How to prove that there is no regular parametrized curve whose image is a semicubical parabola? It is known that a level curve that admits regular parametrizations also admits irregular ones, for example ($t$, $t^2$) and ($t^2$, $t^4$) have both the same image, a parabola, but the first one is regular while the second isn't. A parametrized semicubical parabola ($t^3$, $t^2$) is irregular, because the tangent vector vanishes at the origin. But how could I see that the image of this parametrized curve admits no regular parametrizations (i.e. that there is no regular curve whose image is a semicubical parabola)?
 A: The image lies purely in the right half-plane $\{(x,y) \colon y \ge 0\}$ and touches the origin. If $\gamma$ is a regular curve with $\gamma(t_0) = (0,0)$, then the only way for $\gamma$ to stay in the right half-plane is if the tangent at zero is vertical - otherwise the implicit function theorem would tell us we could write it as a graph over some $(-\epsilon,\epsilon)$. 
But the tangent being vertical means the curve must stay in the double cone $\{ |y| \ge |x| \}$ for some short time, which is clearly untrue - the only solution of $x = y^{2/3}$ in this cone near the origin is $(0,0)$.
A: It suffices to show that the semicubical parabola admits a cusp at some point on the curve.
Using the given parametrization $\gamma(t) = (t^3, t^2)$, we see that the tangent vector at $t$ is given by
\begin{align}
\gamma'(t) = (3t^2, 2t) \ \ \Rightarrow \ \ \mathbf{N}(t) = \left(\frac{3t}{\sqrt{9t^2+4}}, \frac{2}{\sqrt{9t^2+4}}  \right)
\end{align}
where $\mathbf{N}$ is the unit tangent vector at $t$. Observe as $t \rightarrow 0^+$, we see that
\begin{align}
\left(\frac{3t \operatorname{sgn} (t)}{\sqrt{9t^2+4}}, \frac{2 \operatorname{sgn} (t)}{\sqrt{9t^2+4}}  \right) \longrightarrow \left(0, 1  \right)
\end{align}
whereas as $t\rightarrow 0^-$ we see that
\begin{align}
\left(\frac{3t \operatorname{sgn} (t)}{\sqrt{9t^2+4}}, \frac{2 \operatorname{sgn} (t)}{\sqrt{9t^2+4}}  \right) \longrightarrow \left(0, -1  \right).
\end{align}
Hence it follows the semicubical parabola has a cusp at $(0, 0)$ on the $xy$-plane. 
