x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v
|
|
|
- まいえ あくや
- 5 years ago
- Views:
Transcription
1 c /(13)
2 x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v 1 v 2 v 1 v 2 v 1 v 2 = v 1 v 2 cos θ = b 1 b 2 + c 1 c 2, i j k v 1 v 2 = 1 b 1 c 1 = (b 1 c 2 c 1 b 2, c c 2, 1 b 2 b 1 2 ) 2 b 2 c 2 θ v 1 v 2 0 θ π v 1 v 2 v 1, v 2 0 = (0, 0, 0) v 1, v 2 v 1 v 2 v 1, v 2 v 1 v 2 sin θ v 1, v 2, v 1 v 2 v i = ( i, b i, c i ) i = 1, 2, 3 [v 1 v 2 v 3 ] v 1 v 3 1 b 1 c 1 [v 1 v 2 v 3 ] = v 1 (v 2 v 3 ) = 2 b 2 c 2 3 b 3 c 3 v 1 (v 2 v 3 ) v 1 (v 2 v 3 ) = (v 1 v 3 ) v 2 (v 1 v 2 ) v 3 c /(13)
3 nbl, = (,, ) f (x,, z) v(x,, z) = (u(x,, z), v(x,, z), w(x,, z)) f, grd f v, div v ( f, f, f ) + + w v, rot v ( w, w, ) 2 f, ( f ), f 2 f + 2 f f 2 2 (x,, z) n = (n x, n, n z ) f (x,, z) f n f = n x + n f + n f z v(x,, z) (x,, z) v v(x,, z) v (x,, z) 1 2 v f (x,, z) k k 2 f (x,, z) ( f g) = ( f ) g + ( g) f, ( f A) = ( f ) A + f ( A), ( f A) = ( f ) A + f ( A), (A B) = (B ) A + (A ) B + A ( B) + B ( A), (A B) = B ( A) A ( B), (A B) = (B ) A (A ) B + A ( B) B ( A), ( f ) = 0, ( A) = 0, ( A) = ( A) 2 A c /(13)
4 e r, e θ (r, θ) x = r cos θ, = r sin θ f = f r e r + 1 f r θ e θ, 2 f = 1 ( f ) 1 2 f r + r r r r 2 θ, 2 A = R e r + Θ e θ A = 1 (rr) + 1 Θ r r r θ (r, θ, ϕ) x = r sin θ cos ϕ, = r sin θ sin ϕ, z = r cos θ f = f r e r + 1 f r θ e 1 f θ + r sin θ ϕ e ϕ, 2 f = 1 ( r 2 f ) 1 ( f ) 1 2 f + sin θ + r 2 r r r 2 sin θ θ θ r 2 sin 2 θ ϕ, 2 A = R e r + Θ e θ + Φ e ϕ A = 1 (r 2 R) 1 (Θ sin θ) 1 Φ + + r 2 r r sin θ θ r sin θ ϕ, 1 ( (Φ sin θ) A = Θ ) er + 1 ( 1 R r sin θ θ ϕ r sin θ ϕ (rφ) ) eθ + 1 ( (rθ) R ) eϕ r r r θ (r, θ, z) x = r cos θ, = r sin θ, z = z f = f r e r + 1 f r θ e θ + f e z, 2 f = 1 ( f ) 1 2 f r + r r r r 2 θ + 2 f 2, 2 A = R e r + Θ e θ + Z e z A = 1 (rr) + 1 Θ r r r θ + Z, A = ( 1 Z r θ Θ ) er + ( R Z ) eθ + 1 ( (rθ) R ) ez r r r θ θ r e θ e e r r z eϕ r θ x x ϕ e θ x z r θ e z eθ e r c /(13)
5 t r(t) = (x(t), (t), z(t)) r() r(t) s(t) s(t) = t t r (τ) dτ = x (τ) 2 + (τ) 2 + z (τ) 2 dτ s s r(s) = (x(s), (s), z(s)) s t(s) = r (s) = (x (s), (s), z (s)) n(s), b(s) t (s) = κ(s)n(s), b(s) = t(s) n(s) κ(s) t (s) n, b t, n, b τ(s) t (s) = κ(s)n(s), n (s) = κ(s)t(s) + τ(s)b(s), b (s) = τ(s)b(s) c /(13)
6 u, v r(u, v) = (x(u, v), (u, v), z(u, v)) dr = r r du + dv E = r 2 = ( ) 2 ( ) 2 ( ) 2, r + + F = r = + +, G = r 2 = ( ) 2 ( ) 2 ( ) dr dr = E(du) 2 + 2Fdudv + G(dv) 2 n n = r r / r r L = 2 r 2 n, M = 2 r n, N = 2 r 2 n dr dn = L(du) 2 + 2Mdudv + N(dv) 2 L, M, N L = N = 1 EG F 2 1 EG F 2 2 x 2 2 x z 2 2 z 2, M = 1 EG F 2 2 x P P P κ 1, κ 2 H = 1 2 (κ 1 + κ 2 ) K = κ 1 κ 2 H = 1 2 EN 2FM + GL LN M2, K = EG F 2 EG F z, c /(13)
