1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:

    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.


    [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


    User Avatar
    Science Advisor
    Homework Helper

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

  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 &
    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


    User Avatar
    Science Advisor
    Homework Helper

    Nope,direct proportionality means

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

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

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


    User Avatar
    Science Advisor
    Homework Helper

    The real positive semiaxis.

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

  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


    User Avatar
    Science Advisor
    Homework Helper

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

  16. Apr 1, 2005 #15


    User Avatar
    Science Advisor
    Homework Helper

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


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

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


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


    User Avatar
    Science Advisor
    Homework Helper

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

  19. Apr 2, 2005 #18


    User Avatar
    Science Advisor

    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 "voilĂ ".
  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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook