|
|
|
- まとも ことじ
- 7 years ago
- Views:
Transcription
1 2010 4
2
3 ( )
4 (Magnet) (Displacement current) (Electromagnetic Induction) Motional Induction Transformer Induction Motional induction Transformer induction A 87 A A A
5 % 2 20% 30% 30% 60% 1, III III 7
6 ( ) D., R., J. /., 2002 ( :3). 427 H-18.., 2008 ( : ; 3). 427 M [ / ], 2005, 427 T-11.., 1999 ( - ). 427 H-14.., 1997 ( ). 427 K-33. :., , 1987 ( :4) B , 1987 ( :3). 427 K-27., Edward M.Purcell [ ]/., ( :2). 420 B ,, /., 1969 ( :3). 420 F-5. :., W-8. ( ), 2004 ( :2) E-3 2.., 1998 ( One Point:2) O-2. /., 2002 ( ). 427 K-41.., 2005 ( :B-1483). 408 B ( 2)., H-2. ( 3)., T-1 y. 427 T-7 b. :, 1, 2 V. D., M. G. /
7 0.2 7,., B-9 1., 1, 2., O , ( ), J.D. /., 1994 (. 67.) J-3 2,., G-3.., 2002 ( :3) A-4. :,., , 1987 ( :5) B /,., 1981 ( :21) K c.
8
9 9 1 7 *1 600 *2 1 p * *4 elektron electricity 1773 * * * * * *1 * * * * * * * J.J *
10
11 11 r[m] q[c] Q[C] k qq ( k = Nm 2 C 2 r 2 [N] ε 0 k = 1 4πε 0 µ 0 = 4π 10 7 N/A 2 c = 1 ε0 µ 0 c = m/s k ε 0 = 1 c 2 µ 0 = C 2 /N m m (1 )
12 12 1 ~10-2 m ~10-10 m ~10-6 m + + ~10-9 m Å (= ) =,,,, + + m e me 1840me e +e 0 e = C
13 mA 1.5V 1Ω cm ( ) ( ) ( ) ( ) ( ) ( 10 4 m 3 ), ( )., m 3. ( ) 10
14
15 15 2 ( ) *1 (1) ( ) (2) ( ) (3) ( ) (4) ( ), r (x, y, z) r = (x, y, z) r t r(t) t,. r *1 material, object
16 16 2 (scalar),,,,,,,, 2 (x y ) a = (a x, a y ) = a x î + a y ĵ î = (1, 0), î = (0, 1) 2 ( r θ) a = (r cos θ, r sin θ) = r cos θî + r sin θĵ = rˆr ˆr = (cos θ, sin θ), ˆθ = ( sin θ, cos θ) a y y a a ^ ^ r j^ ^ i a x x 2 2 (source) (field) (particle) Q r q F = k qq r 3 r = q E E = k Q r 3 r r E qe (probe) q
17 17 *2 R 1 Q 1 R 2 Q 2 r Q E 1 = k r R 1 ( r R 1 ), Q E2 = k 3 r R 2 ( r R 2 ), 3 q Q 2 Q 1 R 1 r O R 2 *
18 18 2 q r q E 1 ( r) q E 2 ( r) F = q E 1 + q E 2 r E = E 1 + E 2
19 19 4 ( Kane and ternheim, )
20 A( r) = (x, y) A( r) = r r r = (x, y) r = x 2 + y A( r) = ( y, x) xy (0, a) +q (0, a) q r = (x, y) (1) y E x = 0, E y 2k P y 3 P = qd (2) x E x = 0, E y k P x 3 Mathematica
21 A( r) = (x, y) A( r) = r r
22 A( r) = ( y, x)
23 23 3 (global) (local) t t + dt v x(t + dt) x(t) = vdt v = dx dt. ( ) 1 ( ) d
24 24 第3章 電場のフラックス 立体角 ガウスの法則 面積 d を取り出す 球の中心から微小領域までの距離を r とする このとき ドームの壁に映る 微小領域の面積 dω と 微小領域本来の面積 d との関係は d cos α = r2 dω となる ここで 板 の法線ベクトル 面に垂直な単位ベクトル を n として α は n と r のなす角 である ここに現れた dω を 微小領域d を見込む立体角という 立体角は参照球面の一部分の面積 なので これらを全部足し合わせれば必ず 4π になる このことを ZZ dω = 4π と表現する つまり 立体角の積分は 4π である 積分記号に を添えることで ガウス面 上を掃 きつくすように積分を実行することを明示する ガウス面 r 法線ベクトル 立体角 dω ᵼ n α d cosα フラックス 放射状に水を放出できるスプリンクラーを思い浮かべよう スプリンクラーを網で囲む 網面がガウス面 網上の単位面積を毎秒 v リットルの水流が貫くとする このとき 網全体を毎秒 通過する フラックスと呼ぶ は ZZ ZZ Φ= v cos αd = vr2 dω と書けるだろう α は水流が 網を貫く点で網の法線となす角である ここで は網の表面 ガウ ス面 を表す フラックのイメージと数学的な表現が理解できるだろうか 電場のフラックス 点電荷 q の作る電場は = E 1 q r 4πε0 r3
25 25 Φ = E cos αd α E cos αd = E d d = ˆnd d cos α = r 2 dω ( ) q 1 Φ = 4πε 0 r 2 r 2 dω 1 r 2 r2 r Φ = q 4πε 0 dω = q 4πε 0 4π = q ε 0 E d = q ε 0 q q 1, q 2,, q N q N j=1 q j E d = 1 ε 0 N q j j=1
