# Ps02 cmth03 unit 1

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- 1. CHAPTER 1Curve Theory 1.1. What is a curve? Denition 1.1.1. A parametrized curve in Rn is a continuous function : I Rn , where I is an interval in R. Examples 1.1.2. Parametrization of Cartesian curves. (i) A parametrization of a parabola y = x 2 is (t) = (t, t 2 ), t R. The curve 1 (t) = (t 2 , t 4 ), t R is not a parametrization of y = x 2 . The curve (t) = (t 3 , t 6 ), t R is a parametrization of y = x 2 . 2 y2 (ii) The curve (t) = (a cos t, b sin t), t R is a parametrization of the ellipse x 2 + b2 = a 1. (iii) The curve (t) = (a sec t, b tan t), t ( , ) is a parametrization of the hyper2 2 2 y2 bola x 2 b2 = 1. a (iv) The curve (t) = (a cos3 t, a sin3 t), t (0, 2] or t R is a parametrization of the 2 2 2 astroid x 3 + y 3 = a 3 Examples 1.1.3. (i) Let : R R2 be dened by (t) = (a cos t, b sin t), where a, b R{0}. Then 2 y2 the image of is an ellipse in R2 . Its Cartesian equation is x 2 + b2 = 1. In a particular when a = b, then the image of is a circle in R2 . (ii) Let : R R2 be dened by (t) = (t, t 2 ). Then the image of is a parabola in R2 . Its Cartesian equation is y = x 2 . (iii) Let : R R2 be dened by (t) = (a cosh t, b sinh t), where a, b R{0}. Then 2 y2 the image of is a part of the hyperbola x2 b2 = 1. a (iv) Let : R R2 be dened by (t) = (et cos t, et sin t). Then the image of is a logarithmic spiral in R2 . (v) Let : R R3 be dened by (t) = (a+lt, b+mt, c+nt), where a, b, c, l, m, n R and l2 + m2 + n2 = 0. Then the image of is a line in R3 passing through (a, b, c) having direction (l, m, n). Its Cartesian equation is xa = yb = zc . l m n

2. 1. What is a curve?i(vi) Let : R R3 be dened by (t) = (a cos t, b sin t, ct), where a, b, c R{0}. 2 y2 Then the image of is a helix in R3 . It is a helix with the base ellipse x 2 + b2 = 1. a When a = b, it is called circular helix. Denition 1.1.4. A parametrized curve in R3 is called planar if it is contained in some plane of R3 . It is called non-planar or twisted if it is not planar. Let : (a, b) Rn be a parametrized curve. Then = (1 , 2 , . . . , n ), where each i is a mapping from (a, b) to R. The symbol is the derivative of and it means = (1 , 2 , . . . , n ). Throughout it is assumed that all curves are smooth, i.e., all the derivatives of exist. Examples 1.1.5. (i) The curve (t) = (t, t 2 , t 3 ) is not planar. Suppose that is planar. Then the exist a, b, c, d R, a2 + b2 + c2 = 0, such that at + bt 2 + ct 3 = d for all t. But then a = b = c = 0 (how???). This contradicts the fact a2 + b2 + c2 = 0. (ii) The curve (t) = (cos t, sin t, 3 sin t + 4 cos t) is planar. One can see that the coordinates of satisfy the equation of the plane z = 4x+3y. Hence is planar. 3 (iii) The curve 1 (t) = 4 cos t, 1 sin t, 5 cos t) is a plane curve. 5 Denition 1.1.6. Let be a parametrized curve. Then the vector (t) is called the tangent vector to at the point (t). If (t) = 0, then the equation of the tangent to at the point (t) is R (t) = u (t), u R. Exercise 1.1.7. Find the equation of the tangent to the following curves. (i) (ii) (iii) (iv)(t) = (a cos t, a sin t, bt), (a, 0, 2b). (t) = (t, t 2 , t 3 ), (1, 1, 1). 1 1 (t) = (cos2 t, sin2 t), ( 2 , 2 ). (t) = (et , t 2 ), (1, 0).Example 1.1.8. For a logarithmic spiral (t) = (et cos t, et sin t), show that the angle between (t) and (t) is independent of t. Here (t) = (et cos t, et sin t) and (t) = (et cos t et sin t, et sin t + et cos t). Let (t) be the angle between (t) and (t). Then (t) = cos1 (t) (t) (t) (t) 3. ii1. CURVE THEORY= cos1e2t 2e2t= cos11 2. Therefore the angle between (t) and (t) is independent of t. Proposition 1.1.9. Let be a parametrized curve in Rn . If = a, where a = 0, then is a part of a line. PROOF. Since = a, we have (t) = at + b. Hence is part of a line.1.2. Regular curves Denition 1.2.1. The arc-length of a curve starting at the point (t0 ) is the function s(t) given by ts(t) = (u) du. t0Thus, s(t0 ) = 0 and s(t) is positive or negative according to whether t is larger or smaller than t0 . Example 1.2.2. Compute the arc-length of the logarithmic spiral (t) = (ekt cos t, ekt sin t) starting at the point (1, 0). We have (t) = (ekt sin t + kekt cos t, ekt cos t + kekt sin t). Therefore (t) = ekt ( sin t + k cos t)2 + (cos t + k sin t)2 = ekt 1 + k2 . Since (0) = (1, 0), the arc-length of starting at the point (1, 0) is t 1 + k2 kt (e 1). s(t) = eku 1 + k2 du = k 0Denition 1.2.3. Let : (a, b) Rn be a parametrized curve. Then is called regular at (t) if (t) = 0 (or (t) > 0). A point (t) of the curve is called a singular point if is not regular at that point. A curve is said to be regular if all its points are regular. Denition 1.2.4. Let : (a, b) Rn be a parametrized curve. Then the scalar (t) is called the speed of at the point (t). The curve is called a unit-speed curve if (t) = 1 for all t. It follows from the denitions that every unit-speed curve is regular. The converse is not true. For example (t) = (cos 2t, sin 2t) is regular but not unit-speed. 