Register to reply 
Why variables in directly proportinality are multipiled 
Share this thread: 
#1
Apr1714, 07:52 AM

P: 59

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

Homework
Sci Advisor
HW Helper
Thanks
P: 12,901

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: 59

##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: 131

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

P: 59




#6
Apr1714, 12:28 PM

P: 131

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


#7
Apr1714, 03:49 PM

Mentor
P: 21,291

##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: 131

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

Homework
Sci Advisor
HW Helper
Thanks
P: 12,901

@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

Mentor
P: 21,291




#11
Apr1714, 08:31 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,901

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



#12
Apr1814, 12:52 AM

P: 59

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


#13
Apr1814, 05:46 AM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,901

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


#14
Apr1814, 07:29 AM

P: 59




#15
Apr1814, 07:40 AM

P: 131

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

Homework
Sci Advisor
HW Helper
Thanks
P: 12,901

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: 59




#18
Apr1914, 07:34 PM

P: 131

Very happy I could help! :)



Register to reply 
Related Discussions  
How derive a CDF from MGF directly ?  Calculus  1  
If we arent directly evolved from chimps, how are our dogs directly evolved wolves?  Biology  12  
When the sun was directly overhead in syene, why was it not directly overhead in alex  Introductory Physics Homework  10  
Can these be used directly?  Advanced Physics Homework  2  
Directly proportional?  General Math  7 