Video: Polar Coordinates

In this video, we will learn how to define and plot points given in polar coordinates and convert between the Cartesian and polar coordinates of a point.

12:33

Video Transcript

A coordinate system represents a point in a plane by an ordered pair of numbers called coordinates. We’re most used to working with Cartesian coordinates in an π‘₯𝑦-plane. These are directed distances from two perpendicular axes. But what if this sort of coordinate system isn’t convenient? Well, as far back as 130 BC, there are references to a Greek astronomer and mathematician, named Hipparchus, using angles and chord lengths from a pole to establish the position of stars. This system was formalised by Isaac Newton, though its name polar coordinates is attributed to Gregorio Fontana in the 18th century.

In this video, we’ll learn how to define and plot points given in polar coordinates and convert between the Cartesian and polar coordinates of a point. Let’s have a look at the system itself. We choose a point in the plane that’s called the pole. This is a bit like the origin. And we label it 𝑂. Then, we draw a half line or a ray starting at 𝑂, called the polar axis. This axis is usually drawn horizontally to the right and corresponds to the positive π‘₯-axis in a Cartesian plane. We add this point 𝑃. Then, π‘Ÿ is the distance between 𝑂 and 𝑃. Then, πœƒ is the angle, the line segment 𝑂𝑃 makes with the polar axis, measured in a counterclockwise direction. 𝑃 is then represented by the ordered pair π‘Ÿ and πœƒ. These are the polar coordinates of 𝑃.

What this means is that an angle measured in a clockwise direction will be negative. We also say that if π‘Ÿ is negative, the points π‘Ÿ, πœƒ and negative π‘Ÿ, πœƒ lie on the same line through the origin. And they’re the same distance π‘Ÿ from 𝑂, but on opposite sides. We can see that negative π‘Ÿ, πœƒ can also be described as π‘Ÿ, πœƒ plus πœ‹. And in fact, we try to avoid using negative values for π‘Ÿ wherever possible. So it, of course, describes the length from the origin. It’s also worth noting that, due to the nature of the system, we say that two points are coincident if they lie at the same point. For instance, the points two, 80 degrees and two, 440 degrees are coincident. Their value of π‘Ÿ is the same. And 440 is 360 degrees greater than 80. Essentially, we move from 80 to 440 by completing a full turn, thereby ending up at the same point. Let’s use this information to simply write the polar coordinates of some points.

Consider the points plotted on the graph. Write down the polar coordinates of 𝐢, giving the angle πœƒ in the range πœƒ is greater than negative πœ‹ and less than or equal to πœ‹.

We’re interested in the point 𝐢. And we want to know its polar coordinates. Remember, these are of the form π‘Ÿ, πœƒ. Let’s add a half line or a ray from the pole to point 𝐢. Our job is going to be to work out the value of π‘Ÿ, that’s the length of our half line, and πœƒ, the angle that this half line makes with the positive π‘₯-axis. And since we’re told that πœƒ must be greater than negative πœ‹ and less than or equal to πœ‹, we’re going to travel in a clockwise direction.

Now, π‘Ÿ is quite easy to calculate. We follow the grid around. And we see that the point is located exactly one unit from the pole. So π‘Ÿ must be equal to one. But what about the angle πœƒ? We know that a full turn is two πœ‹ radians. And half a turn is πœ‹ radians. This half a turn is split into 12 subintervals. So each subinterval must represent πœ‹ by 12 radians. Our half line travels three of these subintervals. That’s three lots of πœ‹ by 12, which is πœ‹ by four. But we’re travelling in a clockwise direction. So our value of πœƒ for the polar coordinates of 𝐢 is negative πœ‹ by four. And the polar coordinates of 𝐢 are therefore one, negative πœ‹ by four. Notice that had we travelled in a counterclockwise direction, we’d have, of course, an angle of seven, πœ‹ by four. But that’s outside of the range of πœƒ given.

