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I am following the arc length tutorial found here. To derive the arc length formula, the author uses the Mean Value Theorem to express $\Delta y$ in terms of $x$.

"By the Mean Value Theorem we know that on the interval $[x_{i-1}, x_1]$ there is a point $x^*_i$ so that,"

$$ \Delta y_i = f'(x_i^*)\Delta x \\ L = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + [f'(x^*_i)]^2}\Delta x $$

The formula is then written as a definite integral, like so $$ L = \int_a^b \sqrt{1 + [f'(x)]^2}d x $$

Why does $x_i^*$ get simplified to a single $x$? Is the integral implying that because of the infinitely-small $\Delta x$, that the special point that makes the MVT true converges to $x$? If so, then why is the MVT even needed for the definition of arc length?

Greg M
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  • The Mean value Theorem guarantees that there is (at least) one point on the curve on each interval where the slope of its tangent is equal to the slope of the line connecting the "endpoints" Through that relation we can "bring in" the derivative into the equation. When the limit is taken, $x*$ gets "squeezed" between two endpoints – imranfat May 16 '17 at 20:34
  • No. use Pythagoras $(dl)^2=(dx)^2+(dy)^2$ – hamam_Abdallah May 16 '17 at 20:36
  • @imranfat is that the Squeeze Theorem? – Greg M May 16 '17 at 20:45
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    Hi OP! Are you familiar with the idea of a Riemann sum? The idea here is that, for the function $g(x) = \sqrt{1 + f(x)^2}$, we can write our approximation of the arc length as $\sum_{j = 1}^n g(x_j^)(x_j - x_{j-1})$, which is a Riemann sum for the function $g$. The mean value theorem is used to find the points $x_j^$ which make this true. As we let the number of line segments in our polygonal approximation go to infinity, then $n$ goes to infinity, and this Riemann sum goes to the integral of $g$. – Chris May 16 '17 at 20:57
  • @GregM See Chris' explanation – imranfat May 16 '17 at 21:01

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It is worth recalling Riemann integral as a $$\int_{a}^{b}f(x)dx=\lim_{\max\Delta x_k \rightarrow 0}\sum_{k=1}^nf(x_k^*)\Delta x_k \tag{1}$$ where $\{x_1,x_2,...,x_{n+1}\}$ is a partition of $[a,b]$ and $x_k^* \in [x_k,x_{k+1}]$.

The Euclidean distance between 2 points $(x_1,f(x_1)),(x_2,f(x_2)), x_1<x_2$ is $$\sqrt{\left(x_2-x_1\right)^2+\left(f(x_2)-f(x_1)\right)^2}=\left(x_2-x_1\right)\sqrt{1+\left(\frac{f(x_2)-f(x_1)}{x_2-x_1}\right)^2}$$ or using MVT, $\exists x_1^{*}\in (x_1,x_2)$: $$\sqrt{\left(x_2-x_1\right)^2+\left(f(x_2)-f(x_1)^2\right)}=\left(x_2-x_1\right)\sqrt{1+\left(f'(x_1^{*})\right)^2}=\sqrt{1+\left(f'(x_1^{*})\right)^2}\Delta x_1$$ The arc length is initially approximated with the sum of distances of $(x_1,f(x_1)),(x_2,f(x_2)),...,(x_{n+1},f(x_{n+1}))$, applying MVT for each adjacent pair of points: $$\sum_{k=1}^n \sqrt{\left(x_{k+1}-x_{k}\right)^2+\left(f(x_{k+1})-f(x_{k})\right)^2}=\sum_{k=1}^n \sqrt{1+\left(f'(x_k^{*})\right)^2}\Delta x_k=...$$ thinking of $g(x)=\sqrt{1+\left(f'(x)\right)^2}$ , we have $$...=\sum_{k=1}^n g(x_k^{*})\Delta x_k$$ taking the limit $\max\Delta x_k \rightarrow 0$ we have the Riemann integral $(1)$ for $g(x)$ $$L=\int_{a}^{b}g(x)dx=\int_{a}^{b}\sqrt{1+\left(f'(x)\right)^2}dx$$

rtybase
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