Pre Calculus Parametric Equations

Hi--welcome back to Educator.com.0000

Today, we are going to talk about parametric equations.0002

Up until this point, whenever we have looked at graphing an equation or a function, we thought in terms of input and output.0005

If we put in x for this, what will y come out to be?0010

If we plug in x at 10, y is going to be some number; if we plug in x at 2, y is going to be some number.0014

We always think of x going in and y coming out, and that is how we graph it.0020

We have the x-axis and the y-axis, so we plug in this x, and that goes to a y; we plug in this x; that goes to a y; we plug in this x; it goes to a different y.0024

And that is how we have thought of graphing so far.0031

Parametric equations are a new way to look at graphing.0033

Instead of graphing input versus output, we will base both x and y on a third new variable, a parameter.0037

Parametric equations give us a great new way to look at how time, or some other changing quantity--0044

some changing parameter--some outside thing--affects an object, such as letting us look at motion over time.0049

There are lots of other uses, but motion over time is a really commonly-used one.0055

This way of graphing has a variety of applications in science, calculus, and advanced math.0059

So, let's look at what it is: using this new idea for graphing, we can describe a set of points in the plane, a graph, as a plane curve.0065

As opposed to thinking of a graph as the x's going to y's, we can now just think of it as a bunch of points on the plane.0072

So, how do we get that bunch of points that we call a plane curve onto the plane?0078

A plane curve is created by two functions, a function f(t) and g(t), that are defined on some interval of the real numbers.0083

t is allowed to go from some set of numbers--has some set of numbers like, say, 0 up until 10.0091

And the curve is the set of points, (x,y), equal to (f(t),g(t));0096

that is, as you plug in all of the various t's that we are allowed to have, it creates a bunch of points on the graph.0102

And that is our curve--that is what we get there.0108

The equations x = f(t) and y = g(t) are called parametric equations, and t is the parameter.0110

While we use t most often for parametric equations, since they usually involve time (t for time), we can use any symbol.0119

We could use whatever symbol made sense for the specific parametric equations we were working with.0127

You might see θ show up, or it might be some other letter, depending on what we are working with.0132

The key idea of a parametric graph is that, instead of having x and y be based on each other, we base them on some outside parameter.0137

For example, as time goes on, how does the object move in the x and the y?0145

As this outside thing goes forward or goes to negative values--as this outside thing changes around--how will x and y be affected by this outside thing changing?0151

x and y are no longer directly linked; they are linked to a parameter now.0162

How is that parameter changing, affecting x and y?0166

How do we graph parametric equations?0169

Graphing a parametric equation is very similar to what we think of as normal graphing.0171

You plug in a value; then you see what point you get.0176

It is just that, instead of plugging in x to get y, like we used to, we now plug in t, and we get x and y.0178

We plug in one t, and it gives us (x,y) out of it.0185

So, if we have x = t + 1, y = 2t - 1, and these are our parametric equations,0190

we can just plug in a variety of different t-values, then plot the points that come out for those t-values; and we will have a graph.0195

So, let's start at 0: we plug in 0; then that would give us the point...x is 1; y comes out to be -1.0203

If we plug in 0 for t, we get 0 + 1 (1); it is 2t - 1 for y, so 2(0) - 1 gets us -1.0210

So, that would come out to be: at t = 0, we have (1,-1) as a point.0220

We go and we plot that: (1,-1)--we plot that point.0226

We do the same thing at time = 1: we have 1 + 1, so 2; and 2(1) - 1 comes out to be 1; so we have the point (2,1).0231

So, we go and we plot (2,1), and that is on it.0243

And we have already drawn this; so if t is allowed to vary, just have complete varying, going over everything (that red line that we see there).0246

But we could just keep plotting points like this: (2,3)...3...we would get...2 + 1 comes out to be 3; 2 times 2 minus 1 is 4 - 1, so 3; we would get (3,3).0253

If we went to negative values of t (we aren't required to only have positive values of t--0266

we weren't given any limitations on what t could be, so we have to allow for t to be anything when we are graphing it)--0270

if we plug it t = -1, -1 + 1 comes out to be 0; 2(-1) - 1 comes out to be -3; so we have (0,-3).0277

So, at this point, it becomes pretty clear that we are graphing a line,0288

although in this picture specifically, we had already seen the line.0291

But it would become clear to us that we are graphing a line; and we could graph points in between them,0294

as well, if we wanted to get an even finer idea of what is going on.0298

But it is pretty obviously a line.0301

The same graph with different equations: it is possible to create the same plane curve with different parametric equations.0304

This is kind of surprising, because we think of the equation changing automatically meaning0310

that the graph will change, that our picture, our plane curve, will change.0313

On the previous slide, we graphed x = t + 1; y = 2t - 1.0317

That was what we had on the previous one.0323

On this slide, we have the exact same graph, but these new equations, x = 3t + 1, y = 6t - 1.0324

But the plane curve, the graph that we get out of it, looks exactly the same as the one we just had on the previous slide.0332

How is this possible--what is going on? Let's investigate.0338

We have x = t + 1; y = 2t - 1 as our left side, and x = 3t + 1, y = 6t - 1 as our right side.0342

Both pairs of equations will produce the same plane curve, will produce the same set of points for our graph.0353

But there is a difference: the second pair is moving faster--this set of equations here is moving faster, in a way.0360

It will move three times as far for the same change in t.0368

Let's try some values here: for our red equations here, if we plug in 0 for t, then we end up getting (1,-1), this point right here.0372

