10.1

Parametric Functions

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Topics for today

A little parametric review

dy/dx of parametrics

d2y/dx2 of parametrics

Lengths of parametric graphs

Tangents of parametric graphs

Parametric equations can be used to describe motion that is not a function.

( ) ( ) x f t y g t= =

→

-6

-4

-2

0

2

4

6

-6 -4 -2 2 4 6

This curve is:

( )

( ) ( )

sin 2

2cos 5

x t t t

y t t t

= +

= +

:

min 5 min 6 min 6

max 5 max 6 max 6

step .1

WINDOW

T X Y

T X Y

T

= − = − = −

= = =

=

The airplanes are at the same altitude. Do the planes hit each other?

Maybe, but likely not…they would have to reach the same point at the same TIME.

Independent vs Dependent Variables

Most relationships we study have 1 dependent variable.

2

2 2

dependent independent

y x

z x y

=

=

= +

2

2

dependent independent

cos

sin

x t t

y t t

=

= +

= +

The independent variable is known as “the parameter”

Parametrics allow study of relationships with multiple dependent variables

Parametrics allow you to model problems with more than 2 variables

2 2 1x y+ =

2cos

4sin

x t

y t

=

=

Regular equation can answer “where”

Parametric equations can answer “where” and “when”

( )

( ) 2

Parametrics also tell you WHEN

150cos 18

150sin 18 3 16

x t

y t t

=

= + −2

A simple quadratic will show you

WHERE the bird will go.

48 3 .112y x x= + −

1 sec2 sec

3 sec

Calculator Items

MODE = PARAMETRIC

WINDOW TMin, TMax, Tstep - tells how much of graph to displayUse TABLE to figure out how to set X and Y parameters

TRACE shows X,Y at different T values.

( )

( ) 2150cos 18

150sin 18 3 16

x t

y t t

=

= + −

Be careful about your Trig mode. Make sure you are in degrees or radians depending on what you are analyzing.

Objectives

Eliminate the parameter and classify graphFind “t” for a given pointFind slope and dy/dxFind 2nd derivativeFind length of graphFind equations of tangents

Find y in terms of x and state domain and range

( ) ( ) 2:

1 0 3

Given

x t t y t t t= − =

This is also known as “eliminating the parameter.”Useful to determine what graph looks like.

From the x(t) relationship:x = t – 1x +1 = t

From the y(t) relationship:y = t2

Substitute, and we find:

y = (x+1)2

Domain and range…plug in t to x(t) and y(t).1 2

0 9

x

y

−

( )

3 2

Find at the point 27,9

x t y t

t

= =

=

327

3

t

t

=

=

29

3

t

t

=

=

3t =

What might we want to know about our graph?

-6

-4

-2

0

2

4

6

-6 -4 -2 2 4 6

( )

( ) ( )

sin 2

2cos 5

x t t t

y t t t

= +

= +Position at t

Slope at time dy

tdx

2

2Concavity at time

d yt

dx

Finding dy/dx

2

sin

1

x t

y t

=

= +

But we want dy

dx

We can easily get and .dy dx

dt dt

dy

dy dtdxdx

dt

=

Look familiar?

dy

dy ddxdx

d

= dy

dy dtdxdx

dt

=

Parametric Polar

You try

2

sin

1

x t

y t

=

= +

dy

dy dtdxdx

dt

=

Find dy

dx

cos 2dx dy

t tdt dt

= =

2

cos

dy t

dx t=

2

2

d y

dx

( ) ( )2

cos 2 2 sin

cos

t t t

t

− −=

Chain ruleQuotient rule

( ) ( )2

cos 2 2 sin

cos

t t t dx

t dt

− − =

2

sin

1

x t

y t

=

= +cos 2

dx dyt t

dt dt= =

2

cos

dy t

dx t=

Change to division because we know dx/dt, not dt/dx

( ) ( ) ( ) ( )2 3

cos 2 2 sin cos 2 2 sincos

cos cos

t t t t t tt

t t

− − − −= =

2

cos

d t

dx t

=

dt

dx

2

2

d y

dx

Chain ruleThe derivative of the 1st

derivative

'dydt

dxdt

=

Let’s Generalize

'dy dt

dt dx=

The derivative of the first derivative wrt t

Divided by dx/dt again

cos 2dx dy

t tdt dt

= =2

'cos

ty

t=

( ) ( )2 2

2

cos 2 2 sin'

cos

cos

t t tdy

d y dt tdxdx t

dt

− −

= =( ) ( )

3

cos 2 2 sin

cos

t t t

t

− −=

-1

-0.5

0

0.5

-2 -1.5 -1 -0.5 0.5

Example 2:

22 3

2Find as a function of if and .

d yt x t t y t t

dx= − = −

→

Example 2:2

2 3

2Find as a function of if and .

d yt x t t y t t

dx= − = −

1. Find the first derivative (dy/dx).

dy

dy dtydxdx

dt

= =21 3

1 2

t

t

−=

−

→

2. Find the derivative of dy/dx as a function of t.

21 3

1 2

dy d t

dt dt t

−=

− ( )

2

2

2 6 6

1 2

t t

t

− +=

−

Quotient Rule

( )1 2t −

Divide by dx/dt again (Chain Rule)

( )

2

3

2 6 6

1 2

t t

t

− +=

−

What if we want to knowThe length of the graph?

-6

-4

-2

0

2

4

6

-6 -4 -2 2 4 6

Greg Kelly, Hanford High School, Richland, Washington

L =dx

dt

2

+dy

dt

2

dta

b

The equation for the length of a parametrized curve is:

→

(Notice the similarity to the polar length formula.)

Ex. Find the length of the arc

Graph it:

The curve can be broken into 4 symmetrical sections. So, let’s just focus on 1st quadrant.

3 3cos , sin , 0 2x t y t t = =

3 3cos , sin , 0 2x t y t t = =

2dy

dt

=

2dx

dt

=

2 2

,dx dy

Sodt dt

+ =

4 2 4 29cos sin 9sin cost t t t= +

3 cos sint t=

So, the length of the curve is:

/2

0

4 3 cos sin 6t t dt

= =

/2

0

/2 11

232 0

0 0

3cos sin

(absolute value is not necessary as

sin and cos are positive in quadrant 1)

let sin

cos

3cos sin 3 = 1.5

t t dt

u t

du tdt

t t dt u du u

=

=

= =

Tangents

To find location of horizontal tangents:

To find location of vertical tangents:

set

dy

dt= 0

set

dx

dt= 0

These will give you the t values of the tangents.

Use the original parametric equations to find (x,y).