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## Multivariable calculus

### Course: Multivariable calculus>Unit 1

Lesson 4: Visualizing vector-valued functions

# Parametric curves

When a function has a one-dimensional input, but a multidimensional output, you can think of it as drawing a curve in space.  Created by Grant Sanderson.

## Want to join the conversation?

• What would happen if you have a function that has a parameter "t" and outputs 3 dimensional vectors? I can imagine that this would draw some weird 3D curve like a trail of smoke, but is this always the case or can you get surfaces with only one paremeter too?
• Think of the parameters as the coordinates ON the resulting shape. A curve will have a starting point and an ending point, no matter how many dimensions it takes (a good example of a 3 dimensional curve is a helix). The input parameter (t), tells you how far along the curve have you gone from the starting point. The parameter (t) doesn't care what the shape of the curve is, it sees the curve as an one dimensional object on which it can only move back and forth.
Analogically, a surface (in a 3D space) will always take two parameters. A surface represents a curved two dimensional plane. For example - any place on the Earth can be represented by two parameters (coordinates) - the latitude and longitude.
• how do we even interpret the rate of this function? for unctions in the normal coordinate system, we've got the derivative to tell us how fast something's changing, but what about here?
• Suppose we have a function f(t)=[t*sin(t)]
[t*cos(t)]
x=t*sin(t)
y=t*cos(t)
If you want to do some parametric graphing, use this link. https://www.desmos.com/calculator/ksjcpazwa9
In our example, we've got one-dimensional input, and we get the 2-dimensional output - we plot points as our output or vectors. So, we'll get a nice spiral graph with the above plot.
To find dy/dx,
we first get dy/dt for which we have to use product rule and we'll get (cost - t*sint) and divide that by dx/dt for which we'll get (sint + t*cost).
So,
dy/dx =

(dy/dt)
-------
(dx/dt)
So, let's take a point. For that, I will take the value of t as 1 (you can take any number). For x, I get 0.841, and for y, I get 0.540.
So, I will input the value of t in my derivative equation and I will get -0.301/1.382 = -0.218.
So, at point (0.841,0.540), dy/dx will be -0.218. If you've plotted the graph, you'll see it clearly there. I hope you understood. In case you didn't, Khan academy has videos on it by the name "Parametric equations differentiation".
• Yes, but how do I go from a function to a parametric equation? Everyone seems to assume that you already know what the parametric form is?
• What's is the software program used to get this kind of math animations? Or is it all programmed in python by Grant Sanderson?? Thanks
• I believe he's using the program Grapher, which comes as a stock app on all Mac computers. :)
• Hello, may I just ask in which computer language are all the simulations of transformations written?
Thank you :)
(1 vote)
• What makes the graph slow or fast when it starts?
• Would it be possible to graph the value of t in three dimensions on the z-axis so that you would have an understanding of what the value of t is?
• What is the difference between vector valued functions and parametric curves? Are parametric curves a special type of vector valued function? If so, what defines them?
(1 vote)
• Simply put, a parametric curve is a normal curve where we choose to define the curve's x and y values in terms of another variable for simplicity or elegance. A vector-valued function is a function whose value is a vector, like velocity or acceleration(both of which are functions of time).