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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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  • $\begingroup$ Your last differential in the second part should be outside the radical, not inside it. $\endgroup$ Jun 22, 2012 at 18:21

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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  • $\begingroup$ Hooo thank you :) I didn't know the first rule :) Yeah, I used dy because I don't know how to put Δ. $\endgroup$
    – Andres
    Jun 22, 2012 at 18:26
  • $\begingroup$ @Andres: The first rule is just a consequence of two things: (i) if $a$ and $b$ are both positive, then $\sqrt{ab} = \sqrt{a}\sqrt{b}$; and (ii) $\sqrt{r^2} = |r|$. Putting them both together, if $a$ and $b$ are positive, then $a\sqrt{b} = $\sqrt{a^2}\sqrt{b}=\sqrt{a^2b}$. $\endgroup$ Jun 22, 2012 at 18:32
  • $\begingroup$ Thank you :) Where you learn that kind of things ? I am at first year in University. Not EEUU, Argentina :) $\endgroup$
    – Andres
    Jun 22, 2012 at 18:39
  • $\begingroup$ @Andres This is standard root manipulation, which I believe is taught in second or third year in high school in Argentina. $\endgroup$
    – talmid
    Jun 22, 2012 at 18:41
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    $\begingroup$ @Andres I suggest khanacademy.org/math/algebra/exponents-radicals, for example. Or pick any third-year high school math book. Properties and tricks of algebraic manipulation like these are ubiquitous in higher-level courses, and people often use them without pointing out what they are doing, like in this case. $\endgroup$
    – talmid
    Jun 22, 2012 at 18:55
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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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  • $\begingroup$ It helps if you think about the Pythagorean theorem, first, noting how a right triangle (with legs $dx$ and $dy$) has its hypoteneuse be $\sqrt{(dx)^{2}+(dy)^{2}}$; we want to change this expression into a differential, i.e., something that looks like $f(x)\,dx$. We can do this using the algebraic manipulations shown in this answer... $\endgroup$ Jun 22, 2012 at 18:25
  • $\begingroup$ Thank you talmid :) Like I said in the other answer i didn't know the rule: $$a \times \sqrt{r+s} = \sqrt{a^2 \times (r+s)}$$ $\endgroup$
    – Andres
    Jun 22, 2012 at 18:36
  • $\begingroup$ @Andres Yes, because (assuming $a \geq 0$) $\sqrt{a^2(r+s)} = \sqrt{a^2}\sqrt{r+s} = a \sqrt{r+s}$. $\endgroup$
    – talmid
    Jun 22, 2012 at 18:39

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