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David Terr
Ph.D. Math, UC Berkeley

9.5. Parametric Equations

Thus far, we have looked at many types of graphs. In Chapter 1, we introduced graphs and looked at graphs of circles and simple functions. In Chapter 6, we introduced polar coordinates and looked at graphs in these coordinates. Now we introduce yet another way of representing curves, known as parametric equations, as well as how to graph them. We will graph some familiar curves, such as circles and ellipses using their parametrized forms, as well as some as yet unfamiliar ones, such as cycloids and Lissajous Figures.

A parametrized curve is a curve written represented by the following system of equations:

• (9.5.1a) x = f(t)
• (9.5.1b) y = g(t)

where f and g are specified functions of the variable t, and t has a specified range, i.e. an interval such as [0,1], [0,2π), or (-∞, ∞). The variable t is known as the parameter of the system. Equations (9.5.1a) and (9.5.1b) are known as parametric equations for the coordinates x and y respectively.

Unit Circle

As a first example of a parameterized curve, consider the unit circle. As we have seen, the unit circle is parametrized by the polar angle θ. Specifically, it has the following parametrization:

• (9.5.2a) x = cos θ
• (9.5.2b) y = sin θ

where θ lies in the interval [0,2π).

Example 1: Graph the unit circle using the parametrization given by Equations (9.5.2).

Solution: To make a graph of a parametrized curve, we first need to tabulate x and y for as many values of the parameter as we can. We may use Table 4.5.1, which we reproduce below with x replaced by θ and the last two columns interchanged, for this purpose. Note that for the purpose of graphing, we no longer need to know the value of the parameter θ accurately.)

θ x = cos θ y = sin θ
0
1.000
0.000
π/6
3/2 ≈ 0.866
1/2 = 0.500
π/4
2/2 ≈ 0.707
2/2 ≈ 0.707
π/3
1/2 = 0.500
3/2 ≈ 0.866
π/2
0.000
1.000
2π/3
-1/2 = -0.500
3/2 ≈ 0.866
3π/4
-√2/2 ≈ -0.707
2/2 ≈ 0.707
5π/6
-√3/2 ≈ -0.866
1/2 = 0.500
π
-1.000
0.000
7π/6
-√3/2 ≈ -0.866
-1/2 = -0.500
5π/4
-√2/2 ≈ -0.707
-√2/2 ≈ -0.707
4π/3
-1/2 = -0.500
-√3/2 ≈ -0.866
3π/2
0.000
-1.000
5π/3
1/2 = 0.500
-√3/2 ≈ -0.866
7π/4
2/2 ≈ 0.707
-√2/2 ≈ -0.707
11π/6
3/2 ≈ 0.866
-1/2 = -0.500
1.000
0.000

Now we just plot these points and draw a smooth curve through them, obtaining the following graph:

Figure 9.5.3: Graph of a Parametrized Unit Circle

Ellipse

It is also useful to know how to parametrize an ellipse. As it turns out, an ellipse centered at the origin with semi-major axis a and semi-minor axis b has the following parametrization:

• (9.5.4a) x = a cos t
• (9.5.4b) y = b sin t

where t ranges over the interval [0,2π).

Example 2: Use the parametric equations of an ellipse to graph an ellipse with semi-major axis 5, semi-minor axis 4, and centered at the origin.

Solution: It is easy to tabulate x and y in this case; we simply take our previous table and multiply all values of x by 5 and all values of y by 4. We obtain the following table:

t x = 5 cos t y = 4 sin t
0
5.00
0.00
π/6
4.33
2.00
π/4
3.54
2.83
π/3
2.50
3.46
π/2
0.00
4.00
2π/3
-2.50
3.46
3π/4
-3.54
2.83
5π/6
-4.33
2.00
π
-5.00
0.00
7π/6
-4.33
-2.00
5π/4
-3.54
-2.83
4π/3
-2.50
-3.46
3π/2
0.00
-4.00
5π/3
2.50
-3.46
7π/4
3.54
-2.83
11π/6
4.33
-2.00
5.00
0.00

Once again, we plot a smooth curve through these points, obtaining the following graph:

Figure 9.5.5: Graph of a Parametrized Ellipse

Note that this is the same ellipse we graphed in Example 4 of Section 9.1, using different points.

Now we introduce some more exotic curves, which are nevertheless easy to graph, due to their simple parametrizations.