4. 2. Regular curvesiiiLemma 1.2.5. Let : (a, b) Rn be a parametrized curve. If (t) = 1 for all t, then (t) and (t) are perpendicular for all t. In particular, if is a unit-speed curve, then (t) (t) for all t. PROOF. Since (t) = 1 for all t, we have (t)(t) = 1. Hence (t)(t) = 0 for all t, for all t. i.e., (t) (t) It follows that if is a unit-speed curve, then (t) (t) for all t as (t) = 1 for all t. Lemma 1.2.6. Let : (a, b) Rn be a regular curve. Then the arc-length of , starting at any point of the curve, is a smooth map. t PROOF. We have s(t) = t0 (u) du. Then ds = . Let = (1 , 2 , . . . , n ). Since is dt regular, it follows that 12 + 22 + + n2 > 0. Dene f : (0, ) R and g : (a, b) R by f(t) = t and g(t) = 12 (t) + + n2 (t). Then both f and g are smooth maps. Since the range of g is contained in (0, ), the domain of f, it follows that the map f g is a smooth map. But f g = ds . Therefore ds is a smooth map and hence s is a smooth dt dt map.Denition 1.2.7. A parametrized curve : ( , b) Rn is called a reparametrization a a of : (a, b) Rn if there exits a bijective smooth map : ( , b) (a, b), whose inverse is also smooth, such that = . The map is called the reparametrization map for the above reparametrization. If is a reparametrization of with the reparametrization map , then = 1 . Hence is a reparametrization of . We note that two curves which are reparametrizations of each other have the same image. Hence they have the same geometric properties. Exercise 1.2.8. Let C be the collection of all parametrized curve in Rn . Let , C. We say if is a reparametrization of . Show that the above relation on C is an equivalence relation. Proposition 1.2.9. Any reparametrization of a regular curve is regular. PROOF. Let : ( , b) Rn be a reparametrization of a regular curve : (a, b) Rn . a Then there is a bijective smooth map : ( , b) (a, b), whose inverse is also smooth, a such that = . Let : (a, b) ( , b) be the inverse of . Then (t) = t for all a d d t (a, b). Therefore d dt = 1 and hence d ( = 0 for any ( , b). Since = , t) t a t d t d d d d d we have d = dt d . Since is regular dt = 0. Since d is never vanishing, it follows t t t that d is never zero (vector), i.e., is regular. d t 5. iv1. CURVE THEORYProposition 1.2.10. A parametrized curve has a unit-speed reparametrization i it is regular. PROOF. Let be a unit-speed reparametrization of the curve . Since is a reparametrization of , is a reparametrization of . Since is regular (as it is unit-speed) and any reparametrization of a regular curve is regular, it follows that is regular. Conversely, assume that : (a, b) Rn be regular. Let s be the arc-length of t , starting at the point (t0 ), i.e., s(t) = t0 (u) du. We note that s is a smooth map. Since is regular, ds (t) = (t) > 0 for all t, and hence s is strictly increasing dt ds and so it is one-one. Since dt (t) = 0 for all t, it follows from the inverse function theorem that s1 : ( , b) (a, b) is smooth, where ( , b) is the range of s. Consider a a n the reparametrization : ( , b) R given by = s1 , i.e., s = . Now a d ds d ds = dt . Then d dt = d = ds , i.e., d = 1. Hence has a unit-speed dt dt d dt t t d t reparametrization. Corollary 1.2.11. Let be a regular curve, and let be a unit-speed reparametrization of given by u = , where u is a smooth map. Then u = s + c, where s is the arc-length of , starting at any point, and c is a constant. Conversely, if u = s + c, then is a unit-speed reparametrization of . PROOF. Since u = , we have d du = d . This implies d du = d . Therefore dt dt dt d dt t d t du = ds , as is unit-speed. This proves that u = s + c. dt dt Conversely, assume that u = s + c. Clearly u is bijective, smooth and its inverse is also smooth. We have (s + c) = . Hence d ds = d . This implies d ds = dt d dt t d dt t d = ds . Therefore d = 1, i.e., is unit-speed reparametrization of . dt dt d t1.3. Curvature and Torsion Denition 1.3.1. Let : (a, b) R3 be a unit-speed curve. Then the curvature of (t) of at the point (t) is dened by (t) = (t) . Examples 1.3.2. (i) Let a, b R3 , and let a = 1. Then the curvature of the unit-speed curve (t) = ta + b is zero everywhere. (ii) Let (x0 , y0 ) R2 , and let R > 0. Then the curvature of the unit-speed curve t t (t) = (x0 + R cos( R ), y0 + R sin( R )), which is a circle with center (x0 , y0 ) and 1 radius R, is R everywhere. 