# Constant of proportionality

by autodidude
Tags: constant, proportionality
 P: 333 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
HW Helper
P: 1,391
 Quote by autodidude If a ∝ b and a ∝ c, why do you multiply b and c together to find the constant?
You don't. If $a\propto b$ and $a \propto c$, then that tells you that $a = k_3 bc$, where k3 is some constant of proportionality. This is because:

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

$$1 = \frac{k_1(c)b}{k_2(b)c},$$

or

$$\frac{k_1(c)}{c} = \frac{k_2(b)}{b}.$$

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 $a = k_3 bc$.

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

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