Constant of proportionality

  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


    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

  2. jcsd
  3. mathman

    mathman 6,754
    Science Advisor
    Gold Member

    Your question is very confusing.
  4. Mute

    Mute 1,390
    Homework Helper

    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]


    [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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