Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Parametric equations X equal F of T&Y equal G of T So the curve of the points X, Y defined by the functions F of T&G of T is A is a parametric curve over some interval I of T values. T is a parameter of the curve and I is the parameter interval. If the curve is on a closed interval, then there is an initial point and a terminal point. So this is a way that we can define graphs that aren't currently functions as two separate functions, looking at the X individually and then the Y individually. If we look at an example, X equal the square root of T to the fourth plus 9 and Y equal t ^2 where T is greater than or equal to 0. The first thing that we sometimes want to do is figure out what it's going to actually look like in a Cartesian coordinate system. So if we thought about plotting T values 0123 etcetera. So if T is 0, we stick 0 in here and 0 + 9. Square root of that is 3 and 0 ^2 is 0. So when T was 0, we have a point of 30. That would be our initial point because T was greater than or equal to 0. So we could plot that point on a graph. Right here would be our initial point. If we want to plot a few more points to figure out what's happening, we could say let's T equal 1, so 1 to the fourth plus 9 sqrt 10 bigger than three, less than four and one squared, so bigger than three less than 4, 1. So if we go out three in a little bit up one, we get that point there. If we put in two, two to the 424816 plus 9 is 25 sqrt 25 is five. So when our X values 5, our Y value is 4. So go out five and up four. And remember this is not to scale because I can't draw well when T is 3/3 to the fourth plus 9 sqrt 90. So more than 9, less than 10, 3 ^2 is 9. So we come out here and we get another point. We have a direction. These errors are showing the direction. We had our initial point and we're going to just keep going out and out and out. Another thing that we're frequently asked to do is to change the parametric equation into X's and Y's or Cartesian equations. So if we know T to the 4th, we could think of T to the 4th as t ^2 ^2 and t ^2 is just Y. So we could think of this t ^2 as turning into Y, and then we'd have X equal sqrt y ^2 + 9. Squaring each side, we'd get X ^2 equal y ^2 + 9. If we subtract the Y ^2 over, we'd get X ^2 -, y ^2 equal 9. We know this is a hyperbola. If we divide everything 3 by 9, we'd get a one here. So X ^2 / 9 -, y ^2 / 9 equal 1. Now the parametric gives us more information than the Cartesian the way it is. And the reason is the parametric told us T had to be greater than or equal to 0. So we knew that we wanted X values that were positive because we had X equaling a square root. And we also knew that we wanted the Y values that were positive. And so when we look at this, the values of X that are positive and Y that are positive would only be this portion of the hyperbola. Let's look at another example. In a circle. We know that the generic formula is X ^2 + y ^2 equal a ^2. So if we thought about X equaling a cosine T&Y equaling a sine T, these are parametrics. So if we divided each side by a, we get X / a equal cosine T and y / a equals sine T We know our trigonometric functions and we know that cosine squared of an angle plus sine squared of the angle is 1. So in this case, we could put X / a ^2 + y / a ^2 equaling 1. And that's just the equation of a circle. So X ^2 + y ^2 equal a ^2. Now in this case, we actually have a starting point and an ending point. Let's say that our parameter that they're going to give us is 0 less than or equal to T less than or equal to 2π. If that was the case, we'd think about what happens when our T is 0. So if we have T being 0, where's our X and where's our Y? Well, if T is 0, cosine of 0 is 1, so we know that our X is a sine of 0 would be 0, so our Y is 0. So out here somewhere is our initial point, a 0 for this interval. Now we can change the interval and that will change our initial starting point. If we thought about it, Pi halves, we know that cosine of π halves is 0, so X is going to be 0. Sine of π halves is 1. So our Y is going to be A. So this is going to have a direction that's coming around like this. It's a circle, so it's going to come all the way back. If we wanted it to be half a circle, we would change our parameters and say instead of to 2π, we'd go to π. If we wanted the circle to perhaps start up here, we would make our T say we wanted just half of the circle. We would have, In this case, Pi halves less than or equal to T, less than or equal to three Pi halves. We would have a starting point and an ending point. So our initial point and our ending point. If we had an ellipse that