If we plug in 1 for t, we will get (2,1) out of it, this point right here.0385

Compare that to the blue one: if we plugged in 0 in the blue one, we will have 3(0) + 1, so we will be at 1 in our x;0392

and then, 6(0) - 1...so -1; we will have the same starting location.0400

At time = 0, at t = 0, we will be at the same place as in the red one.0405

However, when we plug in time = 1, t = 1, we end up getting 3(1) + 1 is 4, and our y-value is 6(1) - 1; it comes out to be 5.0410

We have managed to go much farther than we did; look at how much farther the blue graph has gone than the red graph.0421

So, the blue graph has managed to go three times as far.0428

We would have to go out to t = 3 to be able to get the same set of points: we would have 3 + 1 = 4 and y...2(3) is 6, minus 1 is 5.0432

So, if we had plugged in t = 3, we would be at the same point, (4,5).0442

So effectively, the blue graph is moving three times faster.0448

It does the amount that would take one time interval in blue--that would take three time intervals in red.0451

So, three time intervals in red is one time interval in blue.0459

That is one way of looking at it: they have the same picture, when we just look at it;0463

but if we think about how fast the point is moving in regards to t changing around, they are totally different in regards to that.0466

All right, here is another one to consider: x = -t + 1, and y = -2t - 1.0474

We get, once again, the exact same graph; what is going on here--how is this possible?0482

Well, we actually have the same plane curve, but once again, the motion, how the t change shows which point we are at, is completely different.0487

At 0, we are at (1,-1), just like before.0495

However, as we go to positive numbers...if we plug in 1, we get (0,-3), because -1 + 1 gets us to 0, and then -2t - 1 will get us to -3.0499

So, we will be down here.0510

If we plug in +2, we will be at (-1,-5).0511

Previously, as t went up, as our t increased, the graph went up to the top right; we saw it moving up.0517

And it went down to the bottom left as t became negative.0525

But now, if we plug in a negative value, we get (2,1) for plugging in -1; so we see that it is the opposite.0528

Negative points go to the top right, but positive points go down to the bottom left; so it is the opposite of what it was before.0535

If we want to show the direction of motion--if we want to show which way it is moving--we can draw arrows along the curve.0542

In this case, it would be useful to distinguish this graph from the previous one by showing it with arrows.0549

We just place little arrowheads along the curve occasionally, so that we can see which way it is pointing.0554

That is just two things like that, the arrowhead, just on part of the line...0559

And we wouldn't make it that large, unless we were trying to make a point.0564

We just make little arrowheads, so we can see which way the motion is going.0568

All right, a way that we can think about a parametric equations (and plane curves, and all of these ideas):0572

we can think of parametric equations as a way to describe the motion of an object.0578

As the object moves, it leaves a glowing trail behind where it moved.0582

So, the places that it moves through--what we see is the graph; the plane curve is the places that it has been through.0586

The glowing trail is the plane curve graphed by the equations.0592

So, the equations tell us its motion, its location at a specific time.0596

As it passes through various times, it moves through different locations.0600

And what we see as a graph, what we see as the plane curve, is just where it has been.0603

However, this graph can show us where it went (it shows us where it has been); and it can show us the direction of motion by those little arrows on it.0609

But it can't show us the speed.0616

Consider these three graphs: briefly, really quickly: I am human; I won't make exactly the same graph each of the three times I am going to draw this.0618

But it will be pretty close; the idea is just that, if you drew the exact same graph, we could draw it in three different ways.0625

In our first one, we draw it like this--fairly slow-moving; it makes a loop down here, and then goes up like this.0631

So, if we saw this as a picture, we would be able to see its motion, and we would be able to see where it had gone.0643

But we wouldn't have any idea that it went pretty slowly.0649

But we could have the same thing go through the exact same set of locations;0652

and it would show us the same motion, but we would have no idea that that one went so much faster.0660

That red one went so much faster than the blue.0665

My picture isn't exactly the same between the two; but let's pretend it is.0667

So, the red one has the same motion and the same picture as the blue one, but only by having watched it move were we able to see that it went faster.0672

Finally, we could have one that is different than both of those,0681

where it starts slow, and then it speeds up, and then it slows back down, and then it speeds up,0683

and then it slows back down, and then it speeds up...so it is changing around as it goes through it.0690

Once again, my picture is not quite perfect; it is a little bit more jerky than I would like.0696

But what we are seeing is that we can only show where it is being and the direction it went.0700

But once we are looking at a still, 2D picture, we can't see the speed of motion.0706

That is the difference, in many ways, between what we are seeing with the 3t + 1 and 6t - 1, versus the set of parametric equations.0711

We are just seeing how fast it is going.0721

They give us the same picture, but the picture isn't quite everything with parametric equations.0723

There is also this question of how fast it managed to move through that picture.0727

All right, sometimes it is useful to turn a pair of parametric equations into an old-fashioned rectangular equation.0731

A rectangular equation is just a fancy way to say an equation that only uses x and y,0737

where it is the two things related to each other, like we were used to before in math.0742

To do that, we must eliminate the parameter t from the equation.0746

How do we go about eliminating parameters from parametric equations?0749

Well, we do this by solving for t in one equation; and then, since we have t--we know what t is, as a value--we can plug it into the other.0753

We plug that into the other; there is no more parameter--it is gone.0761

For example, the ones that we were working with were x = t + 1, y = 2t - 1.0764

Well, what we do is: we would solve for the t here, and so we get x - 1 = t.0769

It is not too difficult; at that point, we can take this value for t, and we can plug it in here;0774

so we have x - 1 taking the spot of the t over here in the y equation.0779

We have y = 2(x - 1) - 1; we simplify that, and we get y = 2x - 3, at which point we have managed to solve this;0784

and we have a rectangular equation--we have eliminated the parameter from it, because we no longer have t.0794

