 Welcome back everyone. We're gonna continue our discussion. We started here in lecture 30, but we're gonna move from section 10.1 to section 10.2 about calculus with parametric curves. Based upon examples we saw in James Stewart's calculus textbook. So now that I have a good idea of these parametric curves, what we wanna do is start considering doing calculus with these parametric curves. So we're talking about limits, derivatives, integrals, the whole shebang right there, right? And so how does one do a calculation? How does someone do calculus with parametric curves? Well, it turns out that for limits, there's really not a whole lot that's gonna be going on here. So we first also mentioned that our curve is given by X equals F of T and Y equals G of T. These are gonna be our parametric equation. So calculation of limits is gonna be basically the exact same thing we saw in calculus one. We can calculate the limit of X as T approaches A, some value. We can calculate the limit of Y as T approaches B or something like that, right? We can calculate these limits of the function. And so if we wanna calculate the limit, like say X approaches some number A, what happens to Y? Well, we just have to figure out, well, how does T approach this value, right? And so, I mean, this is just gonna be a slight slide of hand here. You have to switch over. Well, we don't wanna describe it in terms of X. So let's say that we have to compute the limit X, the limit of X we call A as say, T approaches some value alpha. And then you plug that in over here. T approaches alpha. You can handle something like that. Derivatives are gonna be the same basic idea, right? We can calculate the derivative of X with respect to T, the derivative of Y with respect to T by the usual derivative rules. But the issue that gets a little bit tricky is when we start to look at slopes of tangent lines, right? What if we wanna compute Y prime? And by Y prime, we really mean DX over, or DY over DX. That is, what's the derivative of Y with respect to X? Because if we wanna describe the slope of tangent lines, then we need to have rise over run, which means a change of Y over a change of X. A small change of Y is what DY is. Small infinitesimal change of X is what DX means. So to find a change of rise over run, we need to get DY over DX. Now it turns out using the chain rule, we can accomplish this. Because notice that DY over DT by the chain rule is gonna equal DY over DX times DX over DT. Notice the DX cancel, you have DY over DT. Now to solve for DY over DX is what we have right here, just divide by the DX over DT on both sides of the equation. And you get this guy right here. DY over DX is equal to the derivative of Y with respect to T and the derivative of X with respect to T. Now because there's multiple variables going on here, one has to be careful when we say things like Y prime or X prime. In this context, we will never ever say X prime. We'll be specific in what we mean here. And when we say Y prime, we will always mean DY DT, sorry, DY DX in this context. If we wanna talk about the derivative of Y with respect to T, we will write it out, DY over DT to avoid confusion right here. So DY over DX or Y prime will equal the derivative of Y with respect to T over the derivative of X with respect to T. Now this is a valid form as long as the derivative with respect, derivative of X with respect to T is not zero. If it were zero, we actually would have a vertical tangent line at the location so we can handle it. And then we get an equation for our tangent line, usual equation, Y equals K is equal to DY over DX times X minus H. And so we could describe tangent lines or parametric curves so long as we remember to adapt the derivative. We need DY over DX. We can also do similar things for higher derivatives but one has to be careful, right? If we take the example of the second derivative, D squared Y over DX squared, for short hand we will call this Y double prime, right? Now the second derivative by its definition is the derivative of the derivative. So we have to take the derivative with respect to X of the first derivative, DY over DX. And so by the principle we saw above, you're gonna take the derivative of Y prime with respect to T and divide that by the derivative of X with respect to T. Or in short hand I like to think of it as DY prime over DT divided by DX over DT. The important thing to realize here is that the second derivative of Y with respect to X is not the quotient of the second derivative Y with respect to T with the second derivative of X with respect to T. That's a very common confusion that students try to do right here. Why is there, so this, so the thing is our second derivative doesn't look like this, it should look like this although I don't wanna put an equal sign there. These two things are what's equal. DY prime over DT over DX DT, that's the correct anti, that's the correct second derivative right there. And so I wanna show you an example using these principles here. So we have a parametric curve C which is given by the formula X equals T squared and Y equals T cubed minus three T. And so you can see an illustration of what this thing looks like with some examples given to you. For example, when T equals zero, you'll have the origin zero zero. When T equals one, you'll be at the point one, comma, negative two. When T equals, let's see, when T equals, well, actually this would be T equals the square root of three, we'll come