Cycloid

A cycloid is a curve obtained by marking the position of a point on the rim of a wheel of a moving vehicle, such as a bicycle. A cycloid generated by a wheel of radius R has the following parametrization:

• (9.5.6a) x = R (θ - sin θ)
• (9.5.6b) y = R (1 - cos θ)

Here, the parameter θ represents the angle by which the wheel has turned since the marked point was touching the ground. Here the parameter θ ranges over all real numbers, i.e. from -∞ to -∞.

Example 3: Graph two full cycles of a cycloid with R = 1, using the above parametrization.

Solution: First we need to compute and plot the values of x and y for θ ranging over an interval of length 4π. The simplest interval to use for this purpose is [-2π, 2π]. The following table gives values of x and y as a function of the parameter θ ranging over this interval:

θ x = θ - sin θ y = 1 - cos θ   θ x = θ - sin θ y = 1 - cos θ
-2π ≈ -6.283
-6.283
0.000
0
0.000
0.000
-11π/6 ≈-5.760
-6.260
0.134
π/6 ≈ 0.524
0.024
0.134
-7π/4 ≈-5.498
-6.205
0.293
π/4 ≈ 0.785
0.078
0.293
-5π/3 ≈ -5.236
-6.102
0.500
π/3 ≈ 1.047
0.181
0.500
-3π/2 ≈ -4.712
-5.712
1.000
π/2 ≈1.571
0.571
1.000
-4π/3 ≈ -4.189
-5.055
1.500
2π/3 ≈2.094
1.228
1.500
-5π/4 ≈ -3.927
-4.634
1.707
3π/4 ≈ 2.356
1.649
1.707
-7π/6 ≈-3.665
-4.165
1.866
5π/6 ≈2.618
2.118
1.866
-π ≈ -3.142
-3.142
2.000
π ≈ 3.142
3.142
2.000
-5π/6 ≈-2.618
-2.118
1.866
7π/6 ≈ 3.665
4.165
1.866
-3π/4 ≈ -2.356
-1.649
1.707
5π/4 ≈3.927
4.634
1.707
-2π/3 ≈ -2.094
-1.228
1.500
4π/3 ≈ 4.189
5.055
1.500
-π/2 ≈ -1.571
-0.571
1.000
3π/2 ≈ 4.712
5.712
1.000
-π/3 ≈ -1.047
-0.181
0.500
5π/3 ≈ 5.236
6.102
0.500
-π/4 ≈ -0.785
-0.078
0.293
7π/4 ≈5.498
6.205
0.293
-π/6 ≈ -0.524
-0.024
0.134
11π/6 ≈5.760
6.260
0.134
0
0.000
0.000
2π ≈ 6.283
6.283
0.000

Below is a graph of two cycles of the cycloid, based on this data.

Figure 9.5.7: Graph of a Cycloid

Lissajous Figures

Lissajous figures are some of the most interesting parametrized curves. A Lissajous figure has the following parametrization:

• (9.5.8a) x = cos at
• (9.5.8b) y = sin bt

where a and b are positive integers and t ranges over the interval [0, 2π). Lissajous figures can be generated on an oscilloscope from two sinusoidal inputs of different frequencies. When a=b, the figure is a circle.

Below are some examples of Lissajous figures with given vales of a and b.

Figure 9.5.9: Lissajous Figures

(Image taken from Wikipedia)

Example 4: Use the parametrization given by Equations (8.5.8) to plot the Lissajous figure with a=3 and b=2.

Solution: First we tabulate x = cos 3t and y = sin 2t against t, for t ranging from 0 to 2π.

t x = cos 3t y = sin 2t   t x = cos 3t y = sin 2t
0
1.000
0.000
π
-1.000
0.000
π/12
0.707
0.500
13π/12
-0.707
0.500
π/6
0.000
0.866
7π/6
0.000
0.866
π/4
-0.707
1.000
5π/4
0.707
1.000
π/3
-1.000
0.866
4π/3
1.000
0.866
5π/12
-0.707
0.500
17π/12
0.707
0.500
π/2
0.000
0.000
3π/2
0.000
0.000
7π/12
0.707
-0.500
19π/12
-0.707
-0.500
2π/3
1.000
-0.866
5π/3
-1.000
-0.866
3π/4
0.707
-1.000
7π/4
-0.707
-1.000
5π/6
0.000
-0.866
11π/6
0.000
-0.866
11π/12
-0.707
-0.500
23π/12
0.707
-0.500
π
-1.000
0.000
1.000
0.000

Below is a graph of this Lissajous figure, based on the data.