Answer
False
Work Step by Step
Because \(\mathbf{r}'(t)\) is undefined for the value of \(t\) such that \(\mathbf{r}'(t)\), let \[ \mathbf{r}(t) = t\mathbf{i} + |t|\mathbf{j} \] Since \(|t|\) is not differentiable at \(t = 0\), then \(\mathbf{r}'(t)\) is not defined at \(t=0\), hence \(\mathbf{r}(t)\) is not smooth. Result: False