To find the length of the curve described by the vector function $\mathbf{r}(t) = 6\,\mathbf{i} + 4t^2\,\mathbf{j} + 3t^3\,\mathbf{k}$, we can use the following formula for the length of a space curve:

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

where $\mathbf{r}'(t)$ is the derivative of the vector function $\mathbf{r}(t)$, and $|\mathbf{r}'(t)|$ is the magnitude of the derivative. The limits of integration are given by $a = 0$ and $b = 1$ in this case.

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

The derivative of the vector function $\mathbf{r}(t)$ is:

$\mathbf{r}'(t) = \frac{d}{dt}(6\,\mathbf{i} + 4t^2\,\mathbf{j} + 3t^3\,\mathbf{k})$ $\mathbf{r}'(t) = 0\,\mathbf{i} + 8t\,\mathbf{j} + 9t^2\,\mathbf{k}$ $\mathbf{r}'(t) = 8t\,\mathbf{j} + 9t^2\,\mathbf{k}$

Step 2: Find $|\mathbf{r}'(t)|$

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

$|\mathbf{r}'(t)| = \sqrt{(0)^2 + (8t)^2 + (9t^2)^2}$ $|\mathbf{r}'(t)| = \sqrt{64t^2 + 81t^4}$ $|\mathbf{r}'(t)| = \sqrt{t^2(64 + 81t^2)}$ $|\mathbf{r}'(t)| = t\sqrt{64 + 81t^2}$

Step 3: Set up the integral

Now, integrate from $t = 0$ to $t = 1$:

$L = \int_{0}^{1} t \sqrt{64 + 81t^2} \, dt$

This integral requires substitution. Let's use the substitution $u = 64 + 81t^2$, hence $du = 162t\, dt$.

Step 4: Solve the integral

Now we can calculate the integral to find the length. Let me do the calculation.The length of the curve is:

$L = \frac{-512}{243} + \frac{145\sqrt{145}}{243}$

This is the exact expression for the curve length. If you'd like, I can also provide the numerical approximation of this value.

Would you like further details or an approximation?

Here are 5 related questions to deepen your understanding:

1. What is the physical significance of the length of a curve in vector calculus?
2. How does the derivative of a vector function relate to the velocity vector in physics?
3. Can you find the curvature of a curve given a vector function?
4. What are some methods for solving more complex integrals involving square roots?
5. How would the solution change if the curve was parametrized differently?

Tip: When dealing with integrals involving square roots, substitution is often the key to simplifying the expression.