Department of Mathematics Education

J. Wilson, EMAT 6680

 

Matthew Tanner

EMAT 6680

Write-up #10

July 10, 2004

 

           

 

The Graph of

x = cos(t)

 

y = sin(t)

 

for 0 [ t [ 2o

 

 

Figure 1

 

 

is a that of a circle.  To explore other graphs, we may vary the weight of the components of the parametric equation or vary the scalar of the parameter t within the component functions. 

 

 

 

 

x = cos(at)

 

y= sin(bt)

 

for 0 [ t [ 2o

 

 

Figure 2.  a = 2.33

 

 

 

 

In Figure 2, a = 2.33.  Note that the graph of the function completes two and third horizontal oscillations within the range of t – one from A to B and the second from B to C.  In the interval D, the graph of the function makes and additional third of a cycle.  Click on the diagram for an animated demonstration.

 

Similarly, values of b greater that 1 induces vertical oscillations.

 

 

Figure 3.  b=3

 

 

 

 

The combined effect of varying a and b is that of inducing the simple ratio of a and b and factoring t by original value of the term (a or b) that reduces to 1.

 

 

Figure 4.  a = 7, b = 3, t varies from 0 to 2p/3

 

 

 

 

Here the ratio a/b = 7/3 = 2.33/1.  The effect is that the two component functions “progress” at the same rate in relationship to each other, but at a rate 3-times faster than t when compared to the graph of the function in Figure 2 (above).  Scaling the domain of t by a like factor produces the same graph as Figure 2.

 

For various a and b

 

x = a cos(t)

 

y= b sin(t)

 

for 0 [ t [ 2o

 

induce an ellipse.

 

Figure 5

 

 

 

 

The scalars of the component functions predictably amplify the graph of the function in a way that is a good illustration of the x components dominance in the graph of the function as t approaches 0 and o and the y component dominates the graph of the function as t approaches o/2 and 3o/2.  Click on the diagram for an animated demonstration.

 

The graph of the parametric equation

 

 

Figure 6

 

 

 

 

is like that of a circle – if we are given any value of t and calculate the hypotenuse

 

 

However, unlike a circle, the y component of equation only has one real zero.  In the intervals [o/2,3o/2], the value of y does not go below the absolute value of .   Likewise, within any interval [-h,h], the y component has only one zero at t=0 and comes only as close as the absolute value of to a second zero.  To observe this we can simply increase the range of the of the function.

Figure 7.

Range = [-10o,10o]

 

 

The graphs of the family of linear equations

 

x = a + t

y = b +kt

 

reveal some important properties.

 

Figure 8

 

Figure 9

>

Observation: it is clear that k represents the slope of the function.  More specifically, the ratio of the scalar of t in the y component to the scalar of t in the x component is the slope as is demonstrated by:

 

 

Graph

x = t + 1

y = 2t –1

for some appropriate range t.

 

Interpret.  Is there anything to vary to help understand the graph?

 

Figure 10

 

 

 

As above, note that the .  Clearly, the y-intercept is fixed by the value of t for which x = 0.š Since, x = 0 when t = -1, the y-intercept is (2*-1)-1=-3.š Similarly, the x-intercept is determined by the value for t for which y=0 or 1.5.

 

Figure 11

 

 

 

 

Alternating the scalars of t in the x and y components (Figure 11) produces a line of reciprocal slope.

 

Note that where

x = ta + b

y = tc + d

the y-intercept is given by

d - bc/a.

 

 

 

 

7.       Write parametric equations of a line segment through (7,5) with slope of 3.  Graph the line segment using your equations.

 

We begin by casting the equation using the point-slope formula:

 

.

 

Next, setting t equal to y and solving for x, we obtain parametric equations that meet the description.

 

> plot1:=plot( [(t+16)/3,t, t=0..10] ):

plot2:=PLOT(POINTS([7,5],SYMBOL(DIAMOND))):

with(plots):

display(plot1, plot2);

 

 

 

 

Note that, given

x = ta + b

y = tc + d

 

c/a=3.