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Why variables in directly proportinality are multipiled

  1. Apr 17, 2014 #1
    Why variables (RHS) in directly proportionality are always multiplied.

    Suppose the newton 2nd law

    ##{F}\propto{m}##

    ##{F}\propto{a}##

    ##{F}\propto{m*a}##
     
    Last edited by a moderator: Apr 18, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    Simon Bridge

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    If you know that for variables Q, a, and b, if you know ##Q \propto a##, and that ##Q \propto b## then you also know that ##Q\propto ab## ...

    This is because that is what "directly proportional to" means.

    Similarly if ##Q\propto a## and ##Q\propto 1/b## then ##Q\propto a/b##
     
  4. Apr 17, 2014 #3
    If
    ##Q \propto a##
    ##Q \propto b##

    then you also know that

    ##Q\propto (a+b)##
    this can aslo be true why multiply.
     
    Last edited by a moderator: Apr 18, 2014
  5. Apr 17, 2014 #4
    This isn't true.

    The constant of proportionality can't depend on whatever's on the right side. So, [itex]Q \propto a[/itex] means [itex]Q = C_1 \cdot f(b) \cdot a[/itex] (where [itex]C_1[/itex] doesn't depend on [itex]a[/itex] or [itex]b[/itex]).

    Similarly, [itex]Q \propto b[/itex] really means [itex]Q = C_2 \cdot f(a) \cdot b[/itex], for some (possibly different) constant [itex]C_2[/itex].

    The only way this can be true simultaneously is if [itex]Q = C \cdot a \cdot b[/itex] for some constant [itex]C[/itex] -- or, more simply, if [itex]Q \propto ab[/itex].
     
  6. Apr 17, 2014 #5
    I didn't understand what are u trying to say.
     
    Last edited by a moderator: Apr 18, 2014
  7. Apr 17, 2014 #6
    Proportional to [itex]x[/itex]
    means the same as
    Equal to [itex]x[/itex], times some constant.

    Right?
     
  8. Apr 17, 2014 #7

    Mark44

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    I didn't either, especially the parts about f(a) and f(b).
    ##Q \propto a## means Q = ka for some constant k. If you form the ratio Q/a, you get (ka)/a, or k, the constant of proportionality.

    BTW, using "textspeak" such as "u" for "you" isn't allowed here.
     
  9. Apr 17, 2014 #8
    Let me start by answering your original question in a different way.

    Suppose [itex]Q[/itex] is some function of [itex]a[/itex] and [itex]b[/itex]. I'll write it as [itex]Q(a, b)[/itex] to emphasize this.

    To say [itex]Q(a, b) \propto a[/itex] means that [itex]Q(ka, b) = kQ(a, b)[/itex] for any constant [itex]k[/itex]. In words: if you scale up [itex]a[/itex], you scale up [itex]Q[/itex] by the same amount, because [itex]Q[/itex] is proportional to [itex]a[/itex].

    You said that [itex]Q(a, b) = a + b[/itex] satisfies [itex]Q(a, b) \propto a[/itex]. Let's check!
    [tex]
    \begin{align}
    Q(ka, b) &= ka + b \\
    kQ(a, b) &= ka + kb \\
    &\ne Q(ka, b)
    \end{align}
    [/tex]
    Therefore, [itex]a + b[/itex] is not proportional to [itex]a[/itex].

    ---

    Now as to my apparently-hard-to-understand notation: the [itex]f(a)[/itex] notation just means "any function of [itex]a[/itex]". Note that Mark44's constant [itex]k[/itex] could well depend on [itex]a[/itex]! For example, if [itex]Q(a, b) = \sin(a)b[/itex], then [itex]Q(a, b) \propto b[/itex] is true. I used the [itex]f(a)[/itex] notation to emphasize this.
     
  10. Apr 17, 2014 #9

    Simon Bridge

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    @22990atinesh (OP): You seem to be confused about what "directly proportional to" means
    Here is the definition:

    If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##.

    (I suspect you've got the first part but not the second part.)

