WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about parametric equations.
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Up until this point, whenever we have looked at graphing an equation or a function, we thought in terms of input and output.
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If we put in x for this, what will y come out to be?
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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.
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We always think of x going in and y coming out, and that is how we graph it.
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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.
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And that is how we have thought of graphing so far.
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Parametric equations are a new way to look at graphing.
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Instead of graphing input versus output, we will base both x and y on a third new variable, a parameter.
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Parametric equations give us a great new way to look at how time, or some other changing quantity--
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some changing parameter--some outside thing--affects an object, such as letting us look at motion over time.
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There are lots of other uses, but motion over time is a really commonly-used one.
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This way of graphing has a variety of applications in science, calculus, and advanced math.
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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.
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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.
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So, how do we get that bunch of points that we call a plane curve onto the plane?
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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.
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t is allowed to go from some set of numbers--has some set of numbers like, say, 0 up until 10.
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And the curve is the set of points, (x,y), equal to (f(t),g(t));
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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.
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And that is our curve--that is what we get there.
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The equations x = f(t) and y = g(t) are called parametric equations, and t is the parameter.
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While we use t most often for parametric equations, since they usually involve time (t for time), we can use any symbol.
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We could use whatever symbol made sense for the specific parametric equations we were working with.
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You might see θ show up, or it might be some other letter, depending on what we are working with.
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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.
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For example, as time goes on, how does the object move in the x and the y?
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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?
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x and y are no longer directly linked; they are linked to a parameter now.
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How is that parameter changing, affecting x and y?
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How do we graph parametric equations?
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Graphing a parametric equation is very similar to what we think of as normal graphing.
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You plug in a value; then you see what point you get.
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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.
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We plug in one t, and it gives us (x,y) out of it.
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So, if we have x = t + 1, y = 2t - 1, and these are our parametric equations,
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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.
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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.
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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.
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So, that would come out to be: at t = 0, we have (1,-1) as a point.
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We go and we plot that: (1,-1)--we plot that point.
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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).
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So, we go and we plot (2,1), and that is on it.
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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).
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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).
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If we went to negative values of t (we aren't required to only have positive values of t--
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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)--
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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).
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So, at this point, it becomes pretty clear that we are graphing a line,
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although in this picture specifically, we had already seen the line.
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But it would become clear to us that we are graphing a line; and we could graph points in between them,
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as well, if we wanted to get an even finer idea of what is going on.
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But it is pretty obviously a line.
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The same graph with different equations: it is possible to create the same plane curve with different parametric equations.
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This is kind of surprising, because we think of the equation changing automatically meaning
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that the graph will change, that our picture, our plane curve, will change.
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On the previous slide, we graphed x = t + 1; y = 2t - 1.
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That was what we had on the previous one.
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On this slide, we have the exact same graph, but these new equations, x = 3t + 1, y = 6t - 1.
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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.
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How is this possible--what is going on? Let's investigate.
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We have x = t + 1; y = 2t - 1 as our left side, and x = 3t + 1, y = 6t - 1 as our right side.
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Both pairs of equations will produce the same plane curve, will produce the same set of points for our graph.
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But there is a difference: the second pair is moving faster--this set of equations here is moving faster, in a way.
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It will move three times as far for the same change in t.
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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.
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If we plug in 1 for t, we will get (2,1) out of it, this point right here.
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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;
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and then, 6(0) - 1...so -1; we will have the same starting location.
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At time = 0, at t = 0, we will be at the same place as in the red one.
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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.
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We have managed to go much farther than we did; look at how much farther the blue graph has gone than the red graph.
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So, the blue graph has managed to go three times as far.
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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.
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So, if we had plugged in t = 3, we would be at the same point, (4,5).
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So effectively, the blue graph is moving three times faster.
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It does the amount that would take one time interval in blue--that would take three time intervals in red.
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So, three time intervals in red is one time interval in blue.
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That is one way of looking at it: they have the same picture, when we just look at it;
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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.
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All right, here is another one to consider: x = -t + 1, and y = -2t - 1.
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We get, once again, the exact same graph; what is going on here--how is this possible?
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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.
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At 0, we are at (1,-1), just like before.
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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.
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So, we will be down here.
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If we plug in +2, we will be at (-1,-5).
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Previously, as t went up, as our t increased, the graph went up to the top right; we saw it moving up.
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And it went down to the bottom left as t became negative.
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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.
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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.
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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.
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In this case, it would be useful to distinguish this graph from the previous one by showing it with arrows.
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We just place little arrowheads along the curve occasionally, so that we can see which way it is pointing.
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That is just two things like that, the arrowhead, just on part of the line...
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And we wouldn't make it that large, unless we were trying to make a point.
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We just make little arrowheads, so we can see which way the motion is going.
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All right, a way that we can think about a parametric equations (and plane curves, and all of these ideas):
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we can think of parametric equations as a way to describe the motion of an object.
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As the object moves, it leaves a glowing trail behind where it moved.
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So, the places that it moves through--what we see is the graph; the plane curve is the places that it has been through.
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The glowing trail is the plane curve graphed by the equations.
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So, the equations tell us its motion, its location at a specific time.
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As it passes through various times, it moves through different locations.
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And what we see as a graph, what we see as the plane curve, is just where it has been.
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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.
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But it can't show us the speed.
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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.
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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.
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In our first one, we draw it like this--fairly slow-moving; it makes a loop down here, and then goes up like this.
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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.
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But we wouldn't have any idea that it went pretty slowly.
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But we could have the same thing go through the exact same set of locations;
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and it would show us the same motion, but we would have no idea that that one went so much faster.
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That red one went so much faster than the blue.
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My picture isn't exactly the same between the two; but let's pretend it is.
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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.
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Finally, we could have one that is different than both of those,
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where it starts slow, and then it speeds up, and then it slows back down, and then it speeds up,
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and then it slows back down, and then it speeds up...so it is changing around as it goes through it.
