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Kreyszig by YHLee;100510; 9-1 Chapter 9 Vector Differential Calculus, Grad, Div, Curl
9.1 Vectors in 2‐Space and 3‐Space Two kinds of quantities used in physics, engineering and so on. A scalar : A quantity representing magnitude. A vector : A quantity representing magnitude and direction. A vector is represented by an arrow. The tail is called initial point. The head (or the tip) is called terminal point The distance between the initial and the terminal points is called the distance, magnitude, or norm. A velocity is a vector, v , and its norm is v .
A vector of length 1 is called a unit vector. Definition Equality of Vectors
Two vectors and a b are equal, a b= , if they have the same length and the same direction. A translation does not change a vector.
Components of a Vector A vector a is given with initial point ( )1 1 1: , ,P x y z and terminal point ( )2 2 2: , ,Q x y z .
The vector a has three components along x, y, and z coordinates. [ ]1 2 3, , a a a a=
where 1 2 1a x x= − , 2 2 1a y y= − , 3 2 1a z z= −
The length a is given by
2 2 21 2 3a a a a= + +
Initial point
Terminal point
Kreyszig by YHLee;100510; 9-2 Example 1 Components and Length of a Vector A vector a with initial point ( ): 4, 0, 2P and terminal point ( ): 6, 1, 2Q − .
Its components are 1 2 36 4 2, 1 0 1, 2 2 0a a a= − ⇒ = − − ⇒ − = − ⇒
Hence [ ]2, 1, 0a = −
Its length is ( )22 22 1 0 5a = + − + ⇒
• Position vector A point A:(x, y, z) is given in Cartesian coordinate system. The position vector of the point A is a vector drawn from the origin to the point A. [ ], ,r x y z=
Theorem 1 Vectors as Ordered Real Triple Numbers
A vector in a Cartesian coordinate system can be represented by three real numbers ( )1 2 3, , a a a ,
which corresponds to three components.
Hence a b= means that 1 1 2 2 3 3, and a b a b a b= = = .
Vector Addition, Scalar Multiplication Definition
Addition of two vectors [ ]1 2 3, ,a a a a= and [ ]1 2 3, ,b b b b=
[ ]1 1 2 2 3 3, , a b a b a b a b+ = + + +
Parallelogram rule Head‐To‐Tail rule
Kreyszig by YHLee;100510; 9-3 Basic Properties of Vector Addition
a b b a+ = + : commutative
( ) ( )a b c a b c+ + = + + : associative
0 0a a a+ = + = ( ) 0a a+ − =
A vector a− has the same length a as a but with the opposite direction.
Definition
In scalar multiplication, a vector a is multiplied by a scalar c.
[ ]1 2 3, , ca ca ca ca=
ca has the same direction with a with increased length for c >0,
“ opposite direction “ “ for c <0 Basic Properties of Scalar Multiplication
( )c a b ca cb+ = +
( )c k a ca ka+ = +
( ) ( )c ka ck a cka= =
1a a= , 0 0a = , ( )1 a a− = − → ( )b a b a+ − = −
Unit Vectors ˆˆ ˆ, , i j k A vector can be represented as
[ ]1 2 3 1 2 3ˆˆ ˆ, , a a a a a i a j a k= = + +
using three unit vectors ˆˆ ˆ, , i j k along x, y, z axes
[ ]ˆ 1,0,0i = , [ ]ˆ 0,1,0j = , [ ]ˆ 0,0,1k =
Kreyszig by YHLee;100510; 9-4 Example 2 Vector Addition. Multiplication by Scalars Let
[ ]4, 0,1a = and 132, 5,b = −⎡ ⎤⎣ ⎦
Then [ ]4, 0, 1a− = − − ,
[ ]7 28, 0, 7a =
436, 5,a b+ = −⎡ ⎤⎣ ⎦
( ) 232 2 2, 5, 2 2a b a b− = = −⎡ ⎤⎣ ⎦
Example 3 ˆˆ ˆ, , i j k Notation for Vectors The two vectors in Example 2 are
ˆˆ4a i k= +
13ˆˆ ˆ2 5b i j k= − +
9.2 Inner Product (Dot Product) Definition
Inner Product (Dot Product) of Vectors is defined as
1 1 2 2 3 3cosa b a b a b a b a b• = γ = + +
0 ≤ γ ≤ π , the angle between and a b
Orthogonality
a is orthogonal to b if 0a b• = .
