Answer
$\frac{7\pi}{3}$
Work Step by Step
Recall: The speed of the particle is formulated by $v(t)=\sqrt{(dx/dt)^2+(dy/dt)^2}$.
Given: $x=2+5\cos (\pi t/3)$ and $y=-2+7\sin(\pi t/3)$
Find $dx/dt$ and $dy/dt$:
$dx/dt=0+5\cdot \frac{\pi}{3}\cdot (-\sin (\pi t/3))=-\frac{5\pi}{3}\sin (\pi t/3)$
$dy/dt=0+7\cdot \frac{\pi}{3}\cdot \cos (\pi t/3)=\frac{7\pi}{3}\cos(\pi t/3)$
Evaluate the speed of the particle at time $t=3$:
$v(t)=\sqrt{(-\frac{5\pi}{3}\sin(\pi t/3))^2+(\frac{7\pi}{3}\cos(\pi t/3))^2}$
$v(t)=\sqrt{\frac{25\pi^2}{9}\sin^2(\pi t/3)+\frac{49\pi^2}{9}\cos^2(\pi t/3)}$
$v(3)=\sqrt{\frac{25\pi^2}{9}\sin^2\pi+\frac{49\pi^2}{9}\cos^2\pi}$
$v(3)=\sqrt{\frac{25\pi^2}{9}\cdot 0^2+\frac{49\pi^2}{9}\cdot (-1)^2}$
$v(3)=\sqrt{\frac{49\pi^2}{9}}$
$v(3)=\frac{7\pi}{3}$
Thus, the speed of the particle is $\frac{7\pi}{3}$.
