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Direct proportion

  • Thread starter Reshma
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This seems to be a simple question. But I'm rather dubious at a particular step.
The question is: Prove
[tex]\frac{1}{x}-\frac{1}{y}[/tex] is directly proportional to [tex]\frac{1}{x}[/tex]

This how I went about it:
[tex]\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}[/tex]
[tex]\frac{y-x}{xy}=[\frac{y-x}{y}]\frac{1}{x}[/tex]

Can we say the quantity on the right hand side is proportional to [tex]\frac{1}{x}[/tex]?
 
998
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Well, what does proportional mean?
 
743
4
I'll give a more practical definition: If the quantity on the right hand side increases or decreases then the quantity on the left hand side also increases or decreases.There is a constant which can be any real number which relates the two quantites.

Now, in context to this question: Can you call the term in the parenthesis a 'constant' since both 'x' and 'y' are variables?
 
998
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So when you say

[tex] a \propto b [/tex]

you mean that there's some [itex]k[/itex] independent of [itex]a[/itex] and [itex]b[/itex] such that

[tex] a = k b[/tex]?

or to be more precise, [itex]a \propto b[/itex] iff [itex]\frac{da}{db}[/itex] is constant.

Then

[tex] \frac{1}{x} - \frac{1}{y} \propto \frac{1}{x}[/tex]

is indeed false.

For example, just let [itex] u = \frac{1}{x}[/itex], and [itex]v = \frac{1}{y}[/itex]. Then your question is to prove

[tex]u - v \propto u[/tex]

which is silly unless [itex]v \propto u[/itex] (easy to prove), and presumably [itex]y[/itex] is independent of [itex]x[/itex] in your question~

I imagine that your teacher isn't using that definition for proportionality, because otherwise he/she is being very silly~
 
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743
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I know this a silly sum. But I think more precise definition of proportionality would be the geometrical one:

[tex] y \propto x [/tex]

[tex] y = m x[/tex]
m is the slope. With variable y on y-axis and variable x on x-axis.
 
998
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Yeah, that's the same as the one I gave in a different form :smile:

[tex] y \propto x \Longleftrightarrow \frac{dy}{dx} \ \mbox{is constant} [/tex]
 

dextercioby

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Nope,in the way it's formulated,it definitely doesn't match our understanding of direct proportionality.

Daniel.
 
743
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Of course the x and y above are different from the question.
The result of this question seems obivious by your second method and you can get a straight line on a graph considering :
[tex] \frac{1}{x} - \frac{1}{y}[/tex]
on y-axis &
[tex]\frac{1}{x}[/tex]
on x-axis.
 
998
0
Actually, my definition doesn't work! Ignore it! That's what I get for staying up all night writing lab reports.

Your geometrical one is fine though. They aren't the same, I'm just silly when I haven't slept in a day :wink:
 

dextercioby

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Nope,direct proportionality means

[tex] a \propto b \leftrightarrow a=k b,k\in R^{*}_{+} [/tex]

Daniel.
 
743
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What's [tex]R^{*}_{+}[/tex]?

'k' can be any real number, right?
 

dextercioby

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The real positive semiaxis.

[tex] \mathbb{R}_{+}^{*}=:\{x\in \mathbb{R}| x>0 \} [/tex]

Daniel.
 
743
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But the spring constant 'k' in the restoring force of an oscillating body is given by:
F = -kx , x-displacement
 
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dextercioby

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That's not a correct formula.F=-kx.It's not a direct proportionality.As "x" increases,the force decreases & viceversa.

Daniel.
 

BobG

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Reshma said:
This seems to be a simple question. But I'm rather dubious at a particular step.
The question is: Prove
[tex]\frac{1}{x}-\frac{1}{y}[/tex] is directly proportional to [tex]\frac{1}{x}[/tex]

This how I went about it:
[tex]\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}[/tex]
[tex]\frac{y-x}{xy}=[\frac{y-x}{y}]\frac{1}{x}[/tex]

Can we say the quantity on the right hand side is proportional to [tex]\frac{1}{x}[/tex]?
No, because [tex]\frac{y-x}{y}[/tex] is not a constant. A change in x changes both terms.
 
743
4
The solution

I have found the solution to this question. The problem was the crucial 'missing' component. Now, don't blame me, the source of this question is a very bad text. Let me demonstrate how 'excruciatingly' simple the solution is.

The missing component:

[tex] y \propto x [/tex]

Now the proof:

[tex] y \propto x [/tex]
Hence;
[tex]\frac{1}{y} \propto\frac{1}{x} [/tex]
So;
[tex]\frac{1}{y} \ = k \frac{1}{x} [/tex]
k= constant

Consider,
[tex]\frac{1}{x}-\frac{1}{y}[/tex]

[tex]\frac{1}{x}-\frac{1}{y} = \frac{1}{x} - k \frac{1}{x}[/tex]

Hence;
[tex]\frac{1}{x}-\frac{1}{y} = (1-k)\frac{1}{x} [/tex]

Finally,
[tex](\frac{1}{x}-\frac{1}{y})\propto\frac{1}{x}[/tex]

Voila.... :biggrin:!
 

dextercioby

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You mean "voilà"...:wink: Yeah,hopefully you'll be more careful in the future.You had us all puzzled...

Daniel.
 

HallsofIvy

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Or as teenagers nowadays like to say "Viola"! I figure about 10% of them know it's a joke, the other 90% really think that's the correct pronunciation of "voilà".
 
743
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Well, I don't have a French PC.
 
77
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What is direct proportionality?
In ordinary life I understand it.For example if I buy more tickets i will more pay.But in physical formula I don t understand it.For example in formula of momentum we have p=m*v.But why is there " * "?Why not "+"?That means p=m+v?I know that is stupid question,but I m down.
From what is this formula,p=m*v derived?
Thanks
 

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