TN span the osculating plane. T N is to osculating plane B will suffice as a normal vector.

Lecture 4

Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-08

Math objects made using MathType; figures taken from Advanced Engineering Math 4th Edition (Zill); graphs made with Winplot.

Announcements

• Test will be July 17th for sure

• No formula sheet will be given on the test

• As much as French hates our textbook, still use it. Also, use your first year Stewart Calculus textbook.

9.3 Review

e.g. 1)

Find a_N,a_T.

( )

t = cosh

( )

t , sinh

( )

t

r

Recall:

( )

( )

( )

( )

( )

( )

cosh

2

sinh

2 d

cosh sinh d

d

sinh cosh

d

t t

t t

e e t

e e t

t t

t

t t

t

−

−

+ =

− =

=

 

 

=

 

 

Back to the question:

( )

( )

( )

( )

( )

( )

( )

( )

sinh , cosh cosh , sinh

t t t t

t t t t

′ = =

′′ = =

r v

r a

( )

( )

2 sinh t cosh t ′ ′′

⋅ = ⋅ =

v a r r

( )

( )

2 2

0, 0, sinh t cosh t 0, 0, 1

× = − = −

v a

1 ⇒ × =v a

( )

( )

( )

( )

2 2

2 sinh cosh sinh cosh T

T a

t t

a

t t

⋅ =

′ ′′⋅ =

′ =

+

v a v

r r r

( )

( )

2 2

1

sinh cosh N

a

t t

× =

′ ′′× =

′

=

+

v a v r r

r

e.g. 2)

Find the equation of the osculating plane,r

( )

t = 2 cos

( )

t , 2 sin

( )

t , 3t at 4 t=π .

We need a point and a normal vector. The point is given by:

( )

4

( )

4

( ) ( )

4 4

3 4

2 cos , 2 sin , 3 2, 2,

π π π π

π =

=

r

T, N span the osculating plane.∴ ×T Nis⊥to osculating plane

B will suffice as a normal vector.

( )

( )

( )

( )

2

( )

2

( )

2 sin , 2 cos , 3 4 sin 4 cos 9 13

t t t

t t t

′ = −

′ = + +

=

r

r

( )

( )

_{( )}

1

( )

( )

2 sin , 2 cos , 3 13

t

t t t

t ′

= = −

′

r T

r

( )

1

( )

( )

2 cos , 2 sin , 0 13

t t t

′ = − −

T

( )

4 2

3 13

t

′ = =

T

( )

t = −2 cos

( )

t , 2 sin−

( )

t , 0

N

( )

( )

( )

( )

4 4 4

1 1

4 2 2

2 cos , 2 sin , 0 , , 0

π π π

π

= − −

= − −

N

N

3 3 2

, ,

4 4 4 26 26 13

π π π −

 ₌  _×  ₌

     

     

B T N

Take an arbitrary point on the osculating plane,

(

x y z, ,

)

. 3

2, 2,

4

x y z π

⇒ − − − is in the plane 3

2, 2,

4

3 3 2 3

, , 2, 2, 0

4 26 26 13

x y z

x y z

π

π

⊥ − − −

−

⋅ − − − =

B

3

3 3 2 2

2 x− y+ z= π

9.4 – Partial Derivatives

( )

,

z= f x y

2

:

f  →

The graph of f consists of points in3like

(

x y f x y, ,

( )

,

)

and describes a surface in3.

Level Curves

Imagine cutting graph of the f by successive horizontal planes.

Pick some k in the range of f, such thatz=k.

Every point on the intersection of surface and the plane looks like

(

x y k₀, ₀,

)

, where

(

0, 0

)

f x y =k.

This is the basis of contour mapping.

Level Curves of f are the curves with equations f x y

( )

, =k, where k ϵ Range of f.

e.g. 3)

Sketch level curves of f x y

( )

, = 9−x2−y2 To find the range, find the domain.

