A parameter is merely a variable.

Given    x = 3 + 2t
             y = 2 -   t

The given equations are parametric equations with parameter t.

Example:  Find parametric equations of the line through (-1,3) and (1,1)

  1. Choose one of the endpoints as the starting point.  We'll select (-1,3).
  2. Consider the x values.  How far does -1 have to move to get to 1?  Obviously, 1 - (-1) or 2 units.  So x = -1 + 2t.
  3. Consider the y values.  How far does 3 have to move to get to 1?  Obviously, 1 - 3 or -2 units.  So y = 3 - 2t
A vector is a ray with direction and magnitude (length).

Example:  Find a vector equation of the line through (-1,3) and (1,1).

  1. It helps to draw the line first so you can see what it looks like.
  2. Choose one of the endpoints as the starting point.  We'll select (-1,3)
  3. The direction vector of the line is the vector (represented by an ordered pair) that tells you how to move from one point to another.  We know we have to move the x 2 units and the y -2 units, so our direction vector is (2,-2).
  4. The vector equation is (-1,3) + (2,-2)t.

    5. (-1,3) is the starting point and (2,-2) is the direction vector.
Example:  Find the Cartesian equation of the line passing through (-1,3) and (1,1).
  1. Find the slope of the line.  m = -2/2 = -1
  2. Choose one of the points to substitute into the point-slope form of an equation.
  3. So y - 3 = -1(x + 1)   or    y - 1 = -1(x - 1)
  4. Simplifying would give y = -x + 2
OR
  1. Take the above parametric equations:  x = -1 + 2t

    2.                                                            y =  3  - 2t
  2. Solve the first equation for t.  x + 1 = 2t

    3.                                          x + 1 = t
                                                   2
  3. Substitute that value of t into the second equation.

    4.                                               y = 3 - 2( x + 1 )
                                                                        2
                                                     y = 3 - x - 1
                                                     y = -x + 2

Problems:

(a)  Eliminate the parameter t and write it as a rectangular equation.
       x = t2 + 3
       y = t – 1
          for  -2 < t < 4

     Solve the first equation for t and plug that value into the second equation.
      t2 = 3 - x
      t = \/(3 - x)

     y = \/(3 - x) - 1

(b)  Write the parametric equations as a rectangular equation
       x = 2 cos q
       y = 5 sin q

      Since cos2q + sin2q = 1, solve for sin qand cos q in the above equations and substitute
      into the identity statement.
         x/2 = cos q
         y/5 = sin q

         cos2q + sin2q = 1
        (x/2)2 + (y/5)2 = 1
        x2  +   y2 = 1
        4        25

        We have an ellipse.

(c)  Find a set of parametric equations to represent the graph of y = x2 + 2x – 1
      Let x = t
            y = t2 + 2t  - 1

(d)  Find a set of parametric equations to represent the graph of   x2 + y2 = 4.
       Divide by 4,  x2 + y2 = 1
                            4      4
       Considering   cos2q + sin2q = 1,  cos q = x/2  and sin q = x/2
        so  x = 2 cos q
             y = 2 sin q

(e)  A baseball is hit when it is 3 feet above the ground and leaves the bat with initial velocity of
      150 ft per second and at an angle of elevation of 20o.  A 6 mph wind is blowing in the
      horizontal direction against the batter.  A 20 ft high fence is 400 ft from home plate.  Will the
      hit be a home run?

     Change 6 mph to ft/sec:  6 mi      1 hr     1 min     5280 ft   = 8.8 ft
                                              hr      60 min     60 sec      mi              sec
     Set up the situation:
       x represents the horizontal motion:  x = 150t cos 20 o - 8.8t (distance obtained by the
                                                                        horizontal velocity and windspeed)
       y represents the vertical motion:     y = -16t2 + 150t sin 20o + 3  (distance obtained by
                                                 gravity, vertical velocity, and  original height)
      The question now is, how high is the ball when it reaches the fence 400 ft out?  Set x = 400
      and solve for t.
           400 = 150t cos 20 o - 8.8t
           400 = (150 cos 20 o - 8.8)t
                  400                =  t
          150 cos 20 o - 8.8

              3.027 = t

         Now, without rounding, put this value for t into the y equation to see how the ball was at
         this time:
         y = -16( 3.027)2 + 150( 3.027) sin 20o + 3 = 11.701 ft
         No, the hit was not a home run.  It didn't make it over the fence.

         (To simulate this in parametric mode in the calculator, set X1 = 400 and
          Y1 = 20 t / tmax   to show the fence.  Then X2 and Y2 are the equations of
          motion above.)