Solution manual for classical dynamics of particles and systems 5th edition by thornton

Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file at CHAPTER Matrices, Vectors, and Vector Calculus 1-1 x2 = x2′ x1′ 45˚ x1 45˚ x3 x3′ Axes x′1 and x′3 lie in the x1 x3 plane The transformation equations are: x1′ = x1 cos 45° − x3 cos 45° x2′ = x2 x3′ = x3 cos 45° + x1 cos 45° x1′ = 1 x1 − x3 2 x2′ = x2 x3′ = 1 x1 − x3 2 So the transformation matrix is:        2 − 1  2      Full file at Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 2at./ CHAPTER 1-2 a) x3 D E γ O β x2 B α A C x1 From this diagram, we have OE cos α = OA OE cos β = OB (1) OE cos γ = OD Taking the square of each equation in (1) and adding, we find 2 OE cos α + cos β + cos γ  = OA + OB + OD (2) But 2 2 2 OA + OB = OC (3) and OC + OD = OE (4) Therefore, 2 OA + OB + OD = OE (5) Thus, cos α + cos β + cos γ = (6) b) x3 D E D′ O A A′ x1 E′ θ B′ B C x2 C′ First, we have the following trigonometric relation: OE + OE′ − 2OE OE′ cos θ = EE′ Full file at 2 (7) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ But, 2       EE′ = OB′ − OB  + OA′ − OA  + OD′ − OD        2     = OE′ cos β ′ − OE cos β  + OE′ cos α ′ − OE cos α        + OE′ cos γ ′ − OE cos γ    2 (8) or, 2 EE′ = OE′ cos α ′ + cos β ′ + cos γ ′  + OE  cos α + cos β + cos γ  − 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′  = OE′ + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′  (9) Comparing (9) with (7), we find cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ (10) 1-3 x3 e3 O e1 e2 e2′ e3 A x2 e2 e1′ e3 ′ e1 x1 Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 Denote the new axes by x′1 , x′2 , x′3 and the corresponding unit vectors by e1′ , e2′ , e3′ The effect of the rotation is e1 → e3′ , e2 → e1′ , e3 → e2′ Therefore, the transformation matrix is written as:  cos ( e1′ , e1 ) cos ( e1′ , e2 ) cos ( e1′ , e3 )  0    λ =  cos ( e2′ , e1 ) cos ( e2′ , e2 ) cos ( e2′ , e3 )  = 0    1 0    cos ( e3′ , e1 ) cos ( e3′ , e2 ) cos ( e3′ , e3 )   1-4 a) Let C = AB where A, B, and C are matrices Then, Cij = ∑ Aik Bkj (1) k (C ) t Full file at ij = C ji = ∑ Ajk Bki = ∑ Bki Ajk k k Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 4at./ CHAPTER ( ) Identifying Bki = Bt ik ( ) and Ajk = At , kj (C ) = ∑ (B ) ( A ) t t ij t ik k (2) kj or, C t = ( AB) = Bt At (3) ( AB) B−1 A−1 = I = ( B−1 A−1 ) AB (4) ( AB) B−1 A−1 = AIA−1 = AA−1 = I (5) (B (6) t b) To show that ( AB) = B−1 A−1 , −1 That is, 1-5 −1 ) A−1 ( AB) = B−1 IB = B−1B = I Take λ to be a two-dimensional matrix: λ = λ11 λ12 = λ11λ 22 − λ12 λ 21 λ 21 λ 22 (1) Then, 2 2 2 λ = λ112 λ 22 − 2λ11λ 22 λ12 λ 21 + λ12 λ 21 + ( λ11 λ 21 + λ122 λ 22 ) − ( λ112 λ 212 + λ122 λ 222 ) ( ) ( ) ( 2 2 = λ 22 λ112 + λ122 + λ 21 λ112 + λ122 − λ112 λ 21 + 2λ11λ 22λ12 λ 21 + λ12 λ 22 ( )( ) 2 2 = λ11 + λ12 λ 22 + λ 21 − ( λ11λ 