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- こおが そめや
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2 r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2
3 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) dt 2 = (ẋ(t), ẏ(t), ż(t)) ) = (ẍ(t), ÿ(t), z(t)) d dt r(t) 2 =2r(t) v(t) r(t) 2 = x(t) 2 + y(t) 2 + z(t) 2 d dt r(t) 2 = 2(x(t)ẋ(t)+y(t)ẏ(t)+z(t)ż(t)) =2r(t) v(t)
4 t = 0 (1, 0, 0) t = 1 ( 1, 1, 1) t = 1 ( 1, 1, 1) r(t) = (1 t)(1, 0, 0) + t( 1, 1, 1) = (1 2t, t, t) t = 0 (1, 0, 0) r(t) O v(t) =( 2, 1, 1) a(t) = (0, 0, 0)
5 t =0 (1, 0) ω v(t) a(t) O r(t) r(t) = (cos ωt, sin ωt) v(t) =( ω sin ωt, ω cos ωt) = ωr π/2 r(t) R θ = [ cos θ sin θ sin θ cos θ a(t) =( ω 2 cos ωt, ω 2 sin ωt) = ω 2 r(t) v(t) = ω a(t) = ω 2 ] θ
6 ma = f Isaac NEWTON ( ) m dv dt = f m d2 r dt 2 = f
7 y v 0 mg 0 x v 0 =(u 0,v 0 ) m d2 r dt 2 = f f = (0, mg) r(0) = 0 g dr dt (0) = v 0 { mẍ = 0 mÿ = mg { ẋ = u0 ẏ = v 0 gt x(0) = 0, ẋ(0) = u 0 y(0) = 0, ẏ(0) = v 0 x(t) = u 0 t y(t) = v 0 t g 2 t 2
8 x kv mg t = 0 v g v(t)
9 m dv dt = mg kv dy dx P(x) y = 0 log y = dy y = P(x) y = ce P(x)dx P(x)dx () dy dx αy = 0 y = ce αdx + c = ce αx
10 dy dx m dv dt P(x) y = Q(x) y = c(x)e = mg kv ( ) P(x )dx if Q(x) = 0 y = ce P(x)dx c(x) = e P(x )dx Q(x)dx + c y = e P(x )dx ( P(x )dx e Q(x)dx + c) dy dx αy = β y = β α + ce αx c:
11 x kv mg v = mg k t = 0 v (1 e k m t) m dv dt mg k v g = mg kv v = gt v(t) v(0) = 0 v = mg k t
12 v ( ) v --- dv/dt
13 du dt = ru(1 u/k) du/dt 0 < u < K u u > K u 0 K u dv dt = g k m v v < mg/k v > mg/k v v dv du/dt 0 mg/k uv
14 v v v(0) > mg/k mg k mg k v(0) = 0 t t dv dt = g k m v v < mg/k v > mg/k v v dv du/dt 0 mg/k uv
15 0 x(t) x x(t)
16 kx m m k x(t) d 2 x dt 2 + ω2 0x = 0 m d2 x dt 2 = kx ω 0 = k/m
17 d 2 x dt 2 + ω2 0x = 0 (1) x = cos ω 0 t x = sin ω 0 t A cos ω 0 t + B sin ω 0 t = 0 A = B = 0 x 2 (t) x 1 (t) x(t) = c 1 x 1 (t) + c 2 x 2 (t)
18 K k=0 a k (t) dk u dt k = f(t) f(t) =0 d 2 x dt 2 + ω2 0x = 0
19 d 2 x dt 2 + ω2 0x = 0 (1) A cos ω 0 t + B sin ω 0 t = 0 A = B = 0 t = 0 A = 0 t = π/(2ω 0 ) B = 0
20 d 2 x dt 2 + ω2 0x = 0 (1) x 2 (t) x 1 (t) x(t) = c 1 x 1 (t) + c 2 x 2 (t) d 2 x dt 2 + ω2 0x = d2 dt 2 (c 1x 1 + c 2 x 2 ) + ω0(c 2 1 x 1 + c 2 x 2 ) = c 1 ( d 2 x 1 dt 2 + ω2 0x 1 ) + c 2 ( d 2 x 2 dt 2 + ω2 0x 2 ) = 0
21 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t dx dt (0) = 0 x(0) = 1, dx dt (0) = 1 = C sin (ω 0 t + φ) x(0) = 0, dx x(0) = a, dt (0) = b
22 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t dx dt (0) = 0 = C sin (ω 0 t + φ) x(0) = 1, x = cos ω 0 t
23 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ) dx dt (0) = 1 x(0) = 0, x(t) = 1 ω 0 sin ω 0 t
24 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ) dx x(0) = a, dt (0) = b x(t) =a cos ω 0 t + b ω 0 sin ω 0 t
25 K k=0 a k (t) dk u dt k = f(t) f(t) =0
