# 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. Sin

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