Vector Calculus Preliminaries

picture to proof at the drop of a hat

S. Gill Williamson

1

2

pythagorean theorem

(a + b)2 = 4((1/2)(ab) + c2 =⇒ a2 + 2ab + b2 = 2ab + c2

=⇒ a2 + b2 = c2

law of cosines

(a sinx)2 + (b − a cosx)2 = c2

(a sinx)2 + (a cosx)2 + b2 − 2ab cosx = c2

a2 + b2 − 2ab cosx = c2

cosine of x − y

12 + 12 − 2 · 1 · 1 cos(x − y) = | |(cos(x), sin(x)) − (cos(y), sin(y))| |2

2 − 2 cos(x − y) = (cos(x) − cos(y))2 + (sin(x) − sin(y))2

2 − 2 cos(x − y) = 2 − 2(cos(x) cos(y) + sin(x) sin(y)) =⇒

(1) cos(x − y) = cos(x) cos(y) + sin(x) sin(y) [y B −y] =⇒

(2) cos(x + y) = cos(x) cos(y) − sin(x) sin(y) [y B y − π/2] =⇒

(4) sin(x + y) = cos(x) sin(y) + sin(x) cos(y) [y B −y] =⇒

(3) sin(x − y) = − cos(x) sin(y) + sin(x) cos(y)

From these, ei(x+y) = eixeiy .

dot product de�nitions

|A − B |2 = |A|2 + |B |2 − 2|A| |B | cos(x)

(1)n∑i=1(ai − bi )

2 =

n∑i=1

a2i +n∑i=1

b2i − 2|A| |B | cos(x)

(2)n∑i=1(ai − bi )

2 =

n∑i=1

a2i +n∑i=1

b2i − 2n∑i=1

aibi

=⇒

n∑i=1

aibi = |A| |B | cos(x)

3

determinants and volumes

Let A ∈ M3(R) where A = (A(1),A(2),A(3)) in terms of row vectors.We show det(A) = ±vol((A(1),A(2),A(3)). See �gure—>Let e1, e2, e3 be an orthonormal basis for R3.Assume w.l.o.g. that A1 = |A1 |e1, A2 = c1e1 + |Q |e2, andA3 = d1e1 + d2e2 + |P |e3.

Thus, A =©­«|A1 | 0 0c1 |Q | 0d1 d2 |P |

ª®¬ and det(A) = |A1 | |Q | |P |,

the volume of the parallelepiped de�ned by A1, A2, A3,its sign + if RHS, - if LHS.NOTE: If (f1, f2, f3) is any o.n. basis it transforms to (e1, e2, e3) by an orthogonal matrix, det 1.

orthogonal matrices and bases e = (e1, e2, e3), f = (f1, f2, f3)

A = [I ]fe =

©­«e1 · f1 e2 · f1 e3 · f1e1 · f2 e2 · f2 e3 · f2e1 · f3 e2 · f3 e3 · f3

ª®¬ satis�es AT = A−1 (orthogonal).

A geometric algorithm involves rotations about a �xed axis.T (e1) = e1, T (e2) = e ′2, T (e3) = e ′3 =⇒

A = [T ]ee =©­«1 0 00 cos(x) − sin(x)0 sin(x) cos(x)

ª®¬ where A is orthogonal.

We call such a rotation in R3 a simple rotation.Indicate the plane generated by f1, f2 by P =< f1, f2 >.We transform basis e into f with three simple rotations.

e1 e2 e3

e1 e ′2 e ′3

e ′1 e ′2 f3

f1 f2 f3

Rotate e system about e1 until e2 ∈ P .Call the rotated e2 e ′2 ∈ P .Rotate the new system about e ′2 ∈ Puntil e1 ∈ P . Call the rotated e1, e

′1 ∈ P .

Since e ′1 ∈ P and e ′2 ∈ P , the third vector is f3.Now a simple rotation about f3 yields f .

