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. |