Show that $B(t): \mathbb{R} \rightarrow \mathbb{R}^{2} := (t^{3}, |t^{3}|)$ is not an example of a $1-$manifold in $\mathbb{R}^{2}$

Question

Munkres defines a $k$-manifold in $\mathbb{R}^{n}$ as follows:

As an example (according to Munkres):

$$B(t): \mathbb{R} \rightarrow \mathbb{R}^{2} := (t^{3}, |t^{3}|)$$

is not an instance of a $1-$manifold in $\mathbb{R}^{2}$, since although $B(t)$ has a continuous inverse and is $C^{2}$, the rank of $DB(t)$ at $t = 0$ is not $1$.

I understand everything except how $B(t)$ is $C^{2}$ however:

If $U'$ is an open set in $\mathbb{R}$ containing $0$, then $$\frac{\partial B_{2}}{\partial t} = \frac{\partial|t^{3}|}{\partial t} $$

does not exist at $t = 0$. If the first partial does not exist, how can the function even be considered of class $C^{2}$?

Answer 1

The first derivative of $\lvert t^3\rvert$ exists at $t=0$; it is $0$, since the left-hand and right-hand limits of the Newton quotient are both $0$, because

$$ \frac{\lvert t^3\rvert-0}{t-0}=\text{sgn}(t)t^2\to 0 $$

The second derivative also exists and is $0$ at $t=0$. It is only the third derivative that fails to exist, making this a function of class $C^2$.

Note for comparison that $\lvert t\rvert$ is a function that doesn't have a first derivative.

Answer 2

Because the first two derivatives of $t^3$ are zero at $t-0$, $|t^3|$ is in fact twice differentiable at $t=0$. (It is not thrice differentiable.)