Plane Geometry - Bruce E. Shapiro
Plane Geometry - Bruce E. Shapiro Plane Geometry - Bruce E. Shapiro
308 SECTION 52. ARC LENGTHProof. Suppose that γ(t) = (x(t), y(t)) for some parameterization, and letv = (p, q) be any unit vector in the plane, and define the functionf(t) = px(t) + qy(t) = γ(t) · v (52.2)f ′ (t) = px ′ (t) + qy ′ (t) = γ ′ (t) · vFrom the fundamental theorem of calculus,But sincef(b) − f(a) =∫ baf ′ (t)dt =∫ baγ ′ (t) · v dtγ ′ (t) · v = |γ ′ (t)||v| cos θ(t) ≤ |γ ′ (t)|because |v| = 1 (v is a unit vector), and θ(t) is the angle between v andγ ′ (t). HenceFrom equation 52.2,From equation 52.1,f(b) − f(a) ≤∫ bγ(b) · v − γ(a) · v ≤a|γ ′ (t)|dt∫ ba|γ ′ (t)|dtγ(b) · v − γ(a) · v ≤ L(γ) (52.3)Since equation 52.3 holds for all unit vectors v, we can apply it to any onein particular we want. In particular, we can choosev =γ(b) − γ(a)|γ(b) − γ(a)|(52.4)Substituting equation 52.4 into 52.3 givesL(γ) ≥ (γ(b) − γ(a)) · v= (γ(b) − γ(a)) ·=|γ(b) − γ(a)|2|γ(b) − γ(a)|= |γ(b) − γ(a)|= d(γ(a), γ(b))γ(b) − γ(a)|γ(b) − γ(a)|« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
SECTION 52. ARC LENGTH 309Thus in the Real Euclidean plane the shortest path between two points isgiven by a straight line segment. Now we want to ask the same questionabout the surface of a sphere: What is the path of shortest distance betweentwo points on the surface of a sphere embedded in Euclidean 3-space? Wecan use the same sort of approach to answer this question.Let us confine ourselves to the surface of a sphere, centered at the origin,that has radius 1; this is not really a limitation because any result we derivecan ultimately be transformed to a different radius or center. Then we candefine spherical coordinatesx(t) = cos θ(t) sin ϕ(t)y(t) = sin θ(t) sin ϕ(t)z(t) = cos ϕ(t)⎫⎬⎭(52.5)Now we let γ : [a, b] ↦→ S be any curve on the surface of the unit sphere,connecting the points A = γ(a) and B = γ(b). Differentiating equation52.5x ′ (t) = −θ ′ (t) sin θ(t) sin ϕ(t) + ϕ ′ (t) cos θ(t) cos ϕ(t)y ′ (t) = θ ′ (t) cos θ(t) sin ϕ(t) + ϕ ′ (t) sin θ(t) cos ϕ(t)z ′ (t) = −ϕ ′ (t) sin ϕ(t)⎫⎬⎭(52.6)Dispensing with the t to save space,|γ ′ (t)| 2 =(−θ ′ sin θ sin ϕ + ϕ ′ cos θ cos ϕ) 2 + (θ ′ cos θ sin ϕ + ϕ ′ sin θ cos ϕ) 2 + (−ϕ ′ sin ϕ) 2= (θ ′2 sin 2 θ sin 2 ϕ − 2θ ′ ϕ cos θ sin θ cos ϕ sin ϕ + ϕ ′2 cos 2 θ cos 2 ϕ)+ (θ ′2 cos 2 θ sin 2 ϕ + 2θ ′ ϕ ′ cos θ sin θ sin ϕ cos ϕ + ϕ ′2 sin 2 θ cos 2 ϕ)+ ϕ ′2 sin 2 ϕ= θ ′2 (sin 2 θ sin 2 ϕ + cos 2 θ sin 2 ϕ) + φ ′2 (cos 2 θ cos 2 ϕ + sin 2 θ cos 2 ϕ + sin 2 ϕ)= θ ′2 sin 2 ϕ(sin 2 θ + cos 2 θ) + φ ′2 (cos 2 ϕ(cos 2 θ + sin 2 θ) + sin 2 ϕ)= θ ′2 sin 2 ϕ + ϕ ′2Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.
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308 SECTION 52. ARC LENGTHProof. Suppose that γ(t) = (x(t), y(t)) for some parameterization, and letv = (p, q) be any unit vector in the plane, and define the functionf(t) = px(t) + qy(t) = γ(t) · v (52.2)f ′ (t) = px ′ (t) + qy ′ (t) = γ ′ (t) · vFrom the fundamental theorem of calculus,But sincef(b) − f(a) =∫ baf ′ (t)dt =∫ baγ ′ (t) · v dtγ ′ (t) · v = |γ ′ (t)||v| cos θ(t) ≤ |γ ′ (t)|because |v| = 1 (v is a unit vector), and θ(t) is the angle between v andγ ′ (t). HenceFrom equation 52.2,From equation 52.1,f(b) − f(a) ≤∫ bγ(b) · v − γ(a) · v ≤a|γ ′ (t)|dt∫ ba|γ ′ (t)|dtγ(b) · v − γ(a) · v ≤ L(γ) (52.3)Since equation 52.3 holds for all unit vectors v, we can apply it to any onein particular we want. In particular, we can choosev =γ(b) − γ(a)|γ(b) − γ(a)|(52.4)Substituting equation 52.4 into 52.3 givesL(γ) ≥ (γ(b) − γ(a)) · v= (γ(b) − γ(a)) ·=|γ(b) − γ(a)|2|γ(b) − γ(a)|= |γ(b) − γ(a)|= d(γ(a), γ(b))γ(b) − γ(a)|γ(b) − γ(a)|« CC BY-NC-ND 3.0. Revised: 18 Nov 2012
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