Why does these curves are different at the origin.

Question

Let $\alpha:\mathbb R\to \mathbb R^2$ be given by $\alpha(t)= (t^3,t^2)$. The trace of $\alpha$ is drawn below:

Since $\alpha'(t)=(3t^2,2t)$, we have in $t=0$: $\alpha (0)=(0,0)$ and $\alpha'(0)=(0,0)$, then at the origin the tangent vector is zero.

Now let $\beta:\mathbb R\to \mathbb R^2 $ be given by $\beta(t)=(t,|t|)$. The trace of $\beta$ is drawn below:

Note that $\beta(0)=(0,0)$ and the curve is not differentiable at this point (we can see this by elementary calculus).

I would like to know intuitively why these curves are so different at the origin, for me they are pretty much the same (a corner).

Thanks in advance

Answer 1

To find the difference, don't look in $(0,0)$ (that's just one point and it cannot tell you anything about the local behavior), but look in a neighborhood, at points where $|t|$ is small.

If $t$ is small, the tangent to $\alpha$ is $(3t^2,2t)$ while the tangent to $\beta$ is $(1,\pm 1)$. As $t$ goes to $0$ we have that $3t^2$ is much smaller than $1$ so the horizontal part of the tangent vector to $\alpha$ goes to zero faster than the horizontal part of $\beta$. But the tangent vector shows the direction in which the curve "goes" at time $t$. Intuitively we see that for $t$ very small $\alpha$'s tangent vector is almost vertical. That's why the curve $\alpha$ gets into $(0,0)$ vertically and the curve $\beta$ gets into $(0,0)$ with an angle.