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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 at , is it possible to isolate t on one of the equations, plug it into the other, and solve for the derivative?
• 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
• 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!
• this video is very unclear. it's the first time that I did not understand sal explanation
• Which part did you first get confused at? There were two confusing parts for me:

1) when he just says that you can do
``dy/dx = (dy/dt)/(dx/dt)``
as if they were just fractions, even though they are not. It does make sense if you think of
``dy/dx``
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/dt``
and
``dx/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.

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
``y=2x+3``
). 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:
``(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
• At , Sal is just using a modified version of the Chain rule right?
• 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....!
• 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.
• 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
• 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".
• 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!
• 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?
• 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
• 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`?