미적분학 해설
미적분학 해설서
김동화,이성종,이영회,진영훈 지음
미적분학 해설서
21 개정판
김동화, 이성종, 이영회, 진영훈
2022/03/03
서문
지은이: 수학과 17 김동화, 17 이성종, 17 이영회, 18 진영훈.
17 김동화의 말 : 일반수학은 모든 이공계열 학생들이 기본으로 배우는 학문임과 동시에 다른 학문을 익히기 위해서
필수적인 학문입니다. 이 해설서를 통하여 많은 학생들이 기초를 다지는 데에 큰 도움이 되었으면 하며, 공부에 자신감을
더해줄 수 있었으면 합니다. 기초가 튼튼해야 심화된 전공 개념을 배울 때에도 더 쉽게 이해할 수 있기에, 수강생 여러분
들은 일반수학 과목을 꼭 열심히 수강하시길 바랍니다.
17 이성종의 말 : 2020년을 맞아 일반수학 교재가 새롭게 집필되었다는 소식을 처음 들었을 때, 저는 순전히 호기심
으로 책을 사서 구매하였습니다. 하지만 아직 연습문제의 답이 없는 것을 발견, ’ 내가 한 번 해설을 만들어보면 어떨까?’
라고 생각하여 이 해설서를 작성하게 되었습니다. 저희 집필진들은 평소에도 같이 전공을 공부하는 스터디메이트들인데,
LATEX를 이용하여 풀이를 만들어놓는 공부를 자주 하곤 했습니다. 그래서 이렇게 일반수학 해설서가 만들어진 것이죠.
비록 아직 학부생들이 만든 풀이에 불과하기 때문에, 계산 실수도 잦고 개념 오류도 군데군데 있으리라 짐작합니다. 답
들을 너무 맹신하지는 마시고 그저 참고서로만 사용하시길 바랍니다.
17 이영회의 말 : 새 책이 나온다는 소식을 듣자마자 뜻 맞는 동기들과 모여 해설서를 만들기 시작했습니다. 신입생들,
그리고 미적분학 책을 가지고 공부할 학우들에게 도움이 됐으면 하는 바람을 가지고 집필하였습니다. 물론 이런 비교적
사소한? 동기와는 다르게 많은 시간과 노력을 들였습니다. 그렇기에 이 해설서가 많은 친구들에게 큰 도움이 되었으면
합니다.
공동작업을 했기 때문에 각자의 스타일이 달라 검토를 많이 했음에도 불구하고 오타나 오류가 있을 수 있습니다. 너
그러운 마음으로 이해하여 주시고 만약 오탈자나 다른 좋은 의견이 있다면 밑에 적어둔 이메일로 알려주신다면 아주 큰
도움이 될 것 같습니다.
특히 수학과 학우분들은 증명 문제에 대해 해설서에 적혀있는 내용뿐만 아니라 다른 풀이는 없는지 다양하게 생각하
는 습관을 가졌으면 합니다. 미적분학은 앞으로 전공을 공부할 때 가장 기본적으로 쓰이는 기둥입니다. 기둥을 단단히
해야 그만큼 더 앞으로 나아갈 수 있다고 생각합니다. 보잘것없어 보이는 이 해설서가 그래도 수학과 학우분들께는 큰
동기부여가 되었으면 하고 바랍니다.
18 진영훈의 말: 혼자 공부할 때 또 주변에 물어볼 사람이 없을 때 이 해설서를 잘 활용하시면 조금이라도 도움을 받을
수 있을 것입니다. 저는 이 해설서가 단지 답만 보는 용도가 아닌 적힌 풀이들을 이해하면서 여러분들의 수학적 사고력을
향상할 수 있는 수단으로 활용되었으면 좋겠습니다. 비록 저희들의 풀이가 완벽하고 깔끔한 풀이는 아닐 수 있지만, 개
정을 거듭하면서 더 좋은 풀이들로 발전해나가겠습니다. 일반수학 열심히 공부하시고 좋은 성적 받으시길 바랍니다.
