Find the length of the loop of the following curve.
x = 6t − 2t3,    y = 6t2

To find the length of the loop of the given parametric curve \(x(t) = 6t - 2t^3\) and \(y(t) = 6t^2\), we need to follow the **arc length formula for parametric curves**. 

The arc length \(L\) is given by:

\[
L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
\]

### Step 1: Compute the derivatives
Given:
\[
x(t) = 6t - 2t^3, \quad y(t) = 6t^2
\]

- Derivative of \(x(t)\):
  \[
  \frac{dx}{dt} = 6 - 6t^2
  \]

- Derivative of \(y(t)\):
  \[
  \frac{dy}{dt} = 12t
  \]

### Step 2: Substitute into the arc length formula
\[
L = \int_{t_1}^{t_2} \sqrt{\left(6 - 6t^2\right)^2 + \left(12t\right)^2} \, dt
\]

Simplify the expression inside the square root:
\[
\left(6 - 6t^2\right)^2 + \left(12t\right)^2 = 36 - 72t^2 + 36t^4 + 144t^2 = 36t^4 + 72t^2 + 36
\]

\[
36(t^4 + 2t^2 + 1) = 36(t^2 + 1)^2
\]

Thus, the integrand becomes:
\[
\sqrt{36(t^2 + 1)^2} = 6(t^2 + 1)
\]

### Step 3: Find the limits of integration
The loop occurs when \(x(t) = 0\), i.e.,
\[
6t - 2t^3 = 0 \quad \Rightarrow \quad t(3 - t^2) = 0
\]
This gives \(t = 0\) and \(t = \pm\sqrt{3}\). The loop corresponds to the segment between \(t = -\sqrt{3}\) and \(t = \sqrt{3}\).

### Step 4: Evaluate the integral
Now we integrate:
\[
L = \int_{-\sqrt{3}}^{\sqrt{3}} 6(t^2 + 1) \, dt
\]

This simplifies to:
\[
L = 6 \int_{-\sqrt{3}}^{\sqrt{3}} (t^2 + 1) \, dt
\]

Since the integrand is even, we can write:
\[
L = 12 \int_0^{\sqrt{3}} (t^2 + 1) \, dt
\]

Evaluate the integral:
\[
\int_0^{\sqrt{3}} (t^2 + 1) \, dt = \left[\frac{t^3}{3} + t\right]_0^{\sqrt{3}} = \left(\frac{(\sqrt{3})^3}{3} + \sqrt{3}\right) - (0) = \left(\frac{3\sqrt{3}}{3} + \sqrt{3}\right) = 2\sqrt{3}
\]

Thus:
\[
L = 12 \cdot 2\sqrt{3} = 24\sqrt{3}
\]

### Final Answer
The length of the loop is:

\[
L = 24\sqrt{3}
\]

---

Do you want further details, or have any questions?  

Here are some related questions for further practice:
1. How do you compute the length of a curve defined implicitly?
2. What is the general arc length formula for polar coordinates?
3. How does symmetry simplify curve length calculations?
4. What happens if you swap the parametric equations for \(x(t)\) and \(y(t)\)?
5. How do you determine the area enclosed by a parametric curve?

**Tip:** When integrating symmetric curves, it is often easier to integrate over half the range and multiply by two!

Find the length of the loop of the following curve. x = 6t − 2t^3, y = 6t^2

Learn how to compute the length of the loop for a parametric curve defined by x = 6t − 2t^3 and y = 6t^2 using the arc length formula. This problem includes detailed steps, from finding derivatives to evaluating the integral. Suitable for undergraduate calculus students.

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