Video: Conversion between Parametric and Rectangular Equations

In this video, we will learn how to convert from the parametric form of an equation to its equivalent rectangular form and vice versa.

12:31

Video Transcript

In this video, we’re going to learn how to convert between parametric and rectangular or Cartesian equations. We’ll learn how this process can help us to sketch the curve given by a pair of parametric equations and how to parameterize circles with centres at the origin and line segments between two points.

Let’s begin by recalling that a Cartesian equation or a rectangular equation is one given in the variables π‘₯ and 𝑦. It’s usually given as 𝑦 is equal to some function of π‘₯, although that’s not always the case. A pair of parametric equations are given in terms of a third variable, usually 𝑑, such that π‘₯ is equal to some function of 𝑑 and 𝑦 is equal to some other function of 𝑑. And one method we have for converting from parametric to rectangular form is rearranging to eliminate the variable 𝑑.

Let’s see what that might look like.

Convert the parametric equations π‘₯ equals 𝑑 squared plus two and 𝑦 equals three 𝑑 minus one to rectangular form.

Here, we have a pair of parametric equations. We have π‘₯ is equal to some function of 𝑑 and 𝑦 is equal to some other function of 𝑑. To convert parametric equations to rectangular form, we need to find a way to eliminate the 𝑑. So looking at our equations, we can see that we can rearrange the equation in 𝑦 to make 𝑑 the subject. We begin by adding one to both sides. And then, we divide through by three. So we see that 𝑑 is equal to 𝑦 plus one all over three.

Now, we go back to our equation for π‘₯. We replace 𝑑 with 𝑦 plus one over three. And we find that π‘₯ equals 𝑦 plus one over three all squared plus two. And there will be certain circumstances where we’re required to distribute the parentheses and simplify. In this case, that’s not necessary. And so, we’re finished. We’ve converted the parametric equations π‘₯ equals 𝑑 squared plus two and 𝑦 equals three 𝑑 minus one into rectangular or Cartesian form. It’s π‘₯ equals 𝑦 plus one over three all squared plus two.

In our next example, we’re going to look at how trigonometric identities can help us to eliminate the variable 𝑑.

Find the Cartesian equation of the curve defined by the parametric equations π‘₯ equals two plus cos 𝑑 and 𝑦 equals four cos of two 𝑑.

Remember, a Cartesian equation is one which contains only the variables π‘₯ and 𝑦. So we’re going to need to find a way to eliminate our third variable 𝑑 from our parametric equations. And at first glance, it doesn’t seem to be a nice way to do so. But we can begin by recalling some trigonometric identities. We have cos of two 𝑑 in our second parametric equation. And we know that cos of two 𝑑 is equal to two times cos squared 𝑑 minus one. This means we can rewrite our equation for 𝑦 as four times two cos squared 𝑑 minus one.

Next, we’ll look to rearrange our equation for π‘₯ to make cos of 𝑑 the subject. Once we’ve done that, we’ll be able to find an expression for cos squared 𝑑 in terms of π‘₯. We can subtract two from both sides. And we see that π‘₯ minus two equals cos of 𝑑. Then, by squaring both sides of this equation, we find that cos squared 𝑑 is equal to π‘₯ minus two all squared. And so, we’re now able to replace cos squared 𝑑 with π‘₯ minus two squared. That gives us 𝑦 equals four times two times π‘₯ minus two all squared minus one.

We distribute this first pair of parentheses. And we find that π‘₯ minus two all squared is equal to π‘₯ squared minus four π‘₯ plus four. We distribute again by multiplying each of these terms by two and then simplifying: eight minus one is seven. Well, finally, we distribute one more time by multiplying each term of two π‘₯ squared minus eight π‘₯ plus seven by four. And we find the Cartesian equation of the curve defined by parametric equations π‘₯ equals two plus cos 𝑑 and 𝑦 equals four cos of two 𝑑 is 𝑦 equals eight π‘₯ squared minus 32π‘₯ plus 28.

Now, we might be able to see that this process can be really useful in helping us to sketch curves given in parametric form. For example, in this case, we were given a pair of parametric equation. It might now be instantly obvious what the curve defined by these parametric equations looks like. But by writing it in Cartesian form, we can see that we have a quadratic curve with a positive leading coefficient of π‘₯. So we know we’re going to have that typical parabola shape. And we can use our usual methods for sketching quadratic curves to sketch the curve defined by these parametric equations.

Let’s see what the whole process might look like.

Convert the parametric equations π‘₯ equals three cos 𝑑 and 𝑦 equals three sin 𝑑 to rectangular form.

