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The Beginning of Parametric Equations

The Beginning of Parametric Equations. (Sec. 6.3a). 2. y = –16t + 420. Consider a rock dropped from the top of a 420-foot tower…. The rock’s height y above the ground t seconds later can be modeled with the following equation:. Since the horizontal position of the rock never changes,

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The Beginning of Parametric Equations

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  1. The Beginning ofParametric Equations (Sec. 6.3a)

  2. 2 y = –16t + 420 Consider a rock dropped from the top of a 420-foot tower… The rock’s height y above the ground t seconds later can be modeled with the following equation: Since the horizontal position of the rock never changes, it can be modeled by an equation such as: x = 2.5 These are examples of parametric equations, with a parameter of t (often, t represents time)

  3. Definition: Parametric Curve, Parametric Equations The graph of the ordered pairs (x, y) where x = f(t), y = g(t) are functions defined on an interval I of t-values is a parametric curve. The equations are parametric equations for the curve, the variable t is a parameter, and I is the parameter interval.

  4. Look back at our first equations: 2  Let’s graph these in our calculator, and see if we can interpret this graph… y = –16t + 420 x = 2.5

  5. More examples For each of the given parameter intervals, use your calculator to graph the parametric equations (A) (B) (C) How do the graphs differ from each other???

  6. More examples Complete the table for the parametric equations and then plot each of the points. t x 1 0 –1 0 1 y 0 1 0 –1 0 Where’s the graph???

  7. More examples t x 1 0 –1 0 1 y 0 1 0 –1 0 Now, let’s explore these functions on the calculator… • What happens when we manipulate the range for t, or the “t-step” value???

  8. More examples Sketch a graph of the following parametric equations by hand, then verify your work using a calculator.

  9. t x y Verify with a calculator!!!

  10. Eliminating the Parameter,Parametrization of Curves

  11. In some situations, we can eliminate the parameter from parametric equations, obtaining a rectangular equation that represents the curve… 8 8 Ex: x = 1 – 2t, y = 2 – t, – < t < Solve the first equation for t: Substitute for t in the second equation: y = .5x + 1.5  Identify the curve!!!

  12. Eliminate the parameter and identify the graph: 2 x = t – 2, y = 3t y = + 3 x + 2 – • Identify the curve!!! Can we graph the curve in both forms???

  13. Eliminate the parameter and identify the graph: First, check the graph… 2 2 x + y = 4

  14. Using vectors, we can also find parametric equations for a line or a line segment: Find a parametrization of the line through the points A(–2, 3) and B(3, 6). OA + AP = OP y P(x, y) AP = OP – OA B(3, 6) OA + AB = OB AB = OB – OA A(–2, 3) AP must be a scalar multiple of AB (let the scalar be “t”) x O

  15. Using vectors, we can also find parametric equations for a line or a line segment: Find a parametrization of the line through the points A(–2, 3) and B(3, 6). AP = tAB y P(x, y) OP – OA = t(OB – OA) B(3, 6) x + 2, y – 3 = t 5, 3 x + 2 = 5t y – 3 = 3t A(–2, 3) x = –2 + 5t y = 3 + 3t • Try graphing these parametrics!!! x O

  16. Now, how do we find the equation of the segment through the same two points??? y x = –2 + 5t y = 3 + 3t B(3, 6) What happens when we plug in t = 0 and 1??? A(–2, 3) t = 0 produces point A, t = 1 produces point B x O So, use the same equations, but restrict t!!! x = –2 + 5t y = 3 + 3t 0 < t < 1

  17. Find a parametrization for the line segment with endpoints (5,2) and (–2,–4). One possibility:

  18. Find a parametrization for the circle with center (–2,–4) and radius 2. One possibility:

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