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# Parametric equations differentiation

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.G (LO)
,
CHA‑3.G.1 (EK)

## Video transcript

- [Voiceover] So what we have here is x being defined in terms of t and y being defined in terms of t, and then if you were to plot over all of the t values, you'd get a pretty cool plot, just like this. So you try, t equals zero, and figure out what x and y are, t is equal to one, figure out what x and y are, and all of the other ts, and then you get this pretty cool-looking graph. But the goal in this video isn't just to appreciate the coolness of graphs or curves, defined by parametric equations. But we actually want to do some calculus, in particular, we wanna find the derivative, we wanna find the derivative of y, with respect to x, the derivative of y with respect to x, when t, when t is equal to negative one third. And if you are inspired, I encourage you to pause and try to solve this. And I am about to do it with you, in case you already did, or you just want me to. (chuckles) All right, so the key is, is well, how do you find the derivative with respect to x, derivative of y with respect to x, when they're both defined in terms of t? And the key realization is the derivative of y with respect to x, with respect to x, is going to be equal to, is going to be equal to, the derivative of y with respect to t, over the derivative of x with respect to t. If you were to view these differentials as numbers, well this would actually work out mathematically. Now, it gets a little bit non-rigorous, when you start to do that, but, if you though of it that, it's an easy way of thinking about why this actually might make sense. The derivative of something versus something else, is equal to the derivative of y with respect to t, over x with respect to t. All right, so how does that help us? Well, we can figure out the derivative of x with respect to t and the derivative of y with respect to t. The derivative of x with respect to t is just going to be equal to, let's see, the derivative of the outside, with respect to the inside, it's going to be two sine whoops, the derivative of sign is cosine, two cosine of one plus three t, times the derivative of the inside with respect to t. So that's going to be, derivative of one is just zero. Derivative of three t with respect to t is three. So times three, that's the derivative of x with respect to t I just used the Chain Rule here. Derivative of the outside two sine of something, with respect to the inside, so derivative of this outside, two sine of something with respect to one plus three t, is that right over there. And the derivative of the inside with respect to t, is just our three. Now, the derivative of y with respect to t is a little bit more straight-forward. Derivative of y with respect to t, we just apply the Power Rule here, three times two is six, t to the three minus one power, six t squared. So this is going to be equal to six t squared, six t squared, over, well, we have the two times the three, so we have six times cosine of one plus three t, and then our sixes cancel out, and we are left with, we are left with t squared over cosine of one plus three t. And if we care when t is equal to negative one third, when t is equal to negative one third, this is going to be equal to, well, this is going to be equal to negative one third squared. Negative one third squared, over, over, over the cosine of one plus three times negative one third is negative one. So it's one plus negative one, so it's a cosine of zero. And the cosine of zero is just going to be one. So this is going to be equal to positive, positive one ninth. Now let's see if we can visualize what's going on here. So let me draw a little table here. So, I'm gonna plot, I'm gonna think about t, and x, and y. So t, and x, and y. So when t is equal to negative one third, well our x is going to be, this is going to be sine of zero, so our x is going to be zero, and our why is going to be, what, two over, or negative two over 27. So, we're talking about, we're talking about the point zero comma negative two over 27. So that is that point right over there. That's the point where we're trying to find the slope of the tangent line. It's telling us that that slope is one ninth, that slope is one ninth, so if we move, I guess one way to think about it is if we move four, one, two, three, four and a half, we're gonna move up half. So, if I wanted to draw the tangent line right there, it would look something like, something like that. Something, something, something like that. Let's see if we've got, one, two, three, four, and a half, that's what we got, just like that is pretty close. So that's what we just figured out. We figured out that the slope of the tangent line, right at that point is one ninth. So, it's not only neat to look at, but I guess somewhat useful.
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