26 n n n n u 1,u 2,,,,u n α 1 u 1 + α 2 u 2 +,,,,+α n u n
27
28 K a k (t) d k k = 0 dt k L L(u) = f (t) L : L(u + v) = L(u) + L(v) L(αu) = αl(u) α :
29 m c dx dt x(t) m d2 x dt 2 = kx cdx dt
30 γ = c 2m, ω 0 = k m d 2 x dx + 2γ dt2 dt + ω2 0x = 0 x(t) = e λt x(t) = e λt λ λ 2 + 2γλ + ω 2 0 = 0
31 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 = λ 2 + 2γλ + ω 2 0 = 0 γ 2 ω 2 0 γ = c/(2m) γ = 0 0 < γ < ω 0 γ = ω 0 γ > ω 0 x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ)
32 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ > ω 0 λ 2 + 2γλ + ω 2 0 = 0 λ = γ ± γ 2 ω 2 0 x 1 = e ( γ+ γ 2 ω0 2)t x 2 = e ( γ γ 2 ω0 2)t x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t
33 γ > ω 0 x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t v γ = 5.0 ω 0 = 1.0 x x t
34 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 0 < γ < ω 0 λ 2 + 2γλ + ω 2 0 = 0 λ = γ ± iω ω = ω 2 0 γ2 x 1 = e ( γ+iω)t = e γt e iωt = e γt (cos ωt + i sin ωt) x 2 = e ( γ iω)t = e γt e iωt = e γt (cos ωt i sin ωt) x 1, x 2 (x 1 + x 2 )/2 = e γt cos ωt (x 1 x 2 )/(2i) = e γt sin ωt x = e γt (A cos ωt + B sin ωt)
35 0 < γ < ω 0 x = e γt (A cos ωt + B sin ωt) v γ = 0.05 ω 0 = 1.0 x x t
36 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ = ω 0 λ 2 + 2γλ + γ 2 = 0 λ = γ x 1 = e γt x 2 = te γt x = e γt (A + Bt)
37 γ = ω 0 x = e γt (A + Bt) v γ = 1.0 ω 0 = 1.0 x x t
38 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ = 0 0 < γ < ω 0 γ = ω 0 x = A cos ω 0 t + B sin ω 0 t x = e γt (A cos ωt + B sin ωt) ω = ω 2 0 γ2 x = e γt (A + Bt) γ > ω 0 x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t
39 γ = 0 v γ = 0.0 ω 0 = 1.0 x x t 0 < γ < ω 0 v γ = 0.05 ω 0 = 1.0 x x t
40 γ = ω 0 v γ = 1.0 ω 0 = 1.0 x x t γ > ω 0 v γ = 5.0 ω 0 = 1.0 x x t
41 0 γ < ω 0 γ > ω 0 γ = ω 0 x = e γt (A cos x = e γt (A + Bt) γ γ = ω 0 ) ω0 2 γ2 t + B sin ω0 2 γ2 t x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t λ ω 0 0 ω 0 γ λ 2 + 2γλ + ω0 2 = 0 iω 0 iω 0
42 d 2 x dt 2 + 3dx dt + 2x = 0 d 2 x dt 2 + 2dx dt + 2x = 0 d 2 x dt 2 + 2dx dt + x = 0 d 2 x dt 2 + x = 0 d 2 x dt 2 + dx dt + x = 0 d 2 x dt 2 + 3dx dt + x = 0
43 x(0) = 1, ẋ(0) = 3 x(0) = 1, ẋ(0) = 0 x(0) = 1, ẋ(0) = 0 x(0) = 3, ẋ(0) = 1 x(0) = 2, ẋ(0) = 1 x(0) = 2, ẋ(0) = 3
44
45 ma cos ωt (A > 0, ω > 0) m d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt 0 < γ < ω 0 ω
46 z(t) = x(t) + iy(t) d 2 z dz + 2γ dt2 dt + ω2 0z = Ae iωt z(t) x(t) d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt y(t) d 2 y dy +2γ dt2 dt + ω2 0y = A sin ωt
47 =) d 2 z dz + 2γ dt2 dt + ω2 0z = Ae iωt z s (t) =R(ω)Ae iωt R(ω) {(ω 2 0 ω 2 )+2γωi}R(ω)Ae iωt = Ae iωt R(ω) = 1 (ω 2 0 ω2 )+2γωi
48 2 au 1 + bu 2 2 2
49 K a k (t) d k k = 0 dt k L(u) = f (t) L S 1 S 2 S 1 S 2 proof L(S 1 ) = f (t) L(S 2 ) = f (t) L(S 1 S 2 ) = L(S 1 ) L(S 2 ) = f (t) f (t) = 0