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Once again, my picture is not quite perfect; it is a little bit more jerky than I would like.
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But what we are seeing is that we can only show where it is being and the direction it went.
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But once we are looking at a still, 2D picture, we can't see the speed of motion.
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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.
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We are just seeing how fast it is going.
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They give us the same picture, but the picture isn't quite everything with parametric equations.
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There is also this question of how fast it managed to move through that picture.
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All right, sometimes it is useful to turn a pair of parametric equations into an old-fashioned rectangular equation.
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A rectangular equation is just a fancy way to say an equation that only uses x and y,
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where it is the two things related to each other, like we were used to before in math.
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To do that, we must eliminate the parameter t from the equation.
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How do we go about eliminating parameters from parametric equations?
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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.
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We plug that into the other; there is no more parameter--it is gone.
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For example, the ones that we were working with were x = t + 1, y = 2t - 1.
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Well, what we do is: we would solve for the t here, and so we get x - 1 = t.
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It is not too difficult; at that point, we can take this value for t, and we can plug it in here;
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so we have x - 1 taking the spot of the t over here in the y equation.
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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;
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and we have a rectangular equation--we have eliminated the parameter from it, because we no longer have t.
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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.
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But because it no longer has the parameter, it no longer has a direction, and it no longer has this idea of speed.
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So, when we turn it into a rectangular equation, it is just a question of what its graph looks like.
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These ideas of speed and direction disappear once we get rid of the parameter,
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because it is the parameter changing that allows us to have the idea of speed;
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it is the parameter changing that allows us to have the idea of which way we are moving.
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Are we moving from left to right? Are we moving from right to left?
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I want you to be careful when you are eliminating parameters.
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You will sometimes need to alter domains to keep the graph the same.
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Furthermore, it is not always possible to solve for t directly, so occasionally we will have to be clever.
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Sometimes, you will have to come up with an identity or some other relationship that is going on.
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You won't be able to get this nice, clean t = something involving a variable.
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It will be something a little bit more thoughtful, a little bit more difficult.
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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.
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We will see a little bit of this in the examples.
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We can also do the reverse of this, where we start with a rectangular equation, and then we want to parameterize it.
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We want to get a parameter into that, so that we can express this rectangular equation, instead, using parametric equations.
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This is really easy to do if the original graph is in this form: y = f(x).
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That is to say, y is simply a function of x; pretty much all of the things that we are used to
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of y = stuff involving x...stuff involving x...stuff involving x.
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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.
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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.
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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."
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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.
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So, at this point, we have x = t, y = t³ - 2t + 3; we have parametric equations.
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And it is really, really easy, if it is in this nice, clean form, y = f(x).
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Also, if it was in the form x = function of y, where it is x = stuff involving y, then you just set y = t.
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You can do the same thing in the other direction.
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However, it won't always be that easy.
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Sometimes, the original rectangular equation can be a little more complicated, where it has x² + y² = 1, or something like that.
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In those cases, it is going to take more thought.
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If you want to try to come up with identities or relationships that will help you create the appropriate parametric equations,
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how can you get these two things to connect to each other in a way where you can bring a parameter to bear?
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There is no one easy way to do this; it will depend on the specific thing you are working on.
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So, just try to think of things that look similar or that might be connected to this.
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Just try to play around and use whatever you know to be able to come up with some way to turn it into something
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where you can get a parameter to show up, at which point you can easily turn it into parametric equations.
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But it can require some playing around and cleverness that won't just be immediately apparent.
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So, just think of things that look similar, and see if there is any way to use them.
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Parametric equations allow us to make really interesting graphs very easily.
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And by "interesting," I mean things that are crazy, bizarre, cool, strange...they are pretty unlike any other graph that we are used to.
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Let's start with this red one first, because it is considerable more tame.
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If x is equal to 3cos(πt) and y = t, well, as t increases, our y is just going to increase, as well.
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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.
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So, we would be out here at 3; as t goes to 1, we are going to get cos(π) eventually.
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The cosine of π is -1, so we will end up going out to -3.
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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.
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But other than that, it is just going to end up oscillating back and forth.
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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.
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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.
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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.
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And we can keep that going down this way, as well.
00:17:59.100 --> 00:18:02.000
All right, there we go; now let's talk about the blue one.
00:18:02.000 --> 00:18:09.100
This one is considerably more crazy: x = t times cos(2t); y = 5sin(3t).
00:18:09.100 --> 00:18:15.400
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.
00:18:15.400 --> 00:18:21.100
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...
00:18:21.100 --> 00:18:25.600
cosine of 0 would be 1, but it is multiplied by 0 times cosine of stuff.
00:18:25.600 --> 00:18:30.300
So, we start out at (0,0) for the beginning, and then, from there, we work our way out.
00:18:30.300 --> 00:18:43.600
It goes up and around, and around some more, like that.
00:18:43.600 --> 00:18:50.400
It is a pretty unusual thing; there is no way that we would be able to write that easily with rectangular coordinates.
00:18:50.400 --> 00:18:53.700
There is no way that we could create a function--that clearly fails the function test.
00:18:53.700 --> 00:18:58.800
But as a parametric equation, creating it parametrically is totally fine, and not very difficult to do.
00:18:58.800 --> 00:19:07.100
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.
00:19:07.100 --> 00:19:09.100
We can create these really strange-looking things.
00:19:09.100 --> 00:19:15.300
Parametric equations give us a lot of power to make really interesting-looking stuff without that much difficulty.
00:19:15.300 --> 00:19:19.500
This brings us to the idea that graphing calculators are nice to have; why?
00:19:19.500 --> 00:19:23.500
because working with parametric equations is a great time to use a graphing calculator.
00:19:23.500 --> 00:19:25.800
It is a totally new way of looking at graphing.
00:19:25.800 --> 00:19:31.000
Even if you have seen it just a little bit before in previous classes, parametric equations takes a little while to understand--
00:19:31.000 --> 00:19:36.700
this idea of something that...you are not seeing t on the graph; t doesn't ever show up on the graph.
