大一经济数学微积分06-07(2)_B__answer

∇f =< 2 xy + 1, x 2 + 1,1 > . (2’) At the point P0 (1,0, −1) . ∇f =< 1, 2,1 > . (2’) The equation for the tangent plane is ( x − 1) + 2 y + ( z + 1) = 0 . (2’) The parametric equations for the normal line is x = 1 + t , y = 2t , z = −1 + t (2’)
=∫
1 0
1
∫
2− x
x
( x + 3)dydx (4’)
= ∫ (3 + (4’) ============================ 12. (14pts) Solve differential equations.
(a)
x y ' = e y+ x , x > 0 ;
∫∫
0
8
2
x
3
f ( x, y )dydx
Solution: The graph of the region (3’) y-limits of integration (2’) x-limits of integration (2’)
∫∫
0
2
y3
0
f ( x, y ) dxdy (1’)
==================================== 10 (8pts) Change the Cartesian integral into an equivalent polar integral. Then, evaluate the polar integral.
sin[ ( x 2 + y 2 )]dxdy 2 Solutions: x = r cos θ , y = r sin θ (1’) dxdy → rdrdθ (1’)
0 0
∫∫
1
1− y 2
π
∫ ∫
0
π /2 1
0
sin
π r2
2
drdθ (4’)
第 4 页（共 4 页）
1 ⎡ π r2 ⎤ 1 (2’) ⎢ cos ⎥ = π⎣ 2 ⎦0 π ======================================= 11. (8pts) Find the mass of a thin triangular plate bounded by the y-axis and the lines y = x and y = 2 − x if the density δ ( x, y ) = x + 3 . Solution: =−
======================================== 6. (6pts) Find the critical point of the function f ( x, y ) = x 2 − 2 xy + 2 y 2 − 2 x + 2 y + 1 , and determine if a local extremum occurs at the critical point. Solution: f x = 2 x − 2 y − 2 = 0 , f y = −2 x + 4 y + 2 = 0 , (1’) critical point is (1,0) (2’) f xx = 2, f yy = 4, f xy = −2 ; f xx f yy − ( f xy ) 2 = 4 > 0 , (2’) At the critical point is a saddle point, the function has a local minimum. (1’) ====================================== 7. (12pts) Find the absolute maximum and absolute minimum of the function f ( x, y ) = 6 xy − 4 x3 − 3 y 2 on the rectangular plate 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 . Solution: f x = 6 y − 12 x 2 = 0 , f y = 6 x − 6 y = 0 . The critical points are (0,0)
1 1 and ( , ) . (3’) 2 2
第 3 页（共 4 页）
1 1 f (0,0) = 0 , f ( , ) = 1/ 4 . (1’) 2 2 On the line x = 0,0 ≤ y ≤ 1 , f (0, y ) = −3 y 2 , min= −3 at (0,1) , max=0 at (0,0) ; (1’)
第 2 页（共 4 页） 4. (6pts) Let f ( x, y ) = e (a) Find
2 y − x2
2
.
∂f ∂f ∂ f , and ; (b) Find the linearization of f at (1,1) . ∂x ∂y ∂x∂y 2 ∂f = −2 xe 2 y − x , (1’) Solution: (a) ∂x 2 ∂f = 2e 2 y − x (1’) ∂y
2 ∂2 f = −4 xye 2 y − x (1’) ∂x∂y (b) L( x, y ) = e − 2e( x − 1) + 2e( y − 1) (3’) =============================
5. (8pts) Find the equations for the tangent plane and the normal line of the surface x2 y + x + y + z = 0 at the point P0 (1,0, −1) Solution: Let f ( x, y , z ) = x 2 y + x + y + z
dy e x = ey dx x
(b) xy '+ y = − sin x , x > 0 , y ( ) = 0 . 2 (2’)
π
Solutions: (a)
∫
e − y dy = 2∫ e x d x (2’)
−e − y = 2e x + C (2’) 1 sin x (b) y '+ y = − (2’) x x v( x) = e ∫ =x (1) 1 y = [ ∫ (− sin x)dx + C ] x 1 = (cos x + C ) (3’) x
On the line y = 0,0 ≤ x ≤ 1 , f ( x,0) = −4 x3 , min= −4 at (1,0) , max=0 at (0,0) ; (1’) On the line x = 1,0 ≤ y ≤ 1 , f (1, y ) = 6 y − 4 − 3 y 2 , min= −4 at (1,0) , max= −1 at (1,1) . (2’) On the line y = 1,0 ≤ x ≤ 1 , f ( x,1) = 6 x − 4 x 3 − 3 , max= 2 2 − 3 at (1/ 2,1) , min= −1 at (1,1) (2’) The abs. max value is 1/4, and the abs. min value is −4 . (2’) ================================= 8. (8pts) Use the method of Lagrange multipliers to find the points on the sphere x 2 + y 2 + z 2 = 16 where f ( x, y , z ) = x + y + 2 z has its maximum and minimum values.
南 京 航 空 航 天 大 学
第 1 页 （共 4 页）
二○○六
～ 二○○七
学年
第 一
学期
课程名称： 《 命题教师：
Calculus
》参考答案及评分标准
试卷类型： B 试卷代号：
曹荣美
1. (6pts) Let v = 4 j − 3k , u = i + j + k . (a) Find v ⋅ u , v , and u . (b) Find the cosine of the angle between v and u . (c) Find the vector proj v u . Solution: (a) v ⋅ u =< 0, 4, −3 >i< 1,1,1 >= 1 , v = 5 , u = 3 (2’) (2’) 5 3 u ⋅ v v 1 4j − 3k 4 3 (c) = ⋅ = = j − k (2’) v v 5 5 25 25 (b) cos θ =
(2’)
Solve the system, we obtain the critical points (2, 2, 2 2) and. (−2, −2, −2 2) (2’)
f (2, 2, 2 2) = 8 and. f (2, 2, 2 2) = −8 Max=8, min=-8 (2’) ===================================== 9. (10pts) Sketch the region of integration and write an equivalent double integral with the order of integration reversed.