7.1 Parametric Equations

[STARTS AT 8:15]

Instructor: All right, example 2, eliminate the parameter for each of the plane curve described by the following parametric equations and then described the resulting graph. So what we're going to do is actually create a function of y in terms of x. 

How we're going to do that is we're going to solve for t in one of these equations. I'm going to go ahead and solve the x equation, my x of t. I'm going to solve that for t, so this would be x squared equals 2t plus 4. Subtracting 4, x squared minus 4 equals 2t, which then gives me t equals x squared minus 4 over 2. 

Nothing crazy complicated about that. I've just isolated my t. And when I go back to my y of t, that is 2t plus 1. Replacing my new value of t in terms of x, x squared minus 4 over 2. The 2's will reduce, and so I'll get x squared minus 4, plus 1. 

So x squared minus 3 is my equation without a t, so just in terms of y and x. So this would be-- I'll go ahead and sketch-- this'll look like. This would be a parabola opening up. 

However, we have a restriction on t. So my beginning value of t is negative 2, which would make that at 0. All right, so if I have t of negative 2, it gives me an x value of 0 if I want an accurate picture here. 

If I plug in negative 2 there, I get negative 3. So my beginning value is actually the vertex of that parabola. Evaluating at 6 will give me a 4. I have a 6, I get 13. 

So actually, this end of my graph does not work, because my starting value's over here and my ending value is somewhere along here and it is traveling in that direction. So effectively it is a piece of a parabola. 

All right, next, we are going to eliminate the variable with these. Now this is actually very similar to what you would do on that last question on 1c to see that it's actually a circle. 

But let's go and go with this. First, I'm going to isolate my t value. So this would be x over 4 equals cosine t, which means t equals arccosine of x over 4. 

Well, taking my y value here, my y function, 3 sine t-- well, rather than putting in t, I'm going to have arccosine of x over 4. Now I'm going to need a right triangle for this. 

Well, if the arccosine-- or if cosine, actually, we'll just go back to that-- if cosine t equals x over 4, that means the adjacent over the hypotenuse, so x, 4. So my opposite side here is the square root of 16 minus x squared. That's the Pythagorean theorem there. 

All right, so if I want to find the sine of that, the sine of that would be the opposite over the hypotenuse. So this is 3 times square root of 16 minus x squared over 4. So that is my function. 

All right, now one thing is, when we take the square root, we actually are implicitly specifically saying the positive root. When in reality as a graph, we have a plus or minus. So we have something along these lines, which is an ellipse. 

So I want to take just a side note here. And if I take this identity-- so this is a separate way to realize the same concept-- cosine squared plus sine squared t equals 1. If I square my x over 4 equals cosine t and my y over 3 sine t, and my x over 4 equals cosine 7, this is x squared over 16, plus y squared over 9, equals 1, which is the graph of a hyperbola with a major axis of 4 and a minor axis of 3. 

So that exactly matches what this is. And it will, in fact, be oriented this way. Because it's going from 0 to 2 pi, we should make a full ellipse there. 

[ENDS AT 14:06]