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)
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1 2011 I 2 II III 17, 18, : : : x = a x = a
2 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)
3 2 3 (a, b) f(a, b) f f 1 x = a f f (a) = 0. x = a (a r, a + r) f(a) f(x) f(x) f(a) 0 f(x) f(a) a r < x < a = 0 x a f(x) f(a) a < x < a + r = 0 x a f x a f (a) f (a) = x a 0 x a+0 f(x) f(a) 0 x a f(x) f(a) 0 x a f (a) = 0 0 x 2 xy 2 f (a, b) f f (a, b) = (a, b) = 0 x y. f (a, b) y b x 1 f(x, b) x = a f(x, b) x x = a 0 f x (a, b) = 0 f y (a, b) = 0 xy 0 f(x) = x 3 0
4 : 1 1 f (a) = 0 a 2 f (a) > 0 f (a) < 0 f (a) = 0 f(x) = x 4, x 4, x 3 x = 0 f 0 f 0 2 f (a) = 0 f (a) > 0 f C 2 f f (a) = (f ) (a) > 0 f (x) a f (a) = 0 a r < x < a f (x) < 0 f(x) a < x < a + r f (x) > 0 f(x) x = a f f > 0 f f (a) > 0 f (x) a f (a) = 0 1 a
5 x = a f(x) x = a f (a) = f(a) + f (a) (x a) 2 2 f (a) > 0 x = a f (a) < 0 x = a x = a 2 f(x) f (a) > 0 x = a f (a) < 0 x = a f(x) = f(a) + f (a)(x a) f (a + θ(x a))(x a) 2 θ 0 1 a f (a) = 0 f(x) = f(a) f (a + θ(x a))(x a) 2 x a f f C 2 f (a) f (a + θ(x a)) f (a) > 0 f(x) = f(a) f (a + θ(x a))(x a) 2 > f(a) f(a) f (a) < 0 f(x) = f(a) f (a + θ(x a))(x a) 2 < f(a) f(a) 1.3 : f(x, y) (a, b) f x (a, b) = f y (a, b) = 0 f x (a, b) = f y (a, b) = 0 f (a, b) n n
6 f(x, y) f f (a, b) = (a, b) = 0 x y (a, b) f f(x, y) = x 2 + y 3 (0, 0) f(0, y) y y (0, 0) f(0, 0) = f (a) = P (x) = f(a) + p(x a) 1 f(x) x = a 1 f(x) P (x) = 0 x a x a p = f (a) P (x, y) = f(a, b) + p(x a) + q(x a) 2 f(x, y) (a, b) 1 (x,y) (a,b) f(x, y) P (x, y) (x a)2 + (y b) 2 = P (x, y) p q p = f (a, b), x q = f (a, b) y 1 2 Q(x) 1 f(x) x = a 2 f(x) Q(x) x a (x a) 2 = 0
7 2 7 Q(x) Q(x) = f(a) + f (a)(x a) + f (a) (x a) Q(x, y) f(x, y) (a, b) 2 (x,y) (a,b) f(x, y) Q(x, y) = 0 (1) (x a)2 + (y b) 22 (1) Q(x, y) 2 (x, y) (a, b) (a, b) x = a + r cos θ, y = b + r sin θ (x, y) (a, b) r 0 r 0 θ r 0 r 0 θ (1) Q(x, y) x a y b 2 c + p(x a) + q(y b) + A(x a) 2 + 2B(x a)(y b) + C(y b) 2 c, p, q, A, B, C (1) x = a + r cos θ, y = b + r sin θ r 0 1 r 2 { f(a + r cos θ,b + r sin θ) c pr cos θ qr sin θ Ar 2 cos 2 θ 2Br 2 cos θ sin θ Cr 2 sin 2 θ } = 0 (2) r 0 r 2 0 r 2 0 r = 0 0 f(x, y) c = f(a, b) r 2 r r 0 r 2 r r 0 r 2 0 f(a + r cos θ, b + r sin θ) f(a, b) pr cos θ qr sin θ = 0 r 0 r x, y f(x, y) f(a, b) p(x a) q(y b) = 0 (x,y) (a,b) (x a)2 + (y b) 2 (a, b) p = f (a, b), x q = f (a, b) y