γ should be / 2π when 0 and 0a b≠ ≠
Theorem 1 Orthogonality
0a b• = if and only if and a b are perpendicular to each other Length and Angle
a a a= •
Then
cosa b a b
a b a a b b
• •γ = =
• •
Kreyszig by YHLee;100510; 9-5
Basic Properties of Inner Product
( )1 2 1 2q a q b c q a c q b c+ • = • + • : linearity
a b b a• = • : symmetry 0a a• ≥ : positive definiteness 0 if and only if 0a a a• = = : positive definiteness
( )a b c a c b c+ • = • + • : distributive
a b a b• ≤ : Cauchy‐Schwarz inequality
a b a b+ ≤ + : triangle inequality
( )2 2 222a b a b a b+ + − = + : parallelogram equality
Example 1 Inner Product. Angle between Vectors
Two vectors [ ]1, 2, 0a = and [ ]3, 2, 1b = − are given
1 3 2 ( 2) 0 1 1a b• = ⋅ + ⋅ − + ⋅ ⇒ −
2 2 21 2 0 5a a a= • ⇒ + + ⇒
( )22 23 2 1 14b b b= • ⇒ + − + ⇒
1 1arccos cos 1.69061 96.865
5 14oa b
a b−• −
γ = = = =
Example 2 Work Done by a Force A constant force p is exerted on a body.
But the body is displaced along a vector d . Then the work done by the force in the displacement of the body is
( )cosW p d p d= α ⇒ •
↑ Inner product is used nicely here. Example 3 Component of a Force in a Given Direction What force in the rope will hold the car on a 25o ramp. The weight of the car is 5000 lb. Since the weight points downward, it can be represented by a vector as [ ]0, 5000, 0a = −
a can be given by a sum of two vectors a c p= + ↑ ↑ Force exerted to the rope by the car Force exerted to the ramp by the car
From the figure, ( )cos 65 2113 op a lb= ⇒
• A vector in the direction of the rope 1, tan25 , 0ob ⎡ ⎤= −⎣ ⎦
The force on the rope 2113 b
p a lbb
⎛ ⎞⎜ ⎟= • − ⇒⎜ ⎟⎝ ⎠
Kreyszig by YHLee;100510; 9-6
• Projection (or component) of a in the direction of b
cosa b
p ab
•= γ =
p is the length of the orthogonal projection of a onto b .
• Orthonormal Basis The orthogonal unit vectors in Cartesian coordinates system form an orthonormal basis for 3‐space. ↑ ( )ˆˆ ˆ, , i j k
An arbitrary vector is given by a linear combination of the orthonormal basis. The coefficients of a vector can be determined by the orthonormality.
1 2 3ˆˆ ˆv l i l j l k= + + → 1 2 3
ˆˆ ˆ, , l v i l v j l v k= • = • = •
↑ 1 2 3
ˆˆ ˆ ˆ ˆ ˆl i i l j i l k i= • + • + •
Example 5 Orthogonal Straight Lines in the Plane Find the straight line L1 passing through the point P: (1, 3) in the xy‐plane and perpendicular to the straight line L2: x‐2y+2=0 The equation of the straight line 2 1 2: bL x b y k+ =
↑ b r k• = , in vector form.
[ ] [ ]= =1, ‐2, 0 , , y, 0b r x
From the next Example 6, b is normal to L2.
A normal vector to [ ]= −1, 2, 0b is [ ]= 2, 1, 0a
A line normal to a is 1 : 2L x y c+ =
L1 passes through P: (1, 3)
→ 2+3=c → 1 : 2 5L x y+ =
Kreyszig by YHLee;100510; 9-7 Example 6 Normal Vector to a Plane Find a unit vector normal to the plane 4 2 4 7x y z+ + = − Express the plane in vector form : a r c• =
The unit vector : ˆ an
a=
Rewriting
a r c• = → • = ≡ˆ cn r p
a : /p c a= , a constant.
↑ Projection of r onto n̂ A position vector r is from the origin to a point on the plane. The same projection p for any r → n̂ should be a surface normal.
Since [ ]4, 2, 4a = is given, the surface normal is obtained as 1ˆ6
an a
a= ⇒
9.3 Vector Product (Cross Product, Outer Product) Definition
The vector product of and a b is defined as
v a b= × : another vector Its magnitude
sinv a b a b= × = γ : , angle between and a bγ
Its direction is perpendicular to both and a b conforming to right‐handed triple( or screw).
Note that a b× represents the area of the parallelogram
formed by and a b .
