For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Parametric Equations
 Parametric equations are a new way to look at graphing. Instead of graphing input versus output, we'll base both x and y on a third, new variable: a parameter.
 Using this new idea for graphing, we can describe a set of points in the plane (a graph) as a plane curve. A plane curve is created by two functions f(t) and g(t) defined on some interval of the real numbers. The curve is the set of points (x,y)
= ( f(t), g(t) ). The equations
x = f(t) and y=g(t)  Graphing with parametric equations is very similar to "normal" graphing. You plug in a value, then see what point you get. Just instead of plugging in x to get y, you plug in t to get x and y.
 If we want to show the direction of motion in the plane curve, we can draw arrows along the curve to show which way it moves.
 Sometimes it's useful to turn a pair of parametric equations into an old fashioned rectangular equation (one using just x and y). To do that, we must eliminate the parameter t from the equations. We do this by solving for t in one equation, then plugging it into the other. POOF! No more parameter. [Caution: Be careful when eliminating parameters. We will sometimes need to alter domains to keep the same graph. Furthermore, it is not always possible to solve for t directly, so occasionally we'll have to be clever.]
 If you have access to a graphing calculator, it's great to try graphing some parametric equations with it. It's a new way of looking at graphing, so it helps just to play around. For more information, check out the appendix on graphing calculators.
Parametric Equations

 Parametric equations are based on some outside parameter. As this parameter changes, it causes the x and y values to change. Unlike a normal equation, where the change in one variable causes a change in the other, we have the change in some third parameter variable (t) causing changes in both x and y. A table of values is very similar for parametric equations. Plug in the given input value to get the outputs. The only difference is that instead of one input producing one output, we have one input (t) producing two different outputs (x and y).
 For example, to find the values of x and y at t=−3, simply plug in to the x and yequations:
t=−3 ⇒ x = 2(−3)^{2}+1 = 19 t=−3 ⇒ y = −(−3)^{3}+5 = 32  Repeat the above process for each of the t values given in the table, then fill in the corresponding x and y values.


 


 


 


 


 


 


 




 The domain is the allowed set of input values. If a problem does not specify or imply a specific set of allowed inputs, then we assume the domain is the set of all inputs that do not "break" the function(s)/equation(s). Thus, for this problem, the domain of t is the set of all input values that do not cause the equations for x or y to "break" (be undefined).
 Looking at x = 3 √{t+7}, we see the only part that can "break" is the square root function. Square root works as long as what is underneath the square root is not negative, so we have
Looking at y = ln(42−t) + 4, w see the only part that can "break" is the natural log function. The natural log works as long as its argument is positive (0 is not allowed), so we havet+7 ≥ 0 ⇒ t ≥ −7 42−t > 0 ⇒ 42 > t  Since a value for t must work in both the xequation and the yequation, both restrictions on the domain of t apply simultaneously. Combining these restrictions, we have
We can write this in interval notation as [−7, 42).−7 ≤ t < 42.

 Graphing a set of parametric equations is very similar to graphing a normal function/equation. Simply plug in various input values, take down the outputs, then plot the points. The only difference is that instead of plotting input vs. output, we plot the (x, y) values created for various values of t.
 Create a table of values, filling it out as necessary to get a good sense for how the graph will behave. If you already feel comfortable with the equations and think you have a good idea for what the graph will look like, calculate just a few points. If you're unsure about what it will look like, calculate more until the graph becomes clear to you.
We create a table along the lines of the one below:
t x y −4 6 −10 −3 5 −3 −2 4 2 −1 3 5 0 2 6 1 1 5 2 0 2 3 −1 −3 4 −2 −10  Finally, once you have enough points to understand how the graph works, plot the points and connect them with curves. When graphing parametric equations, we often also put little arrows along the graph to show which direction the graph "moves" as the tvalues move to larger and larger (positive) values.

 Sometimes we want to turn a pair of parametric equations into a rectangular equation: an equation using just x and y, like we normally use. To do this, we must remove t from the equations and "merge" the x and yequations together.
 We can do this through substitution. Solve for t in one of the equations, then plug that in to the other equation. In so doing, we will have eliminated t from existing in either equation.
Now that we have t solved for in one equation, plug that in to the other.x = 2−t ⇒ t = 2−x y = −t^{2} + 6 ⇒ y = − (2−x)^{2} + 6  The above equation of y = − (2−x)^{2} + 6 is perfectly acceptable as an answer. It describes the same parabolic curve that the original parametric equations gave, but does so without using t.
Still, depending on the problem, we might be expected to give an alternative, equivalent form. For example, we might want to be able to easily identify the vertex of the parabola:
Or we might be expected to expand the right side into a clear polynomial:y = − (2−x)^{2} + 6 ⇒ y = − ⎛
⎝− (x−2) ⎞
⎠2
+ 6 ⇒ y = − (x−2)^{2} + 6
All of these are equivalent equations, so they are all acceptable answers for the problem.y = − (2−x)^{2} + 6 ⇒ y = −(4−4x+x^{2}) + 6 ⇒ y = −x^{2} + 4x + 2