was defined as X equal a cosine T&Y equal B sine T for 0 less than or equal to T less than or equal to 2π, I'm going to do the same skills that we did with the circle. If we divide by A and we divide by B, we get X / a equal cosine T and y / b equals sine T by our Pythagorean identity. Cosine squared T plus sine squared T equal 1, so X / a ^2 + y / b ^2 equal 1, so X ^2 / a ^2 + y ^2 / b ^2 1. Now once again, we could change those conditions. They're fairly standard, but we really can change them depending on where we want our starting point to be. So if we thought about looking at this graph, if T is 0, we'd have a zero. If T is π halves, we'd have 0B. If PT is π, we'd have negative A0. If T is 3 Pi halves, we'd get zero, negative B. So this graph is going to be a graph of an ellipse. Now, it didn't tell us whether A was bigger or B was bigger, but somewhere out here we're going to have a point a 0. Somewhere up here we're going to have a .0, B, etcetera, and we're going to go in an elliptical fashion. I realize that looks like a circle, but it's an ellipse because A&B currently we think probably aren't equal. So in this case my initial point is the same as my final point. If we look at another one, how about X equals sine of 2π of the quantity 1 -, t and Y equal cosine of the quantity 2π of 1 -, t where T is going from zero to 1. Back to our Pythagorean identity, we know that sine squared of an angle plus cosine squared of an angle equal 1. In this case our angle is just going to be this 2π times the quantity 1 - t So we get sine squared plus cosine squared equal 1 or X ^2 + y ^2 equal 1. Now this time if we think about putting AT value of 0, if I stick in zero I get 1 - 0 and the sine of 2π * 0 or the sine of 0 is 0 cosine of 1 - 0 is one 2π * 1. Did I just do that wrong up here? 2 Pi 1 - 0 so 2π 1 - 0 is 1 cosine of 2π is 0 cosine of 2π of 1 - 0 the so the cosine of 2π is 1. So our starting point in this case is the .01 for the X&Y. If I stuck in something like 1/6, we'd have 1 - 1/6 or five sixths. 5 sixths times 2π would be 5 Pi thirds. Sine of five Pi thirds would be negative root 3 / 2. If we have cosine of five Pi thirds, we know that's a half. So now we have another point. If we looked at when T is a 4th, 1 -, 1/4 is 3/4, 2π * 3/4 would be 3 Pi halves. Sine of three Pi halves is -1 cosine of three Pi halves is 0. So we know this is a circle and we currently know our starting point of 01. And now we know our X is going to be negative for the next point and our Y is positive. So that gives us our direction. We're coming this way. We're going counter clockwise, and if we thought about at one, 1 -, 1 is zero, 2π times 00, sine of 00 cosine of 1 - 1 is 00 times 2π, cosine of 0 is 1. So we're back to our starting and finishing point being the same point. In this example, what if X equaled 4, sine T&Y equal 5, cosine T0 less than or equal to T less than or equal to π? If we got our sine and cosines by themselves and used our Pythagorean identity, we could see we'd have X ^2 / 16 + y ^2 / 25 equal 1, which we know is an ellipse. But what's really happening as far as our starting and our ending points? So if we looked at when T is 0, sine of 0 is 0, so our X is 0 and our Y is going to be 5. Let's choose another value in between 0 and π maybe π halves and then maybe π and see if that's enough to get us an idea of what the graph is going to be looking like. So sine of π halves is 1 and 1 * 4 is 4. Cosine π halves is 0. So we get the .40 sine of π is 00 times 4 cosine of π is -1. So in this example, we're going to have 05 as our starting point. We're going to go then to 40, and then we're going to end at zero -5. So we're going to have it coming like this. It's going to be half of an ellipse, and it definitely has a starting point and an ending point and direction. So here's our initial point, and here's our ending point. If we looked at X equals secant T&Y equal tangent T, well, we know that we have a value or we have a Pythagorean identity that says tangent squared t + 1 equal secant squared T, Or we could think of that as secant squared t -, 1 equal tangent squared T. Those are just rewriting the same formula. Secant T is X, so secant squared T is X ^2 -, 1, tangent T is Y, so tangent squared T is y ^2. If we take everything to one side, we can see this as a hyperbola. And this example, they didn't give us what our values were. So we wouldn't know a starting point and an ending point. Let's change it a little bit and let's say we're going to go to π halves to three Pi halves. So if we're doing this, if we have T is π halves, secant of π halves is going to give me, oh, undefined, isn't it? Tangent of π halves