Geometry and Calculus of Variations by
???
2011
Notes of Paul Boeck
Last changes: March 14, 2011
Contents
Planar Curve 2
Tangents of planar curves . . . 2
Implicitly Defined Curves . . . 3
Orientation . . . 3
Arc–length parametrization . . . 4
Frenet frame . . . 5
Rate of change ofT(t) . . . 5
The Rotation–Index . . . 6
Families of Curves . . . 7
Envelope of implicitly defined curves . . . 7
Calculus of Variations 8 Minimal area of surface of revolution . . . 8
Several variables . . . 9
Special Cases ofF− −L . . . 10
Applications . . . 10
Space Curves 10 Curvature for Space curves . . . 11
Structural Equations . . . 11
Local behavior of curves . . . 12
Implicit Space Curves . . . 12
Vector fields . . . 14
Arc length . . . 14
Isometry . . . 15
Index 16
PLANAR CURVE
Notations
• Rnn–dimensional Euclidean space. Usuallyn=2 orn=3.
• Points inRnx∈Rn:x=
x1
... xn
= x1 . . . xn⊤ x1. . .xnare calledcoordinatesofx.
• Inner product∀x,y∈Rnthe inner product is defined as(x,y) =x·y=x1y1+· · ·+xnyn
• Magnitudeofx∈Rnis|x|=√ x·x=
s n
∑
i=1
x2i
• Distanceofx,y∈Rnis|x−y|= s n
∑
i=1
(xi−yi)2
Planar Curve
Definition 1
• AcurveinR2is a smooth mapx:I→R2where I= (a,b), a,b∈R, I∋t→x(t) = x1(t)
x2(t)
• t is calledparameter
• Thetraceofxis the image setx(t)∈R2={z∈R2, st.∃t∈I:z=x(t)}
Remark • Smooth map means thatx1(t)andx2(t)are continuously differentiable as many times as we need.
Therefor the trace ofx(t)cannot have jumps.
Definition 2
• Atangentvector toR2is a pair(p,v)where p∈R2is the point of application andv∈R2is the vector acting at p.
• Equality of Tangent Vectors(p,v)and(q,w)is given iff
p=q and v=w
• For fixed p∈R2the set of all tangent vectors(p,v)is thetangent spacetoR2at p and denoted as TpR2.
• Vector fieldVonR2is a function what assigns a tangent vectorv(p) = (p,V(p))to p∈R2.
Tangents of planar curves
Definition 3
For a curvexinR2thevelocityofxat time t is the tangent vector x′(t) =
x′1(t) x′2(t)
at x(t)
Example 4 • x(t) = cost
sint
witht∈(0,π). By definition velocityx′(t) =
−sint cost
=
cos(t+π2) sin(t+π2)
. and we conclude thatx(t)andx′(t)are orthogonal.
Considerx·x′=cost(−sint) +sintcost=0.
• Straight line x(t) = a+tu =
a1+tu1
a2+tu2
, a is a fixed vector, u is a fixed unitvector, t ∈ R parameter.
x′(t) = u1
u2
=u.
Definition 5 (Speed)
Letx:I→R2, thespeedv:I→Rdefined as v(t) =|x′(t)| ≥0
Implicitly Defined Curves PLANAR CURVE
Definition 6 (Regular Curves) A curvex:I→R2is
• regularat t0if v(t0) =|x′(t0)| 6=0
• regularif v(t)6=0for all t∈ I.
Example 7 x(t) =
t3
|t|3
, t ∈ R. x2(t). x1(t) = t3smooth. x2(t) = |t|3 =
(t3 ift≥0
−t3 ift<0 =
(x1 ift≥0
−x1 ift<0. Therefore x2=|x1|and has no tangent at 0. The speed there is 0 and therefore the curve is not regular.
x(t) = t3
t2
. Again bothx1(t)andx2(t)are smooth. But the tangent again is not defined at the origin because the curve is not regular there.