21 + λ12 λ 22 ) But since λ is an orthogonal transformation matrix, ∑λ λ ij ) (2) kj = δ ik j Thus, 2 λ112 + λ122 = λ 21 + λ 22 =1 (3) λ11λ 21 + λ12 λ 22 = Therefore, (2) becomes λ =1 1-6 The lengths of line segments in the x j and x ′j systems are L= ∑x j Full file at (4) j ; L′ = ∑ x′ i i (1) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ If L = L′ , then ∑ x = ∑ x′ j i (2) xi′ = ∑ λ ij x j (3) j i The transformation is j Then,  ∑ x = ∑  ∑ λ j j i ik k   xk   ∑ λ iA xA    A (4)   = ∑ x k x A  ∑ λ ik λ iA   i  k ,A But this can be true only if ∑λ ik λ iA = δ k A (5) i which is the desired result 1-7 x3 (0,0,1) (1,0,1) (0,0,0) x1 (0,1,1) (1,1,1) (1,0,0) (0,1,0) x2 (1,1,0) There are diagonals: D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ; D , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D ; D , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D ; and D , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D The magnitudes of the diagonal vectors are D1 = D = D = D = The angle between any two of these diagonal vectors is, for example, D1 ⋅ D (1,1,1) ⋅ ( −1,1,1) = cos θ = = D1 D 3 Full file at Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 6at./ CHAPTER so that θ = cos−1   = 70.5°  3 Similarly, D1 ⋅ D D ⋅D D ⋅D D ⋅D D ⋅D = = = = =± D1 D D1 D D2 D3 D2 D D3 D 1-8 Let θ be the angle between A and r Then, A ⋅ r = A2 can be written as Ar cos θ = A2 or, r cos θ = A (1) This implies QPO = π Therefore, the end point of r must be on a plane perpendicular to A and passing through P 1-9 a) A = i + 2j − k B = −2i + 3j + k A − B = 3i − j − 2k 2 A − B = ( 3) + ( −1) + (−2)2  12 A − B = 14 b) B θ A component of B along A The length of the component of B along A is B cos θ A ⋅ B = AB cos θ B cos θ = A ⋅ B −2 + − = = or A 6 The direction is, of course, along A A unit vector in the A direction is ( i + 2j − k ) Full file at (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ So the component of B along A is ( i + 2j − k ) cos θ = c) A⋅B 3 = = ; θ = cos −1 AB 14 7 θ  71° i j k −1 −1 A × B = −1 = i −j +k −2 −2 −2 d) A × B = 5i + j + k A − B = 3i − j − 2k e) A + B = −i + 5j i j k ( A − B ) × ( A + B ) = −1 − −1 ( A − B) × ( A + B) = 10i + 2j + 14k 1-10 r = 2b sin ω t i + b cos ω t j v = r = 2bω cos ω t i − bω sin ω t j a) a = v = −2bω sin ω t i − bω cos ω t j = −ω r speed = v =  4b 2ω cos ω t + b 2ω sin ω t  = bω  cos ω t + sin ω t  12 speed = bω  cos ω t + 1 b) At t = π 2ω , sin ω t = , cos ω t = So, at this time, v = − bω j , a = −2bω i So, θ  90° Full file at 12 12 Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 8at./ CHAPTER 1-11 a) Since ( A × B) i = ∑ ε ijk Aj Bk , we have jk (A × B) ⋅ C = ∑∑ ε ijk Aj Bk Ci j,k i = C1 ( A2 B3 − A3 B2 ) − C2 ( A1B3 − A3 B1 ) + C3 ( A1B2 − A2 B1 ) C1 = A1 C2 A2 C3 A1 A3 = − C1 A2 C2 A3 A1 C3 = B1 A2 B2 A3 B3 = A ⋅ ( B × C ) B1 B2 B3 B1 