Selected solutions to exercises from Pavel Grinfeld’s Introduction to Tensor Analysis and the Calculus of Moving Surfaces David Sulon 9/14/14 ii Contents I Part I 1 1 Chapter 1 3 2 Chapter 2 7 3 Chapter 3 13 4 Chapter 4 17 5 Chapter 5 33 6 Chapter 6 39 7 Chapter 7 47 8 Chapter 8 49 9 Chapter 9 51 II Part II 57 10 Chapter 10 59 11 Chapter 11 67 12 Chapter 12 77 III Part III 89 13 Chapter 16 101 14 Chapter 17 109 iii iv CONTENTS Introduction Included in this text are solutions to various exercises from Introduction to Tensor Analysis and the Calculus of Moving Surfaces, by Dr. Pavel Grinfeld. v vi CONTENTS Part I Part I 1 Chapter 1 Chapter 1 Ex. 1: We have x = 2x0, y = 2y0. Thus F 0(x0; y0) = F (2x0; 2y0) 2 2y0 = (2x0) e 2 2y0 = 4 (x0) e : 1 1 Ex. 2: Note that the above implies x0 = 2 x, y0 = 2 y. We check @F 0 2y0 (x0; y0) = 8 (x0) e @x0 2 1 y 1 ( 2 ) = 8 x e 2 = 4xey @F (x; y) = 2xey: @x @F 0 @F Thus, (x0; y0) = 2 (x; y) as desired. @x0 @x Ex. 3: Let a; b R, a; b = 0, and consider the "re-scaled" coordinate basis 2 6 a i = 0 0 0 j = ; 0 b where each of the above vectors is taken to be with respect to the standard 2 basis for R . Thus, given point (x; y) in standard coordinates, we have x = ax0, y = by0, where (x0; y0) is the same point in our new coordinate system. Now, let T (x; y) be a di¤erentiable function. Then, @F @F T = (x; y); (x; y) r @x @y 3 4 CHAPTER 1. CHAPTER 1 in standard coordinates @F @x @F @y = (x ; y ) 0 (x; y); (x ; y ) 0 (x; y) @x 0 0 @x @y 0 0 @y 0 0 (think of x0 as a function of x). @F 1 @F 1 = (x ; y ) ; (x ; y ) @x 0 0 a @y 0 0 b 0 0 1 @F 1 @F = ; pa2 + 0 @x b2 + 0 @y 0 0 1 @F 1 @F = ; ; pi i @x0 pj0 j0 @y0 0 0 as desired. Ex. 4: Assume x0 a cos sin x = + : y0 b sin cos y Then, x0 = a + (cos ) x (sin ) y y0 = b + (sin ) x + (cos ) y Also, x0 a cos sin x = y0 b sin cos y 1 x cos sin x0 a = y sin cos y0 b cos sin cos2 +sin2 cos2 +sin2 x0 a = sin cos 2 2 2 2 y0 b cos +sin cos +sin cos sin x0 a = sin cos y0 b Thus, x = (cos )(x0 a) + (sin )(y0 b) y = (sin )(x0 a) + (cos )(y0 b) : 5 Further notice that we obtain i0; j0 from the standard basis [Note: this "basis" would describe points be with respect to this point (a; b)] cos sin 1 i = 0 sin cos 0 = cos i+ sin j cos sin 0 j = 0 sin cos 1 = sin i+ cos j Now, we have, given a function F , we compute @F @F @F @x @F @y @F @x @F @y (x; y) i + (x; y) j = (x ; y ) 0 (x; y) + (x ; y ) 0 (x; y) i + (x ; y ) 0 (x; y) + (x ; y ) 0 (x; y) j @x @y @x 0 0 @x @y 0 0 @x @x 0 0 @y @y 0 0 @y 0 0 0 0 @F @F @F @F = (x0; y0) cos + (x0; y0) sin i+ (x0; y0) sin + (x0; y0) cos j @x @y @x @y 0 0 0 0 @F @F @F @F = (x0; y0) cos i (x0; y0) sin j + (x0; y0) sin i + (x0; y0) cos j @x0 @x0 @y0 @y0 @F @F = (x0; y0) (cos i sin j) + (x0; y0) (sin i + cos j) @x0 @y0 @F cos @F sin = (x ; y ) i j + (x ; y ) i j @x 0 0 sin @y 0 0 cos 0 0 @F @F = (x0; y0) i0 + (x0; y0) j0 @x0 @y0 [NOT SURE - I will ask about this one tomorrow] (a; b) cooresponds to a shifted "origin," corresponds to angle for which the whole coordinate system is rotated. Ex. 5: We may obtain any a¢ ne orthogonal coordinate system by rotating the "standard" Cartesian coordinates via (1.7) and then applying a rescaling. 