26 26 3 N j=1 q j ( dv ) ρ ρdv V ρdv V E d = 1 ε 0 V ρdv ε 0
27 a σ 3.2 λ 3.3 a Q 3.4, 5, ( N/C), λ = C/m,, ) ε 0 = C 2 /N m a, (1) F = k qq r 2 F = k qq r 2.1 (2) J.J. ( ) a ( Q) e (1) (2) a = m, e = C, m = kg
28 [ / ], p [ / ],
29 Q x m q v = dx/dt m dv dt = k qq x 2 mv dv dt = k qq dx x 2 dt = d ( ) 1 dt 2 mv2 = d ( k qq ) dt x = d ( 1 dt 2 mv2 + k qq ) = 0 x = 1 2 mv2 + k qq x = E ( ) E E U (x) = k qq x ( ) F (x) x U (x) du (x) F (x) = dx mv dv dt = F (x)dx dt = d ( ) 1 dt 2 mv2 = d ( U (x)) dt = d ( ) 1 dt 2 mv2 + U (x) = 0 = 1 2 mv2 + U (x) = E ( U (x) F U x U(x) = F (x )dx x 0 x 0
30 (1) (2) 4.2 F (x) x 1 x 2., F W x1 x 2 = x2 x 1 F (x )dx du (x). F F (x) = dx x2 du (x ) W x1 x 2 = x 1 dx dx = U (x 1 ) U (x 2 ) 4.3 Q x E(x) = k Q x 2 U(x) = q x x 0 E(x )dx x U(x)/q = V (x) = E(x )dx x 0 dv (x) E(x) = dx V (x) x 0 x x 0 V (x) = x k Q x 2 dx = [ k Q ] x x = k Q x U(x) = qv (x) [V] 4.2 +Q A +Q B B v 0
31 V ( r) ( ) V ( r) V ( r) V ( r),, x y z V ( r) grad V ( r) ( r) ( V V = x, V y, V ). z V = 1 r r = x 2 + y 2 + z 2 V = r r 3 Q r E = k Q r 3 r q r Q V = r r 3 V = k Q r E = V 3 q m d v dt = q V v = d r dt m v d v dt d r = q dt V d ( ) 1 dt 2 m v2 = q dx V dt x q dy V dt y q dz V dt z d ( ) 1 dt 2 m v2 = q dv dt d ( ) 1 dt 2 m v2 + qv = 0 = 1 2 mv2 + qv = E (
32 32 4 qv V =?? r V ( r) = E( r ) d r r 0 r 0 r 4.3 r 4.4 r 0 Q r V ( r) V ( r) = 1 Q 4πε 0 r r 0 E( r) 4.5 r 1, r 2, r N, Q 1, Q 2, Q N, r V ( r) V ( r) = 1 4πε 0 N j=1 Q j r r j E( r) 4.6 p V ( r) = 1 p r 4πε 0 r (1) (2) E Q (3) E P
33 r 0 r w = N 1 j=0 f( r j ) ( r j+1 r j ) r r0 r 0 r N w = r f( r) d r r 0 r C r0 C C f = (x, y) f( r ) d r r 0 r 0 = (0, 0), r = (1, 1) 1: y = x 2: y = x 2 3: y = x y (1,1) r (0,0) x 1 f = (x, y) = (x, x) d r = (dx, dy) = (dx, dx) f d r = 2xdx 1 0 2xdx = 1
34 f = (x, y) = (x, x 2 ) d r = (dx, dy) = (dx, 2xdx) f d r = xdx + 2x 3 dx 1 0 (x + 2x 3 )dx = 1 3 f = (x, y) = (x, x) d r = (dx, dy) = (dx, 1 2 x 1/2 dx) f d r = xdx + 1 dx = (x + 1/2)dx (x + 1/2)dx = 1 3 V ( r) = 1 2 (x2 + y 2 ) f x = V x, f y = V y, z ( V f = x, V ) V y r 0 = (0, 0), r = (1, 1) f d r = dv = V ( r) + V ( r 0 ) = 1 x y dv (x, y) dv = V V dx + x y dy V f = V 4.9 V ( r) = mgz m, g V
35 (x ) x v(x) v(x x/2) y z x v(x) x v(x+ x/2) y zn y zn 3 x, y, z ( ) (x, 0, 0) 1 2 (x x/2, 0, 0), (x+ x/2, 0, 0) 1 2 y z 1 2 v(x + x/2) y z } {{ } 2 v(x x/2) y z } {{ } 1 v(x) x x y z 1 2 ˆn, +ˆn 1 = y zˆn, 2 = + y zˆn,, 1 2 v 1, v 2, v v 2 2 = v x(x) x x y z 3, r = (x, y, z) v( r) 6 v v v v v v 6 6 [ vx (x) = + v y(x) + v ] z(x) x y z x y z = v( r) x y z
36 v( r) d = v( r)dv V, ( ), v( r) = (y 2, 2xy + z 2, 2yz) , V Q, ρ [C/m 3 ] Q = ρ( r)dv, E( r) d = 1 ρ( r)dv ε 0 V V, E( r) d = E( r)dv V E( r)dv = 1 ρ( r)dv ε 0, V V E( r) = 1 ε 0 ρ( r), 4.8, [ V ( r)] = 1 ρ( r) ε 0 ( ) 2 = x y z 2 V ( r) = 1 ρ( r) ε 0 = 2 V ( r) = 1 ε 0 ρ( r)
37 ( r) V = 1 ε 0 ρ 4.11, x V = V 0 (x/d) 4/3..
