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## Calculus 2

### Course: Calculus 2 > Unit 5

Lesson 1: Parametric equations intro# Parametric equations differentiation

Sal finds the derivative of the function defined by the parametric equations x=sin(1+3t) and y=2t³, and evaluates it at t=-⅓.

## Want to join the conversation?

- Instead of using the equation dy/dx = (dy/dt)/(dx/dt) Sal mentioned at1:20, is it possible to isolate t on one of the equations, plug it into the other, and solve for the derivative?(16 votes)
- You can, but it would be much messier. It will have different form but is equivalent (at least in this problem).

Given x = 2sin(1+3t) and y = 2t³. We want to find dy/dx so we want a function y(x). This mean we need to find t in term of x.

x = 2 sin(1+3t)

x/2 = sin(1+3t)

arcsin(x/2) = 1 + 3t

[arcsin(x/2) - 1]/3 = t

Now substitute that in for t in y equation.

y = 2t³ = 2[(arcsin(x/2) -1)/3]³

dy/dx = 6 [(arcsin(x/2) -1)/3]² * [1/3 * 1/√(1 - (x/2)²)] * 1/2

We are asked to find dy/dx when t = -1/3. We have dy/dx in term of x so we need to find what x is when t = -1/3, so we use x = 2sin(1+3t) = 2sin(1+3(-1/3)) = 2sin(0) = 0. So x=0 when t=-1/3

6 [(arcsin(0/2) -1)/3]² * [1/3 * 1/√(1 - (0/2)²)] * 1/2

= 6 * [(0-1)/3]² * [1/3 * 1/√(1-0)] * 1/2

= 6 * 1/9 * 1/3 * 1/2

= 6/54 = 1/9(30 votes)

- I didn't get why dy/dx = (dy/dt)/(dx/dt). I have an intuition why this works, but I need a better explanation. This would be make me happier!(14 votes)
- Try checking out this short article: http://abe-research.illinois.edu/faculty/dickc/mathematics/parametric2a.htm

I hope it helps!(11 votes)

- this video is very unclear. it's the first time that I did not understand sal explanation(3 votes)
- Which part did you first get confused at? There were two confusing parts for me:

1) when he just says that you can do

as if they were just fractions, even though they are not. It does make sense if you think of`dy/dx = (dy/dt)/(dx/dt)`

as "the infinitely small change in y over the infinitely small change in x", because that just means "slope, but over a tiny (infinitely small) interval". Then`dy/dx`

and`dy/dt`

just mean a change in some function y and a change in some function x, which are each in terms of a variable t. If you did that with a normal slope, you'd cancel out the third variable, like Sal does here.`dx/dt`

2) The other area of confusion might have to do with how parametric functions work, since Sal actually doesn't do a quick review here at all. Normally, the y-value of a function is determined by the x value of the function (like a line

). In a parametric function, the y and the x values of the function are broken out and defined separately, then put together after they have been defined. You could think of it like your regular (x,y) coordinates, except that the x and the y are being defined by another set of function, like this:`y=2x+3`

`(x,y)=( 2sin(3t) , 2t^3 )`

Another way to think about it is that the parametric equation tells you where you pencil should be, in x,y coordinates, at any time after you start drawing the graph.

This allows you to have a graph that violates the vertical line test, as this one does.

check out this video for an introduction to parametrics: https://www.khanacademy.org/math/algebra-home/alg-trig-functions/alg-parametric/v/parametric-equations-1(25 votes)

- At1:38, Sal is just using a modified version of the Chain rule right?(6 votes)
- Yes, the two parametric equation can be seen as composite functions..... sal's this example can also be seen as composite function but they are a little messy to seperate. so i have a less messy example for you........... see........... y=cost and x= arcsin(t) .............these are parametric equations but if we seperate t in 2nd equation as t=sinx then this function can be substitute in 1st equation ......... y=cos(sinx) or y=cos(t(x)).......... then there is question why sal divides like (dy/dt)/dx/dt ............. and it is because chain rule say first take derivative of outside function with respect to inside function and we exactly do that...... dy/dt.......... and then inside funtion with respect to x and that is what we don't do........ we take derivative of x with respect to inside function...look......we take derivative of x= arcsin(t) that is dx/dt but that should be dt/dx ...... therefore we divide..... hope that helps....!(5 votes)

- Does the graph shown in the video represents a functions?

I think it doesn't as for x = 0 there seem more than 1 value for y.(4 votes)- y isn't a function of x using your logic. however, x is a function of the parameter t, and y is also a function of t(3 votes)

- dy/dx is normally defined as the derivative of the continuous function y of x but in this case y is clearly not a function of x. So how do you
**rigorously**define dy/dx in the general case?

Maybe it's silly but my brain wants to simply reject this as a "syntax error".(4 votes)- That's why, when dealing with parametric functions, we have to redefine dy/dx as dy/dt/dx/dt. I believe you can use the chain rule to derive the new definition.

I hope this helps!(2 votes)

- Just out of curiosity, how does the graph used in the video work? Does the horizontal axis represent x and the vertical represent y? Also, how do you graph this on only two axises if there also a t variable?(3 votes)
- You get the x- and y-values (for the graph) by plugging in various t values. For example, when t=1, x=sin (4) and y=2. You then plot the sine value as the x and the 2 as the y. It didn't show it in the video, but parametric graphs also indicate direction along the curve over time. Hope this helps(3 votes)

- It seems that there are two derivatives on the graph, how come there's only one answer ? Shouldn't there be another derivative at =-1/3 other than 1/9?(3 votes)
- At4:58, how come did he move four and a half?(1 vote)
- Sal is trying to pickup another point in the tangent line to draw it properly. The slope is 1/9, so if you move 1 to the right along x axis, you need to move 1/9 up in the y axis. He moved 4.5 to the right along x axis so he needs to move 0.5 (4.5*1/9 = 4.5/9 = 0.5) up along y axis. The dot 4.5 right and 0.5 up was chosen because it's easier to plot than the dot 1 right and 1/9 up.(5 votes)

- In one of the practice questions for parametric functions differentiation, you need to get the derivative of
`4e^(6t)`

, which the hints show to be equal to`24e^(6t)`

. If the derivative of`e^x`

is`e^x`

, isn't the derivative of`e^(6t)`

also`e^(6t)`

? If not, how is it that you only bring down the`6`

and not also the`t`

?(1 vote)- This is an application of the chain rule. (f(g(x)))' = f'(g(x)) · g'(x). Or d/dx (f(g(x))) = df/dg · dg/dx

Let f(x) = e^x, and g(x) = 6x

Giving e^6x · 6

There are a few videos on the chain rule, starting with this one: https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/chain-rule-ab/v/chain-rule-introduction(3 votes)

## 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.