이 해설서는 2020년도 정규 일반수학 강의가 끝나기도 전에 만들어진 해설서이기 때문에 문제에 접근하는 관점이 배
우실 것들과 다를 수 있음을 알려드립니다. 참고용으로만 사용하여 주시고, 오탈자나 틀린 풀이, 혹은 다른 좋은 의견이
있다면 이성종 : lovekrandleesj1004@gmail.com 로 메일을 남겨주신다면 개정 시 적극적으로 반영하겠습니다.
해설서 작성을 흔쾌히 허락해주신 미적분학 집필진 수학과 교수님 세 분께 감사의 말을 드립니다. 좋은 교재를 집필
해주시고, 좋은 공부 기회를 주심에 다시 한번 감사의 말을 드립니다. 또 이 해설서로 공부하실 독자분들에게도 감사의
말을 드립니다.
ii
SOLUTION
CONTENTS
CONTENTS
Contents
1 —i¸ h⇠
4
2 ˘\
10
3 ¯Ñ
16
4
Ñ
36
5 4\ ⇠
48
6 q ⇠@ L|Ï
55
¨
7 å\ƒ
76
8 °0@ â,
83
9 ıå⇠
96
10 ·
99
11 ‰¿⇠h⇠@ ¯Ñ
106
12 ∏¯ÑX \©
122
13 ‰⌘ Ñ
137
14 ·t
152
15 °0•
162
16 °0•X
172
Ñ ¨
3
SOLUTION
1
—i¸ h⇠
¯¨t ‰L¸ ⇡‰.
p
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F
µ8⌧ 1.1 .
1. ‰L Ö⌧ 8ÑD Ùt‹$.
a) p ! (q ! r)¸ p ^ q ! r @ ŸXt‰.
proof ) P : p ! (q ! r), Q : p ^ q ! r \ ì‡ ƒ¨\| ¯¨t
‰L¸ ⇡‰.
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
q!r
T
F
T
T
T
F
T
T
p^q
T
T
F
F
F
F
F
F
P
T
F
T
T
T
T
T
T
Q
T
F
T
T
T
T
T
T
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
p!q
T
T
F
F
T
T
T
T
p!r
T
F
T
F
T
T
T
T
q^r
T
F
F
F
T
F
F
F
P
T
F
F
F
T
T
T
T
q
T
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F
r
T
T
F
F
T
T
F
F
T
T
F
F
T
T
F
F
s
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
p!r
T
T
F
F
T
T
F
F
T
T
T
T
T
T
T
T
q!s
T
F
T
F
T
T
T
T
T
F
T
F
T
T
T
T
p^q
T
T
T
T
F
F
F
F
F
F
F
F
F
F
F
F
r_s
T
T
T
F
T
T
T
F
T
T
T
F
T
T
T
F
P
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
Q
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
e) ((p ! q) _ (p ! r)) ! (q _ r)¸ p _ q _ r@ ŸXt‰.
proof ) P : ((p ! q) _ (p ! r)) ! (q _ r), Q : p _ q _ r\ ì‡
ƒ¨\| ¯¨t ‰L¸ ⇡‰.
b) (p ! q) ^ (p ! r)¸ p ! q ^ r@ ŸXt‰.
proof ) P : (p ! q) ^ (p ! r), Q : p ! q ^ r\ ì‡ ƒ¨\|
¯¨t ‰L¸ ⇡‰.
p
T
T
T
T
F
F
F
F
1 —i¸ h⇠
p
T
T
T
T
F
F
F
F
Q
T
F
F
F
T
T
T
T
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
p
q
T
T
F
F
T
T
T
T
p
r
T
F
T
F
T
T
T
T
(p
q) _ (p ! r)
T
T
T
F
T
T
T
T
q_r
T
T
T
F
T
T
T
F
P
T
T
T
T
T
T
T
F
2. ‰L¸ ⇡t ¸¥ƒ Ö⌧X Ä Ö⌧| ‹$.
a) a < x  b.
sol) x  a ⇣î x > b.
c) p ! (q _ r)@ (p^ ⇠ q) ! r@ ŸXt‰.
proof ) P : p ! (q _ r), Q : (p^ ⇠ q) ! r\ ì‡ ƒ¨\| b) ®‡ ‰⇠ x– XÏ x2 6x + 9 0t‰.
sol) ¥§ ‰⇠ x– XÏ x2 6x + 9 < 0t‰.