00:19:36.700 --> 00:19:41.800
You see what its effects are through x and y, but t itself never shows up on the graph.
00:19:41.800 --> 00:19:47.300
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?
00:19:47.300 --> 00:19:49.900
You are thinking in terms of this thing that never shows up.
00:19:49.900 --> 00:19:52.200
It is a totally new way of thinking about graphing.
00:19:52.200 --> 00:19:58.300
It really helps to just play around; if you have a graphing calculator, just plot random things.
00:19:58.300 --> 00:20:01.600
Plot down some equation that you think might be interesting.
00:20:01.600 --> 00:20:06.300
And then, once you have an understanding of an equation, alter that equation if you already understand it;
00:20:06.300 --> 00:20:11.400
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;
00:20:11.400 --> 00:20:15.400
how you can get it to move left by 2; how you can get the thing to squish down.
00:20:15.400 --> 00:20:19.200
What can you do to it to get different stuff to happen--how can you play with the thing?
00:20:19.200 --> 00:20:25.300
Just do weird stuff to it; play with your graphing calculator, and just get a sense for how parametric equations work.
00:20:25.300 --> 00:20:30.100
There is pretty much no better way to learn this sort of thing than just playing around for a while with it.
00:20:30.100 --> 00:20:35.200
You just do weird things, and eventually you realize, "Oh, this all makes sense!"
00:20:35.200 --> 00:20:38.600
And it will make sense eventually, but you have to get experience with it.
00:20:38.600 --> 00:20:42.700
And the easiest way to get experience quickly with graphing things isn't by graphing it by hand.
00:20:42.700 --> 00:20:46.300
That takes a long time; but if you use a graphing calculator, you can get really the best of both worlds.
00:20:46.300 --> 00:20:49.100
You can see your graphs quickly, but you can also think about what is going on.
00:20:49.100 --> 00:20:54.100
If you want more information on graphing calculators, there is an appendix on graphing calculators at the end of the course.
00:20:54.100 --> 00:20:56.400
So, just go and check that out; there is a lot more information there
00:20:56.400 --> 00:20:59.700
if you don't know much about graphing calculators and if you are interested in getting one.
00:20:59.700 --> 00:21:04.700
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.
00:21:04.700 --> 00:21:09.400
In the very first lesson on graphing calculators, I talk about some of the free options that you can have
00:21:09.400 --> 00:21:14.400
for ways that you can have the function of a graphing calculator without actually needing to go out and buy one.
00:21:14.400 --> 00:21:20.100
If you are watching this video right now, there are graphing calculators that you can use for free right now on the Internet.
00:21:20.100 --> 00:21:27.400
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.
00:21:27.400 --> 00:21:32.500
There is really nothing better that you can do for understanding this stuff than just getting the chance to play around.
00:21:32.500 --> 00:21:37.800
Also, when you are using a graphing calculator, pay attention to the interval that the parameter is using.
00:21:37.800 --> 00:21:43.300
Most will only start with t going from 0 to 2π or t going from -10 to 10.
00:21:43.300 --> 00:21:48.500
But because the interval is limited at the beginning, it might end up cutting off some of your graph.
00:21:48.500 --> 00:21:52.700
So, you want to pay attention to what interval it starts by giving your t.
00:21:52.700 --> 00:21:55.400
Set the interval as you need for whatever you are plotting.
00:21:55.400 --> 00:22:01.300
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.
00:22:01.300 --> 00:22:07.800
Or maybe go crazy, like -100 to 100; and that will very, very likely catch anything that you would end up wanting to graph.
00:22:07.800 --> 00:22:14.200
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.
00:22:14.200 --> 00:22:17.300
So, that is something to think about as you are working with the graphing calculator.
00:22:17.300 --> 00:22:22.400
Also, if you don't quite understand how to get a graphing calculator to work with parametric equations,
00:22:22.400 --> 00:22:28.100
you can check out the lesson on graphing parametric and polar stuff in the appendix.
00:22:28.100 --> 00:22:35.600
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.
00:22:35.600 --> 00:22:42.300
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.
00:22:42.300 --> 00:22:44.800
First, let's see what this thing graphs out as.
00:22:44.800 --> 00:22:53.200
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.
00:22:53.200 --> 00:22:57.800
Instead of giving out one value, it now gives out a pair of values, which is our point.
00:22:57.800 --> 00:23:04.900
Let's consider if we plugged in 0: if we plugged in 0 into x, 0² - 3, we would have -3.
00:23:04.900 --> 00:23:09.800
And plug in 0 into y; 0 - 2 gets us -2.
00:23:09.800 --> 00:23:14.600
If we plug in +1 into x, that gets us 1² - 3, so that will be -2.
00:23:14.600 --> 00:23:17.600
1 into y...that is 1 - 2, which is -1.
00:23:17.600 --> 00:23:29.300
2: 2² - 3, 4 - 3, -1; 2 - 2 is 0; 3: 3³ is 9 - 3 is positive 6; 3 - 2 is positive 1.
00:23:29.300 --> 00:23:36.200
If we went in the other direction and we plugged in -1, (-1)² - 3 is positive 1 minus 3; that gets us -2.
00:23:36.200 --> 00:23:47.900
-1 - 2 is -3; -2 into t² - 3...(-2)² becomes positive 4, minus 3 becomes positive 1; -2 - 2 is -4.
00:23:47.900 --> 00:23:55.300
-3 squared is positive 9; 9 - 3 is positive 6; -3 - 2 is -5.
00:23:55.300 --> 00:24:07.000
That gives us a pretty good set to plot--let's plot this out; OK.
00:24:07.000 --> 00:24:28.500
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.
00:24:28.500 --> 00:24:30.800
So, let's start plotting some of these points.
00:24:30.800 --> 00:24:36.100
We see, at 0, when we plug in time = 0, notice: 0 is not going to show up at all on our graph.
