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

Similar documents
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

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

1

高等学校学習指導要領

高等学校学習指導要領

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

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

応力とひずみ.ppt

6. Euler x

.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 =,

() 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 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

sekibun.dvi

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a

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.

Microsoft Word - 触ってみよう、Maximaに2.doc

i

mugensho.dvi

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

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

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,

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

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

II 2 II

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

DVIOUT

(2000 )

limit&derivative

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π


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

( ) 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) (

高校生の就職への数学II

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

koji07-02.dvi


A

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

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

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,

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

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 :

a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552

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,,

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

CALCULUS II (Hiroshi SUZUKI ) f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b)

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 =

Chap10.dvi

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

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

,2,4

(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

untitled

数学Ⅱ演習(足助・09夏)

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

2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

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

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

II Time-stamp: <05/09/30 17:14:06 waki> ii

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

< 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)

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)

極限

a a s d f g h j a s d f g h a s

40 6 y mx x, y 0, 0 x 0. x,y 0,0 y x + y x 0 mx x + mx m + m m 7 sin y x, x x sin y x x. x sin y x,y 0,0 x 0. 8 x r cos θ y r sin θ x, y 0, 0, r 0. x,


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 { (

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

Chap9.dvi

( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,


JKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n =

di-problem.dvi

II (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%


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

高等学校学習指導要領解説 数学編

p q p q p q p q p q p q p q p q p q x y p q t u r s p q p p q p q p q p p p q q p p p q P Q [] p, q P Q [] P Q P Q [ p q] P Q Q P [ q p] p q imply / m

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

x ( ) x dx = ax

2000年度『数学展望 I』講義録

f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f

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

Part () () Γ Part ,

( [1]) (1) ( ) 1: ( ) 2 2.1,,, X Y f X Y (a mapping, a map) X ( ) x Y f(x) X Y, f X Y f : X Y, X f Y f : X Y X Y f f 1 : X 1 Y 1 f 2 : X 2 Y 2 2 (X 1

( ) 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 +

untitled

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 I

1 I p2/30

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb


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

1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1


Transcription:

013 4 16 5 54 (03-5465-7040) nkiyono@mail.ecc.u-okyo.ac.jp hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y = + 1 y = + + 1 6: ac b 0 a + by + cy + p + qy + r = a( 0 ) + b( 0 )(y y 0 ) + c(y y 0 ) + r

0, y 0, r ( 0, y 0 ) 1 p = a 0 by 0 q = b 0 cy 0 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 1 y = y = z = f(, y) a a z = ( + ba ) y + ac b a y z = ( + ba ) y ac b a y 3 y 1 X, Y X = + b a y 0, y 0, r 3 z = f(, y) y

3 Y ac b ac b b ac Y = a y Y = a y ac b = 0 Y = y X Y a > 0, ac b > 0 z = X + Y a > 0, ac b = 0 z = X a > 0, ac b < 0 z = X Y a < 0, ac b > 0 z = X Y a < 0, ac b = 0 z = X a < 0, ac b < 0 z = X + Y a 0 a = 0 c 0 y a = c = 0 b 0 by = b ( ( + y) ( y) ) X = + y, Y = y b z = X Y z = X + Y a 0 z = X Y 1 Y z = X z = X 7 ac b = 0 1 z = X z = X 7:

4 X Y 1 z = X + AX + BY + C z = X + AX + BY + C Y 1 X 7 zx z = X + AX + C z = X + AX + C X = A X = A Y Y z z = BY + C 1 z = X + Y XY (, y) = (0, 0) XY z = k X + Y = k k z = X + Y 1 z 1 Xz Xz Y = 0 z = X + Y z = X z z = X Y 8 X, Y, y z = X + Y z = X Y 8: y z = X Y z = X + Y X Y z = X Y XY X Y = k

5 Y z = 0 z = O X z = z = 3 z = 1 z = 3 z = 1 9: z = X Y k X k Y k = 0 Y = ±X 9 XY Xz Y = k z = X k X = 0 k z = k Y z Xz z = X Y z z = Y z = X Y? 10 z = X Y 10: z = X Y 4. (1) + 4y + 5y + + 3y + 1 () + 6y + 5y + + 3y + 1

