高等学校学習指導要領

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高等学校学習指導要領

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

, 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,,,,

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) 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 {

() 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)

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

春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,

( ) 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 +

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高等学校学習指導要領解説　数学編

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

, 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

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(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

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

arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =

18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (

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

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1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :

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

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

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

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y

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

< 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)

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

(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n

( ) x y f(x, y) = ax

(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z

.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 =,

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5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4

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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

dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp

But nothing s unconditional, The Bravery R R >0 = (0, ) ( ) R >0 = (0, ) f, g R >0 f (0, R), R >

Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

2010 IA ε-n I 1, 2, 3, 4, 5, 6, 7, 8, ε-n 1 ε-n ε-n? {a n } n=1 1 {a n } n=1 a a {a n } n=1 ε ε N N n a n a < ε

f : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y

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

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

B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:

[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29

重力方向に基づくコントローラの向き決定方法

3 1 1 BCA ACD HP A AB BC ABC ONP x AM, CN x 30 DM DM! CN CN AM AMD 10 1 AB AC

2 x x, y, z x,, z F c : x x x cos y sin z z 8 F F F F F x x x F x x F 9 F c J Fc J Fc x x x y y y cos sin 0 sin cos 0 0 0, J Fc 0 J Fc t x /x J Fc,, z

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

Transcription:

sin cos tan

i

lim

e