C:/KENAR/0p1.dvi
|
|
|
- さあしゃ おおふさ
- 5 years ago
- Views:
Transcription
1 2{ {3 [ ] ,15 m 10,10 m 2 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
3 2{3. 55 (b) x 1 kgf 1 g = m=s 2 1kg 1 kgf 1 kgf = 1 kg 9:80665 m=s 2 =9:80665 N 2 3 N kgf 2 3 (i) 2.5 mg N N 1 kgw kg. 2 kg
4 56 2 (ii a ma = N, mg (iii) a a = N m, g (a) ; N =1:20 10 kgf = 1:20 10 kg g; m =1:00 10 kg N ( )+( ) 1:20 10 kg g a =, g 1:00 10 kg =0:20g =0:20 9:8 m=s 2 2:0m=s 2 (b) ; N =8:0 kgf=8:0 kg g; m =1:00 10 kg a = 8:0 kg g 1:00 10 kg, g =,0:20g,2:0m=s 2 a <0 a =0 N = mg a =,g N =0 0 kgf 1 a C M m N C N C g (1)(a) (b) (2)(a) (b) (3) N ; N C M;m;g;a (1)(a) N (b) mg (2)(a) N C (b) N ;Mg (3) N = m(g+a); N C =(m+m)(g+a)
5 2{3. 57 a b c C D 2.6: a (a) (b) b c C 2.7: a b C c D (a) 2 (b) (a) 2 (b)
6 N R 05 R< 0 N (2.29) 0 N N 2 2 R R = N (2.30) 2 6 I{2{1 3 < 0 (2.31) 4 5 F F Reibung 6
7 2{ m N = c R g (1) =0 mg sin (2) < c mg cos mg (3) 2.8: (4) > c 4 mg N R 2.8 x y x +mg sin y,mg cos (1),(2) 0 ( x :0=mg sin, R (2:32) y :0=N, mg cos (2:33) (2.32) (2) R = mg sin =0 (1) R =0 (3) 0 (2.33) N = mg cos (2.29) R 0 N; mg sin <mgcos! c mg sin c! mg cos c 0 =tan c (4) R = N = mg cos x ma = mg sin, mg cos
8 60 2 a = g(sin, cos ) tan >tan c (3) sin cos > 0 (2.31) sin > 0 cos >cos a>0 2 l = 176 cm h =93cm s = 168 cm t =2:0 s 7 2, : 2.9 1m N s 2 = tan, =0:521 gt 2 cos sin = h l
9 2{ m m F F F g t =0 1 C D D E C D E 2.11 m g F N F C N C m g E N C x C 2.10: T CD N F m g D N C C N DF T DE F E D E N E y N E m g T DE T CD 2.11:
10 62 2 N E C N C CD D C T CD E N E D E T DE 4 1 D T CD T DE N DF C D E x y O t =0 t x y y = x dy dx dt = dt d2 y dt 2 = d2 x dt 2 8 F v a C D E v a D 4 3 C D E F (x ) m a = N C (2.34) (y ) m 0=N F, m g (2.35) m a = m g, N E (2.36) C 0 a = T CD, N C (2.37) D(x ) 0 a cos 4 = N DF cos 4, T CD (2.38) D(y ) 0 a sin 4 = T DE, N DF sin 4 (2.39) E 0 a = N E, T DE (2.40) C E D 8
11 2{ (2.34) (2.40) (2.35) (m + m )a = m g (2.41) 1 a = m m + m g (2.42) (2.42) a (2.34)(2.40) N F = m g; N C = T CD = T DE = N E = p 1 N DF = 2 m m m + m 5 t =0 t t 0 t =0 x = y =0 v =0 t v = at; x = y = 1 2 at2 6 : v t x y I{1{2 a = dv (2.41) dt (m + m ) dv dt = m g v v = dy dt dy dt (m + m )v dv dt = m g dy dt