2 2 2 2

9−x −x ≥ ⇔ ≥0 9 x +y

Thus, the domain is a disk with equationx2+y2 =9

( )

( )

2 2

2 2

0 , 3

9 , 0

x y f x y

x y f x y

+ = ⇒ =

+ = ⇒ =

Range:

[ ]

0, 3 Try k = 0

( )

2 2

2 2

, 0

9 9 f x y

x y x y

=

= − −

+ =

k = 1

( )

2 2

2 2

, 1

9 8 f x y

x y x y

=

= − −

+ =

k = 2

( )

2 2

2 2

, 2

9 5 f x y

x y x y

=

= − −

+ =

k = 3

( )

( ) ( )

2 2

2 2

, 3

9 0

, 0, 0 f x y

x y x y

x y =

= − −

+ =

⇔ =

Note: example is from Stewart’s 14.1, example 11

(

)

3

, , :

w f x y z F

= → 

Graph is a set of points in 4

, _

(

)

independent _dependent

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

 

 

 

 

Graph is 4D

(

, ,

)

,

F x y z =k k∈Range of F This set forms a surface in3.

Setting w = k gives the level surface of F.

e.g. 4)

(

)

2 2 2

, , 3 6

F x y z =x + y + z Draw some level surfaces. Any k ≥ 0 makes sense. k = 0

(

)

(

) (

)

2 2 2

, , 0

3 6

, , 0, 0, 0

F x y z

x y z

x y z

=

= + +

⇔ =

k = 1

(

)

2 2 2

, , 1 3 6

F x y z = =x + y + z ←ellipsoid k = 2

(

)

2 2 2

, , 2 3 6

F x y z = =x + y + z ←ellipsoid Partial Derivatives

Recall:z= f x y

( )

,

(

)

( )

0

, ,

lim x

f x x y f x y z

x ∆ → x

+ ∆ −

∂ =

∂ ∆

= partial derivative with respect to x

(

)

( )

0

, , ,

lim y

f x y y f x y z

y ∆ → y

∆ −

∂ ₌

∂ ∆

= partial derivative with respect to y

Find the derivative with respect to one variable and hold the others constant.

e.g. 5)

( )

3 2 3 2

,

6 5

f x y z

z x x y y

=

= − + +

2 3

2 2

3 12 0

0 18 10

z

x xy

x z

x y y

y

∂ _{= − + +}

∂

∂ _{= + +}

∂

e.g. 6)

( )

( )

2_tan1 2

,

x y

f x y z z e

−

= =

( )

_{( )}

( )

2 1 2

2 1 2

tan _{1 2}

tan ₂

4

2 tan

2 1

x y

x y

z

e x y

x

z y

e x

y y

−

−

−

∂ _{=  }

 

∂

 

∂

= _{ }

∂ _ + _

x x z f

f z x x

∂ ₌ ∂ _{= =}

∂ ∂

( )

,

f x y =z ,

f f x y ∂ ∂

∂ ∂ are also functions of 2 variables

2 _→

 

So you may differentiate f_x,f_yagain with respect to either x or y.

e.g. 7)

3 2 3 2

6 5

x x y y

− + +

2 3

2

2

2 3

3

3 12

3 12

6 12 z

x xy x

z z

x x x

x xy x

x y

∂ _{= − +} ∂

∂ ₌ ∂ ∂    ∂ ∂ ∂ 

∂  

= _− + _

∂ = − + OR

2

2 3

2

3 12

36

z x

y x y x

x xy

y xy

∂ ₌ ∂ ∂ 

 

∂ ∂ ∂ ∂ 

∂  

= _− + _

∂ =

(2nd mixed partial) Notation

 _nd st 3

2

1 2

z y x ∂ ∂ ∂

  rd _nd st 3

3 ₂ 1

z x y z

∂ ∂ ∂ ∂



st 1 2

x y z

f y x ∂

= ∂ ∂

The order is because it is intuitive to read f_xyand differentiate it in order that it appears and the order for the other notation is read right-left because you split up your derivatives from right-left, like so:

2

z z

y x x y  

∂ ∂ ∂

= _{ }

∂ ∂ ∂ ∂_{ }.

e.g. 7 continued

Continued from example 7.

