Vector Algebra and Calculus

1. Revision of vector algebra, scalar product, vector product

2. Triple products, multiple products, applications to geometry
3. Dierentiation of vector functions, applications to mechanics
4. Scalar and vector elds. 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. Dierentiating Vector Functions of a Single Variable
Your experience of dierentiation and integration has extended as far as scalar functions of single and multiple
variables
d
dx
f (x) and

x
f (x, y, t)
No surprise that we often wish to dierentiate vector functions.
For example, suppose you were driving along a wiggly road with
position r(t) at time t.
Dierentiating r(t) should give velocity v(t).
Dierentiating v(t) should yield acceleration a(t).
Dierentiating a(t) should yield the jerk j(t).
r
o
Dierentiation of a vector 3.2
Dierentiation of a vector 3.3
By analogy with the denition for a scalar function, the derivative of a vector function a(p) of a single parameter
p is
da
dp
(p) = lim
p0
a(p + p) a(p)
p
.
If we write a in terms of components relative to a FIXED coordinate system (,,

k constant)
a(p) = a
1
(p) + a
2
(p) + a
3
(p)

k
then
da
dp
(p) =
da
1
dp
+
da
2
dp
+
da
3
dp

k .
To dierentiate a vector function dened wrt a xed coordinate system,
dierentiate each component separately
All the familiar stu works ... 3.4
This means that
All the familiar rules of dierentiation apply
they dont get munged by operations like scalar product and vector products.
For example:
d
dp
(a b) =
da
dp
b + a
db
dp
d
dp
(a b) =
da
dp
b + a
db
dp
.
NB! (obvious really): da/dp has
a dierent direction from a
a dierent 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.
r(t) = x(t) + y(t) + z(t)

k
Then the instantaneous velocity is
v(t) =
dr
dt
=
dx
dt
+
dy
dt
+
dz
dt

k
and the acceleration is
a(t) =
dv
dt
=
d
2
r
dt
2
.
Chain rule: more good news 3.6
Likewise, the chain rule still applies.
If u = u(p):
da(p)
dp
=
da
du
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
dr
dt
=
dr
du
du
dt
Its acceleration is
d
2
r
dt
2
=
d
2
r
du
2
_
du
dt
_
2
+
dr
du
d
2
u
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
dt
(a a) = a
da
dt
+
da
dt
a = 2a
da
dt
.
But (a a) = a
2
= const.
So
d
dt
(a a) = 0 2a
da
dt
= 0 (QED)
Integration of a vector function 3.9
As with scalars, integration of a vector function of a single scalar variable is the reverse of dierentiation.
In other words
_
p
2
p
1
_
da(p)
dp
_
dp = a(p
2
) a(p
1
)
Eg, from dynamics-ville
_
t
2
t
1
a dt = v(t
2
) v(t
1
)
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 elds.
See lecture 4!
Geometrical interpretation of derivatives 3.10
Position vector r(p) traces a space curve.
Vector r is a secant to the curve
r/p lies in the same direction as r(p)
Take limit as p 0
dr/dp is a tangent to the space curve
r
r
(p) r
p) (p +
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

dr
ds

= 1
and
dr/ds is a unit tangent to r at s
For s arc-length and p some other parametrization, we have
dr
dp
=
dr
ds
ds
dp
and

dr
dp

dr
ds

ds
dp
=
ds
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.
ds
1
r
r
r
dr
(s +
r (s)
r
r s)
(t)
(t + t)
r d
dt
ds
dt
SPEED
Example 3.13
Question: Draw the curve
r = a cos(
s

a
2
+ h
2
) + a sin(
s

a
2
+ h
2
) +
hs

a
2
+ h
2

k
where s is arc length and h, a are constants.
Answer
Example ctd 3.14
r = a cos(
s

a
2
+ h
2
) + a sin(
s

a
2
+ h
2
) +
hs

a
2
+ h
2

k
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
dr
ds
=
a

a
2
+ h
2
sin () +
a

a
2
+ h
2
cos () +
h

a
2
+ h
2

k
Projection on the xy plane has magnitude a/

a
2
+ h
2
Projection in the z direction h/

a
2
+ h
2
So the elevation angle is tan
1
(h/a), a constant.
We are expecting |dr/ds| = 1, and indeed it is!
2 2
a/ a + h
d
ds
r
2
h/ a + h
2
x
y
z
Length
Length
Why cant 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 =
_
p
1
p
0

dr
dp

dp .
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
ds
dp
=

_
dx
dp
_
2
+
_
dy
dp
_
2
+
_
dz
dp
_
2
.
z
y
x

y
z
x

s
Arc length is special /ctd 3.16
Suppose we had parameterized our helix as
r = a cos p + a sin p + hp

