Q:

Find the length of the loop of the given curve: $x=3t-t^3$ $y=3t^2$

I used the arc length formula (where you take the integral of square root of x' squared + y' squared $\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$) to get $t^3 + 3t + C$ which seems to be the wrong answer. Not sure what I did wrong. Please tell me the correct answer and why my solution was incorrect. Thanks

A:

If the path described makes a loop, then there will be 2 values $t_1$ and $t_2$ such that $
\begin{bmatrix} x(t_1) \\ y(t_1) \end{bmatrix} = \begin{bmatrix} x(t_2) \\ y(t_2) \end{bmatrix}
$ , so:
$$\begin{bmatrix} 3t_1 - t_1{}^3 \\ 3t_1{}^2 \end{bmatrix} = \begin{bmatrix} 3t_2 - t_2{}^3 \\ 3t_2{}^2 \end{bmatrix}$$
From $3t_1{}^2 = 3t_2{}^2$, we get $t_1 = -t_2$, and from $3t_1 - t_1{}^3 = -3t_1 + t_1{}^3$, we get that 
$$t_1 = -\sqrt{3} \text{ and }t_2 = \sqrt{3}$$
Now the formula for parametrized path length is similar to the pythagorean formula, explicitly:
$$L = \int_{t_1}^{t_2} { \sqrt{ \left(\frac{{\rm d}y}{{\rm d}t} \right)^2 + \left(\frac{{\rm d}x}{{\rm d}t} \right)^2 } {\rm d}t }$$
And filling in:
$$\begin{align}
%
L &= \int_{-\sqrt{3}}^{\sqrt{3}} { \sqrt{ \left(6t \right)^2 + \left(3 - 3t^2 \right)^2 } {\rm d}t }
%
\\ &= \int_{-\sqrt{3}}^{\sqrt{3}} { \sqrt{ 9t^4 + 18t^2 + 9 } {\rm d}t }
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\\ &= \int_{-\sqrt{3}}^{\sqrt{3}} { \left \vert 3t^2 + 3 \right \vert {\rm d}t }
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\\ &= t^3 + 3t ~\bigg\vert_{t = -\sqrt{3}}^{t = \sqrt{3}} 
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\\ &= 12 \sqrt{3}
%
\end{align}$$