Kreyszig by YHLee;100510; 9-8 • In components
[ ]2 3 3 2 3 1 1 3 1 2 2 1, , a b a b a b a b a b a b a b× = − − −
v a b= × can be calculated as follows
2 3 1 3 1 21 2 3
2 3 1 3 1 21 2 3
ˆˆ ˆ
ˆˆ ˆi j k
a a a a a aa b a a a i j k
b b b b b bb b b
× = ⇒ − +
Example 1 Vector Product
[ ]1, 1, 0a = , [ ]3, 0, 0b =
The vector product
Example 2 Vector Products of the Standard Basis Vectors
Theorem 1 General Properties of Vector Products
( ) ( )ka b k a b a kb× = × = × : for every scalar k
( ) ( ) ( )a b c a b a c× + = × + × : distributive
( ) ( ) ( )a b c a c b c+ × = × + × : distributive
( )b a a b× = − × : anticommutative
( ) ( )a b c a b c× × ≠ × × : not associative
Kreyszig by YHLee;100510; 9-9 Example 3 Moment of Force
A force p is exerted on a point A. Point Q and point A is connected by a vector r . The moment m about a point Q is defined as m p d= ,
where d is the perpendicular distance from Q to L. → sinm p r= γ
In vector form m r p= × : Moment vector Example 5 Velocity of a Rotating Body
A vector w can describe a rotation of a rigid body. → Its direction the rotation axis (right‐hand rule) Its magnitude = angular speed ω (radian/sec) The linear speed at a point P
→ sinv d w r w r= ω ⇒ γ⇒ ×
In vector form v w r= ×
Kreyszig by YHLee;100510; 9-10 Scalar Triple Product The scalar triple product is defined as
( ) ( ) a b c a b c= • ×
It can be calculated as
( )1 2 3
1 2 3
1 2 3
a a a
a b c b b b
c c c
• × =
Theorem 2 Properties and Applications of Scalar Triple Products
(a) The dot and cross can be interchanged
( ) ( )a b c a b c• × = × •
(b) Its absolute value is the volume of the parallelepiped formed by , and a b c . (c) Any three vectors are linearly independent if and only if their scalar tipple product is nonzero.
Proof:
(a) It can be proved by direct calculations
(b) ( ) ( )cos cosa b c a b c a b c• × = × β ⇒ β ×
↑ ↑ area of the base height of the parallelepiped (c) If three vectors are in the same plane or on the same straight line,
either the dot or cross product becomes zero in ( )a b c• × .
( ) 0a b c• × ≠ → Three vectors NOT in the same plane or on the same straight line.
They are linearly independent. Example 6 Tetrahedron A tetrahedron is formed by three edge vectors,
[ ]2, 0, 3a = , [ ]0, 4, 1b = , [ ]5, 6, 0c =
Find its volume. First, find the volume of the parallelepiped using scalar triple product.
The volume of tetrahedron is 1/6 of that of parallelepiped. Therefore, the answer is 12.
Kreyszig by YHLee;100510; 9-11 9.4 Vector and Scalar Functions and Fields. A vector function gives a vector value for a point p in space
( ) ( ) ( ) ( )1 2 3, , v v p v p v p v p= = ⎡ ⎤⎣ ⎦ ( ) ( ) ( ) ( )In Cartesian Coord.1 2 3 , , , , , , , , , , v x y z v x y z v x y z v x y z⎯⎯⎯⎯⎯⎯→ = ⎡ ⎤⎣ ⎦
A scalar function gives scalar values : ( )f f p=
A vector function defines a vector field. A scalar function defines a scalar field. In Engineering Meaning of field = Meaning of function. The field implies spatial distribution of a quantity.
Example 1 Scalar function The distance from a fixed point op to any point p is a scalar function, ( )f p .
( )f p defines a scalar field in space → It means that the scalar values are distributed in space.
( ) ( ) ( ) ( ) ( )2 2 2, , o o of p f x y x x x y y z z= = − + − + −
In a different coordinate system → p and op have different forms, but ( )f p has the same value.
→ ( )f p is a scalar function.
Direction cosines of the line from op to p depend on the choice of coordinate system.
→ Not a scalar function.
Kreyszig by YHLee;100510; 9-12 Example 3 Vector Field (Gravitation field) Newton's law of gravitation
2
cF
r= : ( ) ( ) ( )2 2 2
o o or x x y y z z= − + − + −
The direction of F is from p to po. ( ), ,F x y z defines a vector field in space.
• In vector form Define the position vector,
( ) ( ) ( ) ˆˆ ˆo o or x x i y y j z z k≡ − + − + − : Its direction is from po to p.
Then
3
cF r
r= −
Vector Calculus Convergence An infinite sequence of vectors (1) (2) (3), , ,...a a a converges to a if
( )lim 0nna a
→∞− = .
