Vector Algebra and Calculus 1. Revision of vector algebra, scalar product, vector product 2. Triple products, multiple products, applications to geometry 3. Differentiation of vector functions, applications to mechanics 4. Scalar and vector fields. Line, surface and volume integrals, curvilinear co-ordinates 5. Vector operators — grad, div and curl 6. Vector Identities, curvilinear co-ordinate systems 7. Gauss’ and Stokes’ Theorems and extensions 8. Engineering Applications

3. Differentiating Vector Functions of a Single Variable • Your experience of differentiation and integration has extended as far as scalar functions of single and multiple variables d ∂ f (x) and f (x, y , t) dx ∂x • No surprise that we often wish to differentiate vector functions.

• For example, suppose you were driving along a wiggly road with position r(t) at time t. • Differentiating r(t) should give velocity v(t). • Differentiating v(t) should yield acceleration a(t).

r

• Differentiating a(t) should yield the jerk j(t).

o

Differentiation of a vector

3.2

Differentiation of a vector

3.3

• By analogy with the definition for a scalar function, the derivative of a vector function a(p) of a single parameter p is a(p + δp) − a(p) da (p) = lim . δp→0 dp δp ˆ constant) • If we write a in terms of components relative to a FIXED coordinate system (ˆı,ˆ, k ˆ a(p) = a1(p)ˆı + a2(p)ˆ + a3(p)k then

da da1 da2 da3 ˆ k . (p) = ˆı + ˆ + dp dp dp dp

To differentiate a vector function defined wrt a fixed coordinate system, differentiate each component separately

All the familiar stuff works ...

3.4

• This means that – All the familiar rules of differentiation apply – they don’t get munged by operations like scalar product and vector products. • For example:

d da db (a × b) = ×b+a× dp dp dp da db d (a · b) = ·b+a· . dp dp dp

• NB! (obvious really): da/dp has – a different direction from a – a different magnitude from a.

Position, velocity and acceleration

3.5

• Suppose r(t) is the position vector of an object moving w.r.t. the orgin. ˆ r(t) = x(t)ˆı + y (t)ˆ + z (t)k • Then the instantaneous velocity is • and the acceleration is

v(t) =

dr dx dy dz ˆ k = ˆı + ˆ + dt dt dt dt

d 2r dv = 2 . a(t) = dt dt

Chain rule: more good news

3.6

• Likewise, the chain rule still applies. • If u = u(p):

da(p) da du = dp du dp

• This follows directly from the fact that the vector derivative is just the vector of derivatives of the components.

♣ Example of chain rule

3.7

• The position of vehicle is given by r(u) where u is amount of fuel used by time t, so that u = u(t). • Its velocity must be • Its acceleration is

dr dr du = dt du dt  2 dr d 2u d 2r du d 2r + = dt 2 du 2 dt du dt 2

♣ Example: direction of the derivative

3.8

Question 3D vector a has constant magnitude, but is varying over time. What can you say about the direction of da/dt? Answer Using intuition: if only the direction is changing, then the vector must be tracing out points on the surface of a sphere. So da/dt is orthogonal to a??? To prove this write d da da da (a · a) = a · + · a = 2a · . dt dt dt dt But (a · a) = a2 = const. So d da (a · a) = 0 ⇒ 2a · = 0 (QED) dt dt

Integration of a vector function

3.9

• As with scalars, integration of a vector function of a single scalar variable is the reverse of differentiation. • In other words

Z

p2 p1



Eg, from dynamics-ville Z

 da(p) dp = a(p2 ) − a(p1 ) dp

t2 t1

a dt = v(t2 ) − v(t1 )

• However, other types of integral are possible, especially when the vector is a function of more than one variable. • This requires the introduction of the concepts of scalar and vector fields. See lecture 4!