38
39 B At
40 40 5 n[ /m 3 ] Na 2.65 Cu 8.47 Ag 5.86 Au 5.90 Fe 17.0 Mn σ σ/ε (1) σ/ε 0 (2) σ 2 /2ε 0 electric tension electric stress
41 (condenser) (capacitor) [ / ],
42 Q Q V Q = CV C (capacitance) F( ) 1F 1µF= 10 6 F 1pF= F Q q A q B q A q B Q q C Q q D q C q D q A q B Q E Q V Ed Q q A q B q C q D q A q B q C q D q A q B q C q D C d q A q B q C q D 2. (a) Q Q Q Q b a Q Q r(a<r<b) Q E 4 r Q b V ae dr 4 a C 4 ab b a b a :C = 4πε 0 a 5.4 a, ( 2a, 3a),, 1, 2 ε 0 [A] 1 ( ), Q [A-1] [A-2] [B] 1 2 ( ), Q [B-1]
43 Q [B-2] [B-3] 2a a 3a d ±Q d U F E = E = E + E = σ ε 0 σ 2ε 0 E = σ 2ε 0 σ = Q/ F F = QE = σ2 2ε 0 = 1 2 QE U = F d = σ2 2ε 0 d = 1 2 QEd = 1 2 QV = 1 2 CV 2 = Q2 2C U = σ2 2ε 0 d
44 44 5 d u = U d = σ2 2ε 0 = 1 2 ε 0E ε 0E V (1) Q r (2) Q (3) Q V (4) C (5) ( ) 1 U = 2 ε 0E 2 dτ V dτ dv V (6) U = 1 2 CV 2 L b a 5.8 W = 1 2 N q i V ( r i ) i=1 W = ε 0 E 2 dτ 2 V 5.6
45 v (steady current) j A/m 2 n j = en v 6.1 j = en v 6.2 n /m 3. A = 2mm 2 1A, [ v m/s 1/300cm] ], 21 [ /
46 E J. J. Thomson Drude Drude τ τ 1 m a = e E a = e E/m τ v = ee m τ j = en v = ne2 τ m E j = σ E σ = ne2 τ m (electrical conductivity) (electrical resistivity) ρ=1/σ
47 [ / ], ρ 10 8 Ω m = ρ Ω m = ρ Ω m = (thermal conductivity) C V = 3 2 k B κ σ = nc V v 2 τ 3 κ = nc V v 2 τ 3 m ne 2 τ =k B m v 2 e 2 = k B 3 2 e 2 2 k BT = 3 2 ( ) 2 kb m v 2 = k BT L = 3 ( kb 2 e ) 2 = 3 2 κ σ = 3 2 ( kb e ) 2 T ( ) 2 [J/K] [V 2 /K 2 ] [C] e
48 48 6 κ σt = L (Wiedemann-Franz) 6.3 ρ L R = ρ L (resistance) G = 1 R (conductance) E = σ 1 j L LE = Lσ 1 j = V = L j = V = RI }{{} σ }{{} =I =R 6.3 ε σ C R = ε σc 6.4 a ( l), b, σ. V,.
49 b a V 6.5 ( ) 6.4 R Q J = I 2 R 1840 J.P.Joule Joule 6.6 m F µ
50 t = 0 V C R (1) Q(t) (2) Q(t) (3) I(t) (4) R dq(t) dt + Q(t) C (5) (6) (7) (6) = V 6.8 ( ) 70mV 6.0 nm Benjamin Crowell, Electricity and Magnetism
51 (Wheatstone bridge) R3 Rx V R1 R ρ R = ρ ( ) h π ab
52 V j d j. d dv t, j d } {{ } + V ρ t dv } {{ } j d = V = 0 jdv V jdv + V ( ρ dv = 0 = t V j+ ρ ) dv = 0 t, j+ ρ t = 0 (continuity equation) j d = I V ρ t dv = d ( ) ρdv = dq dt V dt
53 6.7 ( ) 53 I dq dt I + dq dt = ( 1/e ). t = 0 ρ 0, t ρ. (1) j+ ρ t = 0 E = ρ ε j = σ E ρ t = σ ε ρ. (2) τ. (3) ( ), σ = Ω 1 m 1. ε = C 2 /N m 2. (4) 6.7 ( ) ( ) ( ) ( ) ( ) ( ) l x 1. de(x) = en(x) dx ε dv (x) 2. E(x) = dx 3. m dv dt = ee(x) mv2 = ev (x) 5. j = env
54 54 6 n(x) ( ) j x d2 V (x) dx 2 dv dx dv d 2 V (x) dx dx 2 = j ε = dv dx = = en(x) ε = j εv = j ( ) 1/2 m ε 2eV (x) { m dv 2eV (x) dx = d ( ) } 2 1 dv = j ( m ) 1/2 d dx 2 dx ε 2e dx ( ) 1/2 4j m V 1/4 = dv ( ) 1/2 4j m ε 2e V = dx 1/4 ε 2e ( ) 1/2 4j m x = dv V 1/4 = = V = ε ( 9j m 4ε 2e 2e ) 2 3 x 4 3 ) 1/2 dx = 4 3 V 3/4 = ( 4j ε m 2e (2V 1/2) x = l V = ( ) 9j m 2 3 l 4 3 4ε 2e j = 4ε 2e 9l 2 m V 3/2
55 a b ( ) a b, a b sin φ, a b a x a y a z a x b x b y b z b x a x b y a y b x a y b z a z b y a z b x a x b z a b z a b x a b y a b a b sinφ a b b a φ b a 7.2 B T T=N/A m q B v F L = q v B 7.1 B v 0 q
