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Parametric Equations. ordered pairs (x , y) are based upon a third variable, t, called the parameter t usually represents time 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 ]. Ex. . Ex. .
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Parametric Equations • ordered pairs (x , y) are based upon a third variable, t, called the parameter • t usually represents time • 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 [ tmin , tmax ]
Graphing Parametric Eq. • Graphs can be made without a calculator by plotting the points found in the previous table
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)
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
The graph of the parametric curve from the example above looks like:
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.
Ex. #1 x = 2t + 1 y = t – 1 solution: solve the 2nd equation for t and then substitute the expression into the 1st equation t = y + 1 x = 2(y + 1) + 1 x = 2y + 3 x – 3 = 2y .5x – 1.5 = y or y = .5x – 1.5 ( the equation of a line)
Ex #2 y = 3t t = y/3 (solve 2nd equation for t) x = (y/3)2 – 2 (substitute into the 1st equation) x = y2/9 – 2 (simplify) y2 = 9(x + 2) (the equation of a parabola)
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. x2 + y2 = 4cos2t + 4sin2t = 4(cos2t + sin2t) = 4(1) = 4 so x2 + y2 = 4 (the equation of circle)
Practice Problem #1 Eliminate the parameter: and describe the graph Solution: t = x + 1 y = 2(x + 1)2 or y = 2x2 + 4x + 2 (parabola which opens up)
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 = (- , )
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 + 7t or x = 3 – 7t y = 5 – 7t y = –2 + 7t I = [0 , 1] I = [0 , 1]