2012 Parametric Functions
AP Calculus : BC BONUS
Parametric vs. Cartesian Graphs
yx
2
 y1  t

 x1  t
2
Addsinitial
position
and
(x , y ) a position graph
and  are called
parameters.
t
orientation
x = f (t) adds time,
y = g (t) motion, and
change
( f (t), g (t) ) is the
ordered pair
Parametric vs. Cartesian graphs (by hand)
f (t )  t 2  4
t  [2,3]
t
g (t ) 
2
t
-2
-1
0
1
2
3
x
y
Parametric vs. Cartesian graphs (calculator)
t  [0, 2 ]
x(t )  3cos(t )
y (t )  3sin(t )
t
x
y
0
MODE: Parametric
/2
ZOOM: Square

3/2
2
Try this.
x(t )  3cos(t )
y (t )  3sin(t )
t  [0, 2 ]
Parametric graphs are never
unique!
Eliminate the Parameter
Algebraic: Solve for t and substitute.
x  3t  1
y  2t  1
Eliminate the Parameter
Trig: Use the Pythagorean Identities.
Get the Trig function alone and square both sides.
x  3cos(t )
y  2sin(t )
Insert a Parameter
Easiest:
Let t equal some degree of x or y and plug in.
y  x  7 x  8x
4
3
x  y  5y
2
4
Calculus!
The Derivative finds the RATE OF CHANGE.
x  f (t )
y  g (t )
dy

dx
Words!
dy
dt
dx
dt
Example 1:
x  2t
y  3t  4
2
dy

dt
dx

dt
dy


dx
2
Eliminate the parameter. y  3  x   4
2
and
dy

dx
Calculus!
The Derivative finds the RATE OF CHANGE.
x = f (t) then
y = g (t) then
dx
dt
finds the rate of horizontal change
dy
dt
finds the rate of vertical change
with respect to time.
with respect to time.
(( Think of a Pitcher and a Slider.))
dy
dx
still finds the slope of the tangent at any time.
dy
dx

dy
dt
dx
dt
Example 2:
x  24t
2
y  16t 2  24t
2
dy
a) Find and interpret
dt
dy
b) Find and interpret
dx
dx
and
dt
at t = 2.
at t = 2
Example 3:
x  2 cos(t )
y  3sin(t )

Find the equation of the tangent at t = 4 ( in terms of x and y )
Find the POINT.
Find the SLOPE.
m
Graph the curve and its tangent

dy
dx
Example 4:
x  t  3t  24t  5
3
2
y  2t 2  12t
Find the points on the curve (in terms of x and y) , if any, where
the graph has horizontal and/or vertical tangents
m

dy
dx
Horizontal Tangents Slope = 0
therefore, numerator = 0
Vertical Tangent Slope is Undefined
therefore , denominator = 0
The Second Derivative
Find the SECOND DERIVATIVE of the Parametric Function.
2
d y

2
dx
 dy 
d 
 dx 
dt
dx
dt
1). Find the derivative of the
derivative w/ respect to t.
dx
2). Divide by the original
.
dt
x  t t  2
2
Example 1:
y  t  3t
3
Find the SECOND DERIVATIVE of the Parametric Function.
dy

dt
dx

dt
dy

dx
 dy 
d 
 dx  
dt
dx

dt
=
Last Update:
• 10/19/07