→ ( )lim nna a
→∞= : a , limit vector
Similarly, a vector function ( )v t has the limit l at ot if
( )lim 0ot tv t l
→− = .
→ ( )limot tv t l
→=
Continuity ( )v t is continuous at ot t= if
( ) ( )limo
ot tv t v t
→=
( )v t is continuous at ot t= if and only if its three components are continuous at ot
( ) ( ) ( ) ( )1 2 3ˆˆ ˆv t v t i v t j v t k= + +
Kreyszig by YHLee;100510; 9-13 Definition Derivatives of a Vector Function
( )v t is differentiable at t if the limit exists
( ) ( ) ( )0
' limt
v t t v tv t
tΔ →
+ Δ −=
Δ
↑ Called derivative of ( )v t
( )v t is differentiable at t if and only if its three components are differentiable at t.
→ ( ) ( ) ( ) ( )1 2 3' ' , ' , 'v t v t v t v t= ⎡ ⎤⎣ ⎦
• Differentiation rules ( )' 'cv cv= : c, constant
( )'u v u v′ ′+ = +
( )'u v u v u v′ ′• = • + •
( )'u v u v u v′ ′× = × + ×
( ) ( ) ( ) ( ) w ' ' w ' w w'u v u v u v u v= + +
Example 4 Derivative of a Vector Function of Constant Length
Let ( )v t be a vector function with a constant length, ( )v t c= .
( ) ( ) ( )2 2v t v t v t c= • =
The total derivative ( )' 2 ' 0v v v v• = • =
↑ ' 0 or 'v v v= ⊥ Partial Derivatives of a Vector Function Let the components with two or more variables be differentiable
( ) ( ) ( )1 1 2 2 1 2 3 1 2ˆˆ ˆ, , ... , , ... , , ... n n nv v t t t i v t t t j v t t t k= + +
The partial derivative of v with respect to mt
31 2 ˆˆ ˆm m m m
vv vvi j k
t t t t
∂∂ ∂∂= + +
∂ ∂ ∂ ∂
The second partial derivative
22 22
31 2 ˆˆ ˆl m l m l m l m
vv vvi j k
t t t t t t t t
∂∂ ∂∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
Example 5 Partial derivatives
Kreyszig by YHLee;100510; 9-14 9.5 Curves. Arc length. Curvature. Torsion
A major application of vector calculus concerns curves and surfaces. A curve C can be represented by a vector function with a parameter t.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ˆˆ ˆ, , r t x t y t z t x t i y t j z t k= = + +⎡ ⎤⎣ ⎦ : parametric representation
The direction of the curve is determined by increasing values of t • Another representation of a curve ( ) ( ), , x y f x z g x= =
↑ ↑ projection of C onto xy plane projection of C onto xz plane Another representation of C can be given by an intersection of two surfaces ( ) ( ), , 0, , , 0F x y z G x y z= =
Example 1 Circle The circle 2 2 4x y+ = , 0z = is given In parametric representation ( ) [ ] [ ]( ), ( ), ( ) 2cos , 2sin , 0r t x t y t z t t t= ⇒ : 0 2t≤ ≤ π
Check
( ) ( )2 22 2 2cos 2sin 4x y t t+ ⇒ + =
For t=0 : ( ) [ ]0 2, 0, 0r =
For / 2t = π : ( ) [ ]/ 2 0, 2, 0r π =
… As t increases, ( )r t moves counterclockwise.
