A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
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1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = (4, ) = 1 ( 4) 4 =
2 A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
3 = log (1) e () (a, log a) a = log = 1 (a, log a) log a = 1 ( a) a 1 (1) 1 e 1 a = e a = 1 e = log 1 log 1 ( e = e 1 ) e log a a 1 = e () 1 (0, 0) 0 log a = 1 (0 a) a log a = 1 log a a = e 1 1 a 1 = 1 ( e) e = 1 e = log
4 = e (1) 1 () e a A = e 1 a
5 B = 4 ( 3, 1) m + = 4 + = 0 0 = = 3 = 1 ( 3, 1) m = 3 1 = = 1 (, 1) 8 + = = = 4 (, 1) 4 1 = 1 1 = 1 ( ) = 1 +
6 A (1) + 8 = 1 A( 1, ) A 1 () = 1 A(, 1) 1 A
7 C A A A A(a, f(a)) = f() A(a, f(a)) f (a) f (a) 0 A f 1 (a) =f() = f() (a, f(a)) f (a) 0 f(a) = f 1 0 ( a) (a) f (a) = 0 = a 4. = e (1, e) f() = e =e f () = e f (1) = e 1 a (1, e) e = 1 ( 1) e = 1 e + e + 1 e
8 A (1) = e A( 1, e) A( 1, e) =e 1 ( ) π () = sin A 6, 1 A ( π, ) 1 6 =sin
9 A f() = f() A(a, f(a)) B(b, f(b)) AB f() (a, b) =f() A B AB B 1 f(b) f(a) AB b a c f A (c) f(b) f(a) b a = f (c), a < c < b f() [a, b] (a, b) a c b f(b) f(a) b a = f 0 (c), a < c < b c 4.3 f() = + 1 [0, ] (0, ) 5 = + 1 f() f(0) 0 f (c) = c = 5 1 = c = c = 1 1 1
10 c (1) f() = + 4 a = 0 b = 3 c 3 () f() = 3 a = 1 b = 1 c
11 B 4. a > 0 1 a + 1 < log(a + 1) log a < 1 a log(a + 1) log a log(a + 1) log a = (a + 1) a f() = log f() = log > 0 f () = 1 [a, a + 1] log(a + 1) log a (a + 1) a a < c < a + 1 = 1 c 1 c a > 0 1 a + 1 < 1 c < 1 a 1 1 a + 1 < log(a + 1) log a < 1 a 4.6 a < b e a < eb e a b a < eb
12 A f() [a, b] (a, b) 1 1 (a, b) f () > 0 f() [a, b] (a, b) f () < 0 f() [a, b] 3 (a, b) f () = 0 f() [a, b] 1 [a, b] a 1 < b 1 f( ) f( 1 ) = f (c)( 1 ), 1 < c < c (a, b) f () > 0 1 f (c) > 0 1 > 0 f( ) f( 1 ) > 0 f( 1 ) < f( ) 1 1 I 1 1 < f( 1 ) < f( ) f() I 1 < f( 1 ) > f( ) f() I
13 f() g() (a, b) g () = f () f() g() [a, b] g() = f() + C C 4.8
14 140 4 f () f() 4.3 f() = f() 0 f () = = f () = 0 =f() = f() f () 0 + f() 0 1 f() (1) f() = e = 1 () f() = log 1 1
15 (3) f() = + sin (0 π) π π B f() = a f() = a f(a) f() = b f() = b f(b) b a f() f() = a = a f (a) = 0 f() f (a) = 0 = a f () = a a f () + 0 f() a f () 0 + f() f (a) = 0 = a f () f(a) f (a) = 0 f(a) ()
16 (1) f() = e () f() = + 4 (1) f () = 1 e + ( e ) = (1 )e f () = 0 = 1 f() e > 0 1 f () + 0 f(1) = 1 e 1 = f() 1 e 1 e f(1) = 1 e () 0 f () = 1 4 = 4 = ( + )( ) f () = 0 =, f() > 0 0 f () f() 4 4 f( ) = 4 f() = (1) () 4 1 e 1 4 =
17 (1) f() = () f() = e 4 e
18 144 4 (3) f() = log 1 e 1 e 1 (4) f() = + =
19 f() = a = a 4.3 f() = = < 0 = 1 0 f() = + 1 f () = = ( + 1) = > 0 f () > 0 1 < 0 f() = + 1 f () = f () = 0 = 3 f() f () f() ( f 3 ) = 3 9 f(0) = 0 3 9
20 (1) f() = () f() =
21 4.4 f() = + + a = 1 a f() = 1 f ( 1) = 0 a f () = ( + 1)( 1) ( + + a) = 1 a ( 1) ( 1) f() = 1 f() = 1 f ( 1) = 0 a 4 = 0 f ( 1) = a 4 a = f() = f () = 3 ( 1) = ( + 1)( 3) ( 1) f() ( 1) > f () f() 1 7 a = = 1 f( 1) = 1 f(3) = 7 ( ) a = f() = + a = 1 a f()
22 148 4 C 4.5 = cos + sin (0 π) = cos ( sin ) + cos = cos (1 sin ) 0 < < π = 0 cos = 0 = π, 3 π sin = 1 0 π cos = 0 = π, 3 π sin = 1 = π 0 π 3 π π = π, = 3 π 1 =cos + sin 3 π π π
23 (1) = (1 + cos ) sin (0 π) () = (1 4)
