From your first related question, you asked:
Of two variables, which affects $y$ more? I have two equations:
- $\displaystyle y = \frac {0.0060}{k}$
- $\displaystyle y= \frac{0.00016}{m}$
How do I determine which variable has a greater affect on $y$?
E.g., how do I determine if doubling $k$ as a greater effect on $y$ than doubling $m$ and so forth?
Note also that since $\displaystyle y =\frac{0.0060}{k}\quad$ and $\quad\displaystyle y =\frac{0.00016}{m}$, then
$\frac{0.0060}{k} = \frac{0.00016}{m} \implies \frac{k}{m} = \frac{0.0060}{0.00016} = 37.5$
We call $37.5$ the proportionality constant (the value of the ratio $\displaystyle \frac km$). That is, $k$ and $m$ are directly proportional. Another way to express this proportionality is that $k$ is a multiple of $m$: $k = 37.5m$.
(It would be good to take a look at Wikipedia-proportionality).
In either of these forms, it is easy to see that multiplying both $k$ and $m$ by the same factor does nothing to change the proportionality constant: and doing so for each of $k$ and $m$ results in the same effect on $y$.
E.g. if we look at $k = 37.5 m$, multiply both $m$ and $k$ by, say, $3$, we have:
$(3k) = 37.5(3m) \implies 3(k) = 3(37.5m)$ which simplifies back to our original equation: $k = 37.5m$.
In relation to $y$, $y$ and $k$, $y$ and $m$ are inversely proportional, hence:
$\displaystyle y = \frac{0.0060}{k} = \frac{0.00016}{m}$ then $\displaystyle \frac{0.0060}{3k} = \frac {1}{3} y = \frac{0.00016}{3m}$
Going back to the ratio of $k$ to $m$: $\displaystyle \frac{k}{m} = 37.5$ Now, we have seen that multiplying both $k$ and $m$ by the same factor (of any value) does not change the proportionality of $k$ to $m$, any any such change in the values of $k$ and $m$ impact the value of $y$ in exactly the same manner.
Similarly, with your post today: as others have shown, multiplying both $m$ and $k$ by the same number has the same impact on $y$, regardless of the fact that you now have $y$ being expressed as functions of the square-root of $k$ and of $m$, respectively, and for similar reasons:
In this case, we have that $\displaystyle y = \frac{0.0029}{\sqrt{k}} = \frac{0.0060}{\sqrt{m}}$ and so we have the proportion of $\displaystyle \frac{\sqrt{k}}{\sqrt{m}}=\frac{0.0029}{0.0060} \approx 0.5$
Multiplying both $k$ and $m$ by $2$ is equivalent to multiplying both $\sqrt{k}$ and $\sqrt{m}$ by $\sqrt{2}$, thus cancelling out, leaving the proportionality constant unchanged.
However (I suspect this is what your intuition was with respect to your post:
IF (as in the first post, but choosing simpler numerators) $\displaystyle y = \frac{4}{k} = \frac{1}{m}$, then we have that $\displaystyle \frac{k}{m} = 4/1 = 4$, or equivalently, $k = 4m$.
NOW if you were to ask: what is the impact on $y$ if you were to take the square root of both $k$ and $m$, then, indeed, one of the two will have a greater impact on $y$ than the other, under a few conditions.
In this case, let $\displaystyle y_k = \frac{4}{\sqrt{k}}$ and $\displaystyle y_m = \frac{1}{\sqrt{m}}$, then squaring both sides of each equation gives us: $y_{k}^{2} = \frac{4^2}{k} = 4\left(\frac{4}{k}\right) = 4y$ $y_{m}^{2} =\frac{1^2}{m}= \frac{1}{m} = y$
Hence we have that replacing $k$ from with the square root of $k$ results in a value of $y$ which is $4$ times greater than the value of $y$ when $m$ is replaced with the square root of $m$.
Similarly, for any exponent $p \neq 1$ and any equations of the form $\displaystyle y = \frac xk$ and $\displaystyle y = \frac ym, \;x \neq y$, replacing $k$ with $k^p$ and $m$ with $m^p$ (where $p \neq 0$) will result in differing impacts on the original value of $y$. The magnitude of the difference in impact depends on the given values for the numerators, and the value of the exponent $p$.
Try playing around with this a bit!