Let’s just see what this might have looked like for, say, point 𝐸. This time, our point lies two units from the pole. So π‘Ÿ must be equal to two. Measuring in a counterclockwise direction from the positive π‘₯-axis, we travel eight lots of πœ‹ by 12, which is two πœ‹ by three radians. And therefore, the polar coordinates are two, two πœ‹ by three. And that’s all fine and well. But now, suppose we want to convert between polar and Cartesian coordinates. What then?

Imagine we have a point 𝑃, given by Cartesian coordinates π‘₯, 𝑦 and polar coordinates π‘Ÿ, πœƒ. We can form a right angle triangle with this information. We know that the length of the hypotenuse is π‘Ÿ-units. The height of this one is 𝑦-units. And its width is π‘₯-units. We’ll use right angle trigonometry to form expressions for π‘₯ and 𝑦 in terms of π‘Ÿ and πœƒ. We know that cos of πœƒ is equal to adjacent over hypotenuse. In this case, it’s equal to π‘₯ over π‘Ÿ. Multiplying both sides of this equation by π‘Ÿ and we find π‘₯ to be equal to π‘Ÿ cos πœƒ. Similarly, sin πœƒ is equal to opposite over hypotenuse. This time, when we multiply by π‘Ÿ, we find that 𝑦 is equal to π‘Ÿ sin πœƒ. These are the equations we use to convert from polar coordinates to Cartesian. And although they have been deduced from our diagram, where π‘Ÿ is a positive value and πœƒ is greater than zero and less than πœ‹ by two, these equations are valid for all values of π‘Ÿ and πœƒ.

For instance, suppose we wanted to convert the point with polar coordinates two, negative πœ‹ by three into Cartesian coordinates. We see that π‘Ÿ is equal to two. And πœƒ is equal to negative πœ‹ by three. This means π‘₯ must be equal to two cos of negative πœ‹ by three, which is equal to one. And 𝑦 is equal to two sin of negative πœ‹ by three, which is negative root three. In Cartesian form then, our point is given by the coordinates one, negative root three.

But what about converting in the other direction? If we want to go the other way, we use the Pythagorean theorem to find an expression for π‘Ÿ in terms of π‘₯ and 𝑦. Remember, the sum of the squares of the smaller two sides in a right angle triangle is equal to the square of the larger side. So π‘Ÿ squared is equal to π‘₯ squared plus 𝑦 squared. Or π‘Ÿ is equal to the square root of π‘₯ squared plus 𝑦 squared. We know that tan of πœƒ is equal to opposite over adjacent. In this case, that’s 𝑦 divided by π‘₯. To solve for πœƒ, we find the inverse tan of both sides. So πœƒ is equal to the inverse tan of 𝑦 over π‘₯. We do, however, need to be a little bit careful when converting from polar to Cartesian form. This formula works beautifully for coordinates plotted in the first quadrant. But for the other quadrants, a calculator will give us an incorrect value. And so we have a couple of options. We could plot the coordinates out and go from there. Or we can quote the following standard results for values of πœƒ between negative πœ‹ and πœ‹ or negative 180 and 180.

For points plotted in the first or fourth quadrant, we use the value of πœƒ generated by a calculator. That’s πœƒ is equal to the inverse tan of 𝑦 divided by π‘₯. For coordinates plotted in the second quadrant, remember, that’s the top left quadrant, we add πœ‹ radians or 180 degrees to the value of πœƒ we get from our calculator. And for points plotted in the third quadrant, we subtract πœ‹ radians or 180 degrees from the calculator value. Let’s see what this might look like.

Convert negative two, five to polar coordinates. Give the angle in radians and round to three significant figures throughout.

Remember, polar coordinates are given in the form π‘Ÿ, πœƒ. There are some general formulae that we can use to evaluate these. But we do need to be extra careful of our value for πœƒ. This is because we usually measure in a counterclockwise direction from the positive π‘₯-axis. And if we draw a sketch of our Cartesian coordinates, we see we end up with a point plotted in the second quadrant. But we have a set of rules that can help us. For points plotted in the first quadrant and the fourth quadrant, we use the calculator values. For points plotted in the second quadrant, we add πœ‹ to the value we get on our calculator. And for points plotted in the third quadrant, we subtract πœ‹ from our calculator value.

