1. **State the problem:** Find the angle $\theta$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$ where the curve given by $r = \theta + \sin 2\theta$ is farthest from the origin. 2. **Recall the formula:** The distance from the origin in polar coordinates is $r$. To find where $r$ is farthest, we find critical points by setting the derivative $\frac{dr}{d\theta}$ to zero. 3. **Calculate the derivative:** $$\frac{dr}{d\theta} = 1 + 2\cos 2\theta$$ 4. **Set the derivative equal to zero to find critical points:** $$1 + 2\cos 2\theta = 0$$ 5. **Solve for $\cos 2\theta$:** $$2\cos 2\theta = -1$$ $$\cos 2\theta = -\frac{1}{2}$$ 6. **Find $2\theta$ values where $\cos 2\theta = -\frac{1}{2}$:** Within $0 \leq \theta \leq \frac{\pi}{2}$, $2\theta$ ranges from $0$ to $\pi$. The solutions for $\cos x = -\frac{1}{2}$ in $[0, \pi]$ are: $$x = \frac{2\pi}{3}, \frac{4\pi}{3}$$ Only $\frac{2\pi}{3}$ lies in $[0, \pi]$, so: $$2\theta = \frac{2\pi}{3} \implies \theta = \frac{\pi}{3}$$ 7. **Evaluate $r$ at critical point and endpoints:** - At $\theta = 0$: $$r = 0 + \sin 0 = 0$$ - At $\theta = \frac{\pi}{3}$: $$r = \frac{\pi}{3} + \sin \left(2 \times \frac{\pi}{3}\right) = \frac{\pi}{3} + \sin \frac{2\pi}{3} = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.047 + 0.866 = 1.913$$ - At $\theta = \frac{\pi}{2}$: $$r = \frac{\pi}{2} + \sin \pi = \frac{\pi}{2} + 0 = 1.571$$ 8. **Conclusion:** The maximum distance from the origin is approximately $1.913$ at $\theta = \frac{\pi}{3} \approx 1.047$ radians. **Note:** The answer you provided, 8.378, does not match the interval $0 \leq \theta \leq \frac{\pi}{2}$ or the function given. The correct farthest angle in the given interval is $\theta = \frac{\pi}{3}$.