I have that $f(x,y) = (-y,x)$, and $\gamma(t)=(\cos(t),\sin(t)),$ where $t\in I=[0,2\pi]$. The function $f$ maps a vector to each point in the $(x,y)$ plane. The result is a vector field, which curls clockwise and is such that the magnitude of each vector increases as we move away from the origin.
The continuous mapping $\gamma$ is a parametrization for the unit circle ( a curve ), which lies in this same $(x,y)$ plane. I think this is correct. But what is the relationship/significance between $f$ and $f\circ \gamma$? ( They should form a vector field of sorts ) The relevant path integral is $$\int_\gamma f dt=\int_I\langle(-\sin(t),\cos(t)),(-\sin(t),\cos(t)\rangle dt.$$
A path integrals determines the area of the function under the curve, or something. What does this path integral tell us?
The path is a unit circle $\vec{\gamma}(t) = (\cos t, \sin t)$ or $(x^2 + y^2 = 1)$
Vector field $\vec{F} = (-y, x)$ or $(-\sin t, \cos t)$.
Let's take a point on the circle, $(1, 0)$. Our vector field at that point is $(0,1)$ pointing upward and tangent to the circle. You can check similarly at other points that the vector field is tangent to the circle. So it is a rotational vector field (say some velocity field for example).
Now $\vec{\gamma} \,'(t) = (-\sin t, \cos t)$ which is the gradient of the curve and slope of the tangent in $2D$. You can see that our vector field is working in the same direction.
So if you see the line integral, it becomes
$\displaystyle \int_\gamma \vec{F}(\gamma(t))\cdot \vec{\gamma} \,'(t) dt = \int_{0}^{2\pi} (-\sin t, \cos t) \cdot (-\sin t, \cos t) dt = \int_{0}^{2\pi} 1 \, dt$
and which is the magnitude of our rotational vector field (unit) working on the unit circle.