00:24:36.100 --> 00:24:50.200
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.
00:24:50.200 --> 00:24:56.800
Let's go to positive 1 for time; that is (-2,-1), so it will be here.
00:24:56.800 --> 00:25:10.200
We plug in time = 2; then we are at -1 for x and...oops, that won't end up being the case.
00:25:10.200 --> 00:25:20.700
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.
00:25:20.700 --> 00:25:24.900
It makes sense, because that matches up to the -2 value up here; I'm sorry about that.
00:25:24.900 --> 00:25:32.800
So, plug in time = 2; we are at (1,0) (the 0 is for our y-value).
00:25:32.800 --> 00:25:39.300
Plug in time = 3; we are at 6: 1, 2, 3, 4, 5, 6; and 1 height.
00:25:39.300 --> 00:25:41.900
So, we could draw in a curve; let's also think about it a bit.
00:25:41.900 --> 00:25:51.100
x is basically behaving as a parabola, compared to time; the speed of motion in x, our horizontal motion,
00:25:51.100 --> 00:25:55.700
is going to speed up as time increases, because it has that t² factor.
00:25:55.700 --> 00:26:00.800
So, it will start at -3; but as time gets bigger and bigger, it is going to move faster and faster and faster.
00:26:00.800 --> 00:26:07.200
What about y? y, on the other hand, is t - 2; it is linear, so it just maintains the same constant rate of increase.
00:26:07.200 --> 00:26:09.200
It never gets faster; it never gets slower.
00:26:09.200 --> 00:26:15.500
That is why we end up seeing this top curving out like this, because as time increases more and more,
00:26:15.500 --> 00:26:21.600
our y-value stays at the same amount of increase, but our x-value moves more and more to the right.
00:26:21.600 --> 00:26:26.000
So, we move faster horizontally, and that is why we are seeing it curve out like that.
00:26:26.000 --> 00:26:29.700
If we went to the negatives, we would plug those in as well.
00:26:29.700 --> 00:26:43.800
At time = -1, we are at (-2,-3); at time = -2, we are at (1,-4); at time = -3, we are at (6,-5).
00:26:43.800 --> 00:26:54.600
So, we are going to end up seeing the same thing, because it is curving out like this.
00:26:54.600 --> 00:26:56.700
It basically curves a lot like a parabola.
00:26:56.700 --> 00:27:07.600
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.
00:27:07.600 --> 00:27:11.800
All right, so now that we have the graph, let's see about eliminating that parameter.
00:27:11.800 --> 00:27:15.000
We have the graph part done; how can we eliminate that parameter?
00:27:15.000 --> 00:27:25.200
Well, notice: we have y = t - 2; so if y equals t - 2, then we have y + 2 = t--that is nice.
00:27:25.200 --> 00:27:34.000
So, at this point, we can plug that in for x = t² - 3; we have x = y + 2,
00:27:34.000 --> 00:27:39.400
what we are swapping out for our t; and then we just go back to what it had been, squared, minus 3.
00:27:39.400 --> 00:27:44.600
So, that this point, we have x = (y + 2)² - 3, which we could expand if we wanted.
00:27:44.600 --> 00:27:49.700
But that doesn't really help us understand what is going on any better, so that is a fine answer, right there.
00:27:49.700 --> 00:27:53.800
And if you wanted to, if you had to, you could expand (y + 2)² - 3.
00:27:53.800 --> 00:27:57.200
But for my purposes, that actually makes it easier to understand, because then,
00:27:57.200 --> 00:28:04.400
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.
00:28:04.400 --> 00:28:09.900
So, it has been shifted left by three, and it has been shifted down by 2, because it has y + 2.
00:28:09.900 --> 00:28:21.400
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.
00:28:21.400 --> 00:28:27.000
All right, let's graph x = 2cos(θ) and y = sin(2θ).
00:28:27.000 --> 00:28:31.500
The first thing to do here is to think about where the interesting stuff happens.
00:28:31.500 --> 00:28:37.700
θ is our parameter here, so θ is allowed to change freely; so where would we want to start?
00:28:37.700 --> 00:28:41.200
Where do we want to stop? How often do we need to have points?
00:28:41.200 --> 00:28:51.600
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.
00:28:51.600 --> 00:28:55.700
Those are the absolute most interesting parts to look at on a trigonometric function.
00:28:55.700 --> 00:29:02.800
Notice, though, that we have 2θ; 2θ isn't going to have its most interesting stuff happen at 0 and then π/2,
00:29:02.800 --> 00:29:06.600
because if you plug in π/2 into 2θ, it is going to put out π.
00:29:06.600 --> 00:29:13.000
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.
00:29:13.000 --> 00:29:22.600
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.
00:29:22.600 --> 00:29:31.200
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.
00:29:31.200 --> 00:29:45.300
We have θ, x, and y; let's plot a bunch of points, so that we can see what is going on here.
00:29:45.300 --> 00:29:53.800
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.
00:29:53.800 --> 00:29:59.700
Plug 0 into sin(2θ); sin(0) is just 0; so that is our first point.
00:29:59.700 --> 00:30:05.800
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.
00:30:05.800 --> 00:30:10.600
And cos(θ)...well, it has to pay attention to what 2θ's most interesting stuff is.
00:30:10.600 --> 00:30:17.000
So, π/4 into 2cos(θ): well, 2cos(π/4)...we use a calculator to come up with what that is approximately.
00:30:17.000 --> 00:30:20.600
That comes out to be around 1.41.
00:30:20.600 --> 00:30:27.100
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.
00:30:27.100 --> 00:30:33.600
Continue on with this process: we plug in π/2; π/2 into cos(θ) is going to come out as 0.
00:30:33.600 --> 00:30:41.500
π/2 into sin(2θ) is going to come out as sin(π); so we get 0 here, as well.
00:30:41.500 --> 00:30:52.900
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.
