Video: Conversion between Rectangular and Polar Equations

In this video, we will learn how to convert equations from polar to rectangular form and vice versa.

10:47

Video Transcript

In this video, we’ll learn how we can use our understanding of polar and Cartesian coordinates to change between equations in polar and Cartesian form. We’ll consider how these techniques might help us to recognise graphs of equations written in polar form by converting to Cartesian or rectangular form and interpreting from there.

Remember, the polar coordinate system is a way of describing points in a plane by their distance from the origin or the pole and the angle the line joining this point of the origin makes with the positive horizontal axis measured in a counterclockwise direction. These are of the form π‘Ÿπœƒ, where π‘Ÿ is the distance from the origin to that point and πœƒ is that angle.

We 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 these equations are suitable for all values of π‘Ÿ and πœƒ. The converse formulae are π‘Ÿ squared equals π‘₯ squared plus 𝑦 squared and tan πœƒ is equal to 𝑦 divided by π‘₯.

Now in this case, we need to be a little bit careful establishing the value of πœƒ because it works beautifully for coordinates plotted in the first quadrant. But for other quadrants, the calculator can give us an incorrect value. And we do have a set of rules that we can follow for calculating the exact value of πœƒ. However, we don’t really need this formula in this video. We’re looking to establish how we can convert between polar equations, one where π‘Ÿ is some function of πœƒ, and Cartesian or rectangular equations, where 𝑦 is some function of π‘₯. We do, however, use the other three formulae to perform these conversions. Let’s see what that might look like.

Convert the equation π‘₯ squared plus 𝑦 squared equals 25 into polar form.

Remember, we convert polar coordinates to Cartesian or rectangular coordinates using the formulae π‘₯ equals π‘Ÿ cos πœƒ and 𝑦 equals π‘Ÿ sin πœƒ. And these are suitable for all values of π‘Ÿ and πœƒ. In our original equation, we’ve got π‘₯ squared and 𝑦 squared. So let’s use our formulae for π‘₯ and 𝑦 to generate expressions for π‘₯ squared and 𝑦 squared in terms of π‘Ÿ and πœƒ.

Since π‘₯ is equal to π‘Ÿ cos πœƒ, it follows that π‘₯ squared must be π‘Ÿ cos πœƒ all squared, which we can distribute and say that π‘₯ squared is equal to π‘Ÿ squared times cos squared πœƒ. Similarly, we see that 𝑦 squared must be equal to π‘Ÿ sin πœƒ all squared, which is equal to π‘Ÿ squared sin squared πœƒ.

Now our original equation says that the sum of these is equal to 25. So we can say that π‘Ÿ squared cos squared πœƒ plus π‘Ÿ squared sin squared πœƒ equals 25. Our next step is to factor π‘Ÿ squared on the left-hand side of this equation. So π‘Ÿ squared times cos squared πœƒ plus sin squared πœƒ equals 25. But why did we do this?

Well, here is where it’s useful to know some of our trigonometric identities by heart. We know that cos squared πœƒ plus sin squared πœƒ is equal to one for all values of πœƒ. So we can replace cos squared πœƒ plus sin squared πœƒ in our equation with one. So π‘Ÿ squared times one equals 25. Well, we don’t need this one. π‘Ÿ squared is simply equal to 25. We solve this equation by taking the square root of both sides. And we find that π‘Ÿ is equal to five.

Remember, we would usually take both the positive and negative square root of 25. But since π‘Ÿ represents a length, we don’t need to. π‘₯ squared plus 𝑦 squared equals 25 is the same as π‘Ÿ equals five in polar form.

Now if we think about what we know about the equation π‘₯ squared plus 𝑦 squared equals 25 and polar coordinates, this makes a lot of sense. The equation π‘₯ squared plus 𝑦 squared equals 25 represents a circle whose centre lies at the origin and whose radius is the square root of 25, or five. We can also think about what the equation π‘Ÿ equals five means in polar form. Well, it must be all points whose distance from the origin is five units.

Now of course, if we think right back to what we know about locus or loci, it follows that this form is a circle whose centre is the origin and whose radius is five. Let’s now look at converting a polar equation into rectangular form.

Convert the polar equation π‘Ÿ equals four cos πœƒ minus six sin πœƒ to the rectangular form.

Remember, we convert from polar coordinates to Cartesian coordinates or rectangular coordinates using the following formulae. π‘₯ is equal to π‘Ÿ cos πœƒ and 𝑦 is equal to π‘Ÿ sin πœƒ. And these are suitable for all values of π‘Ÿ and πœƒ. Our aim is going to be to manipulate each of our equations so that we have an equation for cos πœƒ and sin πœƒ.

Well, if we divide both sides of our first equation by π‘Ÿ, we see that cos πœƒ is equal to π‘₯ over π‘Ÿ. Similarly, dividing through by π‘Ÿ in our second equation, and we find sin πœƒ equals 𝑦 over π‘Ÿ. We can then replace cos πœƒ with π‘₯ over π‘Ÿ and sin πœƒ with 𝑦 over π‘Ÿ in our original polar equation. And we see that π‘Ÿ is equal to four times π‘₯ over π‘Ÿ minus six times 𝑦 over π‘Ÿ. This simplifies to four π‘₯ over π‘Ÿ minus six 𝑦 over π‘Ÿ.