Geometrical interpretation of derivatives

3.10

δr • Position vector r(p) traces a space curve. • Vector δr is a secant to the curve δr/δp lies in the same direction as δr(p) • Take limit as δp → 0

r (p)

r (p + δ p)

dr/dp is a tangent to the space curve

• Nothing special about the parameter p – may be various ways of parametrizing a particular curve. • Consider helix aligned with z -axis. Could parametrize by for example: z , the “height” up the helix, or s, the “length” along the curve

Geometrical interpretation of derivatives /ctd

3.11

• If the parameter s is arc-length or metric distance, then we have: |dr| = ds so

and

dr =1 ds

dr/ds is a unit tangent to r at s • For s arc-length and p some other parametrization, we have

dr dr ds = dp ds dp

and dr = dr ds = ds dp ds dp dp

Geometrical interpretation of derivatives /ctd

3.12

• To repeat, the derivative dr/dp is a vector • Its direction is always a tangent to curve r(p) • Its magnitude is ds/dp, where s is arc length • If the parameter is arc length s, then dr/ds is a unit tangential vector. • If the parameter is time t, then magnitude |dr/dt| is the speed.

δr

r (s)

δr

r (s + δ s) dr ds

1

r (t)

dr dt

r (t + δ t) ds SPEED dt

♣ Example

3.13

Question: Draw the curve hs s s ˆ √ √ k )ˆ  + r = a cos( √ 2 )ˆ ı + a sin( 2 2 2 2 2 a +h a +h a +h where s is arc length and h, a are constants. Answer

♣ Example ctd

3.14

s s hs ˆ √ √ r = a cos( √ 2 k )ˆ ı + a sin( )ˆ  + 2 2 2 2 2 a +h a +h a +h Show that the tangent dr/ds to the curve has a constant elevation angle w.r.t the xy -plane, and determine its magnitude. Answer a dr a h ˆ √ √ = −√ 2 k sin ()ˆ ı + cos ()ˆ  + 2 2 2 2 2 ds a +h a +h a +h

Length

z

2

√ Projection on the xy plane has √ magnitude a/ a2 + h2 Projection in the z direction h/ a2 + h2 So the elevation angle is tan−1 (h/a), a constant. We are expecting |dr/ds| = 1, and indeed it is!

2

h/ a + h

dr ds

y

x 2

2

Length a/ a + h

Why can’t we say any old parameter is arc length?

3.15

• Arc length s parameter is special because ds = |dr|, • Or, in integral form, whatever the parameter p, Accumulated arc length =

Z

p1 p0

• Using Pythagoras’ theorem on a short piece of curve. In the limit as ds tends to zero ds 2 = dx 2 + dy 2 + dz 2 . So if a curve is parameterized in terms of p s   2  2 2 dx dy dz ds + + . = dp dp dp dp

dr dp . dp

z

δz

y δx

δs

δy x

Arc length is special /ctd

3.16

• Suppose we had parameterized our helix as ˆ r = a cos pˆı + a sin pˆ + hp k • p is not arc length because s 2  2  2 q dr dy dz dx = a2 sin2 p + a2 cos2 p + h2 + + = dp dp dp dp p = a2 + h2 6= 1 √ • So if we want to parameterize in terms of arclength, replace p with s/ a2 + h2.

Curves in 3D

3.17

• Let’s look more closely at parametrizing a 3D space curve in terms of arclength s. • Introduce – orthogonal coord frames for each value s – each with its origin at r(s).

• To specify a coordinate frame we need – three mutually perpendicular directions – should be intrinsic to the curve – NOT fixed in an external reference frame.

O

r (s)

Curves in 3D

• Rollercoaster will help you see what’s going on ... • But it has a specially shaped rail or two rails that define the twists and turns.

• We are thinking about a 3D curve – just a 3D wire. Does the curve itself define its own twist and turns?

3.18

The Fr´enet-Serret Local Coordinates

3.19

Yes: method due to French mathematicians F-J. Fr´enet and J. A. Serret

n

1. Unit tangent ˆt Obvious choice is ˆt = dr(s)/ds 2. Principal Normal n ˆ Proved earlier that if |a(t)| = const then a · da/dt = 0. So ˆt = ˆt(s), |ˆt| = const ⇒ ˆt · dˆt/ds = 0 Hence the principal normal n ˆ is defined from κˆ n = dˆt/ds where κ ≥ 0 is the curve’s curvature.

