Answer
Not a smooth function
Work Step by Step
Step 1 : Given \[ \mathbf{r}(t) = t e^{-t} \mathbf{i} + (t^2 - 2t) \mathbf{j} + \cos(\pi t) \mathbf{k} \] Step 2 : Since \[ \mathbf{r}'(t) = (1 - t) e^{-t} \mathbf{i} + (2t - 2) \mathbf{j} - \pi \sin(\pi t) \mathbf{k} \] 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 \((1 - t)e^{-t}\), \(2t - 2\), and \(-\pi \sin(\pi t)\) are continuous functions, and \(\mathbf{r}'(t) = 0\) at \(t = 1\). Hence, \(\mathbf{r}'(t) = 0\) and \(\mathbf{r}(t)\) is not a smooth function.
