User:Nitsujbrownie/Taxicab geometry

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.

$$\begin{align} s &= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \Delta x_i(1+|f'(x^*_i)|) \\ & = \int_{a}^{b} 1+|f'(x)| \,dx \end{align} $$

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{align} s &= \int_{0}^{r} 1+|x\sqrt{r^2-x^2}|dx\\ &= x+\sqrt{r^2-x^2} \bigg|^{r}_{0}\\ &= r-(-r)\\ &= 2r

\end{align} $$

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, if a function $$f$$ is monotonic and differentiable with a continuous derivative over an interval $$[a,b]$$, then the arc length of $$f$$ over $$[a, b]$$ is $$L = (b-a) + \mid f(b)-f(a) \mid$$.