1. We are asked to find the unit tangent vector to the curve $$\mathbf{r}(t) = k t \ln(t) \mathbf{i} + 4 t^3 \mathbf{j} - 2 t \mathbf{k}$$ at the point \(t=2\). 2. First, find the derivative \(\mathbf{r}'(t)\), which gives the tangent vector: $$\mathbf{r}'(t) = \frac{d}{dt}\big(k t \ln(t)\big) \mathbf{i} + \frac{d}{dt}(4 t^3) \mathbf{j} + \frac{d}{dt}(-2 t) \mathbf{k}$$ 3. Differentiate each component: - For \(k t \ln(t)\), use product rule: $$\frac{d}{dt} (k t \ln(t)) = k \big( \ln(t) + 1 \big)$$ - For \(4 t^3\): $$\frac{d}{dt} (4 t^3) = 12 t^2$$ - For \(-2 t\): $$\frac{d}{dt} (-2 t) = -2$$ 4. Thus, $$\mathbf{r}'(t) = k (\ln(t) + 1) \mathbf{i} + 12 t^2 \mathbf{j} - 2 \mathbf{k}$$ 5. Evaluate at \(t = 2\): - Calculate \(\ln(2)\) (natural logarithm), approximately 0.693. $$\mathbf{r}'(2) = k (0.693 + 1) \mathbf{i} + 12 \times 2^2 \mathbf{j} - 2 \mathbf{k} = k (1.693) \mathbf{i} + 48 \mathbf{j} - 2 \mathbf{k}$$ 6. The tangent vector at \(t=2\) is: $$\mathbf{v} = k (1.693) \mathbf{i} + 48 \mathbf{j} - 2 \mathbf{k}$$ 7. To find the unit tangent vector, divide \(\mathbf{v}\) by its magnitude: $$|\mathbf{v}| = \sqrt{\big(k \times 1.693\big)^2 + 48^2 + (-2)^2} = \sqrt{k^2 (1.693)^2 + 2304 + 4} = \sqrt{k^2 \times 2.866 + 2308}$$ 8. Therefore, the unit tangent vector is: $$\hat{\mathbf{T}} = \frac{1}{|\mathbf{v}|} \big( k (1.693) \mathbf{i} + 48 \mathbf{j} - 2 \mathbf{k} \big) = \frac{k (1.693)}{\sqrt{k^2 \times 2.866 + 2308}} \mathbf{i} + \frac{48}{\sqrt{k^2 \times 2.866 + 2308}} \mathbf{j} - \frac{2}{\sqrt{k^2 \times 2.866 + 2308}} \mathbf{k}$$ Final answer depends on the constant \(k\). For a specific \(k\), substitute its value to get numerical components.