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2. Vector Calculus Richard E. Haskell This chapter begins our discussion of vector calculus. A prerequisite for this chapter is knowledge of differential and integral calculus as is generally obtained in the first two college-level calculus courses. Differentiation of a vector with respect to a scalar (in particular, with respect to time) with applications to particle motion is treated.

Differentiation of a Vector Frame 2.1

Answer (Frame 2.1)

magnitude direction

The derivative of the scalar function Φ(t) with respect to t is given by

0

( ) limt

d tdt t∆ →

Φ ∆Φ=

∆

where ( ) ( )t t t∆Φ = Φ + ∆ − Φ . In a similar way the derivative of the vector function A(t) with respect to t is given by

0

( ) limt

d tdt t∆ →

∆=

∆A A

where ( ) ( )t t t∆ = + ∆ −A A A . ∆A can occur as the result of a change in the magnitude of A, the direction of A, or both. It is assumed that A(t) is a well-behaved function, that is, that 0∆ →A as 0t∆ → . If r is the displacement vector of a particle, the velocity of the particle is given by ( )

0lim

td dt t

∆ →= = ∆ ∆v r r . For linear motion in a

straight line through the origin, ∆r is due to a change in the _________________________of r. For uniform circular motion with the origin at the center of the circle, ∆r is due to a change in the ______________________________of r.

2

Frame 2.2

Answer (Frame 2.2)

0

( )lim yx z

t

dAdA dAd tt dt dt dt dt∆ →

∆= = + +

∆A A i j k

Frame 2.3

Answer (Frame 2.3)

22 22

2 2 2 2yx z

d Ad A d Addt dt dt dt

= + +A i j k

In a Cartesian coordinate system the vector A can be written as

x y zA A A= + +A i j k

Since the unit vectors i, j and k are fixed (and therefore not a function of t), it follows that x y zA A A∆ = ∆ + ∆ + ∆A i j k If we divide by ∆t and take the limit as 0t∆ → , we obtain

0

( )lim ______ ______ ______t

d tt dt∆ →

∆= = + +

∆A A i j k

From the previous frame we can write 2 2d dtA in terms of the Cartesian components as

2

2 ________________________ddt

=A

3

Frame 2.4

Answer (Frame 2.4)

( )d d daa adt dt dt

= +AA A

Frame 2.5

Answer (Frame 2.5)

( )d d ddt dt dt

• = • + •B AA B A B

Frame 2.6

Answer (Frame 2.6)

( ) ( )d d d ddt dt dt dt

• × = • × + • × + • ×C B AA B C A B A C B C

We can write the derivative of the sum of two vectors as

( )d d ddt dt dt

+ = +A BA B

Similarly, the derivative of a scalar times a vector is given by

( ) _________________d adt

=A

The derivative of a dot product can be written as

( ) ___________________ddt

• =A B

When writing the derivative of the cross product ( )d dt×A B it is important to maintain the order or the cross product. Thus,

( )d d ddt dt dt

× = × + ×B AA B A B

Similarly,

( ) ( ) ( )d d ddt dt dt

• × = • × + • ×AA B C A B C B C

= _______________________________

4

Particle Motion Frame 2.7

Answer (Frame 2.7)

rdd rdt dt

= =λrv

In all the cases considered so far the unit vectors have been fixed and therefore constant. This is not always the case. Consider a particle moving in uniform circular motion in the x-y plane. It is convenient to introduce a pair of orthogonal unit vectors λr and λθ in the directions of increasing r and θ that move around with the particle as shown in the figure. The position vector r can then be written as r = r λr. Since r is a constant for circular motion, the velocity v is given by

__________ddt

= =rv

x

y

θ∆θ

1P

2P

rλθλ

r r+ ∆λ λθ θ+ ∆λ λ

+ ∆r r

rt

t t+ ∆

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Frame 2.8

Answer (Frame 2.8) λθ Frame 2.9

Answer (Frame 2.9)

drdt θθ

=v λ

In moving from P1 to P2 along the circle in the previous frame, r changes to r + ∆rand the unit vector λr changes to λr + ∆λr as shown in the figure. If ∆θ is small, the magnitude of ∆λr is approximately ∆θ and approaches this value exactly as 0θ∆ → .Also as 0θ∆ → , the direction of ∆λr approaches the direction of the unit vector ___________. Therefore we can write r θθ∆ = ∆λ λ

∆θ

rλ

r r+ ∆λ λr∆λ

Using the result of the previous frame we can write

0 0

lim limr r

t t

d ddt t t dtθ θ

θ θ∆ → ∆ →

∆ ∆= = =

∆ ∆λ λ λ λ

Combining this result with the result in Frame 2.7 we obtain

_______________rdrdt

= =λv

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Frame 2.10

Answer (Frame 2.10) v = ωr Frame 2.11

Answer (Frame 2.11)

The angular displacement θ in radians is defined as s rθ = , where s is the arc length measured along the circle. Since r is a constant for circular motion,

( )( )1d dt r ds dtθ = . The linear speed v of the particle is just v ds dt= and the angular speed ω is defined by d dtω θ= . Thus, d dt v rω θ= = . We can therefore write v θ=v λ where v = __________

In the previous frame we found that for uniform (v = constant) circular motion v θ=v λ . The acceleration of the particle is given by

dda vdt dt

θ= =λv

Now θλ changes to θ θ+ ∆λ λ because the velocity vector is changing direction.Sketch the vectors θλ , θ θ+ ∆λ λ , and θ∆λ .

