equation I am suppose to solve is y/(t*(sqrt(y^2+b^2))-1 = 0

    Is it correct now?
     
  7. Jan 26, 2014 #6

    Dick

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    No, the p is gone. Your first step of putting it over a common denominator is wrong. It happened when you expressed the term 1/p with that common denominator.
     
    Last edited: Jan 26, 2014
  8. Jan 26, 2014 #7
    So I am lost here :( What would you surgest I do as a first step with original equation? I can add 1/p on both sides I see that and next add the denominator on both sides. But what next?
     
  9. Jan 26, 2014 #8

    Dick

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    The strategy is to do what you were doing. Express X(y) over a common denominator. But do it right.
     
  10. Jan 26, 2014 #9
    I get that now but the only commen denominator I can deduce is my pee size brain is :)

    p*t*(b^2+y^2))

    But what I get from what you are saying is that denominator is wrong?
     
  11. Jan 26, 2014 #10

    Dick

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    I would use p*t*sqrt(b^2+y^2) as a common denominator. Actually that's what I thought you were using and just forgetting the 'sqrt' part. If you have two fraction a/b-c/d you can always use c*d as a common denominator. What do you get expressing a/b-c/d over a common denominator?
     
  12. Jan 26, 2014 #11
    Okay, Glad I am not totally stupid then :)

    So any if I take the original equation : y/(t*(sqrt(b^2+y^2)) - 1/p

    and use the commen denominator p*t*sqrt(b^2+y^2)

    I arrive at

    py/(p*t*sqrt(b^2+y^2) - (t*sqrt(b^2+y^2)/(p*t*sqrt(b^2+y^2) =

    y/(t*sqrt(b^2+y^2)) - 1 = 0 ?

    But if that the right cause of action I still have y twice with the term y^2 and y, and no matter what I do I can't arrive at the result that

    y = b/(sqrt(p/t +1) * sqrt(p/t-1)) :( So I understand I must be doing something wrong and there possibly a term which can be rewritten to arrive that the right result for y, but I cant see which one :(
     
  13. Jan 26, 2014 #12

    Dick

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    Sorry, I've been assuming that some of the things you were writing were typos, when they are actually algebra mistakes. I revised the post you quoted. I'm asking you to show me how to put a/b-c/d over a common denominator.
     
  14. Jan 26, 2014 #13

    haruspex

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    You confused me dropping the '0' at the end of the first of those two lines.
    You have dropped a p in getting to the last line.
    Don't collect everything over on the left like that. Leave it as py = (t*sqrt(b^2+y^2)).
    You next step is to get rid of the square root. How can you do that?
     
  15. Jan 26, 2014 #14
    Thats easy a/b - c/d = ad/bd - bc/bd = (ad-bc)/bd

    But regarding the other expression I attempted now 10 times if I take my equation

    X(y) = y/(t*(sqrt(b^2+y^2)) - 1/p = 0

    and use the commen denominator (p*t(b^2+y^2))

    I end up with the expression:

    py/(pt*sqrt(b^2+y^2)) - (t(b^2+y^2))
    /(p*t(b^2+y^2)) = ?

    I must have messed up somewhere :( ?
     
  16. Jan 26, 2014 #15

    Dick

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    a/b-c/d=(ad-bc)/bd is correct. You are messing up the second part. Actually, looking at it, it's not messed up - but you don't have a common denominator. Just apply the correct pattern with a=y, b=t*(sqrt(b^2+y^2)), c=1 and d=p.
     
    Last edited: Jan 26, 2014
  17. Jan 26, 2014 #16

    Ray Vickson

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    Your equation is
    [tex] \frac{y}{t\sqrt{b^2+y^2}}-\frac{1}{p} = 0 \Rightarrow \frac{y}{t\sqrt{b^2+y^2}} = \frac{1}{p}[/tex]
    Square both sides to get
    [tex] \frac{y^2}{t^2(b^2+y^2)} = \frac{1}{p^2}[/tex]
    Solve for ##y^2##; in other words, let ##y^2 = z## and solve for ##z## from the simple equation
    [tex] \frac{z}{b^2 t^2 + t^2 z} = \frac{1}{p^2}[/tex]
     
  18. Jan 26, 2014 #17

    Dick

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    Yes, I know there is an easier way to attack this. You can also do it by putting everything over a common denominator and going from there. I was trying to diagose what was going so wrong with the OP's attempt to put things over a common denominator. Sometimes you have to do that. Here, it's optional.
     
  19. Jan 27, 2014 #18

    thank and then I should be able to arrive at the solution ?



    But if I solve the rewritten equation with respect to z I get z=-b^2*t^2/(t^2-p^2)?

    I am no closer to y = b/(sqrt(p/t +1) * sqrt(p/t-1)) which is suppose to be the solution for original equation.

    Which is still no closer to y = b/(sqrt(p/t +1) * sqrt(p/t-1))
     
    Last edited: Jan 27, 2014
  20. Jan 27, 2014 #19

    haruspex

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    Yes you are... almost there in fact. Just a little more juggling.
     
  21. Jan 27, 2014 #20
    I got it now :D
     
    Last edited: Jan 27, 2014
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