2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

Similar documents
高等学校学習指導要領

高等学校学習指導要領

() 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)

Gmech08.dvi

all.dvi

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

sec13.dvi

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

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

Chap10.dvi

dvipsj.8449.dvi

重力方向に基づくコントローラの向き決定方法

表1-表4_No78_念校.indd

Gmech08.dvi

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

Gmech08.dvi

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

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

c y /2 ddy = = 2π sin θ /2 dθd /2 [ ] 2π cos θ d = log 2 + a 2 d = log 2 + a 2 = log 2 + a a 2 d d + 2 = l

微分積分 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. このサンプルページの内容は, 初版 1 刷発行時のものです.

untitled

TOP URL 1

数学演習:微分方程式

Chap9.dvi

untitled

LLG-R8.Nisus.pdf

untitled

no35.dvi

Note.tex 2008/09/19( )

3 filename=quantum-3dim110705a.tex ,2 [1],[2],[3] [3] U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x

物性基礎

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)) ( )

1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ

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)

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e

φ s i = m j=1 f x j ξ j s i (1)? φ i = φ s i f j = f x j x ji = ξ j s i (1) φ 1 φ 2. φ n = m j=1 f jx j1 m j=1 f jx j2. m

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

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

2 1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a 2 + b 2 α (norm) N(α) = a 2 + b 2 = αα = α 2 α (spure) (trace) 1 1. a R aα =

大野川水系中流圏域

untitled

untitled


all.dvi

1 yousuke.itoh/lecture-notes.html [0, π) f(x) = x π 2. [0, π) f(x) = x 2π 3. [0, π) f(x) = x 2π 1.2. Euler α

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a

( ) ( )

Onsager SOLUTION OF THE EIGENWERT PROBLEM (O-29) V = e H A e H B λ max Z 2 Onsager (O-77) (O-82) (O-83) Kramers-Wannier 1 1 Ons

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


ẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k.

main.dvi

基礎数学I

TOP URL 1

m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d

I

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx

1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2

°ÌÁê¿ô³ØII

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

振動と波動

: , 2.0, 3.0, 2.0, (%) ( 2.

数学の基礎訓練I

TOP URL 1

III,..

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co

KENZOU Karman) x

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

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

クレイによる、主婦湿疹のケア

(3)(4) (3)(4)(2) (1) (2) 20 (3)

untitled

群馬県野球連盟


<82B582DC82CB8E7188E782C48A47967B41342E696E6464>

untitled



<91E F1938C966B95FA8ECB90FC88E397C38B5A8F708A778F7091E589EF8EC08D7388CF88F5837D836A B E696E6464>

Microsoft Word - 入居のしおり.doc

untitled

ESPEC Technical Report 12

syogaku

-26-

untitled

Transcription:

1 16 10 5 1 2 2.1 a a a 1 1 1

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2

5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a h a x y x y 5 3

(3) (4) x y (x + e2 tan φ ) 2 y 2 1 e 2 l + 1 e 2 = l 2 ( ) (1 e 2 ) 2 e 2 sec 2 φ 1 6 φ e φ δ 7 e cos φ (6) sin δ e (5) (5) (1 e 2 )x 2 + 2l e 2 tan φ x + y 2 + l 2 (1 e 2 tan 2 φ) = 0 (7) (5) (7) e 1) e = 0,, x 2 + y 2 = l 2 ( ) 2) 0 < e < 1,, x + e2 tan φ 2 ) l 1 e + y 2 = (e l2 2 1 e 2 (1 e 2 ) 2 sec 2 φ 1 2 3) e = 1,, 2l tan φ x + y 2 + l 2 (1 tan 2 φ) = 0 ( ) 4) 1 < e <,, x e2 tan φ 2 l y 2 e 2 1 5) e =,, x = l tan φ e 2 1 = (e l2 (e 2 1) 2 sec 2 φ 1 2 (6) φ e φ δ e ) (5) 2.4.1 e = 0 (φ = 90, δ > 0 ) (φ = 90, δ < 0 ) e = 0 l 2.4.2 e = (δ = 0 ( e = ) x = l tan φ 6 Appendix 7 4

2.4.3 66.5 φ 66.5 23.5 δ +23.5 e e φ δ 2.4.4 ( φ > 66.5 ) 66.5 66.5 δ 23.5 δ +23.5 8 3 2004 8 12 9 3.1 12 2.0 cm 5.0 cm h 12 (1) cot h 12 = L l = 2.0 = 0.4 (8) 5.0 h 12 = 68.2 3.2 10 8 φ δ φ + δ = 90 180 ( e = 1) e = 1 e = 1 66.5 9 9 10, 10 11, 11 12, 12 13, (13 14, (14 16 10 5

1: 135 3.3 2.4 x y (x i, y i ) i 9 i = 1 16 i = 27 11 11 12 2cm 6

(l = 5cm) i x i (cm) y i (cm) X i (= x i /l) Y i (= y i /l) 1 1.47 4.71 0.294 0.941 2 1.61 4.18 0.321 0.835 3 1.71 3.65 0.341 0.729 4 1.79 3.26 0.359 0.653 5 1.86 2.91 0.372 0.582 6 1.88 2.53 0.376 0.506 7 1.91 2.06 0.382 0.412 8 1.93 1.71 0.386 0.341 9 2.00 1.35 0.400 0.271 10 1.96 0.91 0.392 0.182 11 2.00 0.59 0.400 0.118 12 2.01 0.29 0.402 0.059 13 2.00 0.00 0.400 0.000 14 2.06-0.74 0.412-0.147 15 2.03-1.31 0.406-0.262 16 2.06-1.56 0.412-0.318 17 2.06-2.12 0.412-0.424 18 1.91-2.44 0.382-0.488 19 1.94-2.82 0.388-0.565 20 1.88-3.32 0.376-0.665 21 1.76-3.82 0.352-0.765 22 1.71-4.32 0.341-0.865 23 1.47-4.88 0.294-0.976 24 1.41-5.47 0.282-1.094 25 1.29-5.94 0.259-1.188 26 1.06-6.56 0.212-1.312 27 0.88-7.44 0.177-1.488 3.4 (7) (7) X Y X x l, Y y l (9) (7) X, Y (1 e 2 )X 2 + 2e 2 tan φ X + Y 2 + (1 e 2 tan 2 φ) = 0 (10) l (10) 7

(10) φ φ δ e tan φ = 0.734 e 2 2 12 X i, Y i (i = 1, 2,, 27) e 2 e 2 = 27 i=1 {(X2 i 2X i tan φ + tan 2 φ)(xi 2 + Y i 2 + 1)} 27 i=1 (X2 i 2X i tan φ + tan 2 = 10.89 (11) φ) 2 e e( ) = 3.30 (12) 13 8 12 e = 3.30 (6) (12) δ( ) δ( ) = +14 09 (13) (2004 δ( ) δ( ) = +14 56 (14) 4 2004 8 12 14 12 2 Appendix 13 8 12 δ > 0) 14 8

2: 4.1 l 2004 8 12 12 68.2 8 12 12 L = 3.8cm 15 4.2 x y 12 138.3 15 9

4.3 x y (x i, y i ), i = 1, 2,, 27, l (X i, Y i ) 4.4 (11) (X i, Y i ), i = 1, 2,, 27, e 8 12 δ (δ = +14 56 ) 4.5 4.6 4.7 5 (5) : a : h : δ φ 10

l (x, y) x = l cot z cos a y = l cot z sin a (15) x y z a (δ, H) H δ (h, a) (δ, H) cos h sin a = cos δ sin H, cos h cos a = sin δ cos φ + cos δ sin φ cos H, (16) sin h = sin δ sin φ + cos δ cos φ cos H cos δ sin H cot h sin a = sin δ sin φ + cos δ cos φ cos H, cot h cos a = sin δ cos φ + cos δ sin φ cos H sin δ sin φ + cos δ cos φ cos H (17) X x sin δ cos φ + cos δ sin φ cos H = l sin δ sin φ + cos δ cos φ cos H, (18) Y y l = cos δ sin H sin δ sin φ + cos δ cos φ cos H (19) X Y H (sin φ X cos φ) cos δ sin H = Y sin δ (sin φ X cos φ) cos δ cos H = X sin δ sin φ + sin δ cos φ (20) sin H cos H X 2 (sin 2 δ cos 2 φ) + 2X sin φ cos φ + Y 2 sin 2 δ + sin 2 δ cos 2 φ cos 2 δ sin 2 φ = 0 (21) sin 2 δ cos 2 φ sin 2 δ = cos 2 φ 11

5.1 sin 2 δ cos 2 φ ( X + sin φ cos φ ) 2 sin 2 δ + (sin 2 δ cos 2 φ) (sin 2 δ cos 2 φ) Y 2 = sin 2 δ cos 2 δ (sin 2 δ cos 2 φ) 2 (22) 5.1.1 sin δ = 0 X = tan φ = const. (23) (19) Y = tan H cos φ (24) Y X (18) (X = tan φ) sin δ = 0 δ = 0 λ sin δ = sin λ sin ɛ (25) ɛ sin δ = 0 sin λ = 0 λ = 0, π 5.1.2 sin δ 0 (X + e2 tan φ) 2 Y 2 + 1 e 2 (1 e 2 ) = cot2 δ (1 e 2 ) 2 = 1 ( ) (1 e 2 ) 2 e 2 sec 2 φ 1 (26) e e cos φ sin δ e 2 (27) 5.2 sin 2 δ = cos 2 φ Y 2 + 2X tan φ + (1 tan 2 φ) = 0 (28) 12

6 (10) (X i, Y i ) (1 e 2 )X 2 i + 2e 2 X i tan φ + Y 2 i + (1 e 2 tan 2 φ) = 0 (29) (X i, Y i ) (29) ξ i (1 e 2 )X 2 i + 2e 2 X i tan φ + Y 2 i + (1 e 2 tan 2 φ) = X 2 i + Y 2 i + 1 e 2 (X 2 i 2X i tan φ + tan 2 φ), i = 1, 2,, 27 (30) ξ i (30) φ ξ i e 2 S(e 2 ) S(e 2 27 ) ξi 2 (31) i=1 e 2 e 2 (30) ξ i S(e 2 ) = [ 27 i=1 [ 27 2 (X 2 i 2X i tan φ + tan 2 φ) 2] (e 2 ) 2 ] (Xi 2 2X i tan φ + tan 2 φ)(xi 2 + Y 2 + 1) e 2 27 + (Xi 2 + Yi 2 + 1) (32) 2 i=1 i=1 e 2 (e 2 ) 2 e 2 e 2 = 27 i=1 (X2 i 2X i tan φ + tan 2 φ)(xi 2 + Y 2 + 1) 27 i=1 (X2 i 2X i tan φ + tan 2 (33) φ) 2 S(e 2 ) e 2 (33) 13