Why are Line Integrals of Vector Fields independent of Magnitude of $\vec{r}'$?

Question

If we have a line integral of a vector field $\vec{F}$ along a space curve $\vec{r}(t)$ $$\int\limits_{t=a}^{t=b}\vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt$$ Why doesn't it matter that $\vec{r}'$ doesn't always have unit length? For example, in this visualization of a line integral (https://en.wikipedia.org/wiki/File:Line_integral_of_vector_field.gif), it seems that $r'$ always has unit length, while it usually doesn't. Why does the length/magnitude of $\vec{r}'$ not matter?

Answer 1

The line integral is expressed more univerally as $$\int_C \vec F\cdot d\vec r$$ where $C$ is the curve of integration. This can be considered at the limit of a Riemann sum, where the $P = \{\vec r_i\}_{i=0}^N$ are partition points of on the curve: $$\int_C \vec F\cdot d\vec r = \lim_{|P|\to 0} \sum_{i=1} \vec F(\vec r_i)\cdot (\vec r_i - \vec r_{i-1})$$ When we move to a paramtetrization, $\vec r = \vec r(t)$, then by the vector version of the chain rule, $d\vec r = \frac {d\vec r}{dt}dt$.

The fact that $\frac {d\vec r}{dt}$ is not restricted to be $1$ is exactly what is needed. If we change to a parameter that zooms through the curve twice as fast, $u = 2t$, then $du = 2dt$, so we would have $d\vec r = \frac {d\vec r}{dt}\frac {du}2$. But this works because by the chain rule $$\frac {d\vec r}{dt} = \frac {d\vec r}{du}\frac {du}{dt} = 2\frac {d\vec r}{du}$$ so the differential becomes $$d\vec r = \left(2\frac {d\vec r}{du}\right)\left(\frac {du}2\right) = \frac {d\vec r}{du}du$$ which is the exactly the same form as $t$. If they did not have the same form, then this expression for $d\vec r$ would not be true for any arbitrary smooth parametrization.

Basically, when the parameter runs through the curve faster, we need $\vec r'(t)$ to be larger to compensate for $dt$ being smaller to cover the same ground.