B2 B3 C2 C3 B3 C1 (1) We can also write C1 C2 C3 B1 B2 (A × B) ⋅ C = − B1 A1 B2 B3 = C1 C2 C3 = B ⋅ ( C × A ) A2 A3 A2 A3 A1 (2) We notice from this result that an even number of permutations leaves the determinant unchanged b) Consider vectors A and B in the plane defined by e1 , e2 Since the figure defined by A, B, C is a parallelepiped, A × B = e3 × area of the base, but e3 ⋅ C = altitude of the parallelepiped Then, C ⋅ ( A × B) = ( C ⋅ e3 ) × area of the base = altitude × area of the base = volume of the parallelepiped 1-12 O c C a h a–c b A b–a c–b B The distance h from the origin O to the plane defined by A, B, C is Full file at Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ h= = = a ⋅ ( b − a ) × ( c − b) ( b − a ) × ( c − b) a ⋅ ( b × c − a × c + a × b) b×c−a×c+a×b a⋅b× c a×b+b×c+c×a (1) The area of the triangle ABC is: A= 1-13 1 ( b − a ) × ( c − b) = ( a − c ) × ( b − a ) = ( c − b) × ( a − c ) 2 Using the Eq (1.82) in the text, we have A × B = A × ( A × X ) = ( X ⋅ A ) A − ( A ⋅ A ) X = φ A − A2 X from which X= ( B × A ) + φA A2 1-14 a)  −1    −2       AB =   0 −1  =  −2     1   3  Expand by the first row AB = −2 3 +2 −2 +1 5 AB = −104 b)  −1          AC = 0    = 13         7   AC = 13    Full file at (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 10 a.u/ CHAPTER 5  −1     ABC = A ( BC ) = 0   −2 −3      c)  −5 −5    ABC =  −5   25 14  AB − Bt At = ? d)  −2    AB =  −2   3  (from part a )  1  2  1 5      B A =  −1    =  −2 −2 3    −1 1   3 t t  −3 −4    6 AB − B A =   −6  t 1-15 t If A is an orthogonal matrix, then At A = 1 0 1 0  1 0      0 a a  0 a − a  = 0   − a a  0 a a   0  1  0 2a 0 a= Full file at  1 0     = 0  2a  0  Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 14 a.u/ CHAPTER ∑= ε for i = A = , 213 ε 213 + ε 231 ε 231 = + = For i = A = , jk A = gives ∑= ε 312 ε 312 + ε 321 ε 321 = But i = 1, jk ∑ = Likewise for i = 2, A = ; i = 1, A = ; i = 3, A = ; i = 2, A = ; i = 3, A = jk Therefore, ∑ε ijk ε Ajk = 2δ iA (2) j,k c) ∑ε ijk ε ijk = ε 123 ε 123 + ε 312 ε 312 + ε 321 ε 321 + ε 132 ε 132 + ε 213 ε 213 + ε 231 ε 231 ijk = ⋅ + ⋅ + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + (1) ⋅ (1) or, ∑ε ijk ε ijk = (3) ijk ( A × B)i = ∑ ε ijk Aj Bk 1-21 (1) jk ( A × B) ⋅ C = ∑ i ∑ε ijk Aj Bk Ci (2) jk By an even permutation, we find ABC = ∑ ε ijk Ai Bj Ck ijk 1-22 To evaluate ∑ε ijk ε Amk we consider the following cases: k a) i= j: ∑ε ijk k b) i=A: ∑ε ε Amk = ∑ ε iik ε Amk = for all i , A , m k ijk k ε Amk = ∑ ε ijk ε imk = for j = m and k ≠ i , j k = for j ≠ m c) i = m: ∑ε ijk k ε Amk = ∑ ε ijk ε Aik = for j ≠ A k = −1 for j = A and k ≠ i , j d) j=A: ∑ε k ijk ε Amk = ∑ ε ijk ε jmk = for m ≠ i k = −1 for m = i and k ≠ i , j Full file at (3) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ ∑ε j = m: e) 15 ε Amk = ∑ ε ijk ε Ajk = for i ≠ A ijk k k = for i = A