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Why variables in directly proportinality are multipiled 
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#1
Apr1714, 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}## 


#2
Apr1714, 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## 


#3
Apr1714, 11:02 AM

P: 58

##Q \propto a## ##Q \propto b## then you also know that ##Q\propto (a+b)## this can aslo be true why multiply. 


#4
Apr1714, 12:23 PM

P: 128

Why variables in directly proportinality are multipiled
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]. 


#5
Apr1714, 12:26 PM

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#6
Apr1714, 12:28 PM

P: 128

Proportional to [itex]x[/itex]
means the same as Equal to [itex]x[/itex], times some constant. Right? 


#7
Apr1714, 03:49 PM

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##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. 


#8
Apr1714, 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 apparentlyhardtounderstand 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. 


#9
Apr1714, 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? 


#10
Apr1714, 08:23 PM

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#11
Apr1714, 08:31 PM

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@Mark44: so noted  post #9 edited to reflect your comments :)



#12
Apr1814, 12:52 AM

P: 58

It must be ##k_2=k*a## and ##k_1=k*b## 


#13
Apr1814, 05:46 AM

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Can you say the same for ##Q=k(a+b)##? 


#14
Apr1814, 07:29 AM

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#15
Apr1814, 07:40 AM

P: 128

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.
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. 


#16
Apr1814, 09:04 AM

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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. 


#17
Apr1814, 09:29 AM

P: 58




#18
Apr1914, 07:34 PM

P: 128

Very happy I could help! :)



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