6 1..3 f(, y) z = f(, y) z = y = = f(, y) 1 g() = + y z = f(, y) z = g() 0 z z = X + Y = X + Y 1 1 1 3 4 1 4 3 4 n 1 1 1 1.1 1 a f() f() f(a) lim a a (3) f() = a f() = a f (a) df d (a) d d f() =a

7 f() a f() a f (a) f() f (), df d (), d d f() = a a f() = f () = f(3) = 3 = 6 f f() f(3) 3 (3) = lim = lim 3 3 3 3 = lim ( 3)( + 3) 3 3 = lim 3 ( + 3) = 3 + 3 = 6 = 3 a f f() f(a) a (a) = lim = lim a a a a = lim ( a)( + a) a a = lim a ( + a) = a + a = a a ( ) = 5. 1 { < 0 f() = 3 0. (3) f() f(a) f(a) f() a

8 a f(a) a a 1 (a, f(a)) (, f()) 11 y f() f(a) f() f() f(a) a f(a) a 1 y = f() O a a + 1 11: a = a 1 y = f() (a, f(a)) 1.3 1 1 1 1

9 y f (a) f() f(a) a O a a + 1 1: f() = c = a 0 f() f(a) a = c c a = 0 0. 0 n n = a na n 1 n a n = ( a)( n 1 + n a + + a n + a n 1 ) n a n lim a a = lim a (n 1 + n a + + a n + a n 1 ) = a n 1 + a n a + + aa n + a n 1 = na n 1 n n n 1 (a 0 + a 1 + a + + a n n ) = a 0 (1) + a 1 () + a ( ) + + a n ( n ) = a 1 + a + 3a 3 + + na n n 1

10 d d sin = cos d. d cos = sin d d e = e an log ( f()g() ) = f ()g() + f()g () f()g() f(a)g(a) = f()g() f(a)g() + f(a)g() f(a)g(a) f()g() f(a)g(a) lim = lim a a = (f() f(a))g() + f(a)(g() g(a)) f() f(a) a a = f (a)g(a) + f(a)g (a) ( ) 1 = f () f() (f()) lim a 1 f() 1 f(a) a g() g(a) g() + lim f(a) a a f(a) f() 1 = lim a a f()f(a) = f 1 (a) f(a)f(a) an an = sin cos ( (an ) = sin 1 ) ( ) = (sin ) 1 1 cos cos + sin cos = cos 1 ) ( ( cos + sin (cos ) cos = 1 + sin sin ) cos = 1 + sin cos = 1 + an 1 cos an = 1 + an = cos cos + sin cos = 1 cos

11 (f(g()) = f (g())g () g() f() f(g()) = g () = 1 f (g()) log log e e log = (e ) = e (log ) = 1 e log = 1 > 0 < 0 (log( )) = 1 ( ) = 1 ( 1) = 1 (log ) = 1 r ( r ) = r r 1 > 0 r = e r log ( r ) = (e r log ) = e r log (r log ) = r r = rr 1 6. f() = e sin g() = 3 + + 1 h() = log(cos ) 3 n 1,,..., n f f n 1

1 n = n = n 3 3.1 (a, b) f(, y) y = b 1 4 = a φ (a) φ() := f(, b) f(, y) (a, b) = a y 1 y = b ψ (b) ψ(y) := f(a, y) f(, y) (a, b) y φ() 1. f (, y) (, y) = (a, b) f f(, b) f(a, b) lim a a f (a, b) (a, b) f (a, b) f(a, b) (a, b) f y f(a, y) f(a, b) (a, b) = lim y y b y b. 4 φ() := f(, b) φ() f(, b) := =:

13 1 d MS-IME f(, y) (a, b) f (a, b) (a, b) 1 (a, b) (, y) 1 1 1 f(, y) = y 3 f (, y) = y 3 y f y (, y) = 3 y. y 1 y f (a, b) (a, b) f (a, b) (a, b) f(, y) f (, y). f(, ) () f f(, ) () 1 φ() = f (, () ) f f (, () ) () df (, ) d φ () (, ) 1 7. f(, y) f(, y) = sin ( 3 y ) (, y), (, y) y

