Andrew Pressley-Instructor\'s Solutions Manual to Elementary Differential Geometry-Springer (2012 ) PDF

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Solutions to the Exercises in Elementary Differential Geometry Chapter 1 1.1.1 It is a parametrization of the part of the parabola with x ≥ 0. 1.1.2 (i) γ (t) = (sec t, tan t) with −π/2 < t < π/2 and π/2 < t < 3π/2. Note that γ is defined on the union of two disjoint intervals: this corresponds to the fact that the hyperbola y 2 − x2 = 1 is in two pieces, where y ≥ 1 and where y ≤ −1. (ii) γ (t) = (2 cos t, 3 sin t). 1.1.3 (i) x + y = 1. (ii) y = (ln x)2 . 1.1.4 (i) γ˙ (t) = sin 2t(−1, 1). (ii) γ˙ (t) = (et , 2t). 1.1.5 γ˙ (t) = 3 sin t cos t(− cos t, sin t) vanishes where sin t = 0 or cos t = 0, i.e. t = nπ/2 where n is any integer. These points correspond to the four cusps of the astroid (see Exercise 1.3.3).

1.1.6 (i) The squares of the distances from p to the foci are (p cos t ± ǫp)2 + q2 sin2 t = (p2 − q2 ) cos2 t ± 2ǫp2 cos t + p2 = p2 (1 ± ǫ cos t)2 , so the sum of the distances is 2p. n.γ˙ = 0. Hence the (ii) γ˙ = (−p sin t, q cos t) so if nn = (q cos t, p sin t) then n. distances from the foci to the tangent line at γ (t) are (p cos t ∓ ǫp, q sin t).n n pq(1 ∓ ǫ cos t) = 2 2 kn k (p sin t + q2 cos2 t)1/2 1

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and their product is (iii) It is enough to

p2 q 2 (1−ǫ2 cos2 t) = q2 . (p2 sin2 t+q 2 cos2 t) (p −f2 ).n (p −f1 ).n . = prove that kp −f1 k kp −f2 k

Computation shows that

both sides are equal to q.

1.1.7 When the circle has rotated through an angle t, its centre has moved to (at, a), so the point on the circle initially at the origin is now at the point (a(t − sin t), a(1 − cos t)) (see the diagram above). 1.1.8 Suppose that a point (x, y, z) lies on the cylinder if x2 + y 2 = 1/4 and on the sphere if (x + 21 )2 + y 2 + z 2 = 1. From the second equation, −1 ≤ z ≤ 1 so let z = sin t. Subtracting the two equations gives x + 41 + sin2 t = 43 , so x = 12 − sin2 t = cos2 t − 12 . From either equation we then get y = sin t cos t (or y = − sin t cos t, but the two solutions are interchanged by t 7→ π − t). √ √ 1.1.9 γ˙ = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos 2t) = √ √ 2( 2 − 1, 1) at t = π/4. So the 1 tangent line is y − ( √2 − 1) = (x − 2)/( 2 − 1) and the normal line is √ √ 1 y − ( √ − 1) = −(x − 2)( 2 − 1). 2 1.1.10 (i) Putting x = sec t gives y = ± sec t tan t so γ (t) = (sec t, ± sec t tan t) gives parametrizations of the two pieces of this curve (x ≥ 1 and x ≤ −1). 2 3t , y = 3t . (ii) Putting y = tx gives x = 1+t 3 1+t3 1.1.11 (i) From x = 1 + cos t, y = sin t(1 + cos t) we get y = x sin t so y 2 = x2 (1 − (x − 1)2 ) = x3 (2 − x). (ii) y = tx so x4 = t2 x3 + t3 x3 = y 2 x + y 3 = y 2 (x + y). 1.1.12 (i) γ˙ (t) = (− sin t, cos t + cos 2t) so γ˙ (t) = 0 if and only if t = nπ (n ∈ Z) and (−1)n + 1 = 0, so n must be odd.