k
p is not arc length because

dr
dp

_
dx
dp
_
2
+
_
dy
dp
_
2
+
_
dz
dp
_
2
=
_
a
2
sin
2
p + a
2
cos
2
p + h
2
=
_
a
2
+ h
2
= 1
So if we want to parameterize in terms of arclength, replace p with s/

a
2
+ h
2
.
Curves in 3D 3.17
Lets 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 xed in an external reference frame.
r
(s)
O
Curves in 3D 3.18
Rollercoaster will help you see
whats going on ...
But it has a specially shaped
rail or two rails that dene the
twists and turns.
We are thinking about a 3D curve just a 3D wire.
Does the curve itself dene its own twist and turns?
The Frenet-Serret Local Coordinates 3.19
Yes: method due to French mathematicians F-J. Frenet and J. A. Serret
1. Unit tangent

t Obvious choice is

t = dr(s)/ds
2. Principal Normal n
Proved earlier that if |a(t)| = const then a da/dt = 0. So

t =

t(s), |

t| = const

t d

t/ds = 0
Hence the principal normal n is dened from
n = d

t/ds
where 0 is the curves curvature.
n
t
d
ds
t
s increasing
3. The Binormal

b
The third member of a local r-h set is the binormal,

b =

t n .
Deriving the Frenet-Serret relationships 3.20
Tangent

t, Normal n : d

t/ds = n, Binormal

b =

t n
Since

b

t = 0, if we dierentiate wrt s ...

d

b
ds

t +

b
d

t
ds
=
d

b
ds

t +

b n = 0
from which
d

b
ds

t = 0.
This means that d

b/ds is along the direction of n:

d

b
ds
= (s) n(s)
where is the space curves torsion.
Deriving the Frenet-Serret relationships 3.21
Tangent

t, Normal n, Binormal

b =

t n
d

t/ds = n, d

b/ds = (s) n(s)

Dierentiating n

t = 0:
(d n/ds)

t + n (d

t/ds) = 0
(d n/ds)

t + n n = 0
(d n/ds)

t =
Now do the same to n

b = 0:
(d n/ds)

b + n (d

b/ds) = 0
(d n/ds)

b + n () n = 0
(d n/ds)

b = +
HENCE
d n
ds
= (s)

t(s) + (s)

b(s).
Summary of the Frenet-Serret relationships 3.22
These three expressions are called the Frenet-Serret relationships:
d

t/ds = n
d n/ds = (s)

t(s) + (s)

b(s)
d

b/ds = (s) 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 (s) and (s) for the curve
r(s) = a cos (s/) + a sin (s/) + h (s/)

k
where =

a
2
+ h
2
Answer:
Denote sin, cos(s/) as S and C.
We found the unit tangent earlier as

t = (dr/ds) = [(a/) S, (a/) C, (h/)] .

Hence
n =
_
d

t/ds
_
=
_

_
a/
2
_
C,
_
a/
2
_
S, 0

The curvature must be positive, so

=
_
a/
2
_
n = [C, S, 0] .
So the curvature is constant, and n parallel to the xy-plane.
Example /continued 3.24
Recall

t = [(a/) S, (a/) C, (h/)] n = [C, S, 0] .

So the binormal is

b =

t n =

k
(a/)S (a/)C (h/)
C S 0

=
__
h

_
S,
_
h

_
C,
_
a

__
Hence
_
d

b/ds
_
=
__
h/
2
_
C,
_
h/
2
_
S, 0

=
_
h/
2
_
n
So the torsion
=
_
h/
2
_
again a constant.
Derivative (eg velocity) components in plane polars 3.25
In plane polar coordinates, the radius vector of any point P is given by
r = r (cos + sin ) = re
r
where we have introduced the unit radial vector
e
r
= cos + sin .
The other natural unit vector in plane polars is orthogonal to e
r
and
is
e

= sin + cos
so that e
r
e
r
= e

= 1 and e
r
e

= 0.
e
r
e

r
P
Aside: notation 3.26
Some texts will use the notation
r,

to denote unit vectors in the radial and tangential directions

I prefer the more general notation
e
r
, e

(as used in, eg, Riley).

