tangent plane at Pis the plane that is normal toF evaluated at P

Lecture 5

Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-10

Math objects made using MathType; graphs made using Winplot.

Announcements

• Multiple choice test

• Chapters:

o 12.1−12.4

o 9.1−9.8

Chapter 9.6 – Tangent Planes and Normal Lines

Geometric Interpretation of the Gradient

( )

,

z= f x y differentiable, k, a number in range of f, Level curve:k = f x y

( )

,

P is a point on the curve, c,

(

x y k₀, ₀,

)

Parametrize c:

( )

t = x t

( ) ( )

,y t

r , such that x t

( )

0 =x y t0,

( )

0 = y0

Plug into f: k = f x t

(

( ) ( )

,y t

)

( ) ( )

(

)

d

, dtk= f x t y t 

( ) ( )

( ) ( )

d d

0

d d

0 , ,

0 ,

x y

f x f y x t y t f f x t y t

f x y t

∂ ∂

= +

∂ ∂

′ ′

= ⋅

′

= ∇ ⋅r

At P,t =t₀,∇f x y

(

₀, ₀

) ( )

⋅r′ t₀ =0←since they are⊥ f

∇ is orthogonal to level curve at P.

In 3-D,∇f is normal to the level surface at P.

This is useful because it lets you find the equation of the tangent plane.

Tangent Planes

LetP x y z

(

₀, ₀, ₀

)

be a point on the graph ofF x y z

(

, ,

)

=k, where∇Fis nonzero. The tangent plane at P is the plane that is normal to∇Fevaluated at P.

How to computeF x y z

(

, ,

)

=k?

We know∇ ⊥F tangent plane⇔ ∇ ⊥F to any vector in the tangent plane P has coordinates

(

x y z₀, ₀, ₀

)

, such that P lies on the level surface

(

x y z, ,

)

any point in tangent plane at P. Then, x−x y₀, −y z₀, −z₀ is in the plane

(

)

(

)

(

) (

) (

)

(

)(

)

(

)(

)

(

)(

)

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

, , , ,

, , , , 0

, , , , , , , , , , 0

, , , , , , 0

x y z

x y z

F x y z x x y y z z F x y z x x y y z z

F x y z F x y z F x y z x x y y z z F x y z x x F x y z y y F x y z z z

⇒ ∇ ⊥ − − −

⇔ ∇ ⋅ − − − =

⇔ ⋅ − − − =

⇔ − + − + − =

The equation of a tangent plane to surfaceF x y z

(

, ,

)

=kat P. 2 cases:

i)z= f x y

( )

,

CreateF x y z

(

, ,

)

= f x y

( )

, −z, then

( )

0 ,

F = ⇔ =z f x y

Look at level surfaceF x y z

(

, ,

)

=0

ii) Anything else

1. Group all x’s, y’s, and z’s on one side of the equation and a number on the other. 2. Set the side with the variables as F.

3. Set the side with the number as k and look at the level surface 4. F x y z

(

, ,

)

=k

e.g. 1)

Find the equation of the tangent plane to the surface at indicated point.

(

)

2 2 2

5x −y +4z =8, 2, 4,1

So use case 2.

(

)

2 2 2

, , 5 4 , 8

F x y z = x −y + z k =

Compute the gradient:∇ =F 10 , 2 ,8x − y z Now plug in the point:∇F

(

2, 4,1

)

= 20, 8,8− Use 5, 2, 2−

Input

(

2, 4,1

)

Compute the equation:

5, 2, 2 2, 4, 1 0

5 10 2 8 2 2 0

x y z

x y z

− ⋅ − − − =

− − + + − =

Group stuff like said here. 5x−2y+2z=4

As long as you get a scalar multiple of the final answer, you’re good. Your plane should be linear, i.e. no powers greater than one for x,y,z.

e.g. 2)

(

2 2

)

1 1

ln , , , 0

2 2

z= x +y _ _

 

This is case 1

(

)

(

2 2

)

, , ln

F x y z = x +y −z 0

F = (level surface)

2 2 2 2

2 2

, , 1

1 1

, , 0 2, 2, 1

2 2

x y

F

x y x y F

∇ = −

+ +

 

∇ _{ }= −

 

1 1

2 2

2, 2, 1 , , 0

2 1 2 1 0

2 2 2

x y z

y z

x y z

− ⋅ − − =

× − + − − =

+ − =

Finding Normal Lines

You can also find the equation of a normal line to the surface at any point, using the gradient as a direction vector.

e.g. 3)

Find parametric & symmetric equations of the normal line to the surface,x2+y2−z2 =0, at the given point,

(

3, 4, 5 .

