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Why variables in directly proportinality are multipiled

by 22990atinesh
Tags: directly, multipiled, proportinality, variables
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22990atinesh
#1
Apr17-14, 07:52 AM
P: 58
Why variables (RHS) in directly proportionality are always multiplied.

Suppose the newton 2nd law

##{F}\propto{m}##

##{F}\propto{a}##

##{F}\propto{m*a}##
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Simon Bridge
#2
Apr17-14, 09:49 AM
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If you know that for variables Q, a, and b, if you know ##Q \propto a##, and that ##Q \propto b## then you also know that ##Q\propto ab## ...

This is because that is what "directly proportional to" means.

Similarly if ##Q\propto a## and ##Q\propto 1/b## then ##Q\propto a/b##
22990atinesh
#3
Apr17-14, 11:02 AM
P: 58
Quote Quote by Simon Bridge View Post
If you know that for variables Q, a, and b, if you know ##Q \propto a##, and that ##Q \propto b## then you also know that ##Q\propto ab## ...

This is because that is what "directly proportional to" means.

Similarly if ##Q\propto a## and ##Q\propto 1/b## then ##Q\propto a/b##
If
##Q \propto a##
##Q \propto b##

then you also know that

##Q\propto (a+b)##
this can aslo be true why multiply.

chogg
#4
Apr17-14, 12:23 PM
P: 128
Why variables in directly proportinality are multipiled

Quote Quote by 22990atinesh View Post

If
##Q \propto a##
##Q \propto b##

then you also know that

##Q\propto (a+b)##
this can aslo be true why multiply.
This isn't true.

The constant of proportionality can't depend on whatever's on the right side. So, [itex]Q \propto a[/itex] means [itex]Q = C_1 \cdot f(b) \cdot a[/itex] (where [itex]C_1[/itex] doesn't depend on [itex]a[/itex] or [itex]b[/itex]).

Similarly, [itex]Q \propto b[/itex] really means [itex]Q = C_2 \cdot f(a) \cdot b[/itex], for some (possibly different) constant [itex]C_2[/itex].

The only way this can be true simultaneously is if [itex]Q = C \cdot a \cdot b[/itex] for some constant [itex]C[/itex] -- or, more simply, if [itex]Q \propto ab[/itex].
22990atinesh
#5
Apr17-14, 12:26 PM
P: 58
Quote Quote by chogg View Post
This isn't true.

The constant of proportionality can't depend on whatever's on the right side. So, [itex]Q \propto a[/itex] means [itex]Q = C_1 \cdot f(b) \cdot a[/itex] (where [itex]C_1[/itex] doesn't depend on [itex]a[/itex] or [itex]b[/itex]).

Similarly, [itex]Q \propto b[/itex] really means [itex]Q = C_2 \cdot f(a) \cdot b[/itex], for some (possibly different) constant [itex]C_2[/itex].

The only way this can be true simultaneously is if [itex]Q = C \cdot a \cdot b[/itex] for some constant [itex]C[/itex] -- or, more simply, if [itex]Q \propto ab[/itex].
I didn't understand what are u trying to say.
chogg
#6
Apr17-14, 12:28 PM
P: 128
Proportional to [itex]x[/itex]
means the same as
Equal to [itex]x[/itex], times some constant.

Right?
Mark44
#7
Apr17-14, 03:49 PM
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Quote Quote by chogg View Post
This isn't true.

The constant of proportionality can't depend on whatever's on the right side. So, [itex]Q \propto a[/itex] means [itex]Q = C_1 \cdot f(b) \cdot a[/itex] (where [itex]C_1[/itex] doesn't depend on [itex]a[/itex] or [itex]b[/itex]).

Similarly, [itex]Q \propto b[/itex] really means [itex]Q = C_2 \cdot f(a) \cdot b[/itex], for some (possibly different) constant [itex]C_2[/itex].

The only way this can be true simultaneously is if [itex]Q = C \cdot a \cdot b[/itex] for some constant [itex]C[/itex] -- or, more simply, if [itex]Q \propto ab[/itex].
Quote Quote by 22990atinesh View Post

I didn't understand what are u trying to say.
I didn't either, especially the parts about f(a) and f(b).
##Q \propto a## means Q = ka for some constant k. If you form the ratio Q/a, you get (ka)/a, or k, the constant of proportionality.

BTW, using "textspeak" such as "u" for "you" isn't allowed here.
chogg
#8
Apr17-14, 04:07 PM
P: 128
Let me start by answering your original question in a different way.

Suppose [itex]Q[/itex] is some function of [itex]a[/itex] and [itex]b[/itex]. I'll write it as [itex]Q(a, b)[/itex] to emphasize this.