7 n k r 0 r k = (x k, k, z k ) f (r) k f k r n 1 k rk+1 f k (x k+1 x k ) f k k=0 f x f dx, z r n n 1 f k r k+1 r k f k=0 f ds t r(t) ( t b) t b b f dx = f (r(t))x (t)dt, f ds = f (r(t)) r (t) dt A(r) = (P(r), Q(r), R(r)) k A k n 1 A k (r k+1 r k ) A dr k=0 t(r) r b A dr = Pdx + Qd + Rdz = A tds = A(r(t)) r (t)dt 2 n k k f (r) k f k n f k k f d f (r) k=1 u, v r(u, v) u b, c v d d b f d = f (r(u, v)) r c r dudv c /(13)
8 r n(r) A(r) A nd A d A nd = A d = d b c A(r(u, v)) ( r r ) dudv r n r 3 n k k ϕ(r) ϕ k n ϕ k k ϕd k=1 u, v, w u b, c v d, e w f r(u, v, w) = (x(u, v, w), (u, v, w), z(u, v, w)) f ϕd = e d b c ϕ(r(u, v, w)) J(u, v, w) du dv dw J (x,, z) J(u, v, w) = (u, v, w) = w w w c /(13)
9 x P(x, ), Q(x, ) ( Q P ) dxd = Pdx + Qd x 2 xz A(x,, z) = z n (P(x,, z), Q(x,, z), R(x,, z)) A d = A nd n x ( P + Q + R ) dxddz = Pddz + Qdzdx + Rdxd 3 f (x,, z), g(x,, z) ( f 2 g+ f g) d = f g n d, ( ( f 2 g g 2 g f ) d = f n g f ) d n f n n f c /(13)
10 4 xz z n A = (P, Q, R) n ( A) nd = A t ds t x t ( R Q ) ( P ddz + R ) ( Q dzdx + P ) dxd = Pdx + Qd + Rdz c /(13)
11 dω = ω D D D n D D ω n 1 dω ω k ω = i 1<i 2< <i k f i1i 2 i k dx i1 dx i2 dx ik dx i dx j = dx j dx i dx i dx i = 0 f d f = f 1 dx 1 + f 2 dx f n dx n dω dω = i 1<i 2< <i k d f i1i 2 i k dx i1 dx i2 dx ik D xz 2 1 ω ω = Pdx + Qd + Rdz ( R Q ) ( P d dz+ R ) ( Q dz dx+ P ) dx d = Pdx+ Qd+Rdz f dx d f dxd (x,, z) (x,, z) (u, v, w) dx d dz = du dv dw (u, v, w) c /(13)
12 P p i j (x 1, x 2, x 3 ) (e 1, e 2, e 3 ) ( x 1, x 2, x 3 ) (ẽ 1, ẽ 2, ẽ 3 ) x i = p i j x j, ẽ i = p i j e j A j=1 j=1 A = 1 e e e 3 = ã 1 ẽ 1 + ã 2 ẽ 2 + ã 3 ẽ 3 A i ã i ã i = p i j j j=1 T 11 T 12 T 13 T = T 21 T 22 T 23 T i j = p i p jb T b T 31 T 32 T,b=1 33 T 2 T i j T + T i j + i j ϕt i j 0 T T T = T v = (v 1, v 2, v 3 ) w = Tv ( ) w j = T i j v j w T j=1 T i j = T ji T T i j = T ji T Tv = λv v ( 0) T λ 27 T T i jk P T i jk = p i p jb p kc T bc T 3 n,b,c=1 2 1 c /(13)
13 P i r i 1 i N F i r i F i F i N r i F i = 0 i=1 2 3 t (x,, z) ρ(x,, z, t) v(x,, z, t) ρ + (ρv) = 0 t 3 3 v(x,, z, t) v tds 4 ρ(x,, z) D(x,, z) D = ρ Q D d = ρd = Q 5 ω = (ω 1, ω 2, ω 3 ) P r = (x 1, x 2, x 3 ) ρ M = (M 1, M 2, M 3 ) M = r (ω r) ρ d I i j = ( k=1 xk 2 ρ d) δ i j x i x j ρ d M i = I i j ω j I i j j=1 c /(13)
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................
