2012 Parametric Functions
AP Calculus : BC BONUS
Parametric vs. Cartesian Graphs2
1
1
y tx t
(x , y ) a position graph x = f (t) adds time, y = g (t) motion, and
change
( f (t), g (t) ) is the ordered pair
t and are called parameters.
Adds-
initial position
and
orientation
2y x
Parametric vs. Cartesian graphs (by hand)2( ) 4 [ 2,3]
( )2
f t t ttg t
t x y
-2
-1
0
1
2
3
Parametric vs. Cartesian graphs (calculator)
( ) 3cos( ) [0,2 ]( ) 3sin( )
x t t ty t t
t x y
0
/2
3/2
2
MODE: Parametric
ZOOM: Square
( ) 3cos( ) [0, 2 ]( ) 3sin( )
x t t ty t t
Try this.
Parametric graphs are never unique!
Eliminate the Parameter Algebraic: Solve for t and substitute.
3 12 1
x ty t
Eliminate the Parameter Trig: Use the Pythagorean Identities.
Get the Trig function alone and square both sides.
3cos( )2sin( )
x ty t
Insert a Parameter
4 3
4 2
7 8
5
y x x x
x y y
Easiest:
Let t equal some degree of x or y and plug in.
Calculus!The Derivative finds the RATE OF CHANGE.
dxdt
dydt
dydx
( )( )
x f ty g t
Words!
Example 1:
2
2
3 4
x t
y t
dydtdxdt
dydx
Eliminate the parameter. and dydx
2
3 42xy
Calculus!The Derivative finds the RATE OF CHANGE.
x = f (t) then finds the rate of horizontal change
with respect to time.
y = g (t) then finds the rate of vertical change
with respect to time.
(( Think of a Pitcher and a Slider.))
dxdt
dydt
still finds the slope of the tangent at any time.dydx dy
dy dtdxdxdt
Example 2:
2
24 2
16 24 2
x t
y t t
a) Find and interpret and at t = 2dydt
dxdt
b) Find and interpret at t = 2.dydx
Example 3:2cos( )3sin( )
x ty t
Find the equation of the tangent at t = ( in terms of x and y )
dymdx
4
Find the POINT. Find the SLOPE.
Graph the curve and its tangent
Example 4: 3 2
2
3 24 5
2 12
x t t t
y t t
Find the points on the curve (in terms of x and y) , if any, where the graph has horizontal and/or vertical tangents
dymdx
Vertical Tangent Slope is Undefined therefore , denominator = 0
Horizontal Tangents Slope = 0therefore, numerator = 0
The Second DerivativeFind the SECOND DERIVATIVE of the Parametric Function.
2
2
d ydx
dyddx
dt
dxdt
1). Find the derivative of the derivative w/ respect to t.
2). Divide by the original .dxdt
Example 1:2
3
2
3
x t t
y t t
Find the SECOND DERIVATIVE of the Parametric Function.
dydx
dydtdxdt
dyddx
dt
dxdt
=
Last Update:
• 10/19/07