Kreyszig by YHLee;100510; 9-15 Example 2 Ellipse An ellipse in the form of vector function ( ) [ ] [ ]( ), ( ), ( ) cos , sin , 0r t x t y t z t a t b t= =
→ cos , sinx a t y b t= =
→ ( ) ( )2 2
2 2
2 2cos sin 1
x yt t
a b+ ⇒ + = : Conventional form of an ellipse
Example 3 straight line
A straight line in the direction of a unit vector b . It passes through a point A. a is a position vector. A point on the line is given by a position vector
( ) [ ]1 1 2 2 3 3, , r t a t b a tb a tb a tb= + ⇒ + + +
• Curves Plane curve : a curve in a plane. Twisted curve : the others. Simple curve : a curve without multiple points at which the curve intersects or touches itself. Arc of a curve : a portion between two points in the curve, simply called 'curve'. Example 3 circular helix A circular helix in parametric representation ( ) [ ]= cos , sin , r t a t a t ct
c>0 : right handed screw c<0 : left handed screw c=0 : circle
Kreyszig by YHLee;100510; 9-16 Tangent to a Curve A tangent is a straight line touching a curve. A vector passing through both P and Q is
( ) ( )1r t t r t
t+ Δ −⎡ ⎤⎣ ⎦Δ
Take the limit
( ) ( ) ( )0
1' lim
tr t r t t r t
tΔ →= + Δ −⎡ ⎤⎣ ⎦Δ
↑ Tangent vector of C at P The unit tangent vector
'
ˆ'
ru
r= : Direction of increasing t
• The function for the tangent of C at P ( ) 'q w r wr= + : and 'r r , constant vectors. w, parameter
Example 5 Tangent to an Ellipse
Find the tangent to the ellipse 2 211
4x y+ = at ( ): 2, 1 / 2P
An ellipse, 2 2
2 21
x ya b
+ = → a=2, b=1
The parametric presentation → ( ) [ ]2cos , sin , 0r t t t=
At ( ): 2, 1 / 2P → / 4t = π
The derivative → ( ) [ ]' 2sin , cos , 0r t t t= −
At P → ( )' / 4 2, 1/ 2, 0r ⎡ ⎤π = −⎣ ⎦
The tangent is
( ) ( ) ( )( )2, 1 / 2 2, 1 / 2 2 1 , 1 / 2 1q w w w w⎡ ⎤⎡ ⎤ ⎡ ⎤= + − ⇒ − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Kreyszig by YHLee;100510; 9-17 Length of a Curve The length of the curve
0
1
lim ( ) ( ( 1) )N
tn
l r a n t r a n tΔ →
=
= + Δ − + − Δ∑
0
1
( ) ( ( 1) )lim
N
tn
r a n t r a n tt
tΔ →=
+ Δ − + − Δ⇒ Δ
Δ∑
( )b
a
r t dt′⇒ ∫
Hence
( )b b
a a
l r t dt r r dt′ ′ ′= = •∫ ∫
Arc Length s of a Curve The upper limit is now a variable
( )t
a
s t r r dt′ ′≡ •∫ : Arc length of C. dr
rdt
′ = (11)
Linear element ds Differentiate (11)
2 2 2 2
ds dr dr dx dy dzdt dt dt dt dt dt
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= • ⇒ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
, : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k= + +
→ 2 2 2 2( ) ( ) ( ) ( )ds dx dy dz dr dr= + + ⇒ • : ˆˆ ̂ dr dx i dy j dz k= + + ds is called the linear element of C. Arc Length as Parameter The parameter t can be replaced by the arc length s in the equation of a curve. For example, the unit tangent vector is given as ( ) ( )ˆ 'u s r s=
where
( ) ( ) ( )0
tangent vector ' '' lim
magnitude of the tangent vectors
r s s r sr s
sΔ →
←+ Δ −=
←Δ
Note that
( )' 1ds
r sds
= =
Kreyszig by YHLee;100510; 9-18 Example 6 Circular Helix. Circle. Arc Length as Parameter.
A helix ( ) [ cos , sin , ]r t a t a t ct= → ( ) [ sin , cos , ]r t a t a t c′ = −
2 2r r a c′ ′• = +
The arc length is 2 2 2 2
0
t
s a c dt t a c= + = +∫
Replace t by s in ( )r t
2 2 2 2 2 2 2 2
ˆˆ ˆcos sins s s s
r a i a j c ka c a b a c a c
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
↑ ≡ r s* ( ) , a new function • For a circle, let c=0 then /t s a=
→ 1ˆ ˆ( ) cos sin
s s sr r s a i a j
a a a⎛ ⎞ ⎛ ⎞ ⎛ ⎞≡ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Curves in Mechanics. Velocity. Acceleration A curve C may represent a path of a moving body. ( )r t : t is time in this case ( ) ( )'v t r t= : velocity vector
The magnitude
dr ds ds
v rds dt dt
′= ⇒ ⇒
↑ Speed of the body along the curve, v is parallel to the tangent of C. ( v is the velocity of the moving body along C ) ( ) ( )' ( )"a t v t r t= = : acceleration vector
Tangential and Normal Acceleration In general tan norma a a= + : tangential and normal acceleration vectors
Proof:
( )
( )dr t dr ds
v tdt ds dt
= ⇒ : s, arc length
↑ = ˆ( )u s , unit tangent vector on C Differentiate the both sides
2 2
2
ˆˆ ˆ( ) ( ) ( )
d ds du ds d sa t u s u s
dt dt ds dt dt⎡ ⎤ ⎛ ⎞= ⇒ +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
: ˆd ( )
ˆ( ), tangential vector. , normal vectordu s
u ss
↑ ↑ Normal acceleration. Tangential acceleration.