¯¨t ‰L¸ ⇡‰.
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
q_r
T
T
T
F
T
T
T
F
⇠q
F
F
T
T
F
F
T
T
p^ ⇠ q
F
F
T
T
F
F
F
F
P
T
T
T
F
T
T
T
T
c) ®‡ ‰⇠ x– XÏ x2 + 4 0t‡ x2  0t‰.
sol) ¥§ ‰⇠ x– XÏ x2 + 4 < 0tpò x2 > 0t‰.
Q
T
T
T
F
T
T
T
T
d) ¥§ ‰⇠ x– XÏ x < 3tpò x > 6t‰.
sol) ®‡ ‰⇠ x– XÏ x 3t‡ x  6t‰.
e) ¥§ ‰⇠ x– XÏ x3 = 2tt x = 1t‰.
sol) ®‡ ‰⇠ x– XÏ x3 = 2t‡ x 6= 1t‰.
f) ®‡ ¨å@ ⇠Y ¨‡| \‰.
d) (p ! r) _ (q ! s)@ p ^ q ! r _ sî ŸXt‰.
sol) ¥§ ¨å@ ⇠Y ¨‡| X¿ Jî‰.
proof ) P : (p ! r) _ (q ! s), Q : p ^ q ! r _ s\ ì‡ ƒ¨\|
4
Q
T
T
T
T
T
T
T
F
SOLUTION
3. ‰L <L– ıX‹$.
a) ‰⇠ x–
XÏ P Ö⌧ p, q
‰L¸ ⇡D L,
p:
q : |x
10 < x < a,
1 —i¸ h⇠
b) A \ B = {3, 4, 5}
c) 1 62 A B
d) 2 62 B A
6| < 2,
sol) A = {2, 3, 4, 5}, B = {1, 3, 4, 5}\ Pt ⌅X $ ptD ®P
Ãq\‰.
Ö⌧ p !⇠ q 8t ⇠ƒ] Xî ê⇠ a 1X ⌧⇠| lX‹
$.
sol) 1  a  4tt x  4 Ãq⇠¿\ Ö⌧ p !⇠ q î 8t‰. 3. —i A = {13k + 10|k 2 Z}, B = {5m|m 2 Z}, C =
a 5x Ω∞, x = 4.5| › Xt xî Ö⌧ p| ÃqX¿Ã Ö⌧ {13(5n + 1) 3|n 2 Z}– XÏ A \ B = CÑD Ùt‹$.
⇠ q| ÃqX¿ ªX¿\ Ö⌧ p !⇠ q p”t ⌧‰. 0|⌧
•\ ê⇠ aî 1, 2, 3, 4\, ¯ ⌧⇠î 4⌧t‰.
proof ) A \ B = CÑD Ùt0 ⌅t A \ B ✓ Ct‡ C ✓ A \ B
ÑD Ùtt ⌧‰.
b) ‰⇠ x– XÏ 8 Ö⌧ p, q, r ‰L¸ ⇡D L,
p:x
10,
q:x<
6 + a,
r : |x| 
i) A \ B ✓ C : x 2 A \ Btt ¥§
a
2
⇠ k, m 2 Z
t¨XÏ
x = 13k + 10 = 5m
Ö⌧ ⇠ p ! q@ q !⇠ rt ®P 8t ⇠ƒ] Xî ‰⇠ aX \◆ t‰. ⇣\ 13k = 5m 10t¿\ 13k = 5(m 2)t‰. t L, 13
✓¸ \ü✓D lX‹$.
