Video: Parametric Equations and Curves in Two Dimensions

In this video, we will learn how to graph curves given by a pair of parametric equations in 2D.

12:06

Video Transcript

By this stage, you should feel confident in sketching curves to find by the equation 𝑦 equals 𝑓 of π‘₯. You might look for asymptotes, intercepts, critical points, and intervals of increase and decrease when doing so.

In this lesson, we’re going to learn how to graph curves defined using a third parameter 𝑑. We see that these curves are defined parametrically by the equations π‘₯ equals 𝑓 of 𝑑 and 𝑦 equals 𝑔 of 𝑑. Imagine that a particle moves through the π‘₯𝑦-plane along a curve 𝑐, as shown. It would be impossible to define this curve in the usual way, 𝑦 as some function of π‘₯, as it fails the vertical line test. That is to say, each unique input can have more than one output. In other words, for a value of π‘₯ equals π‘Ž, we can have more than one output 𝑦.

So instead, we look to find alternative ways to describe curves of this form. And we do so by introducing a new parameter for time 𝑑. And our π‘₯- and 𝑦-coordinates are functions of time such that π‘₯ is 𝑓 of 𝑑 β€” π‘₯ is a function in time β€” and 𝑦 is 𝑔 of 𝑑 β€” 𝑦 is a different function in time. Each point on our curve will now be described by an ordered pair, 𝑓 of 𝑑, 𝑔 of 𝑑. And as 𝑑 varies, we trace our curve 𝑐, which we call a parametric curve. A key skill is being able to sketch such curves. And we do so by plotting key points. There are, of course, a few curves whose shape we can learn. And we’ll consider those too.

Consider the parametric equations π‘₯ of 𝑑 equals 𝑑 squared plus two and 𝑦 of 𝑑 equals three 𝑑 minus one, which is greater than negative two and less than one. Which of the following is the sketch of the given equation?

Here we’ve been given a pair of parametric equations and asked to sketch the curve over the open interval for 𝑑 from negative two to one. We’re going to need to begin by finding some coordinate pairs that satisfy our parametric equations. And whilst we don’t want to include 𝑑 equals negative two and 𝑑 equals one, we know that 𝑑 approaches both of these values. So we’ll use direct substitution to find the coordinate pair that our curve approaches at the end points of our open interval.

Now we’re just looking to identify which of the graphs is the sketch of our equation. So I’ve chosen seven values of 𝑑. If we were looking to sketch the graph from scratch, we may choose more values of 𝑑, perhaps choosing to go up in intervals of 0.25 as supposed to 0.5.

To find our π‘₯- and 𝑦-coordinates, we’re going to substitute each value of 𝑑 into the parametric equations. For example, when 𝑑 is negative two, π‘₯ is equal to negative two squared plus two, which is six. And when 𝑑 is equal to negative two, 𝑦 is equal to three times negative two minus one, which is negative seven. When 𝑑 is equal to negative 1.5, π‘₯ is equal to negative 1.5 squared plus two, which is 4.25. And when 𝑑 is equal to negative 1.5, 𝑦 is equal to three times negative 1.5 minus one, which gives us a 𝑦-value of negative 5.5.

We continue this process, substituting 𝑑 equals negative one into π‘₯ to get three and into 𝑦 to get negative four. When we substitute 𝑑 equals negative 0.5, we get π‘₯ equals 2.25 and 𝑦 equals negative 2.5. When 𝑑 is zero, our coordinate pair is two, negative one. And the final two values of 𝑑 give us coordinate pairs of 2.25, 0.5 and three, two.

We could now plot each ordered pair on a pair of axes. But of course, we’re looking to find which of the given sketches represents our pair of parametric equations. The first coordinate that we’re going to look to find is six, negative seven. That’s here. We then look to find 4.25, negative 5.5, which is around here. And is in fact, as we would expect, the remaining five coordinates do indeed lie on this line sketched.

Notice how the arrows describe the direction in which the graph is sketched. So our curve is sketched through increasing values of 𝑑 from negative two to 𝑑 equals one. The sketch representing the parametric equations π‘₯ of 𝑑 equals 𝑑 squared plus two and 𝑦 of 𝑑 equals three 𝑑 minus one is C.

