Solutions for Math 348 Assignment #4 1

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1 Solutions for Math 348 Assignment #4 1 (1) Do the following: (a) Show that the intersection of two spheres S 1 = {(x, y, z) : (x x 1 ) 2 + (y y 1 ) 2 + (z z 1 ) 2 = r 2 1} S 2 = {(x, y, z) : (x x 2 ) 2 + (y y 2 ) 2 + (z z 2 ) 2 = r 2 2} in R 3 is a curve with constant positive curvature and zero torsion, assuming that S 1 and S 2 meet at more than one point and S 1 S 2. (b) Show that a smooth curve γ : I R 3 has constant positive curvature and zero torsion if and only if γ(i) is contained in a circle. Proof. For (a), note that S 1 S 2 = {(x, y, z) :(x x 1 ) 2 + (y y 1 ) 2 + (z z 1 ) 2 = r 2 1, 2(x 2 x 1 )x + 2(y 2 y 1 )y + 2(z 2 z 1 )z = (r 2 1 r 2 2) (x y z 2 1) + (x y z 2 2)} is contained in a plane. So it has torsion 0. Since the curvature of a curve is invariant under isometry, we may use a translation to move (x 1, y 1, z 1 ) to the origin. That is, we assume that x 1 = y 1 = z 1 = 0. Let u = (x 2 x 1, y 2 y 1, z 2 z 1 ). Using Gram-Schmidt, we may expand {u/ u } to an orthonormal basis of R 3. That is, we have an orthogonal matrix A whose first column is u/ u. Let f : R 3 R 3 be the orthogonal transformation given by f(v) = A 1 v. So f(u) = ( u, 0, 0) and f(s 1 S 2 ) is { x 2 + y 2 + z 2 = r 2 1 (x u ) 2 + y 2 + z 2 = r 2 2 x = r2 1 r u 2 2 u = a y 2 + z 2 = ((r 1 + r 2 ) 2 u 2 )( u 2 (r 1 r 2 ) 2 ) = b 2 4 u 2 x = a y = b cos t z = b sin t 1 xichen/math34815f/hw4sol.pdf 1

2 2 which has constant curvature 1/b. For (b), it is obvious that a circle in R 3 has constant nonzero curvature and zero torsion. On the other hand, let γ be a curve parameterized by its arc length with constant curvature k > 0 and torsion zero. Note that the curve α : I R given by ( cos(ks) α(s) = k, sin(ks), 0 k has curvature k and torsion 0. By the Fundamental Theorem of the Local Theory of Curves, there exists an isometry f : R 3 R 3 such that γ = f α. Therefore, γ(i) is contained in a circle. (2) Find the curvature and torsion of the curve γ : R R 3 given by γ(t) = (t, t 2, t 3 ) and write down the Frenet equations associated to γ. Solution. The curvature and torsion of γ are κ = γ γ γ 3 = t 2 + 9t 4 (1 + 4t 2 + 9t 4 ) 3/2 τ = det [ γ γ γ ] γ γ 2 ) = (1, 2t, 3t2 ) (0, 2, 6t) (1, 2t, 3t 2 ) 3 3 = 1 + 9t 2 + 9t 4 respectively. So the Frenet-Serret equations associated to γ are d T N = γ T N dt B B t 2 +9t t 2 +9t 4 = 2 1+9t 2 +9t t 2 +9t 4 1+4t 2 +9t 4 1+9t 2 +9t T N t 2 +9t 4 0 B 1+9t 2 +9t 4 (3) (MPDC p. 25 Ex. 13) Let α : I R 3 be a regular smooth curve parameterized by its arc length satisfying that κ(s) 0, τ(s) 0 and κ (s) 0. Show that α(i) lies on a sphere if and only if R 2 + (R ) 2 T 2 c

3 for some constant c R, where R = 1/κ, T = 1/τ and R is the derivative of R with respect to s. Proof. Let e 1, e 2, e 3 be the unit tangent, normal and binormal vectors of α, respectively. Note that α(i) lies on a sphere if and only if there exists p R 3 such that α(s) p 2 const d ( ) α(s) p 2 0 ds (α(s) p).α (s) 0 (α(s) p).e 1 (s) 0 α(s) p f(s)e 2 (s) + g(s)e 3 (s) α (s) (f(s)e 2 (s) + g(s)e 3 (s)) e 1 (s) (f(s)e 2 (s) + g(s)e 3 (s)) for some smooth functions f and g on I. So α(i) lies on a sphere if and only if e 1 = f e 2 + g e 3 + fe 2 + ge 3 on I for some smooth functions f, g : I R. By Frenet-Serret equations, e 1 e 2 = 0 κ 0 κ 0 τ e 1 e 2 0 τ 0 we have e 3 e 1 = f e 2 + g e 3 + fe 2 + ge 3 e 1 = f e 2 + g e 3 + ( fκe 1 fτe 3 ) + gτe 2 e 3 3 on I. e 1 = fκe 1 + (f + gτ)e 2 + (g fτ)e 3 1 = fκ 1 = fκ 0 = f ( ) + gτ f 0 = g = fτ = g fτ τ 1 ( ( ) ) 1 1 κ τ = R = T (T R ) τ κ RR = T R (T R ) (R 2 ) = ((T R ) 2 ) R 2 + (R ) 2 T 2 const