56 B v 0 q Physics for cientists and Engineers 6th ed. (College Text), erway and Jewett
57 7.3 (Magnet) (Magnet) 1 N
58 N N N B d = 0, B = 0
59 K q E v K, K v c v c K B = 1 c 2 v E. *1 c = 1 ε0 µ 0. *1 11-3
60 (1)., λ[c/m]. r E. (2) v. r P B., P θ. 8.2 E = q r (q = e) v 4πε 0 r3 B = 1 c 2 v E = µ 0q v r 4π r I d l r d B q = endl v v d l db = µ 0 ( endl) v r 4π r 3 = µ 0 ( en v) dl r 4π r 3 ( en v) dl = ( env) d l = Id l d B = µ 0I 4π d l r r
61 db I r d` d l r θ d B db = µ 0I 4π sin θdl r 2 * 2 *2 11-3
62 (1) z (0, 0, z) B z z B db z = µ 0I 4π a (a 2 + z 2 ) 3/2 dl z db O I a Id` (2) a Benjamin Crowell, Electricity and Magnetism
63 ( a n ) I B z z z + dz ndz z B db = µ 0I 2 a 2 { a 2 + (z z ) 2} 3/2 ndz 8.4 (1) AB I. P. (2) P A 1 2 r B a I. (1) O. (2) d P. P d 2a 2a O
64 a 8.7 a b a b
65 M = µ 0 I r m q I = q T = qv 2πr (T = 2πr/v) ( qv ) M = µ 0 πr 2 ẑ 2πr ẑ l = mrvẑ ( qv ) M = µ 0 πr 2 1 2πr mrv l = ( µ0 q ) l = g l 2m g = µ 0q 2m ` M
66 H = 1 µ 0 B µ 0 = 4π 10 7 N/A 2 H d r = I C., C.. C I H dr,, j( r)., C, I I = j d j r d C,. C H d r = j d I 1, I 2 ( r) F = µ 0I 1 I 2 2πr
67 r =1m I 1 = I 2 = I 1m N/m I µ ,. (1) (a) R i. xy.., x Q(x, 0) B 1 (x). 0 < x < R R < x. (2) (b) r i. xy. (a, 0)., a r < x < a + r, Q(x, 0) B 2 (x). (3) (c), R, a b. i. a r < x < a + r, Q(x, 0) B(x). (a) y (b) y (c) y R RR R O Q x O Q P r x O Q r P x a a
68
69 f = (f x, f y, f z ) ( f fz = y f y z, f x z f z x, f y x f ) x y f, rot f f = i j k x y z f x f y f z i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) 9.1 f (1) f = (x, y, 0) (2) f = ( y, x, 0) 9.2 (A) (F) y y y x x x y y y x x x
70 f = (f x, f y, f z ) xy A B C D A 1 f(x, y, z) W A B C D A. x y+ y D C y y y A x B x x+ x W A B C D A = W A B + W B C + W C D + W D A W A B = f x (x, y, z) x, [ W B C = f x (x + x, y, z) y f y (x, y, z) + f y(x, y, z) x [ W C D = W D C = f x (x, y + y, z) x W D A = W A D = f y (x, y, z) y. ] x y, f x (x, y, z) + f x(x, y, z) y y ] x,, [ W A B C D A = f x (x, y, z) x + [ f x (x, y, z) + f x(x, y, z) [ fy (x, y, z) = x f y (x, y, z) + f y(x, y, z) x ] y y f x(x, y, z) y ] x y x f y (x, y, z) y ] x y. W A B C D A = A B C D A B C D f d r. [ fy (x, y, z) f d r = f ] x(x, y, z) x y x y ABCD yz zx [ fz (x, y, z) f ] y(x, y, z) y z, y z [ fx (x, y, z) f ] z(x, y, z) z x z x. xy ABCDEF 1 2
71 D C F E A D B C F A E G B E G G E [A B C D E F A ] A G E F A ] + G B C D G ] G E E G A B C D E F A A G E F A G B C D G C xy, xy C C [ fy (x, y, z) f d r = f ] x(x, y, z) dxdy = [ C x y f] x dxdy = ( f) d d xy C xy C C C f d r = ( f) d
72 f = (0, 2xz + 3y 2, 4yz 2 ) 1 1 [ (0, 0, 0) (0, 1, 0), (0, 1, 1), (0, 0, 1)] 9.3 H d r = j d C ( H) d = j d, H = j 9.4 { E = ρ ε 0 E = 0 E = V ( ) V = 0 { B = 0 B = µ 0 j
73 73 10 (Displacement current) I(t)., ±Q(t).,., E(t) = Q(t) ε 0 I(t) = dq(t) dt,., I(t) ( ).,.,, I(t), I(t),, I D (t). E t + Q t Q t I t I t I t ID t I t } } }, I D = I I D (t) = ε 0 de(t) dt, ( d I D (t) = ε 0 dt ) E d.,,.,,.,,.
