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## Defining and differentiating parametric equations

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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.