We’re next going to multiply everything through by π‘Ÿ. And we find that π‘Ÿ squared equals four π‘₯ minus six 𝑦. Now we’re obviously not quite done. We want to convert to rectangular form. This is usually of the form 𝑦 is equal to some function of π‘₯, although we’re essentially looking for an equation with π‘₯ and 𝑦 as its only variables. So we recall the other conversion formulae that we use to convert Cartesian to polar coordinates. That’s π‘Ÿ squared equals π‘₯ squared plus 𝑦 squared. We should be able to see now that we can replace π‘Ÿ squared with π‘₯ squared plus 𝑦 squared. So π‘₯ squared plus 𝑦 squared equals four π‘₯ minus six 𝑦.

We’re almost there. You might recognise this equation. We’re going to manipulate it by completing the square. We subtract four π‘₯ from both sides and add six 𝑦. Then we’re going to complete the square for π‘₯ and 𝑦. We halve the coefficient of π‘₯ β€” that gives us negative two β€” and then subtract negative two squared. So we subtract four. Similarly, we halve the coefficient of 𝑦 to get three and then subtract three squared, which is nine. And of course, this is all equal to zero. Negative four minus nine is negative 13. So we add 13 to both sides of our equation. And in rectangular form, our equation is π‘₯ minus two all squared plus 𝑦 plus three all squared equals 13.

Now this is actually a really useful technique as it tells us something about the shape of the graph. We can’t easily spot what the shape of the graph given by π‘Ÿ equals four cos πœƒ minus six sin πœƒ looks like. But we do know that a circle whose centre lies at π‘Ž, 𝑏 and whose radius is π‘Ÿ has the equation π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared equals π‘Ÿ squared. So our polar equation, which also has the rectangular form π‘₯ minus two all squared plus 𝑦 plus three all squared equals 13, must be a circle whose centre is at two, negative three and whose radius is the square root of 13. Let’s have a look at a similar example.

Consider the rectangular equation π‘₯ squared minus 𝑦 squared equals 25. Convert the given equation to polar form. The second part of this question asks us, which of the following is the sketch of the equation?

So we begin by recalling that we can convert from polar to Cartesian coordinates using the formulae π‘₯ equals π‘Ÿ cos πœƒ and 𝑦 equals π‘Ÿ sin πœƒ. Our equation contains π‘₯ squared and 𝑦 squared. So let’s square each of these formulae. And when we do, we find that π‘₯ squared equals π‘Ÿ squared cos squared πœƒ and 𝑦 squared equals π‘Ÿ squared sin squared πœƒ. We know the difference of these is equal to 25. That’s our rectangular equation. So π‘Ÿ squared cos squared πœƒ minus π‘Ÿ squared sin squared πœƒ equals 25.

We can then factor π‘Ÿ squared. So π‘Ÿ squared times cos squared πœƒ minus sin squared πœƒ equals 25. But we know that cos two πœƒ is equal to cos squared πœƒ minus sin squared πœƒ. So let’s replace cos squared πœƒ minus sin squared πœƒ with cos two πœƒ. That tells us that π‘Ÿ squared times cos two πœƒ equals 25. And we can then divide both sides of this equation by cos two πœƒ. Now of course, one over cos πœƒ is sec πœƒ. So we find that π‘Ÿ squared is equal to 25 sec two πœƒ.

For part two, we need to identify which of the following is a sketch of the equation. Now we wouldn’t be particularly easy to sketch the graph of π‘Ÿ squared equals 25 sec two πœƒ. But we do know the general form of the graph whose equation is π‘₯ over π‘Ž all squared minus 𝑦 over 𝑏 all squared equals one. It’s a standard hyperbola, centred at the origin with vertices at plus or minus π‘Ž, zero and covertices at zero, plus or minus 𝑏.

Let’s rearrange our equation to equate it to one. To do that, we divide everything through by 25. And since 25 is five squared, we can write this as π‘₯ over five all squared minus 𝑦 over five all squared equals one. We know we have a standard hyperbola with vertices at plus or minus five, zero. And in fact, there’s only one graph that satisfies this criteria. It’s graph A. Of course, it is useful to know that if we were struggling, we could even try substituting some values of π‘₯ or 𝑦 in and sketching the order pairs.

Let’s now have a look at another example which involves sketching the graph.

Sketch the graph of π‘Ÿ equals two csc πœƒ.

Here we have a polar equation. And it’s not particularly easy to spot what the graph of this function might look like. So instead, we’re going to convert into rectangular form first. We recall that csc πœƒ is one over sin πœƒ. We also know that one of the formulae we use to convert from polar to Cartesian form is the formula 𝑦 equals π‘Ÿ sin πœƒ. Dividing through by π‘Ÿ, and we obtain the second formula is equivalent to saying sin πœƒ equals 𝑦 divided by π‘Ÿ. So csc πœƒ must be equivalent to one over 𝑦 over π‘Ÿ.

Well, when we divide by a fraction, we multiply by the reciprocal of that fraction. So we can actually say that csc πœƒ must be equal to π‘Ÿ over 𝑦. By replacing csc πœƒ with π‘Ÿ over 𝑦 in our original equation, we find that π‘Ÿ is equal to two times π‘Ÿ over 𝑦. Let’s divide through by π‘Ÿ. We get one equals two over 𝑦. Next, we’ll multiply by 𝑦. And we find in rectangular form the equation is 𝑦 equals two. And of course, we can now easily sketch this. It’s simply the horizontal line that passes through the 𝑦-axis at two. This is a great demonstration of when converting to rectangular form can make sketching a graph given in polar form much more simple.

In this video, we’ve seen that, by using the formulae for converting between polar and Cartesian coordinates, we can convert quite easily between polar and rectangular equations. We also saw that this process can help us sketch more complicated graphs given in polar form.

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