74 74 10 (Displacement current), C ( H d r = j d d + ε 0 dt ) E d., C H d r = ( j + ε ) E 0 d t.,. J = j + ε 0 E t
75 75 11 (Electromagnetic Induction) IH 11.3 Motional Induction l B v. V m = v B l = vbl sin θ, motional induction ( ). ( ) C V m = v B d r C
76 76 11 (Electromagnetic Induction) B ` v 10.1 Motional induction, V m = v B l = B l v = B d (t) dt.,, ( ) Transformer Induction 2 C( ), B(t). Φ(t) = B(t) V t = dφ(t) dt, transformer induction. n B, E.. C E d r = dφ(t) dt. ( r B( r, t) ),. V t = C B( r, E d r = t) d t
77 11.4 Transformer Induction z B(t) = (0, 0, B 0 sin(ωt)). xy r E(r).. ( z ) I. 1 a ABCD, AB CD. AB x. (1) x B(x). µ 0. (2) ABCD Φ(x). (3) x I, ABCD. di/dt. (4), I ABCD v. AB x ABCD. v = dx/dt. (5) I ABCD v. AB x ABCD. I A a a D O B x C
78 78 11 (Electromagnetic Induction) 10.4 IH N (1) B(t) = B 0 cos(ωt) AC [ V = 1 3 πb 0ωh 2 N 3 sin (ωt)] Φ = 1 2πN 2 B(t) {r(θ)} 2 dθ r(θ) = hθ/2π. (2) ρ Q J (3) = 0.1cm 2, ρ = 10 8 Ω m h = 0.1cm N = 100 Q J 0
79 11.5 Motional induction Transformer induction Motional induction Transformer induction 10.5( ) l (θ =90) B t B(t) = B 0 cos(ωt) v., x. ` x v motional induction transformer induction V emf = V m + V t = d dt ( (t) B(t)), Φ Φ(t) = B(t) (t)., V emf = dφ(t) dt.. motional induction transformer induction C., C. C. C, ( )., C., C B( r), Φ, Φ(t) = B d., Φ C V = dφ(t) dt., C ( ) E( r). V emf = E d r C
80 80 11 (Electromagnetic Induction) Br d = nd C Ein r. C E d r = d dt ( ) B d ( ) B(r) = { B0 sin (ωt) (0 < r a) B 1 sin (ωt) (a < r) t = 0 r = 2a m e r = 2a (1) B 0 B 1 B 0 = 5B 1 (2) a = 6cm B 0 = 0.02T ω = 100s 1 28% 11.7 E d r = ( E) d, C ( E) d = d dt ( E = B t. ) B d
81 E ( r) = 1 4πε 0 Q r 3 r ρ ( r) Q 2. Q = ρdv E d = 1 ρdv ε 0 V V
82 E d = EdV V 4. E = 1 ε 0 ρ B d = 0 B = B = µ 0 H 2. dh = I d l r 4π r 3 3.
83 83 1. j ( r) I 2. I = j d H d r = j d C 3. H d r = H d C 4. H = j
84 I 2. ( ) d I D = ε 0 E d dt E = ε 0 t d 3. H d r = I + I D C C H d r = ( j + ε ) E 0 d t 4. 5 H = j + ε 0 E t 1. Motional Induction 2. Transformer Induction 3. Φ = B d V = dφ(t) dt C E d r = d dt ( ) B d C E d r = E = B t E d E 0
85 E d = 1 ρdv ε 0 V B d = 0 2. E = 1 ε 0 ρ B = 0 C H d r = C E d r = d dt ( H = j + ε 0 E t E = B t ( j + ε ) E 0 d t ) B d
86
87 87 A A (1) e = n C n (2) 2 10 (3) 3 Q r E (4) λ r 2 xy ( a, 0) q (a, 0) +q r = (x, y) E( r) y r A Ο B x (1) = 2 d E( r) r d k (2) (a, a) x y (3) y (0, y) 1 y 3 3 (1) (5) (A) (F) (1) A( r) = (x, x) (2) A( r) = (x, 1)
88 88 A (3) A( r) = (x, y) (4) A( r) = (y, x) (5) A( r) = ( y, x) y y y x x x y y y x x x 4 a Q r E(r) r } (1) r > a E(r) (2) 0 < r < a E(r) (3) r (4) 1904 J.J. ( ) a ( e) e m T a ε 0 e m 5 6
89 A a λ 7 ρ d
90 90 A A.2 2 A1 A3 B1 B3 1 A1 [9 ] (1) ρ( r) E( r) E( r) V ( r) E( r) = V ( r) ρ( r) (2) σ (3) e n v j = n = /m 3 A = 2mm 2 1A, v = n m/s n n = (4) ε σ ±Q σ ε Q I = σ ε C R = A2 [3 ] (1) p r = (x, y, z) ( p r) = ( ) 1 r 3 = r = x 2 + y 2 + z 2 1 p r r 2 (r n ) n r r (2) p V ( r) = 1 p r 4πε 0 r 3 E( r) = 1 4πε 0 ( f ( r) g ( r) (fg) = f g + g f A3 [4 ] v = (x + y, x y, z) 1 1 vdv = v d V )
91 A vdv = V v d B1 [4 ] [1] a b V b a V (1) C (2) ( ) 1 U = 2 ε 0E 2 dτ V U = 1 2 CV 2 dτ dv
92 92 A V r r + dr [2] ε σ b a V (3) r I(r) a < r < b r r (4) (3) A1 B2 [4 ] R V t = 0 V C R (1) t Q(t) Q(t) (2) (3) (4) (3) B3 [4 ] R Q (1) O x x (2) x (3) Q M x
93 A O x
94 94 A A.3 1 [8 ] E d = 1 ρdv ε 0 V (A.1) a ρ (a) r 0 < r < a a < r (b) r V 0 < r < a a < r (c) (b) V V E V E = V (A.2) (d) E = ρ ε 0 (A.3) 2 [4 ] B B = 0 (A.4) B B 0 (A.5) (a) (A.4) (b) (A.4) (A.5) 3 [8 ] I dh = I d l r 4π r 3 (A.6) (a) (A.6) d l, r, d H (b) (b-1) a z H(z) (b-2) ( a n ) I H z z z + dz ndz z dh = I 2 a 2 ndz
95 A.3 95 (b-3) (b-2) H = ni z 4 [8 ] z a j z r 0 < r < a H = 1 2 j r r = (x, y, 0), r = x 2 + y 2 j = (0, 0, j) xy a/2 C (a) H H x H y H z j, x, y (b) H z (c) H d (d) H d r C C ( a r = 2 cos θ, a ) 2 sin θ, 0 θ 0 2π (c) (d) H d = H d r C 5 [6 ] a d V = V 0 sin (ωt)
96 96 A ε 0 (a) t I (b) (c) r 0 < r < a 6 [16 ] 1831 (a) Motional Induction Transformer Induction (b) Motional Induction Transformer Induction (c) Φ V V = dφ dt (A.7) E B E = B t (A.8) (d) B = B 0 sin (ωt) (B 0 ω ) l R t = 0 x a (d-1) t (d-2) v x t (d-3) (d-2)
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
i 0 1 0.1 I................................................ 1 0.2.................................................. 2 0.2.1...........................
