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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

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PLANAR CURVE

Notations

Rnn–dimensional Euclidean space. Usuallyn=2 orn=3.

• Points inRnxRn:x=

 x1

... xn

= x1 . . . xn x1. . .xnare calledcoordinatesofx.

• Inner product∀x,yRnthe inner product is defined as(x,y) =x·y=x1y1+· · ·+xnyn

MagnitudeofxRnis|x|=√ x·x=

s n

i=1

x2i

Distanceofx,yRnis|xy|= s n

i=1

(xiyi)2

Planar Curve

Definition 1

AcurveinR2is a smooth mapx:IR2where I= (a,b), a,bR, Itx(t) = x1(t)

x2(t)

t is calledparameter

Thetraceofxis the image setx(t)∈R2={zR2, st.tI: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 pR2is the point of application andvR2is the vector acting at p.

Equality of Tangent Vectors(p,v)and(q,w)is given iff

p=q and v=w

For fixed pR2the 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 pR2.

Tangents of planar curves

Definition 3

For a curvexinR2thevelocityofxat time t is the tangent vector x(t) =

x1(t) x2(t)

at x(t)

Example 4x(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, tR parameter.

x(t) = u1

u2

=u.

Definition 5 (Speed)

Letx:IR2, thespeedv:IRdefined as v(t) =|x(t)| ≥0

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Implicitly Defined Curves PLANAR CURVE

Definition 6 (Regular Curves) A curvex:IR2is

• regularat t0if v(t0) =|x(t0)| 6=0

• regularif v(t)6=0for all tI.

Example 7 x(t) =

t3

|t|3

, tR. x2(t). x1(t) = t3smooth. x2(t) = |t|3 =

(t3 ift0

t3 ift<0 =

(x1 ift0

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 hyperbolay2x2=1 All these curves have the formF(x,y) =0.

Definition 9

Let f :R2R. We say that pR2is aregular pointof f iff(p) =∂pf

1,∂p∂f

2

6=0

Let C = {pR2 : f(p) =0}the0–level set of f . Assume that C 6= . Then if f is regular at pC, then C is a regular implicitly defined curve.

Example 10 f(x,y) =

(0 ifx2+y21

x2+y21 ifx2+y2>1.C={pR2,f(p) =0}=unit disc.∇f =0 in the unit disc.

Definition 11

Given f(x,y)andL={pR2: f(p) =ℓ},ℓ∈Ris the–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 neighborhoodx0α,x0+α)×(y0β)×(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) =zF(x,y). (Consider the intersection ofS and{z=0}. ϕ(x,y, 0) =−F(x,y) =0 if x,ycurve

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.

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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,h1are continuous function and h is one–to–one.

Definition 17

Letx:IRbe a curve. If h:JI is a bijection and J interval and h is smooth, thenx1=xh is called areparametriza- tionof the curve.

If h is smooth (C)

x1(t) =xh(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

(x1(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 : IR,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 AC.

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) =s1(t). Hencex1(s) =x(h(s))gives the arc–length parametrized.

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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)3/21) =s(t) measured from 0=x(0). Findtin terms ofs:

s

2 = (1+t)3/21 =⇒ 2s +1= (1+t)3/2 =⇒ (s2+1)2/3=t+1 =⇒ t= (2s +1)2/31=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 tI. 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(ta1(t) +2a2(t)a2(t)

=2 a1(t)

a2(t)

· a1(t),a2(t)=2a(ta(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):JRarc–length parametrization and v(s) =|x(s)|=1. Thecurvatureκis the coefficient in T(s) =κ(s)N(s)

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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(tJ 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 rt

withrthe radius.x(t) =

rsint rcost

. x1(s) =

rcossr rsinsr

andx1(s) =

sinsr

cossr

=T(s).

v(t) =pr22t+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(tv(t) =T1(sv(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

Z L 0 κ(s)ds withκthe curvature ofx.

Proof: 1 2π

Z L

0 κ(s)ds= 1

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

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).

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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

=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:λX(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|| du = d

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 = (c2T(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

. We have ∂X∂λ = 1

+ 0

2t

, ∂X∂t = 1

. 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.

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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∂ydxd(∂y∂F) =0.

Extremas: dJ[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

∂ydxd ∂y∂F =0− dxd

∂y q

1+ [y(x)]2

=0 d

dx

∂y q

1+ [y(x)]2

=0 =⇒ ∂yq1+ [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+y2)

Asy=constant we havey(x) =Ax+B. Using the boundary dataAa+B=uandAb+B=v. This leads to y(x) = uv

abx+ua(uv) ab

Minimal area of surface of revolution

J[y] = Z b

a 2πy q

1+ [y(x)]2dx

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Several variables CALCULUS OF VARIATIONS

Find the extrema of J[y]. E–L equation: ∂F∂ydxd ∂y∂F = 0, F(x,y,y) = 2πyp

1+ [y(x)]2. ThenFy∂y∂F = kfrom previous lecture.

