Answer
$$L = 12$$
Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(3cos(t) - cos(3t))}{dt} = -3sin(t) + 3sin(3t)$
$\frac{dy}{dt} = \frac{d(3sin(t) - sin(3t))}{dt} = 3cos(t) -3cos(3t)$
3. Substitute the values into the definite integral:
$L = \int_{0}^{\pi} \sqrt{(-3sin(t) + 3sin(3t))^2 + (3cos(t) - 3cos(3t))^2} \space dt$
$L = \int_{0}^{\pi} \sqrt{(9sin^2(t) - 18sin(t)sin(3t) + 9sin^2(3t)) + (9cos^2(t) - 18cos(t)cos(3t) + 9cos^2(3t))} \space dt$
$L = \int_{0}^{\pi} \sqrt{(9(sin^2(t) + cos^2(t)) - 18(sin(t)sin(3t) + cos(t)cos(3t)) + 9(cos^2(3t) + sin^2(3t)))} \space dt$
$L = \int_{0}^{\pi} \sqrt{(9(sin^2(t) + cos^2(t)) - 18(sin(t)sin(3t) + cos(t)cos(3t)) + 9(cos^2(3t) + sin^2(3t)))} \space dt$
$L = \int_{0}^{\pi} \sqrt{(9 - 18(cos(3t - t)) + 9)} \space dt$
$L = \int_{0}^{\pi} \sqrt{(18 - 18(cos(2t)) )} \space dt$
$L = \int_{0}^{\pi} \sqrt{18(1 - (cos(2t)) )} \space dt$
** Notice: $cos(2t) = 1 - 2sin^2(t)$
$L = \int_{0}^{\pi} \sqrt{18(1 - (1 - 2sin^2(t)) )} \space dt$
$L = \int_{0}^{\pi} \sqrt{36 sin^2(t)} \space dt$
$L = \int_{0}^{\pi} 6 sin(t) \space dt$
$L = 6(-cos(t))]^{\pi}_0$
$L = -6cos(\pi) - (-6cos(0))$
$L = -6(-1) + 6(1)$
$L = 12$
