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Parametric Equations Lesson 6.7
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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
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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
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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
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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
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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
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Plotting the Points Go to the Y= screen Clear out (or toggle off) any other functions Choose F2, Zoom Data
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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)
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Parametric Plotting on the TI Press the Mode button For Graph, choose Parametric Now the Y= screen will have two functions for each graph
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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
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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
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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?
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Assignment A Lesson 6.7A Page 440 Exercises 1 – 9 odd 27, 29
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Parametric Equations The Sequal Lesson 6.7B
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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
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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
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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
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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?
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Assignment B Lesson 6.7B Page 440 Exercises 11 – 17 odd 31, 33
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