Answer
Smooth function
Work Step by Step
Step 1 : Given \[ \mathbf{r}(t) = \cos^2(t)\mathbf{i} + \sin^2(t)\mathbf{j} + e^{-t}\mathbf{k} \] Step 2 : Since \[ \mathbf{r}'(t) = -2t\sin^2(t)\mathbf{i} + 2t\cos^2(t)\mathbf{j} - e^{-t}\mathbf{k} \] The components are continuous functions on \(\mathbb{R}\), and there is no value of \(t\) for which all three of them are zero. It is clear that \(-2t\sin^2(t)\), \(2t\cos^2(t)\), and \(-e^{-t}\) are continuous functions, and there is no \(t\) that satisfies \(\mathbf{r}'(t) = \mathbf{0}\), hence \(\mathbf{r}'(t) \neq \mathbf{0}\) and \(\mathbf{r}(t)\) is a smooth function.
