As we can remember when we are wanted to skecth the graph of a line given in the form y = ax + b, we make a table of the several values of x and y by giving particular values for them. For exaple, for x = 0, y takes the value b and for y = 0, x takes the value -b/a. In fact, in this sense, the values and the range of the values of the x and y depend on each other. Or, if we think y = f(x) (i.e., y as a function of x) then y varies as y varies and the range of y depends on the range of x. It may be more clear if we look at the table below for a simple equation

y = x + 2

	Range
x	-1	0	1
y (= x + 2)	1	2	3

In this case, we can think x as an independen variable and y as a dependent variable. One immediate question may aries, can we make both x and y as dependent and still get the same equation and same graph. The answer simply is: YES. We can do. Now, let's introduce a definition of parametric curve:

Definition: A parametric curve in the plane is a pair of functions:

where the two continuous functions define ordered pairs (x,y). The two equations are usually called the parametric equations of a curve. The extent of the curve will depend on the range of t and your work with parametric equations should pay close attention the range of t . In many applications, we think of x and y "varying with time t " or the angle of rotation that some line makes from an initial location. For example in physics, position of a particle at a time found by parametric equations of x(t), y(t), z(t).

And graph several sets of curves for for selected values of a, b, and k in an appropriate range for t.

As we may guess, for different values of t, there will be different pairs of (x,y) in R*R. But when we think x nad y seperately (i.e. x = a + t and y = b + kt), the common value in each pair (x,y) is t. So, I can solve and equate both x and y in terms of t. More precisely:

x = a + t implies t = x - a and y = b + kt implies t = (y - b)/k . And if we equate both t's, we get,

x - a = (y - b)/k which imply y = kx + (b-ka). Therefore, when we are dealing with x = a + t and y = b + kt, in fact we are dealing with the equation y = kx + (b-ka) which is the equation of a line with slope k. Now, we are ready to see some graphs.

If we choose a=b=0 and t is between -10 and 10, we get the graphs in Figure 1 below for several values of k. Actually when a=b=0, we are dealing with y = kx which follows from the equation y = kx + (b-ka) we found before. Therefore, all of the graphs passes through the origin whatever the value of k.

Figure 1

When we take a=b=0, x (= a + t) has max value 10 when t= +10 and min value -10 when t= -10. On the other hand, y (= b + kt) has max value 10k and min value -10k when t= +10 and t= -10 respectively. Of course as may expect, the shapes of the graphs for a=b=0 and for the above values of k can be different depending on the range of t. Figure 2 shows this fact.

Figure 2a. t is between 0 and 10

Figure 2b. t is between -25 and -10

Figure 2c. t is between 3 and 5

As we may observe, although slopes of the graphs remain the same, the locations are changed according to the max and min values of x and y. But if we extend the segements in both Figure 1 and Figure 2 so that they become lines (i.e. infinite segment), each corersponding segment lies on the same line.

We may get some different pairs of graphs of x (= a + t) and y (= b + kt) where a and b are not zero. For example if a=1 and b= -1/2 we may get the graphs in Figure 3

Figure 3a t is between -3 and +3

Figure 3b t is between -7 and -2

Figure 3c t is between +2 and +7

As we may observe although the slopes are unchanged, the max and min values of (x,y) are changed according to the different values of t. In other words the boundaries of each line segment are changed.

Note: If we select the range of t big enough, the graph is a line.

Now, let's investigate another pair of parametric equation:

From x = t+ 1, t = x - 1, from y = 2t - 1, t = (y + 1)/2. Equating both t's we may get y = 2x - 3. Therefore, as in the above graphs of the parametric equations, the graphs of x = t+ 1 and y= 2 t - 1 for different finite ranges of t are line segments completely lye on the graph of the line y = 2x - 3.

If you see the demonstration, the only thing varying is the range of t. According to the range of t, the lenght of the segments change but they completely lye on the line y = 2x - 3.

Above we used lines to understand the graphs of parametric equations of the form x = a + t and y = b + kt. Now, let's reverse the event s. t. for a given line we try to parametrize it.

For this purpose, let's try to write parametric equations of a line through (7, 5) with slope of 3. Graph the line using your equations.

The equation of a line passing through (7, 5) with slope of 3 can be found the fact that the general form of a line is y = mx + n where m is the slope. Then our equation in this case is y = 3x + n. Since (7'5) satisfies the equation then 5 = 3.(7) + n and solving for n we found that the equation of the line passing through (7, 5) with slope of 3 is y = 3x -16.

1st Way:

We found above that a parametric equation of the form

implies that t = x - a and t = (y - b)/k so that y = kx + (b-ka).

If we equate y = kx + (b-ka) and the equation y = 3x -16 I can find the coefficients a, b, and k. Thus:

3x -16 = kx + (b-ka) implies k = 3 and b - 3a = -16. So, the parametric equation of the line y = 3x -16 is:

x = a + t

y = b + 3t where a and b satisfies b - 3a = -16 (for example a = 1 and b = -13).

Important Note: The graph of x = a + t and y = b + 3t where a and b satisfies b - 3a = -16 depends on the range of t. So, to get the same graph woth y = 3x - 16 we should choose t large enough (most probably minus ininity to infinity)

2nd Way:

In the equation y = 3x -16, we define a variable t and linear function f(t) of t s.t. x = f(t). Then plugging x = f(t) in the equation of the line we can find another function g(t) of t s.t. y = g(t).

Let's show this by an example:

Let's say x = 3t, then y = 3.(3t) - 16 = 9t -16. Thus our patrametric equation is: x = 3t and y = 9t - 16.

If we choose x = t - 1, then y = 3(t - 1) - 16 = 3t - 19. thus the parametric equation is: x = t-1 and y = 3t - 19.

This page created October 25, 1999

This page last modified October 31, 1999