I am trying to represent the circle of radius 2 in polar coordinates.
$$\gamma(t)=(2\cos(t),2\sin(t))$$
where $t\in [0,2\pi]$. The derivative would be
$$\gamma'(t)=-2\sin(t)\partial_x +2\cos(t)\partial_y.$$
Now if we convert the tangent basis vectors to polar coordinates,
$$\partial_x = \cos(t)\partial_r - \frac{1}{r}\sin(t) \partial_\theta.$$
$$\partial_y = \sin(t)\partial_r + \frac{1}{r}\cos(t) \partial_\theta.$$
Then under the substitution, noting that $r=2$, we have
$$\gamma'(t) = -2\sin(t)\partial_x + 2\cos(t) \partial_y\\= -2\sin(t)\left[\cos(t)\partial_r - \frac{1}{2}\sin(t) \partial_\theta\right] + 2\cos(t)\left[\sin(t)\partial_r + \frac{1}{2}\cos(t) \partial_\theta\right]\\=2\partial_\theta.$$
Thus, in polar coordinates, $\gamma'(t) = 2\partial_\theta$.
Now, what I thought is that in polar coordiantes,
$$\gamma_{\text{polar}}(t) = (2, t)$$ where $t\in [0,2\pi]$.
Then we would have $$\gamma'_{\text{polar}}(t)=\partial_\theta.$$
But this does not align with what I found when I started with cartesian coordiantes. What am I doing wrong?