4 4 (4) (MPDC p. 25 Ex. 14) Let α : I R 2 be a regular smooth curve and p be a fixed point in R 2. Suppose that α(t) p achieves a local maximum at a point t 0 I. Show that κ(t 0 ) 1 α(t 0 ) p. Proof. Suppose that α is parameterized by its arc length. Let f(t) = α(t) p 2. Since f(t) has a local maximum at t 0, f (t 0 ) = 0 and f (t 0 ) 0. Therefore, f (t 0 ) = 0 2(α(t 0 ) p).α (t 0 ) = 0 (α(t 0 ) p).α (t 0 ) = 0 f (t 0 ) 0 α (t 0 ).α (t 0 ) + (α(t 0 ) p).α (t 0 ) (α(t 0 ) p).α (t 0 ) 0. Since both α (t 0 ) and α(t 0 ) p are orthogonal to α (t 0 ), they are linearly dependent. So (α(t 0 ) p).α (t 0 ) = ± α(t 0 ) p α (t 0 ). And since (α(t 0 ) p).α (t 0 ) 1 < 0, we must have and (α(t 0 ) p).α (t 0 ) = α(t 0 ) p α (t 0 ) 1 + (α(t 0 ) p).α (t 0 ) 0 1 α(t 0 ) p α (t 0 ) 0 κ(t 0 ) = α 1 (t 0 ) α(t 0 ) p. (5) Let α : I R 3 and β : I R 3 be two regular smooth curves with nowhere vanishing curvatures and satisfying α(t) β(t) for all t I. Show that the osculating planes of α and β agree for all t I if and only if both α(i) and β(i) lie in the same plane. Proof. Clearly, if α(i) and β(i) lie in the same plane Λ, Λ is the osculating plane of α and β for all t I and hence they have the same osculating planes for all t I. Suppose that α and β have the same osculating planes for all t I. At t 0 I, the osculating planes of α and β are

5 5 parameterized by x y = α(t 0 ) + s 1 α (t 0 ) + s 2 α (t 0 ) and z x y = β(t 0 ) + s 1 β (t 0 ) + s 2 β (t 0 ) z respectively. So they agree if and only if α(t 0 ) β(t 0 ), α (t 0 ), α (t 0 ) Span{β (t 0 ), β (t 0 )}. Therefore, α and β have the same osculating planes for all t if and only if α = β + f 12 β + f 13 β α = + f 22 β + f 23 β α = + f 32 β + f 33 β on I for some smooth functions f ij. It follows that α = (β + f 12 β + f 13 β ) = f 22 β + f 23 β α = (f 22 β + f 23 β ) = f 32 β + f 33 β and hence (1 + f 12 f 22 )β + (f 12 + f 13 f 23 )β + f 13 β = 0 (f 22 f 32 )β + (f 22 + f 23 f 33 )β + f 23 β = 0. We want to show that β (t), β (t), β (t) are linearly dependent for all t I, i.e., det [ β (t) β (t) β (t) ] = 0 for all t I. Let S be the subset of I defined by { S = t I : 1 + f 12(t) f 22 (t) = f 22(t) f 32 (t) = f 12 (t) + f 13(t) f 23 (t) = f 22 (t) + f 23(t) f 33 (t) = } f 13 (t) = f 23 (t) = 0. If t 0 S, β (t 0 ), β (t 0 ), β (t 0 ) are linearly dependent. Suppose that t 0 S. If f 13(t 0 ) 0, then there exists r > 0 such that f 13 (t) 0 for t (t 0 r, t 0 + r) and t t 0 and hence t S for t (t 0, t 0 + r) det [ β (t) β (t) β (t) ] = 0 for t (t 0, t 0 + r). Then by the continuity of det [ β (t) β (t) β (t) ], we conclude that det [ β (t 0 ) β (t 0 ) β (t 0 ) ] = 0.

6 6 Suppose that f 13(t 0 ) = 0. Then } t 0 S f 12 (t 0 ) + f 13(t 0 ) f 23 (t 0 ) = f 13 (t 0 ) = f 23 (t 0 ) = 0 f 12 (t 0 ) = f 13 (t 0 ) = 0 α(t 0 ) = β(t 0 ) f 13(t 0 ) = 0 Contradiction. This proves that β (t), β (t), β (t) are linearly dependent for all t I. So β(i) is contained in some plane Λ β R 3. Since { α = f 22β + f 23β α = f 32 β + f 33 β α = f 22 β + f 23 β α = f 32 β + f 33 β α = f 32β + (f 32 + f 33)β + f 33 β α (t), α (t), α (t) Span{β (t), β (t), β (t)}, α (t), α (t), α (t) are also linearly dependent for all t I. So α(i) is contained in some plane Λ α R 3. Finally, since Λ α and Λ β are the osculating planes of α and β, respectively, we must have Λ α = Λ β.