8 2 8 A, B, C 2 (2) r 1 θ (2) h(r) r 2 g(r) θ h(0) = g(0) = 0 h(r) h(r) h(0) = g(r) g(r) g(0) = h (ρ) g (ρ) ρ 0 r ρ r θ g (ρ) = 2ρ h (ρ) h (ρ) = f f (a + ρ cos θ, b + ρ sin θ) cos θ + (a + ρ cos θ, b + ρ sin θ) sin θ x y f f (a, b) cos θ x y (a, b) sin θ 2Aρ cos2 θ 4Bρ cos θ sin θ 2Cρ sin 2 θ h (r) r 0 g (r) = 0 (f x (a + r cos θ, b + r sin θ) f x (a, b)) cos θ + (f y (a + r cos θ, b + r sin θ) f y (a, b)) sin θ r 0 2r = A cos 2 θ + 2B cos θ sin θ + C sin 2 θ (3) θ (2) (3) (2) f x f y φ(r) = f x (a + r cos θ, b + r sin θ), ψ(r) = f y (a + r cos θ, b + r sin θ) 1 f x (a + r cos θ, b + r sin θ) f x (a, b) = φ (r) = f x r 0 r x (a, b) cos θ + f x (a, b) sin θ y f y (a + r cos θ, b + r sin θ) f y (a, b) = ψ (r) = f y r 0 r x (a, b) cos θ + f y (a, b) sin θ y (4) φ(r) ψ(r) (4) (3) f 2 x 2 (a, b) cos2 θ + 1 ( 2 ) f 2 y x (a, b) + 2 f (a, b) cos θ sin θ f x y 2 y 2 (a, b) sin2 θ = A cos 2 θ + 2B cos θ sin θ + C sin 2 θ θ θ = 0 θ = π/2 A = 1 2 f 2 x 2 (a, b) C = 1 2 f (a, b) 2 y2
9 2 9 2B = 1 2 ( 2 ) f y x (a, b) + 2 f (a, b) x y f(x, y) (a, b) f x (x, y) f y (x, y) (a, b) (a, b) f(x, y) 2 Q(x, y) Q(x, y) =f(a, b) + f f (a, b)(x a) + x y (a, b)(y b) ( 2 ) f 2 y x (a, b) + 2 f (a, b) (x a)(y b) + 1 x y 2 2 f (a, b)(x a)2 x2 2 f (a, b)(y b)2 y2 f, f x, f y C 1 C 1 f x f y f f x f y C 1 f C 2 C C 2 4 C 2 2 f xy = f yx (x a)(y b) : C 2 2 f(x, y) C 2 f (a, b) 2 2 f(a, b) + f f (a, b)(x a) + (a, b)(y b) x y f 2 x 2 (a, b)(x a)2 + 2 f x y (a, b)(x a)(y b) f (a, b)(y b)2 2 y2
10 (a, b) 2 f(x, y) 2 f 2 f 1 1 f(x) x = a f (a) = 0 x = a 2 f(a) + f (a) (x a) 2 2 f (a) > 0 f f (a) < 0 f 2 (a, b) f f x (a, b) = f y (a, b) = 0 f (a, b) 2 f(a, b) x 2 (a, b)(x a)2 + 2 f y x (a, b)(x a)(y b) f (y b)2 2 y2 f 2 (a, b) f(a, b) 2 1/2 u = x a, v = y b u v 2 (0, 0) A = 2 f (a, b) x2 B = 2 f (a, b) y x C 2 Q(u, v) Q(u, v) = Au 2 + 2Buv + Cv 2 = 2 f (a, b) y2 z = Q(u, v) (0, 0, 0) u v u v 2 A, B, C 0 Q(u, v) 0
11 2 11 A, B, C 0 A 0 Q(x, y) = A ( ( u + B ) ) 2 A v AC B2 + A 2 v 2 1 v = u 2 v = 2u 2 z = Q(x, y) A A z = (u + BA ) 2 v + AC B2 A 2 v 2 z = (u + BA ) 2 v AC B2 A 2 v 2 2 u v 1 X, Y X = u + B A v AC B 2 AC B 2 Y = v AC B 2 A 2 B2 AC Y = v AC B 2 = 0 Y = v A 2 A > 0, AC B 2 > 0 z = X 2 + Y 2 A > 0, AC B 2 = 0 z = X 2 A > 0, AC B 2 < 0 z = X 2 Y 2 A < 0, AC B 2 > 0 z = X 2 Y 2 A < 0, AC B 2 = 0 z = X 2 A < 0, AC B 2 < 0 z = X 2 + Y 2 A 0 A = 0 C 0 x y A = C = 0 B 0 2Buv = B 2 ( (u + v) 2 (u v) 2) 2 z = Q(u, v) uv