∆θ θλ

θ θ+ ∆λ λ

θ∆λ

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Frame 2.12

Answer (Frame 2.12) ∆θ −λr Frame 2.13

Answer (Frame 2.13)

2

r rvvr

ω= − = −a λ λ

Referring to your sketch in the previous frame, in the limit as 0θ∆ → the magnitude of θ∆λ is __________ and the direction of θ∆λ is in the direction _________.

From the previous frame rθ θ∆ = −∆λ λ Thus, from Frame 2.11,

0

lim rt

dv vdt t

θ θ∆ →

∆= = −

∆λa λ

which can be written as (see Frame 2.10)

a = _________________________________ Thus the acceleration of the particle is in the r−λ direction; that is, it is always directed toward the center of the circle. It is called the centripetal acceleration and has a magnitude given by

2

2va v rr

ω ω= = =

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Frame 2.14

Answer (Frame 2.14)

sr

θ ∆∆ =

Let us consider uniform circular motion more closely. The figure of Frame 2.7 has been redrawn showing the z axis and the unit vector zλ , which is orthogonal to both

rλ and θλ . We know that s rθ = so that ∆θ equals ____________. The question arises, Can the angular displacement θ be a vector quantity? The answer is no since finite angular displacements do not, in general, commute. That is, the result of two successive rotations of an object in general will depend on the order in which the rotations were carried out. Thus the commutative law for vector addition will not hold for finite angular displacements. However, infinitesimal angular displacements do commute so perhaps θ∆ (in the limit dθ θ∆ → ) can be considered to be a vector quantity. Let us see how this might happen.

x

y

θ∆θ

1P

2P

rλθλ

r r+ ∆λ λθ θ+ ∆λ λ

+ ∆r r

rt

t t+ ∆

z

zλ

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Frame 2.15

Answer (Frame 2.15) r θθ∆ = ∆λ λ Frame 2.16

Answer (Frame 2.16)

zλ

From the figure in the previous frame we can write rr=r λ rr∆ = ∆r λ From Frame 2.8 we know that __________________r∆ =λ so that r θθ∆ = ∆r λ

Consider the vector product ( ) ( )rr r θθ× ∆ = × ∆r r λ λ 2

rr θθ= ∆ ×λ λ The cross product of the unit vectors _______r θ× =λ λ . Thus we can define a vector

zθ= ∆∆θ λ whose magnitude is equal to ∆θ and which is directed along the axis of rotation in a right-handed sense. We can then write

2z rθ × ∆

= ∆ =r r∆θ λ

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Frame 2.17

Answer (Frame 2.17) v

vr

Frame 2.18

Answer (Frame 2.18) θλ

The angular velocity vector ω is defined by

0

limtdt t∆ →

= =∆

dθ ∆θω

Using the results of the previous frame,

2 20lim

t

t dtr r∆ →

× ∆ ×= =

r ∆r r drω

2

____r

×=

r

The direction of ω is the same as that of ∆θ, that is, along the axis of rotation in a right-handed sense. The magnitude of ω is _________ω = =ω Compare this result with Frame 2.10.

In Frame 2.10 we found that v θ=v λ and v rω= . Consider the cross product

0

limt t

θ∆ →

∆× = ×

∆ω r r

Recall from Frame 2.16 that we can write zθ= ∆∆θ λ and rr=r λ . Thus

( )0

lim z r z rt

dr rt dtθ θ

∆ →

∆× = × = ×

∆ω r λ λ λ λ

______rω= Since we already know that r θω=v λ , it follows that = ×v ω r .

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Frame 2.19

Answer (Frame 2.19) T = ×a α r R = ×a ω v centripetal

If ω is a function of time, then the circular motion will not be uniform but will have an angular acceleration d dt=α ω . The total linear acceleration of the particle will be

( )d ddt dt

= = ×va ω r

d ddt dt

= × + ×

ω rr ω

( ) ( )= × + ×α r ω v If only the magnitude of ω changes with time (i.e., the particle motion remains in the x-y plane), then α (which is in the direction of ∆ω) will be in the same direction as ω. The total acceleration a can then be written as the sum of a tangential component and a radial component in the form T R= +a a a , where ______T =a and ______R =a .Sketch , , and T Rα a a in the figure. The component Ra is called the __________________ acceleration.