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(ii) γ˙ (t) = (2t+3t2 , 3t2 +4t3 ). This vanishes ⇐⇒ t(2+3t) = 0 and t2 (3+4t) = 0, i.e. ⇐⇒ t = 0.

1.1.13 (i) Let OP make an angle θ with the positive x-axis. Then R has coordinates γ (θ ) = (2a cot θ, a(1 − cos 2θ )). (ii) From x = 2a cot θ, y = a(1 − cos 2θ), we get sin2 θ = y/2a, cos2 θ = cot2 θ sin2 θ = x2 y/8a3 , so the Cartesian equation is y/2a + x2 y/8a3 = 1. 1.1.14 Let the fixed circle have radius a, and the moving circle radius b (so that b < a in the case of the hypocycloid), and let the point P of the moving circle be initially in contact with the fixed circle at (a, 0). When the moving circle has rotated through an angle ϕ, the line joining the origin to the point of contact of the circles makes an angle θ with the positive x-axis, where aθ = bϕ. The point P is then at the point γ (θ) = ((a + b) cos θ − b cos(θ + ϕ), (a + b) sin θ − b sin(θ + ϕ))      a+b a+b = (a + b) cos θ − b cos θ , (a + b) sin θ b b in the case of the epicycloid,

and γ (θ ) = ((a − b) cos θ + b cos(ϕ − θ ), (a − b) sin θ − b sin(ϕ − θ ))      a−b a−b θ , (a − b) sin θ = (a − b) cos θ − b cos b b in the case of the hypocycloid. 1.1.15 γ˙ (t) = (et (cos t − sin t), et(sin t + cos t)) so the angle θ between γ (t) and γ˙ (t) is given by 1 e2t (cos2 t − sin t cos t + sin2 t + sin t cos t) γ .γ˙ = = , cos θ = k γγ kk γ˙ k e2t ((cos t − sin t)2 + (sin t + cos t)2 ) 2

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so θ = π/3. 1.1.16 γ˙ (t) = (t cos t, t sin t) so a unit tangent vector is t = (cos t, sin t). The distance of the normal line at γγ (t) from the origin is γ (t)...tt| t = | cos2 t + t sin t cos t + sin2 t − t sin t cos t| = 1. |γ Rt 1.2.1 γ˙ (t) = (1, sinh t) so k γ˙ k = cosh t and the arc-length is s = 0 cosh u du = sinh t. 1.2.2 (i) k γ˙ k2 = 14 (1 + t) + 14 (1 − t) + 21 = 1. 2 2 2 9 2 2 (ii) k γ˙ k2 = 16 25 sin t + cos t + 25 sin t = cos t + sin t = 1. 1.2.3 Denoting d/dθ by a dot, γ˙ = (r˙ cos θ − r sin θ, r˙ sin θ + r cos θ) so k γ˙ k2 = r˙2 + r2 . Hence, γ is regular unless r = r˙ = 0 for some value of θ. It is unit-speed if and only if r˙2 = 1 − r2 , which gives r = ± sin(θ + α) for some constant α. To see that this is the equation of a circle of radius 1/2, see the diagram in the proof of Theorem 3.2.2. 1.2.4 Since u is a unit vector, |γ˙ .u | = k γ˙ k cos θ, where θ is the angle between γ˙ and Rb Rb u, so γ˙ .u ≤k γ˙ k. Then, (q q − p )..u = (γ γ(b) γ − γ (a))..u = a γ˙ ..uu dt ≤ a k γ˙ k dt. Taking uu = (q q − p)/ k q − p k gives the result. √ k = 36t2 + 1 − 18t2 + 81t4 = 1 + 9t2 . So the ˙ 1.2.5 γ˙ (t) = (6t, 1 − 9t2 ) so k γγ(t) arc-length is Z t

(1 + 9u2 )du = t + 3t3 .