24 A f() f () C = f() C f () f () f () f () f () f () > 0 f () C f () < 0 f () C C = f() f 00 () = f() f() f () 1 f () > 0 = f() f () < 0 = f()
25 = = 3 6 = 6 6 = 6( 1) 4 = (1, ) = f() f (a) = 0 = a f () (a, f(a)) = f() f (a) = 0 (a, f(a)) = f() f() = 4 = 4 f () = 4 3, f () = 1 f (0) = 0 = 0 f () (0, 0) = (1) =
26 15 4 () = e (, e ) (3) = cos (0 < < π) π + 1 π 1 π π (4) = 4
27 B 4.6 = e = 4e = 4{e + ( 4e )} = 4(4 1)e = 0 = 0 = 0 = ± e 1 1 e ( 1 ) ( ), 1 1 e, 1 e lim = 0 lim = e
28 = 1 = = 1 = 1 ( 1) = ( 1) 1 = ( 1) ( 1) 3 ( ) ( 1) lim = lim = = 1 lim { ( + 1)} = 0 4 { ( + 1)} = 0 = + 1 lim = =1 =
29 (1) = e
30 156 4 () = 4 + 1
31 (3) = + 1 1
32 158 4 C f() f () f () 1 f (a) = 0 f (a) > 0 f(a) f (a) = 0 f (a) < 0 f(a) 1 f (a) > 0 a f () > 0 f () f (a) = 0 =f() < a f () < 0 > a f () > 0 a f ()<0 f () 0 + f () + f() f(a) a f ()>0 f (a)=0 f (a)>0 4.5 f() = f () = f () = 6 f () = 0 =f() f(1) = 1, 1 f ( 1) = 6 > 0 f (1) = 6 < 0 f( 1) f(1) 1 1 f( 1)
33 f() = + cos (0 π) f () = 1 sin f () = cos 0 < < π f () = 0 = π 6, 5 6 π sin = 1 ( π ) f = ( ) 5 3 < 0 f 6 6 π = 3 > 0 ( π ) f = π ( ) 5 3 f 6 π = 5 6 π (1) f() =
34 160 4 () f() = + sin (0 π) f() f (a) = 0 f (a) = 0 f(a) 4.6 (1) f() = 4 () f() = 3 f () = 4 3 f () = 1 f () = 3 f () = 6 f (0) = 0 f (0) = 0 f (0) = 0 f (0) = 0 f(0) f(0) = 4 = 3
35 p 0 = 4p ( 1, 1 ) 1 = p( + 1 ) (1) = 4 ( 1 )
36 16 4 () = = 4 + a 3 + 3a + 1 a [ ] 1 p = 4p 0 (1) = = 1 3 () = = 3 a < 0 8 < a = 0 + a + a = 0 1
37 A 4.5 > 0 e > 1 + > 0 f() = e (1 + ) f() = e (1 + ) f () = e 1 > 0 e > 1 f () > 0 f() 0 > 0 f() > f(0) = 0 f(0) = e 0 (1 + 0) = 0 e > > 0 (1) e >
38 164 4 () log( + 1) < > (1) e > e >, e < 0 e lim =, lim e = 0 n e lim =, n lim n e = 0
39 B 4.6 a e = a = 0 = e = a = 0 0 f() = e 0 1 f ( 1)e f () = () 0 + f() f() e lim f() =, lim f() =, lim +0 f() = 0 = t t ( e t lim f() = lim t t = lim 1 ) = 0 t te t = f() = a = e a =a e a > e a = e a < a < e 0 1
40 a 3 = a( 1)
41 A P P t =f(t) = f(t) t t + t f(t + t) f(t) t t 0 t P v v = d dt = f (t) P v > 0 v < 0 v v v t α α = dv dt = d dt = f (t) P t = f(t) t P v α v = d dt = f 0 (t), α = dv dt = d dt = f 00 (t) 4.9 P t = sin(πt a) t v α a v v = d = π cos(πt a) dt α α = dv dt = π sin(πt a) α = π P
42 v 0 m/ t m = v 0 t 1gt g t v m/ α m/ B P t P (, ) t P PQ PR Q R P(, ) R t d Q dt d dt Q P v ( ) R d v = dt, d P dt Q t P ( ) P(, ) v = d d, dt dt θ tan θ = d dt d dt = d d v P P d dt P θ d dt v
43 v v P d d dt dt ( ) α = d d dt, dt t P α α P P t (, ) t t P v v α α ( v = d, d ) (d ) ( ) d v = + dt dt dt dt ( ) (d ) ( ) α = d dt, d dt α = d dt + dt 4.10 P t = a cos ωt, = a sin ωt (a, ω ) P t P t v α v d = aω sin ωt d = aω cos ωt dt dt v v = ( aω sin ωt) + (aω cos ωt) = a ω (sin ωt + cos ωt) = aw α d dt = aω cos ωt d dt = aω sin ωt α = aω (cos ωt, sin ωt) α α = aω cos ωt + sin ωt = aω
44 P + = a aω a v P α ωt 4.10 α = ω (, ) a a P α P a 4.0 t P t = 3 P (1) = t + 1, = t 4t () = cos πt, = sin πt
45 A f(a + h) f() = a f (a) f(a + h) f(a) lim h 0 h = f (a) h 0 f(a + h) f(a) h f (a) f(a + h) f(a) + f (a)h h 0 f(a + h) h 1 1 h 0 f(a + h) f(a) + f 0 (a)h = f() A(a, f(a)) =f() f(a + h) P f(a) + hf (a) T = f(a) + f (a)( a) f(a) A h 0 h hf (a) Q P T a a + h 4.7 h 0 sin(a + h) 1 (sin ) = cos h 0 sin(a + h) sin a + h cos a