00:30:52.900 --> 00:31:01.800
For 3π/4 into 2θ, that gets us 3π/2; 3π/2 is now on the bottom of the unit circle,
00:31:01.800 --> 00:31:08.700
so it is pointing down at the bottom of the unit circle; so that gets us a -1 in here.
00:31:08.700 --> 00:31:15.000
Plug in π; π into cos(θ) gets us -1; 2 times that will be -2.
00:31:15.000 --> 00:31:23.500
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.
00:31:23.500 --> 00:31:31.100
5π/4...at this point, sine has managed to make one entire arc of the unit circle, because it is doubled up.
00:31:31.100 --> 00:31:33.700
It has 2θ, so it moves twice as fast.
00:31:33.700 --> 00:31:40.100
So normally, it is 0 to π for cos(θ), but cos(2θ) does double that, because it is 2θ.
00:31:40.100 --> 00:31:46.300
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.
00:31:46.300 --> 00:31:57.000
5π/4: plug that in for cos(θ); then we are down to slightly negative: we end up being in -1.41.
00:31:57.000 --> 00:32:04.900
5π/4 times 2 gets us 5π/2, which is equivalent to π/2, so we have sin(π/2), or 1.
00:32:04.900 --> 00:32:16.500
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.
00:32:16.500 --> 00:32:18.400
We are repeating there, remember.
00:32:18.400 --> 00:32:28.100
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.
00:32:28.100 --> 00:32:41.100
7π over...let's just write it out...4, times 2, becomes 7π/2, which is the same thing as 3π/2,
00:32:41.100 --> 00:32:46.500
because we can subtract by 4π/2, and it will still be the same, because that is just one whole unit circle rotation.
00:32:46.500 --> 00:32:50.200
That is the exact same thing, which explains why we get -1 out of there.
00:32:50.200 --> 00:32:57.900
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.
00:32:57.900 --> 00:33:06.800
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.
00:33:06.800 --> 00:33:12.900
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).
00:33:12.900 --> 00:33:16.600
If we were to keep going up with θ, we would just end up seeing completely repeating values.
00:33:16.600 --> 00:33:19.900
If we were to have gone down with θ, we would see repeating going in the other direction.
00:33:19.900 --> 00:33:22.200
So, this is actually enough for us to have.
00:33:22.200 --> 00:33:43.000
All right, let's draw some axes here: OK, let's make a unit of 1, 2, 1, 2, 1, 1.
00:33:43.000 --> 00:33:54.200
Here is 1, 2, 1, -1, -2, -1; OK.
00:33:54.200 --> 00:33:59.100
We plot these points; let's do just the first three, so we can understand what is going on there.
00:33:59.100 --> 00:34:04.400
The first θ = 0: we have x at 2 and y at 0.
00:34:04.400 --> 00:34:13.600
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...
00:34:13.600 --> 00:34:17.900
And then, at θ = π/2, we have managed to get to (0,0).
00:34:17.900 --> 00:34:19.800
Let's try to think about what is going on here.
00:34:19.800 --> 00:34:26.500
2 times cos(θ) is just going to be 2 multiplied on cos(θ)--that is kind of obvious.
00:34:26.500 --> 00:34:31.100
But cos(θ)--how is cos(θ) going to move from 0 to π/2?
00:34:31.100 --> 00:34:35.400
Well, that is a question of how the x changes as we spin up to the top.
00:34:35.400 --> 00:34:42.600
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.
00:34:42.600 --> 00:34:45.900
It is going to move faster as it gets closer to being at the top.
00:34:45.900 --> 00:34:49.800
It will move a little slower at the first, which is why it hasn't gotten very far by π/4.
00:34:49.800 --> 00:34:55.700
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.
00:34:55.700 --> 00:34:59.600
What about sin(2θ)? Well, sin(2θ) is doubling the speed.
00:34:59.600 --> 00:35:07.000
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.
00:35:07.000 --> 00:35:10.300
So, it manages to flip up to here and then flip down to here.
00:35:10.300 --> 00:35:23.100
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.
00:35:23.100 --> 00:35:32.300
What we end up seeing is it going like this: it cuts through and cuts down like this.
00:35:32.300 --> 00:35:36.500
So, if you end up having difficulty understanding how we are figuring out that the curve looks like that,
00:35:36.500 --> 00:35:40.400
and it is not just straight lines going together, just try plotting more points.
00:35:40.400 --> 00:35:45.900
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.
00:35:45.900 --> 00:35:50.300
If you are not sure how all of the curves connect, just plot down more points, and the things will start to make sense.
00:35:50.300 --> 00:35:53.600
The same basic structure is going to go on with the rest of these, so I will move a little faster now.
00:35:53.600 --> 00:36:05.200
The next one: at angle 3π/4, we are at (-1.41,-1), and then at π, we are at -2 and 0.
00:36:05.200 --> 00:36:13.600
At 5π/4, we are at -1.41 and positive 1; at 3π/2, we are back to (0,0).
00:36:13.600 --> 00:36:21.800
At 7π/4, we are at 1.41 and -1; and then we are back to (2,0), and from there, it just wraps.
00:36:21.800 --> 00:36:37.900
But what we end up seeing is that it does the same sort of curving thing, like this.
00:36:37.900 --> 00:36:47.000
So, we get this hourglass figure on its side, and it looks kind of like an infinity.
00:36:47.000 --> 00:36:57.200
And we can see that the direction it is moving is this way; cool.
00:36:57.200 --> 00:37:01.000
All right, any time you have one of these, and you are really not sure how these things work,
00:37:01.000 --> 00:37:03.800
in the worst-case scenario, you just plot a bunch of points.
00:37:03.800 --> 00:37:08.900
Plot really small things for your angle θ; break it into even smaller chunks, and just try it.
00:37:08.900 --> 00:37:16.800
You could always have just plotted in θ = 0, θ = 0.1, 0.2, 0.3, 0.4, 0.5, 0.6...
00:37:16.800 --> 00:37:19.700
and just use a calculator to get decimal approximations for your x and y.