s=

0

1.2.6 We have γ˙ (t) = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos 2t) so k γ˙ (t) k =

p

8(1 − sin t sin 2t − cos t cos 2t =

So the arc-length is s=

Z

x

0

p

8(1 − cos t) = 4 sin

t . 2

 x x t = 16 sin2 . 4 sin dt = 8 1 − cos 2 2 4

1.2.7 The cycloid is parametrized by γγ(t) = a(t − sin t, 1 − cos t), where t is the angle through which the circle has rotated. So t γ˙ = a(1 − cos t, sin t), k γ˙ k2 = a2 (2 − 2 cos t) = 4a2 sin2 , 2 and the arc-length is Z

0

2π

 t t=2π t = 8a. 2a sin dt = −4a cos  2 2 t=0

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1.2.8 The curve intersects the x-axis where cosh t = 3, say at t = ±a where a > 0. We have k γ˙ (t) k2 = 2 cosh2 t − 2 cosh t so the arc-length is Z ap Z ap 2 2 cosh2 t − 2 cosh t dt. s= 2 cosh t − 2 cosh t dt = 2 −a

0

To evaluate the integral put u = cosh t. Then, √ Z 3r hp i3 √ √ √ √ 2+ 3 u √ . u(u + 1) − ln( u + u + 1) = 2 3− 2−ln du = s=2 u+1 1 1+ 2 1

... b + t2c, where aa, b, b c are constant vectors. This implies 1.2.9 γ = 0 =⇒ γ = aa + tb that γ is contained in the plane passing through the point with position vector a and parallel to the vectors bb and cc (i.e. perpendicular to b × cc). (If b and c are parallel there are infinitely-many such planes.)

1.3.1 (i) γ˙ = sin 2t(−1, 1) vanishes when t is an integer multiple of π/2, so γ is not regular. (ii) γ is regular since γ˙ 6= 0 for 0 < t < π/2. (iii) γ˙ = (1, sinh t) is obviously never zero, so γγ is regular. 1.3.2 x = r cos θ = sin2 θ, y = r sin θ = sin2 θ tan θ, so the parametrization in terms of θ is θ 7→ (sin2 θ, sin2 θ tan θ). Since θ 7→ sin θ is a bijective smooth map (−π/2, π/2) → (−1, 1), with smooth inverse t 7→ sin−1 t,√ t = sin θ is a reparametrization map. Since sin2 θ = t2 , sin2 θ tan θ = t3 / 1 − t2 , so the reparametrized curve is as stated. 1.3.3 (i) γ˙ = 0 at t = 0 ⇐⇒ m and n are both ≥ 2. If m > 3 the first components of ... ¨ and γ are both 0 at t = 0 so ¨γ and γ¨ are linearly dependent at t = 0; similarly γ ... ¨ or γ if n > 3. So there are four cases: if (m, n) = (2, 2) or (3, 3) then either γ is zero at t = 0, so the only possibilities for an ordinary cusp are (m, n) = (2, 3) ... and (3, 2) and then γ¨ and γ are easily seen linearly independent at t = 0.   to be 3 t 2 (ii) Using the parametrization γ (t) = t , √1−t2 , we get γ˙ = 0, γ¨ = (2, 0), ... γ = (0, 6) at t = 0 so the origin is an ordinary cusp. (iii) Let γ˜(˜t) be a reparametrization of γγ(t), and suppose γ has an ordinary γ /dt˜ = (dγ cusp at t = t0 . Then, at t = t0 , d˜ γ /dt)(dt/dt˜) = 0, d2γ˜/dt˜2 = 2 3 ˜ 2 , d3γ˜/dt˜3 = (d3γ /dt )(dt/dt˜)3 + 3(d2γ/dt (d2γ /dt )(dt/dt) γ 2 )(dt/dt˜)(d2 t/dt˜2 ). ˜ /d˜t2 and d3γ˜/dt˜3 are Using the fact that dt/dt˜ 6= 0, it is easy to see that d2 γ linearly independent when t = t0 . 1.3.4 (i) If ˜γ (t) = γ (ϕ(t)), let ψ be the inverse of the reparametrization map ϕ. Then γ ˜ (ψ(t)) = γ (ϕ(ψ (t))) = γ(t). γ (ii) If γ˜(t) = γ (ϕ(t)) and γˆ(t) = ˜γ (ψ(t)), where ϕ and ψ are reparametrization maps, then γˆ(t) = γ ((ϕ ◦ ψ)(t)) and ϕ ◦ ψ is a reparametrization map because it d (ϕ(ψ (t)) = ϕ(ψ(t)) ˙ 6= 0 as ϕ˙ and ψ˙ are both 6= 0. ψ(t) ˙ is smooth and dt