14 3. 1 1 1 y = f() = a f (a) y = a 1 z = f(, y) (, y) = (a, b) (, y) = (a, b) (a, b) (, y) = (a, b) z 1 y (a, b) y y f(, y)? y 1. 5 T V 6 T V U f(, y) = T y = V U = f(t, V ) f(, y) V 0 T 0 7 (T 0, V 0 ) (T 0, V 0 ) 5 6 7

15. f(, y) y f T T T V f U = f(t, V ) T V V 0 T T = T 0 T (T 0, V 0 ) U f f U f U T V U U = U(T, V ) 1. 0 (cm) f(, ) 13: f(, )

16 1 ( ) f (, ) := (, ) 1 f(, ) (, ) (, ) = C f (, ) C 8. > 0 f(, ) = 1 e f (, ) = 4 (, ) 13 1 5 3. g(, ) 14: g (, ) = c g (, ) (4) c 9. c f(, ) = sin( + c) + sin( c) (4) 14

17 1 1 f (a, b) φ() = f(, b) 1 φ = a f (a, b) φ() (a, φ(a)) φ() φ() f(, y) f(, y) yz z = f(, y) z = φ() φ() f(, b) z = φ() z = f(, y) y = b y = b yz y = b z b = 0 z b 0 z y y b φ() z = f(, y) ( a, b, f(a, b) ) z f (a, b) φ (a) z y = b z = φ() z = φ() = f(, b) b a 15: ( a, b, f(a, b) ) 15 f y (a, b)

18 z = f(, y) ( a, b, f(a, b) ) yz ( a, b, f(a, b) ) f (a, b) z = f(a, b) (a, b, f(a, b)) f y (a, b) y 16 = a f y (a, b) z y = b f (a, b) b a 16: y

19 4 1 a y b y c (1) a = 1, b =, c = 5 a = 1 > 0 ac b = 1 5 = 5 4 = 1 > 0 z = X + Y () a = 1, b = 3, c = 5 ac b = 1 5 3 = 5 9 = 4 < 0 z = X Y 5 < 0 f() f () = < 0 > 0 f() 3 f () = 3 > 0 = 0 f() 0 = 0 f() f(0) lim 0 0 f (0) f() 0 0 0 0 +0 f (0) f() f(0) lim = lim 0 0 0 = lim = 0 0 f() f(0) 3 lim = lim +0 0 +0 = lim +0 = 0

0 f (0) = 0 { < 0 f () = 3 0 0 f() = 0 = 0 0 f() = 3 +0 3 = 0 6 f () = (e ) sin + e (sin ) = e sin + e cos = e (sin + cos ) g () = ( ) ( 3 + + 1) ( 3 + + 1) ( 3 + + 1) = 4 + 4 + 3 4 ( 3 + + 1) = 4 + + ( 3 + + 1) h () = log (cos ) (cos ) = 1 ( sin ) = an cos 7 y 1 1 y g() = 3 y 1 f(, y) sin z z = g() sin z = cos z g () = ( 3) y = 3 y (, y) = ( sin g() ) g () = ( cos ( 3 y )) 3 y = 3 y cos ( 3 y )

1 h(y) = 3 y y 1 f(, y) sin z z = h(y) y (, y) = ( sin h(y) ) h (y) = ( cos ( 3 y )) 3 y = 3 y cos ( 3 y ) 8 1 f(, ) (, ) = e 1 f (, ) = ( e ) = e e = 4 e 1 f(, ) 1 (, ) = e f (, ) = 4 + 1 e e = 4 e e = = 4 (, ) 9 (, ) = cos( + c) + cos( c) ( ) f (, ) = (, ) = sin( + c) sin( c) = f(, ) f(, ) (, ) = c cos( + c) c cos( c) f (, ) = ( ) (, ) = c sin( + c) c sin( c) = c f(, ) f (, ) = c f(, ) = c ( f(, )) = c f (, ) (4)