( ) ( )
20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
II 1 II 2012 II Gauss-Bonnet II
II 1 II 212 II Gauss-Bonnet II 1 1 1.1......................................... 1 1.2............................................ 2 1.3.................................. 3 1.4.............................................
24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x
![24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x](/thumbs/94/118744578.jpg "24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x")
24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),
1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
II 2 II
II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................
t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
notekiso1_09.dvi
39 3 3.1 2 Ax 1,y 1 Bx 2,y 2 x y fx, y z fx, y x 1,y 1, 0 x 1,y 1,fx 1,y 1 x 2,y 2, 0 x 2,y 2,fx 2,y 2 A s I fx, yds lim fx i,y i Δs. 3.1.1 Δs 0 x i,y i N Δs 1 I lim Δx 2 +Δy 2 0 x 1 fx i,y i Δx i 2 +Δy
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
meiji_resume_1.PDF
β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E
Gmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
![18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb](/thumbs/93/114079295.jpg "18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb")
r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
![II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re](/thumbs/94/118770263.jpg "II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re")
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
曲面のパラメタ表示と接線ベクトル
L11(2011-07-06 Wed) :Time-stamp: 2011-07-06 Wed 13:08 JST hig 1,,. 2. http://hig3.net () (L11) 2011-07-06 Wed 1 / 18 ( ) 1 V = (xy2 ) x + (2y) y = y 2 + 2. 2 V = 4y., D V ds = 2 2 ( ) 4 x 2 4y dy dx =
II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
Gmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
i 18 2H 2 + O 2 2H 2 + ( ) 3K
i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................
1 nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC
1 http://www.gem.aoyama.ac.jp/ nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC r 1 A B B C C A (1),(2),, (8) A, B, C A,B,C 2 1 ABC
1 B () Ver 2014 0 2014/10 2015/1 http://www-cr.scphys.kyoto-u.ac.jp/member/tsuru/lecture/... 1. ( ) 2. 3. 3 1 7 1.1..................................................... 7 1.2.............................................
Kroneher Levi-Civita 1 i = j δ i j = i j 1 if i jk is an even permutation of 1,2,3. ε i jk = 1 if i jk is an odd permutation of 1,2,3. otherwise. 3 4
[2642 ] Yuji Chinone 1 1-1 ρ t + j = 1 1-1 V S ds ds Eq.1 ρ t + j dv = ρ t dv = t V V V ρdv = Q t Q V jdv = j ds V ds V I Q t + j ds = ; S S [ Q t ] + I = Eq.1 2 2 Kroneher Levi-Civita 1 i = j δ i j =
I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1
4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1
1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
![1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1](/thumbs/94/121802164.jpg "1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1")
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 2 N(ε 1 ) N(ε 2 ) ε 1 ε 2 α ε ε 2 1 n N(ɛ) N ɛ ɛ- (1.1.3) n > N(ɛ) a n α < ɛ n N(ɛ) a n
http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 1 1 1.1 ɛ-n 1 ɛ-n lim n a n = α n a n α 2 lim a n = 1 n a k n n k=1 1.1.7 ɛ-n 1.1.1 a n α a n n α lim n a n = α ɛ N(ɛ) n > N(ɛ) a n α < ɛ
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (
II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )
p = mv p x > h/4π λ = h p m v Ψ 2 Ψ
II p = mv p x > h/4π λ = h p m v Ψ 2 Ψ Ψ Ψ 2 0 x P'(x) m d 2 x = mω 2 x = kx = F(x) dt 2 x = cos(ωt + φ) mω 2 = k ω = m k v = dx = -ωsin(ωt + φ) dt = d 2 x dt 2 0 y v θ P(x,y) θ = ωt + φ ν = ω [Hz] 2π
2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =
1 1 1.1 I R 1.1.1 c : I R 2 (i) c C (ii) t I c (t) (0, 0) c (t) c(i) c c(t) 1.1.2 (1) (2) (3) (1) r > 0 c : R R 2 : t (r cos t, r sin t) (2) C f : I R c : I R 2 : t (t, f(t)) (3) y = x c : R R 2 : t (t,
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
ii
ii iii 1 1 1.1..................................... 1 1.2................................... 3 1.3........................... 4 2 9 2.1.................................. 9 2.2...............................