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1.3.5 (i) γ˙ (t) = (2t, 3t3 ) which vanishes ⇐⇒ t = 0. So γ is not regular. (ii) γ˙ (t) = (− sin t − sin 2t, cos t + cos 2t) so k γ˙ (t) k = 2 cos 2t . Hence, γ˙ (t) = 0 ⇐⇒ t is an odd multiple of π. Thus, when t is retricted to the interval −π < t < π, γ is regular.   cos t 4t γ ˙ is obviously never zero, so γ is regular. Let ϕ(t) = 1+sin (t) = 2, − 1.3.6 t. (1+t2 )2 1 Then, dϕ dt = − 1+sin t is never zero so ϕ is a reparametrization map. Setting u = ϕ(t) we get

γ (u) =

2 cos t , 1 + sin t 1 +

2 cos2 t (1+sin t)2

!

=



2 cos t 2(1 + sin t)2 , 1 + sin t 2 + 2 sin t



= γ˜(t),

so γ is a reparametrization of γ˜ . 1.3.7 γ˙ (t) = (aω cos ωt, b cos t) = 0 ⇐⇒ cos ωt = cos t = 0 ⇐⇒ ωt = (2k +1) 2π and t = (2l + 1) π2 , where k, l ∈ Z. This means that ω = 2k+1 2l+1 is the ratio of two odd integers. Rt R˜ γ /d˜ 1.3.8 We have s = t0 k dγ γ /du k du, s˜ = ˜tt0 k d˜ u k d˜ u. By the chain rule, Rt γ /du = (d˜ γ/d˜ u k (d˜ u/du) du = ±˜ s, the sign u)(d˜ u/du), so s = ± t0 k d˜γ /d˜ dγ being that of d˜ u/du. 1.3.9 We can assume that the curve γγ is unit-speed and that the tangent lines all pass through the origin (by applying a translation to γ ). Then, there is a scalar λ(t) ˙ γ + λ γ˙ = (λ˙ + λ2 )γ. γ This implies γ (t) for all t. Then, γ¨ = λγ such that γ˙ (t) = λ(t)γ that γ¨ is parallel to γ˙ . But these vectors are perpendicular since γγ is unit-speed. aa+bb where aa, bb are constant vectors, i.e. γγ is a straight Hence, ¨γ = 0, so γ (t) = ta line. If all the normal lines are parallel, so are all the tangent lines. If γ is assumed to be unit-speed, this means that γ˙ is constant, say equal to a. a Then γ (t) = taa + bb as before. 1.4.1 It is closed because γγ(t + 2π) = γ (t) for all t. Suppose that γ (t) = γ (u). Then cos3 t(cos 3t, sin 3t) = cos3 u(cos 3u, sin 3u). Taking lengths gives cos3 t = ± cos3 u so cos t = ± cos u, so u = t, π − t, π + t or 2π − t (up to adding multiples of 2π). The second possibility forces t = nπ/3 for some integer n and the third possibility is true for all t. Hence, the period is π and for the self-intersections we need only consider t = π/3, 2π/3, giving u = 2π/3, π/3, respectively. Hence, there is a unique self-intersection at γ (π/3) = (−1/8, 0). 1.4.2 The curve ˜γ (t) = (cos(t3 + t), sin(t3 + t)) is a reparametrization of the circle γ (t) = (cos t, sin t) but it is not closed. 1.4.3 If γ is T -periodic then it is kT -periodic for all k 6= 0 (this can be proved by induction on k if k > 0, or on −k if k < 0). If γ is T1 -periodic and T2 -periodic

