Parametrically Defined Curves
D. P. Morstad, University of North Dakota
Objectives of Assignment
1.
2.
3.
4.
To learn how X(PLORE) graphs parametrically defined curves.
To examine parametric representations of circular and elliptical paths.
To examine Lissajous figures.
To investigate a variety of other parametrically defined curves.
**** IF YOU ARE NOT USING A COMPUTER IN THE MATH LAB: This unit
requires the use of a special file, slowpara.xpl. You can get a copy from the
C:\mathstuff\x(plore) subdirectory of any computer in the Math Lab, or you can download it
from our website at:
http://www.und.nodak.edu/dept/math/mathlab/labunits.html#CALCII.
Put it in the directory of your computer that contains X(PLORE).*****
I.
Using X(PLORE) to Graph Parametrically Defined Curves
For this unit, you must first load slowpara.xpl. To do this, start X(PLORE), then press
*. The computer should ask the name of the file you want to load. Type in slowpara.xpl and
press E. X(PLORE) will say that slowpara is now loaded.
With X(PLORE), graphing parametrically defined curves is very similar to graphing
standard functions of one variable. You need to declare a window, you need to define two
parametric functions x(t) and y(t), and then you use a special command to graph the curve.
For example, suppose you want to graph the curve given by:
x( t ) = t 2 and y ( t ) = t 3 for t = −1 to 2 ,
in the window –8 ≤ x ≤ 8, –6 ≤ y ≤ 6. Simply type the following four lines:
window(–8, 8, –6, 6)
grid(1,1)
x(t) = t^2
y(t) = t^3
slowpara(x(t), y(t), –1, 2, 3)
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Notice that the range of the parameter t, from –1 to 2, is declared inside the command
“slowpara(x(t), y(t), –1, 2, 3)”. The last number, 3, slows the graphing so you can see the
direction and relative speed as the curve is drawn – any integer from 1 to 100 can be used, but
anything above 10 will make it excruciatingly slow.
IMPORTANT: For this assignment, always use windows which have an x:y
proportion of 4:3. That is, use windows such as (–4, 4, –3, 3), (–8, 8, –6, 6), and (–12, 12, –
9, 9). This proportion compensates for the fact that the computer screen is not a square, and it
will guarantee that squares will look like squares instead of like arbitrary rectangles, and that
circles will look like circles instead of ellipses. Also, the command grid(1, 1) provides a grid
on your graph.
It can be helpful to think of parametric equations in terms of an Etch–a–Sketch toy.
Imagine that the x(t) function describes how you are turning the left–right knob and that the y(t)
function describes how you are turning the up–down knob. For example, the equations x(t) = 1
and y(t) = t would be the same as leaving the left–right button alone and turning the up–down
knob at a constant rate. You should get a vertical line. However, with the equations x(t) = t
and y(t) = t, you would be simultaneously turning both knobs at the same rate and this would
give you a line at a 45 degree angle.
Recall that if parameterizing equations are used to describe the path of a particle, then
the initial and final locations of the particle and the rate at which it traverses the path are as
important as the curve of the path. While graphing the examples in this assignment, pay close
attention to how these features can be manipulated.
II.
Circles, Ellipses, and Variations
Graph the following curves in the window –8 ≤ x ≤ 8, –6 ≤ y ≤ 6.
1. Graph x(t) = t, y(t) = 4sin(t), for t = 0 to 2π.
(If you were to draw this curve using an Etch–a–Sketch toy, you would have to turn
the left/right knob at a constant rate and turn the up/down knob back and forth.) You
should see a typical sine curve with amplitude 4. Notice that as it was drawn, the height
went from 0 up to 4, then down to –4, and then back up to 0.
2. Graph x(t) = 4sin(t), y(t) = 4sin(t), for t = 0 to 2π.
(This corresponds to moving both knobs back and forth exactly the same.) You should
get a 45 degree line segment because x(t) = y(t), and the line y = x has a slope of 1.
Notice that as it was drawn, its height also went from 0 up to 4, then down to –4, and
then back up to 0.
3. Graph x(t) = 4cos(t), y(t) = 4sin(t), for t = 0 to 2π.
(Now the knobs would be moving back and forth, but 90 degrees out of phase. That
is, one is half–way through being turned in one direction exactly when the other is
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starting to be turned back the other way.) This should give you a circle of radius 4.
The reason circles are hard to draw on a real Etch–a–Sketch is because it's hard to
coordinate your left and right hands exactly while moving them at true sinusoidal rates.
Again, as it was drawn, its height went from 0 up to 4, then down to –4, and then back
up to 0.
4. Graph x(t) = 4cos(t) + 1, y(t) = 4sin(t) – 2, for t = 0 to 2π.
Notice the translation.
5. Graph x(t) = 2cos(t), y(t) = 4sin(t), for t = 0 to 2π.
(This corresponds to moving the up–down knob twice as far back and forth as the
other knob, but maintaining exactly the same frequency.) Parametric equations provide
a quick and easy way to represent ellipses . . . . . . . . . . . .
6. To graph a hyperbola, simply switch to hyperbolic functions: graph x(t) = 2cosh(t),
y(t) = 4sinh(t), for t = –1 to 1, which should give you part of the right half; and then
change the x(t) to x(t) = –2cosh(t) for part of the left half.
7. Now for the pretty stuff, but first erase the screen . Try to predict what will happen if
you graph x(t) = 6cos(t), y(t) = 4sin(3t), for t = 0 to 2π. This will all be inside –8
≤ x ≤ 8 and –6 ≤ y ≤ 6, but it won’t be an ellipse any more because the curve has to go
up and down 3 times while it only goes back and forth only once. That is, the vertical
frequency is 3 times the horizontal frequency.
