五 4. 方向導數 : 梯度

五 _4. 方向導數 _: 梯度

4.3 因為曲線f (x, y) = C為_{f (x, y)}過(x₀, y₀)的等高線_,

所以由定理_4.1(2)知_{,∇f (x0, y0)}與過_{(x0, y0)}點的等高線垂直_, 令_{(x, y)}為曲線f (x, y) = C過_{(x0, y0)}的切線上之一點_, 所以

∇f (x₀, y₀) · (x − x₀, y − y₀) = 0 ⇒ (^∂f_∂x|_(x0,y0),^∂f_∂y|_(x0,y0)) · (x − x₀, y − y₀) = 0

⇒ ^∂f_∂x|_(x0,y0)(x − x₀) + ^∂f_∂y|_(x0,y0)(y − y₀) = 0

4.7 (1)

∂f

∂x = ye^xy,^∂f_∂y = xe^xy 令_{~u =} ^√1

2(1, 1) = (^√1

2,^√1

2)(~u要單位長), 所以

∂f

∂~u|_(0,0) = ∇f (0, 0) · (^√1

2,^√1

2) = (0, 0) · (^√1

2,^√1

2) = 0

(3)

∂f

∂x = ¹

1+^x2

y2

(_y¹) = _y2+x^{y 2},^∂f_∂y = ¹

1+^x2

y2

(^−x_y2) = _y2^−x+x²

令_{~u =} ¹

2(1,√

3) = (¹₂,

√3

2 ),所以

∂f

∂~u|_(1,1) = ∇f (1, 1) · (¹₂,

√ 3

2 ) = (¹₂,⁻¹₂ ) · (¹₂,

√ 3

2 ) = ¹₄ −

√ 3 4 = ¹⁻

√ 3 4

(4)

∂f

∂x = _x2+y^{1 4}(2x) = _x2^2x+y⁴,^∂f_∂y = _x2+y^{1 4}(4y³) = _x2^4y+y³⁴

令~u = ^√1

5(−2, 1) = (^−2√

5,^√1

5), 所以

∂f

∂~u|_(1,0) = ∇f (1, 0) · (^−2√

5,^√1

5) = (2, 0) · (^√−2

5,^√1

5) = ^−4√

5

(6)

∂f

∂x = z cos xz,^∂f_∂y = ¹_z,^∂f_∂z = ^−y_z2 + x cos xz 令_{~u =} ^√1

14(2, 1, 3) = (^√2

14,^√1

14,^√3

14), 所以

∂f

∂~u|_(π,2,2) = ∇f (π, 2, 2) · (^√2

14,^√1

14,^√3

14) = (2,¹₂, −¹₂ + π) · (^√2

14,^√1

14,^√3

14) = ^3+3π√

14

4.8 令∇f (a, b) = (x, y), ~u = ^√1₂(1, 1) = (^√1₂,^√1₂), ~v = ^√1₂(−1, 1) = (^√−1₂,^√1₂)

∂f

∂~u = 1 ⇒ ∇f · (^√1

2,^√1

2) = 1 ⇒ ^√1

2x + ^√1

2y = 1 · · · (1)

∂f

∂~v = 2 ⇒ ∇f · (^−1√

2,^√1

2) = 2 ⇒ ^−1√

2x + ^√1

2y = 2 · · · (2) (1)(2) ⇒ x = ⁻

√2 2 , y = ³

√2

2 , 所以∇f (a, b) = (⁻

√2 2 ,³

√2 2 )

1

4.9 (3)

∂f

∂x = y cos xy,^∂f_∂y = x cos xy + z cos yz,^∂f_∂z = y cos yz 所以過點_(π,¹

6, π)切面方程式為

∇f (π,¹₆, π) · (x − π, y −¹₆, z − π) = 0

⇒ (¹₆ cos^π₆, 2π cos^π₆,¹₆cos^π₆) · (x − π, y − ¹₆, z − π) = 0 ⇒ x + 12πy + z = 4π

(4)

∂f

∂x = _1+x2¹y²z²(yz) = _1+x^yz2y²z²,^∂f_∂y = _1+x^xz2y²z²,^∂f_∂z = _1+x^xy2y²z²

所以過點_(2,¹

3,³₂)切面方程式為

∇f (2,¹₃,³₂) · (x − 2, y − ¹₃, z −³₂) = 0

⇒ (¹₄,³₂,¹₃) · (x − 2, y − ¹₃, z − ³₂) = 0 ⇒ 3x + 18y + 4z = 18

2