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CHAPTER 4 DIFFERENTIAL VECTOR CALCULUS
4.1 Vector Functions
4.2 Calculus of Vector Functions
4.3 Tangents
REVIEW: Vectors Scalar – a quantity only with its magnitude
Example: temperature, speed, mass, volume
Vector – a quantity with its magnitude and its direction
Example: velocity acceleration, force
Vector is denoted by A, or A A. OA - position vector AB - displacement vector
x
y
z
k
j
i
Vector Form of a Line Segment
If 0r is vector in 2-space or 3-space with its initial point at the origin, then the line that passes through the terminal point of 0r and is parallel to the vector v can be expressed in vector form as 0r r tv= + .
0 1 0( )r r t r r= + − or 0 1(1 )r t r tr= − +
Vector Algebra
• Addition
• associative law ( ) ( )A B C A B C+ + = + +
• commutative law A B B A+ = +
• multiplication by scalar
• distributive law
( ) )k B C kB kC+ = +
• unit vectors: , ,i j k
1 2 3 1 2 3, ,A A A A A i A j A k=< >= + +
1, 0, 0 , 0, 1, 0 , 0, 0, 1i j k=< > =< > =< >
Unit vector, u of v : vuv
=
BkB
• scalar product (dot product)
cos ABA B A B θ=i
A B B A=i i
( )A B C A B A C+ = +i i i
1 1 2 2 3 3A B AB AB A B= + +i
• vector product (cross product)
sin ABA B A B θ× =
B
A
A B×
A B B A B A× =− × ≠ ×
( ) ( )A B C A B C× × ≠ × ×
1 2 3
1 2 3
i j k
A B A A A
B B B
× =
1 2 3
1 2 3
1 2 3
( )
A A A
A B C B B B
C C C
× =i
4.1 Vector Functions
4.1.1 Vector-valued function
Definition A vector-valued function ( or simply vector function) is a function whose domain is a set of real numbers and whose range is a set of vectors.
Vector function, ( )r t :
( ) ( ), ( ), ( )r t f t g t h t=< >
where f, g, and h are real-valued functions called the component functions of r ; t is the independent variable (time).
Note
• If domains are intervals of real numbers, the vector functions represent a space curve
• If domains are regions in the plane, the vector functions represent surfaces in space.
4.1.2 Graph of a Vector Function
Consider a particle moving through space during a time interval I. The coordinates are seen as functions defined as:
( ) , ( ) , ( ) ,x f t y g t z h t t I= = = ∈ (1)
The points ( , , ) ( ( ), ( ), ( ))x y z f t g t h t= make up the curve in space, called the particle’s path.
Eqn. (1) parameterize the curve. A curve in space can also be represented in vector form. The vector
( ) ( ) ( ) ( )r t f t i g t j h t k= + +
is the particle’s position vector.
Definition
Let F be a vector function, and suppose the initial point of the vector ( )F t is at the origin. The graph of F is the curve traced out by the terminal point of the vector ( )F t as t varies over the domain set D.
4.1.3 Vector Functions Operations
Theorem
Let F and G be vector functions of the real variable t, and let f (t) be a scalar function. Then
i. ( ) )(ˆ)(ˆ)(ˆˆ tGtFtGF +=+
ii. )(ˆ)())(ˆ( tFtftFf =
iii. ( ) )(ˆ)(ˆ)(ˆˆ tGtFtGF ×=×
iv. ( ) )(ˆ)(ˆ)(ˆˆ tGtFtGF •=•
Question 1 Sketch the graph of the vector function
2( ) ( 3) , 2 2r t ti t j t= + + − ≤ ≤
Label the position of ( 2)r − , (1)r and (2)r .
Question 2 Sketch the graph of the vector function
( ) ( )( ) 3 2 3 2r t t i t j t k= − + + −
Label the position of (0)r .