This type of curve is called a Lissajous figure. The motion of simple pendulums
trace out Lissajous figures. For example, consider someone on a swing in a playground
swinging back and forth. If you were to come up and push the swinger sideways a little
bit, he/she would start swinging all over the place instead of just back and forth. This
motion would closely resemble a Lissajous path. (It might not be exact in this case
because swings are supported by two ropes rather than one. Therefore, the swing
might behave as a compound pendulum rather than as a simple pendulum.)
8. Try a batch of different Lissajous figures by altering the frequencies of both x(t) and
y(t). Try to determine what it takes to make the curve look like a closed loop instead
of just a curve that starts in one corner and ends in another corner.
III.
Standard Curves and Paths
Standard functions of one variable such as f(x) = x 2 and curves such as x = y2 are
easily represented by parametric equations. (Erase before graphing the next four.)
1. Graph x(t) = t, y(t) = t 2, for t = –2 to 2.
2. Graph x(t) = t 2, y(t) = t, for t = –2 to 2.
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3. Graph x(t) = sin2(t), y(t) = sin(t), for t = 0 to 2π.
4. Graph x(t) = tan2(t), y(t) = tan(t), for t = –π/2 to π/2.
For the next several graphs, notice how small changes affect both the direction of the
path and its initial and final positions.
5.
Erase. Graph x(t) = a*cos(2πt), y(t) = b*sin(2πt), for t = 0 to 1, with:
a) a = 6, b = 6. Where did it start and end? Did it go clockwise or counterclockwise?
b) a = –6, b = 6. Same questions, different answers?
c) a = –6, b = –6. Same questions, different answers?
d) a = 6, b = –6. Same questions, different answers?
6.
Graph x(t) = a*sin(2πt), y(t) = b*cos(2πt), for t = 0 to 1, with:
a) a = 6, b = 6. Same questions, different answers?
b) a = –6, b = 6. Same questions, different answers?
c) a = –6, b = –6. Same questions, different answers?
d) a = 6, b = –6. Same questions, different answers?
7.
Graph x(t) = 6cos(t), y(t) = 4sin(t), for t = a to b, with:
a) a = 0, b = 3π/4. From the origin, what angle does the curve subtend?
b) a = –2π/3, b = 5π/3. Same questions, different answers?
c) a = –π, b = π. Same questions, different answers?
d) a = π/6, b = 5π/6. Same questions, different answers?
IV.
Some Fancier Curves
All of the following curves should be smooth except at a few specific points. If you
graph something and it looks like the scribbling of a small child, use a smaller range for t or
change your values of r, b, and c.
1.
Cycloids. Suppose you are driving along some warm summer night and a bicycle rider
crosses your path about a block ahead of you. If he has a reflector in his spokes, you will
notice a very distinctive path that the reflector is taking. The closer the reflector is to the tire, the
more distinctive the path and movement will be. If you have ever seen this, the reflector seems
to kind of hop along very frenetically. If the reflector was mounted right on the side of the tire,
the path it will follow is called a cycloid.
The parametric equations for a cycloid are:
x(t) = r*(t – sin(t)), y(t) = r*(1 – cos(t)),
where r is the radius of the tire. Graph a few of these using t = –3π to 3π with r = 0.5, 1, 2,
and 3. Try a few others.
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2.
Trochoids. If the reflector happened to be somewhere between the tire and the axle,
the path it will follow is called a trochoid. It is somewhat different from a cycloid.
The parametric equations for a trochoid are:
x(t) = r*(t – b*sin(t)), y(t) = r*(1 – b*cos(t)),
where r is the radius of the tire and r*b is the distance from the reflector to the axle. Try a few
of these using the same values as in #1 along with some values of b which are between 0 and r.
(Try r = 0.5 and b = 0.4 first.)
3.
Epicycloids. If the bicycle wheel happened to be rolling around the outside of a large
circle, then the path of the reflector is called an epicycloid. The equations are:
r +b 
r + b 
x( t ) = ( r + b ) cos( t ) − b * cos
t , y( t ) = ( r + b ) sin(t ) − b * sin
t
 b 
 b 
where r is now the radius of the large circle, and b is the radius of the wheel. When r = b, the
resulting epicycloid is called a cardiod. Use t = 0 to 4π, and: r = 2, b = .25; r = 2, b = .6;
r = 2, b = 2; r = 1, b = 2. Try a few others.
4.
Hypocycloids. If the wheel is rolling around inside a larger circle, the path of the
reflector is called a hypocycloid. These equations are:
r−b 
r−b 
x( t ) = ( r − b ) cos( t ) + b * cos
t , y( t ) = ( r − b ) sin(t ) − b * sin
t ,
 b 
 b 
where r is the radius of the larger circle and b is the radius of the wheel. You don’t need to
type these in, just modify the equations you used for the last problem. Graph a few of these
using t = 0 to 8π, and: r = 4, b = 0.5; r = 3, b = 1; r = 2, b = 1; r = 5, b = 3; r = 5, b =
4; r = 3, b = 4.
5.
Spirograph patterns. The Spirograph toy is just a simple mechanical device which
allows you to trace epicycloids and hypocycloids where the pencil (or the reflector, if you want
to use the bicycle analogy) is not right on the edge of the wheel, but is somewhere between the
edge of the wheel and its center. These curves can be represented parametrically as well.
These equations are just slight modifications of the hypocycloid equations, simply multiply the
denominators by a factor c:
r −b 
r − b 
x( t ) = ( r − b ) cos( t ) + b * cos
t , y( t ) = ( r − b ) sin(t ) − b * sin
t .
 b× c 
 b×c 
You don’t need to type these in, just modify the equations you typed in for #4. Try some of
these using the same numbers as in the last problem, but then use 0.5, 0.9, 1, 2, and 3 for c.
Erase between graphs.
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