ˆ 3. The Binormal b ˆ = ˆt × ˆ n. The third member of a local r-h set is the binormal, b

t s increasing

dt ds

Deriving the Fr´enet-Serret relationships

3.20

ˆ = ˆt × n Tangent ˆt, Normal ˆ n : dˆt/ds = κˆ n, Binormal b ˆ ˆ · ˆt = 0, if we differentiate wrt s ... • Since b ˆ ˆ db dˆt db ˆ ˆ ·t+b· = · ˆt + ˆ b · κˆ n=0 ds ds ds from which

ˆ db · ˆt = 0. ds

ˆ • This means that d b/ds is along the direction of n ˆ: ˆ db = −τ (s)ˆ n(s) ds where τ is the space curve’s torsion.

Deriving the Fr´enet-Serret relationships ˆ = ˆt × ˆ Tangent ˆt, Normal n ˆ, Binormal b n ˆ dˆt/ds = κˆ n, d b/ds = −τ (s)ˆ n(s) • Differentiating n ˆ · ˆt = 0: (d n ˆ/ds) · ˆt + n ˆ · (dˆt/ds) = 0 (d n ˆ/ds) · ˆt + n ˆ · κˆ n = 0 (d n ˆ/ds) · ˆt = −κ • Now do the same to n ˆ·ˆ b = 0: ˆ+n ˆ (d n ˆ/ds) · b ˆ · (d b/ds) = 0 ˆ+ˆ (d n ˆ/ds) · b n · (−τ )ˆ n = 0 ˆ (d n ˆ/ds) · b = +τ • HENCE

dn ˆ ˆ = −κ(s)ˆt(s) + τ (s)b(s). ds

3.21

Summary of the Fr´enet-Serret relationships

3.22

These three expressions are called the Fr´enet-Serret relationships: • dˆt/ds = κˆ n

ˆ • dn ˆ/ds = −κ(s)ˆt(s) + τ (s)b(s)

ˆ • d b/ds = −τ (s)ˆ n(s)

• They describe by how much the intrinsic coordinate system changes orientation as we move along a space curve.

♣ Example

3.23

Question Derive κ(s) and τ (s) for the curve ˆ r(s) = a cos (s/β)ˆı + a sin (s/β)ˆ + h (s/β) k √ where β = a2 + h2 Answer: • Denote sin, cos(s/β) as S and C.

We found the unit tangent earlier as ˆt = (dr/ds) = [− (a/β) S, (a/β) C, (h/β)] .

• Hence



2 ˆ κˆ n = d t/ds = − a/β C,

• The curvature must be positive, so

κ = a/β

2




n = [−C, ˆ

• So the curvature is constant, and n ˆ parallel to the xy -plane.

− a/β

2




S, 0

− S, 0] .



♣ Example /continued • Recall • So the binormal is

• Hence • So the torsion again a constant.

3.24

ˆt = [− (a/β) S, (a/β) C, (h/β)]

n ˆ = [−C,

− S, 0] .

ˆ     ˆı ˆ k   h h a ˆ S, − C, b = ˆt × n ˆ = (−a/β)S (a/β)C (h/β) = β β β −C −S 0

ˆ d b/ds = h/β 2 C,



h/β 2 S, 0 = −h/β 2 n ˆ

τ = h/β 2




Derivative (eg velocity) components in plane polars

3.25

In plane polar coordinates, the radius vector of any point P is given by

ˆ eθ

r = r (cos θˆı + sin θˆ) = r ˆ er

ˆ er

where we have introduced the unit radial vector ˆ er = cos θˆı + sin θˆ . The other “natural” unit vector in plane polars is orthogonal to ˆ er and is ˆ eθ = − sin θˆı + cos θˆ

so that ˆ er · ˆ er = ˆ eθ · ˆ eθ = 1 and ˆ er · ˆ eθ = 0.