Remember, the rectangular form of an equation is one which contains the variables π‘₯ and 𝑦 only. So we’ll need to find a way to eliminate the third variable 𝑑 from our parametric equations. And at first glance, it doesn’t seem to be a nice way to do so. But let’s recall some trigonometric identities. We know that cos squared πœƒ plus sin squared πœƒ is equal to one. So let’s begin by simply squaring our expressions for π‘₯ and 𝑦.

With π‘₯, we get π‘₯ squared equals three cos 𝑑 all squared, which is equal to nine cos squared 𝑑. And so, we can say that cos squared 𝑑 must be equal to π‘₯ squared over nine. Similarly, we can say that 𝑦 squared is equal to three sin 𝑑 all squared. We distribute the parentheses and we find that 𝑦 squared is nine times sin squared 𝑑. And then, we divide through by nine. And we see that sin squared 𝑑 is equal to 𝑦 squared over nine.

Then, if we replace πœƒ with 𝑑 in our identity, remember that doesn’t change the identity. We see that we can replace cos squared 𝑑 with π‘₯ squared over nine. We can replace sin squared 𝑑 with 𝑦 squared over nine. And then, this is all equal to one. We next multiply through by nine. And we find that the rectangular form of our equation is equal to π‘₯ squared plus 𝑦 squared equals nine.

Hence sketch the curve described by this pair of parametric equations.

Well, we’ve just seen that this pair of parametric equations π‘₯ equals three cos 𝑑 and 𝑦 equals three sin 𝑑 is π‘₯ squared plus 𝑦 squared equals nine in rectangular form. And so, we recall that the equation of a circle whose centre is at π‘Ž, 𝑏 and whose radius is π‘Ÿ is π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared equals π‘Ÿ squared. We compare our equation β€” that’s π‘₯ squared plus 𝑦 squared equals nine. And in doing so, we see that π‘Ž and 𝑏 simply have to be zero. We also find that π‘Ÿ squared is equal to nine. Well, if we square root this, we find that π‘Ÿ is equal to three. And so, we see that the parametric equations π‘₯ equals three cos 𝑑 and 𝑦 equals three sin 𝑑 give a circle centred at zero, zero β€” that’s the origin β€” with a radius of three. That might look a little something like this.

We do, however, need to work out the direction which our curve is being sketched. So let’s work out the value of our π‘₯- and 𝑦-coordinates when 𝑑 is equal to zero. When 𝑑 is equal to zero, π‘₯ is equal to three times cos of zero, which is simply three. And when 𝑑 is equal to zero, 𝑦 is equal to three sin of zero, which is zero. So we would begin by plotting the point with Cartesian coordinates three, zero. Let’s next choose 𝑑 is equal to πœ‹ by two.

Now, we could, of course, choose 𝑑 equals one. But since we’re working with trigonometric expressions, πœ‹ by two seems to make a little more sense. So when 𝑑 is equal to πœ‹ by two, π‘₯ is equal to three cos of πœ‹ by 𝑑, which is zero and 𝑦 is equal to three times sin of πœ‹ by two, which is equal to three. So the second point we plot has Cartesian coordinates zero, three. We can, therefore, see that we move in a counterclockwise direction while sketching the circle.

Next, we’re going to look at how we can go up from rectangular into parametric form.

Convert the rectangular equation π‘₯ squared plus 𝑦 squared equals 25 to parametric form.

Let’s begin by recalling what we actually know about this equation. We know that a circle whose centre is at the origin and whose radius is π‘Ÿ can be given by the Cartesian equation π‘₯ squared plus 𝑦 squared equals π‘Ÿ squared. By rewriting our rectangular equation as π‘₯ squared plus 𝑦 squared equals five squared, we see that we have a circle whose centre is at the origin and whose radius is five units. And so, I’ve sketched that on the π‘₯𝑦-plane, as shown.

We’re looking to convert this equation to parametric form. So we know that a pair of parametric equations describe the π‘₯- and 𝑦-coordinates in terms of a third parameter, 𝑑. So let’s pick a general point π‘₯, 𝑦. I’m going to choose this one in the first quadrant. We can drop in a right triangle, whose height is 𝑦 units and whose width is π‘₯ units. And then, we can label the included angle 𝑑. Since the radius of the circle is five units, we know that the hypotenuse of our triangle is five. Then, using standard conventions, we label the sides of our triangle. We have the adjacent; that’s π‘₯. We have the opposite; that’s 𝑦. We have the hypotenuse; that’s five.

