1. **Problem statement:** We have a particle moving along a curve described by $$x=e^{-3t},\quad y=5\cos(3t),\quad z=-5\sin(3t)$$ where $t$ is time. (i) Find velocity $\vec{v}$ and acceleration $\vec{a}$ at any time $t$. 2. **Velocity components:** Velocity is the derivative of position with respect to time $t$. $$v_x=\frac{dx}{dt} = \frac{d}{dt} e^{-3t} = -3e^{-3t}$$ $$v_y=\frac{dy}{dt} = \frac{d}{dt} 5\cos(3t) = 5(-3\sin(3t)) = -15 \sin(3t)$$ $$v_z=\frac{dz}{dt} = \frac{d}{dt} (-5\sin(3t)) = -5(3\cos(3t)) = -15 \cos(3t)$$ Hence the velocity vector: $$\vec{v} = (-3e^{-3t}, -15 \sin(3t), -15 \cos(3t))$$ 3. **Acceleration components:** Acceleration is the derivative of velocity. $$a_x = \frac{dv_x}{dt} = \frac{d}{dt}(-3e^{-3t}) = 9 e^{-3t}$$ $$a_y = \frac{dv_y}{dt} = \frac{d}{dt} (-15 \sin(3t)) = -15 (3 \cos(3t)) = -45 \cos(3t)$$ $$a_z = \frac{dv_z}{dt} = \frac{d}{dt} (-15 \cos(3t)) = -15(-3 \sin(3t)) = 45 \sin(3t)$$ So acceleration vector: $$\vec{a} = (9 e^{-3t}, -45 \cos(3t), 45 \sin(3t))$$ 4. **Magnitude of velocity $||\vec{v}||$:** $$||\vec{v}|| = \sqrt{(-3 e^{-3t})^2 + (-15 \sin(3t))^2 + (-15 \cos(3t))^2}$$ $$= \sqrt{9 e^{-6t} + 225 (\sin^2(3t) + \cos^2(3t))}$$ $$= \sqrt{9 e^{-6t} + 225}$$ 5. **Magnitude of acceleration $||\vec{a}||$:** $$||\vec{a}|| = \sqrt{(9 e^{-3t})^2 + (-45 \cos(3t))^2 + (45 \sin(3t))^2}$$ $$= \sqrt{81 e^{-6t} + 2025 (\cos^2(3t) + \sin^2(3t))}$$ $$= \sqrt{81 e^{-6t} + 2025}$$ 6. **At time $t=0$:** - Velocity magnitude: $$||\vec{v}|| = \sqrt{9 e^0 + 225} = \sqrt{9 + 225} = \sqrt{234} = 3 \sqrt{26}$$ - Acceleration magnitude: $$||\vec{a}|| = \sqrt{81 e^{0} + 2025} = \sqrt{81 + 2025} = \sqrt{2106} = 3 \sqrt{234}$$ **Final answers:** (i) $\vec{v} = (-3e^{-3t}, -15\sin(3t), -15 \cos(3t))$, $\vec{a} = (9 e^{-3t}, -45 \cos(3t), 45 \sin(3t))$ (ii) $||\vec{v}|| = \sqrt{9 e^{-6t} + 225}$, $||\vec{a}|| = \sqrt{81 e^{-6t} + 2025}$ (iii) At $t=0$, $||\vec{v}|| = 3 \sqrt{26}$, $||\vec{a}|| = 3 \sqrt{234}$