00:37:19.700 --> 00:37:22.300
And then, just plot all of those, and that will help you see the curve.
00:37:22.300 --> 00:37:27.900
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,
00:37:27.900 --> 00:37:31.700
as many time values, for later parts of the curve.
00:37:31.700 --> 00:37:36.100
Always just plot more points if you are not sure how the curve looks.
00:37:36.100 --> 00:37:41.800
All right, the third example: Eliminate the parameter, but make sure that the resulting rectangular equation gives the same graph.
00:37:41.800 --> 00:37:44.600
The hint here is to control the domain in the rectangular equation.
00:37:44.600 --> 00:37:51.000
Before we try to eliminate this, let's look at what x = e^t, y = e^2t + 7 would look like.
00:37:51.000 --> 00:37:54.000
It would actually look a lot like a line, surprisingly...
00:37:54.000 --> 00:37:57.800
Actually, no: it won't look a lot like a line--it will look a lot like a parabola, surprisingly.
00:37:57.800 --> 00:38:00.500
But there is this strange thing that is going on.
00:38:00.500 --> 00:38:08.600
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.
00:38:08.600 --> 00:38:14.300
But if you plug in really, really negative values for t, you can only get close to 0.
00:38:14.300 --> 00:38:21.300
e to the -100 is 1 divided by e to the 100, which is really, really small, but it is not actually 0.
00:38:21.300 --> 00:38:30.100
You can never get actually to 0; so that means the range for our x = e^t is everything from 0
00:38:30.100 --> 00:38:35.300
(but not including 0, because it can never actually get there) up until positive infinity.
00:38:35.300 --> 00:38:41.800
Similarly, the range for y = e^2t + 7...well, that +7 will always be added in;
00:38:41.800 --> 00:38:45.100
so the question is how small e^2t can get.
00:38:45.100 --> 00:38:48.600
Once again, it can only get close to 0; it can't even actually make it up to 0.
00:38:48.600 --> 00:38:55.800
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.
00:38:55.800 --> 00:39:00.600
But e^2t can become arbitrarily large, so we can get all the way up to positive infinity.
00:39:00.600 --> 00:39:04.400
With that idea in mind, let's get rid of this parameter--let's eliminate this parameter.
00:39:04.400 --> 00:39:12.700
We know that x equals e^t, so we can rewrite y = e^2t as e^t times 2, plus 7.
00:39:12.700 --> 00:39:17.100
We are still not quite sure how to plug into that; we can write that as (e^t)²,
00:39:17.100 --> 00:39:22.000
which we remember from what we learned about exponents: (e^t)² + 7 =...
00:39:22.000 --> 00:39:31.100
at this point, we see x = e^t, so we swap out, and we have x² + 7.
00:39:31.100 --> 00:39:35.300
So now, we have y = x² + 7.
00:39:35.300 --> 00:39:42.400
However, what we figured out about range, right from the beginning, was that x can never get below 0.
00:39:42.400 --> 00:39:50.700
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.
00:39:50.700 --> 00:39:56.200
Similarly, the y-value can't actually drop below 7; it can't even get quite to 7.
00:39:56.200 --> 00:40:02.800
So, y = x² + 7 is kind of a problem, because while that will never get below 7 for its range,
00:40:02.800 --> 00:40:07.200
we will certainly be allowed to put in x-values that are different than 0 to infinity.
00:40:07.200 --> 00:40:10.800
We will be able to put negative infinity into this; so we have to restrict the domain.
00:40:10.800 --> 00:40:18.900
We restrict the domain, and we say that the domain for x is going to be from 0 up until infinity.
00:40:18.900 --> 00:40:23.600
If we allow 0 up until infinity, well, then, we have satisfied the range that was allowed for x.
00:40:23.600 --> 00:40:27.100
What our x-values are allowed to be is only between 0 and infinity.
00:40:27.100 --> 00:40:32.000
And we have also satisfied from 7 to infinity, because if we can't ever actually plug in 0 for x²,
00:40:32.000 --> 00:40:36.300
y (which is 0² + 7) will never come out to be...we will never be able to get y = 7 out of it.
00:40:36.300 --> 00:40:40.900
We will only get really close to it, which is going to satisfy the range for our y, as well.
00:40:40.900 --> 00:40:44.900
So, that also will have to have this domain before our answer is truly right.
00:40:44.900 --> 00:40:52.100
The parameter, eliminated, gives us the rectangular equation y = x² + 7.
00:40:52.100 --> 00:40:59.100
But we have to have this restriction, that the domain has to be x going from 0 up until infinity, not including 0.
00:40:59.100 --> 00:41:04.800
The fourth example: Notice that x = cos(θ), y = sin(θ), gives a circle centered at (0,0) with radius 1.
00:41:04.800 --> 00:41:08.300
We won't show this precisely; if you are not sure about this, try graphing it--it is pretty cool.
00:41:08.300 --> 00:41:11.500
It comes out to be pretty clear, if you just go through a few points.
00:41:11.500 --> 00:41:21.500
So, x = cos(θ); y = sin(θ) gives us (0,0) with radius 1; we end up getting a circle with radius 1.
00:41:21.500 --> 00:41:28.200
All right, let's think about if we wanted to give a parametric equation for a circle centered at (3,-2) with radius 5.
00:41:28.200 --> 00:41:30.000
How could we do each one of those things?
00:41:30.000 --> 00:41:36.200
Well, we could move the center by adding things to x and y.
00:41:36.200 --> 00:41:42.800
If we have x = cos(θ), y = sin(θ), well, if we just add 3 to x (it is 3 + cos(θ)),
00:41:42.800 --> 00:41:49.500
well, then, all of our x-values will have shifted the entire thing horizontally to the right by 3.
00:41:49.500 --> 00:41:53.000
We will have shifted the entire thing horizontally, because we just added the 3 to x.
00:41:53.000 --> 00:41:56.100
So, every x point just moved over 3.
