7.2 Calculus of Parametric Curves

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Instructor: Calculate the derivative dy dx for each of the following parametrically defined plane curves. And then we're going to look for critical points on the graphs. I'm going to try to keep that graph hidden for the time being. Well, from what we have just seen dy dx is the derivative of y divided by the derivative of x. So the derivative of y is 2. The derivative of x is 2t. And so that will be 1 over t. 

Right, now our critical points are going to occur whenever t equals 0 because it is undefined. Well if t is 0, the point for that, evaluating each of those, is negative 3 and negative 1. So negative 3, negative 1 is where we're going to have a critical point on our graph. 

So here is the graph in very small font. And here's our point negative 3, negative 1. If you notice, the slope of that is undefined. So we can find that. And that is the only critical point here. 

And if you notice, this graph is actually not a function. That's one thing about parametric equations, we can actually have equations and do calculus with functions that do not have-- well that aren't functions, at all. 

So part b, again dy dx is the derivative of y with respect to t, which is 3t squared minus 3 over dx dt, the derivative of x with respect to t. So that is 2. Now where that's going to be 0, this is not going to be undefined at any point. But where that's going to be 0, is when 3t squared minus 3 equals zero factoring out a 3, t squared minus 1 equals 0, which means t is plus or minus 1. 

So our t being equal to negative 1, and our t being equal to positive 1, are going to produce two points. Evaluating at negative 1. We have a negative 1 for x. And evaluating at negative 1 is 6. Negative 1,6 is our first critical point. Evaluating at 1, we get a point of 3 comma. Evaluating at 1, 1 minus 3 plus 4, that should be positive 2. So 3,2. There are our critical points. And checking that with our graph, we see the point negative 1, 6 a local maximum. OK absolutely an extrema. An absolute-- eh, not quite absolute. But it's absolutely a locum extrememum. Extrema. And the point 3, 2 same concept we have a local minimum there. 

All right, 1 c. x of t is 5 cosine t. And y of t is 5 sine t from 0 to 2 pi. OK. So let's go ahead and find dy dx. The derivative of y with respect to t would be 5 cosign t. Derivative of x with respect to t will be 5, negative 5 sine t. 

Now, where are we going to have critical points here? Well, wherever five cosine t is equal to 0. Or wherever negative 5 sine t is equal to 0, because this would be undefined. Well cosine is 0 at pi over 2 and 3 pi over 2. And sine is 0 at 0 pi and 2 pi, in the interval 0 to 2 pi. So we have critical points at four places. 2 pi and 0 being identical. 

So let's go ahead and graph this. We've got our value of t, x, and y. Let's go ahead and use those intervals there. All right. So at 0, we have 5,0. Pi over 2, 0, 5. We actually had something like this in the last section. Negative 5, 0. 0, Negative 5. And back at 5, 0 as we were at 0. 

So if we graph this, what we get-- and I'm not going to do it justice, I'm sure. As we are going to have a circle. We have a circle where we have a slope that is 0, at pi over 2. OK. We have a slope 0 at 3 pi over 2. Undefined at zero, and at pi. So we do, in fact, have critical points at all of those places. And those points are given here in our chart. 

OK, so cosine and sine actually uniquely define some circles. In this particular, it has a radius of 5. So there we have it. Going in the counterclockwise direction. 

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