)

(

)

2 2 2

, ,

F x y z =x +y −z

Similarly to when finding the planes, take the gradient:∇F

(

3, 4, 5

)

= 6,8, 10− This is the line to the desired vector.

Use 3, 4, 5−

( )

t = 3, 4, 5 +t 3, 4, 5− r

3 3

4 4 parametric 5 5

x t

y t

z t

= +   = +   = − 

3 4 5

3 4 5

x− ₌ y− ₌ z− _←

− parametric

9.7 – Curl & Divergence

Vector Fields

Vector fields are vector-valued functions of 2 or 3 variables.

( )

( )

( )

scalar scalar

, , ,

F x y =P x y i+Q x y j

 

(

, ,

)

(

, ,

)

(

, ,

)

(

, ,

)

F x y z =P x y z i+Q x y z j+R x y z k

e.g. 4) 9.7 example 1

Note: this is an example from the textbook, Advanced Engineering Math 4 (Zill). And example 1 from Chapter 16 from the textbook, Stewart’s Calculus Early Transcendentals 7th edition.

Graph the vector field,F x y

( )

, = − +yi xj.

( )

( )

(

)

( )

( )

0, 0

1,1 1,1 1, 1 1, 1 0,1 1, 0 1, 0 0,1 F

F F

F F

= = −

− − = −

= − =

0

( )

2 2

,

F x y = y +x ←radius of a circle

All points on a circle of a given radius result in a vector of the same length.

You may be required to look at a vector field and determine details about it.

The gradient of a scalar function results in a vector field, so you have experience with vector fields.

ϕ is a scalar function,

φ

∇ is a vector field

How does∇interact with vector fields? 2 ways:

i)

(

, ,

)

(

, ,

)

(

, ,

) (

, , ,

)

F x y z =P x y z i+Q x y z j R x y z k

curlF R Q P R Q P

y z z x x y

F

x y z

P Q R

∂ ∂  ∂ ∂  ∂ ∂ 

=_ − _ +_ − _ +_ − _

∂ ∂ ∂ ∂  ∂ ∂

   

= ∇×

∂ ∂ ∂

=

∂ ∂ ∂

i j k

i j k

ii)

(

, ,

)

F x y z =Pi+Qj+Rk

Divergence of F: divF P Q R

x y z

∂ ∂ ∂

= + +

∂ ∂ ∂

This gives a scalar function

, , , ,

F

P Q R x y z

= ∇ ⋅

∂ ∂ ∂

= ⋅

∂ ∂ ∂

Vector fields can model force fields, fluid flow, etc.

If F is a velocity field, then at point, P, divF models flux per unit volume. If divF at P > 0, then P is a source for F.

If divF at P < 0, then P is a sink for F; net inward flow.

Use divergence to model the rate of change of density of a fluid at a point. 0

F

∇ ⋅ = , incompressible/solenoidal

Curl

Fluid flow, modeled by F. Insert a paddle into the flow; measure tendency of paddle to be turned about vertical axis.

f is a scalar function, with continuous 2nd partials

e.g. 5) 9.7, exercise 29

Compute curl

( )

∇ = ∇ ∇f x

( )

f

(

)

(

)

(

)

, ,

x y z

x y z

zy yz zx xz yx xy

f f f f

x f

x y z f f f

f f f f f f

∇ =

∂ ∂ ∂

∇ ∇ =

∂ ∂ ∂

= − + − + −

=

i j k

i j k

0

e.g. 6

F is a vector field with continuous 2nd order partials. Compute div curl

(

( )

F

)

= ∇ ⋅ ∇×

(

F

)

F =Pi+Qj+Rk

, ,

R Q P R Q P

F

y z z x x y

∂ ∂ ∂ ∂ ∂ ∂

∇× = − − −

∂ ∂ ∂ ∂ ∂ ∂

(

)

2 2 2 2 2 2

, , *

0 F

x y z

R Q P R Q P

x y x z y z y x z x z y

∂ ∂ ∂

∇ ⋅ ∇× = ⋅

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

= − + − + −

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

=

9.8 – Line Integrals

( )

t = f t

( ) ( )

,g t ,a≤ ≤t b, r

A, B are starting / ending points of C, respectively i) Smooth

C is smooth if f', g' are continuous on [a,b], and not simultaneously zero on (a,b) ii) Piecewise-smooth

C is piecewise-smooth if it consists of a finite number of smooth curves,C C₁, ₂,...,C_n, joined together end-to-end

1 2 ... n

C=C ∪C ∪ ∪C iii) Closed

C is closed if A = B

iv) Simple closed curve

C is a simple closed curve if it is closed, and does not intersect itself.

