Let’s work through the problems step by step, starting with the solutions to Question 1. I’ll provide detailed explanations for each part.

Question 1: Find the polar coordinates for the given Cartesian points:

The polar coordinates $(r, \theta)$ can be calculated from Cartesian coordinates $(x, y)$ as:

• $r = \sqrt{x^2 + y^2}$
• $\theta = \arctan\left(\frac{y}{x}\right)$ (adjusted for the quadrant of the point)

(a) $M_1(-3, 3)$:

1. Calculate $r$: $r = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

2. Calculate $\theta$: $\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{3}{-3}\right) = \arctan(-1)$ Since the point is in the second quadrant, add $\pi$ to the angle: $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$

3. Polar coordinates: $M_1 = (3\sqrt{2}, \frac{3\pi}{4}).$

(b) $M_2(-4\sqrt{3}, -4)$:

1. Calculate $r$: $r = \sqrt{(-4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8.$

2. Calculate $\theta$: $\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-4}{-4\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.$ Since the point is in the third quadrant, add $\pi$: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.$

3. Polar coordinates: $M_2 = (8, \frac{7\pi}{6}).$

(c) $M_3(0, -5)$:

1. Calculate $r$: $r = \sqrt{(0)^2 + (-5)^2} = 5.$

2. Calculate $\theta$: The point lies on the negative y-axis, so: $\theta = \frac{3\pi}{2}.$

3. Polar coordinates: $M_3 = (5, \frac{3\pi}{2}).$

(d) $M_4(-2, -1)$:

1. Calculate $r$: $r = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}.$

2. Calculate $\theta$: $\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-1}{-2}\right) = \arctan\left(\frac{1}{2}\right).$ Since the point is in the third quadrant, add $\pi$: $\theta = \pi + \arctan\left(\frac{1}{2}\right).$

3. Polar coordinates: $M_4 = (\sqrt{5}, \pi + \arctan(\frac{1}{2})).$

(e) $M_5(4, -2)$:

1. Calculate $r$: $r = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}.$

2. Calculate $\theta$: $\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-2}{4}\right) = \arctan\left(-\frac{1}{2}\right).$ Since the point is in the fourth quadrant, add $2\pi$: $\theta = 2\pi + \arctan\left(-\frac{1}{2}\right).$

3. Polar coordinates: $M_5 = (2\sqrt{5}, 2\pi + \arctan(-\frac{1}{2})).$

Would you like to proceed to Question 2, or should I clarify any steps above?

5 Related Questions:

1. How can you determine the correct quadrant for the angle $\theta$?
2. What are the formulas to convert from polar to Cartesian coordinates?
3. Why do we add $\pi$ or $2\pi$ to the angle for certain quadrants?
4. How would you calculate polar coordinates if $x = 0$ or $y = 0$?
5. Can polar coordinates have negative values for $r$?

1 Tip:

When working with polar coordinates, always sketch the point’s approximate location on the plane to avoid errors in determining the angle’s quadrant.