西南大学2020级《高等数学IB》英文课程考核试卷C及答案

西南大学计算机与信息科学学院
《 高等数学IB 》课程试题 【C 】卷
阅卷须知：阅卷用红色墨水笔书写，得分用阿拉伯数字写在每小题题号前，用正分表示，不得分则在题号前写0；大题得分登录在对应的分数框内；统一命题的课程应集体阅卷，流水作业；阅卷后要进行复核，发现漏评、漏记或总分统计错误应及时更正；对评定分数或统分记录进行修改时，修改人必须签名。

PLEASE ANSWER IN CHINESE OR IN ENGLISH OR BILINGUALISM!!
1. Fill the correct answer in the blanks (3 points each ，15 points in all)
(1)The general solution to the differential equation )0(1
12d d >-=+x x
y x y x is __________
________ .
(2)The convergence set of 1
211)
12()1(+∞
=-∑--n n n x
n n is _______________ __ . (3))sec(xy z =, so the total differential dz=_____________ ___ .
(4)The tangent plane of the surface x y z arctan = at the point )4
,1,1(π
is . ____________________ __ __
(5)Reversing the order of integration:
⎰⎰⎰⎰-+2
120
1
00),(),(x
x
dy y x f dx dy y x f dx ＝_______ __ ________ __ __.
特别提醒：学生必须遵守课程考核纪律，违规者将受到严肃处
2. Choose the corresponding letter of the best answer that best completes the statement or answers the question among A, B, C, and D, and fill in the blanks (3 points each ，15 points in all).
(6).)1(1
2
∑∞
=-n n n
e n is absolutely converges. converges.
(A) absolutely converges (B) diverges.
(C) conditionally converges. (D) All of the above are wrong
(7)The derivative of z xy x z y x f --=2
3
),,( at the point (1,1,0) in the direction of v=2i-3j+6k is () (A)
21 (B) 73 (C) 74 (D) 5
4 (8)At the point (0, 0), For the description of the function xy y x f =),( , the following
is correct. （ ）
(A)It has no partial derivatives.
(B)It has partial derivatives but is not differentiable.
(C)It is differentiable but the partial derivatives are not continuous. (D)The partial derivatives are continuous.
(9)The principal unit normal for the circular motion j t i t t r )2(sin )2(cos )(+= is ( )
(A) j t i t )2(sin )2(cos +- (B) j t i t )2(sin )2(cos - (C) j t i t )2(sin )2(cos -- (D) j t i t )2(sin )2(cos + (10)The curl of xyk j xe i z x F z
++-=)(2
is
(A)k e j y i e x z
z
-+-+)1()1( (B) k e j y i e y z
z
+---)1()1( (C) k xe j x i e x z
z
++--)1()1( (D) k e j y i e x z
z
++--)1()1(
3. Find the solutions for following problems by computing (8 points each ，40 points
in all)
(11)Show that 222)0,0(),(lim y x xy x y x ++→ does not exist.
（12）Find y x z
∂∂∂2 if 01=-+-z xy e z .
(13)
2(2sin 19)D
x x y d σ+++⎰⎰, where D is 221x y +≤.
(14)⎰--+L
y
dx y dy e
x )2
1
()(sin , where L is consisted of line segment AB a nd arc ⋂BC .
1:=+y x AB , from )0,1(A to )1,0(B . )0(12≤-=⋂
x x y BC ： , from )1,0(B to
)0,1(-C .
(15)⎰⎰++
∑dS z y x )342(, where ∑is the part of plane 14
32=++z
y x in the first octant.
4. Solve the following comprehensive problems (10 points each ，30 points in all)
(16)Find the point on the paraboloid 2
2
y x z += that is closest to (0,0,2). What is the
minimum distant?
(17)Find the sum for ∑∞
=--11
21
2n n n x .
(18)⎰⎰∑
dxdy z 2
, where
∑ is the part of the sphere )0(222>--=a y x a z t hat is
inside the cylinder ax y x =+22, taking n
to be upward normal.