2πy q

1+y2y·2πy·12p 1 1+y22y

| {z }

∂F

∂y

=2πy q

1+y22πy y

2

p1+y2 =2πy q

1+y2y

2

p1+y2

!

=2πy 1

p1+y2 =k =⇒ p y

1+y2 =k y =

ry2

k21ODE solve=⇒ xx0= 1

kcosh1y

k =⇒ y=kcosh(k(xx0))

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

∂xidtd ∂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[x1(t)]2+ [x2(t)]2 depends only on derivatives.

0− dtd ∂x∂F

1

=0 ∂F

∂x1 =

∂x1 q

[x1(t)]2+ [x2(t)]2= q x1

[x1(t)]2+ [x2(t)]2 =k1

0− dtd ∂x∂F

2

=0 ∂F

∂x2 = q x2

[x1(t)]2+ [x2(t)]2

=k2

Since dxd ∂x∂F

i =0 =⇒ ∂x∂F

i =ki. Assume thatk1=0 =⇒ x1=0 =⇒ x1is linear =⇒ x2is linear. Ifk16=0, then k2

k1 = x2

x1 = slope If the slope is constant,x(t)is linear.

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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

∂xi =0 1 equation of 1st order

Applications

Hamiltons principle J[x] = Z b

a

1

2|x|2V(x)

E− −Ldxd 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|=pr2+r2θ2=F(t,(),(r,θ))

Space Curves

x:IR3,I= [a,b]⊂R

(i) For space curves we will see that curvaturek0

(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.

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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 TTsince|T|=1. Differentiate 1=|T|=T·T =⇒ 0=T·T. Definition 46

N= |TT| is the normal for biregular curves.NTby definition.

Definition 47

TheBinormalisB=T×N Definition 48

The Frenet frame for space curves is(T,N,B).

With a given curvex(t),tI, 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′′=vT+vT =vT+v2N|Tv| andx′′=vT+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|=

=⇒ T=·N (1) Step 2: Equation forB

B=T×N

d

=dtB = (T×N)

Ifa(t),b(t),tI = (a,b), e.g. I=R, then dtd(a(tb(t)) =a(tb(t) +a(tb(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

=dtN=B×T+B×T = (−rv)N×T

| {z }

B

+vkB×N

| {z }

T

N= (−rv)(−B) +vk(−T) =−vkT+rvB

(12)

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′′=vT+vT=vT+v2κN

x′′′ =v′′T+vvκN+2vvκN+v2κN+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: IR3arc 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 (κN(0) +κ(−κT+rB))

| {z }

from structural equations.

=T(0)ss3!3κ2+N(0)s2!2κ+s3!3κ

+B(0)s3!3

=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) : R3Rthen C is animplicit space curve. G(x,y,z) : R3Rif 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 ifF× ∇G6=0.

Lemma 54

If C is regular then it can be arc–length parametrised.

Definition 55 (Regularity)

f :R3R, f(x,y,z) =c, cR,Σ=(x,y,z)∈R3: f(x,y,z) =c . If

f =6=0 for all (x,y,z)∈Σ

(13)

Implicit Space Curves SPACE CURVES

Example 56f(x,y,z) =|x|21,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) = z2x2y2. Σ: f(x) = 0,z2x2y2 =0. ∇f =0 ⇐⇒ (x,y,z) = 0∈Σ. Therefore the surface is irregular.

Given a regular surfaceΣ, two pointsA,BΣ. 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:IR3is a curve inΣ, ifα(t)∈Σ:{α(t)}tIΣ. Lemma 58

Letαbe a curve inΣ-regular local surface. Then there exists a unique smooth curveu:ID such thatα(t)∈x(u(t)). Proof: Σ=x(D).α(t) =x(u(t)).α=xx1ff=x◦(x1ff)

| {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 pΣsurface. 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=x1z. Assume thatu0=u(t0) =⇒

d

dtz(t) = d

dt(xu(t)) =xu(u(t))·dudt +xv(u(t))du dv whereu(t) = (u(t),v(t)). Taket=t0

v=xu(u0u(t0)

| {z }

α

+xv(u0v(t0)

| {z }

β

=αxu(u0) +βxv(u0)∈span{xu,xv}

Step 2 spanp{xu,xv} ⊂TpΣ.

vspanp{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}.

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