1. Parametric Equations ordered pairs (x, y) are based upon a third variable, t, called the parameterordered pairs (x, y) are based upon a third variable, t, called the parameter t usually represents timet usually represents time there are two equations, one for x and one for y, each in terms of t.there are two equations, one for x and one for y, each in terms of t. An interval, I, is provided to define the values for t [ t min, t max ]An interval, I, is provided to define the values for t [ t min, t max ]

2. Ex. txy -3 -2 0 1 2 3

3. Ex. txy -38 -230 01 0 2 103 234 385

4. Graphing Parametric Eq. Graphs can be made without a calculator by plotting the points found in the previous table

5. Graphing Parametric Eq. Graphs can be made without a calculator by plotting the points found in the previous table Graphs can be made with a calculator by changing the mode MODE  PAR (instead of FUNC)

6. Graphing Parametric Eq. Graphs can be made without a calculator by plotting the points found in the previous table Graphs can be made with a calculator by changing the mode MODE  PAR (instead of FUNC) go to y = to see how that screen has changed

7. The graph of the parametric curve from the example above looks like:

8. Eliminating the Parameter Parametric equations can be changed into a rectangular equation with x and y by “eliminating the parameter” Solve one equation for t and then substitute it into the other equation and simplify.

9. Ex. #1 x = 2t + 1 y = t – 1 y = t – 1 solution: solve the 2 nd equation for t and then substitute the expression into the 1 st equation t = y + 1 x = 2(y + 1) + 1 x = 2(y + 1) + 1 x = 2y + 3 x = 2y + 3 x – 3 = 2y x – 3 = 2y.5x – 1.5 = y or y =.5x – 1.5.5x – 1.5 = y or y =.5x – 1.5 ( the equation of a line) ( the equation of a line)

10. Ex #2 y = 3t  t = y/3 (solve 2 nd equation for t) x = (y/3) 2 – 2 (substitute into the 1 st equation) x = y 2 /9 – 2 (simplify) y 2 = 9(x + 2) (the equation of a parabola)

11. Ex #3 (special case) x = 2 cos t y = 2 sin t I = [0, 2π] If you graph this it looks like a circle, let’s see why. x 2 + y 2 = 4cos 2 t + 4sin 2 t = 4(cos 2 t + sin 2 t) = 4(1) = 4 so x 2 + y 2 = 4 (the equation of circle)

12. Practice Problem #1 Eliminate the parameter: and describe the graph and describe the graph Solution: t = x + 1 y = 2(x + 1) 2 or y = 2x 2 + 4x + 2 (parabola which opens up)

13. Finding the Parametrization of a Line Given two points, finding parametric equations to describe the line or segment containing them Use the following formula (not found in the book) Notes: it doesn’t matter which point is #1 or #2 for a segment, set I = [0, 1] for a line, set I = (- ,  )

14. Practice Problem #2 Find the parametrization of the line segment between ( – 4, 5) to (3, – 2). Solution: x = – 4 + (3 – (– 4))t y = 5 + (– 2 – 5)t x = – 4 + 7tor x = 3 – 7t y = 5 – 7t y = –2 + 7tI = [0, 1]