Definition: Parametric Curve, Parametric Equation
If x and y are given as functions: x
  
, y
  
Over an interval of t-values, then the set of points

,
 
     
defined by these equations is
 a parametric curve. The equations are parametric equations for the curve.
The variable t is a parameter for the curve and its domain D is the parameter interval. If D is a closed interval, a t b , the point
    
is the initial point of the curve and the point


    is the terminal point of the curve.
When the equation of a curve is written as parametric equations, we say we have parametrized the curve
Example
Prove that the parametrized equations for a circle of standard equation x
2  y
2  a
2 is x
 a cos
  and y
 a sin
 
where 0 t 2

Solution:
Proof: x
2  y
2 

 a a
2 cos cos
2
 
 

2  t
 a

2 a sin sin
2
 
 

2
 a
2
 cos
2  sin
2

 a
2 remember trig identity
Example
A parametrization is given for a curve is: x
  y

2 ,
 
0 1 a) Graph the curve. What are the initial and terminal points, if any. b) Find a Cartesian equation for a curve that contains the parametrized curve.
Solution:

a)
The initial point is

The terminal point is

    
  
    
  
Note that the graph is drawn from (3, 0) to (0, 2) b) Solve x 3 3 t for t x
  t
3 t
  
3 t

3
3
Now substitute this into equation y

2 t y

2 t

3

 
3
2 x 6

3
 
2 x

3
Example
2
Find a parametrization for the line segment with endpoints

Solution:
Pick one of the points and create the parametric equations x
  at y bt
1, 3

and
 
.

The line goes through the point
 
when t=0
Now set t to any other number, so lets make t=1
Now using the point

1, 3
 this gives us:    b
This then gives us the parametric equations: x y
  t
  t a a 5 b 4
where 0 t 1
Example
For x
 tan
 
, y
 
2sec
 
,

2 t

2 a) Graph the curve. State the initial and terminal points and indicate the direction of the curve. b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the
Cartesian equation is traced by the parametrized curve?
Solution: a) initial point:
  

 tan 


2
 


2



 
terminal point :
  

 tan
   


 
Therefore we do not have an initial or terminal point
We will need to create a table of values to assist us in drawing this curve.

b) We are looking for an algebraic relationship that relates tan(t) and sec(t) to a constant. The most common relationship is sec
2
   tan
2 
1 x
 tan
 x
2  tan
2
Therefore y
 
2sec


  y
2


 sec

 
2
 sec
2
This gives us : sec
2  tan
2 
1 y x 2 1
This is a hyperbola, and the parametrized curve draws the lower branch of the hyperbola. y x
2
1 y
1 x
2 y x
2
2 y
 
2 1

Example
For x 3 3 ,

2 , 0 x
2 t 1 a) Graph the curve. State the initial and terminal points and indicate the direction of the curve. b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the
Cartesian equation is traced by the parametrized curve?
Solution: a) initial point:
      
terminal point :

      
  

  

b) The graph looks like a line, therefore we will try to write y in terms of x y

2 t
 
2 t

2
 t
 
2
 
2
3
  t
 
2
 
2 x

2
3
We could have also derived this by the two point on a line.