4

cross productA,B,C ∈ R3, n a unit vector, E = (e1, e2, e3) an o.n.basis.Cross product def: A × B B area(A,B)n where n ⊥< A,B > (RHS).Note area(A,B) = |A| |B | sin(x) so A × B = |A| |B | sin(x)n.Symbolically, let DET(A,B,E) be the determinant of a matrix withrows A,B,E, where E = (e1, e2, e3);

DET (A,B,E) =

������a1 a2 a3b1 b2 b3e1 e2 e3

������ = (a2b3 − a3b2)e1 − (a1b3 − a3b1)e2 + (a1b2 − a2b1)e3.Note DET(A,B,E) · A = det(A,B,A) = 0 and DET(A,B,E) · B = det(A,B,B) = 0.Thus, DET(A,B,E) is perpendicular to the plane P =< A,B >, and(A × B) · n = area(A,B) = vol(A,B,n) = det(A,B,n) = DET(A,B,E) · n implies(A × B) = DET(A,B,E).

cross product DET rulesThese rules follow from the DET(A,B,E) = A × B formulation:(1) (A + B) ×C = A ×C + B ×C

(2) A × (B +C) = A × B +A ×C

(3) A × sB = sA × B = s(A × B)(4) (A × B) ·C = DET(A,B,E) ·C = det(A,B,C) := [A,B,C] (the triple product)(6) (A × B) ×C , A × (B ×C) in general

cross product hard rules(1) A × (B ×C) = (A ·C)B − (A · B)C . Note A × (B ×C) = KBB − KCC for some KB , KC is clear.(2) (A × B) ×C = (A ·C)B − (B ·C)A. From (1) use A × (B ×C) = −(B ×C) ×A.Assume wlog E = (e1, e2, e3) is such that A = α1e1, B = β1e1 + β2e2, C = γ1e1 + γ2e2 + γ3e3.

A × B =

������α1 0 0β1 β2 0e1 e2 e3

������ = α1β2e3 and (A × B) ×C =

������ 0 0 α1β2γ1 γ2 γ3e1 e2 e3.

������ = α1β2(γ1e2 − γ2e1).Computing, (A ·C)B − (B ·C)A = (α1γ1)(β1e1 + β2e2) − (β1γ1 + β2γ2)α1e1 = α1β2(γ1e2 − γ2e1).Alternative proof of (1): Let д(A,B,C) = (A ·C)B − (A · B)C and f (A,B,C) = A × (B ×C).Note f ,д ∈ M3(×

3R3 : R3), 3-linear functions from ×3R3 to R3. We have f (A,C,B) = −f (A,B,C)

and д(A,C,B) = −д(A,B,C). A basis for ×3R3 is {eγ | γ ∈ 33}, where eγ = (eγ (1), eγ (2), eγ (3)).Show f = д on this basis. Note f (eγ ) = д(eγ ) = 0 if |{γ (1),γ (2),γ (3)}| = 3 or |{eγ (2), eγ (3)}| = 1.Note f (e1, e1, e2) = e1 × (e1 × e2) = e1 × e3 = −e2 and д(e1, e1, e2) = (e1 · e2)e1 − (e1 · e1)e2 = −e2.All other cases are basically the same.

5

cosine law sphereA spherical triangle is de�ned by unit vectors A,B,C. A is thetangential angle, a the arc angle (e.g. B · C = cos(a)).Let A = (0, 0, 1),B = (sin(c), 0, cos(c)),C = (sin(b) cos(A), sin(b) sin(A), cos(b)). From B · C = cos(a)(1) cos(a) = cos(A) sin(b) sin(c) + cos(b) cos(c).

(2) cos(A) =cos(a) − cos(b) cos(c)

sin(b) sin(c)(a law of cosines).

sin2(A) = 1 − cos2(A) = 1 −(cos(a) − cos(c) cos(b)

sin(b) sin(c)

)2sin2(A)sin2(a)

=sin2(b) sin2(c) − (cos(a) − cos(b) cos(c))2

sin2(a) sin2(b) sin2(c)sin2(A)sin2(a)

=(1 − cos2(b))(1 − cos2(c)) − (cos(a) − cos(b) cos(c))2

sin2(a) sin2(b) sin2(c)

(3)sin2(A)sin2(a)

=

(1 − cos2(a) − cos2(b) − cos2(c) + 2 cos(a) cos(b) cos(c)

)2sin2(a) sin2(b) sin2(c)

But (3) symmetric in a, b, c implies

(4)sin(A)sin(a)

=sin(B)sin(b)

=sin(C)sin(c)

for 0 < a, A < π , the law of sines.