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then it is k1 T1 - and k2 T2 -periodic for all non-zero integers k1 , k2 , so γ (t + k1 T1 + k2 T2 ) = γγ (t + k1 T1 ) as γ is k2 T2 -periodic, which = γ (t) as γ is k1 T1 -periodic. 1.4.4 If γ is T -periodic write T = kT0 + T1 where k is an integer and 0 ≤ T1 < T0 . By Exercise 1.4.3 γ is T1 -periodic; if T1 > 0 this contradicts the definition of T0 . 1.4.5 (i) Choose T1 > 0 such that γγ is T1 -periodic; then T1 is not the smallest positive number with this property, so there is a T2 > 0 such that γγ is T2 -periodic. Iterating this argument gives the desired sequence. (ii) The sequence {Tr }r≥1 in (i) is decreasing and bounded below, so must converge to some T∞ ≥ 0. Then γγ is T∞ -periodic because (using continuity of γ ) γ (t + T∞ ) = limr→∞ γ (t + Tr ) = limr→∞ γ (t) = γ (t). By Exercise 1.4.3, γ is (Tr − T∞ )-periodic for all r ≥ 1, and this sequence of positive numbers converges to 0. (iii) If {Tr } is as in (i) and Tr → 0 as r → ∞, then by the mean value theorem 0 = (f (t + Tr ) − f (t))/Tr = f˙(t + λTr ) for some 0 < λ < 1. Letting r → ∞ gives f˙(t) = 0 for all t, so f is constant. 1.4.6 Following the hint, since T0 = (ki /k)Ti is an integer multiple of Ti , each γi is T0 periodic. Let T be the union of the finite sets of real numbers {Ti , 2Ti , . . . , ki Ti } over all i such that γi is not constant, and let P = {T ′ ∈ T | γγ is T ′ -periodic}. Then P is finite (because T is) and non-empty (because T ∈ P). The smallest element of P is the smallest positive number T 0′ such that γ is T0′-periodic (since if γ is T ′ -periodic either T ′ > T or T ′ ∈ P). By Exercise 1.4.4, T0 = k ′ T ′0 for some integer k ′ and then there are integers k ′i such that T0′ = ki′ Ti for all i such that γi is not constant. Then, ki Ti /k = k ′ ki Ti so kk ′ divides each ki . As k is the largest such divisor, k ′ = 1, so T0 = T0′. 2

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u −3 2 2 1.4.7 If γ (t) = γ (u) then tt2 −3 +1 = u2 +1 . This implies t = u . This shows that γ is not periodic and that for a self-intersection we must have t = −u. The √ t(t2 −3) equation γ (t) = γ (−t) implies t2 +1 = 0, so t = 0 or ± 3. Hence, the unique √ √ γ self-intersection is at γγ( 3) = γ(− 3) = 0.

1.4.8 If γ (t) = γ (u) then sin t = sin u. Also, k γγ (t) k2 = k γ (u) k2 , which gives 5 + 4 cos t = 5 + 4 cos u, hence cos t = cos u. So t = u + 2nπ for some integer n. We must have cos ω(t + 2nπ) = cos ωt so ω(t + 2nπ) = ωt + 2mπ for some integer m. Hence, ω = m/n is rational. The period T of γγ is the smallest positive number such that γ(t γ + T ) = γ (t) for all t. From the first part we know that T = 2N π for some integer N . Then, N is the smallest positive integer N such that ω(t+2N π) = ωt +2M π for some integer M , i.e. the smallest positive integer such that mN/n is an integer. If m and n have no common factor, N = n.

