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 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.

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 yxt

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 = t 2 + 4t + 6 yxt

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

Eliminating the parameter we have: t = x – 1 And so y = (x-1) 2 + 4(x-1) + 6 = x 2 + 2x + 3 = (x+1) 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 = 2t 2, y = 2t yxt

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

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 ).

Example 4

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

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]

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

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

Example 5

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

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]

Eliminating the parameter x 2 + y 2 = 9 why? What does this equation represent?

X 2 +y 2 =9

Example 6

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

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]

Eliminating the parameter x 2 + y 2 = 9 ; y ≥0 What does this equation represent?

X 2 +y 2 =9, 0≤y

Example 7

Graph the curve having the parametric equations: x = 2cost, y = 3sint yxt π/2 0-2π

Graph the curve having the parametric equations: x = 2cost, y = 3sint yxt -303π/2 022π2π

Plot the points indicated in the table. Join these points. What do you get? The range of x is [-2,2] and the range of y is [-3,3]

Eliminating the parameter X 2 / 4 + y 2 /9 = 1 How?

What does this equation represent? X 2 / 4 + y 2 /9 = 1 Is the equation for an ellipse centered at the origin and with the vertices: ( 2, 0), ( - 2, 0), ( 0, 3), ( 0, - 3)

X 2 / 4 + y 2 /9 = 1

Example 8

Graph the curve having the parametric equations: x = 2cost, y = 3sint; t ε [π, 2 π] yxt 0- 2π - 303π/2 022π2π

Plot the points indicated in the table. Join these points. What do you get? The range of x is [-2,2] and the range of y is [-3,0]

Eliminating the parameter X 2 / 4 + y 2 /9 = 1 ; y ≤ 0 How?

What does this equation represent? X 2 / 4 + y 2 /9 = 1 It is the equation for an lower half ellipse centered at the origin and with the vertices: ( 2, 0), ( - 2, 0), ( 0, - 3)

Homework