1. **State the problem:** We want to find the arc length $L$ of the curve $$\mathbf{C}: x=\sin(3t)-3t\cos(3t),\quad y=3t\sin(3t)+\cos(3t),\quad z=4t^2$$ between the points parameterized by $t=0$ and $t=\frac{\pi}{2}$. 2. **Recall the formula for arc length of a parametric curve:** $$L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\, dt$$ 3. **Compute the derivatives:** - $\frac{dx}{dt} = \frac{d}{dt}\left(\sin(3t) - 3t \cos(3t)\right)$ = $3\cos(3t) - 3\cos(3t) + 9t \sin(3t) = 9t \sin(3t)$ - $\frac{dy}{dt} = \frac{d}{dt}\left(3t \sin(3t) + \cos(3t)\right)$ = $3 \sin(3t) + 9t \cos(3t) - 3 \sin(3t) = 9t \cos(3t)$ - $\frac{dz}{dt} = \frac{d}{dt}(4t^2) = 8t$ 4. **Compute the integrand inside the square root:** $$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 = (9t \sin(3t))^2 + (9t \cos(3t))^2 + (8t)^2$$ Simplify: $$= 81t^2 \sin^2(3t) + 81t^2 \cos^2(3t) + 64 t^2 = 81 t^2 (\sin^2(3t) + \cos^2(3t)) + 64 t^2 = 81 t^2 (1) + 64 t^2 = (81 + 64) t^2 = 145 t^2$$ 5. **So the integrand is:** $$\sqrt{145 t^2} = \sqrt{145} |t|$$ On $[0, \frac{\pi}{2}]$, $t \ge 0$, so $|t| = t$. 6. **Set up the integral for arc length:** $$L = \int_0^{\frac{\pi}{2}} \sqrt{145} t \, dt = \sqrt{145} \int_0^{\frac{\pi}{2}} t \, dt$$ 7. **Evaluate the integral:** $$\int_0^{\frac{\pi}{2}} t \, dt = \left[ \frac{t^2}{2} \right]_0^{\frac{\pi}{2}} = \frac{(\frac{\pi}{2})^2}{2} = \frac{\pi^2}{8}$$ 8. **Final answer:** $$L = \sqrt{145} \cdot \frac{\pi^2}{8} = \frac{\pi^2 \sqrt{145}}{8}$$