Vector Calculus ReviewSects. 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 & AccelerationSect. 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

• Start with Position: r = r er

• 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

Scalar!

Computing velocity & acceleration is now tedious!

Cylindrical CoordinatesResults Summary

Spherical CoordinatesResults 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 CoordinatesResults 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!)