1. **State the problem:** Given the vector function $\vec{F} = (-\sin(2t))\vec{i} + \cos(t)\vec{j} + 2t\vec{k}$, find: (i) The magnitude of the first derivative $\left|\frac{d\vec{F}}{dt}\right|$ (ii) The magnitude of the second derivative $\left|\frac{d^2\vec{F}}{dt^2}\right|$ 2. **Calculate the first derivative $\frac{d\vec{F}}{dt}$:** $$\frac{d}{dt}[-\sin(2t)] = -2\cos(2t)$$ $$\frac{d}{dt}[\cos(t)] = -\sin(t)$$ $$\frac{d}{dt}[2t] = 2$$ So, $$\frac{d\vec{F}}{dt} = (-2\cos(2t))\vec{i} + (-\sin(t))\vec{j} + 2\vec{k}$$ 3. **Calculate the magnitude of $\frac{d\vec{F}}{dt}$:** $$\left|\frac{d\vec{F}}{dt}\right| = \sqrt{(-2\cos(2t))^2 + (-\sin(t))^2 + 2^2}$$ Simplify: $$= \sqrt{4\cos^2(2t) + \sin^2(t) + 4}$$ 4. **Calculate the second derivative $\frac{d^2\vec{F}}{dt^2}$:** $$\frac{d}{dt}[-2\cos(2t)] = -2 \cdot (-2\sin(2t)) = 4\sin(2t)$$ $$\frac{d}{dt}[-\sin(t)] = -\cos(t)$$ $$\frac{d}{dt}[2] = 0$$ So, $$\frac{d^2\vec{F}}{dt^2} = 4\sin(2t)\vec{i} + (-\cos(t))\vec{j} + 0\vec{k}$$ 5. **Calculate the magnitude of $\frac{d^2\vec{F}}{dt^2}$:** $$\left|\frac{d^2\vec{F}}{dt^2}\right| = \sqrt{(4\sin(2t))^2 + (-\cos(t))^2 + 0^2}$$ Simplify: $$= \sqrt{16 \sin^2(2t) + \cos^2(t)}$$ **Final answers:** (i) $$\left|\frac{d\vec{F}}{dt}\right| = \sqrt{4\cos^2(2t) + \sin^2(t) + 4}$$ (ii) $$\left|\frac{d^2\vec{F}}{dt^2}\right| = \sqrt{16\sin^2(2t) + \cos^2(t)}$$