Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”)

1. PARAMETRIC
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0
NS

2. ( ) ( )
( ) ( ) ( )( )
( ) ( )
.acalledisvariableThe
curve.theofcalledare
equationsThe.acalledis
,,
bydefinedpointsofcollectionThe
.intervalsomeisdomaincommonwhosefunctionstwoare
andwhere,andLet
parameter
equationsparametric
curveplane
t
tgytfx
tgtfyx
I
gftgytfx
==
=
==
The variable t (the parameter) often represents time. We
can picture this like a particle moving along and we know
its x position over time and its y position over time and
we figure out each of these and plot them together to see
the movement of the particle.

3. 0;4,2 ≥== ttytx
Graph the plane curve represented by the parametric equations
We'll make a chart and
choose some t values
and find the
corresponding x and y
values.
t x y
0 ( ) 002 = ( ) 004 =
The t values we pick must
be greater than or equal to
0. Let's start with 0.
( )yx,
( )0,0
1 ( ) 4.112 ≈ ( ) 414 = ( )4,2
( )0,0
( )4,2
2 ( ) 222 = ( ) 824 = ( )8,2
3 ( ) 4.232 ≈ ( ) 1234 =
( )8,2
( )12,6
( )12,6
We see the "path" of the
particle. The orientation is the
direction it would be moving
over time (shown by the arrows)

4. 0;4,2 ≥== ttytx
We could take these parametric equations and find an
equivalent rectangular equation with substitution. This
is called "eliminating the parameter."
Solve for the parameter t in one
of equations (whichever one is
easier).
( )0,0
( )4,2
( )8,2
( )12,6
4
y
t = Substitute for t in the other
equation.






=
4
2
y
x 2
2 y
x =2 2
yx =2
2
We recognize this as a
parabola opening up. Since
our domain for t started at 0,
it is only the right half.

5. π20;sin4,cos2 ≤≤== ttytx
Graph the plane curve represented by the parametric equations
t x y
0 20cos2 = 00sin4 =
The t values we
pick must be
from 0 to 2 π
( )yx,
( )0,2
2
4
cos2 =
π
22
4
sin4 =
π
( )22,2
( )4,0
( )0,2−
Make the orientation
arrows based where the
curve was as t increased.
4
π
2
π
0
2
cos2 =
π
4
2
sin4 =
π
π 2cos2 −=π 0sin4 =π
2
3π
0
2
3
cos2 =
π
4
2
3
sin4 −=
π
( )4,0 −
4
5π
2
4
5
cos2 −=
π
22
4
5
sin4 −=
π
( )22,2 −−
You could fill in with more
points to better see the curve.

6. π20;sin4,cos2 ≤≤== ttytx
Let's eliminate the parameter. Based on our curve we'd expect to
get the equation of an ellipse.
When you want to eliminate
the parameter and you have
trig functions, it is not easy to
solve for t. Instead you solve
for cos t and sin t and
substitute them in the
Pythagorean Identity:
1cossin 22
=+ tt
22 4 4
t
x
t
y
cos
2
sin
4
:aboveFrom ==
1
24
22
=





+




 xy 1
416
22
=+
xy
Here is the rectangular
version of our ellipse.
You can see it matches!

7. When you then enter a
graph, it will have t for
the variable and you
can enter more than
one equation.
Your Casio graphic calculator can plot parametric equations.
Select “Graph” mode and check that “Type” is set to Parm.
If you watch as it draws the
graph, you will see the
orientation (direction) of the
curve.

8. If an object is dropped, thrown, launched etc. at a
certain angle and has gravity acting upon it, the
equations for its position at time t can be written as:
( )tvx o θcos= ( ) htvgty o ++−= θsin
2
1 2
horizontal position initial velocity angle measured from horizontal
time gravitational constant which
is 9.8 m/s2
initial heightvertical position

9. ( )tvx o θcos= ( ) htvgty o ++−= θsin
2
1 2
Adam throws a tennis ball off a cliff, 300 metres high
with an initial speed of 40 metres per second at an angle
of 45° to the horizontal. Find the parametric equations
that describe the position of the ball at time t.
( ) ( ) 30045sin408.9
2
1 2
+°+−= tty( )tx °= 45cos40
How long is the ball in the air? When the ball hits the ground, the
vertical position y will be 0.
30028.289.40 2
++−= tt
30028.289.4 2
++−= ttytx 28.28=
use the quadratic formula
sec23.11or45.5−=t
The negative time value doesn't make sense so we throw it out.

10. Adam throws a tennis ball off a cliff, 300 metres high
with an initial speed of 40 metres per second at an angle
of 45° to the horizontal. Find the parametric equations
that describe the position of the ball at time t.
When is the ball at its
maximum height?
The motion is parabolic (opening down)
so maximum will be at the turning point.
30028.289.4 2
++−= ttytx 28.28=
a
b
t
2
TPofvalue −=
( )
sec89.2
9.42
28.28
≈
−
−=
What is the maximum height?
( ) ( ) 30089.228.2889.29.4
2
++−=y metres8.340≈

11. Adam throws a tennis ball off a cliff, 300 meters high
with an initial speed of 40 meters per second at an angle
of 45° to the horizontal. Find the parametric equations
that describe the position of the ball at time t.
Determine the horizontal distance the ball traveled.
Use time in air from first part of problem.
30028.289.4 2
++−= ttytx 28.28=
( )23.1128.28=x metres6.317≈

12. Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au