and k ≠ i , j f) A = m: ∑ε k ijk ε Amk = ∑ ε ijk ε AAk = for all i , j , k k i ≠ A or m : This implies that i = k or i = j or m = k g) Then, ∑ε ijk ε Amk = for all i , j , A , m k j ≠ A or m : h) ∑ε ijk ε Amk = for all i , j , A , m k Now, consider δ iA δ jm − δ im δ jA and examine it under the same conditions If this quantity behaves in the same way as the sum above, we have verified the equation ∑ε ijk ε Amk = δ iA δ jm − δ im δ jA k a) i = j : δ iA δ im − δ im δ iA = for all i , A , m b) i = A : δ ii δ jm − δ im δ ji = if j = m, i ≠ j , m = if j ≠ m i = m : δ iA δ ji − δ ii δ jA = −1 if j = A , i ≠ j , A c) = if j ≠ A j = A : δ iA δ Am − δ im δ AA = −1 if i = m, i ≠ A d) = if i ≠ m j = m : δ iA δ mm − δ im δ mA = if i = A , m ≠ A e) = if i ≠ A f) A = m : δ iA δ jA − δ il δ jA = for all i , j , A g) i ≠ A , m : δ iA δ jm − δ im δ jA = for all i , j , A , m h) j ≠ A , m : δ iA δ jm − δ im δ iA = for all i , j , A , m Therefore, ∑ε ijk ε Amk = δ iA δ jm − δ im δ jA k Using this result we can prove that A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C Full file at (1) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 16 a.u/ CHAPTER First ( B × C ) i = ∑ ε ijk Bj Ck Then, jk [ A × (B × C) ]A = ∑ ε Amn Am ( B × C ) n = ∑ ε Amn Am ∑ ε njk BjCk mn = mn ∑ε Amn njk ε mn ε jkn Am Bj Ck Am Bj Ck jkmn jk jkmn   = ∑  ∑ ε lmn ε jkn  Am Bj Ck  jkm  n ( ) = ∑ δ jlδ km − δ kAδ jm Am Bj Ck jkm     = ∑ Am BA Cm − ∑ Am BmCA = BA  ∑ AmCm  − CA  ∑ Am Bm   m   m  m m = ( A ⋅ C ) BA − ( A ⋅ B) CA Therefore, A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C 1-23 Write ( A × B) j = ∑ ε jAm AA Bm Am ( C × D) k = ∑ ε krs Cr Ds rs Then, Full file at (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 17    [ ( A × B) × (C × D)]i = ∑ ε ijk  ∑ ε jAm AA Bm   ∑ ε krs Cr Ds  Am jk = ∑ε ijk rs ε jAm ε krs AA BmCr Ds jk Amrs = ∑ε jAm jAmrs =    ∑ ε ijk ε rsk  AA Bm Cr Ds k ∑ ε (δ jAm ir ) δ js − δ is δ jr AA BmCr Ds jAmrs ( = ∑ ε jAm AA BmCi Dj − AA Bm Di C j jAm )     =  ∑ ε jAm Dj AA Bm  Ci −  ∑ ε jAm C j AA Bm  Di  jAm   jAm  = (ABD)Ci − (ABC)Di Therefore, [( A × B) × (C × D) ] = (ABD)C − (ABC)D 1-24 Expanding the triple vector product, we have e × ( A × e) = A ( e ⋅ e) − e ( A ⋅ e) (1) A ( e ⋅ e) = A (2) A = e ( A ⋅ e) + e × ( A × e) (3) But, Thus, e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e Full file at Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 18 a.u/ CHAPTER 1-25 er eφ θ φ eθ The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by eθ = ( cos θ cos φ , cos θ sin φ , − sin θ )    eφ = ( − sin φ , cos φ , )   er = ( sin θ cos φ , sin θ sin φ , cos θ )  (1) Thus, ( e θ = −φ cos θ sin φ − θ sin θ cos φ , φ cos θ cos φ − θ sin θ sin φ , − θ cos θ ) = −θer + φ cos θ eφ (2) Similarly, ( e φ = −φ cos φ , − φ sin φ , ) = −φ cos θ eθ − φ sin θ er (3) e r = φ sin θ eφ + θeθ (4) Now, let