¸ 5 ⌧\åt¿\ ¥§ ⇠ n 2 Z– t
sol) Ö⌧ ⇠ p ! q 8t0 ⌅t⌧î
k = 5n, x = 13k + 10 = 13(5n + 1) 3
10  6 + a =) a
4
t¿\ x 2 Ct‰. 0|⌧ A \ B ✓ Ct‰.
t‡, Ö⌧ q !⇠ r () r !⇠ q 8t0 ⌅t⌧î
⇣
a⌘ ⇣
a⌘
a 6
^ a 6
=) a  4
2
2
ii) C ✓ A \ B : y 2 Ctt ¥§
y = 13(5p + 1)
t¥| X¿\ Ö⌧ ⇠ p ! q@ q !⇠ rt ®P 8t ⇠ƒ] t¿\ y 2 A \ B
Xî ‰⇠ aX î⌅î 4  a  4t‰. 0|⌧ lX‡ê Xî
\ü✓@ 4t‡, \◆✓@ 4t‰.
4. —i A, B, C–
µ8⌧ 1.2 .
1. P —i A = {1, 2}, B = {2, 3}– t⌧ ‰LD lX‹$.
a) A ⇥ B
sol) A ⇥ B = {(1, 2), (1, 3), (2, 2), (2, 3)}.
⇠ p 2 Z–
t
3 = 13(5p) + 10 = 5(13p + 2)
1Ω\‰. 0|⌧ C ✓ A \ Bt‰.
t⌧ ‰L Ò›t 1ΩiD Ùt‹$.
a) (A [ B) \ C = (A \ C) [ (B \ C)
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
x 2 (A [ B) \ C () (x 2 (A [ B)) ^ (x 2 C)
() (x 2 A _ x 2 B) ^ (x 2 C)
() (x 2 A ^ x 2 C) _ (x 2 B ^ x 2 C)
() (x 2 A \ C) _ (x 2 B \ C)
b) B ⇥ A
sol) B ⇥ A = {(2, 1), (2, 2), (3, 1), (3, 2)}.
() x 2 (A \ C) [ (B \ C).
c) (A ⇥ B) \ (B ⇥ A)
sol) (A ⇥ B) \ (B ⇥ A) = {(2, 2)}.
b) (A \ B) [ C = (A [ C) \ (A [ B)
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
x 2 (A \ B) [ C () (x 2 (A \ B)) _ (x 2 C)
d) (A [ B) ⇥ B
sol) (A [ B) ⇥ B = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}.
() (x 2 A ^ x 2 B) _ (x 2 C)
() (x 2 A _ x 2 C) ^ (x 2 B _ x 2 C)
e) (A ⇥ B) [ (B ⇥ B)
sol) (A⇥B)[(B ⇥B) = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}.
() (x 2 A [ C) ^ (x 2 B [ C)
() x 2 (A [ C) \ (B [ C).
2. ‰L $ ptD ®P ÃqXî —i A, B| lX‹$.
a) A [ B = {1, 2, 3, 4, 5}
5. —i A, B–
5
t⌧ ‰L Ò›t 1ΩhD Ùt‹$.
SOLUTION
a) (A [ B)c = Ac \ B c
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
1 —i¸ h⇠
8. —i A, B
c
x 2 (A [ B) () ⇠ (x 2 A [ B)
B ⇢ A| ÃqXt
A ⇥ B = ((A
() ⇠ (x 2 A _ x 2 B)
() ⇠ (x 2 A)^ ⇠ (x 2 B)
B) ⇥ B) [ (B ⇥ B)
t 1ΩhD Ùt‹$.
proof ) B ⇢ At¿\ A = (A B)[B
7X Ö⌧| ©Xt
() (x 2 Ac ) ^ (x 2 B c )
() x 2 Ac \ B c .
A ⇥ B = ((A
b) (A \ B)c = Ac [ B c
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
1Ω\‰. t⌧ µ8⌧
B) [ B) ⇥ B = ((A
B) ⇥ B) [ (B ⇥ B)
t‰.
c
x 2 (A \ B) () ⇠ (x 2 A \ B)
9. P —i A, B– t A@ BX
ence)@ ‰L¸ ⇡t X⌧‰.
() ⇠ (x 2 A ^ x 2 B)
() ⇠ (x 2 A)_ ⇠ (x 2 B)
() (x 2 Ac ) _ (x 2 B c )
6. —i A, B, C–
t⌧ ‰L Ò›t 1ΩhD Ùt‹$.
a) A4(A4B) = B.
proof ) A B = A \ B c t‡,
A4B = (A
(B [ C) () x 2 A ^ x 62 (B [ C)
A)
() x 2 A ^ ((x 62 B) ^ (x 62 C))
() (x 2 A ^ x 62 B) ^ (x 2 A ^ x 62 C)
= ((B [ A) \ (B [ B c )) \ ((Ac [ A)(Ac [ B c ))
() (x 2 A
() x 2 (A
= ((A \ B c ) [ B) \ ((A \ B c ) [ Ac )
B) ^ (x 2 A
B) \ (A
= (B [ A) \ (Ac [ B c )
C)
= (B [ A) \ (A \ B)c
C).
= (A [ B)
b) A (B \ C) = (A B) [ (A C)
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
(A \ B)
ÑD ¨©Xê.
(B \ C) () x 2 A ^ x 62 (B \ C)
A
(A4B) = A
((A [ B)
(A \ B))
((A [ B) \ (A \ B)c )
() x 2 A ^ ((x 62 B) _ (x 62 C))
=A
() (x 2 A
= A \ (A \ B) = A \ B,
() (x 2 A ^ x 62 B) _ (x 2 A ^ x 62 C)
() x 2 (A
7. —i A, B, C–
B) [ (B
= (A \ B c ) [ (B \ Ac )
() x 2 A ^ x 2 A ^ ((x 62 B) ^ (x 62 C))
x2A
A)
tL, ‰L Ò›t 1ΩiD Ùt‹$.
a) A (B [ C) = (A B) \ (A C)
proof ) ‰L– Xt 8⌧X Ò›t 1Ω\‰.
x2A
B) [ (B
A4B = (A
() x 2 Ac [ B c .
m(—i(symmetric differ-
B) _ (x 2 A
B) [ (A
= (A
C)
C).
(A [ B)) [ (A
(A \ B)c )
B) [ (B
A
t‡,
t⌧ ‰Lt 1ΩhD Ùt‹$.
(A4B)
A = ((A
= ((A
(A [ B) ⇥ C = (A ⇥ C) [ (B ⇥ C)
= ((A
proof )
= ((A
(x, y) 2 (A [ B) ⇥ C () x 2 (A [ B) ^ y 2 C
B) [ (B
A))
A)) \ Ac
B) \ Ac ) [ ((B
B)
c
A) [ ((B
c
A) [ Ac )
A)
A)
c
c
= (A \ B \ A ) [ (B \ A \ A )
() (x 2 A _ x 2 B) ^ (y 2 C)
= B \ Ac .
() (x 2 A ^ y 2 C) _ (x 2 B ^ y 2 C)
t‰. A4(A4B) = (A
(A4B)) [ ((A4B)
A4(A4B) = (B \ A) [ (B \ Ac ) = Bt‰.
() (x, y) 2 A ⇥ C _ (x, y) 2 B ⇥ C
() (x, y) 2 (A ⇥ C) [ (B ⇥ C).
6
A)t¿\
SOLUTION
b) (A4B)4C = A4(B4C)
proof ) 忸 ∞¿X ›t
(B [ C)) [ (B
(A
10. —i U = {1, 2, 3, 4, 5, 6}–
Xî —i A, B, C| lX‹$.
(C [ A)) [ (C
(A [ B)) [ (A \ B \ C)
B) [ (B
C = ((A
= ((A
= (A
B)
A))
C
C) [ ((B
A)
(B [ C)) [ (B
C)
(A [ C)),
µ8⌧ 1.3 .
1. 0Ñ ¨ 1.3.3D ùÖX‹$.
proof ) i), ii), iii)î ¯8–⌧ ùÖt DÃ⇠»<¿\ iv)Ã ùÖ
X0\ \‰.
1
a > 0t| Xê.
 0 |‡
Xt 0Ñ ¨ 1.3.3X iv)–
a
Xt
1
1
 0 =) a ·  a · 0 () 1  0
a
a
t‡
C
(A4B) = C
= (C
((A [ B)
= (C
(A \ B))
(A \ B)c )
(A [ B) [ (C
(A [ B)) [ (A \ B \ C),
t¿\
(A4B)4C = ((A4B)
= (A
C) [ (C
(B [ C)) [ (B
[ (A \ B \ C)
t⌧ ‰L 8 ptD ®P Ãq
a) A4C = {1, 2}
b) B4C = {5, 6}
c) A \ B = B \ C = {3, 4}
sol) A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {3, 4}\ Pt ⌅ 8
ptD Ãq\‰.