西南大学计算机与信息科学学院
《 高等数学IB 》课程试题 【C 】卷参考答案和评分标准
阅卷须知：阅卷用红色墨水笔书写，得分用阿拉伯数字写在每小题题号前，用正分表示，不得分则在题号前写0；大题得分登录在对应的分数框内；统一命题的课程应集体阅卷，流水作业；阅卷后要进行复核，发现漏评、漏记或总分统计错误应及时更正；对评定分数或统分记录进行修改时，修改人必须签名。

PLEASE ANSWER IN CHINESE OR IN ENGLISH OR BILINGUALISM!!
4. Fill the correct answer in the blanks (3 points each ，15 points in all)
(1)
2121x C x y +-=
(2)]1,1[-.
(3))()tan()sec(xdy ydx xy xy dz +⋅⋅=
(4)22π
=+-z y x
(5)
⎰⎰
-10
2),(y
y
dx y x f dy
5. Choose the corresponding letter of the best answer that best completes the statement or answers the question among A, B, C, and D, and fill in the blanks (3 points each ，15 points in all). (6)A(7)C(8)B(9)C(10)D
6. Find the solutions for following problems by computing (8 points each ，40 points in all)
特别提醒：学生必须遵守课程考核纪律，违规者将受到严肃处
(11)Show that 2
22)0,0(),(lim y x xy
x y x ++→ does not exist.
[Solution] 00lim lim 2022
20
0==++→→=y
y x xy x y y x , and 1lim lim 22
022200==++→→=x x y x xy x x x y . Thus, we get different values depending on how )0,0(),(→y x .
Therefore, 2
22)0,0(),(lim y x xy
x y x ++→ does not exist.
（12）Find y
x z
∂∂∂2 if 01=-+-z xy e z .
Differentiate both side with respect to x , we obtain:0=∂∂+-∂∂x
z y x z e z
1
+=∂∂⇒
z e y
x z Differentiate both side with respect to y , we obtain:0=∂∂+-∂∂y
z x y z e z
1
+=∂∂⇒
z e x y z )1
()(2+∂∂=∂∂∂∂=∂∂∂z e y
y x z y y x z 2
)1()1(+∂∂-+=z z
z e y
z ye e 3
2)
1()1(+-+=z z
z e xye e (13)
2(2sin 19)D
x x y d σ+++⎰⎰, where D is 221x y +≤. 2
22
(2sin 19)(19)19D D
D
x x y d x d x d σσσπ+++=
+=+⎰⎰⎰⎰⎰⎰对称性
21
220
77cos 1919=
4
4d d π
π
π
θρθρρππ=⋅+=
+⎰⎰
4 points
4 points
2 points
2 points
4 points
2 points
(14)⎰--+L
y
dx y dy e
x )2
1
()(sin , where L is consisted of line segment AB and arc ⋂BC .
1:=+y x AB , from )0,1(A to )1,0(B . )0(12≤-=⋂
x x y BC ：, from )1,0(B to
)0,1(-C .
[Solution] Let )21
(),(--=y y x P , and y e x y x Q sin ),(+= . Then 1-=∂∂y P , and 1=∂∂x
Q
. By Green’s Theorem, ⎰
⋂
--+CA
BC AB y dx y dy e x )21
()(sin ⎰⎰=D
dxdy 2, where D is
the region bounded
by AB , ⋂
BC , and CA .
⎰--+L
y
dx y dy e x )21()(sin ⎰
⋂
--+=
CA
BC AB y dx
y dy e x )2
1
()(sin ⎰--+-
CA
y
dx y dy e x )21()(sin
⎰
⋂
--+CA
BC AB y dx y dy e x )21()(sin ⎰⎰=D
dxdy 221)421(2π
π+=+⋅=
0:=y CA ,
from
1-=x to 1=x .