46 h 0 1 (1) cos(a + h) () tan(a + h) B 0 1 a = 0 h f() f(0) + f 0 (0) 4.11 p 0 (1 + ) p 1 + p f() = (1 + ) p f () = p(1 + ) p 1 f(0) = 1 f (0) = p f() f(0) + f (0) 0 (1 + ) p 1 + p = ( ) =
47 (1) e () log(1 + ) (3) (1) () log 1.01 (3)
48 P t = a(ωt sin ωt), = a(1 cos ωt) (a ω ) 5 8cm 3 t r cm S cm V cm 3 r = dv dr dt dt ds dt
49 π (1) sin 31 () tan 1 4 aω 5 dv dt = 8 (cm3 / ) dr dt = 1 π (cm/ ) ds dt = 8 (cm / ) 6 (1) () [ ( (1) sin 31 = sin π π ) sin π 6 + π 180 cos π 6 f () = cos 1 f (0) = 1 f(0) = 0 tan () f() = tan ]
50 A 1 (1) = + (, 0) ( π ) () = tan 4, 1
51 (1) = log () = 3 3
52 (1) = ( 1) + 1
53 () = sin + cos (0 π)
54 A r θ θ θ l θ r θ B C
55 a = + a (1) a > 0 () a = 0 (3) a < 0
56 > 0 (1) cos > 1 () sin > 3 6
57 B 7 = P Q P PR QR 8 = a + b = log A(e, 1) A a b
58 f() = 3 + 3a + 3b + c C C A C A P A P Q Q C
59 P t = 4 cos t = sin t 0 t π P m A 0.3m B A 5m 4m B A
60 k 1 A B A = B = d dt = (1) = = () = π + 1 = 1 + π (1) = 1 1 e e 3 () = 3, 0, 3 0 = 1, 1 [ (1) = ( 1) ( + 1) = 4( 3) = 1 ( + 1) 3 ] () = ( sin 1)(sin + 1) = cos (4 sin + 1) (1) () = 1 π 6 π 5 π 6 π [ AB = r cos θ BC = r sin θ l = 4r cos θ + r sin θ dl dθ 5 (1) 3 () 0 (3) 1 ] = 4r( sin θ 1)(sin θ + 1) [ = ( 3a) ( + a) 3 ]
61 [ ) (1) f() = cos (1 > 0 f () > 0 f(0) = 0 ) () g() = sin ( 3 6 ] (1) g () > 0 g(0) = 0 7 P t = t + (t + 1) t 8 a = 1 e b = 1 f() = a + b g() = log f(e) = 1 f (e) = g (e) [ 9 : 1 S() h k k = π h S() = π + πh S () = (π3 k) S() = 3 ] k π 10 A ( a, a 3 3ab+c) A C = g() g() = f( a) (a 3 3ab+c) = 3 +3(b a ) g( ) = g() 11 [ (d ) ( ) d 5 + = ( 4 sin t) + ( cos t) dt dt = ] 4{4 sin t + (1 sin t) } [ 1 0.4m/ A B A = B = + = 5 d t d dt + d dt = 0 ] d = 0.3 = 4 = 3 dt dt = 0.4
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... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =
高校生の就職への数学II
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高等学校学習指導要領解説 数学編
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名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト
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No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
Gmech08.dvi
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#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
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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)) ( )
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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
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(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP
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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
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
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4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X
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< 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) =
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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
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Chap10.dvi
=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.