12 2 12 B z = X 2 Y 2 z = X 2 + Y 2 A 0 z = X 2 Y 1 Y z = X 2 z = X 2 1 z = X 2 z = X 2 1: 1 2 z = X 2 + Y 2 XY (x, y) = (0, 0) XY z = k X 2 + Y 2 = k k z = X 2 + Y 2 1 z 1 Xz Xz Y = 0 x = X 2 + Y 2 z = X 2 z z = X 2 Y 2 2 X, Y u, v uv z = X 2 Y 2 z = X 2 + Y 2 X Y z = X 2 Y 2 XY X 2 Y 2 = k k X k Y k = 0 Y = ±X 3
13 2 13 z = X 2 + Y 2 z = X 2 Y 2 2: Y z = 0 z = 2 O X z = 2 z = 3 z = 1 z = 3 z = 1 3: z = X 2 Y 2 XY Xz Y = k z = X 2 k 2 X = 0 k z = k 2 Y z Xz z = X 2 Y z z = Y 2 z = X 2 Y 2 4 z = f(x, y) (a, b) 2 z = Q(u, v) (0, 0) z = Q(u, v) (a, b) z = X 2 + Y 2 X 2 + Y 2 > 0 (0, 0) (a, b) z = X 2 Y 2 (X, Y ) (0, 0) X 2 Y 2 < 0 (0, 0)
14 2 14 z = X 2 Y 2 4: z = X 2 Y 2 (a, b) z = X 2 Y 2 (X, Y ) (0, 0) X > Y X 2 Y 2 > 0 X < Y X 2 Y 2 < 0 (0, 0) z = X 2 Y AC B 2 = 0 2 z = X 2 2 (0, 0) (0, 0) (0, Y ) 0 2 (0, Y ) (0, 0) (a, b) 2 (0, 0) 2 f(x, y) = x 2 + y 4, g(x, y) = x 2 + y 3 (0, 0) 2 x 2 f g (a, b) C 2 f(x, y) A = 2 f (a, b), x2 B = 2 f (a, b), y x C = 2 f (a, b) y2 (1) AC B 2 > 0 A > 0 C > 0 (a, b) (2) AC B 2 > 0 A < 0 C < 0 (a, b) (3) AC B 2 < 0 (a, b) (4) AC B 2 = 0 2 (a, b)
15 : 2 (a, b) (x, y) f(x, y) φ(x) = φ(a) + φ (a)(x a) + φ (a + θ(x a)) (x a) 2 0 < θ < 1 2 (a, b) f (x 0, y 0 ) (a, b) (x 0, y 0 ) 2 ( x y ) = ( a b ) + t ( x 0 a y 0 b t f(x, y) φ(t) φ(t) = f(a + t(x 0 a), b + t(y 0 b)) ) φ(0) = f(a, b), φ(1) = f(x 0, y 0 ) φ(t) t = 1, a = 0 φ(1) = φ(0) + φ (0)1 + φ (θ 1) 1 2 = φ(0) + φ (0) + φ (θ) 2 2 (5) θ (0, 1) φ (t) φ (t) = f x (a + t(x 0 a),b + t(y 0 b))(x 0 a) + f y (a + t(x 0 a), b + t(y 0 b))(y 0 b) φ (0) = f x (a, b)(x 0 a) + f y (a, b)(y 0 b) f C 2 φ (t) f xy = f yx φ (t) = 2 f x 2 (a + t(x 0 a), b + t(y 0 b))(x 0 a) f y x (a + t(x 0 a), b + t(y 0 b))(x 0 a)(y 0 b) + 2 f y 2 (a + t(x 0 a), b + t(y 0 b))(y 0 b) 2
16 (5) (x 0, y 0 ) (x, y) 2 f(x, y) = f(a, b)+ f f (a, b)(x a) + (a, b)(y b) x y f (a + θ(x a), b + θ(y b))(x a)2 2 x2 + 2 f (a + θ(x a), b + θ(y b))(x a)(y b) y x f (a + θ(x a), b + θ(y b))(y b)2 2 y2 (4) f(x, y) = x 2 + y 4 g(x, y) = x 2 + y 3 (1)(2)(3) (1) (a, b) C 2 f(x, y) A = 2 f (a, b), x2 B = 2 f (a, b), y x C = 2 f (a, b) y2 AC B 2 > 0 A > 0 (a, b) (x, y) θ A = 2 f (a + θ(x a), b + θ(y b)) x2 B = 2 f (a + θ(x a), b + θ(y b)) y x C = 2 f (a + θ(x a), b + θ(y b)) y2 (a, b) f x (a, b) = f y (a, b) = 0 f(x, y) = f(a, b) A (x a) 2 + B (x a)(y b) C (y b) 2 u = x a, v = y b 2 2 f C 2 f xx f xx (a, b) = A > 0 (x, y) (a, b) r 1 (a, b) r 1 (x, y) f xx (a + θ(x a), b + θ(y b)) = A > 0 A ( ) 2 A u 2 + 2B uv + C v 2 = A u + B A v + A C B 2 A v 2