 Graphing a set of parametric equations is very similar to graphing a normal function/equation. Simply plug in various input values, take down the outputs, then plot the points. The only difference is that this time instead of plotting input vs. output, we plot the (x, y) values created for various values of t.
 Create a table of values, filling it out as necessary to get a good sense for how the graph will behave. If you already feel comfortable with the equations and think you have a good idea for what the graph will look like, calculate just a few points. If you're unsure about what it will look like, calculate more until the graph becomes clear to you.
For the specific case of this problem, notice that since x and y are based on trigonometric functions, the "interesting" places will occur at the "interesting" locations on the unit circle. Use those locations to help choose which tvalues to use in the table. [Remember, unless indicated otherwise, assume you are working with radians whenever dealing with trig functions.]
Furthermore, notice that the graph will repeat this set of values for tvalues above or below this interval because of the periodic nature of trig functions.t x y 0 3 0 π 42.12 0.71 π 20 1 3π 4−2.12 0.71 π −3 0 5π 4−2.12 −0.71 3π 20 −1 7π 42.12 −0.71 2π 3 0  Finally, once you have enough points to understand how the graph works, plot the points and connect them with curves. When graphing parametric equations, we often also put little arrows along the graph to show which direction the graph "moves" as the tvalues move to larger and larger (positive) values.

 Sometimes we want to turn a pair of parametric equations into a rectangular equation: an equation using just x and y, like we normally use. To do this, we must remove t from the equations and "merge" the x and yequations together.
 Normally, we do this by substitution: solve for t in one equation, then plug it in to the other. However, for this problem, doing so would not work out very well. For example, if we solved for t in the yequation we would get
This is tough, because inverse sine (sin^{−1}) is not very friendly. We could plug it in, then carefully think through the trigonometry involved, and eventually solve for a rectangular equation. But it would be a little bit ugly. Instead of substituting, it can be really useful to remember the most common identity from trigonometry:y = sin(t) ⇒ t = sin^{−1} (y) sin^{2} (θ) + cos^{2} (θ) = 1  The above identity is true for any θ at all, so it works for t as the variable too:
Thus, if we can get the x and yequations to give us sin(t) and cos(t), we can plug into the above instead. Working towards that, we havesin^{2} (t) + cos^{2} (t) = 1
Plugging in to the above, we getx = 3cos(t) ⇒ x 3= cos(t) ⎢
⎢y = sin(t)
[If you're familiar with conic sections, notice that the above equation gives an ellipse that is exactly the same as the one we graphed in the previous problem.]sin^{2} (t) + cos^{2} (t) = 1 ⇒ (y)^{2} + ⎛
⎝x 3⎞
⎠2
= 1 ⇒ x^{2} 9+ y^{2} = 1

 Graphing a set of parametric equations is very similar to graphing a normal function/equation. Simply plug in various input values, take down the outputs, then plot the points. The only difference is that this time instead of plotting input vs. output, we plot the (x, y) values created for various values of t.
 Create a table of values, filling it out as necessary to get a good sense for how the graph will behave. If you already feel comfortable with the equations and think you have a good idea for what the graph will look like, calculate just a few points. If you're unsure about what it will look like, calculate more until the graph becomes clear to you.
We create a table along the lines of the one below:
t x y −4 0.02 −2.96 −3 0.05 −2.90 −2 0.14 −2.73 −1 0.37 −2.26 0 1 −1 1 2.72 2.44 2 7.39 11.78 3 20.09 37.17  Notice that no matter how negative t gets, x will never reach 0, and y will never reach −3. This is because e^{t} only has a range of (0, ∞), so it can never reach 0, causing the previously mentioned effects on x and y. We can show that we approach the point (0, −3) but never actually reach it by using an empty circle at that location on the graph. Other than that, plot the points as normal, then connect them appropriately to create the graph. Since it is a graph of parametric equations, we can also place little arrows to indicate the direction of motion as t grows from negative to positive values.

