In the following definition of the length of a curve which is copied verbatim from this Wikipedia page, I would like to know why the length of a line segment equals $ \big | f(t_i) – f(t_{i-1}) \big | $.
Here is the definition of the arc length from Wikipedia.
Let $f\colon[a,b]\to \mathbb{R}^n$ be a continuously differentiable
function. The length of the curve defined by $f$ can be defined as
the limit of the sum of line segment lengths for a regular partition
of $[a,b]$ as the number of segments approaches infinity. This means$$L(f)=\lim_{N\to\infty}\sum_{i=1}^N \bigg|f(t_i)-f(t_{i-1})\bigg|$$
where $t_i=a+i(b-a)/N=a+i\Delta t$ for $i=0,1,\dotsc,N$.
It seems to me that $ \big | f(t_i) – f(t_{i-1}) \big | $ needs to be replaced with $ \big | (t_i, f(t_i)) – (t_{i-1}, f(t_{i-1})) \big | $ because the latter is the true length of a line segment, not the former.
For instance, for the case where $n = 1$ where we deal with a curve in the plane, if we let $y = f(x)$ then the length of a line segment should be $ \big | (t_i, f(t_i)) – (t_{i-1}, f(t_{i-1})) \big | $ which is equal to $\sqrt{(\Delta x)^2 + (\Delta y)^2}$, not just $ \big | f(t_i) – f(t_{i-1}) \big | $ which is the same as $\Delta y$. Again the latter doesn't make sense to me as it only accounts for the increment in the value of $y$ and does not account for the increment in the value of $x$.
Edit: I understand that $\bf{f}$ is a vector-valued function with $n$ components. But the curve (or the manifold) representing $\bf{f}$ exists in the $(n+1)$-dimensional space, and thus we need to find the distances (namely the line segment lengths) in the $(n+1)$-dimensional space, not in the $n$-dimensional space. Thus $ \big | \big | \textbf{f} (t_i) – \textbf{f} (t_{i-1}) \big | \big | $ should be repalced with $ \big | \big | (t_i, \textbf{f}(t_i)) – (t_{i-1}, \textbf{f}(t_{i-1})) \big | \big | $.
Best Answer
Suppose you have the function $f(t) = (\cos t, \sin t, t).$ This is the parametric equation of a helix. The length of the curve would be $\int_a^b \|f(t)'\| \ dt$
But you might have learned the arc-length formula for the graph of a function e.g. $y = x^2.$
$L = \int \sqrt {1 + (\frac {dy}{dx})^2} \ dx$
But this curve could just as easily be written $f(t) = (t, t^2)$ And we would be back to the formulation in the first paragraph.