Line Integrals How do you integrate?

Divide [a,b] into n equal pieces, length x b a n

−

∆ = .

[

a x, 1

] [

, x x1 2

] [

,..., xn−1,b

]

Choose a sample point,x_i*∈

[

x_i−1,x_i

]

( )

* i

f x ⋅ ∆ =x area of rectangle under the curve on

[

x_i−1,x_i

]

. Area under curve:

( )

* 1 n

i i

f x x

=

≈

∑

∆

( )

( )

*

1

d lim

n b

i

a n

i

f x x f x x

→∞ ₌

=

∑

∆

∫

Curver

( )

t = x t

( ) ( )

,y t ,a≤ ≤t

Letz= f x y

( )

, be a function whose domain includes C.

( )

( )

(

f r t = f C

)

Divide the interval [a,b] into n equal pieces;

[ ] [ ] [

a t,₁ , t t₁, ₂ ,... t_n−1,b

]

This divides C into n sub-arcs:C₁,...Cn.

Let∆s_ibe the length ith sub-arc Choose a sample point

(

* *

)

,

i i

x y fromC_i. Form product:

(

* *

)

,

i i i

f x y ⋅ ∆s

For the area of the rectangle with baseC_i:

Add up for allC_i’s:

(

* *

)

1

,

n

i i i

i

f x y s

=

∆

∑

Line integral with respect to s of f along C.

This is the area of “fence shape”:

Computation: parametrize C. Use: 2 2 d d d d d d y y s t t t     = _{ } +_{ }    

Then, change everything to t’s.

( )

(

( ) ( )

)

(

₂

( )

)

2

(

₂

( )

)

2

, d b , d

a C

f x y s= f x t y t x t + y t t

∫

∫

e.g. 7)

Evaluate

( )

, d

C

G x y s

∫

on the indicated curve

( )

( )

( )

4

, 2 ,

5 cos , 5sin , 0

G x y xy

x t y t t π = = = ≤ ≤

( )

(

( )

)

(

( )

)

(

( )

)

(

( )

)

( ) ( )

( ) ( )

4 4 4 2 2 0 0 0

, d 2 5 cos 5sin 5sin 5 cos d

50 sin cos 5d 250 sin cos d

C

G x y s t t t t t

t t t

t t t π π π = − + = ⋅ =

∫

∫

∫

∫

Other Types

(

* *

)

1 , n

i i i i

i

f x y x x

=

∆ ← ∆

∑

is the change in x from beginning ofC_ito the end ofC_i.

( )

(

)

( )

(

)

* * 1 * * 1

, d lim ,

, d lim ,

n

i i i

n i C

n

i i i

n i C

f x y x f x y x

f x y y f x y y

→∞ ₌ →∞ ₌ = ∆ = ∆

∑

∫

∑

∫

Great definition, but not useful for computation. To compute: parametrize C by r(t), a ≤ t ≤ b Use:

( )

( )

( )

(

( ) ( )

)

( )

( )

(

( ) ( )

)

( )

d d d d

, d , d

, d , d

b a C b a C

x x t t y y t t

f x y x f x t y t x t t f x y y f x t y t yx t t

′ = ′ = ′ = =

∫

∫

∫

∫

e.g. 7 Continued

Continuation from example 7.

Evaluate.

( )

( )

( )

( )

4

, d , , 2

5 cos 5sin 0

C

G x y x G x y xy

x t

y t

t π

=

= = ≤ ≤

∫

( ) ( )

(

( )

)

( ) ( )

2

50 sin cos 5sin d 250 sin cos d

b a b a

t t t t

t t t

= −

= −

∫

∫

( )

( ) ( ) ( )

( )

( )

4

4

0

2 0

, dy 250 sin cos cos d 250 sin cos d

C

G x y t t t t

t t t

π

π

= −

= −

∫

∫

∫

Line Integrals Continued

Line integrals are independent of parametrization.

