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:

$L = 24\sqrt{3}$

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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!