3 2 3 2

6 5

x + x y + y =z

2

2

2 2

2

36

18 10

36 z

xy y x

z x y

x y y x

xy

∂ ₌

∂ ∂

  ∂ ∂

= _{ }

∂ ∂_{ }

∂  

= _ + _

∂ =

f has continuous 2nd order partials => f_xy = f_yx

For 3/+ Variables

Do everything the same, but with more independent variables.

(

, ,

)

w=F x y z 1st _partials

, , F F F

x y z

∂ ∂ ∂

∂ ∂ ∂

Chain Rule

The best way to learn this is to practice like crazy.

Version 1: 2D

Ifz= f x y

( )

, ,u=g x y

( )

, ,v=h x y

( )

, and all have continuous 1st order partials, then :

z z u z v

x u x v x

∂ ₌ ∂ ∂_{⋅ +}∂ ∂_⋅

∂ ∂ ∂ ∂ ∂

Note: x:=ymeans x is defined by y. z z u z v

y u y v y

∂ ∂ ∂ ∂ ∂

= ⋅ + ⋅

∂ ∂ ∂ ∂ ∂

(

1, 2, 3,...

)

, i

z= f u u u u uare functions ofX X₁, ₂,X₃,...,X_k

3

1 2

1 2 3

... n

j j j j n j

u u

u u

z z z z z

x u x u x u x u x

∂ ∂

∂ ∂

∂ ₌ ∂ _{⋅ +} ∂ _{⋅ +} ∂ _{⋅ + +} ∂ _⋅

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

e.g. 8)

2 2

1

2 2

tan u r s

w uv

v r s −  = −

= _

= 

Compute:

(

)

( )

1 1

2 2 2

1 1 1 1

2 2

1 2 1 2

w w u w v

s u s v s

u v s v u r s

uv uv

− −

∂ ₌∂ _⋅∂ ₊∂ _⋅∂

∂ ∂ ∂ ∂ ∂

= ⋅ ⋅ − + ⋅ ⋅

+ +

9.5 – Directional Derivatives

d

& d x x ∂

∂ are not functions, but rather, differential operators think of them as “verbs”

2D

, x y ∂ ∂ ∇ =

∂ ∂

3D , , x y z ∂ ∂ ∂ ∇ =

∂ ∂ ∂

3

:

F  →

3 3

, ,

:

F F F

F

x y z

F

∂ ∂ ∂

∇ =

∂ ∂ ∂

∇ _ →_

∇= “nabla” f

∇ = gradient of f = grad(f) e.g. 9)

(

)

( )

(

)

( )

( )

( )

2

( )

, , cos

, , , ,

cos , cos sin , sin

x y z F x y z xy yz F x y z F F F

y yz x yz xy yz z xy yz =

∇ =

= − −

The gradient is a vector quantity. Directional derivatives

Since the unit vectoruˆ = cos , sinθ θ ,z= f x y

( )

, , the directional derivative of f in direction ˆu

is:

( )

(

)

( )

0

cos , sin ,

, lim

u

h

f x h y h f x y D f x y

h

θ θ

→

+ + −

=

Note: θ = 0, definition yields fx

2

π

θ = , definition yields f_y Theorem:

( )

( )

( )

ˆ

, , cos , sin ˆ

, ,

u

z f x y u D f x y f x y

θ θ

= =

= ∇ ⋅u

ˆ

u

must be of unit length!