Assume A is north pole vector and latitudes and longitudes are given.(5) cos(a) = cos(A) sin(b) sin(c) + cos(b) cos(c) (from (2))Take lat(B) = 45N , lon(B) = 120W (OR-WA border).Take lat(C) = 30N , lon(C) = 75W (east of FL coast).Find the length, a, of a great circle route from B to C.A = c = 45, b = 60, cos(45) = sin(45) = 0.707, sin(60) = 0.866(7) cos(a) = (0.707)(0.866)(0.707) + (0.5)(0.707) = 0.786 from (5).Thus, arccos(0.786) = 0.67 radians.Take RE = 3959 mi. so a = 0.67 × 3959 = 2653 mi.

POLAR TRIANGLE —->

6

curvature and arclengthLet R(t) = (x(t),y(t), z(t)) with T (t) a unit tangent.|∆R | ≈ ∆s (arclength change). Note, ∃ρt 3 ρt |∆T | ≈ ∆s .Thus, |dR | = ds (arclength di�erential) and ∃ρt 3 ρt |dT | = ds .The variable ρt is the radius of curvature at t .∆R is approximately tangent to R(t) as is ∆R

∆t .Thus, dR

dt is a tangent to R(t) and dR/dt|dR/dt | is unit tangent.

If t = s (arclength) then |dR/dt | = |dR/ds | = 1. Thus, dRds is a unit tangent.

Since T (t) is a unit tangent, ddtT (t) is normal to T (t) or parallel to N (see �gure).

For t = s , ddsT (s) = κsN where κs > 0. But |dT (s) |ds =

|dT (s) |ρs |dT (s) |

= κs . Hence, κs = 1/ρs .Apply this to acceleration and circular paths:

Recall that ddt

dRdt is the acceleration if t is time. Vector velocity is V (t) = dR

dt =dsdt

dRds =

dsdtT .

(1)d2R

dt2=

d

dt

(ds

dtT

)=d2s

dt2T +

ds

dt

dT

dt=d2s

dt2T +

ds

dt

dT

ds

ds

dt=d2s

dt2T + (

ds

dt)2κsN .

Apply (1) to a circle of radius r , V (t) = vT = rωT , v the constant speed, ω angular velocity:

(2) A(t) =d

dtvT = v

ds

dt

dT

ds= v2κsN

a curvature computation:

R(x ,y, z) = (sin t − t cos t , cos t + t sin t , t2)(3) R′(x ,y, z) = (t sin t , t cos t , 2t) and ds/dt = |R′ | = 51/2t so d2s/dt2 = 51/2.(4) R′′(x ,y, z) = (sin t + t cos t , cos t − t sin t , 2) and |R′′(t)| = (5 + t2)1/2.From (1) R′′(x ,y, z) = d2R

dt 2 =d2sdt 2T + (

dsdt )

2κsN . Thus, |R′′ |2 = (d2s

dt 2 )2 + (dsdt )

4κ2s .

Substituting values already computed gives 5 + t2 = 5 + 25t4κ2s , κs (t) = 1/5t .In general, let R = (x(t),y(t), z(t)). Compute R′(t) and R′′(t). Note |R′ | = ds/dt and compute d2s

ds2 .We have |R′′ |2 = (x ′′)2 + (y ′′)2 + (z ′′)2 = (d

2sdt 2 )

2 + (dsdt )4κ2s . Solve for κ.

polar coordinatesRecall polar coordinates (r ,θ ) ∈ R2, (r ,θ ) → (r cos(θ ), r sin(θ )) the correspondence is not bijective.Let ur , uθ be the unit vectors. Let R(t) = r (t)ur (θ (t)), (t time). Note dur /dθ = uθ , duθ /dθ = −ur .Let dR/dt B ÛR so ÛR = Ûrur + r Ûur = Ûrur + r (dur /dθ ) Ûθ = Ûrur + ruθ Ûθ . Thus,(1) ÛR = Ûrur + r Ûθuθ .Also, ÜR = Ürur + Ûr Ûθuθ + Ûr Ûθuθ + r Üθuθ + r Ûθ Ûθ (−ur ). Thus,(2) ÜR =

(Ür − r Ûθ 2

)ur +

(2Ûr Ûθ + r Üθ

)uθ .

For circular motion, r is constant and v = r Ûθ :(3) ÛR = vuθ , ÜR = −v2/rur + r Üθuθ .

