Vector Calculus Review Sects. 1. 13, 1. 14. Overview only 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

a = (dv/dt) = d(r er + r θeθ)/dt Computing velocity & acceleration is now tedious! 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! 

Cylindrical Coordinates Results Summary

Spherical 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 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!)