x  f  t  , y  g  t   Parametric Equations
In other words, at time t, the particle is located at the point
c  t    f  t  , g  t    Parametric Curve
Sketch the curve with parametric equations
x  2t  4, y  3  t 2
t
x  2t  4
y  3  t2
2
8
7
0
4
3
2
0
7
Notice, every point has a time
4
4
19
and we also have a direction arrow!
CONCEPTUAL INSIGHT The graph of a function y = f (x) can
always be parametrized in a simple way as c (t) = (t, f (t))
For example, the parabola y = x2 is parametrized by
c (t) = (t, t2) and the curve y = et by c (t) = (t, et). An
advantage of parametric equations is that they enable us to
describe curves that are not graphs of functions. For
example, the curve below is not of the form y = f (x) but it
can be expressed parametrically.
Eliminating the Parameter Describe the parametric curve
c (t) = (2t − 4, 3 + t2)
of the previous example in the form y = f (x).
Solve for y as a function of x:
x4 1
x  2t  4  t 
 x2
2
2
Substitute:
2
1 2
1

y  3   x  2   7  2x  x
4
2

1 2
y  7  2x  x
4
t in terms of x !
y in terms of x !
(We have y in terms of t
and t in terms of x.)
A bullet follows the trajectory
c (t) = (80t, 200t − 4.9t2)
until it hits the ground, with t in seconds and distance in
meters. Find:
(a) The bullet’s height at t = 5s.
(b) Its maximum height.
The height of the bullet at time t is y (t) = 200t − 4.9t2
y  5   877.5 m
The maximum height occurs
at the critical point of y (t):
y '  t   200  9.8t  0
 t  20.408 s  y  20.408   2041 m
Solution
y  mx  b
THEOREM 1 Parametrization of a Line
y  y1  m  x  x1 
(a) The line through P = (a, b) of slope m is parametrized by
x  a  rt ,
y  b  st
t in  ,  
for any r and s (with r  0) such that m = s/r.
(b) The line through P = (a, b) and Q = (c, d) has
x
y
parametrization
x  a  t c  a ,
y  b  t d  b
t in  ,  
The segment from P to Q corresponds to 0 ≤ 1 ≤ t.
 'ly many parametrizations of a line.
These are equivalent expressions for x  t  and y  t  . In (a) we are given point-slope, in (b) two points.
So how do we parametrize a line?
y  y1  m  x  x1 
(a) Use x = a + rt, to write t in terms of x… implies t = (x − a)/r:
 xa
y  b  st  y  b  s 
  b  m x  a
 r 
This is the equation of the line through P = (a, b) of slope m.
The choice r = 1 and s = m yields the parametrization below.
(b) This parametrization defines a
line that satisfies (x (0), y (0)) = (a, b)
and (x (1), y (1)) = (c, d). Thus, it
parametrizes the line through P and
Q and traces the segment from P to
Q as t varies from 0 to 1.
, Q   c, d 
THM 1
Parametrization of a Line Find parametric equations for the
line through P = (3, −1) of slope m = 4.
x  a  rt , y  b  st t in  ,  
s
m   4  r  1 & s  4 P   a, b    3, 1 
r
x  3  t , y  1  4t
Parametric Curve: c  t    3  t , 1  4t 
 'ly many parametrizations of a line.
s
m   4  r  5 & s  20 P   a, b    3, 1 
r
c  t    3  5t , 1  20t 
The circle of radius R with center (a, b) has parametrization
x  R cos  ,
y  R sin 
Let’s verify that a point (x, y) given by the above equation,
satisfies the equation of the circle of radius R centered at
(a, b):
2
2
2
2
 x  a    y  b    a  R cos   a    b  R sin   b 
 R cos   R sin   R
2
2
2
2
2
In general, to translate a
parametric curve horizontally
a units and vertically b units,
replace c (t) = (x (t), y (t)) by
c (t) = (a + x (t), b + y (t)).
Suppose we have a parametrization c (t) = (x (t), y (t)) where
x (t) is an even function and y (t) is an odd function, that is,
x (−t) = x (t) and y (−t) = −y (t). In this case, c (−t) is the
reflection of c (t) across the x-axis:
c (−t) = (x (−t), y (−t)) = (x (t), −y (t))
The curve, therefore, is symmetric with respect to the x-axis.
Parametrization of an Ellipse Verify that the ellipse with
equation
2
2
x  y
    1
a b
is parametrized by
c  t    a cos t , b sin t 
 for    t   
Show that the equation of the ellipse is satisfied with
x = a cos t, y = b sin t:
2
2
2
2
 x   y   a cos t   b sin t 
2
2
     
 +
  cos t  sin t  1
a b  a   b 
Plot the case a = 4, b = 2.
To plot the case a = 4, b = 2,
we connect the points
corresponding to the t-values
in the table. This gives us the
top half of the ellipse
corresponding to 0 ≤ t ≤ π.
Then we observe that
x (t) = 4 cos t is even and
y (t) = 2 sin t is odd. As noted
earlier, this tells us that the
bottom half of the ellipse is
obtained by symmetry with
respect to the x-axis.
c (−t) = (x (−t), y (−t)) = (x (t), −y (t))
The curve, therefore, is symmetric with respect to the x-axis.
Different Parametrizations of the Same Curve Describe the
motion of a particle moving along each of the following paths.
(a) c1(t) = (t3, t6)
Each of these parametrizations
(b) c2(t) = (t2, t4)
satisfies y = x2, so all three
(c) c3(t) = (cos t, cos2 t) parametrize portions of the
parabola y = x2.
c (t) = (t, f (t))
(a) As t varies from   to , t 3 varies from   to .
 c1  t    t 3 , t 6  traces all of y  x 2
(moving left to right and passing through each point once).
Different Parametrizations of the Same Curve Describe the
motion of a particle moving along each of the following paths.
(a) c1(t) = (t3, t6)
Each of these parametrizations
(b) c2(t) = (t2, t4)
satisfies y = x2, so all three
(c) c3(t) = (cos t, cos2 t) parametrize portions of the
parabola y = x2.
(b) x  t 2  0  the path of c2  t    t 2 , t 4  traces only the right
half of the parabola. The particle comes in towards the origin as
t varies from   to 0, and goes back out to the right as t varies
from 0 to .
Different Parametrizations of the Same Curve Describe the
motion of a particle moving along each of the following paths.
(a) c1(t) = (t3, t6)
Each of these parametrizations
(b) c2(t) = (t2, t4)
satisfies y = x2, so all three
(c) c3(t) = (cos t, cos2 t) parametrize portions of the
parabola y = x2.
(c) As t varies from   to , cos t oscilates between 1 and  1.
 the particle following c3  t    cos t , cos 2 t  oscilates between
the points 1,1 and  1,1 on the parabola.