To say [itex]Q(a, b) \propto a[/itex] means that [itex]Q(ka, b) = kQ(a, b)[/itex] for any constant [itex]k[/itex]. In words: if you scale up [itex]a[/itex], you scale up [itex]Q[/itex] by the same amount, because [itex]Q[/itex] is proportional to [itex]a[/itex].

You said that [itex]Q(a, b) = a + b[/itex] satisfies [itex]Q(a, b) \propto a[/itex]. Let's check!
[tex]
\begin{align}
Q(ka, b) &= ka + b \\
kQ(a, b) &= ka + kb \\
&\ne Q(ka, b)
\end{align}
[/tex]
Therefore, [itex]a + b[/itex] is not proportional to [itex]a[/itex].

---

Now as to my apparently-hard-to-understand notation: the [itex]f(a)[/itex] notation just means "any function of [itex]a[/itex]". Note that Mark44's constant [itex]k[/itex] could well depend on [itex]a[/itex]! For example, if [itex]Q(a, b) = \sin(a)b[/itex], then [itex]Q(a, b) \propto b[/itex] is true. I used the [itex]f(a)[/itex] notation to emphasize this.
Simon Bridge
#9
Apr17-14, 08:18 PM
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@22990atinesh (OP): You seem to be confused about what "directly proportional to" means
Here is the definition:

If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##.

(I suspect you've got the first part but not the second part.)

Applying this definition:

(1) ##Q \propto a## means ##Q = k_1 a## where ##k_1## does not depend on ##a##
(2) ##Q\propto b## means ##Q = k_2 b## where ##k_2## does not depend on ##b##

Now consider:

(3) ##Q \propto ab## means that ##Q= k ab## and we can see from (1) and (2) that ##k_1=k/b## does not depend on ##a## and ##k_2=k/a## does not depend on ##b## - so if (1) and (2) are both true, then (3) is also true.


(4) ##Q \propto a+b## means that ##Q=k(a + b)## and we can see from (1) and (2) that: $$k_1=\frac{k(a+b)}{a},\; k_2=\frac{k(a+b)}{b}$$... these expressions are saying that the only way (4) is true is if ##k_1## depends on ##a## and ##k_2## depends on ##b## - which contradicts the definition of "directly proportional to" used to make (1) and (2).

In other words, if (1) and (2) are both true, then (4) is false.

[Note: this is pretty much the argument first appearing in this thread in post #4 (and repeated since)]

You can try this reasoning process yourself for:

(5) ##Q\propto f(a,b)## ... where ##f## is an arbitrary function of ##a## and ##b## together...

... if (1) and (2) are both true, what form(s) can ##f## take so that (5) is also true?
Mark44
#10
Apr17-14, 08:23 PM
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Quote Quote by Simon Bridge View Post
You seem to be confused about what "directly proportional to" means
Who is "you" here?
Quote Quote by Simon Bridge View Post
Here is the definition:

If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##.

Applying this definition:

(1) ##Q \propto a## means ##Q = k_1 a## where ##k_1## does not depend on ##a##
(2) ##Q\propto b## means ##Q = k_2 b## where ##k_2## does not depend on ##b##
I agree, and this is pretty much what I said in post 7.
Quote Quote by Simon Bridge View Post


[Note: this is pretty much the argument in post #4]
Simon Bridge
#11
Apr17-14, 08:31 PM
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@Mark44: so noted - post #9 edited to reflect your comments :)
22990atinesh
#12
Apr18-14, 12:52 AM
P: 58
Quote Quote by Simon Bridge View Post
@22990atinesh (OP): You seem to be confused about what "directly proportional to" means
Here is the definition:

If ##Q \propto P## then ##Q=kP## where ##k## does not depend on ##P##.

(I suspect you've got the first part but not the second part.)

Applying this definition:

(1) ##Q \propto a## means ##Q = k_1 a## where ##k_1## does not depend on ##a##
(2) ##Q\propto b## means ##Q = k_2 b## where ##k_2## does not depend on ##b##

Now consider:

(3) ##Q \propto ab## means that ##Q= k ab## and we can see from (1) and (2) that ##k_1=k/b## does not depend on ##a## and ##k_2=k/a## does not depend on ##b## - so if (1) and (2) are both true, then (3) is also true.


(4) ##Q \propto a+b## means that ##Q=k(a + b)## and we can see from (1) and (2) that: $$k_1=\frac{k(a+b)}{a},\; k_2=\frac{k(a+b)}{b}$$... these expressions are saying that the only way (4) is true is if ##k_1## depends on ##a## and ##k_2## depends on ##b## - which contradicts the definition of "directly proportional to" used to make (1) and (2).