50 2 S 1 (= ) S 2 S 3 (= )
51 K a k (t) d k k = 0 dt k L(u) = 0 L u 1 u 2 αu 1 + βu 2 ( α,β Z) proof L(u 1 ) = 0 L(u 2 ) = 0 L(αu 1 + βu 2 ) = αl(u 1 ) + βl(u 2 ) L(αu 1 + βu 2 ) = αl(u 1 ) + βl(u 2 ) = 0
52 R(ω) R(ω) = 1 (ω 2 0 ω2 )+2γωi R(ω) =R 0 (ω)e iφ(ω) R 0 (ω) R 1 = R 1 0 eiφ R 0 (ω) = φ(ω) 0 < φ(ω) < π 1 (ω 2 0 ω 2 ) 2 + 4γ 2 ω 2 φ(ω) = Arg((ω 2 0 ω 2 ) + 2γωi) R 1 =(ω 2 0 ω 2 )+2γωi φ(ω)
53 x s (t) d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt R 0 (ω), φ(ω) z s (t) =R 0 (ω)ae i(ωt φ(ω)) x s (t) = R 0 (ω)a cos (ωt φ(ω)) A cos ωt x s (t) t R 0 (ω) φ(ω)
54 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 d 2 x dx + 2γ dt2 dt + ω2 0x = f(t) x s x = x s + y x s x y = x x s y
55 d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt x s (t) = R 0 (ω)a cos (ωt φ(ω)) x(t) = x s (t) + e (B γt cos ω 2 0 γ2 t + C sin B, C ) ω0 2 γ2 t
56 γ > 0 d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt x s (t) = R 0 (ω)a cos (ωt φ(ω)) x s (t) x(t) A cos ωt
57 R 0 (ω) ω > 0 R 0 (ω) ω R 0 ( ω) γ < ω 0 / 2 ω = ω0 2 2γ2 R 0 ( ω) = ω 0 = 1, γ = 0.1 ω R 0 ( ω) 1 2γ ω 2 0 γ2 ω ω = = = R 0 ( ω) = = 5 ω
58 ω R(ω) R 0 (ω) π 0 φ(ω) ω 0
59
60 ẍ +ẋ + x = cos t x(0) = 2, ẋ(0) = 0 z +ż + z = e it z s (t) =Re it ir =1 R = i z s (t) = ie it = i(cos t + i sin t) = sin t i cos t x s (t) = Re z s (t) = sin t 3 x(t) = sin t + e t/2 (A cos 2 t + B sin 3 x(t) = sin t +2e t/2 cos 2 t 3 2 t)
数学演習：微分方程式
( ) 1 / 21 1 2 3 4 ( ) 2 / 21 x(t)? ẋ + 5x = 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ
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)
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
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..............
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
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
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 =
(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)
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....
chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
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..........................
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
() (, 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 =
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
) 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
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
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
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
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
() 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 ()
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
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
#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =
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