Review

0 0 0 00 0

0

( , ) ( , ), lim

s

f x sa y sb f x yD f x y

s

u

0 0

0 0

The directional derivative of , in the direction of at ,

is denoted by , : , must a unit vector

f x y x y

D f x y a b u

u

u

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

0 0, where ,x yD f x y f f f f u 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

0 0 0 0( ,y ) is orthogonal to the level (contour) curve ( , ) ( ,y )f x f x y f x

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

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

14.5

Tangent Planes and Differentials

0 0 0

0 0 0

The equation of a plane containing , , with normal vector , ,

is 0

x y z a b c

a x x b y y c z z

0 0 0c z z a x x b y y

0 0 0

a bz z x x y y

c c

0 0 0z z A x x B y y

When we move from a general plane to a tangent plane

to a surface , , and take on very special values.z f x y A B

We can manipulate this algebraically:

Recall:

0 0 0z z A x x B y y

0If we slice the surface with the plane ,y y

1 the trace of the surface on the plane would be C

1 1 the tangent line to is C T

1 is in the tangent planeT

0 0 0z z A x x B y y

0since y y

1 0 is also on the plane T y y

1So the equation for is:T

0 0z z A x x

1The slope of is T A

1The slope of is also the rate of change of the

function in the direction parallel to the axis

T

x

0 0This is , !xf x y 0 0,xA f x y

0

Similarly, if we slice the surface

with the plane :x x

0 0Hence ,yB f x y

0 0 0 0 0 0 0( , ) ( , )x yz z f x y x x f x y y y

0 0 0

This is the equation of the to the

surface , at the point , ,z f x y P x y z

tangent plane

0 0 0z z A x x B y y

0 0z z B y y

2the trace of the surface on the plane is C

2 2the tangent line to is C T

2and the slope of is T B

2 2Find the equation of the tangent plane to 3 2 at 1, 2,1 .z x y x

6 2 xf x

2 2

0 0 0, 3 2 and , , 1, 2,1f x y x y x x y z

2 yf y

1 8 1 4 2z x y

0 8 8 4 8 1x y z

8 4 1 0x y z

Example: 0 0 0 0 0 0 0( , ) ( , )x yz z f x y x x f x y y y

1, 2 8xf

1, 2 4yf

4Find the equation of the tangent plane to at 3,0,2 .yz x e

4, yf x y x e

xf 4

1

2 yx e 3,0xf

4 0

1

2 3 e

1

2 3 1

1

2 2

1

4

yf 4

4

4

2

y

y

e

x e 3,0yf

0

0

4

43

2e

e

2

3 1

1

4

4

2 y

y

e

x e

1

14

2 3 0z x y

0 0 0 0 0 0 0, ,x yz z f x y x x f x y y y

4 2 3 4z x y 4 8 3 4z x y

4 4 5 0x y z

Example:

Let be a surface with equation , ,S F x y z k

is a level surface of a function of three variablesS F

0 0 0Let , , be a point on .P x y z S

Let be any curve that lies on the surface

and passes through the point .

C S

P

Let defined by , ,C t x t y t z tr

0Let be the parameter value corresponding to .t P

0 0 0 0, ,t x y zr

Any point , , on is also on .x t y t z t C S

, ,F x t y t z t k

If , , and are differentiable functions of and is also differentiable, then x y z t F

0F dx F dy F dz

x dt y dt z dt

, , , , 0F F F dx dy dz

x y z dt dt dt

Another way to write this is:

0F t r

F t r

0The tangent vector lies in the tangent

plane to the surface at the point

t

S P

r

0 0 0 00 , , is the normal

vector to the tangent plane to at

F t F x y z

S P

r

Tangent planes of level surfaces

0 0 0 0a x x b y y c z z

0 0 0The equation of the plane with normal , , containing the point , , :a b c x y z

0 0 0, ,xa F x y z 0 0 0, ,yb F x y z 0 0 0, ,zc F x y z

0 0 0 0 0 0 0 0 00 0 0, , , , , , 0x y zF x y z F x y z F x y zx x y y z z

0 0 0When the plane is the tangent plane to the surface , , at the point , , :F x y z k x y z

32 4xF xy z 2

yF x z 212zF xz y

1, 2,1xF 4 4 1, 2,1yF 1 1 1, 2,1zF 12 2

1 20 12 10 0x y z

2 4 10 10 0y z 2 10 6 0y z 5 3 0y z

Example:2 3a) Compute the tangent plane to the surface 4 0 at the point (1,2, 1)x y xz yz

Recall:

Special case: ( , )z f x y or ( , , ) ( , ) 0F x y z f x y z , , 1x yF f f

0 0 0tangent plane ( ) ( ) ( ) 0x yf x x f y y z z 0 0 0or ( ) ( ) ( ) x yz z f x x f y y

As before!

b) What is the straight line normal to the surface at (1,2, 1)? ( ) 1,2, 1 0,2,10t t r

The elliptic paraboloid

appears to coincide with

its tangent plane as

we zoom in toward (1, 1, 3).

