1. **State the problem:** Find the length of the curve given by the vector function $$\mathbf{r}(t) = \langle t^2, 2t, \ln t \rangle$$ for $$1 \leq t \leq e$$. 2. **Formula for curve length:** The length $$L$$ of a curve defined by $$\mathbf{r}(t)$$ from $$t=a$$ to $$t=b$$ is given by: $$ L = \int_a^b \| \mathbf{r}'(t) \| \, dt $$ where $$\mathbf{r}'(t)$$ is the derivative of $$\mathbf{r}(t)$$ and $$\| \mathbf{r}'(t) \|$$ is its magnitude. 3. **Find the derivative:** $$ \mathbf{r}'(t) = \left\langle \frac{d}{dt} t^2, \frac{d}{dt} 2t, \frac{d}{dt} \ln t \right\rangle = \langle 2t, 2, \frac{1}{t} \rangle $$ 4. **Find the magnitude:** $$ \| \mathbf{r}'(t) \| = \sqrt{(2t)^2 + 2^2 + \left(\frac{1}{t}\right)^2} = \sqrt{4t^2 + 4 + \frac{1}{t^2}} $$ 5. **Simplify the expression under the square root:** $$ \sqrt{4t^2 + 4 + \frac{1}{t^2}} = \sqrt{\frac{4t^4 + 4t^2 + 1}{t^2}} = \frac{\sqrt{4t^4 + 4t^2 + 1}}{t} $$ 6. **Recognize the perfect square:** $$ 4t^4 + 4t^2 + 1 = (2t^2 + 1)^2 $$ 7. **Substitute back:** $$ \| \mathbf{r}'(t) \| = \frac{2t^2 + 1}{t} $$ 8. **Set up the integral for length:** $$ L = \int_1^e \frac{2t^2 + 1}{t} \, dt = \int_1^e \left(2t + \frac{1}{t}\right) dt $$ 9. **Integrate term-by-term:** $$ \int_1^e 2t \, dt = \left[t^2\right]_1^e = e^2 - 1 $$ $$ \int_1^e \frac{1}{t} \, dt = \left[\ln t\right]_1^e = \ln e - \ln 1 = 1 - 0 = 1 $$ 10. **Add the results:** $$ L = (e^2 - 1) + 1 = e^2 $$ **Final answer:** The length of the curve is $$\boxed{e^2}$$.