Implicitly Defined Curves
Example 8 (i) circle: x2+y2=1(x,y)∈R2 (ii) Ellipse xa22 +yb22 =1,a,bnon–zero constants.
(iii) rectangular hyperbolay2−x2=1 All these curves have the formF(x,y) =0.
Definition 9
• Let f :R2→R. We say that p∈R2is aregular pointof f if∇f(p) =∂p∂f
1,∂p∂f
2
6=0
• Let C = {p ∈ R2 : f(p) =0}the0–level set of f . Assume that C 6= ∅. Then if f is regular at p∈ C, then C is a regular implicitly defined curve.
Example 10 f(x,y) =
(0 ifx2+y2≤1
x2+y2−1 ifx2+y2>1.C={p∈R2,f(p) =0}=unit disc.∇f =0 in the unit disc.
Definition 11
Given f(x,y)andL={p∈R2: f(p) =ℓ},ℓ∈Ris theℓ–level setof f . Theorem 12
A regular implicitly defined curve islocallythe trace of a regular parametrized curve.
Proof: (Key idea) This follows from the Implicit function theorem. This says that ifF(x,y) =0 andFy(x,y)6=0, then for any(x0,y0)∃a neighborhood near(x0,y0)such that in this neighborhoody= f(x).
In the definition of regularity for implicitly defined curves we said∇F 6= 0. In other words, in suitable co- ordinate system the curve in the neighborhoodx0−α,x0+α)×(y0−β)×(y0+β). If Fy 6= 0, y = f(x). parametrizationx(t) =
t f(t)
,t=x.
Remark Sis the graph ofF(x,y). Introduceϕ(x,y,z) =z−F(x,y). (Consider the intersection ofS and{z=0}. ϕ(x,y, 0) =−F(x,y) =0 if x,y∈curve
ThenN=∂F∂x,∂F∂y,−1is the normal vector onS.
Orientation
Definition 13
Anorientationof a regular implicitly defined curve with no self–intersections is a choice of direction in which the curve is to be traversed.
Definition 14
For parametrized curves the direction of increasing t is defining the natural parametrization.
Arc–length parametrization PLANAR CURVE
Example 15 (Reparameterization) x:(0,π)→R2,x(t) = cost
sint
is a semicircle.
With an arbitraryh(s):(a,b)→(0,π), thenx1(s) =x(h(s))is the same semicircle.
Definition 16 (Bijection)
Let h:(a,b)→R, h is called abijectionif h,h−1are continuous function and h is one–to–one.
Definition 17
Letx:I→Rbe a curve. If h:J→I is a bijection and J interval and h is smooth, thenx1=x◦h is called areparametriza- tionof the curve.
If h is smooth (C∞)
x1(t) =x◦h(t) =x(h(t)) =
x1(h(t)) x2(h(t))
dx1(t)
dt =
dx1 dh dh dx2 dt
dh dh dt
!
(chain rule)
Remark Ifh′ >0,x1(t)has the same orientation.h′<0 gives the reversed orientation.
Definition 18 (Arc length)
Letxbe an oriented curve, A,B be on the curve with A=x(a)and B=x(b) =B, Then thearc lengthfrom A to B is s=
Z b
a v(t)dt= Z b
a
q
(x′1(t))2+ (x2′(t))2dt
The arc length is the Riemann sum ofv(t)withv(t) =|x′(t)|.
Arc–length parametrization
The curvex(s)is arc–length parametrized if the parameter sis the arc length. x : I → R,x(s), s = arc length measured from some fixed point.
Theorem 19
Parameter s arc–length parametrizes curve C if and only if (i) v=1(speed)
(ii) s=0at some point A∈C.
Proof: ⇒ to proof i) we uses=Rb=s
a=0v(t)dt=Rs
0v(t)dt. Differentiating wrt.sgives 1= dsds = dsd Rs
0 v(t)dt= v(s). ii) is clearly satisfied.
