
where the two continuous functions define an
ordered pair (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.
, with
the interval
. The graph of this equation can be seen in Figure
1 below.
.
That is, if the interval was
then only the bottom half of the circle
would be the curve. See Figure 2 below.
, then
the curve would start at (1,0) and create 2½
revolutions and end at (-1,0). The list below shows the values of x and
y that correspond to multiples
of π/4 for the
parameter t.| t | x = cos t | y = sin t |
| 0 | 1 | 0 |
| π/4 | 1/√2 | 1/√2 |
| π/2 | 0 | 1 |
| 3π/4 | -1/√2 | 1/√2 |
| π | -1 | 0 |
| 5π/4 | -1/√2 | -1/√2 |
| 3π/2 | 0 | -1 |
| 7π/4 | 1/√2 | -1/√2 |
| 2π | 1 | 0 |
| 9π/4 | 1/√2 | 1/√2 |
| 5π/2 | 0 | 1 |
| 11π/4 | -1/√2 | 1/√2 |
| 3π | -1 | 0 |
| 13π/4 | -1/√2 | -1/√2 |
| 7π/2 | 0 | -1 |
| 15π/4 | 1/√2 | -1/√2 |
| 4π | 1 | 0 |
| 17π/4 | 1/√2 | 1/√2 |
| 9π/4 | 0 | 1 |
| 19π/4 | -1/√2 | 1/√2 |
| 5π | -1 | 0 |
, with
the interval
. Let b =
1, and let a vary.



| t | x = cos t | x = 2cos t | x = 3cos t | x = -4cos t | y = sin t |
| 0 | 1 | 2 | 3 | -4 | 0 |
| π/4 | 1/√2 | 2/√2 | 3/√2 | -4/√2 | 1/√2 |
| π/2 | 0 | 0 | 0 | 0 | 1 |
| 3π/4 | -1/√2 | -2/√2 | -3/√2 | 4/√2 | 1/√2 |
| π | -1 | -2 | -3 | 4 | 0 |
| 5π/4 | -1/√2 | -2/√2 | -3/√2 | 4/√2 | -1/√2 |
| 3π/2 | 0 | 0 | 0 | 0 | -1 |
| 7π/4 | 1/√2 | 2/√2 | 3/√2 | -4/√2 | -1/√2 |
| 2π | 1 | 1 | 3 | -4 | 0 |
.
Multiplying the equation by a2b2
we get the
equation
. If a
= b then we get the equation x2
+y2 = a2 which is the
equation for the circle.
,
with
the interval
.
This time let a = 1, and let b vary.| t | x = cos t | y = sin t | y = 2sin t | y = -3sin t | y = 4sin t |
| 0 | 1 | 0 | 0 | 0 | 0 |
| π/4 | 1/√2 | 1/√2 | 2/√2 | -3/√2 | 4/√2 |
| π/2 | 0 | 1 | 2 | -3 | 4 |
| 3π/4 | -1/√2 | 1/√2 | 2/√2 | -3/√2 | 4/√2 |
| π | -1 | 0 | 0 | 0 | 0 |
| 5π/4 | -1/√2 | -1/√2 | -2/√2 | 3/√2 | -4/√2 |
| 3π/2 | 0 | -1 | -2 | 3 | -4 |
| 7π/4 | 1/√2 | -1/√2 | -2/√2 | 3/√2 | -4/√2 |
| 2π | 1 | 0 | 0 | 0 | 0 |
Assignment 9
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Assignment 11