 Sometimes we want to turn a pair of parametric equations into a rectangular equation: an equation using just x and y, like we normally use.
To do this, we must remove t from the equations and "merge" the x and yequations together.
Normally, we solve for t in one equation, then plug that in to the other equation. In this case though, we have x=e^{t} and since e^{t} is all that appears in the other equation, we simply substitute based on that. Do so:
y = 2e^{t} − 3 ⇒ y = 2(x) − 3 ⇒ y = 2x−3  However, there is an issue with this. The equation y=2x−3 does not describe the same graph as the original parametric equations. Compare the graph of y=2x−3 to the graph of the parametric equations, which we found in the previous problem. Clearly, they're different. The difference is that y=2x−3 goes to the left and right forever, while the original parametric graph "starts" at x=0 and only goes to the right from there. The reason for this is that, no matter what value we choose for t, e^{t} can never be equal to 0 or a negative number. Thus, since x=e^{t}, it must be that x cannot be equal to 0 or a negative number either.
 This means we need to put a restriction on what numbers x is allowed to be. We restrict x to the interval of (0, ∞). From the parametric equation, there was a similar restriction on y: because y=2e^{t}−3, the value for y can never equal −3 or lower. However, once we put a restriction on the domain allowed for x, this issue is resolved as well. If we can never plug x=0 or lower in to y=2x−3, then we don't have to worry about y ever going into its "forbidden" values either.

 Our pair of parametric equations must be in the below structure
Thus, using that we were given t = x^{3}−7, we must somehow solve for x and y as above.x = stuff involving t ⎢
⎢y = stuff involving t  From the equation t = x^{3}−7, we can solve for x:
We have now found our xequation. Next, we must find y.t = x^{3}−7 ⇒ t + 7 = x^{3} ⇒ x = 3√t+7  We know y = −4x+5, and from what we just solved for, we also have x = ^{3}√{t+7}. Therefore we can solve for y in terms of t by swapping out x based on the equation we just found:
At this point, we now have an xequation and a yequation that are both based on t as the independent variable. We have found a pair of parametric equations.y = −4x+5 ⇒ y = −4 ⎛
⎝3√t+7⎞
⎠+ 5 ⇒ y = −4· 3√t+7+ 5