We also know that in rectangular trigonometry, sin πœƒ is equal to the opposite divided by the hypotenuse and cos πœƒ is equal to the adjacent over the hypotenuse. So we can say that sin 𝑑 equals 𝑦 over five and cos 𝑑 is equal to π‘₯ over five. We can multiply through by five for both of our equations. And we find that 𝑦 is equal to five sin 𝑑 and π‘₯ equals five cos 𝑑. So for any point in our circle, the 𝑦-coordinate is given by five sin of 𝑑 and the π‘₯-coordinate is given by five cos of 𝑑. Now moving in a counterclockwise direction from the positive horizontal axis, we see that as π‘₯ increases from zero, it generates our corresponding π‘₯- and 𝑦-coordinates. Then, our equations are π‘₯ equals five cos 𝑑 and 𝑦 equals five sin 𝑑.

Now, in these previous two examples, we’ve seen that the equation of a circle whose centre is at the origin and whose radius is π‘Ÿ is given by the equation π‘₯ squared plus 𝑦 squared equals π‘Ÿ squared. But then, we also saw we can convert this into parametric form. In this case, we had 𝑦 equals five sin 𝑑 and π‘₯ equals five cos 𝑑, where five is the radius.

In general, we can state this as a result. We define a circle as the locus of the points that satisfy the equations π‘₯ equals π‘Ÿ cos 𝑑 and 𝑦 equals π‘Ÿ sin 𝑑, where π‘₯ and 𝑦 are the coordinates of any point on the circle, π‘Ÿ is the radius, and 𝑑 is the parameter; it’s the angle subtended by the point at the circle centre.

We can also use our knowledge of simple coordinate geometry to parameterize a line segment. Let’s see what that might look like.

Consider the points 𝐴 equals negative one, one and 𝐡 equals four, two. Parameterize the segment 𝐴𝐡, where 𝑑 is greater than or equal to zero and less than or equal to one.

Let’s begin by simply sketching the diagram, showing our points and our line segment 𝐴𝐡. It looks a little something like this. Now, we know we’re going to go from left to right generally. So we’ll begin by letting 𝑑 be equal to zero at the first point, at negative one, one. And that means we need to let 𝑑 be equal to one at the other end of our interval at the point four, two. Now, we’re going to use the vector form for the equation of a line. It’s π‘Ÿ equals π‘₯ nought, 𝑦 nought plus 𝑑 times π‘Žπ‘.

Now, whilst this is an equation of a line in two dimensions, this process can be extended to working with a line in three dimensions. We’ll next say that the vector π‘Ÿ is of the form π‘₯, 𝑦. Then, we’ll combine the vectors on the right to π‘₯ nought plus π‘‘π‘Ž, 𝑦 nought plus 𝑑𝑏. And we see that the only way for the vector on the left to be equal to the vector on the right is if its component parts are equal. That is, if π‘₯ is equal to π‘₯ nought plus π‘‘π‘Ž and 𝑦 is equal to 𝑦 naught plus 𝑑𝑏. In fact, this pair of equations is called the parametric Form of the equation of a line. Now, we can use these along with the information in our question to parameterize the line segment from π‘Ž to 𝑏.

Let’s go back to what we said earlier. We said 𝑑 is equal to zero at negative one, one. When 𝑑 is equal to zero, this gives us the values for π‘₯ nought, 𝑦 nought. So π‘₯ nought must be equal to negative one and 𝑦 nought must be equal to one. So we see the π‘₯ is equal to negative one plus π‘‘π‘Ž and 𝑦 is equal to one plus 𝑑𝑏. We’ll now use the fact that when 𝑑 is equal to one, π‘₯ is equal to four and 𝑦 is equal to two. So our first equation becomes four equals negative one plus one π‘Ž. And our second equation becomes two equals one plus one 𝑏. By adding one to both sides of our first equation, we find π‘Ž to be equal to five. And by subtracting one from both sides of our second, we find 𝑏 is equal to one. And we have our parametric equations to describe the line segment 𝐴 to 𝐡 for values of 𝑑 between zero and one. They are π‘₯ equals five 𝑑 minus one and 𝑦 equals 𝑑 plus one.

In this video, we’ve seen that if we want to convert from a parametric equation to a rectangular equation, we need to find a way to eliminate 𝑑. We also saw that, in general, a circle whose centre is at the origin and whose radius is π‘Ÿ is defined by the parametric equations π‘₯ equals π‘Ÿ cos 𝑑 and 𝑦 equals π‘Ÿ sin 𝑑. And finally, we saw that the parametric equation of a line is π‘₯ equals π‘₯ nought plus π‘‘π‘Ž and 𝑦 equals 𝑦 nought plus 𝑑𝑏.

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