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1.5.1 x(1 − x2 ) ≥ 0 ⇐⇒ x ≤ −1 or 0 ≤ x ≤ 1 so the curve is in (at least) two pieces. The parametrization is defined for t ≤ −1 and 0 ≤ t ≤ 1 and it covers the part of the curve with y ≥ 0. 1.5.2 If γ (t) = (x(t), y(t), z(t)) is a curve in the surface f (x, y, z) = 0, differentiating f (x(t), y(t), z(t)) = 0 with respect to t gives xf ˙ x + yf ˙ y + zf ˙ z = 0, so γ˙ is perpendicular to ∇ f = (fx , fy , fz ). Since this holds for every curve in the surface, ∇f ∇ is perpendicular to the surface. The surfaces f = 0 and g = 0 should intersect in ∇f and ∇g are not parallel at any point of the intersection. a curve if the vectors ∇ ˙ v, ˙ w˙ 1.5.3 Let γ (t) = (u(t), v(t), w(t)) be a regular curve in R3 . At least one of u, is non-zero at each value of t. Suppose that u˙ (t0 ) 6= 0 and x0 = u(t0 ). As in the ‘proof’ of Theorem 1.5.2, there is a smooth function h(x) defined for x near x0 such that t = h(x) is the unique solution of x = u(t) for each t near t0 . Then, for t near t0 , γ (t) is contained in the level curve f (x, y, z) = g(x, y, z) = 0, where f (x, y, z) = y − v(h(x)) and g(x, y, z) = z − w(h(x)). The functions f ∇f = (−vh ˙ ′ , 1, 0), and g satisfy the conditions in the previous exercise, since ∇ ∇g = (−wh ˙ ′ , 0, 1), a dash denoting d/dx.

1.5.4 The equation cannot be satisfied if x = 1, so the line x = 1 separates the curve into at least two connected pieces. The curve γγ is a parametrization because (x − 1)2 (x2 + y 2 ) = cos2 t((1 + cos t)2 + (sin t + tan t)2 )

= cos2 t(1 + cos t)2 (1 + tan2 t) = (1 + cos t)2 = x2 .

But since tan t is undefined when t = ±π/2, ±3π/2, ..., γγ must be defined on one of the intervals (−π/2, π/2), (π/2, 3π/2), etc. The restriction of γγ to the first of these intervals parametrizes the part of the curve with x > 1, the second gives the part with x < 1. This shows that the two parts of the curve with x < 1 and with x > 1 are connected. 1.5.5 Letting f (x, y, z) = x2 + y 2 = 1/4, g(x, y, z) = x2 + y 2 + z 2 + x − 3/4, we find ∂f ∂f ∂f ∂g ∂g ∂g that ( ∂x , ∂y , ∂z ) = (2x, 2y, 0), (∂x , ∂y , ∂z ) = (2x + 1, 2y, 2z). If these vectors are parallel, z = 0; if y 6= 0 the vectors must then be equal and this is impossible; so y = 0 and then we must have x = 1/2 to satisfy both equations. Hence, the vectors are parallel only at the point (1/2, 0, 0). For the parametrization γ in Exercise 1.1.8, γ˙ (t) = (− sin 2t, cos 2t, cos t). This is never zero as k ˙γ (t) k2 = 1 + cos2 t, so γ is regular. 1.5.6 Define Θ(t) = tan πθ(t)/2, where θ is the function defined in Exercise 9.4.3. Then Θ is smooth, Θ(t) = 0 if t ≤ 0, and Θ : (0, ∞) → (0, ∞) is a bijection. The curve  (Θ(t), Θ(t)) if t ≥ 0, γ (t) = (−Θ(−t), Θ(−t)) if t ≤ 0, is a smooth parametrization of y = |x|.

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There is no regular parametrization of y = |x|. For if there were, there would be a unit-speed parametrization γ˜(t), say, and we can assume that ˜γ (0) = (0, 0). ˜˙ would have to be either √1 2 (1, 1) or − √12 (1, 1) when The unit tangent vector γ x > 0, so by continuity we would have γ˜˙ (0) = ± √1 (1, 1). But, by considering the 2

part x < 0 in the same way, we see that γ˙˜(0) = ± √12 (1, −1). These statements are contradictory.