Question 3
If 2ˆ ( ) 2 cosF t t i t j t k= + + and
2ˆ ( ) 5G t t i t j k= + + , find
(a) ( )ˆˆ ( )F G t+ (b) ˆ(sin )( )t F t
(c) ( )ˆˆ ( )F G t× (d) ˆˆ ˆ( ) ( )F G t F t⎡ ⎤• ×⎣ ⎦
4.2 Calculus of Vector Functions
4.2.1 Vector Derivatives
Definition
The derivative F′ˆ of a vector F is defined as:
ttFttFtFdt
Fdt Δ
−Δ+=′=→Δ
)(ˆ)(ˆlim)(ˆˆ
0
where )(),(),()(ˆ thtgtftF =
Theorem The vector function
kthjtgitftF ~)(~)(~)()(ˆ ++= is differentiable whenever the component functions f(t), g(t) and h(t) are all differentiable.
kthjtgitftF ~)(~)(~)()(ˆ ′+′+′=′
Example
(3 sin )d i t jdt
+ =
2(3 cos 4 )td t i t j te kdt
+ + =
4.2.2 Higher Vector Derivatives
Higher derivatives of a vector function F are obtained by successively differentiating the
components of kthjtgitftF ~)(~)(~)()(ˆ ++= .
The second derivative of F is the function
[ ] kthjtgitftFtF ~)(~)(~)()(ˆ)(ˆ ′′+′′+′′=′
′=′′
and the third derivative )(ˆ tF ′′′ is the derivative of
)(ˆ tF ′′ and so forth.
Example
Let ktejeietF ttt ~~~)(ˆ 222 ++= − . Find
(i) unit tangent vector )0(T
(ii) )0(F ′′
(iii) )(ˆ)(ˆ tFtF ′′•′ Theorem: Differentiation rules
Suppose F and G are differentiable vector functions and c is a scalar and f is a real valued function. Then
i. )(ˆ)(ˆ)](ˆ)(ˆ[ tGtFtGtFdtd ′+′=+
ii. )(ˆ)](ˆ[ tFctFcdtd ′=
iii. )(ˆ)()(ˆ)()](ˆ)([ tFtftFtftFtfdtd ′+′=
iv. )(ˆ)(ˆ)(ˆ)(ˆ)](ˆ)(ˆ[ tGtFtGtFtGtFdtd ′•+•′=•
v. )(ˆ)(ˆ)(ˆ)(ˆ)](ˆ)(ˆ[ tGtFtGtFtGtFdtd ′×+×′=×
vi. ))((ˆ)()]((ˆ[ tfFtftfFdtd ′′= , chain rule
Likewise, we can obtained the partial derivatives of a multivariable vector function. Suppose
ˆ( ) ( ) ( ) ( )R t f t i g t j h t k= + +
is a differentiable functions of n variables,
nttt ,,, 21 … . Then, the partial derivative of )(ˆ tR is
1 1 1 1
ˆ( ) f g hR t i j kt t t t∂ ∂∂ ∂= + +∂ ∂ ∂ ∂
ˆ( )n n n n
f g hR t i j kt t t t
∂ ∂∂ ∂= + +∂ ∂ ∂ ∂
and
2 2 2 2
1 1 1 1
ˆ( )m m m m
R t f g hi j kt t t t t t t t∂ ∂ ∂ ∂= + +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ Example
Let 2 2ˆ( , ) 2 ( 2 ) ( )R u v uvi u v j u v k= + − + + .
Find the partial derivatives Ru∂∂ ,
Rv
∂∂ ,
2
2R
u∂∂ ,
2
2R
v∂∂
and 2R
u v∂∂ ∂ .
4.2.3 Vector Integrals Let kthjtgitftF ~)(~)(~)()(ˆ ++= where f, g¸and h are continous functions for )bta ≤≤ . Then,
(i) the definite integral of )(ˆ tF is the vector function
kdtth
jdttgidttfdttF
b
a
b
a
b
a
b
a
~)(
~)(~)()(ˆ
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡=
∫
∫∫∫
(ii) the indefinite integral of )(ˆ tF is the vector function
ˆ( ) ( ) ( )
( )
F t dt f t dt i g t dt j
h t dt k
⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦⎡ ⎤+ ⎣ ⎦
∫ ∫ ∫∫
Question 1
In questions 1(a) - 1(b), find ˆ ˆ ˆ ˆ, , ,F F F F′ ′′ ′ ′′ when
1t = .
