Find the arclength of the curve r(t)=⟨(6sqrt2)t,e^(6t),e^(-6t)⟩
, 0≤t≤1

The arclength \( L \) of a curve given by the vector function \( \mathbf{r}(t) \) over the interval \( [a, b] \) is computed using the following formula:

\[
L = \int_a^b \left| \mathbf{r}'(t) \right| \, dt
\]

### Step 1: Compute the derivative \( \mathbf{r}'(t) \)
Given the vector function \( \mathbf{r}(t) = \left\langle 6\sqrt{2}t, e^{6t}, e^{-6t} \right\rangle \), we first find its derivative:

\[
\mathbf{r}'(t) = \left\langle \frac{d}{dt}(6\sqrt{2}t), \frac{d}{dt}(e^{6t}), \frac{d}{dt}(e^{-6t}) \right\rangle
\]

Computing each component:

1. \( \frac{d}{dt}(6\sqrt{2}t) = 6\sqrt{2} \)
2. \( \frac{d}{dt}(e^{6t}) = 6e^{6t} \)
3. \( \frac{d}{dt}(e^{-6t}) = -6e^{-6t} \)

Thus, the derivative is:

\[
\mathbf{r}'(t) = \left\langle 6\sqrt{2}, 6e^{6t}, -6e^{-6t} \right\rangle
\]

### Step 2: Compute the magnitude \( \left| \mathbf{r}'(t) \right| \)
The magnitude of \( \mathbf{r}'(t) \) is given by:

\[
\left| \mathbf{r}'(t) \right| = \sqrt{\left( 6\sqrt{2} \right)^2 + \left( 6e^{6t} \right)^2 + \left( -6e^{-6t} \right)^2}
\]

Simplifying each term:

1. \( \left( 6\sqrt{2} \right)^2 = 72 \)
2. \( \left( 6e^{6t} \right)^2 = 36e^{12t} \)
3. \( \left( -6e^{-6t} \right)^2 = 36e^{-12t} \)

Thus:

\[
\left| \mathbf{r}'(t) \right| = \sqrt{72 + 36e^{12t} + 36e^{-12t}}
\]

Using the identity \( e^{12t} + e^{-12t} = 2\cosh(12t) \), we get:

\[
\left| \mathbf{r}'(t) \right| = \sqrt{72 + 72\cosh(12t)} = 6\sqrt{2(1 + \cosh(12t))}
\]

### Step 3: Set up the integral
Now, we integrate the magnitude over the interval \( [0, 1] \):

\[
L = \int_0^1 6\sqrt{2(1 + \cosh(12t))} \, dt
\]

This integral does not have an elementary antiderivative, so it can be evaluated numerically.

### Step 4: Numerical Evaluation
To compute the arclength, I'll now calculate the value of the integral.

Let's do that:The arclength of the curve \( \mathbf{r}(t) = \left\langle 6\sqrt{2}t, e^{6t}, e^{-6t} \right\rangle \) over the interval \( 0 \leq t \leq 1 \) is approximately \( 403.43 \).

Would you like more details on any part of the process or have any questions?

Here are 5 related questions you might find interesting:
1. How is the arclength formula derived for vector functions?
2. Can you compute the arclength of a circular arc given in parametric form?
3. How does the hyperbolic cosine function \( \cosh(t) \) arise in this problem?
4. What are some alternative methods for numerically solving integrals?
5. How does the arclength differ for curves in 2D versus 3D?

**Tip:** Always verify whether the integral you are dealing with can be solved analytically or if it requires numerical methods.

Find the arclength of the curve r(t)=⟨(6sqrt2)t,e^(6t),e^(-6t)⟩, 0≤t≤1

Find the arc length of the curve r(t) = ⟨(6√2)t, e^(6t), e^(-6t)⟩ over the interval 0≤t≤1. This problem involves vector calculus, derivatives, and the use of hyperbolic functions. Suitable for students in Grades 11-12 or college level.

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