• Assume that we are given a parametrizationx(t)such that i) v(t) =1
ii) s=s(0) =0 t0=0
=⇒ s=t s= Z t
0=av(t)dt= Z t
0 1dt=t =⇒ s=t
Theorem 20
Aregularcurve can be always arc–length parametrized.
Proof: Letxis regular, thereforev(t) =|x′(t)|>0.s(t) =Rt
av(u)du.
take d
dt =⇒ dsdt = d dt
Z t
a v(u)du=v(t)>0
=⇒ s′(t)>0 hences(t)is a strictly increasing function oft. Therefore the inverse function exists and we get t=h(s) =s−1(t). Hencex1(s) =x(h(s))gives the arc–length parametrized.
Frenet frame PLANAR CURVE
Example 21 x(t) = 3t
2t3/2
. Arc length:x′(t) = 3
3t1/2
,v(t) =|x′(t)|=√
32+32t=3√ 1+t.
s= Z t
0 v(t)dt= Z t
0 3√
1+t dt=3(1+t)3/2·23
t
0
=2(1+t)3/2
t
0
=2((1+t)3/2−1) =s(t) measured from 0=x(0). Findtin terms ofs:
s
2 = (1+t)3/2−1 =⇒ 2s +1= (1+t)3/2 =⇒ (s2+1)2/3=t+1 =⇒ t= (2s +1)2/3−1=h(s)
Frenet frame
Assumexis a regular curve. Theunit tangent vectoris T= x′(t)
|x′(t)| = x′(t)
v(t) =⇒ |T|=1 Theunit normal vectoris
N=JT where J=
0 −1
1 0
90 degree ccw rotation TheFrenet frameatx(t)is(T,N).
Proposition 22
Leta(t)be a unit vector, hence|a(t)|=1for all t∈ I. Thena(t)⊥a′(t). Proof: Leta(t) =
a1(t) a2(t)
. Hence
1= q
a21(t) +a22(t) 1=12=a21(t) +a22(t) take d
dt 0= d
dt
a21(t) +a22(t)=2a1(t)·a′1(t) +2a2(t)a′2(t)
=2 a1(t)
a2(t)
· a′1(t),a2′(t)=2a(t)·a′(t)
In particular witha(t) =T(t)the unit tangent vector of some curve. ThenT(t)andT′(t)are orthogonal.
Given a curve, letθ(t)be the inclination angle. That is the angle of how muchTdiffers from thex–axes.
T(t) =
cosθ(t) sinθ(t)
N(t) =JT(t) =
−sinθ(t) cosθ(t)
Rate of change of T ( t )
dT(t) dt = d
dt
cosθ(t) sinθ(t)
= d
dtcosθ(t)
d
dtsinθ(t)
=
−sinθ′(t) cosθ′(t)
=θ′(t)
−sinθ(t) cosθ(t)
=θ′(t)N(t) θ′(t)measures how fastT(t)turns towardsN
Definition 23
Letx(s):J→Rarc–length parametrization and v(s) =|x(s)|=1. Thecurvatureκis the coefficient in T′(s) =κ(s)N(s)
The Rotation–Index PLANAR CURVE
N′(t)⊥N(t)sinceNis unit, soN′(t) =c(t)T(t). By definitionN(t) =JT(t). d
dtN(t) = d
dtJT(t) =J· dT(t)
dt =J·κ(t)N(t) =J·κ(t)JT(t) =κ(t)·J JT(t) =−κ(t)T(t) becauseJrotates by 90 degrees.
