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