In other words, if (1) and (2) are both true, then (4) is false.

[Note: this is pretty much the argument first appearing in this thread in post #4 (and repeated since)]

You can try this reasoning process yourself for:

(5) ##Q\propto f(a,b)## ... where ##f## is an arbitrary function of ##a## and ##b## together...

... if (1) and (2) are both true, what form(s) can ##f## take so that (5) is also true?
How did ##k_2=k/a## and ##k_1=k/b## come in point (3).
It must be ##k_2=k*a## and ##k_1=k*b##
Simon Bridge
#13
Apr18-14, 05:46 AM
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Quote Quote by 22990atinesh View Post


How did ##k_2=k/a## and ##k_1=k/b## come in point (3).
It must be ##k_2=k*a## and ##k_1=k*b##
##Q=k_1a## and ##Q= kab## true at the same time means that ##k_1 a = kab## or ##k_1=kb## ... so well done: but the argument holds: k_1 does not depend on a - in accordance with the definition given in (1).

Can you say the same for ##Q=k(a+b)##?
22990atinesh
#14
Apr18-14, 07:29 AM
P: 58
Quote Quote by Simon Bridge View Post
##Q=k_1a## and ##Q= kab## true at the same time means that ##k_1 a = kab## or ##k_1=kb## ... so well done: but the argument holds: k_1 does not depend on a - in accordance with the definition given in (1).

Can you say the same for ##Q=k(a+b)##?
Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
chogg
#15
Apr18-14, 07:40 AM
P: 128
Quote Quote by 22990atinesh View Post

Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
Well, several people have already supplied one. But let's try plugging in numbers; maybe examples will help!

You said that [itex]Q = a + b[/itex] can satisfy [itex]Q \propto a[/itex]. If that's true, then for any values of [itex]a[/itex] or [itex]b[/itex], if we scale up [itex]a[/itex], we scale up [itex]Q[/itex] by the same amount.

Let's say [itex]a = 1[/itex] and [itex]b = 2[/itex]. This means that [itex]Q = 3[/itex].

Now let's double [itex]a[/itex], and try predicting what happens to [itex]Q[/itex] in two ways.
  • Using [itex]Q \propto a[/itex], when we double [itex]a[/itex], we double [itex]Q[/itex]. Therefore, we expect [itex]Q = 6[/itex].
  • Using [itex]Q = a + b[/itex], we can just plug in the values. We actually find [itex]Q = 4[/itex].

4 is not the same as 6. Therefore, we were wrong when we said [itex]Q \propto a[/itex] is true when [itex]Q = a + b[/itex].

Proportional means multiply.
Simon Bridge
#16
Apr18-14, 09:04 AM
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Still didn't understand. I dont want a rigorous proof. I just want a simple explanation.
I am at a loss: what education level are you at?

Do you understand that the definition of ##Q\propto P## is
##Q=kP## where k is a number that does not depend on P?
I'm afraid that is as simple as it gets.
22990atinesh
#17
Apr18-14, 09:29 AM
P: 58
Quote Quote by chogg View Post
Well, several people have already supplied one. But let's try plugging in numbers; maybe examples will help!

You said that [itex]Q = a + b[/itex] can satisfy [itex]Q \propto a[/itex]. If that's true, then for any values of [itex]a[/itex] or [itex]b[/itex], if we scale up [itex]a[/itex], we scale up [itex]Q[/itex] by the same amount.

Let's say [itex]a = 1[/itex] and [itex]b = 2[/itex]. This means that [itex]Q = 3[/itex].

Now let's double [itex]a[/itex], and try predicting what happens to [itex]Q[/itex] in two ways.
  • Using [itex]Q \propto a[/itex], when we double [itex]a[/itex], we double [itex]Q[/itex]. Therefore, we expect [itex]Q = 6[/itex].
  • Using [itex]Q = a + b[/itex], we can just plug in the values. We actually find [itex]Q = 4[/itex].

4 is not the same as 6. Therefore, we were wrong when we said [itex]Q \propto a[/itex] is true when [itex]Q = a + b[/itex].

Proportional means multiply.
Thanx for example Chogg. This simple example cleared my doubt.

Quote Quote by Simon Bridge View Post
I am at a loss: what education level are you at?

Do you understand that the definition of ##Q\propto P## is
##Q=kP## where k is a number that does not depend on P?
I'm afraid that is as simple as it gets.
Simon Bridge I appreciate your efforts to explain. But your explanation was a little bit theoretical. If you would have said the same thing with an example, I would have gotten it earlier. Anyway thanx for help
chogg
#18
Apr19-14, 07:34 PM
P: 128
Very happy I could help! :)


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