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Bell Ringer Solve even #’s Pg. 34.

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Presentation on theme: "Bell Ringer Solve even #’s Pg. 34."— Presentation transcript:

1 Bell Ringer Solve even #’s Pg. 34

2 1.4 Parametric Equations Pg. 30

3 There are times when we need to describe motion (or a curve) that is not a function.
We can do this by writing equations for the x and y coordinates in terms of a third variable (usually t or ). These are called parametric equations. “t” is the parameter. (It is also the independent variable)

4 To graph on the TI-89: Graph……. 2 Y= WINDOW GRAPH Example 1: T ) MODE
ENTER PARAMETRIC Y= 2nd T ) ENTER WINDOW GRAPH

5 Hit zoom square to see the correct, undistorted curve.
We can confirm this algebraically: parabolic function

6 Circle: If we let t = the angle, then: Since: We could identify the parametric equations as a circle.

7 Graph on your calculator:
Use a [-4,4] x [-2,2] window. WINDOW GRAPH

8 Ellipse: This is the equation of an ellipse.

9 Converting Between Parametric and Cartesian Equations
We have seen two techniques for converting from parametric to Cartesian: The first method is called eliminating the parameter. It requires solving one equation for t and substituting into the other equation to eliminate t. This is possible when the graph is a function. The second method used the Pythagorean identity to eliminate t by using the fact that Both of these methods only work sometimes. There are many curves that can only be described parametrically.

10 On the other hand, changing from the Cartesian equation for a function to a parametric equation always works and it is easy! The steps are: 1) Replace x with t in the original equation. 2) Let x = t . Example: becomes:

11 In the special case where we want the parametrization for a line segment between two points, we could find the Cartesian equation first and then convert it to parametric, but there is an easier way. We will use an example to illustrate: Find a parametrization for the line segment with endpoints (-2,1) and (3,5). Using the first point, start with: Notice that when t = 0 you get the point (-2,1) . Substitute in (3,5) and t = 1 . The equations become: p

12 Homework: 1.4a 1.4 p34 1,7 1.3 p26 6,15,23,26,36 1.4b 1.4 p34 3,9,13,19,25


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