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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 
2π 
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 semimajor axis a and semiminor 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 semimajor axis 5, semiminor 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 
2π 
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 
2π 
1.000 
0.000 
Below is a graph of this Lissajous figure, based on the data.
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