B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t), y(t), z(t)), a t b.
2009 7 9 1 2 2 2 3 6 4 9 5 14 6 18 7 23 8 25 9 26 10 29 11 32 12 35 A 37 1 B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t),
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
Untitled
II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
F S S S S S S S 32 S S S 32: S S rot F ds = F d l (63) S S S 0 F rot F ds = 0 S (63) S rot F S S S S S rot F F (63)
211 12 1 19 2.9 F 32 32: rot F d = F d l (63) F rot F d = 2.9.1 (63) rot F rot F F (63) 12 2 F F F (63) 33 33: (63) rot 2.9.2 (63) I = [, 1] [, 1] 12 3 34: = 1 2 1 2 1 1 = C 1 + C C 2 2 2 = C 2 + ( C )
ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d
A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2
9 E B 9.1 9.1.1 Ampère Ampère Ampère s law B S µ 0 B ds = µ 0 j ds (9.1) S rot B = µ 0 j (9.2) S Ampère Biot-Savart oulomb Gauss Ampère rot B 0 Ampère µ 0 9.1 (a) (b) I B ds = µ 0 I. I 1 I 2 B ds = µ 0
all.dvi
72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G
II 1 3 2 5 3 7 4 8 5 11 6 13 7 16 8 18 2 1 1. x 2 + xy x y (1 lim (x,y (1,1 x 1 x 3 + y 3 (2 lim (x,y (, x 2 + y 2 x 2 (3 lim (x,y (, x 2 + y 2 xy (4 lim (x,y (, x 2 + y 2 x y (5 lim (x,y (, x + y x 3y
v er.1/ c /(21)
12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,
2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ
1 1 1.1 (Isaac Newton, 1642 1727) 1. : 2. ( ) F = ma 3. ; F a 2 t x(t) v(t) = x (t) v (t) = x (t) F 3 3 3 3 3 3 6 1 2 6 12 1 3 1 2 m 2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t)
A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2.
A A 1 A 5 A 6 1 2 3 4 5 6 7 1 1.1 1.1 (). Hausdorff M R m M M {U α } U α R m E α ϕ α : U α E α U α U β = ϕ α (ϕ β ϕβ (U α U β )) 1 : ϕ β (U α U β ) ϕ α (U α U β ) C M a m dim M a U α ϕ α {x i, 1 i m} {U,
2.4 ( ) ( B ) A B F (1) W = B A F dr. A F q dr f(x,y,z) A B Γ( ) Minoru TANAKA (Osaka Univ.) I(2011), Sec p. 1/30
2.4 ( ) 2.4.1 ( B ) A B F (1) W = B A F dr. A F q dr f(x,y,z) A B Γ( ) I(2011), Sec. 2. 4 p. 1/30 (2) Γ f dr lim f i r i. r i 0 i f i i f r i i i+1 (1) n i r i (3) F dr = lim F i n i r i. Γ r i 0 i n i
7-12.dvi
26 12 1 23. xyz ϕ f(x, y, z) Φ F (x, y, z) = F (x, y, z) G(x, y, z) rot(grad ϕ) rot(grad f) H(x, y, z) div(rot Φ) div(rot F ) (x, y, z) rot(grad f) = rot f x f y f z = (f z ) y (f y ) z (f x ) z (f z )
08-Note2-web
r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)
Fubini
3............................... 3................................ 5.3 Fubini........................... 7.4.............................5..........................6.............................. 3.7..............................
