Joint Proportionality

Question

Although I'm already in calculus and doing good, I don't understand joint variation! Given that $x$ is directly proportional with $y$ and inversely proportional with $z,$ we have that $$\dfrac xy = c_1\quad\text{and}\quad xz = c_2;$$ then why can we combine them to conclude that $$xz = ky \ ?$$ Doesn't $xz$ have a fixed value regardless of what $y$ is?

For example, consider this typical problem:

Given that $x$ is directly proportional with $y$ and inversely proportional with $z:$

if $x = 20, y = 10, z = 5,$

then $x = ? , y = 100, z = 18.$

The given answer: $x = 500/9.$ But $xz = 100$ in the first situation, and $xz = 1000$ in the second situation, and $xz$ is supposed to be a constant; so, how can this be?

Answer 1

It rather depends on what you mean when you say "$x$ is directly proportional with $y$ and inversely proportional with $z$". I suspect it may be intended to mean something like "$x$ is directly proportional to $y$ when $z$ is held constant and $x$ is inversely proportional to $z$ when $y$ is held constant"

Take the example of ideal gases which produce similar results to the problem:

• Charles's law says "when the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be directly related" i.e. $\frac{V}{T}=k_c$

• Boyle's law says "the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system" i.e. $PV=k_b$

• Gay-Lussac's law of pressure–temperature which says "the pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature" i.e. $\frac{P}{T}=k_g$

When all three are combined, allowing all of pressure, temperature and volume to vary together in a closed system, you get $$\dfrac{PV}{T}=k$$ and the suggested result to the problem comes from solving $\dfrac{x \times 18}{ 100} = \dfrac{20 \times 5}{10}$

The alternative approach would be to say that since $x$ is directly proportional to $y$ and inversely proportional to $z$, for consistency you would have to have $y$ inversely proportional to $z$, in which case it would be impossible to have a question which goes from $y=10,z=5$ to $y=100,z=18$. That does not seem to be what was intended

Answer 2

Given that $x$ is directly proportional with $y$ and inversely proportional with $z,$ we have that $$\dfrac xy = c_1\quad\text{and}\quad xz = c_2;$$

The expressions $c_1$ and $c_2$ are not generally constants: instead, $$c_1=f(z)\quad\text{and}\quad c_2=g(y).$$ This is because the proportionality between $x$ and $y$ refers to the value of $z$ being fixed while $x$ and $y$ are varying against each other. Similarly, the value of $y$ is being fixed while $x$ and $z$ are varying against each other.

then why can we combine them to conclude that $$xz = ky \ ?$$ Doesn't $xz$ have a fixed value regardless of what $y$ is?

This result's derivation is at Does $x$ being directly proportional to both $y$ and $z$ imply that $x$ is jointly proportional to $yz?$. It is worth noting that such exercises do assume that $y$ and $z$ are independent of each other (for example, $x=y$ with $xz=1$ satisfies the given statement but violates both this independence assumption and the required result).

Given that $x$ is directly proportional with $y$ and inversely proportional with $z:$

If $x = 20, y = 10, z = 5,$

then $x = ? , y = 100, z = 18.$

The given answer: $x = 500/9.$ But $xz = 100$ in the first situation, and $xz = 1000$ in the second situation, and $xz$ is supposed to be a constant; so, how can this be?

By the previous section, $$xz=ky$$ for some real $k,$ so $xz$ is not actually a constant. Plugging in the first triple gives $$xz=10y,$$ which leads to the given answer.