the Semicubical parabola which has the function
$y^2=x^3$
can be parameterized as $(t^2, t^3)$ but also as $(t, t^{\frac{3}{2}})$. Now if I wanted to calculate the tangent vector
I would get $(2t, 3t^2)$ for the first and $(1, \frac{3}{2}t^{\frac{1}{2}})$ for the second.
I want to calculate the tangent vector at the origin (0,0), I would get
$(2\cdot0, 3\cdot0)=(0,0)$ for the first
and $(1,0)$ in case of the second parameterization.
How can they differ? what mistake did I make?
Because even though the trace of the two parametrizations (the curve they draw) is the same, their velocity is different.
We define the tangent (unit) vector to a curve as $$ \mathbf{\tau}\ (t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} $$ when the velocity magnitude $|\mathbf{r}'(t)|$ is nonzero. So it is clear that you just need to pick the appropriate parametrization.