4 ) 沿著向量 ~u = (1, 3) 的方向導數。

(1)

微乙小考二 (2018/3/29)

1. (5%) 求出 f (x, y) = sin(xy) + cos(x

²

y) 在點 (1, π

4 ) 沿著向量 ~u = (1, 3) 的方向導數。

sol: ∇f (x, y) =

cos(xy) · y − sin(x

²

y) · 2xy, cos(xy) · x − sin(x

²

y) · x

²

∇f (1, π 4 ) =

√ 2 2 · π

4 −

√ 2 2 · π

2 ,

√ 2 2 −

√ 2 2

=

−

√ 2 8 π, 0

Take the unit vector of ~ u ⇒ ~ u

|~ u| =

1 √ 10 , 3

√ 10

∂f

∂~ u (1, π 4 ) =

−

√ 2 8 π, 0

·

1 √ 10 , 3

√ 10

= −π 8 √

5 2. (5%) 求曲面 xyz

²

= 6 在點 (3, 2, 1) 的切平面方程式。

sol: The tangent plane Γ at (3,2,1) can be derived from

f

_x

(3, 2, 1)(x − 3) + f

_y

(3, 2, 1)(y − 2) + f

_z

(3, 2, 1)(z − 1) = 0

Let f (x, y, z) = xyz

²

− 6 = 0

f

x

= ∂f

∂x (x, y, z) = yz

²

f

x

(3, 2, 1) = 2

f

y

= ∂f

∂y (x, y, z) = xz

²

f

y

(3, 2, 1) = 3

f

_z

= ∂f

∂z (x, y, z) = 2xyz f

_z

(3, 2, 1) = 12

Therefore, we have Γ : 2(x − 3) + 3(y − 2) + 12(z − 1) = 0. Namely, 2x + 3y + 12z = 24.

3. (10%) 找出函數 f (x, y) = e

^y

(y

²

− x

²

) 的候選點，並決定它是局部極大值，局部極小值或是鞍點。

sol:

∇f (x, y) = 0 ⇒

( f

_x

(x, y) = −2xe

^y

= 0 (1)

f

y

(x, y) = e

^y

(y

²

− x

²

) + 2ye

^y

= 0 (2) Solving (1) and (2), we do have y(y + 2) = 0. There are two critical points (0, 0) and (0, −2)

Check these points further by calculate D(x) =

f

xx

f

xy

f

yx

f

yy

=

−2e

^y

(−2x)e

^y

(−2x)e

^y

e

^y

(y

²

+ 4y + 2 − x

²

)

∗ For (0, 0) D(0, 0) =

−2 0

0 2

= −4 < 0. Therefore, (0,0) is a saddle point.

∗ For (0, −2) D(0, −2) =

−2e

⁻²

0 0 −2e

⁻²

= 4e

⁻⁴

> 0 and ∂

²

f

∂x

²

_(0.−2)

= −2e

⁻²

< 0.

Hence (0, −2) is a local maximum.

1

4 ) 沿著向量 ~u = (1, 3) 的方向導數。

1. (5%) 求出 f (x, y) = sin(xy) + cos(x

y) 在點 (1, π