12 64 2 t v y t =0 t y =0 y (m + m ) Z v 0 vdv = m g 1 Z y 0 dy 2 (m + m )v 2, m gy =0 (2.43) y =0; v =0 t =0 0 U y =0 U =0 U =,m gy N C W W = N C x N F N F 0 N E W W =,N E y (2.37) (2.40) N E = N C W + W =0 C D E m m 2.12: g t =0 1 [ ] L = 1 (2.41) 2 (m +m )v 2 +m gy =0
13 2{3. 65 y v m m P P m C C g l 1 + l 2 C l 2 t =0 C m = M m =3M m C =4M C 2 t 1 C P 2.14 C P T 1 P P P 3 P l 1 + l 2 m g P l 2 P Q 2.13: T 1 T 2 T 1 T 1 P : Q T 2 T 1 P C Q Q O x Q 2l 1 PC 2l 2 2 T 1 m g C C T 2 m C g 2 3 a = m, m sin m + m g; 1 2 (m + m )v 2, (m, m sin )gy =0
14 66 2 (x, x P )+(x, x P )=2l 1 (2.44) x C + x P =2l 2 (2.45) C P 4 4,2 =2P x C O C x P P x r = x,x P x P x r (2.24) (2.25) x P x P x = x P + x r (2.46) x =2l 1 + x P, (x, x P )=2l 1 + x P, x r (2.47) x C =2l 2, x P (2.48) x x C P v v v C v P ; a a a C a P (2.25) t : x C dt + x P dt =0 v C + v P =0 C v C P v P =,v C (2.24) t x dt, x P x + dt dt, x P =0 dt P v r = v, v P P,v r = v, v P P v r,v r v = v P + v r (2.49) v = v P, v r (2.50) v C =,v P (2.51) (2.49) (2.50) (2.51) t a = a P + a r (2.52) a = a P, a r (2.53) a C =,a P (2.54) 3 P P 0
15 2{ a P =0 g +2T 1, T 2 P C T 2 =2T 1 C m a = m g, T 1 (2.55) m a = m g, T 1 (2.56) C m C a C = m C g, 2T 1 (2.57) 4 (2.55) (2.56) (2.57) (2.52) (2.53) (2.54) m (a p + a r )=m g, T 1 (2.58) m (a p, a r )=m g, T 1 (2.59) C m C (,a P )=m C g, 2T 1 (2.60) T 1 (2.58),(2.59) (2.58)+(2:59), (2:60) (m, m )a P +(m + m )a r =(m, m )g (2.61) (m + m + m C )a P +(m, m )a r =(m + m, m C )g (2.62) 3 m = M m =3M m C =4M (2.61),(2.62) (,2Ma P +4Ma r =,2Mg 8Ma P, 2Ma r =0 a P =, 1 7 g; a r =, 4 7 g (2.52) (2.53) (2.54) a =, 5 7 g; a = 3 7 g; a C = 1 7 g 3 1 [ ] L = 2 m (v p + v r ) 2 + m (v p, v r ) m Cv 2 P + f(m + m, m C )x P +(m, m )x r =0 P r (2.61),(2.62)
16 68 2 C C (2.55) T 1 = 1 2 T 2 = m (g, a )= 12 7 Mg 5 t =0 x = x = l 1 + l 2 x C = l 2 v = v = v C =0 C v =, 5 7 gt; v = 3 7 gt; v C = 1 7 gt C t x = l 1 + l 2, 5 14 gt2 ; x = l 1 + l gt2 ; x C = l gt2 6 :(2.61) (2.62) a P = dv P dt a r = dv r dt v r v P t (v r =) dx r dt (v P =) dx P (2.61) 2 (2.62) 2 dt Z vr (m + m )v P 1dv r =(m + m )v P v r 0 Z vp (m + m )v r 1dv P =(m + m )v P v r 0 v P v r (2.61)+(2.62) Z vp Z vr v r (m + m + m C ) v P dv P +2(m + m )v P v r +(m, m ) 0 0 dv r Z xp Z xr =(m + m, m C )g 1dx P +(m, m )g l 2 0 x P = l 2 x r = x, x P =(l 1 + l 2 ), l (m + m + m C )v 2 P +2(m + m )v P v r (m, m )v 2 r l 1 1dx r
