1. **State the problem:** We have a particle moving along a curve defined by parametric equations: $$x = e^{-t}, \quad y = 3 \sin(3t), \quad z = 2 \cos(3t)$$ We want to: a) Find the velocity and acceleration vectors at any time $t$. b) Find the magnitudes of velocity and acceleration at $t=0$. 2. **Find the velocity vector \( \mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) \):** $$x(t) = e^{-t} \implies \frac{dx}{dt} = -e^{-t}$$ $$y(t) = 3 \sin(3t) \implies \frac{dy}{dt} = 3 \cdot 3 \cos(3t) = 9 \cos(3t)$$ $$z(t) = 2 \cos(3t) \implies \frac{dz}{dt} = 2 \cdot (-3) \sin(3t) = -6 \sin(3t)$$ So, $$\mathbf{v}(t) = -e^{-t} \mathbf{i} + 9 \cos(3t) \mathbf{j} - 6 \sin(3t) \mathbf{k}$$ 3. **Find the acceleration vector \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \):** Differentiate each component again: $$\frac{d}{dt}(-e^{-t}) = e^{-t}$$ $$\frac{d}{dt}(9 \cos(3t)) = 9 \cdot (-3) \sin(3t) = -27 \sin(3t)$$ $$\frac{d}{dt}(-6 \sin(3t)) = -6 \cdot 3 \cos(3t) = -18 \cos(3t)$$ So, $$\mathbf{a}(t) = e^{-t} \mathbf{i} - 27 \sin(3t) \mathbf{j} - 18 \cos(3t) \mathbf{k}$$ 4. **Find magnitudes at $t=0$: ** Velocity magnitude: $$||\mathbf{v}(0)|| = \sqrt{(-e^{0})^{2} + (9 \cos(0))^{2} + (-6 \sin(0))^{2}} = \sqrt{(-1)^2 + 9^2 + 0} = \sqrt{1 + 81} = \sqrt{82}$$ Acceleration magnitude: $$||\mathbf{a}(0)|| = \sqrt{(e^{0})^{2} + (-27 \sin(0))^{2} + (-18 \cos(0))^{2}} = \sqrt{1^{2} + 0 + (-18)^{2}} = \sqrt{1 + 324} = \sqrt{325}$$ **Final answers:** - Velocity vector: $$\mathbf{v}(t) = -e^{-t} \mathbf{i} + 9 \cos(3t) \mathbf{j} - 6 \sin(3t) \mathbf{k}$$ - Acceleration vector: $$\mathbf{a}(t) = e^{-t} \mathbf{i} - 27 \sin(3t) \mathbf{j} - 18 \cos(3t) \mathbf{k}$$ - Magnitude of velocity at $t=0$: $$||\mathbf{v}(0)|| = \sqrt{82}$$ - Magnitude of acceleration at $t=0$: $$||\mathbf{a}(0)|| = \sqrt{325}$$