Therefore we get
T′(s) =κ(s)N(s) N′(s) =−κ(s)T(s) T′(s),N′(s)= (T(s),N(s))
0 −κ(s) κ(s) 0
Example 24 (semi–circle) x(t) =
rcost r∈t
withrthe radius.x′(t) =
−rsint rcost
. x1(s) =
rcossr rsinsr
andx′1(s) =
−sinsr
cossr
=T(s).
v(t) =pr2∈2t+r2cos2t=r s= Z t
a=0v(t)dt=rt (t= sr) T′(s) =
−1rcosrs
−1r sinsr
=−1r cossr
sinsr
=−1rN(s)
Let us consider arbitrary parameterizations of regular curves. Ifx(t)is a regular curve,v(t)>0 andT(t) = xv(t)′(t)is well–defined. Thenx(t) =x1(s(t))withs=Rt
a=0v(u)du.
dx(t) dt = d
dt(x1(s(t))) = dx1
ds · dsdt = dx1(t) ds v(t) T(t)·v(t) =T1(s)·v(t)
The Rotation–Index
TheRotation index is defined as γ(x) = 1
2π (θ(L)−θ(0)) T(s) =
cosθ(s) sinθ(s)
sarc length γ(x)is the number of turns ofOP(s).
Theorem 25
For a closed regular curve
γ(x) = 1 2π
Z L 0 κ(s)ds withκthe curvature ofx.
Proof: 1 2π
Z L
0 κ(s)ds= 1 2π
Z L
0 θ′(s)ds= 1
2π(θ(L)−θ(0)).
Example 26 x(s) =r cossr
sinsr
,T(s) =x′(s) =
−sinsr
cossr
=
cos(sr +π2) sin(sr +π2)
,θ(s) = sr+ π2. We haveκ=θ′(s) = 1r.
γ(x) = 1 2π
Z L 0
1
rds= 1 2πrL=1 T(s) =
cosθ(s) sinθ(s)
,T′(s) =
−sinθ(s)θ′(s) cosθ(s)θ′(s)
=θ′(s)
−sinθ(s) cosθ(s)
=θ′(s)
cos(θ+π2) sin(θ(s) + π2)
=θ′(s)N(s).
Families of Curves PLANAR CURVE
Families of Curves
Definition 27
Afamily of curvesis a smooth map:
X(λ,t)→X(λ,t)
where (λ,t) ∈ D in (x1,x2)–plane, D given as D = (a,b)×(c,d). The idea is that for each fixedλ0 we get a curve xλ0(t) =X(λ0,t).
Example 28 X(λ,t) = λ
λ2
+t 1
2λ
=A+tB=xλ(t). Definition 29
A curveuis anenvelopefor the familyX(λ,t)if at each pointuis tangent to a member of the familybut is not a member of the family.
Notice that the contact point belongs toxλandu.t=T(λ)at the contact points for (unknown) functionT(λ). Theorem 30
Ifu:λ→X(λ,T(λ))is an envelope forX(λ,t)if det
∂X
∂λ,∂X
∂t
=det
∂x1
∂λ
∂x1
∂x2 ∂t
∂λ
∂x2
∂t
!
=0 u(λ) =X(λ,T(λ)).
Proof: If we show ∂X∂λ,∂X∂t are linear dependent, then det = 0. Tλ is the tangent of xλ, Tu of u(λ). Hence Tλ=c1Tu.
u(λ) =X(λ,T(λ)) Tu|| dudλ = d
dλ
X(λ,T(λ))
= ∂X
∂λ+ ∂X
∂tT′(λ) (∗) xλ(t) =X(λ,t)
Tλ|| dxλ(t)
t = ∂
∂tX(λ,t) (∗∗) (∗),(∗∗) =⇒∂X∂λ+∂X
∂tT′(λ) =c2∂
∂tX(λ,t)
t=T(λ)
=⇒ c2=const.
∂X
∂λ = (c2−T′(λ))
| {z }
scalar
∂X
∂t
t=T(λ)
Remark This is a necessary condition for enveloping but we use it to try to locate (recover) envelopes.
Example 31 TakeX(λ,t) = λ
λ2
+t 1
2λ
. We have ∂X∂λ = 1
2λ
+ 0
2t
, ∂X∂t = 1
2λ
. Thus by Theorem we have that
det
1 1 2λ 2λ+2t
=0 ⇐⇒ t=0
Substituting back intoXgives envelope asu(λ)X(λ,T(λ)) = (λ,λ2)⊤which is the parabola.