7

spherical coordinatesspherical coordinates

@

@r

ur

u�

u�

0

0

0

@

@�

ur

u�

u�

u�

�ur

0

@

@�

ur

u�

u�

sin(� )u�

cos(� )u�

�up

8

The vectors {ut | t = x ,y, z,p, r ,θ ,ϕ} are unit vectors.De�ne P to be the projection of ur onto plane < ux ,uy > .

Thus, up ∈< ux ,uy > ∩ < uθ ,ur >.(1) up = cos(ϕ)ux + sin(ϕ)uy in the plane < ux ,up ,uy >.(2) up = cos(θ )uθ + sin(θ )ur in the plane < uθ ,up ,ur ,uz >.(3) ur = sin(θ ) cos(ϕ)ux + sin(θ ) sin(ϕ)uy + cos(θ )uz . Given(x ,y, z),x = r sin(θ ) cos(ϕ), y = r sin(θ ) sin(ϕ), z = r cos(θ ).(4) Note that ∂

∂r ut = 0 for {ut | t = x ,y, z,p, r ,θ ,ϕ}.(5) θ = arctan[(x2 + y2)1/2, z].(6) ϕ = arctan(y,x).

8

partial derivatives commuteLemma: Let f : D ! R be C2, D an open set in R

2. Let [0,a]2 ⇢ D.

There exists {(s1, s2), (t1, t2)} ⇢ (0,a)2, fx� (s1, s2) = f�x (t1, t2) =a

�2(f (a,a) � f (a, 0) � f (0,a) + f (0, 0)).Proof: Apply mean value theorem to f (x ,a) � f (x , 0) : [0,a] ! R.There exists s1,

(1) fx (s1,a) � fx (s1, 0) = a

�1 [f (a,a) � f (a, 0) � f (0,a) + f (0, 0)] .

Apply mean value theorem to fx (s1,�) � fx (s1, 0) : [0,a] ! R.There exists s2,

(2) fx� (s1, s2) = a

�1 [fx (s1,a) � fx (s1, 0)].

Using the expression for fx� (s1,a) � fx� (s1, 0) in (1) gives

(3) fx� (s1, s2) = a

�2 [f (a,a) � f (a, 0) � f (0,a) + f (0, 0)] .

Instead of using f (x ,a) � f (x , 0) : [0,a] ! R, use f (a,�) � f (a, 0) : [0,a] ! R.Thus, there exists (t1, t2),

(4) f�x (t1, t2) = a

�2 [f (a,a) � f (0,a) � f (a, 0) + f (0, 0)] = fx� (s1, s2). QED

If a ! 0, continuity implies fx� (0, 0) = f�x (0, 0).Thus, f�x (u,�) = f�x (u,�) for any (u,�) 2 D (transform problem to case (u,�) = (0, 0)).

�

�

�(s1, s2)

f (s1,a)

f (s1, 0)

9

partial derivatives commute

Lemma: Let f : D → R be C2, D an open set in R2. Let [0,a]2 ⊂ D.

There exists {(s1, s2), (t1, t2)} ⊂ (0,a)2, fxy (s1, s2) = fyx (t1, t2) =

a−2(f (a,a) − f (a, 0) − f (0,a) + f (0, 0)).Proof: Apply mean value theorem to f (x ,a) − f (x , 0) : [0,a] → R.There exists s1,

(1) fx (s1,a) − fx (s1, 0) = a−1 [f (a,a) − f (a, 0) − f (0,a) + f (0, 0)] .

Apply mean value theorem to fx (s1,y) − fx (s1, 0) : [0,a] → R.There exists s2,

(2) fxy (s1, s2) = a−1 [fx (s1,a) − fx (s1, 0)].

Using the expression for fxy (s1,a) − fxy (s1, 0) in (1) gives

(3) fxy (s1, s2) = a−2 [f (a,a) − f (a, 0) − f (0,a) + f (0, 0)] .

Instead of using f (x ,a) − f (x , 0) : [0,a] → R, use f (a,y) − f (a, 0) : [0,a] → R.Thus, there exists (t1, t2),

(4) fyx (t1, t2) = a−2 [f (a,a) − f (0,a) − f (a, 0) + f (0, 0)] = fxy (s1, s2). QED

If a → 0, continuity implies fxy (0, 0) = fyx (0, 0).Thus, fyx (u,v) = fyx (u,v) for any (u,v) ∈ D (transform problem to case (u,v) = (0, 0)).

polar circle arclengthConsider velocity and acceleration for a circle, radius ρ, parameterized by arclength.

