1. Find the rectangular coordinates for the polar point $(3, \frac{\pi}{2})$. - Recall conversion formulas: $x = r \cos \theta$, $y = r \sin \theta$. - Substitute: $x = 3 \cos \frac{\pi}{2} = 3 \times 0 = 0$. - Substitute: $y = 3 \sin \frac{\pi}{2} = 3 \times 1 = 3$. - Answer: $(x, y) = (0, 3)$. 2. Find the polar coordinates for the rectangular point $(1, -1)$. - Radius $r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$. - Angle $\theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{-1}{1} \right) = \tan^{-1}(-1)$. - Since $(1,-1)$ is in the fourth quadrant, $\theta = \frac{7\pi}{4}$. - Answer: $(r, \theta) = \left( \sqrt{2}, \frac{7\pi}{4} \right)$. 3. Convert rectangular equation $3x - y + 2 = 0$ to polar form. - Use $x = r \cos \theta$, $y = r \sin \theta$. - Substitute: $3(r \cos \theta) - (r \sin \theta) + 2 = 0$. - Factor $r$: $r(3 \cos \theta - \sin \theta) + 2 = 0$. - Solve for $r$: $$r = \frac{-2}{3 \cos \theta - \sin \theta}$$. 4. Convert polar equation $\theta = \frac{5\pi}{6}$ to rectangular form. - $\theta = \tan^{-1} \left( \frac{y}{x} \right) = \frac{5\pi}{6}$. - So, $\frac{y}{x} = \tan \frac{5\pi}{6} = -\frac{1}{\sqrt{3}}$. - Multiply both sides by $x$: $y = -\frac{1}{\sqrt{3}} x$. - Answer: $\boxed{ y = -\frac{1}{\sqrt{3}} x }$. 5. Classify the graph of $9x^2 + 4y^2 - 18x + 16y - 119 = 0$. - Group terms: $9x^2 - 18x + 4y^2 + 16y = 119$. - Complete the squares: - For $x$: $9(x^2 - 2x)$; complete square inside: $x^2 - 2x + 1 - 1$. - For $y$: $4(y^2 + 4y)$; complete square inside: $y^2 + 4y + 4 - 4$. - Rewrite: $9(x^2 - 2x + 1) - 9 + 4(y^2 + 4y + 4) - 16 = 119$. $9(x - 1)^2 - 9 + 4(y + 2)^2 - 16 = 119$. - Combine constants: $-9 -16 = -25$, move to right side: $9(x - 1)^2 + 4(y + 2)^2 = 119 + 25 = 144$. - Divide both sides by $144$: $\frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{36} = 1$. - This is the equation of an ellipse centered at $(1, -2)$. 6. Identify surface of $\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1$. - The presence of $+$ terms in $x^2$ and $y^2$, and a $-$ in $z^2$, with constant $= 1$ implies a hyperboloid of one sheet. Final answers: 1. $(0, 3)$ 2. $(\sqrt{2}, \frac{7\pi}{4})$ 3. $r = \frac{-2}{3 \cos \theta - \sin \theta}$ 4. $y = -\frac{1}{\sqrt{3}} x$ 5. Ellipse 6. Hyperboloid of one sheet