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In some notes that I am reading there is the following:

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$$(\delta s)^2=(\delta x)^2+(\delta y)^2 \Rightarrow \left (\frac{\delta s}{\delta x}\right )^2=1+\left (\frac{\delta y}{\delta x}\right )^2$$ When $\delta x \rightarrow 0 $ we get $$(s'(x))^2=1+(y'(x))^2 \Rightarrow s'(x)=\sqrt{1+(f'(x))^2} \Rightarrow s(x)=\int_A^x \sqrt{1+(f'(s))^2}ds$$

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I have understood it as follows:

We have the curve $s$ and $\delta s$ is an approximation of the curve, so we get the triangle $\delta x$, $\delta y$, $\delta s$ and we apply the Pythagorean Theorem to get $(\delta s)^2=(\delta x)^2+(\delta y)^2$. This is equal to $\left (\frac{\delta s}{\delta x}\right )^2=1+\left (\frac{\delta y}{\delta x}\right )^2$.

From the limit $$s'(x)=\lim_{h \rightarrow 0}\frac{s(x+h)-s(x)}{h}=\lim_{h \rightarrow 0}\frac{\delta s}{h}$$ (resp. $y'(x)$) for $h=\delta x$ we get $(s'(x))^2=1+(y'(x))^2$.

Then taking the square root of the last equality we get $s'(x)=\pm \sqrt{1+(y'(x))^2}$.

Why do we take only the positive one, $s'(x)=\sqrt{1+(y'(x))^2}$ ?

After that we take integral to get $s(x)$.

Is everything correct?

Is $s(x)$ the curve or the arc length ?

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After that there is the following:

$\sigma : [0, 1] \rightarrow \mathbb{R}^2 \text{ or } \mathbb{R}^3$ $$I(\sigma )=\int_0^1 ||\sigma '(t)||dt$$ $$d\sigma (t)=\sigma '(t)dt \\ |ds|=||\sigma '(t)||dt \\\ \sigma (t)=(\sigma_1 (t), \sigma_2 (t), \sigma_3 (t)), t \in [0, 1] \\ ||\sigma '(t)||=\sqrt{(\sigma_1' (t))^2, (\sigma_2' (t))^2,( \sigma_3' (t))^2}$$

So when we have a function in $\mathbb{R}$ we use the formula $s(x)$ and when we have a function in $\mathbb{R}^2$ or $\mathbb{R}^3$ we use the last formula $I(\sigma )$ to calculate the arc length?

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