1. Problem Q3: Convert Cartesian equations to polar coordinates using formulas $x = r\cos\theta$ and $y = r\sin\theta$. (a) Given $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Substitute $x = r\cos\theta$, $y = r\sin\theta$: $$\frac{r^2 \cos^2 \theta}{9} + \frac{r^2 \sin^2 \theta}{4} = 1$$ Factor out $r^2$: $$r^2 \left(\frac{\cos^2 \theta}{9} + \frac{\sin^2 \theta}{4}\right) = 1$$ Solve for $r^2$: $$r^2 = \frac{1}{\frac{\cos^2 \theta}{9} + \frac{\sin^2 \theta}{4}}$$ (b) Presumably $y^2 = $ (missing rhs), cannot convert without full equation. (c) Given $x^2 + (y - 2)^2 = 4$. Substitute: $$r^2 \cos^2 \theta + (r \sin \theta - 2)^2 = 4$$ Expand second term: $$r^2 \cos^2 \theta + r^2 \sin^2 \theta - 4r \sin \theta + 4 = 4$$ Combine $r^2$ terms: $$r^2(\cos^2 \theta + \sin^2 \theta) - 4r \sin \theta + 4 = 4$$ Recall $\cos^2 \theta + \sin^2 \theta = 1$: $$r^2 - 4r \sin \theta + 4 = 4$$ Subtract 4 from both sides: $$r^2 - 4r \sin \theta = 0$$ (d) Given $x^2$ (incomplete), no full equation to convert. (e) Given $(x+2)^2 + (y-5)^2 = 16$. Substitute: $$(r \cos \theta + 2)^2 + (r \sin \theta - 5)^2 = 16$$ Expand: $$r^2 \cos^2 \theta + 4r \cos \theta + 4 + r^2 \sin^2 \theta - 10r \sin \theta + 25 = 16$$ Combine $r^2$ terms: $$r^2 (\cos^2 \theta + \sin^2 \theta) + 4r \cos \theta - 10r \sin \theta + 29 = 16$$ Simplify: $$r^2 + 4r \cos \theta - 10r \sin \theta + 29 = 16$$ Subtract 16: $$r^2 + 4r \cos \theta - 10r \sin \theta + 13 = 0$$ 2. Problem Q4: Show point $(r, \theta) = \left(\frac{1}{2}, \frac{3\pi}{2}\right)$ lies on $r = -\sin \theta$. Evaluate right side at $\theta = \frac{3\pi}{2}$: $$-\sin \frac{3\pi}{2} = -(-1) = 1$$ But $r = \frac{1}{2}$ given, which does not equal 1. Hence, point does NOT lie on curve $r = -\sin \theta$. 3. Problem Q5: Identify symmetries of curves. (a) $r = \sin \frac{\theta}{2}$ - Symmetry about polar axis since replacing $\theta$ by $-\theta$ yields same $r$. (b) $r^2 = 4 \cos 2\theta$ - Symmetric about polar axis and line $\theta = \frac{\pi}{2}$ due to cosine even function and double angle. 4. Problem Q6: Find intersections. (a) Equate $r$'s: $$1 + \sin \theta = 1 - \sin \theta$$ $$\sin \theta = 0$$ Solutions: $\theta = 0, \pi, 2\pi, ...$ Substitute back $r = 1$ for these $\theta$ values. Points of intersection are on polar axis with $r=1$. (b) Equate $r$'s: $$\cos \theta = 1 - \cos \theta$$ $$2 \cos \theta = 1$$ $$\cos \theta = \frac{1}{2}$$ $$\theta = \pm \frac{\pi}{3} + 2k\pi$$ Calculate $r = \cos \theta = \frac{1}{2}$. Intersection points $(r,\theta) = \left(\frac{1}{2}, \pm \frac{\pi}{3}\right)$. (c) Given $r = \sqrt{2}$ is a circle radius $\sqrt{2}$. Find intersection with other curves if any (not specified here). Final answers summarize key conversions and identifications above.