Presentation is loading. Please wait.

Presentation is loading. Please wait.

Parameterization. Section 1 Parametrically Defined Curves.

Similar presentations


Presentation on theme: "Parameterization. Section 1 Parametrically Defined Curves."— Presentation transcript:

1 Parameterization

2 Section 1 Parametrically Defined Curves

3 So far, we dealt with relations of the form y = f(x) or F(x,y) = 0 such as: y = 5x or x 2 + y 2 – 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.

4 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 parametric equations.

5 Examples Example 1

6 Let: x = t, y = 2t ; 0 ≤ t To graph this curve, consider the following table yxt 000 211 422 633

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

8 Eliminating the parameter Now, let's examine the situation differently by eliminating the parameter t. Doing that, we get: y = 2x; xε[0,∞) 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 ).

9 Example 2

10 Graph the curve defined by the parametric equations: x = t + 1, y = t 2 + 4t + 6 yxt 18-5-6 11-4-5 6-3-4 3-2-3 2-2 30 610 1121 1832

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

12 Eliminating the parameter we have: t = x – 1 And so y = (x-1) 2 + 4(x-1) + 6 = x 2 + 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!

13 Graph

14 Example 3

15 Graph the curve having the parametric equations: x = 2t 2, y = 2t 2 + 1 yxt 1918-3 98-2 32 100 321 982 19183

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

17 Eliminating the parameter 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 ).

18

19 Example 4

20 Graph the curve having the parametric equations: x = sint, y = 5sint + 2 yxt 200 71π/2 20π -33π/2 202π2π 712π +π/2 202π+ π -32π +3π/2

21 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 = 3 π The range of x is [-1,1] and the range of y is [-3,7]

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

23 Eliminating the parameter y = 5x +2 ; x Є [-1,1] why?

24

25 Example 5

26 Graph the curve having the parametric equations: x = 3cost, y = 3sint yxt 030 30π/2 0-3π 03π/2 032π2π 302π +π/2 0-32π+ π -302π +3π/2

27 Plot the points indicated in the table. Join these points. What do you get? The range of x is [-3,3] and the range of y is [-3,3]

28 Eliminating the parameter x 2 + y 2 = 9 why?

29 Graph

30 Example 6

31 Graph the curve having the parametric equations: x = 3cost, y = 3sint; t ε [0,π] yxt 030 30π/2 0-3π

32 Plot the points indicated in the table. Join these points. What do you get? The range of x is [-3,3] and the range of y is [0,3]

33 Eliminating the parameter x 2 + y 2 = 9 ; y ≥0

34 Graph


Download ppt "Parameterization. Section 1 Parametrically Defined Curves."

Similar presentations


Ads by Google