(1) ur = (cos(s/ρ), sin(s/ρ)) and R(s) = ρur = ρ(cos(s/ρ), sin(s/ρ)).

(2) T (s) = uθ (s) =d

dsur = (− sin(s/ρ), cos(s/ρ)).

(3)d

dsT =

1ρ(− cos(s/ρ),− sin(s/ρ)) =

1ρ(−ur ) =

1ρN = κN .

Thus, the curvature k = 1ρ is constant and −ur = N , the normal.

9

gradient basics Let f (x ,y, z) = ax + by + cz (intuition).

Let f : R3 → R and de�ne ∇f = ∂f∂xux +

∂f∂xuy +

∂f∂xuy .

Consider curve c(s) = (c1(s), c2(s), c3(s)), s ∈ (a,b).

(1)d

dsf (c(s)) =

∂ f

∂x

dc1ds+∂ f

∂y

dc2ds+∂ f

∂z

dc3ds= ∇f ·

dcds

.

If c(s) = w + su then dds f (c(s)) is the directional derivative

of f at w in the direction of u (w, u ∈ R3).Suppose c(s) lies in a surface S = {v | f (v) = c0}.

From (1), s ∈ (a,b) implies ∇f ·dcds= 0.

Thus, for w ∈ S , ∇f (w) is perpendicular to all curves in S that pass through w.Thus, ∇f (w) is perpendicular to the surface S at w and to the tangent plane of S at w.In general, let r̄ = (x ,y, z) be an arbitrary “position vector.” Then

(2) d f =∂ f

∂xdx +

∂ f

∂ydy +

∂ f

∂zdz = ∇f · d r̄ where d r̄ = (dx ,dy,dz).

gradient spherical and cylindrical

Consider f (r ,θ ,ϕ) , f : R3 → R , d f = ∂f∂r dr +

∂f∂θ dθ +

∂f∂ϕdϕ. Let r̄ = rur .

Analogous to gradient basics (2), de�ne the spherical gradient ∇f such that d f = ∇f · d r̄.

(3) d r̄ =∂r̄∂r

dr +∂r̄∂θ

dθ +∂r̄∂ϕ

dϕ =∂(rur )

∂rdr +

∂(rur )

∂θdθ +

∂(rur )

∂ϕdϕ = dr ur + rdθ uθ + r sin(θ )dϕ uϕ .

(4) ∇f =∂ f

∂rur +

1r

∂ f

∂θuθ +

1r sin(θ )

∂ f

∂ϕuϕ .

Similarly, de�ne the cylindrical gradient:Consider f (ρ,ϕ, z) with d f =

∂f∂ρdr +

∂f∂ϕdϕ +

∂f∂z dz.

Let r̄ = ρuρ + zuz so

(5) d r̄ =∂r̄∂ρ

dρ +∂r̄∂ϕ

dϕ +∂r̄∂z

dz = dρ uρ + ρdϕ uϕ + dz uz .

We choose the cylindrical gradient, ∇f , such that d f = ∇f · d r̄.

(6) ∇f =∂ f

∂ρuρ +

1ρ

∂ f

∂ϕuϕ +

∂ f

∂zuz .

euclidean example: f (x ,y, z) = x2 + y2 (cylindrical isotimic surfaces).∀K > 0, ∇f = (2x , 2y, 0) ⊥ {(x ,y, z) | f (x ,y, z) = K}.

spherical example: f (r ,θ ,ϕ) = r sin(θ ) (cylindrical isotimic surfaces).∀K > 0, ∇f = (sin(θ ), cos(θ ), 0) ⊥ {(r ,θ ,ϕ) | f (r ,θ ,ϕ) = K}.

cylindrical example: f (ρ,ϕ, z) = ρ (cylindrical isotimic surfaces)∀K > 0, ∇f = (1, 0, 0) ⊥ {(ρ,ϕ, z) | f (ρ,ϕ, z) = K}.

