Find the length of the loop of the curve $x=3t-t^3,y=3t^2$. 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 ${\int }_a^b\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2}dt$.

Step 3 ：First, we need to find the derivatives of $f(t)$ and $g(t)$. The derivative of $f(t)$ is $f^{\prime }(t)=3-3t^2$ and the derivative of $g(t)$ is $g^{\prime }(t)=6t$.

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

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 $\boxed{{\displaystyle \mathrm{\infty }}}$.