17 2{3. 69 =(m + m, m C )g x P, l 2 )+(m, m )g(x r, l 1 ) 1 2 m (v P + v r ) m (v P, v r ) m Cv 2 P 1 2 m (v P + v r ) m (v P, v r ) m Cv 2 P, m g(x P + x r ), m g(2l 1 + x P, x r ), m C (2l 2, x P ) =,m g(l 1 + l 2 ), m g(l 1 + l 2 ), m C l 2 (2.63) (2.46)(2.51) 1 2 m v m v m Cv 2,m C gx,m gx,m C gx C =,m g(l 1 + l 2 ), m g(l 1 + l 2 ), m C gl 2 C (2.23) x = m P m P Q g 4 P Q 2.16: N N R R C m m : 2.17 N 0 g 4 4 a = 1 2 a = 2m, m g; T =2T = 6m m g m +4m m +4m N C
18 N R N N m g N R C 2.18: C N C R N N m g 2 x y x v v ; a a R a R v = v ; a = a b R v 6= v ; a 6= a 3 a R a = a = a N C (x ) m a = N, R (2.63) (y ) 0=N, m g (2.64) (x ) m a = R (2.65) (y ) 0=N C, N, m g (2.66) b R (2.30) R = N y x (x ) m a = N N (2.67) (x ) m a = N (2.68) 4 a R (2.63)+(2.65) (m + m )a = N (2.69)
19 2{3. 71 N a = N m + m (2.70) N (2.64) (2.66) N = m g (2.71) N C = N + m g =(m + m )g (2.72) (2.65) (2.71) R = m a = m m + m N (2.73) (2.29) R < 0 N (2.70) R < 0 m g (2.74) (2.71) (2.73) m m + m N< 0 m g N< m (m + m ) 0 g m N b R m N < (m + m ) = 0 g m (2.71) R = m g (2.75) (2.67) (2.68) a = N m, g; a = m m g
20 O t =0 x =0 x =0 v = v =0 a b 6 a R (2.69) v dx dt x a = dv dt (m + m )v dv dt = N dx dt t v x 0 v 0 x (m + m ) Z v 0 vdv = N N =const. 1 Z x 0 1dx 2 (m + m )v 2, 0=Nx (2.76) N 0 ) b R (2.67) v dx dt (2.68) v dx dt m v dv dt + m v dv dt =(N, N ) dx dt + N dx dt (2.71) N = m g 1 2 m v m v 2, 0=Nx, m g(x, x ) (2.77) x >x 2 N 100% 2 5 C m m N C 2.19:
21 2{ N 0 N N l g f l 1 x 1 l 2 x f > N = 0 (m + m )g; m gl x 1 > 0 x 1 +x 2 > 0 f>0 l 1 l 2 f<0 x 1 < 0 x 1 +x 2 < :
22 74 2 f x 2 l 2 K f x 1 l 1 f = K x 1 l 1 = K x 2 l 2 n l 1 l 2 l n x 1 x 2 x n l = l 1 + l l n f K x 2 l 2 x =x 1 +x 2 + +x n K x 1 l 1 f K x 1 = K x 2 = = K x n l 1 l 2 l n 2.21: = K x 1 +x 2 + +x n l 1 + l l n K = k l x x >0 x <0 = K x l,t x T x,! T T x,kx (2.78),,! T,T y (,kx < 0 ;,kx > 0 ) 2.22: 6 k 6
23 2{ T m x O x 2.23: k 1 O x x>0 x <0 x,kx 2 a x ma =,kx (2.79) a = d2 x dt 2 m d2 x dt 2 =,kx II 3 t =0 t =0 x v =0
24 (2.79) v dx dt a = dv dt mv dv dt =,kxdx dt t v x 0 v x m Z v 0 vdv =,k 1 Z x xdx 2 mv2 =, 1 2 kx k2 1 2 mv2 1 2 kx2 = 1 2 k2 (2.80) U (x) = 1 2 kx2 (2.81) x =0( ) ( 2.4 (2.80) v x v xy! x 2 a 2 + y2 b 2 =1, O x x = a cos ; y = b sin,! k m =!2 (2.80) x 2 + v2 2 (!) = : x, v x v 2.24!