Notice: if we were to graph y = 2x - 3, it would give us the exact same picture as the x = t + 1, y = 2t -1.0800

But because it no longer has the parameter, it no longer has a direction, and it no longer has this idea of speed.0808

So, when we turn it into a rectangular equation, it is just a question of what its graph looks like.0813

These ideas of speed and direction disappear once we get rid of the parameter,0818

because it is the parameter changing that allows us to have the idea of speed;0822

it is the parameter changing that allows us to have the idea of which way we are moving.0826

Are we moving from left to right? Are we moving from right to left?0829

I want you to be careful when you are eliminating parameters.0835

You will sometimes need to alter domains to keep the graph the same.0837

Furthermore, it is not always possible to solve for t directly, so occasionally we will have to be clever.0841

Sometimes, you will have to come up with an identity or some other relationship that is going on.0847

You won't be able to get this nice, clean t = something involving a variable.0850

It will be something a little bit more thoughtful, a little bit more difficult.0855

Sometimes, it will be as easy as just solving for t and plugging it into the other one; but other times, it will actually take a little bit of thought and creativity.0858

We will see a little bit of this in the examples.0864

We can also do the reverse of this, where we start with a rectangular equation, and then we want to parameterize it.0867

We want to get a parameter into that, so that we can express this rectangular equation, instead, using parametric equations.0873

This is really easy to do if the original graph is in this form: y = f(x).0880

That is to say, y is simply a function of x; pretty much all of the things that we are used to0886

of y = stuff involving x...stuff involving x...stuff involving x.0890

If that is the case, you just set x = t; say x is equal to your parameter t, and you are done--it is as easy as that.0894

So, for example, if we had y = x³ - 2x + 3, that is a function of x; x is the only thing that shows up in there, so it is purely a function of x.0900

At that point, we say, "All right, let's say x is equal to t; let's just set x equal to the parameter t."0910

And then, t...x...they are the same thing, so we go back here, and we swap out t's for x, and we get t³ - 2t + 3 for y.0917

So, at this point, we have x = t, y = t³ - 2t + 3; we have parametric equations.0927

And it is really, really easy, if it is in this nice, clean form, y = f(x).0935

Also, if it was in the form x = function of y, where it is x = stuff involving y, then you just set y = t.0940

You can do the same thing in the other direction.0946

However, it won't always be that easy.0949

Sometimes, the original rectangular equation can be a little more complicated, where it has x² + y² = 1, or something like that.0951

In those cases, it is going to take more thought.0957

If you want to try to come up with identities or relationships that will help you create the appropriate parametric equations,0959

how can you get these two things to connect to each other in a way where you can bring a parameter to bear?0965

There is no one easy way to do this; it will depend on the specific thing you are working on.0970

So, just try to think of things that look similar or that might be connected to this.0973

Just try to play around and use whatever you know to be able to come up with some way to turn it into something0977

where you can get a parameter to show up, at which point you can easily turn it into parametric equations.0982

But it can require some playing around and cleverness that won't just be immediately apparent.0986

So, just think of things that look similar, and see if there is any way to use them.0990

Parametric equations allow us to make really interesting graphs very easily.0995

And by "interesting," I mean things that are crazy, bizarre, cool, strange...they are pretty unlike any other graph that we are used to.0999

Let's start with this red one first, because it is considerable more tame.1007

If x is equal to 3cos(πt) and y = t, well, as t increases, our y is just going to increase, as well.1010

As t goes up, y goes up; however, as t goes up for 3cos(πt), well, at t = 0, cos(π0) would be cos(0), or 1.1017

So, we would be out here at 3; as t goes to 1, we are going to get cos(π) eventually.1027

The cosine of π is -1, so we will end up going out to -3.1035

So, 3cos(πt) is just going to oscillate back and forth; it will end up oscillating faster than normal cos(t), because it has this π factor in there.1038

But other than that, it is just going to end up oscillating back and forth.1048

That is why we see this curve: as it goes up, it oscillates back and forth; let me get that so that it goes your way.1051

That is to the left: so it starts out on the right, and then as it goes up, it bounces back and forth, and we see this oscillation, like that.1058

If we want to draw it in, we can see that from the first two points that we plot about, it is just going to be going back and forth, like this.1067

And we can keep that going down this way, as well.1075

All right, there we go; now let's talk about the blue one.1079

This one is considerably more crazy: x = t times cos(2t); y = 5sin(3t).1082

And we are restricting t to only go from 0 until 6, so it is not allowed to go to anything except 0 until 6.1089

So, this starts...is t is 0 for both of them, then we are going to start out at (0,0), because cosine of 2t...1095

cosine of 0 would be 1, but it is multiplied by 0 times cosine of stuff.1101

So, we start out at (0,0) for the beginning, and then, from there, we work our way out.1105

It goes up and around, and around some more, like that.1110

It is a pretty unusual thing; there is no way that we would be able to write that easily with rectangular coordinates.1123

There is no way that we could create a function--that clearly fails the function test.1130

But as a parametric equation, creating it parametrically is totally fine, and not very difficult to do.1134

We can express this really weird-looking figure in very, very few symbols: t times cos(2t) and 5 times sin(3t); t goes from 0 to 6.1139

We can create these really strange-looking things.1147

Parametric equations give us a lot of power to make really interesting-looking stuff without that much difficulty.1149