Envelope of implicitly defined curves
Similarly extend an implicitly defined curve fλ(x) =0 to a family of implicitly defined curve given byF(x,λ) =0 For each fixed value ofλthus defines an implicitly defined curve fλ(x) =0 where f(x,λ) =F(x,λ).
Theenveloping condition: Analysis similar to above shows that a necessary condition for enveloping is∂F∂λ =0.
CALCULUS OF VARIATIONS
Calculus of Variations
Given a functional J[y](that is a function ofy(x),x ∈ [a,b])ysatisfies boundary conditiony(a) = u,y(b) =v, the numbersu,vare given.
Question:Findythat extremizesJ[y]. Definition 32 (Extremals)
Let J[y]be a functional and y satisfies the boundary conditions as indicated above. Thus y is an extremal of J if d
dεJ[y+εh]
ε=0
=0 for all smooth functions h such that h(a) =h(b) =0.
Lemma 33
If y extremizes J then y must be an extremal of J.
Analogy
function f(ε) functionalsJ[y]
min/max values extremum
critical values f′(ε) =0 extremal (or stationary points) Example 34 Consider E–L equationsJ[y] =Rb
a F(x,y,y′)dxthen ∂F∂y− dxd(∂y∂F′) =0.
Extremas: dεdJ[y+εh]
ε=0
=0 for allh,h(a) =h(b) =0. They satisfy the E–L equation.
Example 35 (Minimal arc length problem) J[y] =
Z b a
q
1+ [y′(x)]2dx y(a) =u, y(b) =v
| {z }
Boundary values
Findythat minimizesJ. Thereforeysolves E–L
F(x,y,y′) =q1+ [y′(x)]2 0= ∂F
∂y − dxd ∂y∂F′ =0− dxd ∂
∂y′ q
1+ [y′(x)]2
=0 d
dx ∂
∂y′ q
1+ [y′(x)]2
=0 =⇒ ∂y∂′q1+ [y′(x)]2=kconstant
∂
∂y′ q
1+ [y′(x)]2= 1 2
1
p1+ [y′(x)]2 ·2y′ = p y′
1+ [y′(x)]2 =k
=⇒ y′(x) =k q
1+ [y′(x)]2 =⇒ (y′)2=k2(1+y′2)
Asy′=constant we havey(x) =Ax+B. Using the boundary dataAa+B=uandAb+B=v. This leads to y(x) = u−v
a−bx+u−a(u−v) a−b
Minimal area of surface of revolution
J[y] = Z b
a 2πy q
1+ [y′(x)]2dx
Several variables CALCULUS OF VARIATIONS
Find the extrema of J[y]. E–L equation: ∂F∂y − dxd ∂y∂F′ = 0, F(x,y,y′) = 2πyp
1+ [y′(x)]2. ThenF−y′∂y∂F′ = kfrom previous lecture.
2πy q
1+y′2−y′·2πy·12p 1 1+y′22y′
| {z }
∂F
∂y′
=2πy q
1+y′2−2πy y′
2
p1+y′2 =2πy q
1+y′2− y′
2
p1+y′2
!
=2πy 1
p1+y′2 =k =⇒ p y
1+y′2 =k y′ =
ry2
k2−1ODE solve=⇒ x−x0= 1
kcosh−1y
k =⇒ y=kcosh(k(x−x0))
Solution depends on 2 parametersk,x0. Uniqueness obtained by applying the boundary conditions.
Several variables
J[x] = Z b
a F(t,x1(t),x2(t),x1′(t),x2′(t))dt= Z b
a F(t,x,x′)dt x= x1
x2
, t∈[a,b] We havex(a) =uandx(b) =v.
Definition 36 (Extremals)d dεJ[x+εh]
ε=0
=0wherehis an arbitrary smooth vector field such thath(a) =h(b) =0.
Theorem 37
Ifxextremises J[x]thenxis an extremal.
Proof: like the single variable case.
Theorem 38
Ifx(t)is an extremal of J[x]than it satisfies
∂F
∂xi − dtd ∂x∂F′
i
i=1, 2
We have two equations with two unknowns x1,x2complemented withx(a) =uandx(b) =v.
Example 39 J[x] =Rb a
q[x1′(t)]2+ [x2′(t)]2dt- arc length.x(a) =u,x(b) =v.
F(t,x,x′) =|x′|=q[x′1(t)]2+ [x′2(t)]2 depends only on derivatives.
0− dtd ∂x∂F′
1
=0 ∂F
∂x′1 = ∂
∂x1′ q
[x′1(t)]2+ [x′2(t)]2= q x′1
[x′1(t)]2+ [x′2(t)]2 =k1
0− dtd ∂x∂F′
2
=0 ∂F
∂x′2 = q x′2
[x′1(t)]2+ [x′2(t)]2
=k2
Since dxd ∂x∂F′
i =0 =⇒ ∂x∂F′
i =ki. Assume thatk1=0 =⇒ x′1=0 =⇒ x1is linear =⇒ x2is linear. Ifk16=0, then k2
k1 = x2′
x1′ = slope If the slope is constant,x(t)is linear.
Applications SPACE CURVES
Special Cases of Euler–Lagrange
(i) Fdoes not depend onx =⇒ ∂x∂F1 = ∂x∂F
2 =⇒ 0 =⇒ ∂x∂F′ i =ki
∂F
∂xi =ki ki constant, 1st order ODE (ii) Fdoes not depend ont
F−
2
∑
i=1
xi′∂F
∂x′i =0 1 equation of 1st order
Applications
Hamiltons principle J[x] = Z b
a
1
2|x′|2−V(x)
E− −L−dxd x+∇xV(x) =0,x′′=∇xV(x): Newtons second law.
Ifx=r(t)(cosθ(t), sinθ(t))⊤then
x′=r′(t)
cosθ(t) sinθ(t)
+r(t)
−sinθ(t) cosθ(t)
θ′(t)
|x′|=pr′2+r2θ′2=F(t,(rθ),(r′,θ′))
Space Curves
x:I→R3,I= [a,b]⊂R
(i) For space curves we will see that curvaturek≥0
(ii) In space, curves have more directions to turn. (2 sorts of deformation: Bending (curvature), twisting (torsion)).
Example 40 (Helix) x(t) =
acost asint
kt
orx2+y2=a2cos2t+a2sin2t=1 andz=kt.
Velocityx′=
−asint acost
k
, speed|x′|=√ a2+k2
Arc length =Rt
0v(u)u=Rt 0
√a2+k2du=t√
a2+k2,t= √ s a2+k2. Arc length parametrisation isx1(s) =
acos√ s a2+k2
asin√ s a2+k2
k√ s a2+k2
.
Example 41 x(t) =
t t2 t3
,x′=
1 2t 3t2
,|x′|=√
1+4t2+9t4,s(t) =Rt 0
√1+4u2+9u4du
x3=x13andx2=x21. The curves lies on these two projections into 2 dimensions.
Example 42 x(t) =
costcos 4t costsin 4t
sint
.x21+x22+x23=1, therefore the curve lies on a unit sphere.
Curvature for Space curves SPACE CURVES
Curvature for Space curves
Definition 43
A space curve is called regular if v(t)>0everywhere.
Remark The unit tangent vectorT= v(t)x′ is well defined for regular curves.
Definition 44
Thecurvatureκis defined asκ= v(t)|T′| ≥0.
Definition 45
ABiregular curveis a curve for which|T′| 6=0.
Remark T′⊥Tsince|T|=1. Differentiate 1=|T|=T·T =⇒ 0=T·T′. Definition 46
N= |TT′′| is the normal for biregular curves.N⊥Tby definition.
Definition 47
TheBinormalisB=T×N Definition 48
The Frenet frame for space curves is(T,N,B).
With a given curvex(t),t∈I, how to findTandN?
