Here, we're asked to convert the point given in rectangular coordinates into polar coordinates, and the point that we're given here is (3, -3). So let's go ahead and get started with our steps so that we can get this to polar coordinates. Now, step 1 tells us to go ahead and plot our point. So here on my rectangular coordinate system, I'm going to plot (3, -3), and I end up right down here in quadrant 4.
Now moving on to step 2, and actually finding our r value by taking the square root of x squared plus y squared. Plugging in my values here, 3 and -3, this gives me that \( r \) is equal to the square root of \( 3^2 + (-3)^2 \). Now \( 3^2 \) and \( (-3)^2 \) are both 9, so this gives me the square root of 9+9 or the square root of 18. Now we can simplify this further using our radical rules to get that \( r \) is equal to \( 3 \sqrt{2} \).
Now, let's go ahead and move on to step number 3 in finding \( \theta \). Now here, we want to pay attention to where our point is located, and it is located in quadrant 4, so that tells me that I want to find \( \theta \) by taking the inverse tangent of \( \frac{y}{x} \). Setting that up and plugging my values in, I get that \( \theta \) will be equal to the inverse tangent of my y value -3 over my x value 3. This simplifies to give me the inverse tangent of -1. If I plug this into my calculator, \( \theta \) value is \( -\frac{\pi}{4} \). So this is my final answer here, \( 3\sqrt{2}, -\frac{\pi}{4} \).
Something that you might be thinking about here is if you chose to calculate your inverse tangent of \( \frac{y}{x} \) differently and perhaps not using a calculator, but rather the unit circle, you may have thought that this angle is also \( 7\frac{\pi}{4} \). That's true. If I represented this point as \( 3\sqrt{2}, 7\frac{\pi}{4} \), that would still be correct. The reason that we got \( -\frac{\pi}{4} \) is because of the restrictions placed on our inverse tangent function. But remember that there are always multiple ways to represent the same point in polar coordinates. So if you were to have \( 3\sqrt{2}, 7\frac{\pi}{4} \), that would still be correct. Even if you tried to take that point back to rectangular coordinates, you would still end up with this point, (3, -3). So let me know if you have any questions. Thanks for watching, and I'll see you in the next one.