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The Beginning of Parametric Equations (Sec. 6.3a).

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Presentation on theme: "The Beginning of Parametric Equations (Sec. 6.3a)."— Presentation transcript:

1 The Beginning of Parametric Equations (Sec. 6.3a)

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

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

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

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

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

9 txy 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… Ex: x = 1 – 2t, y = 2 – t, – < t < 88 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: x = t – 2, y = 3t y = + 3 x + 2  Identify the curve!!! 2 – Can we graph the curve in both forms???

13 Eliminate the parameter and identify the graph: x + y = 4 First, check the graph… 22

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). x y O A(–2, 3) B(3, 6) P(x, y) OA + AP = OP AP = OP – OA OA + AB = OB AB = OB – OA AP must be a scalar multiple of AB (let the scalar be “t”)

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). x y O A(–2, 3) B(3, 6) P(x, y) OP – OA = t(OB – OA) AP = tAB x + 2, y – 3 = t 5, 3 x + 2 = 5ty – 3 = 3t x = –2 + 5ty = 3 + 3t  T T T Try graphing these parametrics!!!

16 Now, how do we find the equation of the segment through the same two points??? x y O A(–2, 3) B(3, 6) x = –2 + 5ty = 3 + 3t What happens when we plug in t = 0 and 1??? restrict t So, use the same equations, but restrict t!!! x = –2 + 5ty = 3 + 3t0 < t < 1 t = 0 produces point A, t = 1 produces point B

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.


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