• Arc length parametrisationv(s) =1 and assume the curve is biregular. ThenT=x′,N= |xx′′′′|andB=x×|xx′′′′|. κ=|x′′|.
In the general case, we havex′=vT,x′′=v′T+vT′ =v′T+v2N|Tv′| andx′′=v′T+v2κ·N.
Structural Equations
(T,N,B)′ = (T,N,B)v·
0 −κ 0 κ 0 −r
0 r 0
withκthe curvature measures bending andrthe torsion that measures twisting.
B′ =−rN.
Recall that for plane curves(T,N)′ = (T,N)
0 κ
−κ 0
·v.
Step 1: Equation forT′.N= |TT′′| =⇒ T′ =N′· |T′|. Furthermoreκ= |Tv′| =⇒ |T′|=vκ
=⇒ T′=vκ·N (1) Step 2: Equation forB′
B=T×N
d
=dt⇒ B′ = (T×N)′
Ifa(t),b(t),t∈I = (a,b), e.g. I=R, then dtd(a(t)×b(t)) =a′(t)×b(t) +a(t)×b′(t).
=⇒ B′=T′×N
| {z }
=0
+T×N′ because N||T′ N′×T=−T×N′||N =⇒ T×N||N =⇒ B′||N
B′ =scalar·N =⇒ B′ =−rvN Step 3: Equation forN.
N=B×T
d
=dt⇒ N′=B′×T+B×T′ = (−rv)N×T
| {z }
−B
+vkB×N
| {z }
−T
N′= (−rv)(−B) +vk(−T) =−vkT+rvB
Local behavior of curves SPACE CURVES
Theorem 49
For biregular curvex(t)
κ= |x′×x′′|
v3 r= (x′×x′′)·x′′′
v6κ2 Proof:
x′=vT
x′′=v′T+vT′=v′T+v2κN
x′′′ =v′′T+v′vκN+2vv′κN+v2κ′N+v2κ(−vκT+rvB) x′×x′′=v3κ(T×N) =v3κB
|x′×x′′|=v3κ|B|=v3κ =⇒ κ= |x′×x′′| v3 (x′×x′′)·x′′′ =v3κB·x′′′ =v3κB(rv3κB) =rv6κ2 Theorem 50
κand r are unchanged under reparametrisation.
Proof: x(t),t= g(u)withua new parameter. Setx1(u) =x(g(u)). κ1 = κ,r1 = r. Ift = g(u)is orientation preserving, than(T,N,B) = (T1,N1,B1)and otherwise−(T1,N1,B1).
The osculating plane, i.e.Π= (T,N)–plane withBthe normal ofΠ. We haveB′ =−rvNor for arc parametrisation (v=1)B′=−rN.κmeasures the bending of the projection inΠ= (T,N).rmeasures the bending of the projection in(B,N).
Local behavior of curves
0∈ I,x: I →R3arc length parametrisationv =1,s= t. Taylor expansion at 0. AssumingT(0) = e1,N(0) =e2, B(0) =e3andx(0) =0.
x(s) =x(0) +sx′(0) + s2!2x′′(0) +s3!3x′′′(0) +· · ·
=0+sT(0) +s2!2κN(0) + s3!3 (κ′N(0) +κ(−κT+rB))
| {z }
from structural equations.
=T(0)s−s3!3κ2+N(0)s2!2κ+s3!3κ′
+B(0)s3!3rκ
=T(0)s+N(0)s22κ+B(0)s63κr
This polynomial is the dominant part in Taylor expansion.x0(s) = (s,s22κ,s63κr)⊤
Implicit Space Curves
Definition 51
F(x,y,z) : R3 → Rthen C is animplicit space curve. G(x,y,z) : R3 → Rif C = {(x,y,z) ∈ R3s.t. F(x,y,z)∧ G(x,y,z) =0}=intersection of0–sets of F and G.
Example 52 S1=x21+x22+x23=1 (unit sphere) andS2=x1+x2+x3=1 plane.