17 2 17 f C 2 g(x, y) g(x, y) = f xx (x, y)f yy (x, y) (f xy (x, y)) 2 g(x, y) g(a, b) = f xx (a, b)f yy (a, b) (f xy (a, b)) 2 = AC B 2 > 0 (x, y) (a, b) r 2 (a, b) r 2 (x, y) g(a + θ(x a), b + θ(y b)) = A C B 2 > 0 A u 2 + 2B uv + C v 2 = ( A )) 2 (u + B A v + A C B 2 v (u, v) (0, 0) 2 (x, y) (a, b) (x, y) (a, b) r 1 r 2 r (a, b) r (x, y) f(x, y) = f(a, b) + > f(a, b) f(a, b) A (1) x2 9 y2 (2) 2x 2 2xy + 5y 2 6x + 12y (3) x 2 + xy + y x + y (4) x 3 9xy + y xy (5) xy(x 2 + y 2 1) (6) (x + y)e xy (7) (x 2 + y 2 )e x2 y (f xy ) 2 f xx f yy = 0 2 2
18 f(x, y) = (y x 2 )(y 2x 2 ) f f x (x, y) = 2x(3y 4x 2 ), f y (x, y) = 2y 3x 2 (0, 0) 2 f xx (x, y) = 6(4x 2 y), f xy (x, y) = 6x, f yy (x, y) = 2 (0, 0) f xx (0, 0) = f xy (0, 0) = 0, f yy (0, 0) = 2 f xx (0, 0)f yy (0, 0) (f xy (0, 0)) 2 = 0 (0, 0) z = f(x, y) f(0, 0) = 0 f(x, y) = 0 (x, y) f(x, y) = (y x 2 )(y 2x 2 ) = 0 y = x 2 y = 2x 2 f 0 x 2 < y < 2x 2 f(x, y) < 0 y < x 2 y > 2x 2 f(x, y) > 0 (0, 0) f(x, y) f(0, 0) = 0 y O x 5: f(x, y) y = x 2 y = 2x 2 0 f(x, y) (0, 0) (0, 0)
19 2 19 (5) x y 0 2. f(x, y) (a, b) z = f(x, y) (a, b) z (a, b) (a, b) x = at, y = bt f(x, y) f(at, bt) = (bt a 2 t 2 )(bt 2a 2 t 2 ) = t 2 (b a 2 t)(b 2a 2 t) t = 0 a, b f(at, bt) b > 0 b/2a 2 < t < b/2a 2 (b a 2 t)(b 2a 2 t) > 0 b = 0 f(at, 0) = 2a 4 t 4
20 (1) X = x/3, Y = y/5 X 2 Y 2 (0, 0) (2) f f x (x, y) = 4x 2y 6 f y (x, y) = 2x + 10y + 12 (1, 1) f x 1 y + 1 f(x, y) = 2(x 1) 2 2(x 1)(y + 1) + 5(y + 1) = 9 > 0, 2 > 0 (1, 1) 7 2 (3) f f x (x, y) = 2x + y 3 x 2 f y (x, y) = x + 2y 3 y 2 (1, 1) 2 f xx (x, y) = x 3 f xy (x, y) = 1 f yy (x, y) = y 3 f xx (1, 1) = f yy (1, 1) = 8, f xy (1, 1) = f yx (1, 1) = 1 f xx (1, 1)f yy (1, 1) f xy (1, 1) 2 = 63 > 0 f xx (1, 1) > 0 (1, 1) 9 (4) f f x (x, y) = 3x 2 9y f y (x, y) = 9x + 3y 2 (0, 0) (3, 3) 2 f xx (x, y) = 6x f xy (x, y) = 9 f xy (x, y) = 6y f xx (0, 0) = f yy (0, 0) = 0, f xy (0, 0) = f yx (0, 0) = 9 f xx (0, 0)f yy (0, 0) f xy (0, 0) 2 = 81 < 0 (0, 0) f xx (3, 3) = f yy (3, 3) = 18, f xy (3, 3) = f yx (3, 3) = 9 f xx (3, 3)f yy (3, 3) f xy (3, 3) 2 = 243 > 0 f xx (3, 3) > 0 (3, 3) 0
21 2 21 (5) f f x (x, y) = y(3x 2 + y 2 1) f y (x, y) = x(x 2 + 3y 2 1) (0, 0), (±1, 0), (0, ±1), (±1/2, ±1/2) 9 2 f xx (x, y) = 6xy f xy (x, y) = 3x 2 + 3y 2 1 f yy (x, y) = 6xy f xx (0, 0) = f yy (0, 0) = 0, f xy (0, 0) = f yx (0, 0) = 1 f xx f yy fxy 2 = 1 < 0 (0, 0) (±1, 0), (0, ±1) 4 f xx = f yy = 0, f xy = f yx = 2 f xx f yy fxy 2 = 4 < 0 ±(1/2, 1/2) f xx = f yy = 3/2, f xy = f yx = 1/2 f xx f yy fxy 2 = 2 > 0 f xx > 0 1/8 ±(1/2, 1/2) f xx = f yy = 3/2, f xy = f yx = 1/2 f xx f yy fxy 2 = 2 > 0 f xx < 0 1/8 (5) 9 f(x, y) 1 x, y 2 x 2 + y 2 1 f(x, y) = 0 x y f(x, y) 6 (0, 0), (±1, 0), (0, ±1) ( 1 2, 1 2 ) ( 1 2, 1 2 ) ( 1 2, 1 2 ) ( 1 2, 1 2 ) (6) f f x (x, y) = (1 xy y 2 )e xy f y (x, y) = (1 x 2 xy)e xy ±(1/ 2, 1/ 2) 2 f xx (x, y) = (xy 2 + y 3 2y)e xy f xy (x, y) = (x 2 y + xy 2 2x 2y)e xy f yy (x, y) = (x 3 + x 2 y 2x)e xy