would give me 0 three Pi force. Maybe Secant of three Pi force is root 2. Tangent of three Pi force. Secant of three Pi force is negative. Root 2. Tangent of three Pi force is -1. Let's look at π. Secant of π is -1 and tangent of π is going to be 0. If we look at 5 Pi force seeking A5 Pi force is going to be negative root 2 again, but tangent of five Pi force is going to be a +1. So this is a hyperbola and it's opening left and right. If we thought about putting in our asymptotes, we would realize we have a box of one in each direction. So our asymptote is going to go like this, and we have negative root 2, negative 1. So we're down here somewhere, and then we're going to go to -1 zero, and then we're going to go to negative root 2, positive 1. So this one's going to look something like this. And officially, I should probably go back. Oh, I did say it's just less than. So it's just that portion of the hyperbola. We started with the negative X and negative Y. We went X being negative when Y was 0, and then we went to X is still negative, but Y turns positive. If we look at another one, X equals secant squared t -, 1 and Y equal tangent T. If we use our Pythagorean identity again, we get one plus tangent squared T equaling secant squared T, so y ^2 equal X. If we thought about subtracting the one, we know this is a parabola, it's going to open to the right. So now what we want to figure out is that the whole parabola or is it a portion? So if we thought about doing our T chart again, so if we have T equaling negative pie force zero and π force, that'll give us a clue as to how it ought to look. Secant squared of negative π force. So secant of negative π force is going to be root 2. Root 2 ^2 is two, 2 - 1 is 1, and the tangent of negative π force is -1 Secant squared T of 0 would be 1 - 1 is 0 and tangent of 0 is 0. Secant squared Pi force root 2 ^2 which is 2 - 1 is 1 and tangent of π force is 1. So in this case, we're going to start when X is positive and Y is negative. So X IS1Y is -1 it's a parabola. So we're starting this direction. We're coming to 00 and we're going off to positive and positive. So our graph is going to look something like that. If we were looking at a parameterization for a line through a point AB having a slope M, we know that y -, b equal m * X -, a. This is just point slope form of a line. So if we thought about letting T be any portion we want, frequently we look for pieces that are inside. So like if I let T equal this quantity that was all inside the ( t = X -, a. So we can see that X would just be t + A and then y -, b would equal m * t because we said T was this X -, a portion. So we could get Y equal Mt plus B and we would have this be for negative Infinity to Infinity because it's a line that would go on forever. Now there are many different generic equations that just happens to be one of them. We could have done something similar by letting T equal y -, b if we had wanted and solved for a different replacement line segment with endpoints. If we have two endpoints, 5 negative 2 and -4 six, we could think about this first point occurring when time is 0 and the second occurring when time is something later. In this case one, because we only have two points. So if we have time is 0 and time is one, we're going to think about when time is 0 we have an X value of five and AY of -2, and when time is one we have an X value of -4 and AY value of 6. So if we thought about we need to have our ending location equaling whatever our start location was, plus some constant times our time value, the time that's changing. So for the XS we'd have -4 equaling 5 + a * 1 or a would equal -9. So if we thought about X = 5 + -9 * t, if T was 0, we'd have X being 5. If T was one, 5 + -9 would be -4. So at T being one we have X equal -4. I'm going to do the same thing for our YS. So 6 equal -2 + b * 1 B in this case is 8, so Y equal -2 + 8 T. If T was 0, we'd get Y equal -2, which is what we wanted. If T is one, we'd get -2 + 8 is 6, which is what we want. And we would let our T be in between zero and one. Because it's a line segment with end points. It's not a continuous line. It doesn't go on and on and on. A wheel of radius A rolls along a horizontal straight line, the path traced by a point P on the wheel. Circumference is called a cycloid parametric equation. For a cycloid X equal A times the quantity T minus sine Ty equal a times the quantity 1 minus cosine TT greater than or equal to 0. This is just a circle with a set point and as the circle is rotating around this point is going to different locations and the arcs that are traced are called the cycloid. Thank you and have a wonderful day.