Chapter 2 2.1.1 (i) γ is unit-speed (Exercise 1.2.2(i)) so 1 1 1 . κ = k γ¨ k = k ( (1 + t)−1/2 , (1 − t)−1/2 , 0) k = p 4 4 8(1 − t2 )

(ii) γ is unit-speed (Exercise 1.2.2(ii)) so κ = k γ ¨ k = k (− 45 cos t, sin t, 35 cos t) k = 1. k(1,sinh t,0)×(0,cosh t,0)k cosh t = sech 2 t using Proposition 2.1.2. (iii) κ = = cosh 3 t k(1,sinh t,0)k3 2 2 (iv) (−3 cos t sin t, 3 sin t cos t, 0)×(−3 cos3 t+6 cos t sin2 t, 6 sin t cos2 t−3 sin3 t, 0) k(0,0,−9 sin2 t cos2 t)k = (0, 0, −9 sin2 t cos2 t), so κ = k(−3 cos2 t sin t,3 sin2 t cos t,0)k3 = 3| sin1t cos t| . This becomes infinite when t is an integer multiple of π/2, i.e. at the four cusps (±1, 0) and (0, ±1) of the astroid.

2.1.2 The proof of Proposition 1.3.5 shows that, if v (t) is a smooth (vector) function of t, then k vv(t) k is a smooth (scalar) function of t provided v(t) is non-zero for all t. The result now follows from the formula in Proposition 2.1.2. The curvature of the regular curve γ (t) = (t, t3 ) is κ(t) = 6|t|/(1 + 9t4 )3/2 , which is not differentiable at t = 0. 2.1.3 Working in R3 , γ˙ (t) = (1 − cosh 2t, 2 sinh t, 0) = 2 sinh t(− sinh t, 1, 0), γ¨(t) = γ˙ ×γ¨ k γ = (0, 0, 4 sinh2 t cosh t) and κ = k kγ˙ k3 = (−2 sinh 2t, 2 cosh t, 0) which gives γ˙ ר 1/2 sinh t cosh2 t. This is non-zero if t > 0, but → 0 as t → ∞.

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2.1.4 We have γ˙ (t) = (sec t tan t, sec t tan2 t + sec3 t, 0), γ¨ (t) = (2 sec3 t − sec t, 6 sec3 t tan t − sec t tan t), which gives γ˙ × γ¨γ = sec4 t(0, 0, 2 sec2 t − 3). Hence, the curvature vanishes where sec2 t = 3/2. If −π/2 < t < π/2, cos t > 0 so the curvature vanishes at the two values of t between −π/2 and π/2 at which cos t = 2/3, i.e. at the point √ (3/2, 5/2). 2.2.1 Differentiate t.n n s = 0 and use t˙ = κsn s . 2.2.2 If γ is smooth, t = γ˙ is smooth and hence so are ˙t and ns (since nns is obtained by applying a rotation to t). So κs = t˙ .n ns is smooth. 2.2.3 For the first part, from the results in Appendix 1 it suffices to show that κ ˜s = −κs if M is the reflection in a straight line l. But this is clear: if we take the fixed angle ϕ0 in Proposition 2.2.1 to be the angle between l and the positive x-axis, then (in the obvious notation) ϕ˜ = −ϕ. Conversely, if γ and γ˜ have the same non-zero curvature, their signed curvatures are either the same or differ in sign. In the first case the curves differ by a direct isometry by Theorem 2.2.5; in the latter case, applying a reflection to one curve gives two curves with the same signed curvature, and these curves then differ by a direct isometry, so the original curves differ by an opposite isometry. 2.2.4 The first part is obvious as the effect of the dilation is to multiply s by a and leave ϕ unchanged. For the second part, consider the small piece of the chain bet...