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′

=′

Uc

Unit tacurve,

angentbut al

t vectolways

ors chahave l

ange dilength

irectio1.

on alonng the

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.