(a) ( ) ( )2 2ˆ ( ) 2 3 2 5F t t i t t j t k= + + + +
(b) ( ) ( ) ( )31 1ˆ ( ) 2 1t tF t e i e j t k− −= + + −
Question 2
In questions 2(a) - 2(b), find ( )ˆˆd F Gdt
• and ( )ˆˆd F Gdt
×
.
(a) 2ˆ ( ) ,tF t e i j t k= + + 3ˆ ( )G t t i j k= + −
(b) ( )2ˆ ( ) 2 1 ,F t t i t j t k= − + +
( )ˆ ( ) 2 3G t t i j t k= − + −
Question 3
Find 2 2
2
ˆ ˆ ˆ ˆ ˆ ˆ, , , ,F F F F F F
x y x y x x y∂ ∂ ∂ ∂ ∂ ∂
×∂ ∂ ∂ ∂ ∂ ∂ ∂
for the given
( ) ( )ˆ , sinxyF x y e i x y j x y k= + − + .
Question 4
Evaluate the integral in questions 4(a) and 4(b).
(a) ( )2
0
26 4 3t i t j k dt∫ − +
(b) ( )3
1
3 lntt i e j t k dt∫ − +
4.3 Tangents
Definition
Suppose )(ˆ tF is differentiable at t0 and that
0)(ˆ0 ≠′ tF . Then )(ˆ
0tF ′ is defined to be a
tangent vector to the graph of )(ˆ tF at the point where t = t0.
Unit tangent vector
If )(ˆ tF is a vector function that defines a smooth graph, then at each point a unit tangent is
ˆ ( )ˆ( ) ˆ ( )F t
T tF t
′=
′
and the principal unit normal vector is ˆ ( )ˆ( ) ˆ ( )T t
N tT t′
=′
Smooth curve
The graph of the vector function defined by )(ˆ tF is said to be smooth on any interval of t where
)(ˆ tF′ is continuous and 0)(ˆ ≠′ tF .
A curve that is smooth has a continuous turning tangent
A curve that is not smooth can have “sharp” points. Note that this graph is piecewise smooth
Arc length
Let ( ) ( ) ( ) ( )r t f t i g t j h t k= + + be a differentiable vector valued function on [a, b]. Then the arc length s is defined by
( ) ( ) ( )
2 2 2
22 2
[ ( )] [ ( )] [ ( )]b
a
b
a
s f t g t h t dt
dydx dz dtdt dt dt
′ ′ ′= + +
= + +⌠⎮⌡
∫
In a more compact form:
( )b
a
s r t dt′= ⌠⌡
By FTC,
( )ds r tdt′=
Example Find the length of the given curve:
2( ) 2 lnr t t i tj tk= + + , 1 t e≤ ≤
Solution
The derivative 1( ) 2 2r t ti j kt
′ = + + has length
1( ) 2r t t t′ = + for 1 t e≤ ≤
Thus,
arc length, ( )1
12e
s t dtt= +⌠⌡
2 2
1
2
ln ( 1) (1 0)e
t t e
e
⎡ ⎤= + = + − +⎣ ⎦=
Binormal Vector
Binormal vector, B T N= × ♦ , and T N B define a moving right handed
vector frame, called Frenet frame or TNB frame ♦ play a role in calculating the paths of particles
moving in space
Question 1
In questions 1(a) - 1(b), find the unit tangent vector ,T
the principal unit normal vector ,N the binormal unit
vector ,B of the given ( )r t at the indicated t.
(a) ( ) cos sin ,
; 0, .
r t a t i a t j btk
ab t π= + +
> =
(b) 2 32( ) , 1.3
r t t i t j t k t= + + =
Question 2
In questions 2(a) - 2(b), the coordinates of a moving particle are given as a function of time t. Find the
speed v, the unit tangent vector ,T as a function of t. (a) cos , sin , 0.t tx e t y e t z= = =
(b) 5 sin 4 , 5 cos 4 , 10 .x t y t z t= = =
Question 3
The position vector of a moving particle is
( ) ( )( ) sin cos sin cos .2tr t t t i t t j k= + + − +
(a) Determine the velocity and speed of the particle.
(b) Determine the acceleration of the particle.
(c) Find a unit tangent to the path of the particle, in
the direction of motion.