Video: Converting Coordinates between Rectangular and Polar Forms

Consider the point 𝐴 with rectangular coordinates (βˆ’4, 7). Calculate the distance π‘Ÿ of this point from the origin. Give your answer in exact form. Find the angle πœƒ that the line segment 𝑂𝐴 makes with the positive π‘₯-axis, giving your answer in radians to two decimal places. Given that the point 𝐴 can be expressed in polar form as (π‘Ÿ, πœƒ), which of the following is also a legitimate polar form for point 𝐴? [A] (π‘Ÿ, πœƒ βˆ’ 2πœ‹) [B] (π‘Ÿ, πœƒ βˆ’ πœ‹) [C] (βˆ’π‘Ÿ, πœƒ βˆ’ 2πœ‹) [D] (π‘Ÿ, πœƒ + 3πœ‹) [E] (βˆ’π‘Ÿ, πœƒ +2πœ‹)

08:39

Video Transcript

Consider the point 𝐴 with rectangular coordinates negative four, seven. Calculate the distance π‘Ÿ of this point from the origin. Give your answer in exact form. Find the angle πœƒ that the line segment 𝑂𝐴 makes with the positive π‘₯-axis, giving your answer in radians to two decimal places. Given that the point 𝐴 can be expressed in polar form as π‘Ÿ, πœƒ, which of the following is also a legitimate polar form for point 𝐴? Option (a) π‘Ÿ, πœƒ minus two πœ‹; option (b) π‘Ÿ, πœƒ minus πœ‹; option (c) negative π‘Ÿ, πœƒ minus two πœ‹; option (d) π‘Ÿ, πœƒ plus three πœ‹; or option (e) negative π‘Ÿ, πœƒ plus two πœ‹.

We’re told that the point 𝐴 has rectangular coordinates negative four, seven. In fact, we’re shown this on a diagram. We need to calculate the distance this point 𝐴 is from the origin. We’ll call this distance π‘Ÿ. And we need to give this in an exact form. In other words, π‘Ÿ is just the length of our line segment 𝑂𝐴. There’s several different methods we could use to find this value. For example, we could use our formula to calculate the distance between two points. However, since we’re given these points in the diagram, we could also do this by using the Pythagorean theorem.

Remember, the point 𝐴 is at the coordinates negative four, seven. So if we drop a vertical line from 𝐴 to the π‘₯-axis, we get the following right-angled triangle. We can see the height of this right-angled triangle will be the 𝑦-coordinate of 𝐴 which is seven. And to find the width of this right-angled triangle, we can see we go from π‘₯ is equal to zero to π‘₯ is equal to negative four. So this right-angled triangle has width four. So now, we have a right-angled triangle with height seven and base four. And we need to find the length of the hypotenuse π‘Ÿ.

We can, of course, do this by using the Pythagorean theorem. The Pythagorean theorem tells us that π‘Ÿ squared will be equal to four squared plus seven squared. And we can calculate four squared plus seven squared to give us 65. And then we can solve for the value of π‘Ÿ by taking the square root of both sides of this equation. Remember, π‘Ÿ represents the length, so it won’t be negative. So we take the positive square root. So by taking the positive square root of both sides of this equation, we get that π‘Ÿ is equal to the square root of 65. And remember, the question wants us to give this length in an exact form. So we can just leave our answer like this.

The next part of this question wants us to find the angle πœƒ. And we’re told that this angle is the angle that the line segment 𝑂𝐴 makes with the positive π‘₯-axis. And we need to give our answer in radians to two decimal places. Once again, we can see that this angle of πœƒ is given to us in the diagram. There’s a lot of different ways of calculating the value of πœƒ. In this video, we’re only going to go through one of them.

Consider the following angle. We can see that this angle that we just marked and πœƒ add together to give us half of a turn. And of course, we know a half turn is πœ‹ radians. So these two angles add together to give us πœ‹. And if these two angles add together to give us πœ‹, then this angle must be equal to πœ‹ minus πœƒ. So instead of finding the angle of πœƒ directly, we’ll instead find the value of this angle, πœ‹ minus πœƒ. To do this, we notice this is an angle in a right-angled triangle, and we know the length of the opposite and adjacent side of this angle. So, in fact, we can just find this angle by using trigonometry.

We need to recall the tangent of an angle in a right-angled triangle will be equal to the length of its opposite side divided by the length of its adjacent side. So by using our right-angled triangle with the angle πœ‹ minus πœƒ, we get the tan of πœ‹ minus πœƒ is equal to seven divided by four. And remember, we’re trying to find the value of πœƒ. So we need to take the inverse tangent of both sides of this equation. Doing this, we get that πœ‹ minus πœƒ is equal to the inverse tan of seven divided by four.

