1. **State the problem:** Find the unit tangent vector to the curve $$\mathbf{r}(t) = k t \ln(t) \mathbf{\hat{k}} + 4 t^3 \mathbf{\hat{i}} - 2 t \mathbf{\hat{j}}$$ at $$t=2$$. 2. **Find the derivative \(\mathbf{r}'(t)\):** - Derivative of \(k t \ln(t)\) with respect to \(t\): $$\frac{d}{dt}[k t \ln(t)] = k \left(\ln(t) + 1\right)$$ by the product rule. - Derivative of \(4 t^3\): $$\frac{d}{dt}[4 t^3] = 12 t^2$$. - Derivative of \(-2 t\): $$\frac{d}{dt}[-2 t] = -2$$. Thus, $$\mathbf{r}'(t) = 12 t^2 \mathbf{\hat{i}} - 2 \mathbf{\hat{j}} + k (\ln(t)+1) \mathbf{\hat{k}}$$. 3. **Evaluate \(\mathbf{r}'(2)\):** - \(12 \times 2^2 = 12 \times 4 = 48\) - \(-2\) remains the same - \(k (\ln(2)+1)\) stays symbolic, write as \(k (\ln(2)+1)\) So, $$\mathbf{r}'(2) = 48 \mathbf{\hat{i}} - 2 \mathbf{\hat{j}} + k (\ln(2)+1) \mathbf{\hat{k}}$$. 4. **Find the magnitude of \(\mathbf{r}'(2)\):** $$\|\mathbf{r}'(2)\| = \sqrt{48^2 + (-2)^2 + \left(k (\ln(2)+1)\right)^2} = \sqrt{2304 + 4 + k^2 (\ln(2)+1)^2} = \sqrt{2308 + k^2 (\ln(2)+1)^2}$$. 5. **Find the unit tangent vector \(\mathbf{T}(2)\):** $$\mathbf{T}(2) = \frac{\mathbf{r}'(2)}{\|\mathbf{r}'(2)\|} = \frac{48 \mathbf{\hat{i}} - 2 \mathbf{\hat{j}} + k (\ln(2)+1) \mathbf{\hat{k}}}{\sqrt{2308 + k^2 (\ln(2)+1)^2}}$$. **Final answer:** The unit tangent vector at \( t=2 \) is $$\boxed{\mathbf{T}(2) = \frac{48 \mathbf{\hat{i}} - 2 \mathbf{\hat{j}} + k (\ln(2)+1) \mathbf{\hat{k}}}{\sqrt{2308 + k^2 (\ln(2)+1)^2}}}$$.