Is the vector function $\mathbf r(t) = \langle t^3, t^3 \rangle$ smooth at $t = 0$?

Question

This was confusing me when learning about curvature and smoothness. The condition for smoothness on interval $I$ is given as:

1. $\mathbf r'$ is continuous;
2. $\mathbf r'(t) \neq \mathbf 0$.

In this particular example, the space curve of this parametrization is just a straight line passing through the origin, but it follows that $$\mathbf r'(t) = \langle 3t^2, 3t^2 \rangle$$ thus $\mathbf r'(0) = \mathbf 0$, meaning that $\mathbf r$ is not smooth at the origin. However, the space curve seems to be perfectly "smooth" and differentiable. What am I missing here?

(Also, by this definition the parametrization $\mathbf s(t) = \langle t, t \rangle$ with the same space curve is smooth at the origin. Why could they be different?)

Answer 1

tl; dr: The term smooth gets used inconsistently. As elsewhere in mathematics, the words-to-concepts relation is not a mapping; the same word can mean multiple things.

To be careful with terminology here, it might be best to say the mapping $r(t) = (t^{3}, t^{3})$ is infinitely differentiable, but not regular at $t = 0$.

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1. In differential topology, the term smooth mapping generally connotes some degree of differentiability, often infinite differentiability.
2. In differential topology and differential geometry, the term smooth manifold connotes a manifold with a smooth atlas (whose overlap maps are smooth). A smooth submanifold is a smooth manifold together with a smooth embedding into, e.g., a Cartesian space.
3. In differential geometry, one sometimes uses the term regular mapping to connote a smooth immersion (a smooth mapping $f:M \to N$ from a smooth manifold into a manifold of equal or larger dimension whose differential $Df$ has rank $\dim M$ at each point). At other times (::cough:: ::cough:: calculus books, and sometimes elementary textbooks on differential geometry), one encounters the term smooth mapping conveying that the mapping is regular, or that the image is a smooth submanifold.