Answer
Not a smooth function
Work Step by Step
Step 1 : Given \[ \mathbf{r}(t) = \sin(\pi t)\mathbf{i} + (2t - \ln(t))\mathbf{j} + (t^2 - t)\mathbf{k} \] Step 2: Since \[ \mathbf{r}'(t) = \pi \cos(\pi t)\mathbf{i} + (2 - \frac{1}{t})\mathbf{j} + (2t - 1)\mathbf{k} \] It is clear that \(\left(2 - \frac{1}{t}\right)\) is not continuous at \(t = 0\) and \(\mathbf{r}(1/2) = 0\). Hence, \(\mathbf{r}(t)\) is not a smooth function.
