Answer
$$
x=3 \cos t-\cos 3 t, \quad y=3 \sin t-\sin 3 t, \quad 0 \leq t \leq \pi
$$
the exact length of the curve is
$$
\begin{aligned}
L &=\int_{0}^{\pi} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=12 .
\end{aligned}
$$
Work Step by Step
$$
x=3 \cos t-\cos 3 t, \quad y=3 \sin t-\sin 3 t, \quad 0 \leq t \leq \pi
$$
then
$$
d x / d t=-3 \sin t+3 \sin 3 t , \quad d y / d t=3 \cos t-3 \cos 3 t
$$
so
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=9 \sin ^{2} t-18 \sin t \sin 3 t+9 \sin ^{2}(3 t)+\\
& \quad \quad +9 \cos ^{2} t-18 \cos t \cos 3 t+9 \cos ^{2}(3 t) \\
&=9\left(\cos ^{2} t+\sin ^{2} t\right)-18(\cos t \cos 3 t+ \\
&\quad \quad + \sin t \sin 3 t)+9\left[\cos ^{2}(3 t)+\sin ^{2}(3 t)\right] \\ &=9(1)-18 \cos (t-3 t)+9(1) \\
&=18-18 \cos (-2 t) \\
& =18(1-\cos 2 t) \\ &=18\left[1-\left(1-2 \sin ^{2} t\right)\right]\\
&=36 \sin ^{2} t \end{aligned}
$$
Thus the exact length of the curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{\pi} \sqrt{36 \sin ^{2} t } d t \\
& =6 \int_{0}^{\pi}|\sin t| d t \\
&=6 \int_{0}^{\pi} \sin t d t \\
&=-6[\cos t]_{0}^{\pi} \\
&=-6(-1-1) \\
&=12 .
\end{aligned}
$$