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
![II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2](/thumbs/101/149159646.jpg "II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
gr09.dvi
.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
all.dvi
5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0
W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =
II 6 ishimori@phys.titech.ac.jp 6.. 5.4.. f Rx = f Lx = fx fx + lim = lim x x + x x f c = f x + x < c < x x x + lim x x fx fx x x = lim x x f c = f x x < c < x cosmx cosxdx = {cosm x + cosm + x} dx = [
dynamics-solution2.dvi
1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n
128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds
127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds
重力方向に基づくコントローラの向き決定方法
( ) 2/Sep 09 1 ( ) ( ) 3 2 X w, Y w, Z w +X w = +Y w = +Z w = 1 X c, Y c, Z c X c, Y c, Z c X w, Y w, Z w Y c Z c X c 1: X c, Y c, Z c Kentaro Yamaguchi@bandainamcogames.co.jp 1 M M v 0, v 1, v 2 v 0 v
S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d
S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T
NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
untitled
20010916 22;1017;23;20020108;15;20; 1 N = {1, 2, } Z + = {0, 1, 2, } Z = {0, ±1, ±2, } Q = { p p Z, q N} R = { lim a q n n a n Q, n N; sup a n < } R + = {x R x 0} n = {a + b 1 a, b R} u, v 1 R 2 2 R 3
tomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.
tomocci 18 7 5...,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. M F (M), X(F (M)).. T M p e i = e µ i µ. a a = a i
Microsoft Word - 11問題表紙（選択）.docx
A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx
Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e
7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z
6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
Holton semigeostrophic semigeostrophic,.., Φ(x, y, z, t) = (p p 0 )/ρ 0, Θ = θ θ 0,,., p 0 (z), θ 0 (z).,,,, Du Dt fv + Φ x Dv Φ + fu +
Holton 9.2.2 semigeostrophic 1 9.2.2 semigeostrophic,.., Φ(x, y, z, t) = (p p 0 )/ρ 0, Θ = θ θ 0,,., p 0 (z), θ 0 (z).,,,, Du Dt fv + Φ x Dv Φ + fu + Dt DΘ Dt + w dθ 0 dz = 0, (9.2) = 0, (9.3) = 0, (9.4)
D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j
![D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j](/thumbs/103/158001152.jpg "D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j")
6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..
http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg
untitled
- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =
Gmech08.dvi
63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ
TOP URL 1
TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................
grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )
2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))
III Kepler ( )
III 20 6 24 3....................................... 3.2 Kepler ( ).......................... 2 2 4 2.................................. 4 2.2......................................... 8 3 20 3..........................................
( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)
( : December 27, 215 CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x f (x y f(x x ϕ(r (gradient ϕ(r (gradϕ(r ( ϕ(r r ϕ r xi + yj + zk ϕ(r ϕ(r x i + ϕ(r y j + ϕ(r z k (1.1 ϕ(r ϕ(r i
8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J
26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK
006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................
(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {
![(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {](/thumbs/94/118399854.jpg "(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {")
7 4.., ], ], ydy, ], 3], y + y dy 3, ], ], + y + ydy 4, ], ], y ydy ydy y y ] 3 3 ] 3 y + y dy y + 3 y3 5 + 9 3 ] 3 + y + ydy 5 6 3 + 9 ] 3 73 6 y + y + y ] 3 + 3 + 3 3 + 3 + 3 ] 4 y y dy y ] 3 y3 83 3
I ( ) 2019
I ( ) 2019 i 1 I,, III,, 1,,,, III,,,, (1 ) (,,, ), :...,, : NHK... NHK, (YouTube ),!!, manaba http://pen.envr.tsukuba.ac.jp/lec/physics/,, Richard Feynman Lectures on Physics Addison-Wesley,,,, x χ,
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
応力とひずみ.ppt
in yukawa@numse.nagoya-u.ac.jp 2 3 4 5 x 2 6 Continuum) 7 8 9 F F 10 F L L F L 1 L F L F L F 11 F L F F L F L L L 1 L 2 12 F L F! A A! S! = F S 13 F L L F F n = F " cos# F t = F " sin# S $ = S cos# S S
1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h
IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r
1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D
1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00
M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -
M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................
4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP
1. 1 213 1 6 1 3 1: ( ) 2: 3: SF 1 2 3 1: 3 2 A m 2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP