User:Nitsujbrownie/sandbox

Arc Length

Let $y=f(x)$ be a continuously differentiable function in $\mathbb {R} ^{2}$. Let $s$ be the taxicab arc length of the planar curve defined by $f$ on some interval $[a,b]$. Then the taxicab length of the $i^{\text{th}}$ infinitesimal regular partition of the arc, $\Delta s_{i}$, is given by:

$\Delta s_{i}=\Delta x_{i}+\Delta y_{i}=\Delta x_{i}+|f(x_{i})-f(x_{i-1})|$

By the Mean Value Theorem, there exists some point $x_{i}^{*}$ between $x_{i}$ and $x_{i-1}$such that $f(x_{i})-f(x_{i-1})=f'(x_{i}^{*})dx_{i}$.

$\Delta s_{i}=\Delta x_{i}+|f'(x_{i}^{*})|\Delta x_{i}=\Delta x_{i}(1+|f'(x_{i}^{*})|)$

Then $s$ is given as the sum of every partition of $s$ on $[a,b]$ as they get arbitrarily small.

${\begin{aligned}s&=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}\Delta x_{i}(1+|f'(x_{i}^{*})|)\\&=\int _{a}^{b}1+|f'(x)|\,dx\end{aligned}}$

Curves defined by monotone increasing or decreasing functions have the same taxicab arc length as long as they share the same endpoints.

To test this, take the taxicab circle of radius $r$ centered at the origin. Its curve in the first quadrant is given by $f(x)=-x+r$ whose length is

$s=\int _{0}^{r}1+|-1|dx=2r$

Multiplying this value by $4$ to account for the remaining quadrants gives $8r$, which agrees with the circumference of a taxicab circle. Now take the Euclidean circle of radius $r$ centered at the origin, which is given by $f(x)={\sqrt {r^{2}-x^{2}}}$. Its arc length in the first quadrant is given by

${\begin{aligned}s&=\int _{0}^{r}1+|x{\sqrt {r^{2}-x^{2}}}|dx\\&=x+{\sqrt {r^{2}-x^{2}}}{\bigg |}_{0}^{r}\\&=r-(-r)\\&=2r\end{aligned}}$

Accounting for the remaining quadrants gives $4\times 2r=8r$ again. Therefore, the circumference of the taxicab circle and the Euclidean circle in the taxicab metric are equal. In fact, for any function $f$ that is monotonic and differentiable with a continuous derivative over an interval $[a,b]$, the arc length of $f$ over $[a,b]$ is $(b-a)+\mid f(b)-f(a)\mid$.