Convert Points Between Polar and Rectangular Coordinates
15. Polar Equations
Convert Points Between Polar and Rectangular Coordinates - Video Tutorials & Practice Problems
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1
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Convert Points from Polar to Rectangular
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6m
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Hey, everyone. Now that we're familiar with both polar and rectangular coordinates, we're going to have to convert between the two and take a point given in polar coordinates as our theta converting it into its rectangular xy equivalent. Now this all just comes back down to triangles, something that you are already an expert on. So here, I'm going to break down for you exactly how to take a point given in polar coordinates and convert it into rectangular coordinates just using this triangle here. So let's go ahead and get started. Now looking at the point that we have on our graph here given in polar cod its five pi over three, we can go ahead and form a triangle with this giving our triangle a hypotenuse of five and an inner angle theta of pi over three. Then our side links here are X and Y the same way that we've seen before on the unit circle. But here, our hypotenuse is no longer just one, the same way that it was there. Now, we can dive a bit deeper here because if I set up a cosine equation here, I would see that the cosine of data of course, is equal to the adjacent side over the hypo nose. And specifically this triangle that would tell me that the cosine of that inner angle pi over three is equal to my side length X over my hypotenuse of five. Then I can do the same thing for a sine function, setting up my sine function sine of theta is equal to the opposite side, which in this case is Y over that hypotenuse value of five. Now I can go ahead and solve each of these for X and Y. So looking at my cosine equation here, if I go ahead and multiply both sides by five to cancel that out on the right side, I end up getting that five times the cosine of pi over three is equal to X. Then doing the same thing for my sine equation. Again, multiplying both sides by five here having that cancel on the right side. I see that five times the sine of pi over three is equal to Y. Now, if I actually multiply these out, I see that I get an X value of 5/2 and A Y value of five root 3/2. So looking at the point that I started with five pi over three in polar coordinates, I now have my point in rectangular coordinates as 5/2 5 root 3/2. Now this will work for any point given in polar coordinates when converting to rectangular, my X value is always going to be equal to R times the cosine of data and Y will always be equal to R times the sine of data. Now, this probably looks really familiar to you because from the unit circle, we saw that X was equal to the cosine of data and Y is equal to the sine of data. Now, we just have this R value to account for, but we're still doing this same exact thing just that R is no longer always equal to one. But now that we know how to find these rectangular points given a polar coordinate. Let's go ahead and get some practice here and work through some examples together. Here, we're given a point in polar coordinates negative three pi over six. Now, here we want to go ahead and plot this point on our polar coordinate system and then convert it into rectangular coordinates. So let's go ahead and start by plotting this on our polar coordinate system negative three pi over six. I locate my angle theta first pi over six along this line. But since I have a negative R value three, I'm going to count in the opposite direction three units and I end up with my point right here. Now that I have that point plotted, let's go ahead and convert this into rectangular coordinates. Now I know that my X value is going to be equal to R times the cosine of theta So X equals R cosine of theta here plugging in my values for R and theta, I get negative three times the cosine of pi over six. Now the cosine of pi over six is the square root of 3/2. So this is negative three times the square root of 3/2. Now I have my X value, let's find our Y value Y is going to, of course, be equal to R times the sine of the. So when we plug these values in here, that ends up giving me a negative three times the sine of that angle pi over six. Now the sign of pi over six is one half. So this is equal to negative three times one half which I can rewrite as being negative 3/2. So now I have my X and Y value negative three root 3/2 and negative 3/2. Now this is my point in rectangular coordinates. Now, in this problem, we're also asked to go ahead and plot this on the xy plane, which is actually going to be easier to do if we have these in their decimal form. So this negative three root 3/2 is going to be about negative 2.6 and negative 3/2 is of course just negative 1.5. So I can go ahead and use those decimals to plot that on my rectangular coordinate system. Now actually locating the X value negative 2.6. And then going down to my Y value of negative 1.5 I end up right here in quadrant three of my rectangular coordinate system. Now, here we see that these points are located in the exact same spot, which makes sense because these are just the exact same point but represented in different ways. One in polar and one in rectangular. Now let's look at one final example here, here we have the point given in polar coordinates, zero negative pi over six. Now again, here we want first plot this on our polar coordinate system. So locating my angle negative pi over six, which will end up being right along this line. Since I know that my R value is zero, I actually just stay right here at the pole. What we think of as being the origin in rectangular coordinates. So here is my point B now that we have that plotted on that polar coordinate system, let's go ahead and calculate our X and Y values and get this point in rectangular coordinates. Now I know that X is going to be equal to R times the cosine of theta which here my R value is zero. So this is zero times the cosine of negative pi over six but zero times anything is just zero. So this ends up giving me an X value of zero. Then for Y, if I also do the same thing here, R times the sine of the plugging in my values of R and theta I get zero times the sign of negative pi over six, which of course, again, zero times anything is still just zero. So I again get a Y value of zero then that gives me my point in rectangular cord zero comma zero, which we know is located at the origin, which I can go ahead and plot right here. So anytime we have an R value zero, that's always going to end up being the 0.00 in rectangular coordinates. And we again see that this is located in the exact same spot as it was on our polar coordinate system. Again, these are the same exact point represented in different ways. Now that we know how to convert points from polar to rectangular. Let's continue practicing. Thanks for watching and I'll see you in the next one.