P ˆ

r θ ˆı

Aside: notation • Some texts will use the notation

3.26

ˆr, ˆ θ

to denote unit vectors in the radial and tangential directions • I prefer the more general notation (as used in, eg, Riley). • You should be familiar and comfortable with either

ˆ er , ˆ eθ

Derivative (eg velocity) components in plane polars

3.27

• Now suppose P is moving so that r is a function of time t. • Its velocity is r˙ =

• Note that

ˆ eθ ˆ er

d dr dˆ er (r ˆ er ) = ˆ er + r dt dt dt dθ dr ˆ er + r (− sin θˆı + cos θˆ) = dt dt dr dθ = ˆ er + r ˆ eθ dt dt = radial + tangential dθ dˆ er = ˆ eθ dt dt

ˆ

P r θ ˆı

d dθ dˆ eθ = (− sin θˆı + cos θˆ) = − ˆ er dt dt dt

Acceleration components in plane polars • Recap ...

dθ dr ˆ er + r ˆ eθ ; dt dt • Differentiating r˙ gives the accel. of P r˙ =

3.28

dˆ er dθ = ˆ eθ ; dt dt

dˆ eθ dθ =− ˆ er dt dt

d 2r dr dθ dθ dθ dr dθ d 2θ ¨r = ˆ e + ˆ e − r ˆ e + ˆ e + r ˆ er r θ θ θ dt 2 dt dt dt dt dt 2 dt dt "   2 #  2 dr dθ dθ d 2θ d r ˆ er + 2 −r eθ +r 2 ˆ = dt 2 dt dt dt dt

Acceleration components in plane polars • We just saw

• Three obvious cases:

"

2

d r −r ¨r = dt 2

3.29



dθ dt

2 #

ˆ er +



 d 2θ dr dθ eθ +r 2 ˆ 2 dt dt dt

d 2r ˆ er θ const : ¨r = dt 2  2 dθ d 2θ r const : ¨r = −r eθ ˆ er + r 2 ˆ dt dt  2 dθ ˆ er r and dθ/dt const : ¨r = −r dt

Fixed, varying, and instrinsic coordinates

3.30

Rotating systems

3.31

ω • Body rotates with constant ω about axis passing through the body origin. Assume the body origin is fixed. We observe from a fixed coord system Oxy z .

ρ

• If ρ is a vector of constant mag and constant direction in the rotating system, then in the fixed system it must be a function of t. dr ˙ = RR ˙ ⊤r r(t) = R(t)ρ ⇒ = Rρ dt * dr/dt will have fixed magnitude; * dr/dt will always be perpendicular to the axis of rotation; * dr/dt will vary in direction within those constraints; * r(t) will move in a plane in the fixed system.

Rotating systems

3.32

˙ ⊤ Consider the term RR • Note that RR⊤ = I, hence

˙ ⊤ is anti-symmetric: • Thus RR

˙ ⊤ + RR˙ ⊤ = 0 RR ˙ ⊤ = −RR˙ ⊤ RR 

 0 −z y ˙ ⊤ =  z 0 −x  RR −y x 0

• Application of a matrix of this form to an arbitrary vector has precisely the same effect as the cross product operator, ω×, where ω = [xy z ]⊤. • Thus

r˙ = ω × r

Rotating co-ordinate systems 2

3.33

• Now ρ is the position vector of a point P in the rotating body, but which is moving too, with respect to the rotating system r(t) = R(t)ρ(t)

P at t+ δ t • Differentiating with respect to time:

dr ˙ + Rρ˙ = RR ˙ ⊤ r + Rρ˙ = Rρ dt

• The instantaneous velocity of P in the fixed frame is dr = Rρ˙ + ω × r dt

δρ ω

δr P at t

(ω

• Second term is contribution from the rotating frame • First term is linear velocity in the rotating frame, referred to the fixed frame

r) δ t r= ρ at t

Rotating co-ordinate systems

3.34

• Now consider second differential:

¨r = ω˙ × r + ω × r˙ + R˙ ρ˙ + R¨ ρ

• If angular velocity constant, first term is zero • Now substituting for r˙ we have ¨r = = = =