Definition 53
C is regular if∇F× ∇G6=0.
Lemma 54
If C is regular then it can be arc–length parametrised.
Definition 55 (Regularity)
f :R3→R, f(x,y,z) =c, c∈R,Σ=(x,y,z)∈R3: f(x,y,z) =c . If
∇f =6=0 for all (x,y,z)∈Σ
Implicit Space Curves SPACE CURVES
Example 56 • f(x,y,z) =|x|2−1,x = (x,y,z)∈R3. f(x,y,z) =0 ⇐⇒ |x|2=1, unit sphere =x2+y2+z2= 1,∇f = (2x, 2y, 2z)
If∇f =0 ⇐⇒ 2(x,y,z) =0 =⇒ (x,y,z) =0. As(x,y,z) =0 /∈Σ, the sphere is regular.
• f(x) = z2−x2−y2. Σ: f(x) = 0,z2−x2−y2 =0. ∇f =0 ⇐⇒ (x,y,z) = 0∈Σ. Therefore the surface is irregular.
Given a regular surfaceΣ, two pointsA,B∈ Σ. Find a curve inΣjoiningAandBsuch that the curve has minimal arc length. Such a curve is called geodesic ofΣ.
Definition 57
LetΣbe a surface. The curveα:I→R3is a curve inΣ, ifα(t)∈Σ:{α(t)}t∈I ⊂Σ. Lemma 58
Letαbe a curve inΣ-regular local surface. Then there exists a unique smooth curveu:I→D such thatα(t)∈x(u(t)). Proof: Σ=x(D).α(t) =x(u(t)).α=x◦x−1◦ff=x◦(x−1◦ff)
| {z }
u(t)
=x(u).
Example 59 x(u,v) =
cosu sinu
v
,(u,v)∈D= (0, 2π)×(0,∞)(cylinder)
(i) u1(t) = 2t
π 4
=⇒ x(u1(t)) =
cos 2t sin 2t
π 4
(ii) u2(t) = π
2t
=⇒ x(u2(t)) =
0 1 t
(iii) u3(t) = 2t
t
=⇒ x(u3(t)) =
cos 2t sin 2t
t
.
Definition 60
Let p∈Σsurface. Atangent vectorvtoR3at p is a tangent toΣifvis the velocity at p of some curve inΣ. The set of all such tangent vectors toΣat p is the tangent space (plane) toΣat p written TpΣ.
Lemma 61
x(D)be a local surface inΣand let p=x(u0)∈x(D). Then TpΣis the subspace of TpR3spanned by partial velocitiesxu,xv. Proof:
Step 1 TpΣ ⊂ span{xu(u0),xv(u0)}. By def. z′(t)curvez(t) inΣand passes through p. z(t)is a curve in Σ =⇒ ∃u(t)such thatz(t) =x(u(t))withu=x−1◦z. Assume thatu0=u(t0) =⇒
d
dtz(t) = d
dt(x◦u(t)) =xu(u(t))·dudt +xv(u(t))du dv whereu(t) = (u(t),v(t)). Taket=t0
v=xu(u0)·u′(t0)
| {z }
α
+xv(u0)·v′(t0)
| {z }
β
=αxu(u0) +βxv(u0)∈span{xu,xv}
Step 2 spanp{xu,xv} ⊂TpΣ.
v∈spanp{xu,xv} =⇒ v=αxu+βxv at the point p=x(u0) αxu(u0) +βxv(u0) = d
dt
x(u0+t(α,β)⊤) t=0
=xu(u0+t(α,β)⊤)
t=0
·α+xv(u0+t(α,β)⊤)
t=0
·β Fir smalltu0+t(α,β)⊤–straight line⊂Dthusx(u0+t(α,β)⊤)≡fl(t)is a curve inΣ.v= dtdfl(t)t=0.
NowTpΣ⊂spanp{xu,xv}from Step 1 and spanp{xu,xv} ⊂TpΣfrom Step 2 bringsTpΣ=spanp{xu,xv}.