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Course: Multivariable calculus>Unit 1

Lesson 6: Visualizing multivariable functions (articles)

Parametric functions, one parameter

Parametric functions give a way to represent functions with a one-dimensional input and a multidimensional output.

What we're building to

• A function with a one-dimensional input and a multidimensional output can be thought of as drawing a curve in space.
• Such a function is called a parametric function, and its input is called a parameter.
• Sometimes in multivariable calculus, you need to find a parametric function that draws a particular curve. This is called parameterizing that curve.

Visualizing vector-valued functions

So you're off, happily reading some math one day, as one does, and you come across a function like this:
$f\left(t\right)=\left[\begin{array}{c}t\cdot \mathrm{cos}\left(2\pi t\right)\\ t\cdot \mathrm{sin}\left(2\pi t\right)\end{array}\right]$
How would you visualize it?
It takes in a single variable, $t$, and outputs a two-dimensional vector. For example, at the input $t=1$, it is evaluated like this.
$f\left(1\right)=\left[\begin{array}{c}1\cdot \mathrm{cos}\left(2\pi \cdot 1\right)\\ 1\cdot \mathrm{sin}\left(2\pi \cdot 1\right)\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$
This output is a vector with length $1$ pointing in the $x$-direction.
But how do you visualize all outputs all at once?
A nice way to do this is to imagine what curve the tip of that vector will trace out as $t$ ranges over some values. For example, the following interactive diagram lets you see what curve the output of $f$ will trace out as $t$ ranges from $0$ to $3$:
This is called a parametric curve. When you choose to interpret the function this way, it is called a parametric function, and the input $t$ is called the parameter.

Only look at output space

Notice, unlike graphs, where we try to depict both the input space and output space of a function at the same time, or contour maps, where we just draw on the input space, interpreting functions parametrically has us looking purely at the output space. The reason this makes sense for the example above is that the output space has more dimensions than the input space.

Input information is lost

The problem with only drawing in the output space is that it is not immediately clear which inputs went to the output values we draw. For instance, consider these two functions:
$\begin{array}{rl}f\left(t\right)& =\left[\begin{array}{c}\mathrm{cos}\left(t\right)\\ \mathrm{sin}\left(t\right)\end{array}\right]\\ \\ g\left(t\right)& =\left[\begin{array}{c}\mathrm{cos}\left(t+\pi \right)\\ \mathrm{sin}\left(t+\pi \right)\end{array}\right]\end{array}$
If we plot these as parametric functions, with $t$ running from $0$ to $2\pi$, they each draw a circle with radius $1$ centered at the origin.
However, they are different functions. For instance, evaluate each one at $t=0$.
Given that $f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)\\ \mathrm{sin}\left(t\right)\end{array}\right]$, what is $f\left(0\right)$?

Given that $g\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t+\pi \right)\\ \mathrm{sin}\left(t+\pi \right)\end{array}\right]$, what is $g\left(0\right)$?

One way is to keep track of the lost input information is to label a few points with the input values
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)\\ \mathrm{sin}\left(t\right)\end{array}\right]$
$g\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t+\pi \right)\\ \mathrm{sin}\left(t+\pi \right)\end{array}\right]$
Alternatively, you could imagine how the curve is drawn over time as $t$ runs from the start value to the end value. This is particularly relevant when the function is meant to model the trajectory of a particle through space.

Parametrization

In multivariable calculus, and especially in a topic called "line integration", it is common to start off with a curve, and search for a parametric function that draws the curve. One example that comes up a lot is the unit circle, meaning the circle with radius $1$ centered at the origin.
Finding a parametric function that describes a curve is called parameterizing that curve. In the previous section I showed two different functions which parameterize the unit circle. The most common one that people use in practice is this one:
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)\\ \mathrm{sin}\left(t\right)\end{array}\right]$
Note: When you are parameterizing a curve, you must not only specify the parametric function, but also the range of input values that will draw the curve. For example, using the function $f\left(t\right)$ to draw the unit circle above, you could let $t$ range from $0$ to $2\pi$.