You should be familiar and comfortable with either
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 =
d
dt
(re
r
) =
dr
dt
e
r
+ r
de
r
dt
=
dr
dt
e
r
+ r
d
dt
(sin + cos )
=
dr
dt
e
r
+ r
d
dt
e

= radial + tangential
e
r
e

r
P
Note that
de
r
dt
=
d
dt
e

de

dt
=
d
dt
(sin + cos ) =
d
dt
e
r
Acceleration components in plane polars 3.28
Recap ...
r =
dr
dt
e
r
+ r
d
dt
e

;
de
r
dt
=
d
dt
e

;
de

dt
=
d
dt
e
r
Dierentiating r gives the accel. of P
r =
d
2
r
dt
2
e
r
+
dr
dt
d
dt
e

+
dr
dt
d
dt
e

+ r
d
2

dt
2
e

r
d
dt
d
dt
e
r
=
_
d
2
r
dt
2
r
_
d
dt
_
2
_
e
r
+
_
2
dr
dt
d
dt
+ r
d
2

dt
2
_
e

Acceleration components in plane polars 3.29

We just saw
r =
_
d
2
r
dt
2
r
_
d
dt
_
2
_
e
r
+
_
2
dr
dt
d
dt
+ r
d
2

dt
2
_
e

Three obvious cases:

const : r =
d
2
r
dt
2
e
r
r const : r = r
_
d
dt
_
2
e
r
+ r
d
2

dt
2
e

r and d/dt const : r = r

_
d
dt
_
2
e
r
Fixed, varying, and instrinsic coordinates 3.30
Rotating systems 3.31
Body rotates with constant about axis passing
through the body origin.
Assume the body origin is xed.
We observe from a xed coord system Oxyz.

If is a vector of constant mag and constant direction in the rotating system, then in the xed system it must be
a function of t.
r(t) = R(t)
dr
dt
=

R =

RR

r
* dr/dt will have xed 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 xed system.
Rotating systems 3.32
Consider the term

RR

Note that RR

= I, hence

RR

+ R

= 0

RR

= R

Thus

RR

is anti-symmetric:

RR

=
_
_
0 z y
z 0 x
y x 0
_
_
Application of a matrix of this form to an arbitrary vector has precisely the same eect as the cross product
operator, , where = [xyz]

.
Thus
r = r
Rotating co-ordinate systems 2 3.33
Now 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)(t)
Dierentiating with respect to time:
dr
dt
=

R + R =

RR

r + R
The instantaneous velocity of P in the xed frame
is
dr
dt
= R + r

P at t+
r=
at t
r
P at t
t

(
r) t
Second term is contribution from the rotating frame
First term is linear velocity in the rotating frame, referred to the xed frame
Rotating co-ordinate systems 3.34
Now consider second dierential:
r = r + r +

R + R
If angular velocity constant, rst term is zero
Now substituting for r we have
r = ( r + R ) +

R + R
= ( r) + R +

RR

R + R
= ( r) + R + R + R
= ( r) + 2 (R ) + R
The instantaneous acceleration is therefore
r = R + 2 (R ) + ( r)
Rotating co-ordinate systems 3.35
The instantaneous acceleration is
r = R + 2 (R ) + ( r)
* Term 1 is Ps acceleration in the rotating frame.
* Term 3 is the centripetal accel: magnitude
2
r 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 xed frame of a projectile red along a line of longitude (with
angular velocity of constant relative to the sphere) if the sphere is rotating with angular velocity .
A In the rotating frame
=
=
= ( )
In xed frame, instantaneous acceleration:
r = ( r) + 2 ( r) + ( r)
In rotating frm + Coriolis + Centripetal
r
t
= m
m
n

Example /ctd 3.37

Repeated: r = ( r) + 2 ( r) + ( r)
1) As =

, = Rcos(t) m+Rsin(t) n acceleration in rotating frame

is
( ) =
2
r
2) Centripetal accel due to rotation of sphere is
( r) =
2
Rsin(t) n
r
t
= m
m
n

3) The Coriolis acceleration is

2 = 2
_
_
0

0
_
_

_
_
_
_

0
0
_
_

_
_
0
Rcos(t)
Rsin(t)
_
_
_
_
= 2Rcos(t)

Example /ctd 3.38

Recap:
Accel in rotating frame
2
r
Centripetal due to sphere rotating
2
Rsin(t) n
Coriolis acceleration: 2Rcos(t)

l
2Rcos(t)

2
Rsin(t) n

2
r
r
r
t
= m
m
n

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 Rcos(t)

+ve -ve +ve +ve

Coriolis force is in the direction opposed to

(i.e. opposing earths rotation).
(NB instantaneously common to earths surface and rocket)
Tangential component of velocity
Rockets velocity in direction of meridian
Tangential velocity of earths surface
Coriolis acceleration 3.40
Because of the rotation of the earth, the Coriolis acceleration is of great importance in meteorology
Coriolis acceleration 3.41
Summary 3.42
We started by dierentiating vectors wrt to a xed 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 Frenet-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.