22 2 22 y 0 x 6: (5) 2 f xx = f yy = 1/ 2e, f xy = f yx = 3/ 2e f xx f yy fxy 2 = 1 2e 9 2e = 4 e < 0 (7) f f x (x, y) = 2x(1 + x 2 + y 2 )e x2 y 2 f y (x, y) = 2y(1 x 2 y 2 )e x2 y 2 (0, 0), (0, ±1) 2 f xx (x, y) = 2(1 + 3x 2 + y 2 + 2x 4 + 2x 2 y 2 )e x2 y 2 f xy (x, y) = 4xy(x 2 + y 2 )e x2 y 2 f yy (x, y) = 2(1 x 2 3y 2 + 2x 2 y 2 + 2y 4 )e x2 y 2 f xx (0, 0) = f yy (0, 0) = 2, f xy (0, 0) = f yx (0, 0) = 0 f xx (0, 0)f yy (0, 0) f xy (0, 0) 2 = 4 > 0 f xx (0, 0) > 0 (0, 0) 0 (0, ±1) f xx = 4/e, f xy = f yx = 0, f yy = 4/e f xx f yy f 2 xy = 16 e 2 < 0
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,
9.. x + y + 0. x,y, x,y, x r cos θ y r sin θ xy x y x,y 0,0 4. x, y 0, 0, r 0. xy x + y r 0 r cos θ sin θ r cos θ sin θ θ 4 y mx x, y 0, 0 x 0. x,y 0,0 x x + y x 0 x x + mx + m m x r cos θ 5 x, y 0, 0,
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(,,
() 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 ()
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
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
009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n
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
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
01 4 17 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
(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)
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 +
( ) x y f(x, y) = ax
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
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II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) 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 x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )
.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(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
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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
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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
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 ),