6 CHAPTER 1. CHAPTER 1 Chapter 2 Chapter 2 Ex. 6: See diagram. Note: Diagrams will be added later for Ex. 7-12 Note: For Ex 7-12, let h denote the distance from P to P , where P is a point arbitrarily close to P along the appropriate direction for which we are taking each directional derivative. De…ne f (h) := F (P ), i.e. parametrize along the unit vector emanating from P in the direction of l (note f (0) = F (P )). Also, for points A,B, AB indicates the (unsigned) length of the vector from A to B. Ex. 7: f (h) = F (P )2 + h2 h f 0 (h) = p F (P )2 + h2 dF (p) = fp(0) dl 0 Ex. 8: We have 1 f (h) = AP h 1 f 0 (h) = (AP h)2 1 = (AP h)2 so dF (p) = f (0) dl 0 1 = (AP )2 7 8 CHAPTER 2. CHAPTER 2 Ex. 9: Let denote the measure of angle OP P . By the Law of Sines, we have sin (F (P )) sin ( ) = AP h OA sin () = OA sin sin (F (P ) F (P )) = OP h From the second equation, we obtain OP sin (F (P ) F (P )) sin = h OP [sin (F (P )) cos (F (P )) cos (F (P )) sin (F (P ))] = h The, from the …rst equation, we have sin (F (P )) OP [sin (F (P )) cos (F (P )) cos (F (P )) sin (F (P ))] = AP h (OA) h (OA) h sin (F (P )) = (OP )(AP h) [sin (F (P )) cos (F (P )) cos (F (P )) sin (F (P ))] (OA) h tan (F (P )) = (OP )(AP h) sin (F (P )) (OP )(AP h) cos (F (P )) tan (F (P )) ((OA) h + (OP )(AP h) cos (F (P ))) tan (F (P )) = (OP )(AP h) sin (F (P )) (OP )(AP h) sin (F (P )) tan (F (P )) = (OA) h + (OP )(AP h) cos (F (P )) (OP )(AP h) sin (F (P )) F (P ) = arctan (OA) h + (OP )(AP h) cos (F (P )) Thus, (OP )(AP h) sin (F (P )) f (h) = arctan (OA) h + (OP )(AP h) cos (F (P )) (OP ) sin (F (P )) [(OA) h + (OP )(AP h) cos (F (P ))] (OP ) sin (F (P )) (AP h) [(OA) (OP )(AP h) cos (F (P ))] 1 f 0 (h) = 2 [(OP )(AP h) sin(F (P ))]2 [(OA) h + (OP )(AP h) cos (F (P ))] 1 + [(OA)h+(OP )(AP h) cos(F (P ))]2 (OP ) sin (F (P )) [(OA) h + (OP )(AP h) cos (F (P ))] (OP ) sin (F (P )) (AP h) [(OA) (OP )(AP h) cos (F (P ))] = [(OA) h + (OP )(AP h) cos (F (P ))]2 + [(OP )(AP h) sin (F (P ))]2 9 dF (p) = f (0) dl 0 (OP ) sin (F (P )) [(OP )(AP ) cos (F (P ))] (OP ) sin (F (P )) (AP ) [(OA) (OP )(AP ) cos (F (P ))] = [(OP )(AP ) cos (F (P ))]2 + [(OP )(AP ) sin (F (P ))]2 (OP ) sin (F (P )) (OP )(AP ) cos (F (P )) (OP ) sin (F (P )) (AP )(OA) + (OP ) sin (F (P )) (AP )(OP )(AP ) cos (F (P )) = (OP )2 (AP )2 cos2 (F (P )) + sin2 (F (P )) (OP ) sin (F (P )) (AP )(OA) = (OP )2 (AP )2 (OA) = sin (F (P )) (OP )(AP ) Ex. 10: Clearly F (P ) = F (P ) for any choice P in such a direction. Thus, f is constant, and we have dF (p) = 0 dl Ex. 11: Put d as the distance between P and the line from A to B. As with the previous problem, the distance from P to line AB! is also d. Thus, 1 F (P ) = (AB) d 2 1 F (P ) = (AB) d; 2 dF (p) and we have F (P ) = F (P ), so dl = 0 as before. Ex. 12: Drop a perpendicular from P to AB!. Let K be this point of intersection. Note that the length AK = F (P ) + h. Then, 1 f (h) = (AB)(F (P ) + h) 2 1 f (h) = AB 0 2 dF (p) = f (0) dl 0 1 = AB 2 Ex. 13: (7) The gradient will point in direction !AP , and will have magnitude 1. (8) [Not sure] 10 CHAPTER 2. CHAPTER 2 (9) The gradient will point in direction !AP (in the same direction as was asked for the directional derivative), and thus will have magnitude (OA) sin (F (P )) (OP )(AP ) (note F (P ) is assumed to satisfy F (P ) ) (10) The gradient will point in direction perpendicular to AB!, and will have magnitude 1. (11),(12) The gradient will point in the direction perpendicular to AB! (in the same direction as was asked for the directional derivative in Ex.12), and thus will have magnitude 1 AB: 2 Ex. 14: The directional derivative in direction L would then correspond to the projection of Of onto L. Ex. 15: [See diagram] R ( + h) R ( ) 2 = 1 + 1 2 cos (h) k k by the Law of Cosines. So, R ( + h) R ( ) 2 = 2 2 cos (h) k k R ( + h) R ( ) = 2 2 cos (h) k k p h = 2 2 cos 2 2 s h = 2 2 sin2 2 s h = 4 sin2 r 2 h = 2 sin : 2 Ex.