Now, in fact, it looks like we have part of a parabola. We could use simultaneous equations to eliminate 𝑑. And when we do, we do indeed obtain the equation for our parabola, or be it a parabola on its side. There are a number of special parametric equations for which we can learn the shape.

Consider the parametric equations π‘₯ of 𝑑 equals two sin 𝑑 and 𝑦 of 𝑑 equals three cos 𝑑, where 𝑑 is greater than zero and less than three πœ‹. Which of the following is the sketch of the given equations?

Here we’ve been given a pair of parametric equations and asked to find the sketch of the curve over the open interval for 𝑑 from zero to three πœ‹. We’re going to need to find some coordinate pairs that satisfy our parametric equations. Now whilst we don’t technically want to include 𝑑 equals zero and 𝑑 equals three πœ‹, we know that 𝑑 approaches both of these values. So we’ll use direct substitution to find the coordinate pair that our curve approaches at the end points on our open interval.

Let’s choose intervals of πœ‹ by two radians. Now if we were actually looking to sketch the curve, we may wish to choose, say, πœ‹ by four as our subinterval. But we’re just simply looking to compare our coordinates to the given graphs. To find the first ordered pair, we substitute 𝑑 equals zero into each of our parametric equations. That gives us π‘₯ equals two sin zero and 𝑦 equals three cos zero. Our first ordered pair is zero, three.

We then substitute 𝑑 equals πœ‹ by two. And we get π‘₯ equals two sin of πœ‹ by two and 𝑦 equals three cos of πœ‹ by two. That gives us π‘₯ equals two and 𝑦 equals zero. We continue the pattern by substituting 𝑑 equals πœ‹ to get two sin πœ‹ for π‘₯ and three cos πœ‹ for 𝑦, giving us an ordered pair of zero, negative three. Substituting 𝑑 equals three πœ‹ by two into each equation, and we get π‘₯ equals negative two and 𝑦 equals zero. For 𝑑 equals two πœ‹, we get the ordered pair zero, three. And our last two ordered pairs occur when 𝑑 equals five πœ‹ by two and three πœ‹. And they are two, zero and zero, negative three, respectively.

Let’s compare these to each of our sketches. And we need to ensure that we move in increasing values of 𝑑. That is, we begin at 𝑑 equals zero and move all the way through to 𝑑 equals three πœ‹. That leaves us with either B or C. And in fact, we move in a clockwise direction. So actually, we’re interested in C. And you might have noticed that the values themselves repeat. It should be quite clear now that, due to the nature of the shape of our curve, this pattern will continue forever. The sketch of the equations π‘₯ of 𝑑 equals two sin 𝑑 and 𝑦 of 𝑑 equals three cos 𝑑 is C.

Now can you predict what our graph would look like if our equation for π‘₯ was in fact three sin 𝑑? The values for 𝑑 is πœ‹ by two, 𝑑 is three πœ‹ by two, and 𝑑 is five πœ‹ by two would change to three, negative three, and three, respectively. Comparing these to our curves, we see that that gives us a circle. Now we can confirm this by eliminating 𝑑 and using the identity sin squared π‘₯ plus cos squared π‘₯ equals one. We square both our equations for π‘₯ and 𝑦. And we see that π‘₯ squared plus 𝑦 squared is equal to nine sin squared 𝑑 plus nine cos squared 𝑑. By factoring nine, we can rewrite this right-hand side as nine sin squared 𝑑 plus cos squared 𝑑. And since sin squared 𝑑 plus cos squared 𝑑 will be equal to one, this becomes nine times one, which is nine.

This corresponds to the general equation for a circle with the center at the origin and a radius π‘Ÿ. And we see that, comparing our graph to our equation, we do indeed have a circle with a center at the origin and a radius of three. In general, we can say that the parametric equations of a circle centered at the origin with the radius π‘Ÿ are π‘₯ equals π‘Ÿ cos 𝑑 and 𝑦 equals π‘Ÿ sin 𝑑. 𝑑 takes values from zero to two πœ‹. And after this, the pattern repeats.