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9 203 6 7 WWW http://www.math.meiji.ac.jp/~mk/lectue/tahensuu-203/ 2 8 8 7. 7 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,
ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D
1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00
III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2 lim. (x,y) (1,0) x 2 + y 2 lim (x,y) (0,0) lim (x,y) (0,0) lim (x,y) (0,0) 5x 2 y x 2 + y 2. xy x2 + y
III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2. (x,y) (1,0) x 2 + y 2 5x 2 y x 2 + y 2. xy x2 + y 2. 2x + y 3 x 2 + y 2 + 5. sin(x 2 + y 2 ). x 2 + y 2 sin(x 2 y + xy 2 ). xy (i) (ii) (iii) 2xy x 2 +
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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
1/5 ( ) Taylor ( 7.1) (x, y) f(x, y) f(x, y) x + y, xy, e x y,... 1 R {(x, y) x, y R} f(x, y) x y,xy e y log x,... R {(x, y, z) (x, y),z f(x, y)} R 3 z 1 (x + y ) z ax + by + c x 1 z ax + by + c y x +
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II 16 16.0 2 1 15 x α 16 x n 1 17 (x α) 2 16.1 16.1.1 2 x P (x) P (x) = 3x 3 4x + 4 369 Q(x) = x 4 ax + b ( ) 1 P (x) x Q(x) x P (x) x P (x) x = a P (a) P (x) = x 3 7x + 4 P (2) = 2 3 7 2 + 4 = 8 14 +
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
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
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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,
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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)
CALCULUS II (Hiroshi SUZUKI ) 16 1 1 1.1 1.1 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) lim f(x, y) = lim f(x, y) = lim f(x, y) = c. x a, y b
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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)
(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )
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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
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
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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)
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
2009 9 6 16 7 1 7.1 1 1 1 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(cos y y sin y) y dy 1 sin
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
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
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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
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