r

v

ω

x

y

z

r

v

ω

x

y

z

aT

aR

α

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Frame 2.20

Answer (Frame 2.20)

rr

d drrdt dt

= +λv λ

Frame 2.21

Answer (Frame 2.21) ∆θ λθ

Consider now the more general case of plane curvilinear motion that is not necessarily circular. As shown in the figure, the magnitude of the radius vector r can now change as the particle moves from point P1 to point P2. Thus both r and λr are now variable in the expression rr=r λ so that the velocity v of the particle can be written as

_________________ddt

= =rv

x

y

r

r + ∆r

∆θ

θ

rλ

r r+ ∆λ λ

θλ

θ θ+ ∆λ λ

1P

2P

∆r

Sketch the vectors λr, λr + ∆λr, and ∆λr. In the limit as 0θ∆ → , the magnitude of ∆λr is _________ and the direction of ∆λr is in the direction of the unit vector _________.

∆θrλ

r r+ ∆λ λ r∆λ

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Frame 2.22

Answer (Frame 2.22)

0

limr

t

ddt t θ

θ∆ →

∆=

∆λ λ

ddt θ θθ ω= =λ λ

Frame 2.23

Answer (Frame 2.23)

From the previous frame we can write r θθ∆ = ∆λ λ Therefore

0 0

lim lim ___________________r rrt t

ddt t∆ → ∆ →

∆= = ∆ =

∆λ λ λ

= ____________________________

∆θθλ

θ θ+ ∆λ λ

θ∆λ

From the previous three frames the particle velocity v can be written as

rdrrdtθω= +v λ λ

The particle acceleration is then

2

2r

rd dd dr d d r drr r

dt dt dt dt dt dt dtθ

θωω ω = = + + + +

λ λva λ λ

To find ddt

θλ , sketch the vectors , , and θ θ θ θ+ ∆ ∆λ λ λ λ .

In the limit as 0θ∆ → , the magnitude of ∆λθ is _________ and the direction of ∆λθ is in the direction of the unit vector _________.

θ∆

r−λ

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Frame 2.24

Answer (Frame 2.24)

r rd ddt dt

θ θ ω= − = −λ λ λ

Frame 2.25

Answer (Frame 2.25)

2

22 2r

d r drr rdt dt θω α ω

= − + +

a λ λ

Frame 2.26

Answer (Frame 2.26)

centripetal

From the previous frame we can write rθ θ∆ = −∆λ λ Therefore

____________ _____________ddt

θ = =λ

Using the expressions for rd dtλ and d dtθλ from Frame 2.22 and 2.24, the particle acceleration given in Frame 2.23 can be written as

2

22r r

dr d r drr rdt dt dtθ θω ω α ω = − + + + +

a λ λ λ λ

where d dtα ω= . Collecting terms, this expression can be rewritten as _______________ _______________r θ= +a λ λ

The expression for the acceleration in the previous frame

2

22 2r

d r drr rdt dt θω α ω

= − + +

a λ λ

should be compared with the cases of circular motion in Frames 2.13 and 2.19. The term 2

rrω− λ above represents the _______________________ acceleration and the term r θα λ represents the tangential acceleration of Frame 2.19. Two additional terms have appeared in the expression for the acceleration as the result of letting r vary. The first is the expected radial acceleration component ( )2 2

rd r dt λ . The other additional

term, ( )2 dr dt θω λ , is in the θλ direction and is called the Coriolis acceleration.

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Frame 2.27

Answer (Frame 2.27) 1

t

In addition to the unit vectors rλ and θλ other sets of unit vectors can be used to describe plane curvilinear motion. A particularly useful set is the tangential and normal unit vectors shown in the figure. In this case t is everywhere tangent to the curve and n is normal to the curve, pointing toward the center of curvature C located a distance R from the curve. R is called the radius of curvature. If s is the arc length of the curve measured from some reference point, then we can write the velocity of the particle as

d d dsdt ds dt

= =r rv

ds dt is just the speed of the particle v. Also

0

lims

dds s∆ →

∆=

∆r r

In this limit the magnitude of s∆ ∆r approaches __________ and the direction of s∆ ∆r approaches the direction of the unit vector _________. Thus we can write

v=v t as expected.