Kreyszig by YHLee;100510; 9-19 Example 7 Centripetal acceleration. Centrifugal Force. The revolving path of a small body
ˆ ˆ( ) cos( ) sin( ) r t R t i R t j= ω + ω : ω, angular speed
The velocity vector : ˆ ˆ( ) sin( ) cos( ) v r t R t i R t j′= ⇒ − ω ω + ω ω
The speed of the body : v R= ω
Linear speed
Angular speed = Distance to the center
: ω
The acceleration vector : 2 2 2ˆ ˆcos( ) sin( ) a v R t i R t j r′= ⇒ − ω ω − ω ω ⇒ −ω ↑ Direction toward the center → 2 2a r R= ω ⇒ω : centripetal acceleration
If the body is to be in the circular orbit during rotation, someone should supply the centripetal force to the body. Otherwise, it will deviate from the orbit due to the centrifugal force. Example 8 Superposition of Rotations
A projectile moves along a meridian of the rotating earth. Find its acceleration. The angular speed of the earth rotation is ω.
The unit vector b also rotates,
( )ˆ ˆ ˆcos( ) sin( )b t t i t j= ω + ω
The projectile on the meridian with an angular speed of γ . Its position vector,
ˆ ˆ( ) cos( ) sin( )r t R t b R t k= γ + γ : R, radius of the earth
The derivatives
ˆ ˆ ˆ( ) cos( ) sin( ) cos( )v r t R t b R t b R t k′ ′= = γ − γ γ + γ γ
2 2ˆ ˆ ˆ ˆcos( ) 2 sin( ) cos( ) sin( )a v R t b R t b R t b R t k′ ′′ ′= = γ − γ γ − γ γ − γ γ
Inserting b̂′ and b̂′′ 2 2ˆ ˆcos( ) 2 sin( )a R t b R t b r′= −ω γ − γ γ − γ
↑ ↑ ↑ Centripetal acceleration by the earth. ↑ Centripetal acceleration by the rotation of P on the meridian. Coriolis acceleration due to interaction of two rotations. Coriolis acceleration is in the direction opposite to the earth’s rotation → The projectile deviates from the meridian toward the earth’s rotation direction in the Northern Hemisphere. (??)
If the projectile is to be in the meridian during rotation, someone should supply the Coriolis acceleration to the projectile. Otherwise, it will deviate from the meridian in the direction opposite to the Coriolis acceleration. Curvature and Torsion (skip)
Kreyszig by YHLee;100510; 9-20 9.6 Calculus Review Chain Rule
Let all the functions be continuous and have continuous first partial derivatives in their domains. Let every point (u, v) in B has the corresponding point [ x(u,v), y(u,v), z(u,v) ] in D. The function defined in B, [ ]( , ), ( , ), ( , )w f x u v y u v z u v=
has first partial derivatives
w w x w y w zu x u y u z u
∂ ∂ ∂ ∂ ∂ ∂ ∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂,
w w x w y w zv x v y v z v
∂ ∂ ∂ ∂ ∂ ∂ ∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂.
Simple examples : If ( ), , w f x y z= and ( ) ( ) ( ), , x x t y y t z z t= = = ,
then
w w x w y w zt x t y t z t
∂ ∂ ∂ ∂ ∂ ∂ ∂= + +
∂ ∂ ∂ ∂ ∂ ∂ ∂.
If ( )w f x= and ( )x x t= ,
then
w w xt x t
∂ ∂ ∂=
∂ ∂ ∂.
