Constant of proportionality

  1. Oct 11, 2011 #1
    If a ∝ b and a ∝ c, why do you multiply b and c together to find the constant?

    I also noticed something, but am not sure of the reason why. If you find the constants individually for each expression and combine them all, you get a the the power of the number of expressions

    e.g.

    a ∝ b
    a ∝ c
    a ∝ d

    So the individual constants would be say, k1, k2 and k3 respectively. If you then multiply it all together

    a ∝ (k1b)(k2c)(k3d)

    You get a^3

    If there're expressions, then a^4 etc. All of the numbers I've tried so far have yielded the result but I'm not sure why that's happening

    Thanks
     
  2. jcsd
  3. Oct 11, 2011 #2

    mathman

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    Your question is very confusing.
     
  4. Oct 11, 2011 #3

    Mute

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    You don't. If [itex]a\propto b[/itex] and [itex]a \propto c[/itex], then that tells you that [itex]a = k_3 bc[/itex], where k3 is some constant of proportionality. This is because:

    Given both [itex]a = k_1(c)b[/itex], where k1(c) is a proportionality factor that you know depends on c, and [itex]a = k_2(b)c[/itex], where k2(b) is a proportionality factor that depends on b, you can divide the two equations to get

    [tex]1 = \frac{k_1(c)b}{k_2(b)c},[/tex]

    or

    [tex]\frac{k_1(c)}{c} = \frac{k_2(b)}{b}.[/tex]

    However, by assumption k1 depends only on c and k2 depends only on b, so the only way this relation can hold is if both sides are equal to the same constant, say k3. It follows then that [itex]a = k_3 bc[/itex].

    I'm not sure what you're talking about here. If a is proportional to all those variables, then [itex]a = k_4bcd[/itex], by similar logic to what I did above. I'm not sure where these powers of a comes from.
     
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