ω × (ω × r + Rρ) ˙ + R˙ ρ˙ + R¨ ρ ˙ ⊤Rρ˙ + R¨ ω × (ω × r) + ω × Rρ˙ + RR ρ ω × (ω × r) + ω × Rρ˙ + ω × Rρ˙ + R¨ ρ ω × (ω × r) + 2ω × (Rρ) ˙ + R¨ ρ

• The instantaneous acceleration is therefore ¨r = R¨ ρ + 2ω × (Rρ) ˙ + ω × (ω × r)

Rotating co-ordinate systems

3.35

• The instantaneous acceleration is ¨r = R¨ ρ + 2ω × (Rρ) ˙ + ω × (ω × r) * Term 1 is P’s acceleration in the rotating frame. * Term 3 is the centripetal accel: magnitude ω 2r and direction −r. * Term 2 is a SURPRISE! It is a coupling of rotation and velocity of P in the rotating frame. It is the Coriolis acceleration.

♣ Examples

3.36

Q Find the instantaneous acceleration as observed in a fixed frame of a projectile fired along a line of longitude (with angular velocity of γ constant relative to the sphere) if the sphere is rotating with angular velocity ω. γt ω = ωm ˆ A In the rotating frame ρ˙ = γ × ρ ρ ¨ = γ × ρ˙ = γ × (γ × ρ)

r m ˆ ℓˆ

In fixed frame, instantaneous acceleration: ¨r = γ × (γ × r) + 2ω × (γ × r) + ω × (ω × r) In rotating frm

+

Coriolis

+

Centripetal

n ˆ

γ = γ ℓˆ

♣ Example /ctd

3.37

Repeated: ¨r = γ × (γ × r) + 2ω × (γ × r) + ω × (ω × r)

ˆ ρ = R cos(γt)m 1) As γ = γ ℓ, ˆ + R sin(γt)ˆ n ⇒acceleration in rotating frame is γ × (γ × ρ) = −γ 2 r 2) Centripetal accel due to rotation of sphere is ω × (ω × r) = −ω 2R sin(γt)ˆ n

γt

ω = ωm ˆ

r

m ˆ ℓˆ n ˆ

3) The Coriolis acceleration is       0 γ 0 2ω × ρ˙ = 2 ω ×  0  × R cos(γt) = 2ωγR cos(γt)ℓˆ 0 0 R sin(γt)

γ = γ ℓˆ

♣ Example /ctd

3.38

Recap:

• Accel in rotating frame −γ 2r

γt

ω = ωm ˆ

• Centripetal due to sphere rotating −ω 2R sin(γt)ˆ n • Coriolis acceleration: 2ωγR cos(γt)ˆl

r m ˆ

2

−ω R sin(γt)ˆ n ℓˆ 2ωγR cos(γt)ˆl −γ 2 r r

n ˆ

γ = γ ℓˆ

♣ Example /ctd

3.39

• Consider a rocket on rails which stretch north from the equator. • As rocket travels north it experiences the Coriolis force exerted by the rails: 2 γ ω R cos(γt) ℓˆ +ve -ve +ve +ve • Coriolis force is in the direction opposed to ℓˆ (i.e. opposing earth’s rotation).

Rocket’s velocity in direction of meridian Tangential velocity of earth’s surface

Tangential component of velocity (NB instantaneously common to earth’s surface and rocket)

♣ Coriolis acceleration • Because of the rotation of the earth, the Coriolis acceleration is of great importance in meteorology

3.40

♣ Coriolis acceleration

3.41

Summary

3.42

• We started by differentiating vectors wrt to a fixed coordinate system. • Then looked at the properties of the derivative of a position vector r with respect to a general parameter p and the special parameters of arc-length s, and time t • considered derivatives with respect to other coordinate systems, in particular those of the position vector in polar coordinates with respect to time. • derived Fr´enet-Serret relationships — a method of describing a 3D space curve by describing the change in a intrinsic coordinate system as it moves along the curve. • discussed rotating coordinate systems; we saw that there is coupled term in the acceleration, called the Coriolis acceleration.