( )

( )

( )

( )

( )

( )

, d , d

, d , d

, d , d

C C

C C

C C

f x y s f x y s f x y x f x y x f x y y f x y y

−

−

− =

− =

− =

∫

∫

∫

∫

∫

∫

In 3D

(

, ,

)

G x y z ,

C r

( )

t = x t

( ) ( ) ( )

,y t ,z t ,a≤ ≤t b

( )

(

* * *

)

1

, d lim , ,

n

i i i i

n i C

G x y s G x y z s

→∞ ₌

=

∑

∆

∫

Theorem: b

(

( ) ( ) ( )

, ,

)

(

( )

)

2

(

( )

)

2

(

( )

)

2d

aG x t y t z t x t′ + y t′ + z t′ t

∫

(

)

(

)

( ) ( ) ( )

(

)

( )

* * * 1

, , d lim , ,

, , d

n

i i i i

n i C

b a

G x y z z G x y z z

G x t y t z t z t t

→∞ ₌ = ∆ ′ =

∑

∫

∫

(

)

( )

( ) ( ) ( )

, , , , , , ,

F x y z

C t f t g t h t P Q R

=

+ +

r

i j k

Line integral of F over C F d

d d d

C

C C C

r

P x Q y R z

= ⋅ = + +

∫

∫

∫

∫

d d C C

F r F r

−

⋅ = −

∫

∫

e.g. 8)

( )

, , F x y = y x

Integrate over C, C is the line segments from (0,0) to (0,1), and from (0,1) to (1,1).

( )

( )

( )

( )

d , d , d

, d , d

C C C

C

F r P x y x Q x y y P x y x Q x y y

⋅ = + = +

∫

∫

∫

∫

1 2 d C C f x ∪

∫

d C F r

∫

parametrizesC₁:r

( )

t = 0, , 0t ≤ ≤t 1

( )

( )

1 1 1 1 1 1 0 0 d d d d

0 d 0 d

0

C C

C C

P x Q y y x x x

t t t

= + = + = + =

∫

∫

∫

∫

∫

∫

2 2 2

d d d

C C C

F⋅ r= y x+ x y

∫

∫

∫

ParametrizeC₂:

( ) (

t = −1 t

)

0,1 +t 1,1 , 0≤ ≤t 1 r

( )

( )

( ) (

)

0,1

1 , 0 1

1,1 P

t t P t Q t

Q

 _{= − + ≤ ≤}

 r

1 1 x t

y t t

= = − + =

( )( )

( )( )

( )

2 2 1 1 0 0 1 0

d d 1 1 d 0 d

1 d 1

C C

y x x y t t t

t + = + = =

∫

∫

∫

∫

∫

Applications

F can be a force field. C a curve, then work done by F over C is given by: d

C

F⋅ r=W

∫

Exercise 19

Evaluate

(

2 2

)

d 2 d

C

x +y x− xy y

∫



Note:

C

=

∫

lazy form of

C

∫



Aside: If

( )

, 2 2, 2 ; d

C

F x y = x +y − xy



∫

F⋅ r

1 2

C=C ∪C Parametrize C₁:

( ) (

)

( )

( )

( )

( )

( )

1 2, 0 2, 0 , 0 1 , 0 , 2 2

2 2 2 4 2

0 4 0

t t t t

t t t

x t t t t

y t x t y t = − − + ≤ ≤ = − ≤ ≤ = − + + = − = ′ = ′ = r r

(

)

(

)

1 1

2 2 2 2

0 1 2 0 1 3 2 16 3 ₀ 16 3

d 2 d 4 2 0 4d 0

4 16 16 4d

4 8 4

4 8 4 star

C

t t

x y x xy y t t

t t t

t t t =

= + − = − + − = − +   = _ − + _ = _ − + ←_

∫

∫

∫



Now, parametrizeC₂.

( )

( )

2 cos

0 , 2 is from the radius 2 sin

x t

t

y t π

= _

≤ ≤ 

= _

( )

( )

( )

( )

2 sin 2 cos

x t t

y t t

′ = −

′ =

(

)

(

( )

)

( ) ( )

( )

( )

( )

( )

2

2 2

0 0

2

0 0

d 2 d 4 2 sin d 2 4 sin cos 2 cos d 8sin d 16 sin cos d

#

C

x y x xy y t t t t t t

t t t t t

π π

π π

+ − = − − ⋅

= − −

=

∫

∫

∫

∫

∫



# + star = Answer