If you want directional derivative in direction v, where v is not unit length, normalize v before applying the theoremuˆ = v

v

Proof

( )

(

cos , sin

)

g t = f x t+ θ y t+ θ , ,

x y θfixed, g is function of 1-variable Computeg′

( )

0 by 2 methods:

Method 1: Limit Definition

( )

(

) ( )

(

)

( )

0

0

0 0

0 lim

cos , sin ,

lim h

h

u

g h g

g

h

f x h y h f x y

h D f

θ θ

→

→

+ −

′ =

+ + −

=

=

( )

1

(

)

[

]

2

(

)

[

]

d d

cos , sin cos cos , sin sin

d d

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

t t

θ θ θ θ θ θ

′ = + + ⋅ + + + + ⋅ +

1, 2

f f are partial derivatives of f x t

(

+ cos ,θ y t+ sinθ

)

with respect to _cos , _sin ,

u v

x t+ θ y+t θ respectively

[

]

[

]

( )

d d

cos sin

d d

cos sin

u v

u v

f x t f y t

t t

f f

g t

θ θ

θ θ

= + + +

= +

′ =

Set t = 0 => u=x v, = y

( )

( )

( )

0 cos sin

, cos , sin ,

,

x y

x y

u

g f f

f f f x y u D f x y

θ θ

θ θ

′ = +

= ⋅

= ∇ ⋅

=

Method 2: 3/+ dimensions

(

)

(

)

₀

(

)

(

)

, ,

cos , sin , cos , ,

, , lim u

h

w F x y z

F x h y h z h F x y z

D F x y z

h

α β γ

→

=

+ + + −

=

α,β,γ, are angles between u and coordinate axes Theorem: D F x y z_u

(

, ,

)

= ∇F x y z

(

, ,

)

⋅uˆ

ˆ

umust be unit length. If not, normalize

e.g. 10)

( )

1

(

)

, tan y @ 2, 2 f x y

x −  

= _{ } −

  and find directional derivative in direction 1, 3−

( )

( )

( )

( ) (

)

(

)

2 2 2

1 1 1

, , ,

1 1

, 2, 2

1 1 1 1 1 1

2, 2 , ,

2 2 2 2 4 4

x y

y y

x x

y f x y f f

x x

x y

f

 

∇ = = ⋅ −_{ } ⋅

 

+ +

= −

 

∇ − = _{ } ⋅ =

 

(

)

1 1

2, 2 , 1, 3

4 4 1 3 4 4

1 2 u

D f − = ⋅ −

= −

− =

Wrong because she didn’t normalize! So let’s normalize first. 1, 3 : 1, 3 1 9 10

1

ˆ 1, 3 10

− − = + =

− =u

(

)

1 1 1 3

2, 2 , ,

4 4 10 10

1 3

4 10 4 10 1

2 10 u

D f − = ⋅ −

= −

− =

What is the directional derivative for a unit vector that is tangent to a level curve? 0 because it is a level curve, so the gradient is not changing.

( )

,

f x y =z, ˆua unit vector

( )

,

( )

, ˆ

( )

, ˆ cos

u

D f x y =Df x y ⋅ = ∇u f x y u φ

ϕ is the angle between ∇f x y

( )

, and ˆuis 0≤ ≤φ π

( )

( )

( )

1 cos 1

, _u , ,

f x y D f x y f x y φ

⇒ − ≤ ≤

⇒ ∇ ≤ ≤ ∇

( )

,

( )

,

u

D f x y = − ∇f x y when ϕ = π; i.e. when ˆuand∇f x y

( )

, are opposite

( )

,

( )

,

u

D f x y = ∇f x y when ϕ = 0, i.e. when ˆuand∇f x y

( )

, point in the same direction

Max of directional derivative is equal to ∇f and occurs when ˆuis in the same direction as∇f Min of directional derivative is equal to− ∇f , and it occurs when ˆuis in the opposite direction of∇f .

Also, holds in higher dimensions e.g. 11)

Find a vector that gives the direction of maximal increase at the given point, and find the rate.

( )

2

( )

, sin @ 0,

4 x

f x y =e y _ π _

 

( )

( )

( )

( )

( )

( )

2 2

0 0

4 4 4

, ,

2 sin , cos

0, 2 sin , cos

1 2,

2 x y

x x

f x y f f

e y e y

f π e π e π

∇ =

=

∇ =

=

Move in that direction.

Maximal rate of increase =

( )

1 5

4 2 2

0, 2

f π

∇ = + =