Parametric Equations: Motion in a Plane
Notes for Section 6.3
In Layman’s terms: Parametric equations allow us to put x and y into terms of a single
variable known as the parameter. Time, t, is a common parameter used in this section. We can
define x and y values each with its own equation in terms of t along with a specified interval
for t values.
Formal Definition: The graph of the ordered pairs  x, y  where x  f (t ) and y  g (t ) are
functions defined on an interval
I (of t -values) is a parametric curve.

x  f (t ) and y  g (t ) are parametric equations for the curve.



The variable t is the parameter.
I is the parameter interval.
Note: If no t interval is indicated, we assume
t   ,   or    t  
When we give parametric equations and a parameter interval, I, for a curve, we have
parametrized the curve. Therefore, parametrization of a curve consists of not only the
parametric equations but also the interval of t -values.
Example 1: a) Find a parametrization of the line AB through points A  2,3 and B  4,6 .
b) What is the t-interval that will generate the segment AB ?
a)First, find the equation for the line through A and B in slope intercept form.
m
1
, so
2
y 3 
…and finally
If we assign
1
1
1
 x  (2)  becomes y  3   x  2  becomes y  3  x  1
2
2
2
y  1 x4
2
t as the parameter and equate it with x, we can then write a set of parametric equations.
1
t 4
2
x=t &y=
b) Letting our parameter interval be
t  2,4
segment.
t
xt
-2
4
-2
4
1
y  t 4
2
3
6
A  2,3
B  4,6
or
2  t  4 , we can generate the endpoints of the
Example 2: Parametrize the circle with center (-2,-4) and radius 2.
We know
x  r cos and y  r sin  defines any point on a circle, so
x  r cos t  h and y  r sin t  k
is the parametric form for a circle.
 x  2cos t  2 and y  2sin t  4, t  0,2 
parametrically defines this circle.
Example 3: a) Graph by hand the parametric equations









x  t 2  2 and y  3t , where 1  t  3 .
t
-1
-1
-3
x t 2
y  3t
2
0
-2
0
1
-1
3
2
2
6
3
7
9
   




y
x








b) Now let us graph the parametric equations on the calculator on the same interval:
Setting the window:
x1T  T  2
2
y1T  3T
Tmin: -1
X-min: -2
Y-min: -5
T max: 3
X-max: 10
Y-max: 10
Tstep: 0.1
X-scl: 1
Y-scl: 1
c) Finally, let’s use the calculator to graph the same equations on the following t intervals.
i)
Example 4:
Let
t  3,1
ii)
t  2,3
2.5 feet away from the side of a 420 foot tower, a rock is dropped.

The vertical path of the falling rock can be represented by x  2.5 .

The height of the rock as it freefalls can be represented by y  16t 2  420 .
t represent seconds, where t  0,5 . See if you can set your window to accommodate this
situation. Use your table to find the height of the rock 1, 3 and 5 seconds after it is dropped.
1 second ___404 ft.____ 3 seconds ___276 ft._____ 5 seconds ____20 ft._______
Eliminating the Parameter
Sometimes we need to eliminate the parameter from a set of parametric equations and be able
to identify the function in its Cartesian (Rectangular) form. This means getting rid of the
parameter, t, and obtaining a single equation in terms of only x and y.

Remember: If no t interval is indicated, we assume
t   ,   or    t  
Here are some examples to practice.
Example 5: Eliminate the parameter and identify the graph of the parametric curves.
x  1  2t & y  2  t
a)
Solve for
t  2,4
where
t in the first equation and substitute for t in the second equation.
1st equation
x  1  2t
x 1  2t
x 1
1 x
 t or t 
2
2
y  2t
2
nd
equation
 1 x 
 
 2 
y  2
1 x
1
3
y  2 
 y  x
2 2
2
2
1
3
 y  x  is a line segment in slope intercept form from  5,4  to  7, 2  .
2
2
b)
x  t 2  2 & y  3t
2nd equation solved for
t: t
Now substitute it for t in the
y
3
1st
equation, and
x  t  2 becomes:
2
And now solving for y . . .
x
y2
2 
9
x2
y2
 9( x  2)  y 2 
9
y   9  x  2
y  3 x2
This is a parabola which opens to the right and has vertex  2,0  .
2
 y
x   2
3
 3 
y  2sin t  where t  0, 
 2 
c) x  2cos  t  &
Hint:  x  h    y  k   r 2
2
x 2  y 2   2cos  t     2sin  t  
2
2
2
x 2  y 2  4cos2 t  4sin 2 t
If x  2cos  t  &
y  2sin  t  then,
x 2  y 2  4(cos2 t  sin 2 t )
x 2  y 2  4(1)
x2  y 2  4
 x  0   y  0
2
This is
3
of a circle from
4
2
 22
 3 
0, 2  with radius 2 and center


