h

bafbhafbaf

hx

,,lim,

0

baxf ,at respect to with of derivative Partial

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

0

, ,, limy

h

f a b h f a bf a b

h

, x

f zf x y D f

x x x

, y

f zf x y D f

y y y

second partial derivatives , , , , , ,xx xy yx yyf x y f x y f x y f x y

yxfz ,dz z dx z dy

dt x dt y dt

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

y

y

z

s

x

x

z

s

z

, x g t y h t

t

y

y

z

t

x

x

z

t

z

Chain Rule:

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

Review:

, 0 implicity defines as a function of :F x y y x

Set , yz F x

2Consider the function 2. Find at 1,4dy

xy x ydx

2 2F x y xy

1

22

xF xy yxy

2 1

2yF x x

xy

2

22

2

yxy

xydy

xdxx

xy

1,4

42 4

2 21

12 2

dy

dx

1 8 28

3 3

4

Implicit differentiation:

solve for : ( )y y f x

Differentiate with respect to : , y , f(x) 0x F x F x

0z F dx F dy

x x dx y dx

Example:

x

y

Fdy

dx F

F F dy

x y dx

x y

dyF F

dx

, , 0w F x y z

yxgz ,

x

z

Fz

x F

y

z

Fz

y F

2 2 2 is defined as a function of and by: 23 . z x y x y y z xz

2 2 2 23 0F x y y z xz

22xF xy z

2 2zF y xz

x

z

Fz

x F

2

2

2

2

xy z

y xz

2

2

1,2,3

2 1 2 3

2 2 1 3

z

x

13

10

Another implicit differentiation:

implicity defines as a function of and z x y

0 x y

w F dx F dy F z F F z zF F

x x dx y dx z x x y x x

similarly ,

Example:

Find at 1,2,3z

x

(Notice 1, 2,3 0 !)F

14.5

Directional DerivativesAnd Gradient Vectors

,z f x y

0 0, is the rate of change of in the direction parallel to the axis.xf x y z x

0 0, is the rate of change of in the direction parallel to the axis.yf x y z y

What about the rate of change of in other directions?z

slope 1,1xf

slope 1,1yf

0 0We want to find the rate of change of at ,

in the direction of an arbitrary vector , .

z x y

unit a bu

The slope of the tangent line to at the point

is the rate of change of in the direction of .

T C P

z u

Project and onto the

plane to get and .

P Q

xy P Q

Surface with equation ,S z f x y 0 0 0,z f x y

0 0 0, ,P x y zThe vertical plane that passes through

in the direction of intersects in a curve .

P

S Cu

Let , , be another point on .Q x y z C

,P Q h ha hb u

0 0 and ha x x hb y y

0 0 and x x ha y y hb

0 0, ,z f x y f x ha y hb

0 0 0 0( , ) ( , )slope of secant line:

f x sa y sb f x y

s

0 0Go along a straight line in the - plane, starting at ( , )

in the direction of at unit speed (arc length).

x y x y

u

2 2, with 1 a b a b u

0 0( ) ( , )t x y s r u 0 0( , ) ,x y s a b

0 0or , yx x sa y sb

0 0 0 0change in : ( , ) ( , ) z z f x sa y sb f x y

0 0 0 0

0

( , ) ( , )rate of change in the direction of lim

s

f x sa y sb f x y

s

u :

0 0

0 0

(Directional Derivative)

The directional derivative of , in the direction of at ,

is denoted by , :

f x y x y

D f x yu

Definition :

u

0 0 0 00 0

0

( , ) ( , ), lim

s

f x sa y sb f x yD f x y

s

u

or cos( ), sin( )a b

0 0,x sa y sb a

a b

b

0 0 0 00 0

0

( , ) ( , ), lim

s

f x sa y sb f x yD f x y

s

u

, with 1a b u u

cos( ),sin( )

a b i j

angle with the axisx

A picture of the directional derivative:

Special cases:

a) ( 1, 0) :a b u i

0 0 0 00 0

0

( s, ) ( , ), lim

s

f x y f x yD f x y

s

u

0 0,xf x y

b) ( 0, 1) :a b ju

0 0 0 00 0

0

( , s) ( , ), lim

s

f x y f x yD f x y

s

u

0 0,yf x y(here )h s

0 0 0 00 0

0

( , ) ( , )A simpler way to compute , lim

s

f x sa y sb f x yD f x y

s

u

0 0Set ( ) ( , )g s f x sa y sb

0 0Then ,D f x y u'(0)g

By the chain rule: '(0)g

0 0 0 0 0 0, ( , ) ( , ) where ,x yD f x y f x y a f x y b a b u u

Example:

