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

enter image description here

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


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 ?


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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.