00:41:56.100 --> 00:42:00.000
If all of the points move over 3 at once, we have just moved the center 3 horizontally.
00:42:00.000 --> 00:42:03.800
If we want to move the center vertically, we just add or subtract the amount to our y.
00:42:03.800 --> 00:42:18.500
We can move the center to (3,-2) with x = 3 + cos(θ), y = -2 + sin(θ).
00:42:18.500 --> 00:42:27.200
And that will give us a circle that is moved over 3 and down 2, and so we will be down here.
00:42:27.200 --> 00:42:38.000
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.
00:42:38.000 --> 00:42:42.700
If we make them bigger by 5 on the whole thing, then that means every point of it just went out by 5.
00:42:42.700 --> 00:42:49.100
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.
00:42:49.100 --> 00:42:59.800
We can change the radius to 5 by having it be x = 5cos(θ), y = 5sin(θ).
00:42:59.800 --> 00:43:10.200
And that will end up giving us a larger circle that now has a radius of 5.
00:43:10.200 --> 00:43:14.300
Notice: if you wanted to make an ellipse, where, instead of having a constant radius everywhere,
00:43:14.300 --> 00:43:19.500
you wanted to maybe make the radius (no longer technically a radius) smaller on the top,
00:43:19.500 --> 00:43:24.400
but then expand out, and then smaller again--if you wanted to have a major axis and a minor axis--
00:43:24.400 --> 00:43:29.600
you could not use the same number multiplied on your cos(θ) and your sin(θ),
00:43:29.600 --> 00:43:33.600
so that one of them will end up being larger out, and then it will shrink down for the other one.
00:43:33.600 --> 00:43:39.000
That is one trick if you want to be able to make an ellipse.
00:43:39.000 --> 00:43:46.200
OK, we take these two ideas, and we put them together; we combine moving the center and expanding the radius out.
00:43:46.200 --> 00:43:51.500
So now, we have a radius of 5 that is 5cos(θ) and 5sin(θ), x and y respectively.
00:43:51.500 --> 00:44:07.200
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).
00:44:07.200 --> 00:44:24.200
That would end up getting us a circle that starts at (3,-2) and has a radius of 5; cool.
00:44:24.200 --> 00:44:29.800
All right, one little idea, before we get to our very last example: projectile motion.
00:44:29.800 --> 00:44:32.700
This is a really good use of parametric equations.
00:44:32.700 --> 00:44:36.800
A projectile that is launched from some starting location, whether that means thrown,
00:44:36.800 --> 00:44:40.900
shot from an arrow, shot from a gun, thrown out of a catapult--whatever it is--
00:44:40.900 --> 00:44:48.400
any projectile--anything that is moving--a human cannonball--whatever it is--anything that is moving from some starting location, (d,h),
00:44:48.400 --> 00:44:55.900
where d is the horizontal location and h is the vertical height, with an initial velocity of "v naught," v₀,
00:44:55.900 --> 00:45:04.900
pronounced "v naught," N-A-U-G-H-T, like "all for naught," and an angle of θ above the horizontal
00:45:04.900 --> 00:45:13.200
(how much above the horizontal--if we wanted to draw that in here, then this here would be our angle θ,
00:45:13.200 --> 00:45:19.400
and this is our v₀, how fast we started out before gravity started to affect us and pull us back down to the earth),
00:45:19.400 --> 00:45:24.800
if we have these ideas here, it can have its motion described by the parametric equations
00:45:24.800 --> 00:45:33.000
x = v₀cos(θ)t + d (our starting horizontal location),
00:45:33.000 --> 00:45:41.300
and y = -1/2gt² + v₀sin(θ)t + height.
00:45:41.300 --> 00:45:44.500
In the equation for y, we are probably wondering what this g is.
00:45:44.500 --> 00:45:49.200
Well, this g is the constant of acceleration for gravity.
00:45:49.200 --> 00:45:56.200
So, on earth, we have an acceleration of gravity of 9.8 meters per second per second.
00:45:56.200 --> 00:46:00.800
Equivalently, in the imperial system, it is 32 feet per second per second.
00:46:00.800 --> 00:46:07.500
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.
00:46:07.500 --> 00:46:15.900
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.
00:46:15.900 --> 00:46:17.600
Let's understand just how this is working.
00:46:17.600 --> 00:46:20.400
We have this initial location that it gets shot out of.
00:46:20.400 --> 00:46:28.100
Then, this velocity ends up going out, and then gravity pulls the thing down; gravity is always pulling down on the object.
00:46:28.100 --> 00:46:33.300
It gets pulled down more and more and more and more and more, until eventually it lands and hits the ground.
00:46:33.300 --> 00:46:42.400
If you take any object, and you toss it up, if you were to carefully graph out what it looked like--
00:46:42.400 --> 00:46:47.400
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.
00:46:47.400 --> 00:46:51.200
Any object ends up having a parabolic arc to it.
00:46:51.200 --> 00:46:56.700
And so, that is what we are seeing, because we are seeing this parabolic arc as a parametric equation.
00:46:56.700 --> 00:46:58.500
All right, we are ready for this final example.
00:46:58.500 --> 00:47:05.300
We have a reminder of these formulas on the top, and our problem is: A marauding horseback archer fires an arrow at a castle.
00:47:05.300 --> 00:47:08.100
And he is on a height of 2 meters, because he is on top of a horse.
00:47:08.100 --> 00:47:11.600
So, he fires, and he starts at a height of 2 meters; and the arrow comes out
00:47:11.600 --> 00:47:17.400
with a speed of 50 meters per second and an angle of 25 degrees above the horizontal.
00:47:17.400 --> 00:47:23.800
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.
00:47:23.800 --> 00:47:30.900
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.
00:47:30.900 --> 00:47:34.900
How far above the wall is the arrow when it flies over?
00:47:34.900 --> 00:47:40.100
Let's draw a little picture: we see our man--he is on horseback, but I will not draw the horse.
