I want to prove that
$\int_C \vec v\cdot d \vec r$ =$\int_C \vec v\cdot \vec t ds$
$\int_C \vec v\cdot \vec N(t) ds$ =$\int_C \vec v\cdot \vec n ds$
where $\vec n$ is the unit normal to C, $\vec t$ is the unit tangent to C, C is a regular curve and s is the arclength parameter of C
Any help is greatly appreciated
$$ \int_C \vec{v} \cdot d\vec{R} = \int_C \vec{v} \cdot \vec{T} ds $$ A position vector $\vec{R}(t)=x(t)i+y(t)j$ traces out a curve in 2D space. If you take $$\frac{d\vec{R}}{dt} = \frac{dx}{dt}i+\frac{dy}{dt}j$$ you get the tangent vector. The unit tangent vector is just the derivative of $R$ divided by its length (sometimes denoted $\frac{d\vec{R}}{ds}$).
The length of the position vector is equal to the step along the curve: $$ ds=\left|\frac{d\vec{R}}{dt}\right| = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} $$
Therefore: $$ \int_C \vec{v}\cdot \vec{T}ds = \int_C \vec{v} \cdot \left(\frac{\frac{dx}{dt}i+\frac{dy}{dt}j}{\sqrt{\left(\frac{dx}{dt^2}\right)^2+\left(\frac{dy}{dt}\right)^2}}\right)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt = \int_C \vec{v} \cdot \left(\frac{dx}{dt}i+\frac{dy}{dt}j\right)dt \\ = \int_C \vec{v}\cdot d\vec{R} $$
The second equation: $$ \int_C \vec{v}\cdot \vec{N}ds=\int_C v\cdot \vec{n} ds $$ is incorrect. The step on the right hand side should be $dt$. $$ \int_C \vec{v}\cdot \vec{N}ds = \int_C \vec{v}\cdot \vec{n} dt $$ This is equation true because the length of the normal vector is the same as the length of the unit tangent vector, and thus the same as $ds$. This is true because for a vector $\vec{v}=xi+yj$, a vector normal to this is $\vec{v}=-yi+xj$ (or $\vec{v}=yi-xj$).
Same as in the previous equation, $ds$ and the length of $\vec{N}$ in the denominator cancel out.