This can be further generalized. And we see that the parametric equations of a circle with the center at π‘₯ nought, 𝑦 nought and a radius π‘Ÿ are π‘₯ equals π‘₯ nought plus π‘Ÿ cos 𝑑 and 𝑦 equals 𝑦 nought plus π‘Ÿ sin 𝑑, for 𝑑 is greater than or equal to zero and less than or equal to two πœ‹. In our final example, we’re going to consider a very special curve represented by parametric equations. It’s called a cardioid.

Sketch the curve defined by the parametric equations π‘₯ equals two cos 𝑑 minus cos two 𝑑 and 𝑦 equals two sin 𝑑 minus sin two 𝑑, where 𝑑 is greater than or equal to zero and less than or equal to two πœ‹.

Here we’ve been given a pair of parametric equations and asked to sketch the curve over the closed interval for 𝑑 from zero to two πœ‹. We’ll begin by finding some coordinate pairs which satisfy our parametric equations. And since we’re sketching the curve this time rather than just identifying the graph, I’ve chosen values of 𝑑 at subintervals of πœ‹ by four radians. So that’s zero, πœ‹ by four, πœ‹ by two, three πœ‹ by four, all the way through to two πœ‹.

We begin by substituting 𝑑 equals zero into π‘₯. That’s two cos of zero minus cos of two times zero, which is equal to one. We then substitute zero into the equation for 𝑦. That’s two sin of zero minus sin of two times zero, which is zero. By substituting 𝑑 equals πœ‹ by four into π‘₯, we get two cos of πœ‹ by four minus cos of two times πœ‹ by four. And two times πœ‹ by four is πœ‹ by two. This is simply root two.

Repeating this process for 𝑦, we get two sin of πœ‹ by four minus sin of πœ‹ by two, which is minus one plus root two. In fact, it will be easier to plot these if they’re in decimal form. So rounded to three decimal places, we have 1.414 and 0.414. The remaining coordinate pairs that we’re interested in are one, two; negative 1.414, 2.414; negative three, zero; and so on.

Plotting these on a pair of coordinate axes gives us something that looks a little like this. We then join the points as shown and then add arrows to show the direction in which the curve is traced. Remember, we do this by following the values of 𝑑 from smallest to largest. And we sketched the curve defined by the parametric equations π‘₯ equals two cos 𝑑 minus cos two 𝑑 and 𝑦 equals two sin 𝑑 minus sin two 𝑑. This curve has a special name. It’s called a cardioid.

You might notice that that name is similar to the word β€œcardiac,” meaning heart. And it’s a fascinating shape that can be created by following the locus of a point on a circle as that circle rolls around another circle of equal radius.

Now we generally use polar form to represent these. But in general, the equation of what we call a horizontal cardioid, like this one, in parametric form is π‘₯ equals π‘Ž times two cos 𝑑 minus cos two 𝑑, 𝑦 equals π‘Ž two sin 𝑑 minus sin two 𝑑. Reversing the π‘₯ and 𝑦 so that π‘₯ was equal to π‘Ž two sin 𝑑 minus sin two 𝑑 and 𝑦 was equal to π‘Ž two cos 𝑑 minus cos two 𝑑 would give us a vertical cardioid. That’s a curve that looks a little bit like a heart in the usual orientation we would expect.

In this video, we’ve learned that we can sketch a parametric curve by making a table of values and drawing the direction in which the curve is traced. We saw that the parametric equations of a circle with center π‘₯ nought, 𝑦 nought and a radius π‘Ÿ are π‘₯ equals π‘₯ nought plus π‘Ÿ cos 𝑑 and 𝑦 equals 𝑦 nought plus π‘Ÿ sin 𝑑 for values of 𝑑 in the closed interval zero to two πœ‹. And finally, we saw how a fascinating shape called a cardioid occurs with equations of the form given. And that if we reverse the equations for π‘₯ and 𝑦, we end up with a cardioid with a different orientation.

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