x

y

rr + ∆r

∆ψ ∆r

R

Reference Point

s

∆s

n t

t + ∆t

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Frame 2.28

Answer (Frame 2.28)

dv dvdt dt

= +ta t

Frame 2.29

Answer (Frame 2.29) ∆ψ ∆t = ∆ψn Frame 2.30

Answer (Frame 2.30)

s R ψ∆ = ∆ sR∆

∆ =t n

From the previous frame the velocity v is v=v t

Thus the acceleration can be written as

______________________ddt

= =va

From the previous frame the acceleration a is given by

dv dvdt dt

= +ta t

To find d dtt , sketch the unit vectors t and t + ∆t (see figure in Frame 2.27). The angle between t and t + ∆t is __________. From the sketch, ∆t = ____________

tt + ∆t

∆t

∆ψ

From the figure in Frame 2.27, ∆s, ∆ψ, and R are related by the expression ∆s = __________________ Using this and the result of the previous frame, we can write

∆t = ____________

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Frame 2.31

Answer (Frame 2.31)

1d dsdt R dt

=t n

Frame 2.32

Answer (Frame 2.32)

2dv v

dt R= +a t n

Frame 2.33

Answer (Frame 2.33)

1dds R

=t n

From the previous frame

1 sR

∆ = ∆t n

Therefore

0

lim _____________________t

ddt t∆ →

∆= =

∆t t

Using the result of the previous frame and the fact that ds dt v= , the acceleration as given in Frame 2.29 can be written as

a = _____________________________

From Frame 2.27 the tangent unit vector, t, is given by dds

=rt

This is often given as the definition of the tangent unit vector. From Frames 2.30 and 2.31 we know that

1 sR

∆ = ∆t n

from which we can write

0

lim ___________________s

dds s∆ →

∆= =

∆t t

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Frame 2.34

Answer (Frame 2.34)

ddsdds

=

t

nt

Frame 2.35

Answer (Frame 2.35) perpendicular

The reciprocal of the radius of curvature, 1/R, is called the curvature, κ, of the particle path. Therefore, from the previous frame we can write

dds

κ=t n

Note that because n is a unit vector, the magnitude of d dst is the curvature κ. Thus,

dds

κ =t

and is often taken as the definition of curvature. Using these two equations, the normal unit vector is sometimes defined as n = ____________________

The binormal unit vector, b, is defined as = ×b t n

Thus, b is perpendicular to both t and n. Since b is a unit vector it follows that 1• =b b Differentiating this equation with respect to s gives

0d dds ds

• + • =b bb b

from which

0dds

• =b b

Therefore, the vectors d dsb and b are ________________________.

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Frame 2.36

Answer (Frame 2.36) perpendicular Frame 2.37

Answer (Frame 2.37) × =b t n Frame 2.38

Answer (Frame 2.38)

dds

τ κ= − × + ×n n t b n

Because the binormal unit vector, b, and the tangent unit vector, t, are perpendicular we can write the dot product as 0• =b t Differentiating this equation with respect to s gives

0d dds ds

• + • =b tt b

The second term is zero because d dst is in the direction of n (see Frame 2.34), which is perpendicular to b. Therefore,

0dds

• =b t

Therefore, the vectors d dsb and t are ________________________.

From Frames 2.35 and 2.36 we see that d dsb is perpendicular to both b and t. It must therefore be parallel to n. We will let

dds

τ= −b n

where τ is called the torsion. Recall that t, n, and b form unit vectors that are all perpendicular to each other, and b was defined as = ×b t n . It follows that ________× =b t

From the previous frame we found that = ×n b t We can therefore write

( )dd d dds ds ds ds

×= = × + ×

b tn b tt b

From Frame 2.37 d ds τ= −b n and from Frame 2.34 d ds κ=t n . Therefore we can write

_____________________dds

=n

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Frame 2.39

Answer (Frame 2.39)

dds

κ τ= − +n t b

Frame 2.40

Answer (Frame 2.40)

0 0

00 0

κκ τ

τ

′ ′ = −

′ −

t tn nb b

From the previous frame we found that

dds

τ κ= − × + ×n n t b n

From Frame 2.35 × = −n t b and × = −b n t . Therefore, we can write

_______________dds

=n

Combining the results from Frames 2.34, 2.39, and 2.37 we can write

dds

κ=t n

dds

κ τ= − +n t b

dds

τ= −b n

These are called the Frenet-Serret formulas. If we define d ds′ =t t , d ds′ =n n ,

d ds′ =b b we can write the Frenet-Serret formulas as the following matrix equation:

′ ′ =

′

t tn nb b

This is also known as the Frenet-Serret theorem.

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Exercises 2-1 A wheel of radius 15 cm is rotating with a constant angular speed of 10 rps. What is

the velocity and acceleration of a point on the outer circumference of the wheel? 2-2 A particle moves in a circle of radius r with an angular displacement given by

2A Btθ = + where A and B are constants. Find the centripetal and tangential accelerations as a function of time.

2-3 A particle moves along a curve described by the position vector 2 2t t= +r i j

(a) Find the velocity v and acceleration a as a function of time. (b) Find the magnitudes of the tangential and normal components of the

acceleration when t = 1.