This brings us to the idea that graphing calculators are nice to have; why?1155

because working with parametric equations is a great time to use a graphing calculator.1159

It is a totally new way of looking at graphing.1163

Even if you have seen it just a little bit before in previous classes, parametric equations takes a little while to understand--1166

this idea of something that...you are not seeing t on the graph; t doesn't ever show up on the graph.1171

You see what its effects are through x and y, but t itself never shows up on the graph.1177

So, it is this new way of thinking: if t moved, how would it cause x to move--how would it cause y to move?1182

You are thinking in terms of this thing that never shows up.1187

It is a totally new way of thinking about graphing.1190

It really helps to just play around; if you have a graphing calculator, just plot random things.1192

Plot down some equation that you think might be interesting.1198

And then, once you have an understanding of an equation, alter that equation if you already understand it;1201

and see how you can get it to move it in some different way--how you can get the whole thing to move up to 3;1206

how you can get it to move left by 2; how you can get the thing to squish down.1211

What can you do to it to get different stuff to happen--how can you play with the thing?1215

Just do weird stuff to it; play with your graphing calculator, and just get a sense for how parametric equations work.1219

There is pretty much no better way to learn this sort of thing than just playing around for a while with it.1225

You just do weird things, and eventually you realize, "Oh, this all makes sense!"1230

And it will make sense eventually, but you have to get experience with it.1235

And the easiest way to get experience quickly with graphing things isn't by graphing it by hand.1238

That takes a long time; but if you use a graphing calculator, you can get really the best of both worlds.1243

You can see your graphs quickly, but you can also think about what is going on.1246

If you want more information on graphing calculators, there is an appendix on graphing calculators at the end of the course.1249

So, just go and check that out; there is a lot more information there1254

if you don't know much about graphing calculators and if you are interested in getting one.1256

Even if you don't own one, and you know for sure that you are not going to buy one, there are lots of free options out there.1260

In the very first lesson on graphing calculators, I talk about some of the free options that you can have1265

for ways that you can have the function of a graphing calculator without actually needing to go out and buy one.1269

If you are watching this video right now, there are graphing calculators that you can use for free right now on the Internet.1274

And if you just go and play around for a little while on one of these free things, it is going to help massively for understanding how parametric equations work.1280

There is really nothing better that you can do for understanding this stuff than just getting the chance to play around.1287

Also, when you are using a graphing calculator, pay attention to the interval that the parameter is using.1292

Most will only start with t going from 0 to 2π or t going from -10 to 10.1298

But because the interval is limited at the beginning, it might end up cutting off some of your graph.1303

So, you want to pay attention to what interval it starts by giving your t.1308

Set the interval as you need for whatever you are plotting.1313

If you have absolutely no idea what kind of interval you want, you might want to just start with a really, really big interval, like -20 to 20.1315

Or maybe go crazy, like -100 to 100; and that will very, very likely catch anything that you would end up wanting to graph.1321

But it is going to take your graphing calculator longer to work through all of that interval, than if it had a small interval to work through.1328

So, that is something to think about as you are working with the graphing calculator.1334

Also, if you don't quite understand how to get a graphing calculator to work with parametric equations,1337

you can check out the lesson on graphing parametric and polar stuff in the appendix.1342

And we will talk a little bit more about what is actually going to be involved in getting a calculator to be able to work with graphing a parametric equation.1348

All right, let's look at some examples: the first one: Graph x = t² - 3; y = t - 2; then go on to eliminate the parameter.1355

First, let's see what this thing graphs out as.1362

What we do is make our normal table of values; we plug in some t-values, and that is going to end up giving out x-values and giving out y-values.1365

Instead of giving out one value, it now gives out a pair of values, which is our point.1373

Let's consider if we plugged in 0: if we plugged in 0 into x, 0² - 3, we would have -3.1378

And plug in 0 into y; 0 - 2 gets us -2.1385

If we plug in +1 into x, that gets us 1² - 3, so that will be -2.1390

1 into y...that is 1 - 2, which is -1.1394

2: 2² - 3, 4 - 3, -1; 2 - 2 is 0; 3: 3³ is 9 - 3 is positive 6; 3 - 2 is positive 1.1397

If we went in the other direction and we plugged in -1, (-1)² - 3 is positive 1 minus 3; that gets us -2.1409

-1 - 2 is -3; -2 into t² - 3...(-2)² becomes positive 4, minus 3 becomes positive 1; -2 - 2 is -4.1416

-3 squared is positive 9; 9 - 3 is positive 6; -3 - 2 is -5.1428

That gives us a pretty good set to plot--let's plot this out; OK.1435

Let's do markings of length 1: 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3; OK.1447

So, let's start plotting some of these points.1468

We see, at 0, when we plug in time = 0, notice: 0 is not going to show up at all on our graph.1471

But when we plug in at time = 0, we have the point (-3,-2); we go to -3: 1, 2, 3; down 2: 1, 2; we plot our first point.1476

Let's go to positive 1 for time; that is (-2,-1), so it will be here.1490

We plug in time = 2; then we are at -1 for x and...oops, that won't end up being the case.1497

Let's go back a little bit: 2² is 4, minus 3, so that gets us positive 1, not -1; I'm sorry about that mistake that I made a while back there.1510

It makes sense, because that matches up to the -2 value up here; I'm sorry about that.1521

So, plug in time = 2; we are at (1,0) (the 0 is for our y-value).1525

Plug in time = 3; we are at 6: 1, 2, 3, 4, 5, 6; and 1 height.1533

So, we could draw in a curve; let's also think about it a bit.1539

x is basically behaving as a parabola, compared to time; the speed of motion in x, our horizontal motion,1542

is going to speed up as time increases, because it has that t² factor.1551

So, it will start at -3; but as time gets bigger and bigger, it is going to move faster and faster and faster.1556

What about y? y, on the other hand, is t - 2; it is linear, so it just maintains the same constant rate of increase.1561

It never gets faster; it never gets slower.1567

That is why we end up seeing this top curving out like this, because as time increases more and more,1569

our y-value stays at the same amount of increase, but our x-value moves more and more to the right.1575

So, we move faster horizontally, and that is why we are seeing it curve out like that.1581

If we went to the negatives, we would plug those in as well.1586

At time = -1, we are at (-2,-3); at time = -2, we are at (1,-4); at time = -3, we are at (6,-5).1590

So, we are going to end up seeing the same thing, because it is curving out like this.1604

It basically curves a lot like a parabola.1614

If we want to know which direction...out here at -3, it is here; and then, as it goes to larger times, it curves in this way; so we can see its direction, like this.1617

All right, so now that we have the graph, let's see about eliminating that parameter.1627

We have the graph part done; how can we eliminate that parameter?1632

Well, notice: we have y = t - 2; so if y equals t - 2, then we have y + 2 = t--that is nice.1635

So, at this point, we can plug that in for x = t² - 3; we have x = y + 2,1645

what we are swapping out for our t; and then we just go back to what it had been, squared, minus 3.1654

So, that this point, we have x = (y + 2)² - 3, which we could expand if we wanted.1659

But that doesn't really help us understand what is going on any better, so that is a fine answer, right there.1664

And if you wanted to, if you had to, you could expand (y + 2)² - 3.1670

But for my purposes, that actually makes it easier to understand, because then,1674

it is in a fairly normal form for a sideways parabola, and so we can see that it has been shifted left by that -3.1677

So, it has been shifted left by three, and it has been shifted down by 2, because it has y + 2.1684

So, if you are used to reading this sort of stuff (conics), then you can see that this ends up coming out to make this picture, just the same.1690

All right, let's graph x = 2cos(θ) and y = sin(2θ).1701

The first thing to do here is to think about where the interesting stuff happens.1707

θ is our parameter here, so θ is allowed to change freely; so where would we want to start?1711

Where do we want to stop? How often do we need to have points?1718

Well, the really interesting, the absolutely most interesting, points to look at on a trigonometry function are 0, π/2, π, 3π/2, and 2π.1721

Those are the absolute most interesting parts to look at on a trigonometric function.1731

Notice, though, that we have 2θ; 2θ isn't going to have its most interesting stuff happen at 0 and then π/2,1736

because if you plug in π/2 into 2θ, it is going to put out π.1743

So, we will have missed the π/2; so the most interesting thing for it, the first interesting thing for it after 0, would be π/4.1746

If we have θ = 0, then we will have a sine of 0; if we plug in θ = π/4, then we will have the sine of 2 times π/4, so the sine of 2π/2.1753

That is a really interesting thing to look at, so we will start at 0, and then work our way down with π/4 chunks.1762

We have θ, x, and y; let's plot a bunch of points, so that we can see what is going on here.1771

The first θ is 0; we plug that into here, into our x: 2 times cos(0)...cos(0) is 1, so we just get 2.1785

Plug 0 into sin(2θ); sin(0) is just 0; so that is our first point.1794

Next, we want to do π/4, because we just talked about how the most interesting things are going to be on the π/4 interval.1800

And cos(θ)...well, it has to pay attention to what 2θ's most interesting stuff is.1806

So, π/4 into 2cos(θ): well, 2cos(π/4)...we use a calculator to come up with what that is approximately.1810

That comes out to be around 1.41.1817

We plug in π/4 into sin(2θ); well, that is going to come out to be sin(π/2): sin(π/2) is 1.1820

Continue on with this process: we plug in π/2; π/2 into cos(θ) is going to come out as 0.1827

π/2 into sin(2θ) is going to come out as sin(π); so we get 0 here, as well.1833

Next, 3π/4: plug in 3π/4 in for 2cos(θ); that is now going to be getting us a negative value--it is going to come out as around -1.41.1841

For 3π/4 into 2θ, that gets us 3π/2; 3π/2 is now on the bottom of the unit circle,1853

so it is pointing down at the bottom of the unit circle; so that gets us a -1 in here.1862

Plug in π; π into cos(θ) gets us -1; 2 times that will be -2.1869

The sine of 2π is just going to come out to be 0; notice that we are starting to see this pattern with how this sin(2θ) is working.1875

5π/4...at this point, sine has managed to make one entire arc of the unit circle, because it is doubled up.1883

It has 2θ, so it moves twice as fast.1891

So normally, it is 0 to π for cos(θ), but cos(2θ) does double that, because it is 2θ.1894

So, it has already hit one entire course around the unit circle; so we are just going to end up seeing a full repeat at this point.1900

5π/4: plug that in for cos(θ); then we are down to slightly negative: we end up being in -1.41.1906

5π/4 times 2 gets us 5π/2, which is equivalent to π/2, so we have sin(π/2), or 1.1917

Next, 3π/2; put that in for cosine; you get 0; 3π/2 into sin(2θ) is sin(3π), effectively, which is the same as sin(π), which is 0.1925

We are repeating there, remember.1936

7π/4...when we plug that in, that comes out to be around 1.41; 7π/4 into sin(2θ) comes out to be -1.1938

7π over...let's just write it out...4, times 2, becomes 7π/2, which is the same thing as 3π/2,1948

because we can subtract by 4π/2, and it will still be the same, because that is just one whole unit circle rotation.1961

That is the exact same thing, which explains why we get -1 out of there.1966

And finally, 2π: now our cosine has managed to make one entire wrap, and it is back to 2, and we are back to 1.1970

Notice...oops, we are not back to 1; we are back to 0; sine of 4π is sine of 2π is sine of 0, which is 0, not 1.1978

At this point, we have managed to make an entire wrap on our cos(θ) and an entire wrap, twice now, on our sin(2θ).1987

If we were to keep going up with θ, we would just end up seeing completely repeating values.1993

If we were to have gone down with θ, we would see repeating going in the other direction.1996

So, this is actually enough for us to have.2000

All right, let's draw some axes here: OK, let's make a unit of 1, 2, 1, 2, 1, 1.2002

Here is 1, 2, 1, -1, -2, -1; OK.2023

We plot these points; let's do just the first three, so we can understand what is going on there.2034

The first θ = 0: we have x at 2 and y at 0.2039

At π/4, we have x at around 1.41, so a little bit under halfway to the 1; and then, at a height of 1...2044

And then, at θ = π/2, we have managed to get to (0,0).2053

Let's try to think about what is going on here.2058

2 times cos(θ) is just going to be 2 multiplied on cos(θ)--that is kind of obvious.2060

But cos(θ)--how is cos(θ) going to move from 0 to π/2?2066

Well, that is a question of how the x changes as we spin up to the top.2071

It is going to move faster, the closer θ gets to π/2; that is what we might be used to from how trigonometric things work.2075

It is going to move faster as it gets closer to being at the top.2082

It will move a little slower at the first, which is why it hasn't gotten very far by π/4.2086

But then, it manages to jump all the rest of the way down to 0 by the time it gets to π/2, only another π/4 forward.2090

What about sin(2θ)? Well, sin(2θ) is doubling the speed.2096

So, it manages, by the time it gets to π/4...π/4 has managed to have the sine effectively feel like it is going to π/2.2099

So, it manages to flip up to here and then flip down to here.2107

Sine is a question of how high we are, like this; that is what sine is measuring--the height of the angle for the unit circle.2110

What we end up seeing is it going like this: it cuts through and cuts down like this.2123

So, if you end up having difficulty understanding how we are figuring out that the curve looks like that,2132

and it is not just straight lines going together, just try plotting more points.2136

Any time you have confusion about how to plot something, just plot more points, and that will tell you the story of what is going on.2140

If you are not sure how all of the curves connect, just plot down more points, and the things will start to make sense.2146

The same basic structure is going to go on with the rest of these, so I will move a little faster now.2150

The next one: at angle 3π/4, we are at (-1.41,-1), and then at π, we are at -2 and 0.2153

At 5π/4, we are at -1.41 and positive 1; at 3π/2, we are back to (0,0).2165

At 7π/4, we are at 1.41 and -1; and then we are back to (2,0), and from there, it just wraps.2173

But what we end up seeing is that it does the same sort of curving thing, like this.2182

So, we get this hourglass figure on its side, and it looks kind of like an infinity.2198

And we can see that the direction it is moving is this way; cool.2207

All right, any time you have one of these, and you are really not sure how these things work,2217

in the worst-case scenario, you just plot a bunch of points.2221

Plot really small things for your angle θ; break it into even smaller chunks, and just try it.2224

You could always have just plotted in θ = 0, θ = 0.1, 0.2, 0.3, 0.4, 0.5, 0.6...2229

and just use a calculator to get decimal approximations for your x and y.2237

And then, just plot all of those, and that will help you see the curve.2240

And then, once you get a sense for how the curve is moving, you will be able to use that to not have to plot as many θ values,2242

as many time values, for later parts of the curve.2248

Always just plot more points if you are not sure how the curve looks.2252

All right, the third example: Eliminate the parameter, but make sure that the resulting rectangular equation gives the same graph.2256

The hint here is to control the domain in the rectangular equation.2262

Before we try to eliminate this, let's look at what x = e^t, y = e^2t + 7 would look like.2264

It would actually look a lot like a line, surprisingly...2271

Actually, no: it won't look a lot like a line--it will look a lot like a parabola, surprisingly.2274

But there is this strange thing that is going on.2278

Think about what values e^t can become; if you plug in really large values for t, you get huge values out of e^t.2280

But if you plug in really, really negative values for t, you can only get close to 0.2288

e to the -100 is 1 divided by e to the 100, which is really, really small, but it is not actually 0.2294

You can never get actually to 0; so that means the range for our x = e^t is everything from 02301

(but not including 0, because it can never actually get there) up until positive infinity.2310

Similarly, the range for y = e^2t + 7...well, that +7 will always be added in;2315

so the question is how small e^2t can get.2322

Once again, it can only get close to 0; it can't even actually make it up to 0.2325

So, we can only make it up to 7, but can't actually get to 7; 7, but not including 7, is the bottom of the range.2328

But e^2t can become arbitrarily large, so we can get all the way up to positive infinity.2336

With that idea in mind, let's get rid of this parameter--let's eliminate this parameter.2340

We know that x equals e^t, so we can rewrite y = e^2t as e^t times 2, plus 7.2344

We are still not quite sure how to plug into that; we can write that as (e^t)²,2353

which we remember from what we learned about exponents: (e^t)² + 7 =...2357

at this point, we see x = e^t, so we swap out, and we have x² + 7.2362

So now, we have y = x² + 7.2371

However, what we figured out about range, right from the beginning, was that x can never get below 0.2375

Our x can't actually get to 0; they can't go more to the left of 0, so we can't actually get to the 0 for x.2382

Similarly, the y-value can't actually drop below 7; it can't even get quite to 7.2391

So, y = x² + 7 is kind of a problem, because while that will never get below 7 for its range,2396

we will certainly be allowed to put in x-values that are different than 0 to infinity.2403

We will be able to put negative infinity into this; so we have to restrict the domain.2407

We restrict the domain, and we say that the domain for x is going to be from 0 up until infinity.2411

If we allow 0 up until infinity, well, then, we have satisfied the range that was allowed for x.2419

What our x-values are allowed to be is only between 0 and infinity.2423

And we have also satisfied from 7 to infinity, because if we can't ever actually plug in 0 for x²,2427

y (which is 0² + 7) will never come out to be...we will never be able to get y = 7 out of it.2432

We will only get really close to it, which is going to satisfy the range for our y, as well.2436

So, that also will have to have this domain before our answer is truly right.2441

The parameter, eliminated, gives us the rectangular equation y = x² + 7.2445

But we have to have this restriction, that the domain has to be x going from 0 up until infinity, not including 0.2452

The fourth example: Notice that x = cos(θ), y = sin(θ), gives a circle centered at (0,0) with radius 1.2459

We won't show this precisely; if you are not sure about this, try graphing it--it is pretty cool.2465

It comes out to be pretty clear, if you just go through a few points.2468

So, x = cos(θ); y = sin(θ) gives us (0,0) with radius 1; we end up getting a circle with radius 1.2471

All right, let's think about if we wanted to give a parametric equation for a circle centered at (3,-2) with radius 5.2481

How could we do each one of those things?2488

Well, we could move the center by adding things to x and y.2490

If we have x = cos(θ), y = sin(θ), well, if we just add 3 to x (it is 3 + cos(θ)),2496

well, then, all of our x-values will have shifted the entire thing horizontally to the right by 3.2503

We will have shifted the entire thing horizontally, because we just added the 3 to x.2509

So, every x point just moved over 3.2513

If all of the points move over 3 at once, we have just moved the center 3 horizontally.2516

If we want to move the center vertically, we just add or subtract the amount to our y.2520

We can move the center to (3,-2) with x = 3 + cos(θ), y = -2 + sin(θ).2524

And that will give us a circle that is moved over 3 and down 2, and so we will be down here.2538

If we want to expand to a radius of 5, if we want to increase our radius, then we are going to end up just multiplying the x and the y by 5.2547

If we make them bigger by 5 on the whole thing, then that means every point of it just went out by 5.2558

So, we can multiply 5 on cos(θ) and 5 on sin(θ), and we will end up having expanded the entire circle by 5.2563

We can change the radius to 5 by having it be x = 5cos(θ), y = 5sin(θ).2569

And that will end up giving us a larger circle that now has a radius of 5.2580

Notice: if you wanted to make an ellipse, where, instead of having a constant radius everywhere,2590

you wanted to maybe make the radius (no longer technically a radius) smaller on the top,2594

but then expand out, and then smaller again--if you wanted to have a major axis and a minor axis--2599

you could not use the same number multiplied on your cos(θ) and your sin(θ),2604

so that one of them will end up being larger out, and then it will shrink down for the other one.2609

That is one trick if you want to be able to make an ellipse.2613

OK, we take these two ideas, and we put them together; we combine moving the center and expanding the radius out.2619

So now, we have a radius of 5 that is 5cos(θ) and 5sin(θ), x and y respectively.2626

So, if we want to move that, we just add 3 and -2; we put these two ideas together, and we get x = 3 + 5cos(θ) and y = -2 + 5sin(θ).2631

That would end up getting us a circle that starts at (3,-2) and has a radius of 5; cool.2647

All right, one little idea, before we get to our very last example: projectile motion.2664

This is a really good use of parametric equations.2670

A projectile that is launched from some starting location, whether that means thrown,2673

shot from an arrow, shot from a gun, thrown out of a catapult--whatever it is--2677

any projectile--anything that is moving--a human cannonball--whatever it is--anything that is moving from some starting location, (d,h),2681

where d is the horizontal location and h is the vertical height, with an initial velocity of "v naught," v₀,2688

pronounced "v naught," N-A-U-G-H-T, like "all for naught," and an angle of θ above the horizontal2696

(how much above the horizontal--if we wanted to draw that in here, then this here would be our angle θ,2705

and this is our v₀, how fast we started out before gravity started to affect us and pull us back down to the earth),2713

if we have these ideas here, it can have its motion described by the parametric equations2719

x = v₀cos(θ)t + d (our starting horizontal location),2725

and y = -1/2gt² + v₀sin(θ)t + height.2733

In the equation for y, we are probably wondering what this g is.2741

Well, this g is the constant of acceleration for gravity.2744

So, on earth, we have an acceleration of gravity of 9.8 meters per second per second.2749

Equivalently, in the imperial system, it is 32 feet per second per second.2756

We have this acceleration; if we went somewhere else, like, say, the moon or Jupiter, we would end up getting a different value of g.2761

But most of the problems we end up ever looking at are on earth, so you will probably end up seeing 9.8 meters per second per second a lot.2767

Let's understand just how this is working.2776

We have this initial location that it gets shot out of.2777

Then, this velocity ends up going out, and then gravity pulls the thing down; gravity is always pulling down on the object.2780

It gets pulled down more and more and more and more and more, until eventually it lands and hits the ground.2788

If you take any object, and you toss it up, if you were to carefully graph out what it looked like--2793

if you were to see what it was, you would see it as an arc of a parabola; this is true for any thrown object.2802

Any object ends up having a parabolic arc to it.2807

And so, that is what we are seeing, because we are seeing this parabolic arc as a parametric equation.2811

All right, we are ready for this final example.2817

We have a reminder of these formulas on the top, and our problem is: A marauding horseback archer fires an arrow at a castle.2818

And he is on a height of 2 meters, because he is on top of a horse.2825

So, he fires, and he starts at a height of 2 meters; and the arrow comes out2828

with a speed of 50 meters per second and an angle of 25 degrees above the horizontal.2831

He starts at a height of 2 meters with a speed of 50 meters per second and an angle of 25 degrees above the horizontal, all for the arrow.2837

The castle walls are 80 meters away; he is firing at a castle, and the walls of the castle are 80 meters away, and they are 10 meters tall.2844

How far above the wall is the arrow when it flies over?2851

Let's draw a little picture: we see our man--he is on horseback, but I will not draw the horse.2855

And there are castle walls out here in the distance.2860

And so, there is a distance of 80 meters between him and them, and the castle walls are 10 meters tall.2866

He fires an arrow, and it flies through the air.2874

And the question we want to know is: Just as it gets above this wall, what is the extra height above that wall?2882

How high is the arrow above the wall?2892

And then, it will end up coming and landing on the other side; hopefully it won't hurt anybody.2894

Well, he is a marauder.2898

OK, let's see how we can figure this out.2900

Our first question is when the arrow is above the wall.2903

We have this great formula here; we can figure out what the height of the arrow is if we know the time,2907

because we know what v₀ is (it is 50 meters per second); we know v₀ = 50;2913

we know θ = 25 (he fired it at an angle of 25 degrees above the horizontal);2919

we know that its starting height was h = 2, and we were given g; so that is everything that we need for y, except for the time.2927

But we don't know what time it is before it manages to make it over those walls.2935

What we need to do is: we first need to figure out when it makes it to the walls--at what point is it at the walls?2938

x = v₀cos(θ)t + d: we know what v₀ is--it was 50; we know what θ is--it is 25.2945

We know how far they are, so we know what our x-value is going to be: it is 80 meters away.2953

So, finally, what is our d? That is the one thing we don't know there.2959

What is the d? Well, let's just say that where the horseback rider starts is 0.2963

We might as well make his horizontal location 0; so he starts at 0 horizontally, so 0 = x here, and then this is 80 = x here.2969

The castle walls are at 80; he starts at 0 in terms of horizontal x location.2979

At this point, we are ready to solve this thing.2984

In general, we have that, for any horizontal location, x is equal to 50 (our initial speed, v₀),2986

times cosine of 25 degrees, times the amount of time that the arrow has flown, plus our initial location (our initial location was 0).2994

At this point, we want to solve: so at x = 80, our time is equal to what?3004

We plug in 80 = 50cosine of 25 degrees, all times time.3011

We divide by 50 times cosine of 25 degrees, so we get t = 80/50, times cosine of 25 degrees.3019

We plug that into a calculator, and we get that t is approximately equal to 1.765 seconds.3029

So, after 1.765 seconds of flight time, the arrow is now at the horizontal location of the walls.3038

So, after 1.765 seconds, we are at the walls; so now we can plug that in; and we can figure out,3045

once it makes it to the walls horizontally, how high up it is--what the arrow's height is once we are at the walls horizontally.3051

So now, we use y = -1/2 times 9.8 (our acceleration due to gravity) t² + v₀...3058

50 times sin(θ), sine of 25 degrees, all times time, plus our initial starting height;3072

our initial starting height was 2, because he fires the arrow--he is on top of a horse,3079

so he is firing it from above the ground; he is not firing it from actually the level of the ground.3086

So, we will start working through that: we want to plug in at time = 1.765 seconds,3090

because that is the time that we are interested in knowing the height.3098

We plug that in here; we plug that in, and we get y =...-1/2 times 9.8 is -4.9, times 1.765 squared, plus 50sin(25) degrees, times 1.765 + 2.3100

It is kind of a lot there; but at this point, we can work this all out.3123

We work it out with a calculator, and we get that it is at 24.03 meters high.3126

So, the arrow is 24.03 meters high when it gets to the walls.3131

However, that is not our answer; we were asked how far above the walls when it gets to it.3137

So, how far above the wall is it?3144

At this point, we take 24.03 minus...the height of the walls is 10 meters tall, so minus 10.3147

That will give us the amount that it is above the wall, so that comes out to be 14.03 meters above the castle walls when it flies over them.3156

All right, that finishes up for parametric equations.3169

The important part is to think about it as describing the motion of an object in terms of its time.3172

Try to think about it as how time would change x, how time would change y...3177

Try to think of both of those together, and you will start to slowly build up a sense of how parametric equations work without even having to graph them.3181

Mainly, experience is a great way to learn how to do these things.3188

But you can really speed up the process of learning and understanding parametric equations3191

by just playing around, honestly, for five or ten minutes with a graphing calculator--just playing around,3194

plugging in random things, and seeing how one thing affects another thing--3199

how changing one constant causes things to move around.3202

Just playing around for five or ten minutes will help you so much more than trying to do 10 graphing problems.3205

All right, we will see you at Educator.com later--goodbye!3210