10

vector �elds and �ows

Let F : R3 → R3. Let r̄ : R→ R3 be a curve, {r̄(t) : t ∈ (a,b)}, e.g., t time. Let r̄ : (a,b) → R.r̄ is a �ow of F if for all t ∈ (a,b), d r̄

dt = F (r̄(t)). Recall the velocity d r̄dt is tangent to the curve r̄(t).

If F (x ,y, z) = (F1(x ,y, z), F2(x ,y, z), F3(x ,y, z)) and r̄(t) = (x(t),y(t), z(t)) then

(1) x ′(t) = F1(x(t),y(t), z(t)) y′(t) = F2(x(t),y(t), z(t)) z

′(t) = F3(x(t),y(t), z(t))

is a system of di�erential equations related to the construction of �ows for F .The general case is di�cult analytically but trivial examples can be constructed:Let r̄(t) = (sin(t),− cos(t), et ) so r̄′(t) = (cos(t), sin(t), et ) is a �ow for F (x ,y, z) = (−y,x , z).

di�erential operators

De�ne (using h ∈ RD for h : D → R) the gradient operator, ∇ : RR3→ (R3)R

3 for f ∈ RR3 by

(1) (∇f )(x ,y, z) =∂ f

∂x(x ,y, z)ux +

∂ f

∂y(x ,y, z)uy +

∂ f

∂z(x ,y, z)uz .

Symbolically: ∇ B ( ∂∂x ,

∂∂y ,

∂∂z ), ∇f = (

∂∂x f ,

∂∂y f ,

∂∂z f ) (scalar multiplication by f )

De�ne the divergence operator ∇· : (R3)R3→ RR3 where F ∈ (R3)R

3 ,F (x ,y, z) = F1(x ,y, z)ux + F2(x ,y, z)uy + F3(x ,y, z)uz , by

(2) ∇ · F =∂F1(x ,y, z)

∂x+∂F2(x ,y, z)

∂y+∂F3(x ,y, z)

∂z.

Symbolically: ∇ · F = ∇ · (F1, F2, F3) = ( ∂∂x ,

∂∂y ,

∂∂z ) · (F1, F2, F2) =

∂∂x F1 +

∂∂y F2 +

∂∂z F3

De�ne the curl operator ∇× : (R3)R3→ (R3)R

3 where F ∈ (R3)R3 ,

F (x ,y, z) = F1(x ,y, z)ux + F2(x ,y, z)uy + F3(x ,y, z)uz , by

(3)(∂

∂yF3 −

∂

∂zF2

)ux +

(∂

∂zF1 −

∂

∂xF3

)uy +

(∂

∂xF2 −

∂

∂yF1

)uz .

Symbolically:

∇ × F =

������∂∂x

∂∂y

∂∂z

F1 F2 F3ux uy uz

������ =(∂

∂yF3 −

∂

∂zF2

)ux +

(∂

∂zF1 −

∂

∂xF3

)uy +

(∂

∂xF2 −

∂

∂yF1

)uz .

De�ne the Laplacian operator ∇2 : RR3→ RR3 for f ∈ RR3 by

(4) (∇2 f )(x ,y, z) =∂2 f

∂x2(x ,y, z) +

∂2 f

∂y2(x ,y, z) +

∂2 f

∂z2(x ,y, z).

Symbolically: ∇2 f = ∇ · ∇f = ( ∂∂x ,

∂∂y ,

∂∂z ) · (

∂∂x f ,

∂∂y f ,

∂∂z f ).

Note: ∇ × (∇f ) = Θ (zero vector in R3) and ∇ · (∇ × F ) = 0.

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Indexcosine law example, 6cosine law sphere, 6cosine of x − y, 3cross product, 5cross product DET rules, 5cross product hard rules, 5curvature and acceleration, 7curvature and arclength, 7curvature computation, 7

determinants volumes, 4dot product defs, 3

gradientbasics, 10cylindrical coordinates, 10spherical coordinates, 10

law of cosines, 3

mean value theorem, 9

orthogonal matrices, bases, 4

partial derivativescommute, 9

polar circlearclength parameter, 9

polar coordinates, 7polar triangle, 6pythagorean theorem, 3

sine law sphere, 6spherical coordinates

de�ned, 8derivatives, 8

vector �elds F : R3 → R3

domain of divergence ∇·, 11domain and range of curl of∇×, 11range of gradient ∇, 11�ows, 11

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