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PARAMETRIC
EQUATIONS
DEFINITION: PARAMETRIC EQUATIONS
   If there are functions f and g with a common domain
T, the equations x = f(t) and y = g(t), for t in T, are
parametric equations of the curve consisting of all
points ( f(t), g(t) ), for t in T. The variable t is the
parameter.
   The equations        x = t + 2 and y = 3t – 1
for example are parametric equations and t is the
parameter. The equations define a graph. If t is assigned
a value, corresponding values are determined for x and
y. The pair of values for x and y constitute the
coordinates of a point of the graph. The complete graph
consists of the set of all points determined in this way
as t varies through all its chosen values. We can
eliminate t between the equations and obtain an
equation involving x and y. Thus, solving either equation
for t and substituting in the other, we get
                      3x – y = 7
    The graph of this equation, which also the graph of
the 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
y




                 t=0
                  ●
         t=-1●         ● t=1

                                       x
  t=-2 ●                 ● t=2




t=-3 ●                         ● t=3
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 + x
Reducing to the standard form,
            y + ¼ = x2 + x + ¼
            y + ¼ = (x + ½)2 , a parabola with V(-½,-¼)
                               opening upward
y




          2

          1

          0    1   2
                       x
-2   -1
          -1

          -2
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 t
And adding the two equations will give us
               x2 + y2 = sin2 t + cos2 t
But
              sin2 t + cos2 t = 1
Therefore
              x2 + y2 = 1 , a circle with C(0, 0) and r = 1
y

          2



          1




          0    1   2
                       x
-2   -1


          -1



          -2
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
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 t
Substitute into x = sin t + cos t , then
                 x = sin t + 1 − sin2 t
But              y = sin t and y2 = sin2 t
Therefore        x=y+      1 − y2
                 x–y=      1 − y2
Squaring both sides
               (x – y)2 = 1 – y2
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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Lesson 14 a - parametric equations

  • 2. DEFINITION: PARAMETRIC EQUATIONS If there are functions f and g with a common domain T, the equations x = f(t) and y = g(t), for t in T, are parametric equations of the curve consisting of all points ( f(t), g(t) ), for t in T. The variable t is the parameter. The equations x = t + 2 and y = 3t – 1 for example are parametric equations and t is the parameter. The equations define a graph. If t is assigned a value, corresponding values are determined for x and y. The pair of values for x and y constitute the coordinates of a point of the graph. The complete graph consists of the set of all points determined in this way
  • 3. as t varies through all its chosen values. We can eliminate t between the equations and obtain an equation involving x and y. Thus, solving either equation for t and substituting in the other, we get 3x – y = 7 The graph of this equation, which also the graph of the 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. y t=0 ● t=-1● ● t=1 x t=-2 ● ● t=2 t=-3 ● ● t=3
  • 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 + x Reducing to the standard form, y + ¼ = x2 + x + ¼ y + ¼ = (x + ½)2 , a parabola with V(-½,-¼) opening upward
  • 6. y 2 1 0 1 2 x -2 -1 -1 -2
  • 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 t And adding the two equations will give us x2 + y2 = sin2 t + cos2 t But sin2 t + cos2 t = 1 Therefore x2 + y2 = 1 , a circle with C(0, 0) and r = 1
  • 8. y 2 1 0 1 2 x -2 -1 -1 -2
  • 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. 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 t Substitute into x = sin t + cos t , then x = sin t + 1 − sin2 t But y = sin t and y2 = sin2 t Therefore x=y+ 1 − y2 x–y= 1 − y2 Squaring both sides (x – y)2 = 1 – y2
  • 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