Unit 1.4 Parametric Equations
Definition: Parametric Curve, Parametric Equation
If x and y are given as functions:
x  f t  ,
y  g t 
Over an interval of t-values, then the set of points
 x, y    f  t  , g t  
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
 f a  , g a 
is the initial point of the curve and the point
 f b , g b  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
and
x 2  y 2  a 2 is x  a cos  t 
y  a sin  t  where 0  t  2
Solution:
Proof:
x 2  y 2   a cos  t     a sin  t  
2
2
 a 2 cos 2  t   a 2 sin 2  t 
 a 2  cos 2  t   sin 2  t  
 a2
remember trig identity
Example
A parametrization is given for a curve is:
x  3  3t ,
y  2t,
0  t 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)
3  30 , 2 0   3, 0
The terminal point is  3  31 , 2 1    0, 2 
The initial point is
Note that the graph is drawn from (3, 0) to (0, 2)
b) Solve
x  3  3t for t
x  3  3t
3t   x  3
x  3
t
3
Now substitute this into equation
y  2t
y  2t
 x  3 
 2

 3 
2 x  6

3
2
 x2
3
Example
Find a parametrization for the line segment with endpoints
Solution:
Pick one of the points and create the parametric equations
x  4  at
y  1  bt
 1,  3
and
 4, 1 .
The line goes through the point
 4, 1
when t=0
Now set t to any other number, so lets make t=1
Now using the point
 1,  3 this gives us:
1  4  a
3  1  b
 a  5
 b  4
This then gives us the parametric equations: x  4  5t where
0  t 1
y  1  4t
Example
For
x  tan  t  , y  2sec  t  , 

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:
  
 , 2sec       ,  
 2 
  2
  
  
terminal point :  x, y    tan   , 2sec      ,  
 2 
 2
a) initial point:

 x, y    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
sec2  t   tan 2  t   1
x  tan  t   x 2  tan 2  t 
Therefore
2
 y 
 y
y  2sec  t      sec  t      sec2  t 
 2 
2
This gives us :
sec2  t   tan 2  t   1
2
 y
2
   x 1
2
 
This is a hyperbola, and the parametrized curve draws the lower branch of the hyperbola.
2
 y
2
   x 1
2
2
 y
2
   1 x
2
 
y
  1  x2
2
y  2 1  x 2
Example
For
x  3  3t , y  2t , 0  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:
 x, y   3  3 0 ,2  0   3,0 
terminal point :  x, y    3  3 1 ,2 1    0,2 
a) initial point:
b) The graph looks like a line, therefore we will try to write y in terms of x
y  2t
  2t  2   2
   2  2t   2
2
 3  3t   2
3
2
 x2
3

We could have also derived this by the two point on a line.