Why do we multiply two directly proportional things?

[Frigus]

when we say one term is directly proportional to something for example if I say x directly proportional to y and I also say x is also directly proportional to z then why we multiply y and z when we say x is directly proportional to something whose value is yz.

 

[tnich]

Directly proportional means that given a relation $x=f(y)$,
$$\frac {x_1} {x_2} = \frac {y_1} {y_2}$$
for $x_1 = f(y_1)$ and $x_2=f(y_2)$. Then if $x=yz$,
$$\frac {x_1} {x_2} = \frac {y_1z} {y_2z}=\frac {y_1} {y_2}$$
so $x$ is directly proportional to $y$, and
$$\frac {x_1} {x_2} = \frac {yz_1} {yz_2}=\frac {z_1} {z_2}$$
so $x$ is directly proportional to $z$.

 

[symbolipoint]

Look directly into the literal meaning of the language for joint & inverse variation!

x directly proportional to y

Let a constant be k.
x=ky

x is also directly proportional to z

let a constant be c.
x=czThose are two formulas for x. The expressions are equal.
ky=cz
or alternatively
y=(c/k)z, where the number c/k is a constant, and this shows y is directly proportional to z.
This did NOT produce your expression yz.

 

[Frigus]

So sir how can we get the expression x=kyz

 

[tnich]

So sir how can we get the expression x=kyz

What are you trying to figure out? Do you have a specific example? Like $PV=nRT$?

 

[symbolipoint]

So sir how can we get the expression x=kyz

We do not automatically get the equation (not expression) x=kyz, unless our numbers are defined or described to show x=kyz. In English worded description, this formula says, "x is directly proportional to y and z."

I may be misunderstanding what you are really try to ask.

 

[Frigus]

What are you trying to figure out? Do you have a specific example? Like $PV=nRT$?

Yes sir,this is the the best example which I can use to tell what I want to say when we say Pv is directly proportional to n,t why do we multiply n and t.

 

[tnich]

I think you are approaching it backwards. If you apply what we have shown you, you will see that $PV$ is directly proportional to $n$ and to $T$. Now that you have learned this pattern, you can apply it in similar situations.

 

[symbolipoint]

Hemant, that example formula, PV=nRT is based on some measurable physical properties which a theory was given, and experimentally found to work well. The variation constant in the formula is R. You could translate the given formula as "The product of P and V is directly proportional to n and T."

 

[tnich]

I think you are approaching it backwards. If you apply what we have shown you, you will see that $PV$ is directly proportional to $n$ and to $T$. Now that you have learned this pattern, you can apply it in similar situations.

Notice in post #2 that when you check the proportionality of $x$ and $y$ by varying $y$, $z$ does not change. You can look at proportionality one pair of variables at a time while holding all other variables constant. So you can multiply $y$ and $z$ together and it does not change their proportionality with $x$.

 

[Frigus]

Sir please explain me this I am very confused if I say gravitational phone directly proportional to m1 and also to m2 and inversely proportional to square of distance between them why do we while combining these term multiply all these terms

 

[symbolipoint]

Sir please explain me this I am very confused if I say gravitational phone directly proportional to m1 and also to m2 and inversely proportional to square of distance between them why do we while combining these term multiply all these terms

A few moments of thought, and my response is,... Basic Physical Science and then expressing the theory as an algebraic formula. The language for describing direct and inverse variation is extremely precise and uncomplicated. Simply learn it, and learn to use it. For the gravitation equation about force, the measurements and testing came first, then someone or some people developed the theory and formula; which probably came as the arithmetic or algebraic formula first (just my guess). Why the multiplication by both masses and then divide by square of distance between them - that is the theory AND the corresponding formula.

You will learn about direct and inverse variation when you study intermediate algebra (depending on where you obtain your mathematical education).

 

[Frigus]

Sir please don't get offend from my reply as I am again and again asking many questions but I am getting through very hard time because I can't understand it.
Please help me to figure out where I am wrong if I say gravitational force is directly proportional to m1 so I can write gravitational force=k1 m1 and also for second mass m2 gravitational force=M2K2 where k1 and K2 are some constants and I can also write gravitational force is inversely proportional to the square of distance between them then I can write gravitational force is equal to k3/r^2 so by combining at by multiplying all the three terms I will get

Fg^3(gravitational force)=(k1)(K2)(k3)(m1)(m2)(m3)/r^2

Then Fg=3√(k1)(K2)(k3)(m1)(m2)/r^2

 

[PeroK]

Sir please don't get offend from my reply as I am again and again asking many questions but I am getting through very hard time because I can't understand it.
Please help me to figure out where I am wrong if I say gravitational force is directly proportional to m1 so I can write gravitational force=k1 m1 and also for second mass m2 gravitational force=M2K2 where k1 and K2 are some constants and I can also write gravitational force is inversely proportional to the square of distance between them then I can write gravitational force is equal to k3/r^2 so by combining at by multiplying all the three terms I will get

Fg^3(gravitational force)=(k1)(K2)(k3)(m1)(m2)(m3)/r^2

Then Fg=3√(k1)(K2)(k3)(m1)(m2)/r^2

First, the gravitational force is proportional to $m_1$ only when $m_2$ and $r$ are fixed. This is what you mean when you say two things are proportional (or inversely proportional): you assume every other variable is fixed.

In this case, if we assume that $r$ is fixed, then:

$F = k_1m_1$ and $F = k_2m_2$ implies $k_1m_1 = k_2 m2$, hence $k_2 = k_1 m_1/m_2$

This gives us: $F = k_2m_2 = (k_1m_1/m_2)m_2 = k_4 m_1m_2$.

Where $k_4 = k_1/m_2 = k_2/m_1$.

Now, if we also have $F = k_3/r^2$, then:

$k_3/r^2 = k_4 m_1 m_2$

Hence:

$k_4 = k/r^2$, where $k = k_3/(m_1 m_2)$

And, finally,

$F = km_1m_2/r^2$

Note that in this equation, if we fix $m_2$ and $r$, say, then the constant of proportionality between $F$ and $m_1$ is $km_2/r^2$ and not just $k$.

 

[Vanadium 50]

Hemant, let me give you two pieces of advice I gave you before.

Hemant, you are replying immediately to the messages you get, I think you will have a better outcome if you think about what people say before responding - it can take a few moments.

If you want to understand physics, you need to put more effort in.

You've disregarded both, and now you're unhappy. I think if you were to take this advice seriously, you'd be happier.

 

[Frigus]

Hemant, that example formula, PV=nRT is based on some measurable physical properties which a theory was given, and experimentally found to work well. The variation constant in the formula is R. You could translate the given formula as "The product of P and V is directly proportional to n and T."

Thanks a lot sir today I understood (after many day) how does this work,

I agained opened the site and started reading from 1st post and when I reached this I understood the thing which you was trying to explain me and also perok and tnich sir.