So let’s begin by working out the value of π‘Ÿ. This is essentially the length of the line segment that joins our point to the pole or the origin. It’s given by the square root of π‘₯ squared plus 𝑦 squared. And our value for π‘₯ is negative two. Our value for 𝑦 is five. So π‘Ÿ is equal to the square root of negative two squared plus five squared, which is root 29. Correct to three significant figures, that gives us a value of π‘Ÿ as 5.39. We calculate πœƒ by evaluating the inverse tan of five divided by negative two, which is negative 1.190 and so on. Our coordinate lies in the second quadrant. So we add πœ‹ radians to this value. And that gives us 1.9513, which correct to three significant figures is 1.95 radians. And so in polar form, our coordinates are 5.39, 1.95.

We’re now going to consider how to find the distance between two points given as polar coordinates. We might recall the distance formula for two Cartesian coordinates π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two. It’s the square root of π‘₯ one minus π‘₯ two squared plus 𝑦 one minus 𝑦 two squared. So how can we adapt this formula to find the distance between two polar coordinates?

Well, for two points 𝑃 one and 𝑃 two given by π‘Ÿ one cos πœƒ one, π‘Ÿ one sin πœƒ one and π‘Ÿ two cos πœƒ two, π‘Ÿ two sin πœƒ two. We can see that the distance between them will be given by the square root of π‘Ÿ one cos πœƒ one minus π‘Ÿ two cos πœƒ two squared plus π‘Ÿ one sin πœƒ one minus π‘Ÿ two sin πœƒ two squared. We distribute the parentheses and recall the fact that cos squared πœƒ plus sin squared πœƒ is equal to one. So we find that the distance is equal to the square root of π‘Ÿ one squared plus π‘Ÿ two squared minus two times π‘Ÿ one times π‘Ÿ two times cos πœƒ one cos πœƒ two plus sin πœƒ one sin πœƒ two. But then, we know that we can say that cos of πœƒ one minus πœƒ two is the same as cos πœƒ one cos πœƒ two plus sin πœƒ one sin πœƒ two. So we can rewrite our formula as the square root of π‘Ÿ one squared plus π‘Ÿ two squared minus two times π‘Ÿ one times π‘Ÿ two times cos of πœƒ one minus πœƒ two. Let’s have a look at the application of this formula.

Find the distance between the polar coordinates two, πœ‹ and three, negative three πœ‹ over four. Give your answer accurate to three significant figures.

Remember, to find the distance between two polar coordinates given by π‘Ÿ one, πœƒ one and π‘Ÿ two, πœƒ two, we use the formula the square root of π‘Ÿ one squared plus π‘Ÿ two squared minus two times π‘Ÿ one π‘Ÿ two times cos of πœƒ one minus πœƒ two. Let’s let π‘Ÿ one be two and πœƒ one be πœ‹. So π‘Ÿ two is three and πœƒ two is negative three πœ‹ by four. Then, we substitute straight into this formula. And we see the distance between them is the square root of two squared plus three squared minus two times two times three times cos of πœ‹ minus negative three πœ‹ by four. That gives us the square root of 13 minus 12 of cos of seven πœ‹ by four, which is equal to 2.124 and so on. So correct to three significant figures, we see the distance between our polar coordinates is 2.12 units.

In this video, we’ve learned that a polar coordinate is one of the form π‘Ÿ, πœƒ, where π‘Ÿ is the distance of that point away from the pole or origin and πœƒ is the angle measured in a counterclockwise direction from the positive π‘₯-axis. We saw that we can convert from polar to Cartesian form using the formulae π‘₯ equals π‘Ÿ cos πœƒ and 𝑦 equals π‘Ÿ sin πœƒ and the given formulae to convert from Cartesian back into polar form. But we also saw that we need to be really careful to establish which quadrant our point lies in to make sure we get the correct value of πœƒ. Finally, we saw that the distance between two points with polar coordinates π‘Ÿ one, πœƒ one and π‘Ÿ two, πœƒ two is given by the square root of π‘Ÿ one squared plus π‘Ÿ two squared minus two times π‘Ÿ one π‘Ÿ two cos of πœƒ one minus πœƒ two.

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