Lesson 14 a - parametric equations

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Lesson 14 a - parametric equations

  1. 1. PARAMETRICEQUATIONS
  2. 2. DEFINITION: PARAMETRIC EQUATIONS If there are functions f and g with a common domainT, the equations x = f(t) and y = g(t), for t in T, areparametric equations of the curve consisting of allpoints ( f(t), g(t) ), for t in T. The variable t is theparameter. The equations x = t + 2 and y = 3t – 1for example are parametric equations and t is theparameter. The equations define a graph. If t is assigneda value, corresponding values are determined for x andy. The pair of values for x and y constitute thecoordinates of a point of the graph. The complete graphconsists of the set of all points determined in this way
  3. 3. as t varies through all its chosen values. We caneliminate t between the equations and obtain anequation involving x and y. Thus, solving either equationfor t and substituting in the other, we get 3x – y = 7 The graph of this equation, which also the graph ofthe parametric equations, is a straight line.Example 1: Sketch the graph of the parametric equations x = 2 + t and y = 3 – t2 . t -3 -2 -1 0 1 2 3 x -1 0 1 2 3 4 5 y -6 -1 2 3 2 -1 -6
  4. 4. y t=0 ● t=-1● ● t=1 x t=-2 ● ● t=2t=-3 ● ● t=3
  5. 5. Example 2: Eliminate the parameter between x = t + 1 and y = t2 + 3t + 2 and sketch the graph.Solution:Solving x = t + 1 for t, we have t = x – 1.Substitute into y = t2 + 3t + 2, then y = (x – 1)2 + 3(x – 1) + 2 y = x2 – 2x + 1 + 3x – 3 + 2 y = x2 + xReducing to the standard form, y + ¼ = x2 + x + ¼ y + ¼ = (x + ½)2 , a parabola with V(-½,-¼) opening upward
  6. 6. y 2 1 0 1 2 x-2 -1 -1 -2
  7. 7. Example 3: Eliminate the parameter between x = sin t and y = cos t and sketch the graph.Solution:Squaring both sides of the parametric equations, we have x2 = sin2 t and y2 = cos2 tAnd adding the two equations will give us x2 + y2 = sin2 t + cos2 tBut sin2 t + cos2 t = 1Therefore x2 + y2 = 1 , a circle with C(0, 0) and r = 1
  8. 8. y 2 1 0 1 2 x-2 -1 -1 -2
  9. 9. Example 4: Find the parametric representation for the line through (1, 5) and (-2, 3).Solution:Letting (1, 5) and (-2, 3) be the first and second points, respectively, of x = x1 + r(x2 – x1)and y = y1 + r(y2 – y1)We then have x = 1 + r(-2 – 1) and y = 5 + r(3 – 5) x = 1 – 3r y = 5 – 2r
  10. 10. Example 5: Eliminate the parameter between x = sin t + cos t and y = sin t.Solution:Solving sin2 t + cos2 t = 1 for cos t, we have cos t = 1 − sin2 tSubstitute into x = sin t + cos t , then x = sin t + 1 − sin2 tBut y = sin t and y2 = sin2 tTherefore x=y+ 1 − y2 x–y= 1 − y2Squaring both sides (x – y)2 = 1 – y2
  11. 11. Exercises:Eliminate the parameter and sketch the curve.• x = t2 + 1, y = t + 1• x = t2 + t – 2 , y = t + 2• x = cos θ , y = cos2 θ + 8 cos θ• x = 4 cos θ , y = 7 sin θ• x = cos θ , y = sin 2θ• x = 1 + cos 2θ , y = 1 – sin θ

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