2008 II 21 1 31 i 0 1 0.1 I................................................ 1 0.2.................................................. 2 0.2.1............................................. 2 0.2.2.............................................
168 13 Maxwell ( H ds = C S rot H = j + D j + D ) ds (13.5) (13.6) Maxwell Ampère-Maxwell (3) Gauss S B 0 B ds = 0 (13.7) S div B = 0 (13.8) (4) Farad
13 Maxwell Maxwell Ampère Maxwell 13.1 Maxwell Maxwell E D H B ε 0 µ 0 (1) Gauss D = ε 0 E (13.1) B = µ 0 H. (13.2) S D = εe S S D ds = ρ(r)dr (13.3) S V div D = ρ (13.4) ρ S V Coulomb (2) Ampère C H =
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.............................................
1 180m g 10m/s 2 2 6 1 3 v 0 (t=0) z max t max t z = z max 1 2 g(t t max) 2 (6) 1.3 2 3 3 r = (x, y, z) e x, e y, e z r = xe x + ye y + ze z. (7) v =
1. 2. 3 3. 4. 5. 6. 7. 8. 9. I http://risu.lowtem.hokudai.ac.jp/ hidekazu/class.html 1 1.1 1 a = g, (1) v = g t + v 0, (2) z = 1 2 g t2 + v 0 t + z 0. (3) 1.2 v-t. z-t. z 1 z 0 = dz = v, t1 dv v(t), v
24.15章.微分方程式
m d y dt = F m d y = mg dt V y = dy dt d y dt = d dy dt dt = dv y dt dv y dt = g dv y dt = g dt dt dv y = g dt V y ( t) = gt + C V y ( ) = V y ( ) = C = V y t ( ) = gt V y ( t) = dy dt = gt dy = g t dt
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
C:/KENAR/0p1.dvi
2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x
96 7 1m =2 10 7 N 1A 7.1 7.2 a C (1) I (2) A C I A A a A a A A a C C C 7.2: C A C A = = µ 0 2π (1) A C 7.2 AC C A 3 3 µ0 I 2 = 2πa. (2) A C C 7.2 A A
7 Lorentz 7.1 Ampère I 1 I 2 I 2 I 1 L I 1 I 2 21 12 L r 21 = 12 = µ 0 2π I 1 I 2 r L. (7.1) 7.1 µ 0 =4π 10 7 N A 2 (7.2) magnetic permiability I 1 I 2 I 1 I 2 12 21 12 21 7.1: 1m 95 96 7 1m =2 10 7 N
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
0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ). f ( ). x i : M R.,,
2012 10 13 1,,,.,,.,.,,. 2?.,,. 1,, 1. (θ, φ), θ, φ (0, π),, (0, 2π). 1 0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ).
5 36 5................................................... 36 5................................................... 36 5.3..............................
9 8 3............................................. 3.......................................... 4.3............................................ 4 5 3 6 3..................................................
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....................................
ii
ii iii 1 1 1.1..................................... 1 1.2................................... 3 1.3........................... 4 2 9 2.1.................................. 9 2.2...............................
1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0
A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1
7 9 7..................................... 9 7................................ 3 7.3...................................... 3 A A. ω ν = ω/π E = hω. E
B 8.9.4, : : MIT I,II A.P. E.F.,, 993 I,,, 999, 7 I,II, 95 A A........................... A........................... 3.3 A.............................. 4.4....................................... 5 6..............................
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 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
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
i
007 0 1 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................ 3 0.4............................................. 3 1
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.................................
E F = q b E (2) E q a r q a q b N/C q a (electric flux line) q a E r r r E 4πr 2 E 4πr 2 = k q a r 2 4πr2 = 4πkq a (3) 4πkq a 1835 4πk 1 ɛ 0 ɛ 0 (perm
1 1.1 18 (static electricity) 20 (electric charge) A,B q a, q b r F F = k q aq b r 2 (1) k q b F F q a r?? 18 (Coulomb) 1 N C r 1m 9 10 9 N 1C k 9 10 9 Nm 2 /C 2 1 k q a r 2 (Electric Field) 1 E F = q
(w) F (3) (4) (5)??? p8 p1w Aさんの 背 中 が 壁 を 押 す 力 垂 直 抗 力 重 力 静 止 摩 擦 力 p8 p
F 1-1................................... p38 p1w A A A 1-................................... p38 p1w 1-3................................... p38 p1w () (1) ()?? (w) F (3) (4) (5)??? -1...................................
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
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
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)
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h f(a + h, b) f(a, b) h........................................................... ( ) f(x, y) (a, b) x A (a, b) x (a, b)
(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)
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,
2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
12 2 E ds = 1 ρdv ε 1 µ D D S S D B d S = 36 E d B l = S d S B d l = S ε E + J d S 4 4 div E = 1 ε ρ div B = rot E = B 1 rot µ E B = ε + J 37 3.2 3.2.