Finally, all we need to do is rearrange this equation to make πœƒ the subject. We’ll add πœƒ to both sides of this equation and then subtract the inverse tangent of seven over four from both sides of the equation. Doing this, we get that πœƒ is equal to πœ‹ minus the inverse tan of seven over four. And remember, we need to set our calculator into radians mode. Doing this to two decimal places, we get that πœ‹ minus the inverse tan of seven over four is approximately equal to 2.09. Therefore, we were able to show to two decimal places the angle the line segment 𝑂𝐴 makes with the positive π‘₯-axis πœƒ is equal to 2.09.

And now, we’re given some information about the point 𝐴. We’re told that the point 𝐴 can be expressed in polar form as π‘Ÿ, πœƒ. And we just calculated the values of π‘Ÿ and πœƒ. And of course, we know that polar forms for a point are not unique. We need to determine which of the following is also a legitimate polar form of our point 𝐴. The first thing we notice is options (a), (b), and (d) all contain a positive π‘Ÿ, whereas options (c) and (e) contain negative π‘Ÿ. So let’s start by recalling what makes two polar coordinates equal.

One way of having two sets of polar coordinates represent the same point is to have the same value of π‘Ÿ. However, they have two different values of πœƒ which represent the same angle. And we know how to do this. For example, instead of using the angle πœƒ, we can instead take one full turn around our π‘₯-axis and then another turn at πœƒ. And we also know that one full turn counterclockwise around from the positive π‘₯-axis is two πœ‹. So this angle will be πœƒ plus two πœ‹. And in fact, we could do any number of full turns around our axis and then an angle of πœƒ. Or alternatively, we could instead take full turns clockwise.

And we know these would instead be represented by negative angles. And this gives us our first relationship between polar coordinates. We know the point π‘Ÿ one, πœƒ one will be the same as the point π‘Ÿ one, πœƒ one plus two πœ‹π‘› for any integer 𝑛. This just tells us we can take 𝑛 full turns clockwise or counterclockwise before we measure our angle. And it won’t change the value of our polar coordinate. In our case, the point 𝐴 has polar coordinates π‘Ÿ, πœƒ. So changing π‘Ÿ one to π‘Ÿ and πœƒ one to πœƒ gives us the following expression.

We know that π‘Ÿ, πœƒ will be equal to π‘Ÿ, πœƒ plus two πœ‹π‘› for any integer 𝑛. In other words, we can add or subtract any even integer multiple of πœ‹ from our angle. It won’t change the point. We can see that option (a) does this. We subtract two πœ‹ from our angle of πœƒ. In other words, our value of 𝑛 is equal to negative one, or we’re taking one extra full turn clockwise around our axis. We can see that option (b) does not do this. We would need to choose 𝑛 equal to negative one-half. Or alternatively, we’re only taking half a turn extra clockwise.

We can also see that option (d) can’t be correct. Our value of 𝑛 would have to be three over two. In other words, we’re taking an extra one and a half turns in the counterclockwise direction. Of course, we can’t use this to determine whether option (c) or option (e) are correct because both of these have negative π‘Ÿ as their first ordinate.

To determine whether options (c) and (e) represent point 𝐴, there’s one more polar relation we need to recall. We need to remember that the point π‘Ÿ, πœƒ will be the same as the point negative π‘Ÿ, πœƒ plus two 𝑛 plus one times πœ‹ for any integer 𝑛. In fact, we can reason this in exactly the same way we did before. This time, our first ordinate is equal to negative π‘Ÿ. And when this value is negative, this just tells us we need to point our line segment 𝑂𝐴 in the opposite direction. But we can then see if we rotate this vector, πœ‹ radians, either clockwise or counterclockwise, we’ll end up back where we started. Then just as we did before, taking full turns clockwise or counterclockwise won’t change the value of our polar coordinate.

So this time, when our value of π‘Ÿ was negative, we instead showed you can add or subtract odd integer multiples of πœ‹ to our value of πœƒ. So we can now use this to determine whether options (c) and (e) represent our point 𝐴. In option (c), we can see that π‘Ÿ is negative. However, we’re subtracting an even integer multiple of point. So option (c) does not represent the same point. Similarly, in option (e), we can see that π‘Ÿ is negative and we’re adding an even integer multiple of πœ‹. So option (e) also can’t be correct. So in actual fact, out of all five of these options, only option (a) was the same as our point 𝐴.

Therefore, given that the point 𝐴 had rectangular coordinates negative four, seven, we were able to show the distance of the point 𝐴 from the origin which we called π‘Ÿ is equal to the square root of 65. The angle that the line segment 𝑂𝐴 makes with the positive π‘₯-axis which we called πœƒ to two decimal places in radians was 2.09. And of the five listed options, only option (a) which was π‘Ÿ, πœƒ minus two πœ‹ had the same representation in polar coordinates as point 𝐴.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.