Example: Parameterizing a loopy curve

Let's say you want to parameterize this loopy pattern:
To parameterize a curve, you should always think about drawing it. In this case, you could imagine sketching it by trying to draw a circle counterclockwise while someone pushes your hand to the right at a steady velocity. To encode this, using formulas, we start with parametric function for a circle:
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)\\ \mathrm{sin}\left(t\right)\end{array}\right]$
This would have us starting at the point $\left(1,0\right)$, and tracing a circle with radius $1$ counterclockwise. Since the loopy curve we are parameterizing starts at $\left(-2,0\right)$, we start tweaking this function by shifting the $x$ value by $-3$.
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)-3\\ \mathrm{sin}\left(t\right)\end{array}\right]$
Being pushed to the right over time corresponds with a steady increase in the $x$-value of your hand with respect to time, unrelated to the motions it's going through for the circle. To encode this, add some constant $c$ times $t$ to the $x$-component of the function.
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)-3+ct\\ \mathrm{sin}\left(t\right)\end{array}\right]$
To figure out what the constant should be, we need to know how far right one has moved after completing one loop. Our current function $f\left(t\right)$ completes one loop as $t$ goes from $0$ to $2\pi$. Looking at the loopy curve, it seems we shift exactly $1$ to the right after a single loop.
This means we must have $2\pi c=1$, and hence $c=\frac{1}{2\pi }$.
$f\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(t\right)-3+\frac{1}{2\pi }t\\ \mathrm{sin}\left(t\right)\end{array}\right]$
Finally, we need to place bounds on $t$. Let's see just how many loops the loopy curve contains:
It looks like it has $6$. Since our chosen function $f\left(t\right)$ completes one loop as $t$ increases by $2\pi$, we should let it range from $0$ to $6\left(2\pi \right)=12\pi$.

Summary

• A function with a one-dimensional input and a multidimensional output can be thought of as drawing a curve in space.
• Such a function is called a parametric function, and its input is called a parameter.
• Sometimes in multivariable calculus, you need to find a parametric function that draws a particular curve. This is called parameterizing that curve.

Want to join the conversation?

• Are there exercises for parameterization? If so, where? If not, when? I would love to work on those. Also, any further advice about how to picture or quickly sketch parametric equations would be greatly appreciated.
• How did we measured it as "2(pi)c" ?
• I think it is because that gives you the shift from the start point of one circle to the start point of another with regards to x. Since x is cos(t) each time t = a 2*pi (with a just being a positive integer) the value of x would reset to the original cos(0). normally you'd just have c*2pi = 0 (since the start point is continuously x = cos(0)) but since you want it to loop your circles start point is going to change as well. Keep in mind that
cos(0) = cos(2pi) = cos(a2pi)
To sum up 2(pi) could be thought of as the time between start points of the circle and c is the constant that determines what that distance between start points will be. I hope that helped.
• On the last example . . . what about the radius? The radius isn't one like it is with a unit circle. Or at least it doesn't look that way. Help!
• CCF the radius is still always one because the centre is always moving to the right at exactly the velocity needed so that it moves exactly one unit to the right for every complete circuit of the line, much as the earth moves while the moon 'circles' it so that the distance between the earth and moon stays (more or less) the same.
• What is the physical meaning of t in parametric equations? For example, x variable has the physical meaning of being in the x-axis, either in the left or in the right. Same with a y variable. This can be physically shown in a Cartesian plane. However, where is t? It is a variable but, what?
• This is a very good question!
The most common meaning t can carry (especially in physics) is time! We can use parametric equations to model the projectile motion. In 2D we would have one equation for the x position, for example x(t) = (v1)t. In this case the projectile was given an initial velocity v1 upon release and moves according to that function in the x direction. The y component may look something like this: y(t) = c1 + (v2) + (g/2)t^2. Where v2 is the initial y velocity and g is acceleration due to gravity. A similar expression can be made for a 3D model as well.
Once you have equations like this you can take time derivatives and get the speed in various directions at whatever time you'd like!

Later on in multivariable you will also be doing parametric surfaces. The time values (generally (u,v) or (s,t)) have less meaning as only the shape of the surface matters and not how "fast" the function outputs it.
• Why does the axis use s and t in the graphs at the end? Wasn't it the idea that the input t is kept in a separate space?
• I think it is just to show you do not need to be confined to using x and y, you can use any two variables, because some functions will also use non x, y variables as well.
• In the paragraph about two parameters it presents the function f(t,s)=(t^3-st+2,s-t,s+t) it parametrizes x as x=t^3-st. Why is it not t^3-st+2?