Vector Calculus Review
Sects. 1.13, 1.14. Overview only. For details, see text!
Differentiation of a Vector with
Respect to a Scalar
– Components of vectors are scalars
 Differentiate a vector component by component.
Derivatives of components are scalars.
– The text proves: The derivative of vector A with
respect to scalar s is a vector:
dA/ds transforms as a vector under orthogonal
coordinate transformations.
Differentiation of Vector with Respect to a Scalar
• Some straightforward identities:
Examples: Velocity & Acceleration
Sect. 1.14
• For the dynamics of point particles (much of this
course!): We use vectors to represent position, velocity
& acceleration, often as functions of time t (a scalar).
• Notation (bold are vectors, without writing the
arrow above):
Position: r(t)
Velocity: v(t)  dr/dt  r
Acceleration: a(t)  dv/dt  d2r/dt2  r
• In Cartesian (rectangular) coordinates:
Position:
r = ∑i xi ei
Velocity:
v = r = ∑i (dxi/dt)ei
Acceleration: a = v = r = ∑i (d2xi/dt2)ei
• This is straightforward in Cartesian
coordinates because the unit vectors ei are
constant in time!
• This is not necessarily so in other coordinate
systems! Taking time derivatives can be messy
because of time dependent unit vectors!
• In non-rectangular coordinate systems: Unit
vectors at the particle position, as it moves
through space, aren’t necessarily constant in t!
 The components of the time derivatives of
position r can be complicated!
• We’ll look at these (briefly) in detail for
cylindrical coordinates (where the xy plane part is
plane polar coordinates) & spherical coordinates.
• Mostly we’ll show results only. For derivations, see
the text!
Cylindrical Coordinates
• In the xy plane, these are plane polar coordinates:
Caution!! In M&T notation, cylindrical coordinate angle 
, & plane polar coordinate angle  θ! (These are really the
same!) See Appendix F!
Coordinates:
x1= r cos, x2= r sin
x3= z, r = [(x1)2+(x2)2]½
 = tan-1(x2/ x1), z = x3
Unit Vectors: er = r/|r|
ez = k (z direction)
e ( direction). er, e, ez: A mutually orthogonal set!
er  e  ez , er  ez
Plane Polar Coordinates
• Consider xy plane motion only! See Fig.    θ
A particle moves along the curve
s(t). In time dt = t2-t1, it moves from
P(1) to P(2). As time passes r & θ
(& r) change, but always er  eθ.
From the figure:
 der= (er)(2) -(er)(1)  er (|| eθ)
or
der= dθ eθ
 deθ = (eθ)(2) - (eθ )(1)  eθ (|| er)
or
deθ = -dθ er
 (der/dt) = (dθ/dt)eθ,
(deθ/dt) = -(dθ/dt)er or
Computing velocity & acceleration is now tedious!
• Start with Position: r = r er
Scalar!
• Compute Velocity:

v = (dr/dt) = (dr/dt)er + r(der/dt)
Or:
v = r er + r er
Using
gives:
v = r er + r θeθ
• Compute Acceleration:
a = (dv/dt) = d(r er + r θeθ)/dt
Or (after manipulation; See Next Page!):
a = [r - r(θ)2]er + [rθ + 2rθ]eθ
Cylindrical Coordinates
Results Summary
Spherical Coordinates
Results Summary. Details left for student exercise!
Unit Vectors: er = r/|r|
eθ in θ direction
e in  direction
er, eθ, e
A mutually orthogonal set!
er  eθ, er  e, eθ  e
They remain  as time passes
& r, θ,  change
Spherical Coordinates
Results Summary. Details left for student exercise!
Position: r = r er
Velocity: v = (dr/dt)
= (dr/dt)er + r(der/dt)
Or: v = r er + r er = ??
Acceleration: a = (dv/dt)
= d(r er + r er)/dt = ??
Student Exercise!!!
(A mess!)
See Problem 25! (Solutions to be posted!)