I have two equations:

1: $\displaystyle y=\frac{0.0029}{\sqrt{k}}$

2: $\displaystyle y=\frac{0.0060}{\sqrt{m}}$

Which variable has a greater effect on $y$? i.e., does doubling $m$ have a greater effect on $y$ than doubling $k$? How do I describe/prove that?


(I asked a similar question previously, here. However I feel that the square root nature of this question makes it an entirely different question, and so I've posted a new question.)

  • $\begingroup$ Well, you want to compare $k$ and $m$, right? $\endgroup$ – Matt Gregory May 22 '11 at 13:09
  • $\begingroup$ Why not try it and see? Pick some $m$, see the effect of doubling it, then do the same with $k$, see if you can work out what's going on. $\endgroup$ – Gerry Myerson May 22 '11 at 13:09
  • $\begingroup$ By the way, the "similar question" referred to was math.stackexchange.com/questions/40475/… $\endgroup$ – Gerry Myerson May 22 '11 at 13:11
  • $\begingroup$ Try and plot some of those graphics for differents $k$ and $m$ at wolframalpha.com, so you can actually see what is happening. $\endgroup$ – Leonardo Fontoura May 22 '11 at 15:13
  • $\begingroup$ you need to start accepting, and/or voting for answers to questions you've asked: people here put a lot of time and effort providing answers to questions. All we ask is that you accept solutions and/or vote up those you find helpful. I realize that you're somewhat new here, but you've asked 5 questions now, all answered, but have provided no feedback. $\endgroup$ – Namaste Jun 2 '11 at 22:49

From your first related question, you asked:

Of two variables, which affects $y$ more? I have two equations:

  1. $\displaystyle y = \frac {0.0060}{k}$
  2. $\displaystyle y= \frac{0.00016}{m}$

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!


If you double $k$ you have

$$y'=\frac{0.0029}{\sqrt{2\cdot k}}=\frac{1}{\sqrt{2}}\cdot y$$

Simillary if you double $m$ you also get

$$y'=\frac{0.0060}{\sqrt{2\cdot m}}=\frac{1}{\sqrt{2}}\cdot y$$

So your value will always change by a constant factor that is $\displaystyle \frac{1}{\sqrt{2}}$. How much the "effect" is depends on how big $y$ was before.


$k$ is linear with $m$. On the other hand, $k$ is linear with $y^2$. So $y$ affects $k$ more than $m$ does.


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