00:47:40.100 --> 00:47:46.400
And there are castle walls out here in the distance.
00:47:46.400 --> 00:47:53.900
And so, there is a distance of 80 meters between him and them, and the castle walls are 10 meters tall.
00:47:53.900 --> 00:48:02.300
He fires an arrow, and it flies through the air.
00:48:02.300 --> 00:48:11.800
And the question we want to know is: Just as it gets above this wall, what is the extra height above that wall?
00:48:11.800 --> 00:48:14.100
How high is the arrow above the wall?
00:48:14.100 --> 00:48:18.500
And then, it will end up coming and landing on the other side; hopefully it won't hurt anybody.
00:48:18.500 --> 00:48:20.000
Well, he is a marauder.
00:48:20.000 --> 00:48:23.000
OK, let's see how we can figure this out.
00:48:23.000 --> 00:48:27.100
Our first question is when the arrow is above the wall.
00:48:27.100 --> 00:48:33.400
We have this great formula here; we can figure out what the height of the arrow is if we know the time,
00:48:33.400 --> 00:48:39.400
because we know what v₀ is (it is 50 meters per second); we know v₀ = 50;
00:48:39.400 --> 00:48:47.100
we know θ = 25 (he fired it at an angle of 25 degrees above the horizontal);
00:48:47.100 --> 00:48:54.700
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.
00:48:54.700 --> 00:48:58.600
But we don't know what time it is before it manages to make it over those walls.
00:48:58.600 --> 00:49:05.200
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?
00:49:05.200 --> 00:49:13.300
x = v₀cos(θ)t + d: we know what v₀ is--it was 50; we know what θ is--it is 25.
00:49:13.300 --> 00:49:18.700
We know how far they are, so we know what our x-value is going to be: it is 80 meters away.
00:49:18.700 --> 00:49:23.600
So, finally, what is our d? That is the one thing we don't know there.
00:49:23.600 --> 00:49:29.000
What is the d? Well, let's just say that where the horseback rider starts is 0.
00:49:29.000 --> 00:49:38.900
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.
00:49:38.900 --> 00:49:43.900
The castle walls are at 80; he starts at 0 in terms of horizontal x location.
00:49:43.900 --> 00:49:45.800
At this point, we are ready to solve this thing.
00:49:45.800 --> 00:49:54.600
In general, we have that, for any horizontal location, x is equal to 50 (our initial speed, v₀),
00:49:54.600 --> 00:50:04.600
times cosine of 25 degrees, times the amount of time that the arrow has flown, plus our initial location (our initial location was 0).
00:50:04.600 --> 00:50:11.200
At this point, we want to solve: so at x = 80, our time is equal to what?
00:50:11.200 --> 00:50:18.700
We plug in 80 = 50cosine of 25 degrees, all times time.
00:50:18.700 --> 00:50:28.800
We divide by 50 times cosine of 25 degrees, so we get t = 80/50, times cosine of 25 degrees.
00:50:28.800 --> 00:50:38.200
We plug that into a calculator, and we get that t is approximately equal to 1.765 seconds.
00:50:38.200 --> 00:50:45.300
So, after 1.765 seconds of flight time, the arrow is now at the horizontal location of the walls.
00:50:45.300 --> 00:50:50.900
So, after 1.765 seconds, we are at the walls; so now we can plug that in; and we can figure out,
00:50:50.900 --> 00:50:57.800
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.
00:50:57.800 --> 00:51:12.100
So now, we use y = -1/2 times 9.8 (our acceleration due to gravity) t² + v₀...
00:51:12.100 --> 00:51:19.200
50 times sin(θ), sine of 25 degrees, all times time, plus our initial starting height;
00:51:19.200 --> 00:51:26.100
our initial starting height was 2, because he fires the arrow--he is on top of a horse,
00:51:26.100 --> 00:51:30.200
so he is firing it from above the ground; he is not firing it from actually the level of the ground.
00:51:30.200 --> 00:51:37.800
So, we will start working through that: we want to plug in at time = 1.765 seconds,
00:51:37.800 --> 00:51:40.600
because that is the time that we are interested in knowing the height.
00:51:40.600 --> 00:52:02.900
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.
00:52:02.900 --> 00:52:06.000
It is kind of a lot there; but at this point, we can work this all out.
00:52:06.000 --> 00:52:11.600
We work it out with a calculator, and we get that it is at 24.03 meters high.
00:52:11.600 --> 00:52:16.700
So, the arrow is 24.03 meters high when it gets to the walls.
00:52:16.700 --> 00:52:24.500
However, that is not our answer; we were asked how far *above* the walls when it gets to it.
00:52:24.500 --> 00:52:26.700
So, how far above the wall is it?
00:52:26.700 --> 00:52:36.500
At this point, we take 24.03 minus...the height of the walls is 10 meters tall, so minus 10.
00:52:36.500 --> 00:52:49.400
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.
00:52:49.400 --> 00:52:52.000
All right, that finishes up for parametric equations.
00:52:52.000 --> 00:52:57.300
The important part is to think about it as describing the motion of an object in terms of its time.
00:52:57.300 --> 00:53:01.600
Try to think about it as how time would change x, how time would change y...
00:53:01.600 --> 00:53:07.800
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.
00:53:07.800 --> 00:53:10.700
Mainly, experience is a great way to learn how to do these things.
00:53:10.700 --> 00:53:14.000
But you can really speed up the process of learning and understanding parametric equations
00:53:14.000 --> 00:53:19.200
by just playing around, honestly, for five or ten minutes with a graphing calculator--just playing around,
00:53:19.200 --> 00:53:22.600
plugging in random things, and seeing how one thing affects another thing--
00:53:22.600 --> 00:53:25.400
how changing one constant causes things to move around.
00:53:25.400 --> 00:53:30.500
Just playing around for five or ten minutes will help you so much more than trying to do 10 graphing problems.
00:53:30.500 --> 00:53:33.000
All right, we will see you at Educator.com later--goodbye!