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I have this:

$\sqrt{(dx)^2 + (dy)^2}$

And my book simplified it as:

$\sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} \times dx$

I don't have even a close idea how he did it. If it helps, is about path lenght whit integration.

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    Your last differential in the second part should be *outside* the radical, not inside it.2012-06-22

2 Answers 2

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$a\sqrt{r} = \sqrt{a^2(r)}\quad\text{if }a\gt 0\text{ and } r\gt 0.$ So, using changes instead of differentials: $\begin{align*} \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x &= \sqrt{\left(\Delta x\right)^2\left(1 + \left(\frac{\Delta y}{\Delta x}\right)^2\right)}\\ &= \sqrt{(\Delta x)^2 + (\Delta y)^2}. \end{align*}$ Taking limits as $\Delta x\to 0$ converts $\Delta x$ to $dx$, $\Delta y$ to $dy$, and $\frac{\Delta x}{\Delta y}$ to the derivative $\frac{dy}{dx}$.

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    Such stupid... It is simply common factor and everything is clear :) Thank you :)2019-01-04
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$\displaystyle \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(1 + \frac{(dy)^2}{(dx)^2}\right)\cdot(dx)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \cdot \sqrt{(dx)^2}= \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\cdot dx$

(Note that we're treating $dx$ and $dy$ as numbers, but that's another issue; see for example this question.)

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    @Andres Yes, because (assuming $a \geq 0$) $\sqrt{a^2(r+s)} = \sqrt{a^2}\sqrt{r+s} = a \sqrt{r+s}$.2012-06-22