⎰⎰==--+-CA
y
dx dx y dy e x 121
)21
()(11sin
2121)21()(sin π
π=-+=--+∴⎰L
y dx y dy e x
(15)
⎰⎰++
∑dS z y x )342( , where ∑ i s the part of plane 14
32=++z
y x in the first 2 points 4 points
4 points
4 points
octant.
[Solution] 1432:=++z y x ∑ . 4342=++∴z y
x
⎰⎰⎰⎰=++∑∑
dS dS z y x 4)342( 3424:y x z --= ∑ , 202
330:),(≤≤-≤≤∈x x y D y x ; .
dxdy dxdy z z dS y x 3
61
12
2=
'+'+= dxdy z z dS z y x D
y x ⎰⎰⎰⎰'+'+=++
2
214)342(∑⎰⎰=D dxdy 3614614=
4. Solve the following comprehensive problems (10 points each ，30 points in all)
(16)Find the point on the paraboloid 2
2y x z += that is closest to (0,0,2). What is the
minimum distant?
[Solution] Let ),,(z y x P be any point on the paraboloid , the square of the distant between (0,0,2)
and P is 222(2)x y z ++-. The
problem
is
to
find
the
minimum
value
of
222(,,)(2)f x y z x y z =++-subject to the constraint 22y x z +=.
Let 22222(,,,)(2)()F x y z x y z z x y λλ=++-+--.
Solving 222202202(2)00
x y
z F x x F y y F z F z x y λ
λλλ=-=⎧⎪=-=⎪⎨
=-+=⎪⎪=--=⎩,
2 points 2 points
4 points
2 points
we have 11132y x z ⎧==⎪⎪⎨⎪=⎪⎩
，22232
y x z ⎧==⎪⎪⎨⎪=⎪⎩，and 33300
y x z ==⎧⎨=⎩.
1112227(,,)(,,)4f x y z f x y z ==
.
333(,,)2
f x y z =
The point on the paraboloid 22y x z
+= that is closest to (0,0,2) is
3()2. The
.
(17)Find the sum for ∑∞
=--11
21
2n n n x .
[Solution] Let 121
2-=-n x u n n . 2211212lim
lim x x n n u u n n n n =⋅+-=∞→+∞→ . So, ∑∞
=--1
1
212n n n x converges when 1<x .
When 1-=x the series,
∑∞
=--1
121
n n , diverges. When 1=x the series, ∑
∞
=-11
21
n n , diverges.
Thus, the convergence set of
∑
∞
=--11
21
2n n n x is )1,1( - . Let
)1,1(1
2)(11
2-∈-=∑∞
=-x n x x s n n .
)1111(2111)12()12()(2
212
2112112x x x x x n x n x x s n n n n n n n n ++-=-==='-='-='∑∑∑∑∞=∞=-∞=-∞
=- ⎰⎰++-='=-x x
dt t
t dt t s s x s 00)1111(21
)()0()(ln
21(= 2 points
4 points
3points
0)0(=s , )(12112x s n x n n =-∴∑∞
=-x x
-+=11ln
21 )1,1(-∈x .
(18)⎰⎰∑
dxdy z 2
, where
∑ is the part of the sphere )0(222>--=a y x a z t hat is
inside the cylinder ax y x =+22, taking n
to be upward normal.
ax y x D y x y x a z ≤+∈--=22222:),(: , ∑, 0cos >γ.
⎰⎰⎰⎰≤+--=
ax
y x dxdy
y x a dxdy z 22)(2
222
∑
⎰⎰⋅-=
-θ
π
πρρρθcos 0
2222)(a d a d 极坐标
⎰--=22cos 0
422)
4121(ππθ
θρρd a a ⎰-=2
0424)cos cos 2(21πθθθd a
⎰+-+=202
4]4)2cos 1(2cos 1[21πθθθd a ⎰+-+=204)8
4cos 122cos 43(21πθθ
θd a 3254a π=
2points
2points
3points
3points
2points
2points。