25 2{3. 77 t t =0 x = cos 0 = ; v =! sin0=0 x v (i) (ii) (iii) j,j (iv) (v) 7 =!t x = cos!t (2.82) T x = cos 2 = cos!t T = 2! =2 r m k (2.83), p v =! 2, x 2 =! sin!t!t < v<0, v =,!sin!t (2.84) 8 (2.80) a = k m x =,!2 x =,! 2 cos!t a, t v, t x, t k m g (1) (2) O 1 2 kx2 (3) v = dx dt = d cos!t =,! sin!t dt 9 6 (1)2r m mg mg,(2) (3) k k k 2.25:
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
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
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
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)
t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
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
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3) 6 y = g(x) x = 1 g( 1) = 2 ( 1) 3 = 2 ; g 0 ( 1) =
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
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 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
sec13.dvi
13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:
() 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 ()
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
1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O
![1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O](/thumbs/93/114442636.jpg "1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O")
: 2014 4 10 1 2 2 3 2.1...................................... 3 2.2....................................... 4 2.3....................................... 4 2.4................................ 5 2.5 Free-Body
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
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0
![1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0](/thumbs/93/114340663.jpg "1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0")
: 2016 4 1 1 2 1.1......................................... 2 1.2................................... 2 2 2 2.1........................................ 2 2.2......................................... 3 2.3.........................................
TCSE4~5
II. T = 1 m!! U = mg!(1 cos!) E = T + U! E U = T E U! m U,E mg! U = mg!(1! cos)! < E < mg! mg! < E! L = T!U = 1 m!! mg!(1! cos) d L! L = L = L m!, =!mg!sin m! + mg!sin = d =! g! sin & g! d =! sin ! = v
Note.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
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
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
f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a
3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (
Acrobat Distiller, Job 128
(2 ) 2 < > ( ) f x (x, y) 2x 3+y f y (x, y) x 2y +2 f(3, 2) f x (3, 2) 5 f y (3, 2) L y 2 z 5x 5 ` x 3 z y 2 2 2 < > (2 ) f(, 2) 7 f x (x, y) 2x y f x (, 2),f y (x, y) x +4y,f y (, 2) 7 z (x ) + 7(y 2)
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,
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
m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2
3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1
http://know-star.com/ 3 1 7 1.1................................. 7 1.2................................ 8 1.3 x n.................................. 8 1.4 e x.................................. 10 1.5 sin
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
![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](/thumbs/95/125540821.jpg "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")
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
A大扉・騒音振動.qxd
H21-30 H21-31 H21-32 H21-33 H21-34 H21-35 H21-36 H21-37 H21-38 H21-39 H21-40 H21-41 H21-42 n n S L N S L N L N S S S L L log I II I L I L log I I H21-43 L log L log I I I log log I I I log log I I I I
DE-resume
- 2011, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 21131 : 4 1 x y(x, y (x,y (x,,y (n, (1.1 F (x, y, y,y,,y (n =0. (1.1 n. (1.1 y(x. y(x (1.1. 1 1 1 1.1... 2 1.2... 9 1.3 1... 26 2 2 34 2.1,... 35 2.2
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 ),
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..........................................
2014 S hara/lectures/lectures-j.html r 1 S phone: ,
14 S1-1+13 http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html r 1 S1-1+13 14.4.11. 19 phone: 9-8-4441, e-mail: hara@math.kyushu-u.ac.jp Office hours: 1 4/11 web download. I. 1. ϵ-δ 1. 3.1, 3..
1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx
1 1 1 1 1. U(x, t) U(x, t) + c t x c, κ. (1). κ U(x, t) x. (1) 1, f(x).. U(x, t) U(x, t) + c κ U(x, t), t x x : U(, t) U(1, t) ( x 1), () : U(x, ) f(x). (3) U(x, t). [ U(x, t) Re u k (t) exp(πkx). (4)
S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
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)
, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
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
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)
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
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....
