Recall from last week:

Length of a curve | '( ) |

b

a

t dt r Arc length ( )

t

a

s t u du r | '( ) | ds

tdt

r

Arc length parametrization ( ) with | '( ) | 1s s r r

'( )Unit tangent vector '(s)

| '( ) |

t

t

rT r

r

Curvature: d

sds

T

r

t

t

T

r

3

t t

t

r r

r

Principal unit normal:

s t

s t

T T N

r r Binormal B T N

Frenet frame (or TNB frame) T,N,B

Torsion : d

ds

BN

d

ds

BN

d

ds

B

2

' '' '''t t t

t t

r r r

r r

( ) ( ), (t), ( )t x t y z t r

Decompose the acceleration vector ''( )t a r a a T NT N

2

2tangential acceleration: = ( ( ) )

d s da t

dt dtT r

22

normal acceleration: ( )ds

a tdt

N r

' ''

'

a v r ra T

v r

2 2' ''

= |'

a

T

v a r ra N a |

v r

constant velocity: ( ) 0d

adt

T r2

a N r

If you travel at constant speed on a circle of radius :a r

21a a

rN0a T

Recall:

13.6

Acceleration in Polar Coordinates

Newton’s law of gravitation (1687):

2

GmM

rF

| r | | r |11 2 2

is the vector from the center of the sun to the planet

is the mass of the sun

is the mass of the planet

is the gravitational constant

6.674 10 G = (from 1798  ) 

M

m

G

N m kg

r

2 ''

GMm

rF = a a r =

| r | | r |

'd

dt r r

hence ' is a constant vector r r C

in particular 0 r C

the planet moves in a plane orthogonal to ! C

Inverse square law

'' r r 0' ' '' r r r r

since is parallel to by Newton's lawr'' r

Recall polar coordinates:

cos( )

sin( )

x r

y r

2 2 2

arctan( )

r x y

y x

0

0 2

r

from Calc I,

area element:

replace , by (perpendicular unit vectors)i j

cos( ) sin( ) , sin( ) cos( )r u = i j u i j

' ' 'rr r r u u

cos( ) sin( ) rr r r r i j u

21

2dA r d

cos( ) sin( )r r r i j

rrr u

'cos( ) 'sin( )

'sin( ) 'cos( )

r r

r r

i j

i j

'

+ '

rr

r

u

u

' ( cos( ) sin( )

r r r i j)'

Kepler’s second law of planetary motion (1609):

The planet sweeps out equal areain equal time

and ' ' 'rr r r u urecall rrr u

recall: ' constant C r r

hence ' ' 'r rr r r C r r u u u ( ') rr r u u

2( ') 'r r r k = k

2since and are constant vectors, ' is constant as wellr C k

21recall:

2dA r d 2 21

hence ' 2

dA dr r

dt dt

Area swept out at constant speed!

Lets show this is true:

http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

i.e., the planet travels in a plane orthogonal to C

but and are unit vectors in the xy plane, hence = (or r r u u u u k k)

we can assume planet travels in xy plane

thus ( ') rr r C u u

14.1

Functions ofSeveral Variables

Recall:

Functions of one variable: ( )y f x

Graph of : the set in the plane ( , ( )) | f x f x x D

Domain : D set of all where ( ) is defined.x f x

Range: set of all ( ) for in the domainf x x

Functions of two (or more) variables: ( , ) (or ( , , ) )z f x y w f x y z

Domain : D region in the plane where ( , ) is defined.f x y

( , ) | f x y x DRange:

2

Find and sketch the domain of the function

, 1 .f x y x y

Example:

21 0x y 2or 1x y

Graph of : the set in 3-space ( , y, ( , )) | ( , ) f x f x y x y D

are curves in the plane given by

( , ) for various constants .

Level curves x y

f x y c c

When lifted to the surface, they are sometimes called contour curves.

Average January sea-level temperatures measured in degrees Celsius

The level curves are called isothermals, they join areas with the same temperature

(a) Contour map (b) Horizontal traces are

raised level curves

2 2The graph of ( , ) 4 is formed by lifting the level curves:z f x y x y

2 2

Level curves and graph of x yz xye

2 23 ( * * ( ), 5..5, 5..5, );plot d x y exp x y x y axes boxed

Maple command:

I II III

IV V VI

yxyxf ),()a( xyyxf ),()b(

221

1),()c(

yxyxf

222),()d( yxyxf

2),()e( yxyxf yxyxf sin),()f(

VI

I

II

V

IV

III

2

4

12sin yxz

2222 yxeyxz

22sin yxz

22 4

1

yxz

23 3xyxz

yxz sinsin

53.

54.

55.

56.

57.

58.

A B C

D E F

I II III

IV V VI

,B III

,C II

,F V

,A VI

,D IV

,E I

14.2

Continuity

Recall:

lim ( ) x a

f x L

the values ( ) approach

more and more as you get closer to .

f x L

x a

Notice: does not have to lie in the domain of definition!

( ( ) may be undefined)

a

f a

We can approach from two sides: both limits lim ( ) and lim ( )

must exists and be the same.

x a x aa f x f x

Examples:

0a) lim

| |x

x

x does not exists since

0 0lim 1 but lim 1

| | | |x x

x x

x x

2

3

9b) lim

x 3x

x

6

2 9since 3

x 3

xx

0

sin( )c) lim

xx

x

1 since sin( ) (sandwich theorem)x x x

is continous at f x a

is continous, if it is continuous for all (domain of definition)f x D

if lim ( ) ( )x a

f x f a

( , ) ( , )lim ( , )

x y a bf x y L

If the values of f(x, y) approach the number L as the point (x, y) approaches the point (a, b) along any path that stays within the domain of f.

Definition:

Examples:2 2

2 2( , ) (1,1)

sin( )a) lim

x y

x y

x y

sin(2)

2

2 2

2 2( , ) (0,0)

sin( )b) lim

x y

x y

x y

2

20

sin(r )limr r

0

sin(z)lim 1z z

2 2

2 2( , ) (0,0)

sin( )c) lim

x y

x y

x y

We can let (x, y) approach (a, b) from an infinite number of directions in any manner whatsoever as long as (x, y) stays within the domain of f.For all of these the limit must be the same.

If 0 then x 2 2 2

2 2 2( , ) (0,0) 0

sin( ) sin(x )lim lim 1

x y x

x y

x y x

If y 0 then 2 2 2

2 2 2( , ) (0,0) 0

sin( ) sin( y )lim lim

x y x

x y

x y y

1

Limit does not exist!

Examples:

2 2( , ) (0,0)a) lim

x y

xy

x y If 0 or 0 then x y

If y then x2

2 2 2( , ) (0,0) 0

1lim lim

2 2x y x

xy x

x y x

Limit does not exist2

2 4( , ) (0,0)b) lim

x y

xy

x y

2 2( , ) (0,0)lim 0

x y

xy

x y

(To be discussed Thursday)