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To compute the length of the curve \( \mathbf{r}(t) = \langle 4 \cos(5t), 4 \sin(5t), t^{3/2} \rangle \) over the interval \( 0 \leq t \leq 2\pi \), we can use the formula for the arc length of a parametric curve:

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

where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \|\mathbf{r}'(t)\| \) is the magnitude of the derivative vector. The limits of integration are \( a = 0 \) and \( b = 2\pi \).

### Step 1: Find \( \mathbf{r}'(t) \)
First, take the derivative of each component of \( \mathbf{r}(t) \):
\[
\mathbf{r}(t) = \langle 4 \cos(5t), 4 \sin(5t), t^{3/2} \rangle
\]

1. \( \frac{d}{dt} [4 \cos(5t)] = -20 \sin(5t) \)
2. \( \frac{d}{dt} [4 \sin(5t)] = 20 \cos(5t) \)
3. \( \frac{d}{dt} [t^{3/2}] = \frac{3}{2} t^{1/2} \)

Thus, \( \mathbf{r}'(t) = \langle -20 \sin(5t), 20 \cos(5t), \frac{3}{2} t^{1/2} \rangle \).

### Step 2: Compute \( \|\mathbf{r}'(t)\| \)
The magnitude of the derivative vector \( \mathbf{r}'(t) \) is:

\[
\|\mathbf{r}'(t)\| = \sqrt{(-20 \sin(5t))^2 + (20 \cos(5t))^2 + \left( \frac{3}{2} t^{1/2} \right)^2}
\]

Simplifying each term:
\[
(-20 \sin(5t))^2 = 400 \sin^2(5t), \quad (20 \cos(5t))^2 = 400 \cos^2(5t), \quad \left( \frac{3}{2} t^{1/2} \right)^2 = \frac{9}{4} t
\]

So,
\[
\|\mathbf{r}'(t)\| = \sqrt{400 (\sin^2(5t) + \cos^2(5t)) + \frac{9}{4} t}
\]

Using \( \sin^2(5t) + \cos^2(5t) = 1 \), we get:
\[
\|\mathbf{r}'(t)\| = \sqrt{400 + \frac{9}{4} t}
\]

### Step 3: Set up the integral
Now, the arc length is given by:
\[
s = \int_0^{2\pi} \sqrt{400 + \frac{9}{4} t} \, dt
\]

### Step 4: Solve the integral
This integral is straightforward but requires a substitution method. Let:
\[
u = 400 + \frac{9}{4} t \quad \Rightarrow \quad du = \frac{9}{4} dt \quad \Rightarrow \quad dt = \frac{4}{9} du
\]

The limits of integration change as follows:
- When \( t = 0 \), \( u = 400 \),
- When \( t = 2\pi \), \( u = 400 + \frac{9}{4} \times 2\pi = 400 + \frac{9\pi}{2} \).

Thus, the integral becomes:
\[
s = \frac{4}{9} \int_{400}^{400 + \frac{9\pi}{2}} \sqrt{u} \, du
\]

The integral of \( \sqrt{u} \) is:
\[
\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}
\]

Thus,
\[
s = \frac{4}{9} \times \frac{2}{3} \left[ u^{3/2} \right]_{400}^{400 + \frac{9\pi}{2}}
\]

Now, compute the bounds:
\[
s = \frac{8}{27} \left[ \left( 400 + \frac{9\pi}{2} \right)^{3/2} - 400^{3/2} \right]
\]

This expression gives the length of the curve. If you would like the numerical evaluation, I can continue with the computation.

Would you like me to continue evaluating it or explain any particular steps in more detail?

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Here are five questions to consider:
1. How does the arc length formula change for different parameterizations?
2. How would the calculation change if the curve was in four dimensions?
3. What is the significance of the trigonometric terms in the curve's components?
4. Can you think of other examples where calculating the arc length is useful?
5. How would you approach this problem if the interval for \( t \) was infinite?

**Tip:** Always simplify trigonometric identities early when computing arc lengths to reduce complexity in integration.

Compute the length of the curve r(t) = <4 cos(5t), 4 sin(5t), t^(3/2)> over the interval 0 <= t <= 2π. (Use symbolic notation and fractions where needed.)

This guide provides a step-by-step solution to compute the length of the parametric curve r(t) = <4 cos(5t), 4 sin(5t), t^(3/2)> over the interval 0 <= t <= 2π. Learn key calculus concepts like vector differentiation and arc length, suitable for undergraduate calculus students.

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