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
r III... IV.. grad, div, rot. grad, div, rot 3., B grad, div, rot I, II ɛ-δ web page (
r 8.4.8. 3-3 phone: 9-76-4774, e-mail: hara@math.kyushu-u.ac.jp http://www.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html Office hours: 4/8 I.. ɛ-n. ɛ-δ 3. 4. II... 3. 4. 5.. r III... IV.. grad, div,
44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
i ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
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
Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
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
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 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
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,
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
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....................................
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
m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
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(,,
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
2 Chapter 4 (f4a). 2. (f4cone) ( θ) () g M. 2. (f4b) T M L P a θ (f4eki) ρ H A a g. v ( ) 2. H(t) ( )
http://astr-www.kj.yamagata-u.ac.jp/~shibata f4a f4b 2 f4cone f4eki f4end 4 f5meanfp f6coin () f6a f7a f7b f7d f8a f8b f9a f9b f9c f9kep f0a f0bt version feqmo fvec4 fvec fvec6 fvec2 fvec3 f3a (-D) f3b
Korteweg-de Vries
Korteweg-de Vries 2011 03 29 ,.,.,.,, Korteweg-de Vries,. 1 1 3 1.1 K-dV........................ 3 1.2.............................. 4 2 K-dV 5 2.1............................. 5 2.2..............................
[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29
![[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29](/thumbs/93/114156928.jpg "[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29")
A p./29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x) + C f(x) A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x)
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
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
(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
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
6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1
6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) (e) Γ (6.2) : Γ B A R (reversible) 6-1 (e) = Clausius 0 = B A: Γ B A: Γ d Q A + d Q (e) B: R d Q + S(A) S(B) (6.3) (e) // 6.2 B A: Γ d Q S(B) S(A) = S (6.4) (e) Γ (6.5)
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
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
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
<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
(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
() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
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
( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
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)
5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (
![5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (](/thumbs/92/107866626.jpg "5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (")
5 5.1 [ ] ) d f(t) + a d f(t) + bf(t) : f(t) 1 dt dt ) u(x, t) c u(x, t) : u(x, t) t x : ( ) ) 1 : y + ay, : y + ay + by : ( ) 1 ) : y + ay, : yy + ay 3 ( ): ( ) ) : y + ay, : y + ay b [],,, [ ] au xx
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
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..........................
mugensho.dvi
1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim
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
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.
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. ttp://www.morikita.co.jp/books/mid/00571 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. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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,
12-7 12-7 12-7 12-7 12-8 12-10 12-10 12-10 12-11 12-12 12-12 12-14 12-15 12-17 12-18 10 12-19 12-20 12-20 12-21 12-22 12-22 12-23 12-25 12-26 12-26 12-29 12-30 12-30 12-31 12-33 12-34 12-3 12-35 12-36
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
untitled
1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β
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
(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)
, 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,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 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,
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)
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
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)
( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED)
rational number p, p, (q ) q ratio 3.14 = 3 + 1 10 + 4 100 ( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED) ( a) ( b) a > b > 0 a < nb n A A B B A A, B B A =
( )
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...............................
[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt
![[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt](/thumbs/101/152334931.jpg "[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt")
3.4.7 [.] =e j(t+/4), =5e j(t+/3), 3 =3e j(t+/6) ~ = ~ + ~ + ~ 3 = e j(t+φ) =(e 4 j +5e 3 j +3e 6 j )e jt = e jφ e jt cos φ =cos 4 +5cos 3 +3cos 6 =.69 sin φ =sin 4 +5sin 3 +3sin 6 =.9 =.69 +.9 =7.74 [.]
) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
![) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4](/thumbs/92/107866707.jpg ") ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4")
1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
A 99% MS-Free Presentation
A 99% MS-Free Presentation 2 Galactic Dynamics (Binney & Tremaine 1987, 2008) Dynamics of Galaxies (Bertin 2000) Dynamical Evolution of Globular Clusters (Spitzer 1987) The Gravitational Million-Body Problem
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
電気電子数学入門 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/073471 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 14 (tool) [ ] IT ( ) PC (EXCEL) HP() 1 1 4 15 3 010 9 ii 1... 1 1.1 1 1.
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..........