14 - University of Pennsylvaniawziller/math114s14/ch14-3-4.pdf · lim ( , ) x y a b f x y L o If...

14.2

Continuity

( , ) ( , )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:

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.

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

If y then mx2 2 3

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

x y x

xy m x

x y x m x

0

2If y then x

2If x then y

2 5

2 4 2 8( , ) (0,0) 0lim lim

x y x

xy x

x y x x

2 4

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

2x y y

xy y

x y y

0

1

2

Limit does not exist

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

x y

xy

x y

2

4 20lim

1x

m x

m x

Example:2

2 2( , ) (0,0) lim

x y

xy

x y

If y then mx2 2 3

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

x y x

xy m x

x y x m x

2

20lim 0

1x

mx

m

2If y then x

2If x then y

2 5

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

x y x

xy x

x y x x

2 4

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

x y y

xy y

x y y y

2

20lim

1y

y

y 0

0

Is the limit 0? Use polar coordinates: cos( ), sin( )x r y r

2 3 2

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

cos( )sin ( )lim lim

x y y

xy r

x y r

2

0lim cos( )sin ( )r

r

0 2

since for all

cos( )sin ( )

"Sandwich theorem"

r r

Limit is 0!

3

20lim

1x

x

x

Also: a) Polynomials ( , ) are continuous (they contain terms like )n mP x y cx y

0 0 0 0

( , )b) Rational functions (quotients of polynomials) are continuous

Q( , )

at ( , ) if Q( , ) 0

P x y

x y

x y x y

2

2 2( , ) (0,0) earlier we showed lim 0

x y

xy

x y

2

2 2a) Where is ( , ) continuous?

xyf x y

x y

2

2 2( , ) (1,1)hence lim

x y

xy

x y

1

2

Examples:

is continuous for all ( , ) (0,0)f x y

can be extended continuously for all ( , ) by defining (0,0) 0f x y f

3 3

b) Can ( , ) tan be extended continuously?x y

f x yx y

3 3

limy x

x y

x y

2 2( )( )limy x

x y x xy y

x y

2Yes, by defining ( , ) tan(3 )f x x x

c) as a last resort, we can try some sample path as in the examples, or use

polar coordinates to decide if a limit exists.

2 2 2lim 3y x

x xy y x

14.3

Partial Derivatives

Recall:

0

'( ) limh

f a h f af a

h

Functions of one variable ( ).y f x

What is the derivative at ?x a

|

( ( ))x a

dy df x

dx dx

h

bafbhafbaf

hx

,,lim,

0

Partial derivatives:

baxf ,at respect to with of derivative Partial

Regard as a constant and differentiate , with respect to y f x y x

Example:2

2 2( , )

xf x y

x y

Compute . xf

2 2

d x

dx x y

2 2

d x

dx x b

2 2

2 2 2

1( ) (2 x)

( )

x b x

x b

2 2

2 2 2( )

x b

x b

2 2

2 2 2( )y

y xf

x y

Regard as if it were a constant : y y b

replace: b y

( , )z f x y

Partial derivative of with respect to at ,f y a b

0

, ,, limy

h

f a b h f a bf a b

h

Regard as a constant and differentiate

, with respect to

x

f x y y

0 0, :xf x y

0 0, :yf x y

Notation:

fDx

zyxf

xx

ffyxf xxx

,,

2

,xy xy

f ff x y f

y x y x

( , )z f x y

Example: 1( , ) arctan tany y

z f x yx x

2

recall:

1arctan( )

1

dx

dx x

2

1arctan

1

y y

x x x xy

x

2 2

1

1

y

xy

x

2 2

y

x y

The derivative with respect to first,

then the derivative with respect to of that

x

y

2

1arctan

1

y y

y x y xy

x

2

1 1

1xy

x

2

1

yx

x

We have four second order derivatives: , , , , , , ,xx xy yx yyf x y f x y f x y f x y

2 2

x

x y

Find ,f f

x y

2 3, lnx

g x y x yy

3 1 12x x

y

g xyy

3 12xy

x

2 2

2

13y x

y

xg x y

y

2 2 13x y

y

3

2

12xxg y

x 2

2

16yyg x y

y

26xyg xy

Mixed partials are equal, or the order of differentiation does not matter.

Example:

26yxg xy

14.4

Chain Rule

Recall: Functions of one variable z (y).f

and depends on : y (x).y x g

so depends on : z ( ) ( (x))z x f y f g

Chain rule:

'( ) '( ) '( ( ) '( )dz dz dy

f y g x f g x g xdx dy dx

yxfz , , x g t y h t

dz z dx z dy

dt x dt y dt

Use the chain rule to find at 1.dz

tdt

2 2cos sin 2xy xyye x e x x

2cosxyxe x

1

2

dx

dt t

1dy

dt t

2 2 21 1cos sin 2 cos

2

xy xy xydzye x e x x xe x

dt tt

If 1,

then 1, 0

t

x y

1

10 cos 1 1 2sin 1 1 cos 1 1

2t

dz

dt

cos1 sin1

Chain rule for a function of two variables:

Example:

z

x

z

y

2Let cos with and ln . xyz e x x t y t

yxfz , , , ,x g s t y h s t

s

y

y

z

s

x

x

z

s

z

t

y

y

z

t

x

x

z

t

z

2, 2 , 7 . z x y x x s t y s t

1

2

zy

x x

2

z x

y y

1x

t

7y

t

4 and 1s t

x

y

9

9

1

1 72 2

z xy

t x y

1 9

3 76 6

18 1 63

6

44

6

22

3

Another chain rule:

Example:

Find if 4 and 1.z

s tt