    Applying this definition:

    (1) ##Q \propto a## means ##Q = k_1 a## where ##k_1## does not depend on ##a##
    (2) ##Q\propto b## means ##Q = k_2 b## where ##k_2## does not depend on ##b##

    Now consider:

    (3) ##Q \propto ab## means that ##Q= k ab## and we can see from (1) and (2) that ##k_1=k/b## does not depend on ##a## and ##k_2=k/a## does not depend on ##b## - so if (1) and (2) are both true, then (3) is also true.


    (4) ##Q \propto a+b## means that ##Q=k(a + b)## and we can see from (1) and (2) that: $$k_1=\frac{k(a+b)}{a},\; k_2=\frac{k(a+b)}{b}$$... these expressions are saying that the only way (4) is true is if ##k_1## depends on ##a## and ##k_2## depends on ##b## - which contradicts the definition of "directly proportional to" used to make (1) and (2).

    In other words, if (1) and (2) are both true, then (4) is false.

    [Note: this is pretty much the argument first appearing in this thread in post #4 (and repeated since)]

    You can try this reasoning process yourself for:

    (5) ##Q\propto f(a,b)## ... where ##f## is an arbitrary function of ##a## and ##b## together...

    ... if (1) and (2) are both true, what form(s) can ##f## take so that (5) is also true?
     
    Last edited: Apr 17, 2014
  11. Apr 17, 2014 #10

    Mark44

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    Who is "you" here?
    I agree, and this is pretty much what I said in post 7.
     
  12. Apr 17, 2014 #11

    Simon Bridge

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    @Mark44: so noted - post #9 edited to reflect your comments :)
     
  13. Apr 18, 2014 #12
    How did ##k_2=k/a## and ##k_1=k/b## come in point (3).
    It must be ##k_2=k*a## and ##k_1=k*b##
     
    Last edited by a moderator: Apr 18, 2014
  14. Apr 18, 2014 #13

    Simon Bridge

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    ##Q=k_1a## and ##Q= kab## true at the same time means that ##k_1 a = kab## or ##k_1=kb## ... so well done: but the argument holds: k_1 does not depend on a - in accordance with the definition given in (1).

    Can you say the same for ##Q=k(a+b)##?
     
  15. Apr 18, 2014 #14
    Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
     
    Last edited by a moderator: Apr 18, 2014
  16. Apr 18, 2014 #15
    Well, several people have already supplied one. But let's try plugging in numbers; maybe examples will help!

    You said that [itex]Q = a + b[/itex] can satisfy [itex]Q \propto a[/itex]. If that's true, then for any values of [itex]a[/itex] or [itex]b[/itex], if we scale up [itex]a[/itex], we scale up [itex]Q[/itex] by the same amount.

    Let's say [itex]a = 1[/itex] and [itex]b = 2[/itex]. This means that [itex]Q = 3[/itex].

    Now let's double [itex]a[/itex], and try predicting what happens to [itex]Q[/itex] in two ways.
    • Using [itex]Q \propto a[/itex], when we double [itex]a[/itex], we double [itex]Q[/itex]. Therefore, we expect [itex]Q = 6[/itex].
    • Using [itex]Q = a + b[/itex], we can just plug in the values. We actually find [itex]Q = 4[/itex].

    4 is not the same as 6. Therefore, we were wrong when we said [itex]Q \propto a[/itex] is true when [itex]Q = a + b[/itex].

    Proportional means multiply.
     
    Last edited by a moderator: Apr 18, 2014
  17. Apr 18, 2014 #16

    Simon Bridge

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    I am at a loss: what education level are you at?

    Do you understand that the definition of ##Q\propto P## is
    ##Q=kP## where k is a number that does not depend on P?​

    I'm afraid that is as simple as it gets.
     
  18. Apr 18, 2014 #17
    Doubt Cleared

    Thanx for example Chogg. This simple example cleared my doubt.

    Simon Bridge I appreciate your efforts to explain. But your explanation was a little bit theoretical. If you would have said the same thing with an example, I would have gotten it earlier. Anyway thanx for help :smile:
     
    Last edited by a moderator: Apr 18, 2014
  19. Apr 19, 2014 #18
    Very happy I could help! :)
     
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