How do you demonstrate that, if $\frac{d f}{d r} = \frac{d g}{d y}$ $(df)/(dr)= (dg)/(dy)$ , then $\frac{d f}{d r} = \frac{d g}{d y} = K$ $(df)/(dr)= (dg)/(dy) = K$ where K is a constant ?

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How do you demonstrate that, if $\frac{d f}{d r} = \frac{d g}{d y}$ $(df)/(dr)= (dg)/(dy)$ , then $\frac{d f}{d r} = \frac{d g}{d y} = K$ $(df)/(dr)= (dg)/(dy) = K$ where K is a constant ?

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Answer 1 · 2017-11-13T16:50:32

EDIT : I'm sorry i allow myself to edit your answer and answer my own question

Explanation:

explanation here :

Let's say that if $r$ $r$ varies $\frac{d f}{d r}$ $(df)/(dr)$ varies too so the equality must be respected so it mean that $\frac{d g}{d y}$ $(dg)/dy$ need to change too, but it doesn't depend of $r$ $r$ so there is a problem.

the only way the equality is respected is that the functions are equal to the same constant.

EDIT 2:

For the math

$\frac{d f}{d r} = \frac{d g}{d y}$ $(df)/(dr) = (dg)/(dy)$

we derivate both side by $y$ $y$

$\frac{d}{d y} (\frac{d f}{d r}) = \frac{d^{2} g}{{d y}^{2}}$ $d/(dy)((df)/(dr)) = (d^2g)/(dy^2)$

$\frac{d^{2} g}{{d y}^{2}} = 0$ $(d^2g)/(dy^2) = 0$

$\frac{d g}{d y} = K = \frac{d f}{d r}$ $(dg)/(dy) = K = (df)/(dr)$

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How do you demonstrate that, if $\frac{d f}{d r} = \frac{d g}{d y}$ $(df)/(dr)= (dg)/(dy)$ , then $\frac{d f}{d r} = \frac{d g}{d y} = K$ $(df)/(dr)= (dg)/(dy) = K$ where K is a constant ?

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Explanation:

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How do you demonstrate that, if dfdr=dgdy(df)/(dr)= (dg)/(dy), then dfdr=dgdy=K(df)/(dr)= (dg)/(dy) = K where K is a constant ?

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Explanation:

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How do you demonstrate that, if $\frac{d f}{d r} = \frac{d g}{d y}$ $(df)/(dr)= (dg)/(dy)$ , then $\frac{d f}{d r} = \frac{d g}{d y} = K$ $(df)/(dr)= (dg)/(dy) = K$ where K is a constant ?