We are tasked with finding the length $L$ of the curve described by the vector function:

$\mathbf{r}(t) = \mathbf{i} + 6t^2 \mathbf{j} + t^3 \mathbf{k}, \quad 0 \leq t \leq \sqrt{33}$

The formula for the length of a curve given by a vector function $\mathbf{r}(t)$ is:

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

Here, $\mathbf{r}'(t)$ represents the derivative of $\mathbf{r}(t)$ with respect to $t$, and $\|\mathbf{r}'(t)\|$ is the magnitude of this derivative.

Step 1: Compute $\mathbf{r}'(t)$

We differentiate each component of $\mathbf{r}(t) = \mathbf{i} + 6t^2 \mathbf{j} + t^3 \mathbf{k}$:

$\mathbf{r}'(t) = \frac{d}{dt}(\mathbf{i} + 6t^2 \mathbf{j} + t^3 \mathbf{k}) = 0 \mathbf{i} + 12t \mathbf{j} + 3t^2 \mathbf{k}$

Thus, the derivative is:

$\mathbf{r}'(t) = 12t \mathbf{j} + 3t^2 \mathbf{k}$

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

The magnitude of $\mathbf{r}'(t)$ is:

$\|\mathbf{r}'(t)\| = \sqrt{(12t)^2 + (3t^2)^2}$

Simplifying:

$\|\mathbf{r}'(t)\| = \sqrt{144t^2 + 9t^4} = \sqrt{9t^2(16 + t^2)} = 3t\sqrt{16 + t^2}$

Step 3: Set up the integral

Now, we can set up the integral for the curve length:

$L = \int_0^{\sqrt{33}} 3t\sqrt{16 + t^2} \, dt$

Step 4: Solve the integral

We'll now proceed with solving this integral. Let me calculate that.The length of the curve is $L = 279$.

Would you like more details or have any questions about the solution?

Here are 5 related questions to consider:

1. How is the formula for the arc length of a curve derived from the vector function?
2. What are the steps for finding the magnitude of a derivative in parametric curves?
3. How would the arc length change if the limits of integration were different?
4. What if one of the components of the curve $\mathbf{r}(t)$ had a trigonometric function?
5. How do we interpret the physical meaning of curve length in different contexts?

Tip: When finding arc length, always check if simplifying the integrand helps before applying more complex integration techniques.