A sailing ship is attacking a citadel at the top of some sea cliffs. The sailing ship is 1000 m from the base of the cliffs, and the cliffs are 100 m tall. The citadel is at the top of the cliffs, situated right next to the edge, and has a height of 20 m. The sailing ship fires a cannonball from a height of 10 m above sea level with a speed of 250 [m/s] and an angle of 10^{°} above horizontal. Does the cannonball hit the citadel? If not, does it strike the cliffs or fly over the citadel? If it does hit the citadel, where does the cannonball make contact relative to the top of the cliffs?
 Figuring out whether or not the cannonball hits the citadel is a matter of knowing the ball's location at the appropriate moment in time. We can set the bottom of the ship as the origin (0, 0). With that in mind, since the cannonball starts 10 m above sea level, the cannonball starts at the location (0, 10). The bottom of the citadel wall starts at (1000, 100) and the top of the wall is at (1000, 120). Thus, we need to know if the cannonball passes through the little line segment defined by the bottom and top of the citadel walls.
 We can model the motion of the cannonball by using the parametric equations above. From the problem, we have
Plugging in to the parametric equations, we getv_{0} = 250, θ = 10^{°}, d=0, h = 10 x = 250 cos(10^{°})·t y = −4.9 t^{2} + 250sin(10^{°}) ·t + 10  Earlier, we realized that the cannonball hits the citadel if it passes through the line segment defined by the points (1000, 100) and (1000, 120). Therefore we want to know the cannonball's height when it makes it to x=1000 (the horizontal edge of the citadel/cliff). However, we can not directly plug x=1000 in to the yequation to find the ball's height at that xlocation. Instead, we must find the time that the ball reaches that horizontal location, then use that to find the height there.
 Let t_{1000} represent the time it takes for the cannonball to reach a horizontal distance of x=1000. Plugging in to the xequation, we can solve for t_{1000}:
Using a calculator, we find that t_{1000} ≈ 4.062.x = 250 cos(10^{°})·t ⇒ 1000 = 250 cos(10^{°}) ·t_{1000} ⇒ t_{1000} = 1000 250cos(10^{°})  Now that we know it takes 4.062 seconds for the cannonball to horizontally reach the citadel/cliff, we need to know what height it has when it gets there. We plug t_{1000}=4.062 in to the yequation to find the height of the cannonball at that time:
Using a calculator, we work out that the height is 105.49 meters above sea level. Therefore, since the bottom of the citadel is at 100 meters above sea level and its top is at 120 meters, the citadel is struck by the cannonball. Furthermore, we see that the citadel is struck at a height of 105.49−100= 5.49 meters above the top of the cliffs.y = −4.9 t^{2} + 250sin(10^{°}) ·t + 10 ⇒ y = −4.9 (4.062)^{2} + 250sin(10^{°}) ·4.062 + 10
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Parametric Equations
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Definition
 Graphing with Parametric Equations
 Same Graph, Different Equations
 How Is That Possible?
 Same Graph, Different Equations, cont.
 Here's Another to Consider
 Same Plane Curve, But Still Different
 A Metaphor for Parametric Equations
 Think of Parametric Equations As a Way to Describe the Motion of An Object
 Graph Shows Where It Went, But Not Speed
 Eliminating Parameters
 Creating Parametric Equations
 Interesting Graphs
 Graphing Calculators, Yay!
 Example 1
 Example 2
 Example 3
 Example 4
 Projectile Motion
 Example 5
 Intro 0:00
 Introduction 0:06
 Definition 1:10
 Plane Curve
 The Key Idea
 Graphing with Parametric Equations 2:52
 Same Graph, Different Equations 5:04
 How Is That Possible?
 Same Graph, Different Equations, cont.
 Here's Another to Consider
 Same Plane Curve, But Still Different
 A Metaphor for Parametric Equations 9:36
 Think of Parametric Equations As a Way to Describe the Motion of An Object
 Graph Shows Where It Went, But Not Speed
 Eliminating Parameters 12:14
 Rectangular Equation
 Caution
 Creating Parametric Equations 14:30
 Interesting Graphs 16:38
 Graphing Calculators, Yay! 19:18
 Example 1 22:36
 Example 2 28:26
 Example 3 37:36
 Example 4 41:00
 Projectile Motion 44:26
 Example 5 47:00
Precalculus with Limits Online Course
Transcription: Parametric Equations
Hiwelcome 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 xaxis and the yaxis, 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 quantity0044
some changing parametersome outside thingaffects 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 commonlyused 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 numbershas 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 curvethat 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 valuesas this outside thing changes aroundhow 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 tvalues, then plot the points that come out for those tvalues; 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 t0266
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 possiblewhat 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 fasterthis 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 yvalue 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 bluethat 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 herehow 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 motionif we want to show which way it is movingwe 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 throughwhat 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 thisfairly slowmoving; 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 oldfashioned 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 twe know what t is, as a valuewe can plug it into the other.0753
We plug that into the other; there is no more parameterit 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 equationwe 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 doneit is as easy as that.0894
So, for example, if we had y = x^{3}  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 t^{3}  2t + 3 for y.0917
So, at this point, we have x = t, y = t^{3}  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 x^{2} + y^{2} = 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 functionthat 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 weirdlooking 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 strangelooking things.1147
Parametric equations give us a lot of power to make really interestinglooking 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 understand1166
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 movehow 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 wayhow 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 happenhow 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 = t^{2}  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 tvalues, and that is going to end up giving out xvalues and giving out yvalues.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, 0^{2}  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 1^{2}  3, so that will be 2.1390
1 into y...that is 1  2, which is 1.1394
2: 2^{2}  3, 4  3, 1; 2  2 is 0; 3: 3^{3} 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 t^{2}  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 plotlet'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: 2^{2} 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 yvalue).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 t^{2} 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 yvalue stays at the same amount of increase, but our xvalue 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 = tthat is nice.1635
So, at this point, we can plug that in for x = t^{2}  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 valueit 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 measuringthe 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 worstcase 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 lineit 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 parameterlet'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 x^{2} + 7.2362
So now, we have y = x^{2} + 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 yvalue can't actually drop below 7; it can't even get quite to 7.2391
So, y = x^{2} + 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 xvalues 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 xvalues 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 x^{2},2427
y (which is 0^{2} + 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 = x^{2} + 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 itit 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 xvalues 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 againif you wanted to have a major axis and a minor axis2599
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 catapultwhatever it is2677
any projectileanything that is movinga human cannonballwhatever it isanything 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," v_{0},2688
pronounced "v naught," NAUGHT, like "all for naught," and an angle of θ above the horizontal2696
(how much above the horizontalif we wanted to draw that in here, then this here would be our angle θ,2705
and this is our v_{0}, 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 = v_{0}cos(θ)t + d (our starting horizontal location),2725
and y = 1/2gt^{2} + v_{0}sin(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 like2793
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 manhe 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 v_{0} is (it is 50 meters per second); we know v_{0} = 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 wallsat what point is it at the walls?2938
x = v_{0}cos(θ)t + d: we know what v_{0} isit was 50; we know what θ isit is 25.2945
We know how far they are, so we know what our xvalue 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, v_{0}),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 iswhat 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) t^{2} + v_{0}...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 arrowhe 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 calculatorjust playing around,3194
plugging in random things, and seeing how one thing affects another thing3199
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 latergoodbye!3210
1 answer
Last reply by: Mohammed Jaweed
Tue Sep 22, 2015 12:35 AM
Post by Mohammed Jaweed on September 22, 2015
You did the math wrong on the first problem. when you plugged in 2 for x you got 1, but it's supposed to be 1.
1 answer
Last reply by: Professor SelhorstJones
Sun May 31, 2015 11:11 AM
Post by Datevig Daghlian on May 26, 2015
Thank you for the lecture! Great explanation!