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Direct proportion

  1. Apr 1, 2005 #1
    This seems to be a simple question. But I'm rather dubious at a particular step.
    The question is: Prove
    [tex]\frac{1}{x}-\frac{1}{y}[/tex] is directly proportional to [tex]\frac{1}{x}[/tex]

    This how I went about it:
    [tex]\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}[/tex]
    [tex]\frac{y-x}{xy}=[\frac{y-x}{y}]\frac{1}{x}[/tex]

    Can we say the quantity on the right hand side is proportional to [tex]\frac{1}{x}[/tex]?
     
  2. jcsd
  3. Apr 1, 2005 #2
    Well, what does proportional mean?
     
  4. Apr 1, 2005 #3
    I'll give a more practical definition: If the quantity on the right hand side increases or decreases then the quantity on the left hand side also increases or decreases.There is a constant which can be any real number which relates the two quantites.

    Now, in context to this question: Can you call the term in the parenthesis a 'constant' since both 'x' and 'y' are variables?
     
  5. Apr 1, 2005 #4
    So when you say

    [tex] a \propto b [/tex]

    you mean that there's some [itex]k[/itex] independent of [itex]a[/itex] and [itex]b[/itex] such that

    [tex] a = k b[/tex]?

    or to be more precise, [itex]a \propto b[/itex] iff [itex]\frac{da}{db}[/itex] is constant.

    Then

    [tex] \frac{1}{x} - \frac{1}{y} \propto \frac{1}{x}[/tex]

    is indeed false.

    For example, just let [itex] u = \frac{1}{x}[/itex], and [itex]v = \frac{1}{y}[/itex]. Then your question is to prove

    [tex]u - v \propto u[/tex]

    which is silly unless [itex]v \propto u[/itex] (easy to prove), and presumably [itex]y[/itex] is independent of [itex]x[/itex] in your question~

    I imagine that your teacher isn't using that definition for proportionality, because otherwise he/she is being very silly~
     
    Last edited: Apr 1, 2005
  6. Apr 1, 2005 #5
    I know this a silly sum. But I think more precise definition of proportionality would be the geometrical one:

    [tex] y \propto x [/tex]

    [tex] y = m x[/tex]
    m is the slope. With variable y on y-axis and variable x on x-axis.
     
  7. Apr 1, 2005 #6
    Yeah, that's the same as the one I gave in a different form :smile:

    [tex] y \propto x \Longleftrightarrow \frac{dy}{dx} \ \mbox{is constant} [/tex]
     
  8. Apr 1, 2005 #7

    dextercioby

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    Nope,in the way it's formulated,it definitely doesn't match our understanding of direct proportionality.

    Daniel.
     
  9. Apr 1, 2005 #8
    Of course the x and y above are different from the question.
    The result of this question seems obivious by your second method and you can get a straight line on a graph considering :
    [tex] \frac{1}{x} - \frac{1}{y}[/tex]
    on y-axis &
    [tex]\frac{1}{x}[/tex]
    on x-axis.
     
  10. Apr 1, 2005 #9
    Actually, my definition doesn't work! Ignore it! That's what I get for staying up all night writing lab reports.

    Your geometrical one is fine though. They aren't the same, I'm just silly when I haven't slept in a day :wink:
     
  11. Apr 1, 2005 #10

    dextercioby

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    Nope,direct proportionality means

    [tex] a \propto b \leftrightarrow a=k b,k\in R^{*}_{+} [/tex]

    Daniel.
     
  12. Apr 1, 2005 #11
    What's [tex]R^{*}_{+}[/tex]?

    'k' can be any real number, right?
     
  13. Apr 1, 2005 #12

    dextercioby

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    The real positive semiaxis.

    [tex] \mathbb{R}_{+}^{*}=:\{x\in \mathbb{R}| x>0 \} [/tex]

    Daniel.
     
  14. Apr 1, 2005 #13
    But the spring constant 'k' in the restoring force of an oscillating body is given by:
    F = -kx , x-displacement
     
    Last edited: Apr 2, 2005
  15. Apr 1, 2005 #14

    dextercioby

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    That's not a correct formula.F=-kx.It's not a direct proportionality.As "x" increases,the force decreases & viceversa.

    Daniel.
     
  16. Apr 1, 2005 #15

    BobG

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    No, because [tex]\frac{y-x}{y}[/tex] is not a constant. A change in x changes both terms.
     
  17. Apr 2, 2005 #16
    The solution

    I have found the solution to this question. The problem was the crucial 'missing' component. Now, don't blame me, the source of this question is a very bad text. Let me demonstrate how 'excruciatingly' simple the solution is.

    The missing component:

    [tex] y \propto x [/tex]

    Now the proof:

    [tex] y \propto x [/tex]
    Hence;
    [tex]\frac{1}{y} \propto\frac{1}{x} [/tex]
    So;
    [tex]\frac{1}{y} \ = k \frac{1}{x} [/tex]
    k= constant

    Consider,
    [tex]\frac{1}{x}-\frac{1}{y}[/tex]

    [tex]\frac{1}{x}-\frac{1}{y} = \frac{1}{x} - k \frac{1}{x}[/tex]

    Hence;
    [tex]\frac{1}{x}-\frac{1}{y} = (1-k)\frac{1}{x} [/tex]

    Finally,
    [tex](\frac{1}{x}-\frac{1}{y})\propto\frac{1}{x}[/tex]

    Voila.... :biggrin:!
     
  18. Apr 2, 2005 #17

    dextercioby

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    You mean "voilà"...:wink: Yeah,hopefully you'll be more careful in the future.You had us all puzzled...

    Daniel.
     
  19. Apr 2, 2005 #18

    HallsofIvy

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    Or as teenagers nowadays like to say "Viola"! I figure about 10% of them know it's a joke, the other 90% really think that's the correct pronunciation of "voilà".
     
  20. Apr 3, 2005 #19
    Well, I don't have a French PC.
     
  21. May 30, 2011 #20
    What is direct proportionality?
    In ordinary life I understand it.For example if I buy more tickets i will more pay.But in physical formula I don t understand it.For example in formula of momentum we have p=m*v.But why is there " * "?Why not "+"?That means p=m+v?I know that is stupid question,but I m down.
    From what is this formula,p=m*v derived?
    Thanks
     
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