1. **State the problem:** Find the curvature $\kappa$ of the curve given by the vector function $$\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + t^2\mathbf{k}.$$\n\n2. **Recall the formula for curvature:** For a space curve $\mathbf{r}(t)$, curvature is given by $$\kappa = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}.$$\n\n3. **Find the first derivative:** $$\mathbf{r}'(t) = \frac{d}{dt}(t, t^2, t^2) = (1, 2t, 2t).$$\n\n4. **Find the second derivative:** $$\mathbf{r}''(t) = \frac{d}{dt}(1, 2t, 2t) = (0, 2, 2).$$\n\n5. **Compute the cross product:** $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2t & 2t \\ 0 & 2 & 2 \end{vmatrix} = \mathbf{i}(2t \cdot 2 - 2t \cdot 2) - \mathbf{j}(1 \cdot 2 - 0 \cdot 2t) + \mathbf{k}(1 \cdot 2 - 0 \cdot 2t) = (0, -2, 2).$$\n\n6. **Calculate the magnitude of the cross product:** $$\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = \sqrt{0^2 + (-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}.$$\n\n7. **Calculate the magnitude of the first derivative:** $$\|\mathbf{r}'(t)\| = \sqrt{1^2 + (2t)^2 + (2t)^2} = \sqrt{1 + 4t^2 + 4t^2} = \sqrt{1 + 8t^2}.$$\n\n8. **Compute the curvature:** $$\kappa = \frac{2\sqrt{2}}{(\sqrt{1 + 8t^2})^3} = \frac{2\sqrt{2}}{(1 + 8t^2)^{3/2}}.$$\n\n**Final answer:** $$\boxed{\kappa(t) = \frac{2\sqrt{2}}{(1 + 8t^2)^{3/2}}}.$$