I'm trying to figure out if the following equality is true:
$$ \int_C \vec{v} \cdot d\vec{r} = \int_C \left\lVert \vec{v} \right\rVert ds$$
I figured that $d\vec{r} = \vec{T}ds$, where $\vec{T}$ is the tangent of the curve at a given point and $ds = \left\lVert d\vec{r} \right\rVert$.
So I get:
$$\int_C \vec{v} \cdot d\vec{r} = \int_C \vec{v} \cdot \vec{T} ds$$
I have been told that the initial equality is true. However, how can it be true if $\vec{v} \cdot \vec{T}$ is only equal to the magnitude of $\vec{v}$ if the vector is completely along the tangent? Can't the vector v point in a different direction.
In general, no.
It may be clarifying to look at the definitions.
Suppose $\vec r:[a,b]\to C$ is a parametrization of the smooth path. Then
$$ \int_C \vec v\cdot d\vec r=\int_a^b \vec v\cdot \vec r'(t)\,dt $$
$$ \int_C \|\vec v\|\,ds=\int_a^b \|\vec v(t)\|\cdot \|\vec r'(t)\|\,dt $$
But the second integral is always nonnegative.
Consider $\vec r(t)=(t,0)$ and $\vec v\equiv(-1,0)$ for $t\in[0,1]$. Then $$ \int_C \vec v\cdot d\vec r=\int_0^1 -1\,dt,\quad \int_C \|\vec v\|\,ds=\int_0^11\,dt $$