2
Problem
Problem
Convert the point to rectangular coordinates.
(−2,−4π)
A
(2,−2)
B
(−1,1)
C
(−2,2)
D
(−22,22)
3
Problem
Problem
Convert the point to rectangular coordinates.
(4,6π)
A
(23,2)
B
(43,4)
C
(2,23)
D
(2,3)
4
Problem
Problem
Convert the point to rectangular coordinates.
(−3,0)
A
(−3,0)
B
(0,−3)
C
(0,0)
D
(3,0)
5
Problem
Problem
Convert the point to rectangular coordinates.
(0,47π)
A
(0,47)
B
(−4,0)
C
(0,0)
D
(0,7)
6
concept
Convert Points from Rectangular to Polar
Video duration:
6m
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Hey, everyone, we just learned how to take points given in polar coordinates and convert them to rectangular coordinates. But we also need to be able to do the opposite take points given in rectangular coordinates and convert them to polar coordinates. Now, looking at this point given in rectangular coordinates here we see this sort of suspiciously familiar looking triangle that we've solved the time and time again. And that's all this is, it all just comes back down to triangles. So here we're going to keep using everything that we already know specific working with the Pythagorean theorem and the tangent function in order to convert points from their rectangular coordinates into polar coordinates. Now, here I'm going to walk you through how to do that step by step. So let's go ahead and get started. Now, looking at the point that we have here in rectangular coordinates, three comma four, we're able to form this triangle and from this rectangular coordinate 0.3 comma four, I know that this side LX is equal to three and my side length Y is equal to four. Now, since I already know X and Y it's just left a find R and theta, which will give me my point in polar coordinates. Now, let's start by finding R. Now looking at my R value here, this is the hypo news of my triangle, but I have my other two side links already. So I can find that hypo news by simply using the Pythagorean theorem, which tells me that C squared is equal to A squared plus B squared or specifically for this triangle R squared is equal to X squared plus Y squared. Now actually plugging in our values here, I get that R squared is equal to three squared plus four squared. Now adding those together gives me 25. So I have R squared is equal to 25 or completely solving that for R simply R is equal to five. Now that I have that R value, let's turn to finding the. Now looking at theta here, I know that I have my opposite side and my adjacent side, which means that I can set up a tangent function because I know that the tangent of the is equal to the opposite over the adjacent side or specifically for this angle for a point given in rectangular coordinates, the tangent of my angle theta is equal to Y over X. Now, again, just plugging in the values that I have here, I end up getting that the tangent of theta is equal to that Y value of four over that X value of three. Now in order to actually solve for theta. Here, I need to take the inverse tangent. So here theta will be equal to the inverse tangent of 4/3. Now, if we actually plug that inverse tangent into our calculator, we end up getting that theta is equal to about 53 degrees. And I now have both my R and my theta value. Now, I originally started with my point in rectangular coordinates, three comma four. And now I have my point in polar coordinates, five comma 53 degrees. Now these are the general equations that we're going to use to convert points from rectangular to polar coordinates. But we have to be really careful here because we end up taking an inverse tangent and the inverse tangent. Remember that this function is only defined over the intervals contained in quadrant one and quadrant four. And even though our point here was located in quadrant one, it won't always be. So I actually have a step by step process for you that will work no matter where your point is located. So let's go ahead and work through these examples together. Now first, we're given at this point here negative 40. And I want to go ahead and plot this point and then convert it into coordinates by following these steps here. Now starting with step one, we just want to go ahead and plot this point on a rectangular coordinate system. Now locating this point negative 40, I end up right here on my X axis for this 1st 0.8. Now with step one done, we move on to step two and calculating R. Now this is the same equation that we saw above but just already solved for R having taken the square root on both sides. So here plugging in my values, I get that R is equal to the square root of negative four squared plus zero squared. Now this ends up being the square root of 16, which is simply equal to four. So I have an R value here of four. Now, with our calculated, we move on to finding theta. Now, in order to find theta, we need to look at the location of our point and looking at where this point A is on my X axis because it's on an axis. I know that theta is going to be one of my quadrant angles and thinking specifically about where this point is located. Thinking about my angles here going from zero to pi radiance or 180 degrees. I know that my angle here is simply pi. So that gives me my value of theta pi and I now have my point in polar coordinates for comma pi. Now let's move on to our next example. Here we are specifically given the point