back to this in a moment. We'll come back to that one in a moment. When T equals negative one, you'll be at the point one, two and you get this sort of loop-de-loop, I guess it goes more like this. You get this loop-de-loop, I look something like this, like you're right in the credit coaster at Disneyland or something like that. So this is what the graph is gonna look like. What I wanna do is I wanna actually describe the two tangent lines at the point three zero. So we can see on the graph that this graph seems to intersect three zero twice, right? Because if you follow the loop-de-loop, you're gonna pass through three zero here and you're also gonna pass through it right here. And as such, you actually get two tangent lines at the same location, two distinct tangent lines. You can see one of them right here in green and then another one right here right there. Clearly, my squiggles aren't too straight there. How are we gonna find them? Well, the first thing to do is how do you get zero? How do you get the point three zero in the first play? What's the parameter T? Now, notice that if X, Y equals three zero, that would imply that X, which is T squared equals three and Y, which is equal to T cubed minus three T is equal to zero, right? And solving these equations. So the first one for X is pretty easy. If you solve for T, you're gonna get T equals plus or minus the square of three. So that's why you get two, you're gonna get two tangent lines because there's two parameters that might work. But be aware that T equaling plus or minus square of three makes X equal to three. Do those simultaneously make Y equal to zero? And if you try to solve this equation, you can factor out a T, you get T squared minus three equals zero. And so you're gonna see that, okay, T could equal zero or plus or minus three, square of three. Okay, so those work. So this point right here, if you come through on this side, that's what happens when you get T equals negative root three. And when you come through on this side, that's when T equals the square of three, which the square of three will be between one and two as a real number. So that does match up with what we see here in the graph. Can we find an equation for this? Well, the first thing to do is we have to find the derivative. We wanna find, we need to find DY over DX, which like we saw before, this is DY over DT divided by DX over DT. All right, and so let's do those calculations. The derivative of X with respect to T, we're gonna be taking the derivative of T squared here, T squared prime with respect to T, you're gonna get a two T, that's nice. And DY over DT take its derivative with respect to T, T cubed minus three T by the usual power, we get three T squared minus three, like so. And so DY over DX then has the form three T squared minus three over two T. And so that is our first derivative right here, which if you want to, you can factor out some of the coefficients, you get three halves, and this will sit above T squared minus one over T. And so we're interested in what is the derivative, what's the slope of the tangent line when T, so if you evaluate this thing, when T equals plus or minus the square root of three, let's evaluate this here, T equals plus or minus the square root of three. So when you evaluate it at plus or minus the square root of three, notice you're gonna end up with three halves. Well, when you square right here, the square root of three or the negative square root of three, you're gonna get a three, three minus one. And on the bottom you're gonna see plus or minus the square root of three. And so of course, three minus one is two, that's gonna cancel with the two that's down here. So we end up with three over plus or minus the square root of three, which if you rationalize that thing, you're gonna get plus or minus the square root of three. So the slopes of these tangent lines are both square root of three, well, plus or minus square root of three, the positive one will be positive, the negative one will be negative. And so once you have the slope here, you can then plug in and find the tangent line. You're gonna get y minus zero is equal to plus or minus the square root of three times x minus three, like so. And then coming back up to our diagram, you see that when you are on the negative approach, you're gonna get this tangent line right here, y equals negative square root of three times x minus three. When you're on the positive approach, you're gonna get this tangent line right here, y equals positive square root of three times x minus three. So we can find tangent lines exactly like we did before, so long as we make sure we're using for our tangent slope, dy over dx. Don't be, we only use dy over dt and dx over dt to help us find dy over dx. We need rise over run. But using the second derivative, we can also make some statements about concavity. Oh, I'll get to that one just a second. I wanna look for horizontal tangent lines and vertical tangent lines. So what we know is a horizontal tangent line coincides when dy over dx equals zero. Now, the only way that a fraction can equal zero is when, let me kind of erase this. The fraction can only equal zero when it's numerator equals zero and that numerator is gonna be dy over dt. So the horizontal tangents are gonna coincide when dy over dt equals zero. And similarly, the vertical tangents is when dy over dx is undefined, but more specifically when its denominator goes to zero. Well, the