If $a$ is directly proportional to $b$ and also directly proportional to $c$,
is it true that $a$ is directly proportional to $bc?$ (It seems like it is true.)
Here is what I did, and I have a feeling I have made a pretty big mistake somewhere:
From the given info, we have$$\frac{a}{b}=x$$ for some constant $x,$ and $$\frac{a}{c}=y$$ for some constant $y.$ Multiplying the two equations gives $${a}^2=bc(xy),$$ where $xy$ is a constant.
So, ${a}^2$ is directly proportional to $bc.$
Update:Now for my real doubt. My physics textbook states the following:
Let two objects $A$ and $B$ of masses $M$ and$m$ lie at a distance $d$ from each other. Let the force of attraction between two objects be $F$. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses; that is,$F ∝ M m.$And the force between two objects is inversely proportional to the square of the distance between them; that is,$F ∝\frac1{d^2}.$Combining the proportionalities we get$F ∝ \displaystyle\frac{Mm}{d^2}$.
So, does it mean that my textbook is wrong? Doesn't the combination of the two proportionalities result in $F^2$ being directly proportional to $\displaystyle\frac{Mm}{d^2}?$
$\endgroup$ 45 Answers
$\begingroup$Revised and updated answer
I did not examine carefully your assertion that ${a}^2$ is directly proportional to $bc$
Actually, a is jointly proportional to $b$ and $c$ which means
$a \propto bc$
Here are two examples:
Distance travelled varies directly with speed for a given time.
Distance travelled varies directly with time for a given speed.
Distance travelled varies jointly with speed and time, distance $\propto$ speed$\times time$
To take another example with both direct variation, and inverse variation,
Time taken for a journey varies directly with the distance for a given time
Time taken for a journey varies inversely as the speed, fpr a given time
Time $\propto$ distance/speed
[ It just happens that in the above two examples, the constant of variation is $1$ ]
Added: Proof for the original example
Let $a$ change to $a_1$ while $c$ is unchanged, and $b$ changes to $b'$, then $a/a_1 = b/b'$
Then let $a_1$ change to $a_2$ while $b'$ is unchanged, and $c$ changes to $c'$, then $a_1/a_2 = $c/c'$
Multiplying the two, $\frac{a}{a_2} = \frac{bc}{b'c'}$
so $a\propto bc$
If $a$ is directly proportional to $b$ and also directly proportional to $c$,
we have $$\frac{a}{b}=x$$ for some constant $x,$
Remember, $a$ depends on both $b$ and $c.$
So, $x$ is actually a function of $c.$ It is really aparameter whose value here is being fixed.
and $$\frac{a}{c}=y$$ for some constant $y.$
Similarly, $y$ is actually a function of $c;$ it is really a parameter whose value here is being fixed.
Multiplying the two equations gives $${a}^2=bc(xy),$$ where $xy$ is a constant.
So, ${a}^2$ is directly proportional to $bc.$
In fact, $xy$ is not a constant, but a function of both $b$ and $c.$Thus, $$a^2=bc\times f(b,c)\not\equiv k(bc).$$ So, it is in fact invalid to conclude that $a^2\propto bc.$
If $a$ is directly proportional to $b$ and also directly proportional to $c$,
is it true that $a$ is directly proportional to $bc?$
This is indeed true (provided that there is no additional dependency between $b$ and $c$).
Here is the proof.
In that answer, I also supplied the counterexample $a=b,\,b=2c.$
$F ∝ M m$ and $F ∝\dfrac1{d^2}.$ Combining the proportionalities we get $F ∝ \dfrac{Mm}{d^2}$.
Doesn't the combination of the two proportionalities result in $F^2$ being directly proportional to $\displaystyle\frac{Mm}{d^2}?$
As explained above, it is $F,$ not $F^2,$ is directly proportional to $\displaystyle\frac{Mm}{d^2}.$
From your first two identities,
$$a=bx=cy.$$
These equalities can hold simultaneously iff $b$ and $c$ are also proportional to each other ! Then $a^2=bcxy$ indeed shows that $a^2$ is proportional to $bc$.
But the usual way to work this out is "with $c$ fixed, $a=bx$, and with $b$ fixed $a=cy$: proportionality works independently for $b$ and $c$.
For this to be possible, you must have
$$a=bx_c=cy_b=bcz$$ or if you prefer, both "constants" are proportional to the other variable
$$x_c=cz,y_b=bz.$$
$\endgroup$ $\begingroup$Consider that $F$ is a function of $M,m,$ and $d.$ That is $F=F(M,m,d).$ Pick some particular non-zero values $M_0,m_0,d_0.$ We have $$F(M,m,d)=F(M,m,d_0)(d/d_0)^{-2}$$ because $F(M,m,d)$ is inversely proportional to $d^2.$ We have $$F(M,m,d_0)=F(M,m_0,d_0)(m/m_0)$$ because $F(M,m,d_0)$ is proportional to $M m,$ and because $M m/(M m_0)=m/m_0.$ We have $$F(M,m_0,d_0)=F(M_0,m_0,d_0)(M/M_0)$$ because $F(M,m_0,d_0)$ is proportional to $M m_0,$ and because $M m_0/(M_0 m_0)=M/M_0.$ Altogether we have $$F(M,m,d)=(M m/d^2)K_0$$ for all $M,m,d$, where $K_0=d_0^2/M_0 m_0.$
$\endgroup$ $\begingroup$The following argumentation has been used in the good old days to teach us about the ideal gas laws of Boyle and Gay Lussac.
- Experiments reveal that the pressure $p$ is proportional to the amount of gas $n$
if temperature $T$ and volume $V$ are held constant. Thus: $\, p \sim n$ . - Experiments reveal that the pressure $p$ is inversely proportional to the volume $V$
if temperature $T$ and amount of gas $n$ are held constant. Thus: $\, p \sim 1/V$ . - Experiments reveal that the pressure $p$ is proportional to the temperature $T$
if volume $V$ and amount of gas $n$ are held constant. Thus: $\, p \sim T$ . - Experiments reveal that the volume $V$ is proportional to the temperature $T$
if pressure $p$ and amount of gas $n$ are held constant. Thus: $\, V \sim T$ .
A next step is to introduce absolute temperature $T$ (Kelvin) instead of relative temperature.
Then from (3) we have $p_1/T_1 = p_2/T_2$ : pressure law of Gay Lussac.
The book proceeds with (4) and the volume law of Gay Lussac : $V_1/T_1 = V_2/T_2$.
And then comes a clue. What happens if we change the volume $V$ as well as the temperature $T$ ?
The book says that we should do this in two subsequent steps:
- Keep the volume $V$ constant and raise the temperature $T$.
Then we have according to Gay Lussac: $\;p_1/T_1 = p_2'/T_2 \quad \Longrightarrow \quad p_2' = p_1\cdot T_2/T_1$ - Keep the temperature $T$ constant and change the volume $V$.
Then we have according to Boyle: $\;p_2'\cdot V_1 = p_2\cdot V_2 \quad \Longrightarrow \quad p_1\cdot T_2/T_1 \cdot V_1 = p_2\cdot V_2$
It is supposed that the same procedure may be applied for Newton's law of gravitation:
- Experiments reveal that the force $F$ is proportional to one of the two masses $m$
if the other mass $M$ as well as the distance $r$ between the masses is held constant: $\, F \sim m$ . - The same argument hold for the other mass if mass $m$ and distance $r$ are held constant.
Thus: $\, F \sim M$ . - Experiments reveal that the force $F$ is inversely proportional to the distance $r$ squared
if the mass $M$ as well as the mass $m$ are held constant. Thus: $\, F \sim 1/r^2$ .
- Keep the distance constant and change only the mass $m$ : $ F_1/m_1 = F_2/m_4 \quad \Longrightarrow \quad F_2 = F_1 m_4/m_1 $
- Keep the distance constant and change only the mass $M$ : $ F_2/M_1 = F_3/M_4 \quad \Longrightarrow \quad F_3 = F_1 m_4 M_4/(m_1 M_1) $
- Keep the masses constant and change only the distance $r$ : $ F_3 r_1^2 = F_4 r_4^2 \quad \Longrightarrow \quad F_4 = F_1 m_4 M_4 r_1^2/(m_1 M_1 r_4^2) $
DR. SCHWEERS EN DRS. P. VAN VIANEN NATUURKUNDE op corpusculaire grondslag DEEL I voor de onderbouw van het v.h.m.o. ZEVENDE DRUK (1960) L.C.G. MALMBERG 's-HERTOGENBOSCH$\endgroup$ 2