any position vector be x Then, x = rer ( (5) ) x = re r + re r = r φ sin θ eφ + θeθ + re r = rφ sin θ eφ + rθeθ + rer ( (6) ( ) )   cos θ + rφ sin θ e + rφ sin θ e + rθ + rθ e + rθe +   x = rφ sin θ + rθφ rer + re r φ φ θ θ ( ( ) )   cos θ + rφ sin θ e + r − rφ sin θ − rθ e = 2rφ sin θ + 2rθφ φ r ( ) + 2rθ + rθ − rφ sin θ cos θ eθ or, Full file at (7) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 19 ( ) 1 d     x = a =  r − rθ − rφ sin θ  er +  r θ − rφ sin θ cos θ  eθ  r dt    d 2 + r φ sin θ  eφ  r sin θ dt  ( 1-26 ) (8) When a particle moves along the curve r = k (1 + cos θ ) (1) we have r = − kθ sin θ       r = − k θ cos θ + θ sin θ   Now, the velocity vector in polar coordinates is [see Eq (1.97)] (2) v = rer + rθ eθ (3) so that v = v = r + r 2θ 2 ( ) = k 2θ sin θ + k + cos θ + cos θ θ = k 2θ 2 + cos θ  (4) and v is, by hypothesis, constant Therefore, θ = v2 2k (1 + cos θ ) (5) v 2kr (6) Using (1), we find θ = Differentiating (5) and using the expression for r , we obtain θ = v sin θ v sin θ = 4r k (1 + cos θ ) (7) The acceleration vector is [see Eq (1.98)] ( ) ( ) a =  r − rθ er + rθ + 2rθ eθ so that Full file at (8) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 20 a.u/ CHAPTER a ⋅ er =  r − rθ ( ) = − k θ cos θ + θ sin θ − k (1 + cos θ ) θ   θ sin θ = − k θ cos θ + + (1 + cos θ ) θ  (1 + cos θ )     − cos θ = − kθ  cos θ + + 1 (1 + cos θ )   = − kθ (1 + cos θ ) (9) or, v2 k (10) v sin θ k + cos θ (11) ( a ⋅ er ) + ( a ⋅ eθ ) (12) v2 k (13) a ⋅ er = − In a similar way, we find a ⋅ eθ = − From (10) and (11), we have a = or, a = + cos θ 1-27 Since r × (v × r) = (r ⋅ r) v − (r ⋅ v) r we have d d r × ( v × r )] = [(r ⋅ r ) v − ( r ⋅ v) r ] [ dt dt = (r ⋅ r) a + (r ⋅ v) v − (r ⋅ v) v − ( v ⋅ v) r − (r ⋅ a) r ( = r 2a + ( r ⋅ v ) v − r v2 + r ⋅ a ) (1) Thus, ( d [ r × ( v × r )] = r 2a + ( r ⋅ v ) v − r r ⋅ a + v dt Full file at ) (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 1-28 21 grad ( ln r ) = ∑ i ∂ ( ln r ) ei ∂x i (1) where ∑x r = i (2) i Therefore, ∂ ln r ) = ( r ∂x i xi ∑x i i = xi r (3) so that grad ( ln r ) =   xe ∑ i i  r  i (4) r r2 (5) or, grad ( ln r ) = 1-29 Let r = describe the surface S1 and x + y + z = describe the surface S2 The angle θ between S1 and S2 at the point (2,–2,1) is the angle between the normals to these surfaces at the point The normal to S1 is ( ) ( grad ( S1 ) = grad r − = grad x + y + z − = ( 2xe1 + ye2 + 2ze3 ) x = 2, y = 2, z = ) (1) = 4e1 − 4e2 + 2e3 In S2 , the normal is: ( ) grad ( S2 ) = grad x + y + z − = ( e1 + e2 + 2ze3 ) = e1 + e2 + 2e3 Therefore, Full file at x = 2, y =−2, z = (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 22 a.u/ CHAPTER cos θ = = grad ( S1 ) ⋅ grad ( S2 ) grad ( S1 ) grad ( S2 ) ( 4e1 − 4e2 + 2e3 ) ⋅ (e1 + e2 + 2e3 ) (3) 6 or, cos θ = (4) 6 from which θ = cos −1 1-30 grad ( φψ ) = ∑ ei i =1 (5)  ∂ψ ∂ φ  ∂ ( φψ ) = ∑ ei φ + ψ ∂x i i  ∂x i ∂x i  = ∑ ei φ i = 74.2° ∂ψ ∂φ + ∑ ei ψ ∂x i ∂x i i Thus, grad ( φψ ) = φ grad ψ + ψ grad φ 1-31 a) 12   ∂rn ∂  n  ∑ xj   grad r = ∑ ei = ∑ ei ∂x i ∂xi  j   i =1  n n 2 n = ∑ ei 2xi  ∑ x 2j  2 j  i n  2 = ∑ ei xi n  ∑ x 2j   j  i −1 −1 = ∑ ei x i n r ( n − 2) (1) i Therefore, grad r n = nr ( n − 2) r Full file at (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 23 b) grad f ( r ) = ∑ ei i =1 ∂f ( r ) ∂ f ( r ) ∂ r = ∑ ei ∂x i ∂ r ∂x i i =1  ∂  = ∑ ei x 2j  ∑  ∂x i  j  i   = ∑ ei xi  ∑ x 2j   j  i = ∑ ei i 12 −1 ∂f ( r ) ∂r ∂f ( r ) ∂r x i ∂f r dr (3) Therefore, grad f ( r ) = r ∂f ( r ) r ∂r (4) c) 12   ∂ ln r ∂2   ∇ ( ln r ) = ∑ = ∑  ln  ∑ x j   ∂xi2 ∂x i   j   i  −1 1     ⋅ 2xi  ∑ x 2j       ∂ j =∑   12   i ∂x i      ∑ xj   j    ∂ =∑ i ∂x i −1      xi  ∑ x 2j     j      = ∑ ( − xi )( 2xi )  ∑ x 2j   j  i ( )( ) = ∑ −2x 2j r i =− −2 −2  ∂x  + ∑ i  ∑ x 2j   i ∂x i  j −1 1 + 3  r  2r + = r r r (5) or, ∇ ( ln r ) = Full file at r2 (6) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 24 a.u/ 1-32 CHAPTER Note that the integrand is a perfect differential: 2ar ⋅ r + 2br ⋅ r = a d d ( r ⋅ r ) + b ( r ⋅ r ) dt dt (1) Clearly, ∫ ( 2ar ⋅ r + 2br ⋅ r ) dt = ar 1-33 + br + const (2) Since d  r  rr − rr r rr = = − dt  r  r2 r r (1) we have ∫ d r   r rr   r − r  dt = ∫ dt  r  dt (2) r  r rr   r − r  dt = r + C (3) from which ∫ where C is the integration constant (a vector) 1-34 First, we note that ( ) d  =A  ×A  +A×A  A×A dt (1) But the first term on the right-hand side vanishes Thus, ∫ ( A × A ) dt = ∫ dt ( A × A ) dt (2) ∫ ( A × A ) dt = A × A + C (3) d so that where C is a constant vector Full file at Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 25 1-35 y x z We compute the volume of the intersection of the two cylinders by dividing the intersection volume into two parts Part of the common volume is that of one of the cylinders, for example, the one along the y axis, between y = –a and y = a: ( ) V1 = π a a = 2π a (1) The rest of the common volume is formed by equal parts from the other cylinder (the one along the x-axis) One of these parts extends from x = to x = a, y = to y = a2 − x , z = a to z = a − x The complementary volume is then a2 − x a V2 = ∫ dx ∫ 0 dy ∫ a2 − x a dz = ∫ dx a − x  a − x − a    a a  x a3 x =  a2 x − − sin −1  a 0  = 16 a − 2π a 3 (2) Then, from (1) and (2): V = V1 + V2 = Full file at 16 a3 (3) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 26 a.u/ CHAPTER 1-36 z d y c2 = x2 + y2 x The form of the integral suggests the use of the divergence theorem ∫ S A ⋅ da = ∫ ∇ ⋅ A dv V (1) Since ∇ ⋅ A = , we only need to evaluate the total volume Our cylinder has radius c and height d, and so the answer is ∫ V dv = π c d (2) 1-37 z R y x To the integral directly, note that A = R3er , on the surface, and that da = daer ∫ S A ⋅ da = R ∫ S da = R3 × 4π R = 4π R (1) To use the divergence theorem, we need to calculate ∇ ⋅ A This is best done in spherical coordinates, where A = r 3er Using Appendix F, we see that ∇⋅A = ∂ r A r = 5r r ∂r ( ) (2) Therefore, ∫ V π 2π 0 ∇ ⋅ A dv = ∫ sin θ dθ ∫ R ( ) dφ ∫ r 5r dr = 4π R5 Alternatively, one may simply set dv = 4π r dr in this case Full file at (3) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file MARS, a.tbaAND nkDVECTOR irect.eCALCULUS u/ 27 1-38 z z = – x2 – y2 y C x2 + y2 = x By Stoke’s theorem, we have ∫ ( ∇ × A) ⋅ da = ∫ S C A ⋅ ds (1) The curve C that encloses our surface S is the unit circle that lies in the xy plane Since the element of area on the surface da is chosen to be outward from the origin, the curve is directed counterclockwise, as required by the right-hand rule Now change to polar coordinates, so that we have ds = dθ eθ and A = sin θ i + cos θ k on the curve Since eθ ⋅ i = − sin θ and eθ ⋅ k = , we have ∫ C A ⋅ ds = ∫ 2π ( − sin θ ) dθ = −π 1-39 a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3) Then AB = (−1, 2, 0) ; AC = (−1, 0, 3) ; and AB × AC = (6, 3, 2) Any vector perpendicular to plane (ABC) must be parallel to AB × AC , so the unit vector perpendicular to plane (ABC) is n = (6 , , ) b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H Then DH = ( x − 1, y − 1, z − 1) is parallel to n given in a); this means x −1 = =2 y−1 and x −1 = =3 z −1 Further, AH = ( x − 1, y , z) is perpendicular to n so one has 6( x − 1) + y + z = Solving these equations one finds H = ( x , y , z) = (19 49 , 34 49 , 39 49) and DH = 1-40 a) At the top of the hill, z is maximum; 0= Full file at ∂z = y − x − 18 ∂x and 0= ∂z = x − y + 28 ∂y (2) Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton Full file 28 a.u/ CHAPTER so x = –2 ; y = 3, and the hill’s height is max[z]= 72 m Actually, this is the max value of z, because the given equation of z implies that, for each given value of x (or y), z describes an upside down parabola in term of y ( or x) variable b) At point A: x = y = 1, z = 13 At this point, two of the tangent vectors to the surface of the hill are t1 = (1, 0, ∂z ) = (1, 0, −8) ∂x (1,1) and t2 = (0,1, ∂z ) = (0,1, 22) ∂y (1,1) Evidently t1 × t2 = (8, −22,1) is perpendicular to the hill surface, and the angle θ between this and Oz axis is cos θ = (0, 0,1) ⋅ (8, −22,1) + 22 + 2 = 23.43 so θ = 87.55 degrees c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is steepest Evidently, dy = (tan α )dx and dz = 2xdy + ydx − xdx − ydy − 18 dx + 28 dy = 22(tan α − 1)dx then tan β = dx + dy dz = dx cos α −1 = 22(tan α − 1)dx 22 cos (α + 45) The hill is steepest when tan β is minimum, and this happens when α = –45 degrees with respect to W-E axis (note that α = 135 does not give a physical answer) 1-41 A ⋅ B = 2a( a − 1) then A ⋅ B = if only a = or a = Full file at