To convert the rectangular point $(3, -4)$ to polar form, we use the following formulas:

1. Magnitude $r$: $r = \sqrt{x^2 + y^2}$

2. Angle $\theta$: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$

Step 1: Calculate $r$

Given $x = 3$ and $y = -4$: $r = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 2: Calculate $\theta$

Using $\theta = \tan^{-1}\left(\frac{-4}{3}\right)$: $\theta = \tan^{-1}\left(-\frac{4}{3}\right)$

The arctangent of $-\frac{4}{3}$ is approximately $-0.93$ radians, but this angle is in the fourth quadrant. To adjust it to the interval $0 \leq \theta \leq 2\pi$, we add $2\pi$: $\theta = 2\pi - 0.93 = 5.36 \, \text{radians} \, (\text{approximately})$

Final Answer:

The polar coordinates are: $(r, \theta) = (5, 5.36)$

Let me know if you would like further details on this conversion!

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Expanding Questions:

1. How do you convert from polar to rectangular coordinates?
2. What is the significance of the interval $0 \leq \theta \leq 2\pi$ in polar coordinates?
3. Can negative $r$ values represent valid points in polar coordinates?
4. How does the arctangent function behave in each quadrant?
5. What would the polar coordinates be if $r < 0$?

Tip: Always ensure angles in polar form align with the correct quadrant based on the signs of $x$ and $y$ in rectangular form.