Constant of proportionality

[autodidude]

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

 

[mathman]

Your question is very confusing.

 

[Mute]

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.