DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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106 Chapter 3. Surfaces: Further Topics4. Use the functional F (u) =∫ ba2πu(t) √ 1+(u ′ (t)) 2 dt to determine the surface of revolution ofleast area with two parallel circles (perhaps of different radii) as boundary. (Hint: You shouldend up with the same differential equation as in Exercise 2.2.19.)5. Prove the analogue of Theorem 4.3 for curves. That is, show that of all closed plane curvesenclosing a given area, the circle has the least perimeter. (Cf. Theorem 3.10 of Chapter 1. Hint:Start with Exercise A.2.5. Show that the constrained Euler-Lagrange equations imply that theextremizing curve has constant curvature. Exercise 1.2.4 and the chain rule will help.)6. Interpreting the integral∫ 10space of continuous functions on [0, 1], prove that iffunctions g with〈f ⊥ , 1〉 = 0.)∫ 10f(t)g(t)dt as an inner product (dot product) 〈f,g〉 on the vector∫ 10f(t)g(t)dt = 0 for all continuousg(t)dt =0,then f must be constant. (Hint: Write f = 〈f,1〉1+f ⊥ , where7. Prove the pedal property of the ellipse: The product of the distances from the foci to the tangentline of the ellipse at any point is a constant (in fact, the square of the semiminor axis).8. The curve α(s) rolls along the x-axis, starting at the point α(0) = 0. Apoint F is fixed relativeto the curve. Let β(s) bethe curve that F traces out. As indicated in Figure 4.4, let θ(s) beFθQF'Q'Figure 4.4[]the angle α ′ cos θ − sin θ(s) makes with the positive x-axis. Denote by R θ =the matrixsin θ cos θthat gives rotation of the plane through angle θ.a. Show that β(s) =(s, 0) + R −θ(s) (F − α(s)).b. Show that β ′ (s) · R −θ(s) (F − α(s)) = 0. That is, as F moves, instantaneously it rotatesabout the contact point on the x-axis. (Cf. Exercise A.1.4.)9. Find the path followed by the focus of the parabola y = x 2 /2asthe parabola rolls along thex-axis. The focus is originally at (0, 1/2). (Hint: See Example 2.)
APPENDIXReview of Linear Algebra and Calculus1. Linear Algebra ReviewRecall that the set {v 1 ,...,v k } of vectors in R n gives a basis for a subspace V of R n if andonly if every vector v ∈ V can be written uniquely as a linear combination v = c 1 v 1 + ···+ c k v k .In particular, v 1 ,...,v n will form a basis for R n if and only if the n × n matrix⎡⎤| | |⎢⎥A = ⎣ v 1 v 2 ··· v n ⎦| | |is invertible, and are said to be positively oriented if the determinant det A is positive. In particular,given two linearly independent vectors v, w ∈ R 3 , the set {v, w, v × w} always gives a positivelyoriented basis for R 3 .We say e 1 ,...,e k ∈ R n form an orthonormal set in R n if e i · e j =0for all i ≠ j and ‖e i ‖ =1for all i =1,...,k. Then we have the followingProposition 1.1. If {e 1 ,...,e n } is an orthonormal set of vectors in R n , then they form a basisfor R n and, given any v ∈ R n ∑,wehavev = n (v · e i )e i .i=1We sayann × n matrix A is orthogonal if A T A = I. Itiseasy to check that the column vectorsof A form an orthonormal basis for R n (and the same for the row vectors). Moreover, from thebasic formula Ax·y = x·A T y we deduce that if e 1 ,...,e k form an orthonormal set of vectors in R nand A is an orthogonal n × n matrix, then Ae 1 ,...,Ae k are likewise an orthonormal set of vectors.An important issue for differential geometry is to identify the isometries of R 3 (although thesame argument will work in any dimension). Recall that an isometry of R 3 is a function f : R 3 → R 3so that for any x, y ∈ R 3 ,wehave‖f(x) − f(y)‖ = ‖x − y‖. Wenowprove theTheorem 1.2. Any isometry f of R 3 can be written in the form f(x) =Ax + c for someorthogonal 3 × 3 matrix A and some vector c ∈ R 3 .Proof. Let f(0) =c, and replace f with the function f − c. Ittoo is an isometry (why?) andfixes the origin. Then ‖f(x)‖ = ‖f(x) − f(0)‖ = ‖x − 0‖ = ‖x‖, sothat f preserves lengths ofvectors. Using this fact, we prove that f(x) · f(y) =x · y for all x, y ∈ R 3 .Wehave‖f(x) − f(y)‖ 2 = ‖x − y‖ 2 =(x − y) · (x − y) =‖x‖ 2 − 2x · y + ‖y‖ 2 ;on the other hand, in a similar fashion,‖f(x) − f(y)‖ 2 = ‖f(x)‖ 2 − 2f(x) · f(y)+‖f(y)‖ 2 = ‖x‖ 2 − 2f(x) · f(y)+‖y‖ 2 .107
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DIFFERENTIALGEOMETRY:A First Course
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4 Chapter 1. Curves2πb2πaFigure 1
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6 Chapter 1. CurvesAlternatively, s
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8 Chapter 1. CurvesAn important obs
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10 Chapter 1. Curvesof the road, st
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12 Chapter 1. CurvesSummarizing,κ(
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14 Chapter 1. Curvescurvature κ. I
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16 Chapter 1. CurvesFigure 2.2Conve
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18 Chapter 1. Curvesalso Exercise 2
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20 Chapter 1. Curvesto the curve β
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22 Chapter 1. Curvesthe osculating
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24 Chapter 1. CurvesWe turn now to
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26 Chapter 1. Curvesintersect C eit
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28 Chapter 1. Curvesh(s,t)α(t)α(s
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30 Chapter 1. Curvesy( x(s) ,y(s))
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32 Chapter 1. Curvesɛ2ɛ 1ɛ 3CFig
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CHAPTER 2Surfaces: Local Theory1. P
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36 Chapter 2. Surfaces: Local Theor
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DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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