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Parametric Equations Lesson 6.7. Movement of an Object  Consider the position of an object as a function of time The x coordinate is a function of time.

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Presentation on theme: "Parametric Equations Lesson 6.7. Movement of an Object  Consider the position of an object as a function of time The x coordinate is a function of time."— Presentation transcript:

1 Parametric Equations Lesson 6.7

2 Movement of an Object  Consider the position of an object as a function of time The x coordinate is a function of time x = f(t) The y coordinate is a function of time y = g(t) time 0

3 Table of Values  We have t as an independent variable Both x and y are dependent variables  Given x = 3t y = t 2 + 4  Complete the table t-4-3-201234 x y

4 Plotting the Points  Use the Data Matrix on the TI calculator  Choose APPS, 6, and Current  Data matrix appears Use F1, 8 to clear previous values

5 Plotting the Points  Enter the values for t in Column C1  Place cursor on the C2 Enter formula for x = f(t) = 3*C1  Place cursor on the C3 Enter formula for y = g(t) = C1^2 + 4

6 Plotting the Points  Choose F2 Plot Setup  Then F1, Define  Now specify that the x values come from column 2, the y's from column 3  Press Enter to proceed

7 Plotting the Points  Go to the Y= screen Clear out (or toggle off) any other functions Choose F2, Zoom Data

8 Plotting the Points  Graph appears Note that each x value is a function of t Each y value is a function of t x = f(t) y = g(t)

9 Parametric Plotting on the TI  Press the Mode button For Graph, choose Parametric  Now the Y= screen will have two functions for each graph

10 Parametric Plotting on the TI  Remember that both x and y are functions of t  Note the results when viewing the Table, ♦ Y Compare to the results in the data matrix

11 Parametric Plotting on the TI  Set the graphing window parameters as shown here Note the additional specification of values for t, our new independent variable  Now graph the parametric functions Note how results coincide with our previous points

12 Try These Examples  See if you can also determine what the equivalent would be in y = f(x) form. x = 2t y = 4t + 1 x = t + 5 y = 3t – 2 x = 2 cos t y = 6 cos t x = sin 4t y = cos 2t x = 3 sin 3 t y = cos t Which one is it?

13 Assignment A  Lesson 6.7A  Page 440  Exercises 1 – 9 odd 27, 29

14 Parametric Equations The Sequal Lesson 6.7B

15 Eliminating the Parameter  Possible to represent the parametric curve with a single (x, y) equation  Example Given x = 1 + t 2 and y = 2 – t 2  Solution: Solve 1 st equation for t 2 in terms of y Substitute into 2 nd equation  Result y = 2 – (x – 1) Verify by graphing

16 Try It  Given x = 3 + 2 cos t and y = - 1 – 3 sin t  Hint – manipulate the equations by using Pythagorean identity sin 2 t + cos 2 t = 1

17 Path of a Projectile  Consider a projectile (such as a pumpkin) launched at a specified angle and initial velocity  Based on vector components and effects of gravity the actual path can be represented by Experiment with Pumpkin Launch Spreadsheet Experiment with Pumpkin Launch Spreadsheet

18 Launch Another!  Graph the path of a pumpkin launched at an angle of 35° with an initial velocity of 195 ft/sec  How far did it go?  How long was it in the air?

19 Assignment B  Lesson 6.7B  Page 440  Exercises 11 – 17 odd 31, 33


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