denominator of dy over dx is dx over dt. So the vertical tangents coincide when the denominator goes to zero right there. All right, let's go back there and I'm gonna erase this line so we have some room to work here. So from what we saw above, dy over dt was three t squared minus three. So if we solve that, three t squared minus three equals zero, we can divide everything by three. We get t squared minus one equals zero. This would tell us that t squared equals one or that is t equals plus or minus one. The horizontal tangents are gonna occur at the parameters t equals one and t equals negative one, which if we come here, they're already labeled here on the graph. When x equals one, we're gonna be at the point x equals one, y equals negative two, that's a horizontal tangent. And then the other ones at t equals negative one right here, which will correspond to the point x equals one, y equals positive two, like so. What about the vertical tangent lines? Well, we have to solve the equation dx over dt equals zero, which as we saw before, dx over dt is equal to two x. So that's actually a fairly simple equation to solve. We have to solve two t, oops, I have to color code this correctly, two t equals zero. Well, that implies that t equals zero. So you'll have a vertical tangent when the parameters equal to zero. So coming up over here, we see that there's a vertical tangent right here when t equals zero, that'll be when x equals zero and y equals zero. And we found this coordinates by plugging into the parametric equations. So we can find and identify tangent lines using this notion of tangent like we did before, but we just have to kind of change a little bit how we handle the derivative. All right, now let's get to this notion of concavity. Concavity, when is the curve concave up and when is it concave down? Well, it's concave upward when the second derivative d squared y over dx squared is positive. It'll be concave downward when the second derivative d squared y over dx squared is negative. So what we have to first do is actually compute the second derivative. So what we saw earlier, y prime, y prime from the above discussion, we had three halves times t squared minus one over t. For the sake of simplicity, I'm gonna write this as three halves times t minus t to the negative one, preparing now for a derivative if we compute dy prime over dt, this is gonna be three halves comma one plus t to the negative two, which if we wanna write this back as a fraction, we can do that, but I think, do I want to, can we just leave it alone for right now? Yeah, I mean, ah, nah, let's take care of this thing. We get three halves times one plus one over t squared. We can tie, well, we wanna write this thing as a t squared over t squared like so. And so then we end up with for our dy prime dt, we're gonna get three times t squared plus one and this sits above two t squared. And so that way, d squared y over dx squared, remember this is dy prime over dt divided by dx over dt. And so we record what we have before, three times t squared plus one over two t squared divided by dx dt, which remember that was before just a two t. And so we're gonna get that, the second derivative are the so-called y double prime. This is equal to three times t squared plus one over four t cubed like so. And so we wanna determine what is this thing positive, what is this thing negative? Well, some things to notice, three is positive. You probably already knew that, four is positive. You might be confused on that one, just kidding there. t squared plus one is always positive. For any real number, t squared will always be greater than equal to zero. If you add one, it's always gonna be greater than equal to one, therefore it's greater than equal to zero. So you're always gonna get a positive here, here and here. The only one in suspect here is t cubed. When is t cubed greater than zero? Well, taking the cube root, that'll be when t is positive, right? And so we see right here that the second derivative will be positive exactly when the parameter is positive. And the second derivative will be negative exactly when the parameter t is negative. Coming back up to our picture, we get this right here. How we wanna interpret this is that when you have a negative, when you have a negative parameter, right? When you have a negative parameter, your curve is over here. So just look at the, well, I guess the direction actually is going like this, but if you look at just that blue card, notice the graph in that location is concave downward, just like we saw with the second derivative. And if we look at the portion which corresponds to the positive parameter, this half of the graph, this actually does form a concave upward curve. And so using the second derivative, d squared y over dx squared, a.k.a. y double prime, we can still determine things like concave, when is it concave up, when is it concave down? So all the principles of curve sketching that we've learned in the past apply to parametric curves. We just have to make sure that when we talk about derivatives by y prime, we mean dy over dx. This will describe tangent lines, this will describe monotonicity. And y double prime, we mean exactly d squared y over dx squared. This will describe concavity. And so one can actually use these principles to describe curves in as variables of times, or we can think of a two dimensional motion can be described as long as we make sure we're using the correct derivatives here.