The tangent plane

and the elliptic paraboloid

become virtually

Indistinguishable the closer

we get to (1, 1, 3).

“locally”

the surface looks linear

The tangent plane is the

linearization of the

function at the point of

tangency.

The tangent plane approximates the surface: 2 22z x y

,

The approximation

, , , ,x y

L x y

f x y f a b f a b x a f a b y b

The of at , is

, , , ,x y

f a b

L x y f a b f a b x a f a b y b

linearization

This can be used to approximate the function at "nearby" points.

0 0 0 0 0 0 0Recall, the tangent plane is ( , ) ( , )x yz z f x y x x f x y y y

is called the of at ,f a blinear approximation

or the of at , .f a btangent plane approximation

0 0 0and ( , )z f x y

2 2Find the linear approximation of the function , 20 7 at 2,1

and use it to approximate 1.95,1.08 .

f x y x y

f

2 2

1

2 20 72 x

x yf x

2 2, 20 7 , 2,1f x y x y a b

2 2

1

2 20 714 y

x yf y

723 3

, 3 2 1f x y x y

2,1 20 4 7 9 3f

723 3

1.95,1.08 3 1.95 2 1.08 1f

723 3

1.95,1.08 3 0.05 0.08f 72 1 23 20 3 25

3

1 1430 75

3

450 5 28150

4271.95,1.08

150f

Example: , , , ,x yL x y f a b f a b x a f a b y b

723 3

Linear approximation: ( , ) 3 2 1L x y x y

If ( , ) is close to (2,1)x y

A calculator shows the

difference is 0.012

2

2,13

xf

7

2,13

yf

Recall the example:2 2

( , ) , (0,0) 0xy

f x y fx y

0,0 0xf

is not continuous at (0,0).f

0,0 0yf

0 0 0 0 0 0 0

Tangent plane would be:

( , ) ( , ) 0x yz z f x y x x f x y y y

0 or z z

Clearly a bad approximation near (0,0)

Recall our definition:

, is differentiable at ( , ) if

z ( , ) ( , ) for small , x y

z f x y a b

f a b x f a b y x y

So this says that is differentiable at ( , ) if the tangent plane approximates

the graph of near ( , ) very well!

f a b

f a b

The "tangent plane" would be the - planex y

(Recall that for ( ) the tangent line approximates

near if '( ) exists!)

y f x f

a f a

Recall the concept of a differential for a function of one variable

|0 0

'( ) lim limx a

h h

f a h f ady yf a

dx h x

hence '( )y f a x

and '( )

Differentials : dx x

dy f a dx

Recall tangent line is

( ) '( )( )y f a f a x a

is the change

along the tangent line

dy

hence along the tangent line

the change in is

( ) '( )( )

y

y f a f a x a

'( ) '( )f a x f a dx

The linear approximation formula

, , , ,x yf x y f a b f a b x a f a b y b

x y

the of xincrement the of yincrement

dx

the of xdifferential

dy

the of ydifferential

x dx

for independent variables, the increment is the same as the differential

y dy

What about the dependent variable ?z

, ,z f x y f a b ?dz

The or the is

, ,x y

dz

dz f x y dx f x y dy

differential total differential

The linear approximation formula

, , , ,x yf x y f a b f a b x a f a b y b

dz

, , , ,x yf x y f a b f a b x a f a b y b

z dz but is much easier to calculatedz

dz zGeometric interpretation of the differential and the incerement

goes along the tangent plane from ( , ) to ( , )dz a b a x b y

1

2

a A xy

x

y

1

2xA y

1

2yA x

5

12

Watch out for units!

x ydA A dx A dy the maximum error

in the measurement

x dx

x

0.002 m. x

0.002dx

similarly 0.002dy 1 1

5 0.002 12 0.0022 2

dA

20.005 0.012 0.017dA m

Relative errorerror

actual

the maximum error in the calculated value of the area

z

2 2 b z x y 2 2 2 2

x ydz dx dy

x y x y

12 50.002 0.002

13 13

.0026 m

Relative error

Problem:

1 1

2 2ydx xdy

0.017

30 .0005 0.05%

0.0026.0002 0.02%

13