If then u i, 0 0 0 0 , ( , )xD f x y f x yu

If then u j, 0 0 0 0 , ( , )yD f x y f x yu

This also shows that it is important that is a unit vector. u

0

( ) (0)lims

g s g

s

0 0 0 0( , ) ( , )x yf x y a f x y b

Example:2 2

0 0

Compute the directional derivative of ( , ) 310 10

at ( , ) (3, 1) in the direction of 1,-1 .

x xyD f f x y

x y

u

2

(3, 1)

23, 1

10 10x

x yf

6 1 5

10 10 10

(3, 1)

23, 1

10y

xyf

6

10

5 1 6 1

3, 110 102 2

D f

u

2

20

0 0 0 0 0 0, ( , ) ( , ) x yD f x y f x y a f x y b u

unit vector u1 1

,2 2

, with 1a b u u

Although decreases in the direction of the axis and axis,

it increases in the direction of !

f x y

u

The gradient vector:

0 0 0 0 0 0, ( , ) ( , ) where ,x yD f x y f x y a f x y b a b u u

0 0 0 0( , ), ( , ) ,x yf x y f x y a b

0 0 0 0 0 0 0 0The gradient vector at ( , ) is f ( , ) ( , ), ( , )x yx y x y f x y f x y

for short: f ,x yf f

If ( , , ), then the directional derivative is:w f x y z

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

where , , is a unit vector

x y zD f x y z f x y z a f x y z b f x y z c

a b c

u

u

0 0 0, , where , ,x y zD f x y z f f f f f u u

0 0,D f x y f u u

0 0 0 0( , ), ( , )x yf x y f x y u

Find the directional derivative of , ,

at the point 1,0, 3 in the direction of 6 3 2

z xf x y z

z y

P

a i j k

6 3 2 36 9 4 49 7 a i j k a

, ,z x

f x y zz y

2 2

1, , , ,

x z y xf x y z

z y z y z y

1,0, 3 1,0, 3D f f u

u

6 3 2, ,

7 7 7

u

, ,xf x y z

2

z x

z y

1

or , ,f x y z z x z y

2

z y z x

z y

2

y x

z y

2

x z

z y

1 4 1

1,0, 3 , ,3 9 9

f

1 4 1 6 3 2

1,0, 3 , , , ,3 9 9 7 7 7

D f f

u u 1 6 4 3 1 2

3 7 9 7 9 7

18 12 2

63

8

63

Example:

1

z y

, ,yf x y z

, ,zf x y z

0 0What is the maximum value of the directional derivative at , ?x y

0 0 0 0, ,D f x y f x y u

u 0 0, cosf x y u

0 0, cosf x y

but 1u

This will be maximized when cos 1 and 0

0 0 0 0The maximum value of , is ,D f x y f x y u

0 0This value is obtained when =0, i.e. in the direction of , .f x y

0 0 0 0

0 0

The minimum value of the directional derivative at , is ,

and occurs when has the direction as , .

x y f x y

opposite f x y

u

0 0Fastest increase is , in the direction of f

f x yf

u

0 0Fastest decrease is , in the direction of f

f x yf

u

Example:2 2

Let ( , ) be the temparature of a hot plate 1 3,1 4.

An ant is sitting on the hot plate at ( , ) (2,3).

In what direction should he run to get off the plate without getting burned?

y xT x y x y

x y

x y

xT yT

3 2 2 3

2 1 1 2,

y xT

x y x y

6 1 1 4(2,3) ,

8 9 4 27T

4882(2,3)

108T

2 1By running in the direction of or about , ,

3 10

the ant can decrease the heat at a rate of about 50%.

T

plot3d({y/x^2+x/y^2,0},x=1..3,y=1..4,axes=boxed);

23 11,

36 108

3 2

2 1y

x y

2 3

1 2x

x y

A geometric interpretation of the gradient vector: f ,x yf f

Consider the level curves ( , ) where is some constant.f x y k k

Assume that ( ) ( ), ( ) is a parametrization of this level curvet x t y tr

Then ( ) '( ), '( ) is tangent to the level curvet x t y tr'

Along this curve we have ( ( ), ( ))f x t y t k

in other words: '( ), '( )f x t y t

is orthogonal to the level curves.f

Hence by the chain rule = df

dt' ' 0 x yf x f y

'( ) 0 f t r

0 0

0 0

, is perpendicular

to the level curve

, that passes

through the point ,

and points in the direction

of fastest increase

f x y

f x y k

P x y

On a topographical map, if , represents the height above sea level

at a point with coordinates , , the path of steepest ascent is perpendicular

to all the contour lines.

f x y

x y