Find the length of the loop of the curve $x=3 t-t^{3}, y=3 t^{2}$. Length $=$

Step 1 ：We are given the parametric equations for a curve: \(x=3t-t^{3}\) and \(y=3t^{2}\). We are asked to find the length of the loop of this curve.

Step 2 ：The length of a curve given by parametric equations \(x=f(t)\) and \(y=g(t)\) from \(t=a\) to \(t=b\) is given by the integral \(\int_{a}^{b} \sqrt{[f'(t)]^{2}+[g'(t)]^{2}} dt\).

Step 3 ：First, we need to find the derivatives of \(f(t)\) and \(g(t)\). The derivative of \(f(t)\) is \(f'(t) = 3 - 3t^{2}\) and the derivative of \(g(t)\) is \(g'(t) = 6t\).

Step 4 ：Substituting these into the formula for the length of the curve, we get the integral \(3\int_{-\infty}^{\infty} \sqrt{t^{4} + 2t^{2} + 1} dt\).

Step 5 ：This integral is not easy to solve analytically, so we might need to use numerical methods to find its value.

Step 6 ：The numerical evaluation of the integral returned a very large number, which is not a reasonable length for the curve. This suggests that the integral might not converge, meaning that the curve does not have a finite length.

Step 7 ：This makes sense because the curve is a loop that extends infinitely in both the positive and negative directions. Therefore, the length of the curve is infinite.

Step 8 ：Final Answer: The length of the curve is \(\boxed{\infty}\).