213 12 1 21 5 524 3-5465-74 nkiyono@mail.ecc.u-tokyo.ac.jp http://lecture.ecc.u-tokyo.ac.jp/~nkiyono/index.html 3 2 1 3.1 ρp, t EP, t BP, t JP, t 35 P t xyz xyz t 4 ε µ D D S S 35 D H D = ε E B = µ H E
(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )
( ) 24 1 ( 26 8 19 ) i 0.1 1 (2012 05 30 ) 1 (), 2 () 1,,, III, C III, C, 1, 2,,, ( III, C ),, 1,,, http://ryuiki.agbi.tsukuba.ac.jp/lec/12-physics/ E104),,,,,, 75 3,,,, 0.2, 1,,,,,,,,,,, 2,,, 1000 ii,
() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
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
Part. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3. 39. 4.. 4.. 43. 46.. 46..
Cotets 6 6 : 6 6 6 6 6 6 7 7 7 Part. 8. 8.. 8.. 9..... 3. 3 3.. 3 3.. 7 3.3. 8 Part. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3.
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H
199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)
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
,. 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,
2 (f4eki) ρ H A a g. v ( ) 2. H(t) ( ) Chapter 5 (f5meanfp) ( ( )? N [] σ e = 8π ( ) e mc 2 = cm 2 e m c (, Thomson cross secion). Cha
![2 (f4eki) ρ H A a g. v ( ) 2. H(t) ( ) Chapter 5 (f5meanfp) ( ( )? N [] σ e = 8π ( ) e mc 2 = cm 2 e m c (, Thomson cross secion). Cha](/thumbs/43/23149233.jpg "2 (f4eki) ρ H A a g. v ( ) 2. H(t) ( ) Chapter 5 (f5meanfp) ( ( )? N [] σ e = 8π ( ) e mc 2 = cm 2 e m c (, Thomson cross secion). Cha")
http://astr-www.kj.yamagata-u.ac.jp/~shibata P a θ T P M Chapter 4 (f4a). 2.. 2. (f4cone) ( θ) () g M θ (f4b) T M L 2 (f4eki) ρ H A a g. v ( ) 2. H(t) ( ) Chapter 5 (f5meanfp) ( ( )? N [] σ e = 8π ( )
/Volumes/NO NAME/gakujututosho/chap1.tex i
2010 4 8 /Volumes/NO NAME/gakujututosho/chap1.tex i iii 1 5 1.1............................... 5 2 9 2.1........................................... 9 2.2................................... 16 2.3...................................
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
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
34 2 2 h = h/2π 3 V (x) E 4 2 1 ψ = sin kxk = 2π/λ λ = h/p p = h/λ = kh/2π = k h 5 2 ψ = e ax2 ガウス 型 関 数 1.2 1 関 数 値 0.8 0.6 0.4 0.2 0 15 10 5 0 5 10
33 2 2.1 2.1.1 x 1 T x T 0 F = ma T ψ) 1 x ψ(x) 2.1.2 1 1 h2 d 2 ψ(x) + V (x)ψ(x) = Eψ(x) (2.1) 2m dx 2 1 34 2 2 h = h/2π 3 V (x) E 4 2 1 ψ = sin kxk = 2π/λ λ = h/p p = h/λ = kh/2π = k h 5 2 ψ = e ax2
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,
Microsoft Word - 計算力学2007有限要素法.doc
95 2 x y yz = zx = yz = zx = { } T = { x y z xy } () {} T { } T = { x y z xy } = u u x y u z u x x y z y + u y (2) x u x u y x y x y z xy E( ) = ( + )( 2) 2 2( ) x y z xy (3) E x y z z = z = (3) z x y
(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
genron-3
" ( K p( pasals! ( kg / m 3 " ( K! v M V! M / V v V / M! 3 ( kg / m v ( v "! v p v # v v pd v ( J / kg p ( $ 3! % S $ ( pv" 3 ( ( 5 pv" pv R" p R!" R " ( K ( 6 ( 7 " pv pv % p % w ' p% S & $ p% v ( J /
x ( ) x dx = ax
x ( ) x dx = ax 1 dx = a x log x = at + c x(t) = e at C (C = e c ) a > 0 t a < 0 t 0 (at + b ) h dx = lim x(t + h) x(t) h 0 h x(t + h) x(t) h x(t) t x(t + h) x(t) ax(t) h x(t + h) x(t) + ahx(t) 0, h, 2h,
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..........
NumRu::GPhys::EP Flux 2 2 NumRu::GPhys::EP Flux 3 2.................................. 3 2.2 EP............................. 4 2.3.....................
NumRu::GPhys::EP Flux 7 2 9 NumRu::GPhys::EP Flux 2 2 NumRu::GPhys::EP Flux 3 2.................................. 3 2.2 EP............................. 4 2.3................................. 5 2.4.............................
d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r
![d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r](/thumbs/85/92098187.jpg "d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r")
2.4 ( ) U(r) ( ) ( ) U F(r) = x, U y, U = U(r) (2.4.1) z 2 1 K = mv 2 /2 dk = d ( ) 1 2 mv2 = mv dv = v (ma) (2.4.2) ( ) U(r(t)) r(t) r(t) + dr(t) du du = U(r(t) + dr(t)) U(r(t)) = U x = U(r(t)) dr(t)
A
A05-132 2010 2 11 1 1 3 1.1.......................................... 3 1.2..................................... 3 1.3..................................... 3 2 4 2.1............................... 4 2.2
応用数学A
応用数学 A 米田 戸倉川月 7 限 1930~2100 西 5-109 V を :x 2 + y 2 + z 2 = 4 で囲まれる内部とする F = ye x xe y + ze z FdV = V e x e y e z F = = 2e z 2e z dv = 2e z 3 23 = 64π 3 e z y x z 4π V n Fd = 1 F nd 2 F nd 法線ベクトル n g x,
/Volumes/NO NAME/gakujututosho/chap1.tex i
2012 4 10 /Volumes/NO NAME/gakujututosho/chap1.tex i iii 1 7 1.1............................... 7 2 11 2.1........................................... 11 2.2................................... 18 2.3...................................