Mean Value Theorem
Let ( , , )f x y z be continuous and have continuous first partial derivative in a domain D. Let two points, 0 0 0: ( , , )oP x y z and 0 0 0: ( , , )P x h y k z l+ + + , be in D
and the line segment between the points be in D. Then,
0 0 0 0 0 0( , , ) ( , , )f f f
f x h y k z l f x y z h k lx y z∂ ∂ ∂
+ + + − = + +∂ ∂ ∂
The partial derivatives are evaluated at a point on the line segment
A simple example: For a function of two variables
0 0 0 0( , ) ( , )f f
f x h y k f x y h kx y∂ ∂
+ + − = +∂ ∂
Kreyszig by YHLee;100510; 9-21 9.7 Gradient of a Scalar Field. Directional Derivative Definition 1 Gradient
The gradient of a scalar function ( , , )f x y z is defined as
ˆˆ ˆgrad f f f
f i j kx y z∂ ∂ ∂
= + +∂ ∂ ∂
: a vector function
"del" operator is defined as
ˆˆ ˆi j kx y z∂ ∂ ∂
∇ = + +∂ ∂ ∂
: differential operator
→ grad f f=∇ Directional Derivative Definition 2 Directional Derivative
Directional derivative of f at P in the direction of b̂ is defined by
( )
0
( )limb s
f Q f PdfD f
ds s→
−= ≡ ,
where Q is a variable point displaced from P by ˆ( )sb
The line L in Cartesian Coordinates
ˆˆ ˆ( ) ( ) ( ) ( )r s x s i y s j z s k= + +
′ = ′ + ′ + ′r s x s i y s j z s k( ) ( ) ( ) ( ) : this is a unit vector
b̂⇒ The function f on C in parametric representation ( )( ), ( ), ( )f x s y s z s
The directional derivative
ˆ ˆˆ ˆ ˆ ˆdf f x f y f z f f f x y zi j k i j k
ds x s y s z s x y z s s s∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + + ⇒ + + • + +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
→ ˆ(grad )df
f bds
⇒ •
Example 1 Directional derivative
Find the directional derivative of 2 2 2( , , ) 2 3f x y z x y z= + + at the point P: (2,1,3) in the direction of ˆˆ 2a i k= −
ˆˆ ˆgrad 4 6 2f xi yj zk= + + → (2,1,3)ˆˆ ˆ(grad ) 8 6 6f i j k= + +
The unit direction vector : 1 2 ˆˆˆ5 5
a i k= −
The directional derivative : ( )1 2 ˆ ˆˆ ˆ ˆ8 6 6 1.7895 5
aD f i k i j k⎛ ⎞
= − • + + = −⎜ ⎟⎝ ⎠
Kreyszig by YHLee;100510; 9-22 Gradient Is a Vector. Maximum Increase Theorem 1 Vector Character of Gradient.
For a scalar function ( ) ( , , )f p f x y z= , its gradient is a vector function and its direction corresponds to the maximum increase of f on P. Its magnitude and direction are independent of the coordinate systems.
proof:
ˆ ˆgrad grad cos grad cosbD f b f b f f= • ⇒ γ ⇒ γ
→ The maximum directional derivative for ˆ grad b f , or γ=0 grad direction of max. Dgf f .
Its max. value is gradf .
• The directional derivative
0
( ) ( )lims
f Q f PsΔ →
−Δ
: independent of coordinates
and b̂ is fixed in space. Therefore, gradf should be independent of a coordinate system
Gradient as Surface Normal Vector A surface S : ( , , )f x y z c=
A curve C : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k= + +
Its derivative : ˆˆ ˆ( ) ( ) ( ) ( )r t x t i y t j z t k′ ′ ′ ′= + + Tangent vector of the curve.
If C is on S, the surface equation becomes [ ]( ), ( ), ( )f x t y t z t c= .
Differentiate the surface function w.r.t t
0f f fx y z
x y z∂ ∂ ∂′ ′ ′+ + =∂ ∂ ∂
: ' /x dx dt= , …
→ (grad ) 0f r ′• = : The tangent vector of C is also tangent to S
↑ ↑ Tangent to C and S. Surface normal vector Theorem 2 Gradient as Surface Normal Vector
When a surface S is given by ( , , )f x y z c= , grad f at a point P on S represents the normal vector of S at P.
Example 2 Gradient as a surface normal vector A cone is given by 2 2 24( )z x y= + . Find a unit surface normal vector at P : (1,0,2)
The surface function is ( )2 2 24 0x y z+ − =
→ At Pˆ ˆˆ ˆ ˆgrad 8 8 2 grad 8 4f x i y j z k f i k= + − ⎯⎯→ = −
The unit surface normal vector is 8 4 ˆˆˆ80 80
n i k= −
Kreyszig by YHLee;100510; 9-23 Vector Fields that are Gradients of Scalar Fields (Potentials) A vector function can be obtained by the gradient of a scalar function. ( ) ( )gradv P f P=
( )f P is called potential.
In this case, the field by ( )v P is a conservative field ( no loss of energy)
9.8 Divergence of a Vector Field A differentiable vector function
1 2 3ˆˆ ˆ( , , ) ( , , ) ( , , ) ( , , )v x y z v x y z i v x y z j v x y z k= + +
The divergence of v is defined as
31 2divvv v
vx y z
∂∂ ∂= + +∂ ∂ ∂
Using del operator
1 2 3ˆ ˆˆ ˆ ˆ ˆdiv ( )v i j k v i v j v k v
x y z∂ ∂ ∂⎛ ⎞
⇒ + + • + + ≡∇•⎜ ⎟∂ ∂ ∂⎝ ⎠ : a scalar function
Theorem 1 Invariance of the divergence
div v is a scalar function independent of coordinate systems
In xyz ‐space with 1 2 3, , v v v In * * *x y z ‐space with * * *1 3 3, , v v v
→ 31 2div vv v
vx y z
∂∂ ∂= + +∂ ∂ ∂
→ ** *31 2
* * *div
vv vv
x y z
∂∂ ∂= + +∂ ∂ ∂
It will be proved in 10.7 • If f is twice differentiable
ˆˆ ˆgradf f f
f i j kx y z∂ ∂ ∂
= + +∂ ∂ ∂
Take the divergence
2 2 2
22 2 2
div(grad )f f f
f fx y z∂ ∂ ∂
= + + ⇒∇∂ ∂ ∂
: 2 2 2
22 2 2x y z
∂ ∂ ∂∇ ≡ + +
∂ ∂ ∂, Laplace operator
→ 2div(grad )f f=∇
Kreyszig by YHLee;100510; 9-24 Example 2 Flow of a Compressible fluid. Physical Meaning of the Divergence A small box B with a volume V x y zΔ = Δ Δ Δ . No source or sink in the volume. Fluid flows through the box B. The fluid velocity vector
1 2 3ˆˆ ˆv v i v j v k= + +
Flux density is defined as
1 2 3ˆˆ ˆu v u i u j u k= ρ = + +
↑ Density of the fluid The flux density is the transfer of certain quantity across a unit area per unit time. ↑ Mass of fluid in this case • Loss of mass by outward flow through surfaces of B (1) Loss of mass due to flow in y direction during a time intervalΔt
{ } { }2 2 2 2 2( ) ( ) ( ) ( )y y y y y y
Vu x z t u x z t u u x z t u t
y+Δ +Δ
ΔΔ Δ Δ + − Δ Δ Δ ⇒ − Δ Δ Δ ⇒ Δ Δ
Δ
↑ 2u= Δ
(2) Loss due to flow in x direction duringΔt
1 1 1( ) { ( ) }x x x
Vu y z t u y z t u t
x+Δ
ΔΔ Δ Δ + − Δ Δ Δ ⇒Δ Δ
Δ
(3) Loss due to flow in z direction during Δt
3 3 3( ) { ( ) }z z z
Vu x y t u x y t u t
z+Δ
ΔΔ Δ Δ + − Δ Δ Δ ⇒Δ Δ
Δ
The total loss of mass in B duringΔt
( )31 2
Vuu uV t t
x y z t
∂ ρΔΔΔ Δ⎛ ⎞+ + Δ Δ = − Δ⎜ ⎟Δ Δ Δ ∂⎝ ⎠
↑ ↑ Reduced mass inside B. Loss by flow through six side surfaces of B.
→ div ut∂ρ
= −∂
: continuity equation, conservation of mass
For a steady flow, 0t
∂ρ=
∂ → div 0u =
For a constant ρ (incompressible) → div 0v = , condition of incompressibility
Kreyszig by YHLee;100510; 9-25 9.9 Curl of a Vector Field
The curl of a vector function 1 2 3ˆˆ ˆ( , , )v x y z v i v j v k= + + is defined as
1 2 3
ˆˆ
curl
i j k
v vx y zv v v
∂ ∂ ∂=∇× =
∂ ∂ ∂
3 2 1 3 2 1ˆˆ ˆv v i v v j v v k
y z z x x y∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎛ ⎞⇒ − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠
Example 1 Curl of a Vector Function
A vector function is given, ˆˆ ˆ3v yzi zxj zk= + + , find the curl of v
ˆˆ
ˆˆ ˆ3 2
3
i j k
v xi yj zkx y zyz zx z
∂ ∂ ∂∇× = ⇒ − + +
∂ ∂ ∂
Example 2 Rotation of a rigid body The rotation is described by the angular speed vector w . → Its direction : right hand rule Its magnitude : angular speed, ω The linear speed of a point on the body : v w r= ×
Let the axis of rotation be z axis, ˆw k= ω
→
ˆˆ
ˆ ˆ0 0
i j k
v w r y i x j
x y z
= × ⇒ ω ⇒−ω +ω
The curl of v
ˆˆ
ˆcurl 2
0
i j k
v kx y zy x
∂ ∂ ∂= ⇒ ω
∂ ∂ ∂−ω ω
→ curl 2v w=
Theorem 2 Grad, Div, Curl
The curl of the gradient of a scalar function is always zero
( ) 0f∇× ∇ = : irrotational
The divergence of the curl of a vector function is always zero
( ) 0v∇• ∇× =
Theorem 3 Invariance of the curl
curl v is a vector independent of coordinate systems
