Constant of proportionality

  • Thread starter autodidude
  • Start date
  • #1
333
0
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
 

Answers and Replies

  • #2
mathman
Science Advisor
7,877
453
Your question is very confusing.
 
  • #3
Mute
Homework Helper
1,388
10
If a ∝ b and a ∝ c, why do you multiply b and c together to find the constant?
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 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 [itex]a = k_4bcd[/itex], by similar logic to what I did above. I'm not sure where these powers of a comes from.
 

Related Threads on Constant of proportionality

  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
12K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
1
Views
3K
Top