Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Plane curves and parametric equations. Suppose that T is a number in an interval. IA. Plane curve is the set of ordered pairs XY where X equal F of Ty equal G of T for T in interval I. The variable T is called a parameter and the equations X equal F of T&Y equal G of T are called parametric equations for the curve. If we wanted to graph a plane curve described by parametric equation, we would select some values of T on the given interval. For each value of T, we use the given parametric equations to compute our X and our Y. We would then plot the points XY in the order of increasing T and connect them with a smooth curve. Parametric equations have starting and ending points. That's an important concept. We have direction and parametric curves. Let's look at some examples. If we have X equal t ^2 + 1 Y equal 5 -, t ^3 at some specific location, let's say at T equal 2. Literally what we're going to do here is we're just going to stick to in for the T and figure out what our X value and our Y value would be. So if T is given to be two, we can see that our X is going to be 5 and our Y is going to be 5 -, 8 or -3. So when T is 2, we have a .5 negative 3:00. What if we were asked to do it on an interval between -3 and 2? So let's have T equaling -3, T equal -2, T equal -1, T equals 0, T equal 1, and T equal 2. If we use the equations X equal t ^2 + 1 and Y equal 5 -, t ^3, we should be able to come up with points at each location. So X is going to equal -3 ^2 + 1 or 10 and Y is going to equal 5 - -3 ^3 -3 ^3 is going to give us -27 minus. The negative is +2727 + 5 is going to give us 32. So the .1032, looking at the next one, we're going to put in -2 for our T, so -2 ^2 + 1 or X is five, Y is 5 - -2 ^3. So -2 ^3 -8 minus the -8 is positive eight, 5 + 8 is 13. So putting in -1 next. Now there's obviously infinitely points on this interval. So if we put in -1 ^2 as 1 + 1 is 2, Y is going to equal 5 - -1 ^3 or the .26. If we go to X or T equaling zero, 0 ^2 + 1 is going to give us one. Y is going to be 5 - 0 ^3, or the .15 X equal 1 ^2 + 1 would give us 2. Again, Y equal 5 - 1 ^3 would give us four. And a moment ago we found that at T equal 2 we have 5 negative 3. So if we wanted to know what this graph was actually going to look like, we would plot those points on that interval. So 1032 somewhere way up there, 513-2615, then 245 negative 3. So if we plotted those, this is absolutely not to scale, we might get something that looked like that. We could actually come up with what the equivalent equation might be by solving for T. We're going to do that in the next example. So if we know that X equal root T&Y equal t -, 1, we could first of all figure out how to graph this. We could give ourselves some values. We can see here that the domain for T would have to be greater than or equal to 0 because we have a square root. So we could say what happens when T is 0 and T is 1 and I'm going to jump to four because I know sqrt 4 and then I'm going to jump to 9:00. So if we had this at t = 0, we'd have 0, -1. At T equal 1, we'd have 10. At T equal 4, we'd have 2-3. At T equal 9, we'd have 3/8, literally sticking the TS in for the X and the Y equations. But what if we actually wanted to know what this equation was without TS? We wanted to come up with the equation without our TS in there. So we would realize that if Y equal t - 1, we could think of T equaling y + 1. So when we have the X equal to square root of T instead of square root of T, we could think of that as sqrt y + 1. So we'd have X equal to sqrt y + 1. That's going to be an equation of a parabola. So at zero, we're at -1. At one, we're at zero, At 2, we're at three, At 3, we're at 8:00. We actually have direction here because T can't be a negative, so we'd have a starting point of zero -1 it's going to be half a parabola going out to Infinity. For this example, we're asked to graph it, but it might help us to know what this is going to look like. So we might actually solve it in terms of getting rid of the TS. To start, we know that X / 2 would equal T, so if we substitute that into this other equation, why would it equal X / 2 instead of the t -, 1? Because it's an absolute value. We're going to have a positive case and a negative case if the inside piece was greater than or equal to 0. So X / 2 greater than or equal to 1X greater than or equal to two. So if the inside piece was greater than or equal to two, we would have it be exactly the way it is without the absolute values. But if the inside is less than two, when we put a number less than two and for X, we're going to get a negative on the inside. Well, how do I get rid of a negative on the inside? We take the opposite of a negative, and an opposite of a negative makes that a positive. So we get negative X / 2 + 1. So these are really going to be two line segments. And what we want to do is we want to put in some T values. Let's put in T equal, 2T equal one, t = 0, T equal -1 negative 2. Literally just choosing some random points. So if T was 2, looking at the original, we'd get X being four and Y being one. If T is one, we'd get 2, 0. If T is 0, we'd have 0, one. If T is -1, we'd get -2, 2. If T is -2, we'd get -4, negative +3. So plotting these points, we'd have 4 12001, negative 2/2, negative 4/3. So we're going to get a graph that looks something like this, which is what we would expect. We had the absolute value of a line if we thought about starting at negative Infinity. That tells me directions going to come in this direction and going off towards the right. The next example is going to involve trig and the easiest way to do this is to use our trig identity that says cosine squared plus cosine squared of an angle plus sine squared of an angle equal 1. So I'm going to start by saying X - 1 equal 3, cosine T divide by 3. So X - 1 / 3 is cosine Ty -2 equal 3, sine Ty -2 / 3 equals sine T. Now we know that cosine squared T plus sine squared T equal 1. So we have X - 1 / 3 ^2 plus y - 2 / 3 ^2 equaling one. The X - 1 quantity squared over 9 + y - 2 ^2 / 9 equaling one. This is actually going to be a circle where we have the radius being 3 and the center being 1/2. So center 12, the radius three, we go right three, we go left 3, we go up three and we go down 3. Not to scale, let's see, we go up 3123 up there somewhere and right three and left three and down 3. So now we actually have a restriction of 0 to 2 zero to π. So we actually don't want this full circle. What happens if we put in T is 0? If T was 0, we'd get 1 + 3, cosine of 0, 2 + 3 sine of 0. Well, cosine of 0 is 1, so we'd have 4:00, Sine of 0 is 0, so we'd have 4 two. So our first point would be right here. This would be our starting location if we thought about putting in T equal π halves. Next, 1 + 3 cosine π halves, 2 + 3 sine π halves. Well, cosine of π halves is 0, so we're going to have one sine of π halves is one, so the .15. So here's our second point. So that's going to give me a direction. If here's my starting, I know I'm going around the circle counterclockwise. My ending point is going to be at π. So we have 1 + 3 cosine π 2 + 3 sine π cosine π is -1 so 1 - 3 negative 2 cosine π is 0 so -2 two. So our ending point is going to be right here. So the portion of the circle we're looking at is only the top half starting at 4 two and ending at -2 two based on the restrictions on our T. If we looked at a generic one, we'd have solving for our cosine again and then solving for our sine. Knowing that our cosine squared plus our sine squared equals 1, we'd get the equation X -, H ^2 + y -, K ^2 equal R-squared, which we know is a circle. Doing another one that's generic solving for our cosine X -, H / a equal cosine Ty minus K / b equals sine T so cosine squared plus sine squared equal 1. So X -, H quantity squared over a ^2 + y - K ^2 / b ^2 equal 1. And this is really just a graph of an ellipse. The last one we're going to look at X -, H / a secant Ty minus K / b equal tangent T. Now what's our identity that relates secants and tangents? We have one plus tangent squared T equaling secant squared T so 1 + y -, K / b ^2 equaling X -, H / a ^2. If we want to have one equal, we get one equaling X -, H quantity squared over a ^2 -, y -, K ^2 / b ^2. And this is just the generic formula for a hyperbola. Thank you and have a wonderful day.