negative one route three. Now we're going to restart our steps here starting from step number one, we're going to start out by plotting our point. Now plotting my point here, it's going to be more useful to know so that the square root of three is about 1.73 as a decimal. So plotting this point at negative 11.73 ends me up right here in quadrant two for point B. Now, at that point plotted, we want to go ahead and find R as we have before taking the square root of X squared plus Y squared. Now plugging these values in here, I end up getting the square root of negative one squared plus the square root of three squared. Now actually adding those together gives me the square root of four or simply two. So I have my R value of two. Now all that's left is to find theta. Again, we're looking at the location of our point and I see that my point B is located in quadrant two. So it's not on an axis. It's not in quadrant one or four as my original example was, but it is in quadrant two. So what do we do now? Well, we're gonna start off the same way that we did for our original angle that was in quadrant one by taking the inverse tangent of Y over X. So let's start there beta is equal to the inverse tangent of my Y value root three over my X value negative one. Now this simplifies to the inverse tangent of the negative square root of three, which from my knowledge of the unit circle or by simply typing this into a calculator, I will end up getting a negative pi over three for that inverse tangent of negative root three. But we're not done yet because let's think about where this point is located or where this angle is located. The negative square root of three is here in quadrant four. And that is not where my point B is located. So that wouldn't make any sense. Now, in order to make sure that this is located in the right place because our point is located here in quadrant two, we need to go ahead and add pi to that angle that we got from the inverse tangent. So this negative pi over three, I need to add pi to it. This ends up giving me two pi over three which thinking about where that angle is located that is located in quadrant two as it should be. So here I have my final value of two pi over three and I have this point. Now in polar coordinates, 22 pi over three. Now when converting points from rectangular to polar coordinates, we're going to start out the same way we're going to plot our point on our graph. And we're going to find R by using this equation here. Then when finding the, we need to pay attention to the location of our point and then we're good to go now that we know how to convert points from a rectangular to polar. Let's continue practicing. Thanks for watching and I'll see you in the next one.
7
example
Convert Points from Rectangular to Polar Example 1
Video duration:
2m
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Here, we're asked to convert the point given in rectangular coordinates into polar coordinates. And the point that we're given here is three negative three. So let's go ahead and get started with our steps so that we can get this to polar coordinates. Now, step one tells us to go ahead and plot our point. So here on my rectangular coordinate system, I'm going to plot three negative three and I end up right down here in quadrant four. Now moving on to step two and actually finding our R value by taking the square root of X squared plus Y squared. Now plugging in my values here three and negative three. This gives me that R is equal to the square root of three squared plus negative three squared. Now three squared and negative three squared are both nine. So this gives me the square root of nine plus nine or the square root of 18. Now we can simplify this further using our radical rules in order to get that R is equal to three times the square root of two. Now let's go ahead and move on to step number three in finding the. Now here we want to pay attention to where our point is located and it is located in quadrant four. So that tells me that I want to find theta by taking the inverse tangent of Y over X. Now actually setting that up and plugging my values in, I get that theta will be equal to the inverse tangent of my Y value negative three over my X value three. Now this simplifies to give me the inverse tangent of negative one. And if I plug this into my calculator, this will give me negative pi over four for my value of data. So here I have my in polar coordinates with that R value three root two comma that beta value negative pi over four. So this is my final answer here three root two negative pi over four. Now something that you might be thinking about here if you chose to calculate your inverse tangent of Y over X differently and maybe not using a calculator. But rather the unit circle you may have thought isn't this angle also seven pi over four. And that's true. If I represented this point as three root to seven pi over four, that would still be correct. The reason that we got negative pi over four 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 three root 27 pi over four, that would still be correct. And even if you tried to take that point back to rectangular coordinates, you would still end up with this 0.3 negative three. So let me know if you have any questions. Thanks for watching and I'll see you in the next one.