1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
(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
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sekibun.dvi
a d = a + a+ (a ), e d = e, sin d = cos, (af() + bg())d = a d = log, cosd = sin, f()d + b g()d d 3 d d d d d d d ( + 3 + )d ( + )d ( 3 )d (e )d ( sin 3 cos)d g ()f (g())d = f(g()) e d e d ( )e d cos d
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九州大学学術情報リポジトリ Kyushu University Institutional Repository 物理工科のための数学入門 : 数学の深い理解をめざして 御手洗, 修九州大学応用力学研究所 QUEST : 推進委員 藤本, 邦昭東海大学基盤工学部電気電子情報工学科 : 教授 http://hdl.handle.net/34/500390 出版情報 : バージョン :accepted
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II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
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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,
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Quiz x y i, j, k 3 A A i A j A k x y z A x A y A z x y z A A A A A A x y z P (x, y,z) x i y j zk P x y z P ( x 1, y 1, z 1 ) Q ( x, y, z ) 1 OP x1i y1 j z1k OQ x i y j z k 1 P Q PQ 1 PQ x x y y z z 1 1
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
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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
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)
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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)
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
2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a
009 I II III 4, 5, 6 4 30. 0 α β α β l 0 l l l l γ ) γ αβ ) α β. n n cos k n n π sin k n π k k 3. a 0, a,..., a n α a 0 + a x + a x + + a n x n 0 ᾱ 4. [a, b] f y fx) y x 5. ) Arcsin 4) Arccos ) ) Arcsin
1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +
6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +
熊本県数学問題正解
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(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37
4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin
limit&derivative
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( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n
1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x
. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +
1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =
arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927
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)
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
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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
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20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
数学Ⅲ立体アプローチ.pdf
Ⅲ Ⅲ DOLOR SET AMET . cos x cosx = cos x cosx = (cosx + )(cosx ) = cosx = cosx = 4. x cos x cosx =. x y = cosx y = cosx. x =,x = ( y = cosx y = cosx. x V y = cosx y = sinx 6 5 6 - ( cosx cosx ) d x = [
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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...............................
B line of mgnetic induction AB MN ds df (7.1) (7.3) (8.1) df = µ 0 ds, df = ds B = B ds 2π A B P P O s s Q PQ R QP AB θ 0 <θ<π
8 Biot-Svt Ampèe Biot-Svt 8.1 Biot-Svt 8.1.1 Ampèe B B B = µ 0 2π. (8.1) B N df B ds A M 8.1: Ampèe 107 108 8 0 B line of mgnetic induction 8.1 8.1 AB MN ds df (7.1) (7.3) (8.1) df = µ 0 ds, df = ds B
no35.dvi
p.16 1 sin x, cos x, tan x a x a, a>0, a 1 log a x a III 2 II 2 III III [3, p.36] [6] 2 [3, p.16] sin x sin x lim =1 ( ) [3, p.42] x 0 x ( ) sin x e [3, p.42] III [3, p.42] 3 3.1 5 8 *1 [5, pp.48 49] sin
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,,
01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 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,
) 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
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
![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](/thumbs/85/91372642.jpg "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")
c 28. 2, y 2, θ = cos θ y = sin θ 2 3, y, 3, θ, ϕ = sin θ cos ϕ 3 y = sin θ sin ϕ 4 = cos θ 5.2 2 e, e y 2 e, e θ e = cos θ e sin θ e θ 6 e y = sin θ e + cos θ e θ 7.3 sgn sgn = = { = + > 2 < 8.4 a b 2
1 2 1 No p. 111 p , 4, 2, f (x, y) = x2 y x 4 + y. 2 (1) y = mx (x, y) (0, 0) f (x, y). m. (2) y = ax 2 (x, y) (0, 0) f (x,
No... p. p. 3, 4,, 5.... f (, y) y 4 + y. () y m (, y) (, ) f (, y). m. () y a (, y) (, ) f (, y). a. (3) lim f (, y). (,y) (,)... (, y) (, ). () f (, y) a + by, a, b. + y () f (, y) 4 + y + y 3 + y..3.
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 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +