 In this video, we'll look at another example of a path described by parametric equations and investigate the behavior of the curve representing that path. Consider the curve described by the parametric equations x of t being negative t cubed minus t squared and y of t being 1 third t cubed minus 2 t squared plus 5. We'll take a look at this graph in our calculator. Okay, that gives us a good sense of what this looks like. We see we have some interesting things happening here. We have some places where the slope is zero and other places where it may be undefined. Let's take a look at the slope of the curve at any time t. Recall that the slope is found by calculating dy dx but since y and x are both functions of t, by chain rule we find that the slope is found by taking the quotient of dy dt and dx dt. Well, since x of t was negative t cubed minus 3 t squared, dx dt is negative 3 t squared minus 6 t and that's our denominator. y of t was 1 third t cubed minus 2 t squared plus 5. So its derivative is t squared minus 4 t. I'm going to factor the numerator and denominator into irreducible factors. Now we know that there exists a horizontal tangent line when dy dx is zero. Now this happens when dy dt is zero but dx dt is not. Now looking at dy dx here, we see that dy dx is zero when this numerator is zero and the denominator is not. dy dt is zero when t is zero and t is four. Now we also notice that the denominator or dx dt is zero when t is zero. So we don't have a horizontal tangent line there but we do when t is four. When t is four, our coordinate, if we substitute t being four into our original equations, we get negative one-twelve, negative five and two-thirds. So looking back at our graph, we have a horizontal tangent line. Somewhere along here, of course the pixelation of the calculator doesn't allow us to really focus exactly where that horizontal tangent line is but we see it's when t is four, we have approximately negative one-twelve. This window is to negative two-hundred. So that should be just about here. So negative one-twelve, negative five and two-thirds is the point at which we have a horizontal tangent line. Now with dy dx, we know that this is undefined when the denominator, which is really dx dt, is zero but dy dt is not. Well what does an undefined slope give us? It gives us when we have a vertical tangent line. We notice that dx dt is zero when t is zero and t is negative two. Now as before, we know that when t is zero, dy dt was also zero which is the numerator here. So we know that we don't have a vertical tangent line there but we do have a vertical tangent line when t is negative two. t is negative two gives us the point, the coordinate, negative four, negative five and two-thirds. Let's take a look at this on our graph. We have a vertical tangent line at the point we just found when t was negative two. That corresponded to negative four, negative five and two-thirds and we see that our graph confirms that this was when t was four and this is when t was negative two. Now we noted that both dy dt and dx dt are zero when t is zero and that corresponds to a point at zero five given our parametric equations. Let's take a look at that on our graph. We notice back here we have a cusp at this point. That's when the particle was actually at rest. dx dt and dy dt are both zero and this is when t was zero. So no tangent line there, neither vertical nor horizontal or any other type. In fact, we don't have a tangent line. That's at the cusp zero five. Let's take a look at concavity. For concavity we need to find the second derivative of y with respect to x. But again, since y and x are functions of t we have a bit of a process here. In order to find the second derivative of y with respect to x we actually split our derivative into the numerator being the first derivative with respect to t of dy dx that we just calculated divided by dx dt. So I'm going to differentiate with respect to t. Now I'm actually doing something a little special. Notice that I'm writing dy dx without the common factor of t that was both in the numerator and the denominator. I'm going to remove that This expression is only valid for non-zero t values. We notice that when t was zero anyway we had a cusp. So we didn't have concavity at that point. So we can actually remove that as essentially a removable discontinuity. And then the denominator is the derivative of x which we saw before and we differentiate the numerator here With respect to t we obtain negative 18 over 9 times t plus 2 squared all over negative 3t times t plus 2 verify that when you differentiate that numerator you do ultimately get this derivative. If we clean all of this up I get a positive 2 divided by 3t times t plus 2 cubed because I have a squared term here I'm dividing that by another factor of t plus 2 giving me t plus 2 cubed. We note that this expression changes sign when the denominator changes sign and that happens when t is zero and t is negative 2 Again verify this using what you know about calculus and this tells us that the concavity of the curve changes at each of those times. Let's confirm this with our graph When t was zero we changed from being concave down to concave up and when t was negative 2 we changed from being concave up here to concave down.