To find the length of the curve $\mathbf{r}(t) = \langle 2t, t^2, 31t^3 \rangle$ from $t = 1$ to $t = 4$, we use the formula for the arc length of a vector-valued function:

$L = \int_{a}^{b} \left| \mathbf{r}'(t) \right| \, dt$

Step 1: Find the derivative $\mathbf{r}'(t)$

The derivative of the vector function $\mathbf{r}(t) = \langle 2t, t^2, 31t^3 \rangle$ is:

$\mathbf{r}'(t) = \langle 2, 2t, 93t^2 \rangle$

Step 2: Compute the magnitude of $\mathbf{r}'(t)$

Now, compute the magnitude of the derivative $\mathbf{r}'(t)$:

$\left| \mathbf{r}'(t) \right| = \sqrt{(2)^2 + (2t)^2 + (93t^2)^2}$ $\left| \mathbf{r}'(t) \right| = \sqrt{4 + 4t^2 + 8649t^4}$

Step 3: Set up the integral

The arc length $L$ is given by:

$L = \int_{1}^{4} \sqrt{4 + 4t^2 + 8649t^4} \, dt$

This integral needs to be computed. Let's proceed with that calculation.

The length of the curve $\mathbf{r}(t) = \langle 2t, t^2, 31t^3 \rangle$ from $t = 1$ to $t = 4$ is approximately $1953.08$ units.

Do you want further details on this process, or do you have any other questions?

Here are some related questions:

1. How does the formula for arc length change in different coordinate systems?
2. How can you compute arc length if given a parametric curve in polar coordinates?
3. How does curvature relate to the arc length of a curve?
4. What are some real-world applications of arc length computations?
5. How does the arc length change if the function is given in higher dimensions?

Tip: When computing arc lengths, simplifying the integrand before solving can sometimes make the problem easier to handle manually.