Find the length of the curve.
r(t) = 6 i + 4t^2 j + 3t^3 k,   0 ≤ t ≤ 1

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.

Find the length of the curve. r(t) = 6 i + 4t^2 j + 3t^3 k, 0 ≤ t ≤ 1

Learn how to find the length of the curve defined by r(t) = 6i + 4t^2j + 3t^3k for t in the range 0 to 1. This problem involves vector calculus, including finding the magnitude of the derivative and solving integrals. Ideal for Calculus III students.

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