 0,0 .
Example 6: Two opposing players in “Capture the Flag” are 100 ft apart. On a signal, they
run to capture a flag that is on the ground midway between them. The faster runner, however,
hesitates for 0.1 sec. The following parametric equations model the race to the flag:
x1=10(t – 0.1), y1= 3
x2= 100 –9t, y2= 3
a. Simulate the game in a [0,100] by [-1,10] viewing window with t starting at 0. Graph
simultaneously.
b. Who captures the flag and by how many feet?
10(t  0.1)  50
t  0.1  5
t  5.1sec.
100  9t  50
 9t  50
t  5.556 sec.
Faster runner still wins by 0.456 seconds. Second runner is still 4.1 feet from the flag at t = 5.1 seconds.
The equation for vertical projectile motion in terms of time:
y  16t 2  v0t  s0 .
Example 7: A baseball is hit straight up from a height of 5 ft with an initial velocity of 80 ft/sec.
a. Write an equation that models the height of the all as a function of time t.
y  16t 2  80t  5
b. Use a parametric mode to simulate the pop-up.
x 3 &
y  16t 2  80t  5
c. Use parametric mode to graph the height against time. [Hint: Let x(t) = t]
x t &
y  16t 2  80t  5
x-min: 0
x-max: 10
y-min: -10
y-max: 120
d. How high is the ball after 4 sec?
y  16(4)2  80(4)  5
y  69 ft.
e. What is the maximum height of the ball? How many seconds does it take to reach its
maximum height?
 b  b  
Use vertex formula.  , f     (2.5 sec, 105ft.)  at 2.5 seconds, the ball 105 feet high.
 2a  2a  
Projectile Motion when looking at horizontal and vertical components.
x  (vo cos )t and y  16t 2  (vo sin )t  so
If a projectile is launched at an initial height of so feet above the ground at an angle of

from
horizontal and the initial velocity is vo feet per second, the path of the projectile is modeled by the
parametric equations shown above.
Example 8: A baseball player is at bat and hits a ball at a height of 4 feet. The ball leaves the bat at
120 ft/sec towards the center field fence, which is 425 feet away. The fence is 12 feet high. If the ball
leaves the bat at an angle of elevation of 39, will the ball be a homerun?
Equations for the ball:
x1  (120cos39 )t and y  16t 2  (120sin39 )t  4
Equations for the fence:
x2  425 and y2  t
Window on Calculator:
T-min: 0
X-min: -10
Y-min: 0
T-max: 12
X-max: 500
Y-max: 100
T-step: 0.1
X-scl: 50
Y-scl: 10
Trace to get an x-value close to 425 ft.
Looking at the table:
Table Setup
TblStart = 4.55
Tbl = .001
At t= 4.558, the ball is 425.07 feet away and 15.807 feet high, so it is
a homerun!
Sinusoidal Problems Using Parametric Equations:
x  r cos  h & y  r sin  k
Example 9: Kristin is riding on a Ferris wheel that has a radius of 30 ft. The wheel is turning
counterclockwise at a rate of one revolution every 10 seconds. Assume the lowest point of the Ferris
wheel (6 o’clock) is 10 feet above the ground and that Kristin is at a point marked A (3 o’clock) at time
t=0. Use parametric equations to find Kristin’s position 22 seconds into the ride.
r = radius = 30 feet
Now we need  as a function of time.
1 revolution 2 


 radians / sec. so,   t
10 sec.
10 5
5
 
t0
5 
Thus, x  30cos 
30'
10'
&
 
t   40
5 
y  30sin 
When t  22, Kristin ' s position is  9.271 ft., 68.532 ft.
This means she is 9.271 feet to the right of the axle ,
and 68.532 feet above ground .