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 +
A B 5 C 9 3.4 7 mm, 89 mm 7/89 = 3.4. π 3 6 π 6 6 = 6 π > 6, π > 3 : π > 3
π 9 3 7 4. π 3................................................. 3.3........................ 3.4 π.................... 4.5..................... 4 7...................... 7..................... 9 3 3. p
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
常微分方程式の局所漸近解析 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/007651 このサンプルページの内容は, 初版 1 刷発行当時のものです. i Leibniz ydy = y 2 /2 1675 11 11 [6] 100 Bernoulli Riccati 19 Fuchs
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) = (, )
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
2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (
(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2
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)
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 ),
Untitled
23 1 11 A 2 A.1..................................... 2 A.2.................................. 4 A.3............................... 5 A.4.................................... 6 A.5.......................
) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
平成18年度弁理士試験本試験問題とその傾向
CBA CBA CBA CBA CBA CBA Vol. No. CBA CBA CBA CBA a b a bm m swkmsms kgm NmPa WWmK σ x σ y τ xy θ σ θ τ θ m b t p A-A' σ τ A-A' θ B-B' σ τ B-B' A-A' B-B' B-B' pσ σ B-B' pτ τ l x x I E Vol. No. w x xl/ 3
1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
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.....................................
0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9
1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),
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
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)
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h 0 f(a + h, b) f(a, b) h............................................................... ( ) f(x, y) (a, b) x A (a, b) x
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
３
- { } / f ( ) e nπ + f( ) = Cne n= nπ / Eucld r e (= N) j = j e e = δj, δj = 0 j r e ( =, < N) r r r { } ε ε = r r r = Ce = r r r e ε = = C = r C r e + CC e j e j e = = ε = r ( r e ) + r e C C 0 r e =
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)
P F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2
1 1 2 2 2 1 1 P F ext 1: F ext P F ext (Count Rumford, 1753 1814) 0 100 H 2 O H 2 O 2 F ext F ext N 2 O 2 2 P F S F = P S (1) ( 1 ) F ext x W ext W ext = F ext x (2) F ext P S W ext = P S x (3) S x V V
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......................
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 =
2 p T, Q
270 C, 6000 C, 2 p T, Q p: : p = N/ m 2 N/ m 2 Pa : pdv p S F Q 1 g 1 1 g 1 14.5 C 15.5 1 1 cal = 4.1855 J du = Q pdv U ( ) Q pdv 2 : z = f(x, y). z = f(x, y) (x 0, y 0 ) y y = y 0 z = f(x, y 0 ) x x =
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
1 1 1 1-1 1 1-9 1-3 1-1 13-17 -3 6-4 6 3 3-1 35 3-37 3-3 38 4 4-1 39 4- Fe C TEM 41 4-3 C TEM 44 4-4 Fe TEM 46 4-5 5 4-6 5 5 51 6 5 1 1-1 1991 1,1 multiwall nanotube 1993 singlewall nanotube ( 1,) sp 7.4eV
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
B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:
![B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:](/thumbs/94/118646867.jpg "B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:")
B. 41 II: ;; 4 B [] S 1 S S 1 S.1 O S 1 S 1.13 P 3 P 5 7 P.1:.13: 4 4.14 C d A B x l l d C B 1 l.14: AB A 1 B 0 AB 0 O OP = x P l AP BP AB AP BP 1 (.4)(.5) x l x sin = p l + x x l (.4)(.5) m d A x P O
1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2
1 1.1 R(x) = 0 y + P (x)y + Q(x)y = R(x)...(1) y + P (x)y + Q(x)y = 0...(2) 1 2 u(x) v(x) c 1 u(x)+ c 2 v(x) = 0 c 1 = c 2 = 0 c 1 = c 2 = 0 2 0 2 u(x) v(x) u(x) u (x) W (u, v)(x) = v(x) v (x) 0 1 1.2
1 1 1 1 1 1 2 f z 2 C 1, C 2 f 2 C 1, C 2 f(c 2 ) C 2 f(c 1 ) z C 1 f f(z) xy uv ( u v ) = ( a b c d ) ( x y ) + ( p q ) (p + b, q + d) 1 (p + a, q + c) 1 (p, q) 1 1 (b, d) (a, c) 2 3 2 3 a = d, c = b
30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
( )
18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)
338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
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,, 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
12 -- 1 4 2009 9 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8 4-9 4-10 c 2011 1/(13) 4--1 2009 9 3 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
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
2012 A, N, Z, Q, R, C
2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)
δ ij δ ij ˆx ˆx ŷ ŷ ẑ ẑ 0, ˆx ŷ ŷ ˆx ẑ, ŷ ẑ ẑ ŷ ẑ, ẑ ˆx ˆx ẑ ŷ, a b a x ˆx + a y ŷ + a z ẑ b x ˆx + b
23 2 2.1 n n r x, y, z ˆx ŷ ẑ 1 a a x ˆx + a y ŷ + a z ẑ 2.1.1 3 a iˆx i. 2.1.2 i1 i j k e x e y e z 3 a b a i b i i 1, 2, 3 x y z ˆx i ˆx j δ ij, 2.1.3 n a b a i b i a i b i a x b x + a y b y + a z b
振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
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..........................................
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................