6. 3. Curvature and TorsionvProposition 1.3.3. If is a regular curve in R3 , then its curvature is given by the formula = . 3 PROOF. Let be a unit-speed reparametrization of given by s = . We note that the curvature of (s(t)) is same as the curvature of at the point (t). We have 2 ds 2 2 = = , and hence ds d s = . Let denote the derivative of with dt dt dt 2 ds respect to s. Since s = , = , i.e., = ds . Dierentiating it again, we have dt= =Hence = =ds dtdtd2 s dt 2 ds 2 dtdt = dsds 2 dtds d2 s dt dt 2 ds 4 dt ( ) ( ) ( ) = . 4 4 ( ) 4 = ( ) 4 = . The last step follows 3 because . Example 1.3.4. Let (t) = (a cos t, a sin t, bt), where a, b > 0. Then computations a shows that (t) = a2 +b2 . Exercises 1.3.5. Compute the curvature of the following curves. (i) (t) = (ekt cos t, ekt sin t). (ii) (t) = (t, cosh t). Also determine a point having maximum curvature. (iii) (t) = (t, t 2 , t 3 ). Now we dene a special type of curvature for plane curves only. Denition 1.3.6. Let : (a, b) R2 be a unit-speed curve. Let t be the unit tangent of , i.e., t = . The unit vector ns obtained by rotating the tangent vector t anticlockwise by an angle is called the signed unit normal of . Since is unit-speed, 2 is perpendicular to t and hence it is parallel to ns . Therefore there exists a map s : (a, b) R such that = s ns . The map s is called the signed curvature of . Since = , it follows that = |s |, i.e., the absolute value of the signed curvature is the curvature Let R : R2 R2 be a rotation operator (it rotates a vector anti-clockwise by an angle ). We note that R is linear and R (1, 0) = (cos , sin ) and R (0, 1) = 7. vi1. CURVE THEORY( sin , cos ). Let (x, y) R2 . Then R (x, y) = R (x(1, 0) + y(0, 1)) = xR (1, 0) + yR (0, 1) = x(cos , sin ) + y( sin , cos ). Hence R (x, y) = (x cos y sin , x sin + y cos ). In fact, R is an isometry of R2 , i.e., it satises R (x1 , y1 ) R (x2 , y2 ) = (x1 , y1 ) (x2 , y2 )((x1 , y1 ), (x2 , y2 ) R2 ).In particular, when = , R (x, y) = (y, x). 2 2 Fix a vector a R2 . Dene Ta : R2 R2 by Ta (x) = x + a. Then Ta is called a translation by a vector a (it translates given vector by a) and it is an isometry. Denition 1.3.7. An isometry of R2 of the form Ta R is called a direct isometry, where a R2 and [0, 2). Example 1.3.8. Compute the signed unit normal of the curve (t) = (ekt cos t, ekt sin t). The curve is not unit-speed. The unit tangent will be the unit vector in the direc tion of the tangent vector, i.e., t(t) = (t) . Now (t) = (ekt sin t+kekt cos t, ekt cos t+ (t) 1 kekt sin t) and (t) = ekt 1 + k2 . Hence t(t) = 1+k2 ( sin t + k cos t, cos t + k sin t). Since ns (t) = R (t(t)), we have 2 ns (t) = 1 1 + k2( cos t k sin t, sin t + k cos t)Exercise 1.3.9. If : (a, b) R2 is a unit-speed curve, then show that both signed unit normal and signed curvature are smooth maps. Proposition 1.3.10. Let : (a, b) R2 be a unit-speed curve, and let s0 (a, b). Let 0 R such that (s0 ) = (cos 0 , sin 0 ). Then there exists a unique smooth map : (a, b) R such that (s) = (cos (s), sin (s)) for all s (a, b) and (s0 ) = 0 . PROOF. Let (s) = (f(s), g(s)) (s (a, b)). Since is smooth, both f and g are smooth maps. Since is unit-speed, f 2 + g 2 = 1 and hence f f + g g = 0. Dene : (a, b) R by s(s) = 0 +(f g fg). s0 8. 3. Curvature and TorsionviiSince f and g are smooth, is a smooth map and (s0 ) = 0 . Let F = f cos + g sin and G = f sin g cos . Then F = f cos f sin + g sin + g cos = (f + g ) cos + (g f ) sin . Now, f + g = f + g(f g fg) = f(1 g 2 ) + fg g = ff 2 + fg g = f(f f + g g) = 0. By similar arguments it follows that g f = 0. Hence F = 0, i.e., F is a constant map. Therefore F(s) = F(s0 ) for all s. Now F(s0 ) = f(s0 ) cos (s0 ) + g(s0 ) sin (s0 ) = cos2 0 + sin2 0 = 1. This shows that F = 1. Similarly, we get G = 0. Hence f cos + g sin = 1 and f sin g cos 0. Solving these equations we get f = cos and g = sin . Therefore (s) = (cos (s), sin (s)) for all s (a, b) and (s0 ) = 0 . For the uniqueness, let : (a, b) R be a smooth map satisfying (s0 ) = 0 and (s) = (cos (s), sin (s)) for all s. Then there is a map n : (a, b) Z such that (s)(s) = 2n(s) for all s. Since both and are smooth, n is smooth and hence it is continuous. Since (a, b) is connected, n is continuous and the non-empty connected subsets of Z are singletons only, it follows that n is a constant map, say, c. Therefore (s) (s) = 2c for all s. Since (s0 ) = (s0 ) = 0 , we get c = 0. Hence (s) = (s) for all s. Denition 1.3.11. Let : (a, b) R2 be a unit-speed curve. Then there is a smooth map : (a, b) R such that (s) = (cos (s), sin (s)) for all s, which is determined uniquely by the condition (s0 ) = 0 . The map is called the turning angle of . Proposition 1.3.12. Let : (a, b) R2 be a unit-speed curve, and let be the turning angle of . Then the signed curvature of can be determined by the formula s = . PROOF. Since is the turning angle of , we have (s) = (cos (s), sin (s)) for all s. Hence = ( sin , cos ). Now ns = R ( ) = ( sin , cos ). Therefore = ns 2 and hence the signed curvature s of is , i.e., s = . It follows from the Proposition 1.3.12 that the signed curvature is the rate at which the tangent vector of the curve rotates. Example 1.3.13. If : (a, b) R2 is a unit-speed curve, then its signed curvature and signed unit normal are smooth maps. 9. viii1. CURVE THEORYThe map ns : (a, b) R2 is given by ns (t) = R/2 (t(t)). The map R/2 : R2 R2 dened by R/2 (x, y) = (y, x) is a smooth map and the map t = is a smooth map and hence the map R/2 t = ns is a smooth map, i.e., the signed unit normal is a smooth map. Now the map s : (a, b) R is given by (t) = s (t)ns (t). Therefore s (t) = (t)ns (t). Since both and ns are smooth maps, it follows that s is a smooth map. Theorem 1.3.14. Let k : (a, b) R be a smooth map. Then there is a unit-speed curve : (a, b) R2 whose signed curvature is k. Moreover, if : (a, b) R2 is any other unit-speed curve whose signed curvature is k, then there is a direct isometry M of R2 such that = M . sPROOF. Let s0 (a, b). Dene : (a, b) R by (s) = s0 k(u)du. Since k is smooth, is a smooth map. Note that (s0 ) = 0 Dene : (a, b) R2 by s s (s) = cos (u)du,s0sin (u)du . s0 Then (s) = (cos (s), sin (s)) for all s. Therefore is unit-speed curve. It follows form the Proposition 1.3.10 that is the turning angle of satisfying (s0 ) = 0 and hence the signed curvature of is . But = k. Hence the signed curvature of is k. Let : (a, b) R2 be a unit-speed curve with signed curvature k. Let be the s turning angle of . Then = k. But then (s) = s0 k(u)du + (s0 ) = (s) + , where = (s0 ) is a constant. So we have = + . Since is the turning angle of , (s) = (cos (s), sin (s)) for all s. Let a = (s0 ). Integrating the above equation, we have s s (s) = cos du,s0s0 ss0sin du s0s= Ta scos( + )du,s0scos du,= Ta sin du + asin( + )du s0s= Ta s0s(cos cos sin sin )du, (sin sin + cos cos )dus0 10. 3. Curvature and Torsionixs= Ta cos scos du sin s0= Ta R sin du, cos s0sscos du,s0sscos du + sin s0 sin dus0sin du s0= Ta R ((s)). Therefore (s) = M((s)) for all s, where M = Ta R is a direct isometry. Hence = M . Example 1.3.15. Let : (a, b) R2 be a regular curve with constant signed curvature. Show that is (part of) a circle. We may assume that is unit-speed. Let k be the curvature of . Let k < 0. Then k = for some > 0. Let : (a, b) R2 be dened by (t) = ( 1 cos( t), 1 sin( t)). Then is a unit-speed and t(t) = (t) = ( sin( t), cos( t)), (t) = ( cos( t), sin( t)). Hence ns (t) = (cos( t), sin( t)). Therefore (t) = ns (t). This implies that the signed curvature of is = k. Since and have the same signed curvature, there exists a direct isometry M of R2 such that = M . Since N is (part of) a circle for any direct isometry N of R2 , it follows that is also a circle. Similarly one can show that is a circle when k > 0. Examples 1.3.16. (i) Compute the signed curvature of (t) = (t, cosh t). We have (t) = (1, sinh t). Let (t) be the angle between positive x-axis and (t). Then (1, 0)(1, sinh t) 1 cos (t) = = . 2 cosh t 1 + sinh t Therefore sin (t) = tanh t and tan (t) = sinh t. Let s be the arc-length of , t i.e., s(t) = 0 (u) du = sinh t. Dierentiating tan (t) = sinh t = s with respect 1 1 to s we have sec2 = 1, i.e, (1 + tan2 ) = 1. Therefore = 1+tan2 = 1+s2 = 1 1 = cosh2 t . Hence s (t) = cosh2 t . 1 (ii) Give an example of a plane curve having signed curvature 1+s2 . It is s s 1 1+sinh2 tcos(tan1 u)du,(s) = 0See the proof of Theorem 1.3.14.sin(tan1 u)du . 0 11. x1. CURVE THEORYDenition 1.3.17. Let : (a, b) R3 be a unit-speed curve with nowhere vanishing curvature. Then the function n = is called the unit normal or principal normal of . t It is clear that n(t) = 1 for all t. Since t(t) = 1 for all t, t = 0, i.e., t = 0. This means t n. The map b dened by b = t n is called the unit binormal of . Thus at each point (t) of a unit-speed curve , we have a right handed system of orthonormal basis {t(t), n(t), b(t)}, i.e, t(t)n(t) = b(t), n(t)b(t) = t(t), b(t)t(t) = n(t), t(t) t(t) = n(t) n(t) = b(t) b(t) = 1. Since b(t) b(t) = 1 for all t, b b = 0, i.e, b b. Also, b = t n. Therefore b = n + t n = (n) n + t n = t n. This implies that b t. Hence b n. t Therefore there is a map such that b = n. The map is called the torsion of . Since n = b t, we have n = b t + b = (n t) + (b n) = b t. t Proposition 1.3.18. Let : (a, b) R3 be a regular curve with nowhere vanishing curvature. Then the torsion of is given by ... ( ) . = 2 PROOF. First assume that is unit-speed. Then = t, = n and ... = n + n = n + (b t). Now ... ( ) = (t n)( n + (b t)) = b( n + (b t)) = 2 , ... 2 2 2 and = b = 2 . Hence ( ) 2 = 2 = . Let be a regular curve with nowhere vanishing curvature (not necessarily unitspeed). Let be its unit-speed reparametrization given by s = , where s is the arc-length of starting at any point of . Note that the torsion of at (s(t)) is same as the torsion of at the point (t). Since s = , we have ds = , i.e,. = ds . dt dt ... 2 2 2 3 Therefore ( ds )2 + d s = and ( ds )3 + 2 ds d s + ds d s + d s = . It follows that dt dt dt 2 dt dt 2 dt 2 dt 3 ... dt = ( )( ds )3 and ( ) = ( ) ( ds )6 . Also 2 = 2 ( ds )6 . dt dt dt ... Hence ( ) 2 = ( ) 2 = , as is a unit-speed curve. Exercises 1.3.19. Compute curvature and torsion of the following curves. (i) (t) = (t, t 2 , t 3 ). (ii) (t) = (a cos t, a sin t, bt) (iii) (t) = (et cos t, et sin t, et )4 (iv) (t) = ( 5 cos t, 1 sin t, 3 cos t) 5 1 1 t (v) (t) = ( 3 (1 + t)3/2 , 3 (1 t)3/2 , 2 ) 12. 3. Curvature and TorsionxiExample 1.3.20. (i) If : (a, b) R3 is a unit-speed curve with nowhere vanishing curvature, then the unit normal, unit binormal, curvature and torsion are smooth maps. We know that n, b : (a, b) R3 and , : (a, b) R. By denition (t) = (t) . 2 2 2 Let = (1 , 2 , 3 ). Since is nowhere vanishing, 1 (t) + 2 (t) + 3 (t) > 0 for all t. Dene f : (0, ) R by f(t) = t and g : (a, b) R by g(t) = 1 2 (t) + 2 2 (t) + 3 2 (t). Then both f and g are smooth maps. Since g(t) > 0 for all t, f g is a smooth map. But f g = and hence is a smooth map. 1 Since is a smooth map and (t) > 0 for all t, is a smooth map. Therefore = n is a smooth map. By denition b = t n and both t and n are smooth maps Therefore b is a smooth map. Now b = n. It means = bn. Since both b (so is b) and n are smooth maps, is a smooth map. (ii) Let be a regular curve in R3 with nowhere vanishing curvature. Then is planar i the torsion of is identically zero. We may assume that is unit-speed. Suppose that is planar. Then there exist a unit vector a R3 and a constant d R such that a = d. Dierentiating, we get ta = 0. Dierentiating it again, we have = 0, i.e., (n a) = 0. This gives ta n a = 0, as is nowhere vanishing. Therefore a t and a n. It means b = a. Since b is continuous, it follows that either b(t) = a for all t or b(t) = a for all t. So, in either case, b = 0. Since b = n, it follows that = 0. Conversely, assume that = 0. Since b = n, we have b = 0, i.e, b is a constant d unit vector. Now dt ( b) = b = t b = 0. This means b is a constant function, say, d, i.e., b = d. Hence is lying on the plane R b = d, i.e., is planar. (iii) Let be a regular curve in R3 with constant curvature and zero torsion. Then is (part of) a line or (part of) a circle. Assume that the curvature = 0. Since = n, = 0. Therefore is (part of) a line. d 1 1 1 Let > 0. Note that = 0. We see that dt ( + n) = t + n = t + (b t) = 0. 1 1 Therefore + n = a for some constant vector a R3 . But then a = k , 1 i.e., is lying on the sphere R a = . Since = 0, is lying on some plane. Therefore is lying on a sphere as well as a plane and hence it is (part of) a circle. (iv) Let be a regular curve in R3 with nowhere vanishing curvature. Show that d is spherical i = dt . 2 13. xii1. CURVE THEORYWe may assume that is unit-speed. Assume that is spherical. Then there exist a constant vector a R3 and a constant R R such that ( a)( a) = a 2 = R2 . Dierentiating it we get ( a)t = 0. Dierentiating it again we 1 have ( a)n = . Again dierentiation gives ( a)b = 2 . Since {t, n, b} is an orthonormal basis of R3 , a = (( a)t)t + (( a)n)n + (( a)b)b. Therefore 1 a = n + 2 b. 2 2 1 2 But then R = a = 2 + 2 4 . Dierentiating it we obtain 2 2 2 4 2 2 (2 4 + 2 42 ) 2 d 2 = 0, i.e., 2 2 = 0, i.e., = dt . 2 4 8 3 4 d 1 Conversely, assume that = dt . Let a = + n 2 b. Then 2 a = t 2n + = t 2n + d = dt1 d n b 2b 2 dt 1 d (b t) b 2 ()n 2 dt b = 0. 2 Hence a is a constant vector. Now a 2 = 12 + 4 . Again using = 2 2 1 2 it follows that 2 + 2 4 is a positive constant, say, R . Therefore a and hence is spherical. 2d dt 2 2 , =R , 2OR 1 1 Let = and = . Assume that is spherical. Then there exist a constant vector a R3 and a constant R > 0 such that ( a)( a) = a 2 = R2 . Dierentiating it we get ( a)t = 0. Dierentiating it again we have ( a)n = 1 = . One more dierentiation gives ( a)b = . Now a = (( a)t)t + (( a)n)n + (( a)b)b = n b. 2Therefore R2 = a 2 = n b = 2 + ( )2 . Dierentiation of the last equation gives + ( ) = 0, which is required. Conversely, assume that + ( ) = 0. Then 2 + ( )2 = d for some constant d > 0(???). Let a = + n + b. Then a = t + n + (b t) + ( ) b n = ( + ( ) )b = 0. Therefore a is a constant vector. Now a 2 = 2 n b = 2 + ( )2 = d. Hence is spherical. (v) Let be a unit-speed curve in R3 with nowhere vanishing curvature. Show that the curve = t is a regular curve and nd the curvature and torsion of . 14. 4. Fundamental Theorem of Space Curvesxiii Here = t. Therefore = = . Since the curvature = > 0, the t = n, = n + n = n + (b t) and curve is regular. Now = t ... 2 3 = n ... (b t) + b + b n 2 t n. + Now = 2 t + 3 b, 3 4 2 ( ) = + , = and = 2 + 2 . Let 1 be the curvature of and 1 be the torsion of . Then 2 2 + 2 2 + 2 1 = = = 3 3 and... ( ) ( ) 1 = = . 2 2 + 2 Theorem 1.3.21. Let be a unit-speed curve in R3 . Let t, n and b be its unit tangent, unit normal and unit binormal respectively. Let and be the curvature and torsion of respectively. Then = t n = t n b = b n The equations in Theorem 1.3.21 are called Frenet-Serret equations or SerretFrenet equations. Example 1.3.22. Verify the Serret-Frenet equations for the curve (a) =a cosa a2 + b2, a sina a2 + b2,b a2 + b2.1.4. Fundamental Theorem of Space Curves Theorem 1.4.1. (Existence) Let k, t : (, ) R be smooth maps with (s) > 0 for all s. Then there exists a unit-speed curve : (, ) R3 whose curvature is k and torsion is t. PROOF. Fix s0 (, ). It follows from the theory of Dierential Equations that the equations T = kN, N = kT + tB and B = tN have a unique solution T, N and B 0 k 0 such that {T(s0 ), N(s0 ), B(s0 )} is a standard basis of R3 . Since the matrix k 0 t 0 t 0 N and B in terms of T, N and B is skew-symmetric, it follows that T, expressing T, N and B are orthogonal unit vectors. Since B is a smooth map orthogonal to both T 15. xiv1. CURVE THEORYand N, there exists a smooth map : (, ) R such that B = T N. Note that (s) is 1 or 1. Since is continuous, either (s) = 1 for all s or (s) = 1 for all s. Since B(s0 ) = T(s0 ) N(s0 ), we have (s) = 1 for all s, i.e., B = T N. Dene : (, ) R3 by s(s) =T(u)du. s0 Then = T. Therefore is a unit-speed curve and T is the unit tangent of . Now = kN. Therefore = k, i.e., k is the curvature of . It follows from = T = T = kN that N is the unit normal of . Since B = T N, B is the unit binormal of . It follows from the equation B = tN that t is the torsion of . Theorem 1.4.2. (Uniqueness) If , : (, ) R3 are unit-speed curves with same curvature and same torsion, then there is an isometry M of R3 such that = M . PROOF. Let t, n and b be the unit tangent, unit normal and unit binormal of , and let t, n and b be those of . Fix s0 (, ). Since {t(s0 ), n(s0 ), b(s0 )} and {t(s0 ), n(s0 ), b(s0 )} are both right handed orthonormal basis of R3 , there is a rotation about origin that maps t(s0 ), n(s0 ) and b(s0 ) to t(s0 ), n(s0 ) and b(s0 ) respectively. Further, there is a translation which takes (s0 ) to (s0 ) (and this has no eect on t, n and b). By applying rotation followed by translation, we can therefore assume that (s0 ) = (s0 ), t(s0 ) = t(s0 ), n(s0 ) = n(s0 ) and b(s0 ) = b(s0 ). Hence to prove the theorem we need to prove = . Dene A : (, ) R by A(s) = t(s)t(s) + n(s)n(s) + b(s)b(s). Then A is a smooth map. Note that A(s) 3 for all s and A(s0 ) = 3. Also observe that A(s) = 0 for all s (you need to verify this). Therefore A is a constant map. Since A(s0 ) = 3, we have A(s) = 3 for all s. Hence it follows that t = t, n = n and b = b. The equation t = t implies that = + c. Since (s0 ) = (s0 ), we get c = 0. Hence = . This proves the theorem. Example 1.4.3. Describe all curves in R3 which have constant curvature > 0 and constant torsion . Let : (a, b) R3 be a unit-speed curve with constant curvature > 0 and constant torsion . Let a = 2 + 2 and b = 2 + 2 . Dene : (a, b) R3 by (t) =a costb2, a sint, bt. + + + b2 a Then is a unit-speed curve. The curvature of is a2 +b2 = and its torsion is b = . Therefore , : (a, b) R3 are unit-speed curves with same curvature a2 +b2 a2a2b2a2 16. 5. Isoperimetric Inequality and Four Vertex Theoremxvand same torsion. So, there is an isometry M of R3 such that = M . But is a helix and hence M is a helix. This implies that is a helix. Hence any curve with constant (non-zero) curvature and constant torsion is a helix.1.5. Isoperimetric Inequality and Four Vertex Theorem Denition 1.5.1. Let a R be a positive constant. A simple closed curve in R2 is a (regular) curve : R R2 such that (t) = (t ) i t t = ka for some k Z. Thus, the point (t) returns to its starting point when t is increased by a, but not before that. By Jordan Curve Theorem any simple closed curve in R2 has an interior and exterior. More precisely, the set of points in R2 that are not on the curve is the disjoint union of two subsets of R2 , denoted by int() and ext(), with the following properties. (i) int() is bounded, (ii) ext() is unbounded, (iii) both int() and ext() are connected, but any curve joining a point of int() to a point of ext() crosses . Proposition 1.5.2. Let be a regular curve with period a. Then its unit-speed reparametrization is ()- periodic, where () is the length of . PROOF. Since is a- periodic, the length, (), of is a (u) du.() = 0Let be a unit-speed reparametrization of given by s = , where s is the arclength of starting at the point (0). Let t, t R, and let t t . Then there is k N {0} such that t t = ka + , where 0 < a. Now ts(t ) s(t) =t (u) du 0 (u) du = (u) du t0t+at+2a (u) du +=ttt+ka (u) du + + t+at+= k () + (u) du. tt+(k1)at+ka+ (u) du + (u) du t+ka 17. xvi1. CURVE THEORYt+ Hence s(t ) s(t) = k () i t (u) du = 0 i = 0 i t t = ka. Now (s(t)) = (s(t )) i (t) = (t ) i t t = ka for some k Z i s(t ) s(t) = k (). Therefore is ()- periodic.Let be a simple closed curve in R2 . Let A(int()) be the area of the interior of . Then A(int()) = int() dxdy. Denition 1.5.3. A simple closed curve in R2 is called positively oriented if its unit signed normal points towards the interior of the curve at ach point of the curve. Theorem 1.5.4. (Greens Theorem) Let be a simple closed curve in R2 , and let be positively oriented. Let f and g be continuous on the closure of int() and smooth on int(). Then g f x y[f(x, y)dx + g(x, y)dy].dxdy = int()Proposition 1.5.5. Let (t) = (x(t), y(t)) be a simple closed curve in R2 with period a. Let be positively oriented. Then a1 A(int()) = 2(x y y x )dt. 01 1 PROOF. Take f(x, y) = 2 y and g(x, y) = 2 x. Then both f and g are smooth functions 2 on R . It follows from the Greens theorem that 1 1 1 ( 1 ( 2 )dxdy = [ 2 xdy 2 ydx]. This implies int() 2 1 1 1 a A(int()) = [ 2 xdy 2 ydx] = 2 0 (x y y x )dt.Proposition 1.5.6 (Wirtingers Inequality). Let F : [0, ] R be a smooth map satisfying F(0) = F() = 0. Then F 2 dt 0F 2 dt, 0with equality holding i F(t) = A sin t for some constant A. F(t) . sin tPROOF. Dene G : [0, ] R by G(t) = Note that G is a smooth map. Then F 2 dt = 0 (G sin t + G cos t)2 dt 0 22G sin tdt + 2= 0G 2 cos2 tdt G G sin t cos tdt + 00 18. 5. Isoperimetric Inequality and Four Vertex Theorem G 2 sin2 tdt = 0G 2 (cos2 t sin2 t)dt + 0 2 22G sin tdt + 0 00 G 2 sin2 tdt 0.2F dt G 2 sin2 tdt.F dt + 02 2G sin tdt = 0HenceG 2 cos2 tdt 0 2=xviiF dt = 0(1.5.6.1)0 0 By equation (1.5.6.1) it follows that F 2 dt = 0 F 2 dt i 0 G 2 sin2 tdt = 0 i G 2 sin2 t = 0 i G sin t i G = 0. But G = 0 implies G(t) = A for all t, for some constant A, i.e., F(t) = A sin t for all t.Example 1.5.7. The length and the are of a simple closed are invariant under isometry. Let : R R2 be a simple closed curve with period , and let M : R2 R2 be an isometry. Let : R R2 be = M . We need to prove that the lengths and the area of interiors of and are same. Let t, t R. Then (t ) = (t) i M((t )) = M((t)) i (t ) = (t) i t t = k for some k Z. Therefore is -periodic. Now (u) du =() = 0DM((u)) du 0 DM((u)) (u) du= 0 M( (u)) du == 0 (u) du = (). 01Since M is an isometry , there are a, b, c, d, e, f R2 such that a2 +b2 = c2 +d2 = a + c2 = b2 + d2 = 1, ac + bd = ab + cd = 0 and M(x, y) = (ax + by + e, cx + dy + f). Let (t) = (x(t), y(t)). Then (t) = (ax(t) + by(t) + e, cx(t) + dy(t) + f). Now 2A(int()) =1 2((ax + by + e)(cx + dy + f) (ax + by + e) (cx + dy + f))dt 01We know that N : R2 R2 is an isometry i there are a, b, c, d, e, f R2 such that a2 + b2 = c2 + d2 = a2 + c2 = b2 + d2 = 1, ac + bd = ab + cd = 0 and N(x, y) = (ax + by, cx + dy) + (e, f) OR N is an isometry of R2 if and only if there is a 2 2 orthogonal matrix A and a R2 such that N(x) = Ax + a 19. xviii1. CURVE THEORY=1 2((ax + by + e)(cx + dy ) (ax + b )(cx + dy + f))dt y 0 =1 (ad bc) 2(x y x y)dt + (ec af)(x( ) x(0)) + (ed bf)(y( ) y(0)) 0=1 2(x y x y)dt = A(int()), 0as ad bc = 1, x(0) = x( ) and y(0) = y( ) (???). Theorem 1.5.8 (The Isoperimetric Inequality). Let be a simple closed curve, let be its length, and let A be the area of interior of . Then 2 4A, with equality holding i is a circle. PROOF. We may assume that is unit-speed. Then the length of is its period, i.e., is - periodic. Consider the curve (t) = (x(t), y(t)) dened by (t) = t , i.e., = s or s1 = , where s(t) = t . [Note that is a unit-speed reparametrization of with the parametrization map s1 (t) = t , i.e., s(t) = t .] Then is a simple closed curve with the same length and period . Also the area of interior of is same as the area of interior of . Since the length and the area of a curve are invariant under isometry, we may assume that (0) = (0) = (0, 0). Let x = r cos and y = r sin . Note that r, : [0, ] R are smooth maps. Since is - periodic, (0) = (). This gives r(0) = r() as (0) = (0, 0). Now x 2 + y 2 = r 2 + r 2 2 and x y y x = r 2 . Since 2 2 ds 2 2 = s, we have = ds . But then r 2 + r 2 2 = x 2 + y 2 = = ( dt ) = 2 . We dt 2 also note that A = A(int()) = 1 (x y y x )dt = 1 r . Now, 0202 21 A = 4 41 ( 2 + r 2 2 )dt r 2 0 r 2 dt 0=1 4 ( 2 + r 2 2 2r 2 )dt r 0 1 = 41 r ( 1) dt + 4 220( 2 r 2 )dt 0. r 0 1 r It follows from the above equation that 4 A = 0 i 4 0 r 2 ( 1)2 dt + 0 ( 2 r 2 )dt = 0 2 2 2 2 i 0 r ( 1) dt = 0 ( r )dt = 0 i = 1 and r = A sin t for some A i = t + r 2 20. 5. Isoperimetric Inequality and Four Vertex Theoremxixfor some constant and r = A sin t for some A i r = A sin( ). But r = A sin( ) is a circle. Hence 2 4A = 0 i is a circle. Example 1.5.9. By applying the isoperimetric inequality to the ellipse (where p and q are positive constants), prove that 2x2 p2+y2 q2= 1x2 p2+y2 q2= 1. p2 sin2 t + q 2 cos2 t dt 2 pq,0with equality holding if and only if p = q. Consider (t) = (p cos t, q sin t). Then the trace of is the ellipse Note the is 2-periodic. Therefore length of is 22p2 sin2 t + q 2 cos2 t dt. (t) dt == 00Also, the area A of the interior of is 21 A= 221 (x y y x )dt = 2 0pq(cos t cos t + sin t sin t)dt =pq 2 = pq. 20It follows from the Isoperimetric inequality that 2 2 p2 sin2 t + q 2 cos2 t dt 4pq = 4 2 pq, 0i.e., 2 p2 sin2 t + q 2 cos2 t dt 2 pq.0Equality holds in above inequality i is a circle i p = q. [Because (t) = (p cos t, q sin t) is a circle i p = q.] Denition 1.5.10. A simple closed curve in R2 is called convex if its interior is a convex subset of R2 . Denition 1.5.11. The vertex of a curve is a point of the curve where its signed curvature s has a stationary point, i.e, s = 0 at that point. Theorem 1.5.12 (Four Vertex Theorem). Every convex, simple closed curve in R2 has at least four vertices. 21. xx1. CURVE THEORYExample 1.5.13. Find the vertices of the curve (t) = (a cos t, b sin t), where a, b > 0 and a = b. Let s be the arc-length of starting at some point of . Denoting the derivative dt of with respect to s by , we have (t) = (a sin t, b cos t) ds . Since = 1, dt = 2 2 1 2 2 . Therefore ds a sin t+b cos t1t(t) = (t) =a2 sin2 t + b2 cos2 t Dierentiating with respect to s, we get (t) = Now ns (t) = R (t(t)) = 2(a2ab (b cos t, a sin t). sin t + b2 cos2 t)2 21 1(a2 sin2 t+b2 cos2 t) 2 = This implies that s (t) =(a sin t, b cos t).(b cos t, a sin t). Hence ab 3(a2 sin2 t + b2 cos2 t) 2ab 3 (a2 sin2 t+b2 cos2 t) 2. Now s (t) = ns . 3ab(a2 b2 ) sin t cos t 5(a2 sin2 t+b2 cos2 t) 2 sin 2t = 0 i t = n 2. Since a, b > 0and a = b, we have s (t) = 0 i sin t cos t = 0 i for some n Z. Since is 2- periodic, it follows that the vertices of are (0) = (a, 0), (/2) = (0, b), () = (0, a) and (3/2) = (0, b) only. y Example 1.5.14. The ellipse x 2 + b2 = 1 (or the curve (t) = (a cos t, b sin t)) is a a convex curve. 2 2 y2 y z2 Let (x, y), (z, w) int(), i.e., x 2 + b2 < 1 and a2 + w2 < 1. Let X = x , b and a a b z Y = a , w . Then X < 1 and Y < 1. Let t [0, 1]. Then tX + (1 t)Y b t X + (1 t) Y < t + 1 t = 1. Now 22(tx + (1 t)z)2 (ty + (1 t)w)2 + = tX + (1 t)Y a2 b2 Therefore t(x, y) + (1 t)(z, w) int() and hence is convex.2< 1.