# Which of these variables affect y more?

I have 2 formulas:

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

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

Which variable has a better result on $y$? i.e., does increasing $m$ have a better result on $y$ than increasing $k$? Just how do I describe/prove that?

Many thanks!

(I asked a comparable inquiry formerly, here. Nonetheless I really feel that the square root nature of this inquiry makes it a totally various inquiry, therefore I've uploaded a new inquiry.)

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2022-06-07 14:32:20
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$k$ is straight with $m$. On the various other hand, $k$ is straight with $y^2$. So $y$ influences $k$ greater than $m$ does.

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2022-06-07 15:08:01
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Of 2 variables, which influences $y$ extra? I have 2 formulas:

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

How do I establish which variable has a better affect on $y$?

E.g., just how do I establish if increasing $k$ as a better result on $y$ than increasing $m$ etc?

Keep in mind additionally that given that $\displaystyle y =\frac{0.0060}{k}\quad$ and also $\quad\displaystyle y =\frac{0.00016}{m}$, after that

$$\frac{0.0060}{k} = \frac{0.00016}{m} \implies \frac{k}{m} = \frac{0.0060}{0.00016} = 37.5$$

We call $37.5$ the symmetry constant (the value of the proportion $\displaystyle \frac km$). That is, $k$ and also $m$ are straight symmetrical . An additional means to share this symmetry is that $k$ is a numerous of $m$: $k = 37.5m$.

(It would certainly be excellent to have a look at Wikipedia-proportionality).

In either of these kinds, it is very easy to see that increasing both $k$ and also $m$ by the very same variable not does anything to transform the symmetry constant: and also doing so for each and every of $k$ and also $m$ cause the very same result on $y$.

E.g. if we consider $k = 37.5 m$, increase both $m$ and also $k$ by, claim, $3$, we have:

$(3k) = 37.5(3m) \implies 3(k) = 3(37.5m)$ which streamlines back to our initial formula: $k = 37.5m$. ¢

In relationship to $y$, $y$ and also $k$, $y$ and also $m$ are vice versa symmetrical , therefore:

$\displaystyle y = \frac{0.0060}{k} = \frac{0.00016}{m}$ after that $\displaystyle \frac{0.0060}{3k} = \frac {1}{3} y = \frac{0.00016}{3m}$

Going back to the proportion of $k$ to $m$: $\displaystyle \frac{k}{m} = 37.5$ Now, we have actually seen that increasing both $k$ and also $m$ by the very same variable (of any kind of value) does not transform the symmetry of $k$ to $m$, any kind of any such adjustment in the values of $k$ and also $m$ influence the value of $y$ in specifically the very same fashion.

In a similar way, with your blog post today: as others have actually revealed, increasing both $m$ and also $k$ by the very same number has the very same influence on $y$, no matter the reality that you currently have actually $y$ being shared as features of the square - origin of $k$ and also of $m$, specifically, and also for comparable factors:

In this instance, we have that $\displaystyle y = \frac{0.0029}{\sqrt{k}} = \frac{0.0060}{\sqrt{m}}$ therefore we have the percentage of $\displaystyle \frac{\sqrt{k}}{\sqrt{m}}=\frac{0.0029}{0.0060} \approx 0.5$

Multiplying both $k$ and also $m$ by $2$ amounts increasing both $\sqrt{k}$ and also $\sqrt{m}$ by $\sqrt{2}$, hence negating, leaving the symmetry constant unchanged.¢

However (I believe this is what your instinct was relative to your blog post :

IF (as in the first blog post, yet picking less complex numerators) $\displaystyle y = \frac{4}{k} = \frac{1}{m}$, after that we have that $\displaystyle \frac{k}{m} = 4/1 = 4$, or equivalently, $k = 4m$.

Currently if you were to ask: what is the influence on $y$ if you were to take the square origin of both $k$ and also $m$, after that, without a doubt, among both will certainly have a better influence on $y$ than the various other, under a couple of problems.

In this instance, allow $\displaystyle y_k = \frac{4}{\sqrt{k}}$ and also $\displaystyle y_m = \frac{1}{\sqrt{m}}$, after that making even both sides of each formula offers 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 changing $k$ from with the square origin of $k$ causes a value of $y$ which is $4$ times more than the value of $y$ when $m$ is changed with the square origin of $m$.

In a similar way, for any kind of backer $p \neq 1$ and also any kind of formulas of the kind $\displaystyle y = \frac xk$ and also $\displaystyle y = \frac ym, \;x \neq y$, changing $k$ with $k^p$ and also $m$ with $m^p$ (where $p \neq 0$) will certainly cause varying influence on the initial value of $y$. The size of the distinction in influence relies on the offered values for the numerators, and also the value of the backer $p$.

Attempt experimenting with this a little bit!

4
2022-06-07 14:55:15
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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 additionally get

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

So your value will certainly constantly transform by a constant variable that is $\displaystyle \frac{1}{\sqrt{2}}$. Just how much the "effect" is relies on just how large $y$ was in the past.

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2022-06-07 14:52:53
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