Parameterization. Section 1 Parametrically Defined Curves.

Parameterization

Section 1

Parametrically Defined Curves

So far, we dealt with relations of the form y = f(x) or F(x,y) = 0

such as: y = 5x or x2 + y2 – 4 = 0, which state a direct relationship between the

variables x and y. However, sometimes, it is more useful to express both x and y in terms of a third variable, which we will call a parameter.

Orientation

Let C be a curve defined by the equations:

x = f(t) , y = g(t) ; a ≤ t ≤ b

The direction of the increasing of the parameter t is called the orientation imposed on C by the parameter.

Examples

Example 1

Let:x = t , y = 2t ; 0 ≤ t

To graph this curve, consider the following table txy

000

112

224

336

Plot the points indicated in the table.Join these points. What do you get?

Eliminating the parameter

Now, let's examine the situation differently by eliminating the parameter t. Doing that, we get:

y = 2x; xε[0,∞) How?

The curve defined by this equation is the line segment situated on the first quadrant of the straight line through the origin and the point ( 1 ,2 ).

Example 2

Graph the curve defined by the parametric equations: x = t + 1 , y = t2 + 4t + 6

txy

-6-518

-5-411

-4-36

-3-23

-2-12

-103

016

1211

2318

Plot the points indicated in the table.

Join these points. What do you get?


we have:t = x – 1And soy = (x-1)2 + 4(x-1) + 6 = x2 + 2x + 3 = (x+1)2 + 2 This is the graph resulting from shifting the

curve of the squaring function one unit to the left and two units upward. Sketch this graph!

Example 3

Graph the curve having the parametric equations: x = 2t2 , y = 2t2+ 1

txy

-31819

-289

-123

001

123

289

31819


we have:y = x +1 ; x Є [0,∞) why?

The curve defined by this equation is the line segment situated on the first quadrant of the straight line which intersects the axes at (0,1) and (-1 , 0 ).

Example 4

Graph the curve having the parametric equations: x = sint , y = 5sint + 2

txy

002

π/217

π02

3π/2-1-3

2π02

2π +π/217

2π+ π02

2π +3π/2-1-3

Notice that the same point (x,y) may be obtained by substituting different values of t.

For example the point (0,2) is obtained by both letting t = π and t = 3π

The range of x is [-1,1] and the range of y is [-3,7]


y = 5x +2 ; x Є [-1,1] why?

7

11

3

2

Example 5

Graph the curve having the parametric equations: x = 3cost , y = 3sint

txy

030

π/203

π-30

3π/20-3

2π30

2π +π/203

2π+ π-30

2π +3π/20-3


x2 + y2 = 9 why?

What does this equation represent?

X2+y2=9

Example 6

Graph the curve having the parametric equations: x = 3cost , y = 3sint; t ε [0,π]

txy

030

π/203

π-30



The range of x is [-3,3] and the range of y is [0,3]


x2 + y2 = 9 ; y ≥0


X2+y2=9, 0≤y

Example 7


txy

020

π/203

π-20


txy

3π/20-3

2π20


X2 / 4 + y2 /9 = 1 How?


X2 / 4 + y2 /9 = 1Is the equation for an ellipse centered at the origin and with

the vertices:

( 2 , 0) , ( - 2 , 0) , ( 0 , 3) , ( 0 , - 3)

X2 / 4 + y2 /9 = 1

Example 8

Graph the curve having the parametric equations: x = 2cost , y = 3sint; t ε [π, 2 π]

txy

π- 20

3π/20- 3

2π20


X2 / 4 + y2 /9 = 1 ; y ≤ 0 How?


X2 / 4 + y2 /9 = 1It is the equation for an lower half ellipse centered at the

origin and with the vertices:

( 2 , 0) , ( - 2 , 0), ( 0 , - 3)

0;194

22

yyx

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Parameterization. Section 1 Parametrically Defined Curves.

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