Path Integral

The path integral is of the real-valued function $f (x, y, z)$ is the integral along the path

\begin{matrix} c : [a, b] \to R^{3} where c (t) = (x (t), y (t), z (t) \\ and c is C^{1} \end{matrix}

\begin{aligned} \int_{c} f d s & = \int_{a}^{b} f (x (t), y (t), z (t)) | | c^{'} (t) | | d t \\ = \int_{a}^{b} f (c (t)) | | c^{'} (t) | | d t \end{aligned}

Basic Idea
Partition $[a, b]$ in order to decompose $c$ into paths on $[t_{i}, t_{i + 1}]$

Therefore each segment on the path will have the length:

Δ s_{i} = \int_{t_{i}}^{t_{i + 1}} | | c^{'} (t) | | d t

(see: arc length)
since these segments are infinitesimal their length is basically constant
by MVT we know

\begin{array}{r} \exists t_{i}^{*} \in [t_{i}, t_{i + 1}] such that \int_{t_{i}}^{t_{i + 1}} | | c^{'} (t) | | d t = | | c^{'} (t) | | (t_{i + 1} - t_{i}) \\ so Δ s_{i} = | | c^{'} (t) | | d t \end{array}

Therefore

\int_{c} f d s = \int_{a}^{b} f (c (t)) | | c^{'} (t) | | d t

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Curvature

if $| | c^{'} (t) = 1 | |$

\begin{matrix} Curvature at t_{0} = κ (c (t_{0})) = | | c^{″} (t_{0}) | | \\ Total Curvature = \int_{c} | | c^{″} (t) | | d s \end{matrix}

If $c (a) = c (b) ⟹ C \geq 2 π$ and equals $2 π$ only if it is a circle
also in that case if $C \leq 4 π$ then it is "unknotted" otherwise "knotted"