\ ⇡LD Ùtê. å¿X Ω∞,
(A4B)
1 —i¸ h⇠
1
> 0t‰. ⇣\ a < 0x Ω∞–î
a
a > 0<\ P‡ ⌘⇡t Xt ⌧‰.
(A4B))
t¿\ tî ®⌧t‰. 0|⌧
(A [ C)) [ (C
(A [ B))
p
©Xt 2. 2îp ¨⇠ DÿD ùÖX‹$.
proof ) 2
¨⇠|‡ Xê. ¯Ït ¥§ ⌧\åx ëX
m, n– t
A (B4C) = (A (B [ C)) [ (A \ B \ C),
p
m
2=
() 2n2 = m2
(B4C) A = (B (C [ A)) [ (C (B [ A))
n
t‰. ∞¿X Ω∞ƒ, A@ CX Ì`D ∏¥ ⌅ Ät|
t¿\
A4(B4C) = (A
= (A
(B4C)) [ ((B4C)
(B [ C)) [ (B
[ (A \ B \ C)
t 1Ω\‰. ⇣\ t ›–⌧ m@ ›⇠ÑD L ⇠ à‡, 0|⌧
m = 2k| ÃqXî 0t Dà ⇠ k t¨\‰. ¯Ït
A)
(A [ C)) [ (C
2n2 = m2 = (2k)2 = 4k 2 =) n2 = 2k 2
(A [ B))
t‡, n ⇣\ ›⇠ÑD L ⇠pà‰. X¿Ã tî m, nt ⌧\å|
î
– ®⌧t‰. 0|⌧ 2î ¨⇠ D»‰.
1ΩX¿\ 忸 ∞¿X ›@ ⌧\ ⇡‰.
c) A \ (B4C) = (A \ B)4(A \ C)
proof )
(A \ B)
p
3. å⇠(prime number) p– XÏ pî ¨⇠ DÿD ùÖ
X‹$.
p
proof ) p
¨⇠|‡ Xê. ¯Ït ¥§ ⌧\åx ëX ⇠
m, n– t
m
p
p=
() pn2 = m2
n
(A \ C) = (A \ B) \ (Ac [ C c )
= (A \ B \ Ac ) [ (A \ B \ C c )
= (A \ B \ C c )
= (A \ (B
C))
ÑD t©Xt
(A \ B)4(A \ C) = ((A \ B)
= (A \ (B
= A \ ((B
(A \ C)) [ ((A \ C)
C)) [ (A \ (C
C) [ (C
⇠
B))
t 1Ω\‰. ⇣\ m, nt ⌧\å|‡
à<¿\ m2 , n2 ⇣\
2
2
⌧\åt‰. X¿Ã pn = m t¿\, pî m2 D ò⌅‡, ⇣\ p
å⇠Ï⌧ mD ò⌅¿\ pk = mD ÃqXî k 2 Z {0}
t¨\‰.
t⌧
(A \ B))
pn2 = m2 = (pk)2 () n2 = k 2
B))
t¿\ kî nD ò ‰. X¿Ã pk = m–⌧ kî mD ò⌅¿\,
p
tî m, nt ⌧\å|î
– ®⌧t‰. 0|⌧ pî ¨⇠
D»‰.
= A \ (B4C)
t¿\ Ò›t 1Ω\‰.
7
SOLUTION
4. ¨⇠ a, b, å⇠(prime number) p–
a = b = 0ÑD ùÖX‹$.
proof ) b 6= 0|
Xt
p
a + b p = 0 () a =
1 —i¸ h⇠
p
XÏ a + b p = 0tt µ8⌧ 1.4 1. A = {1, 2, 3, 4}– t⌧ h⇠ f : A ! R,
f (x) = x2 4x + 2X XÌD lX‡ ¯ò⌅| ¯¨‹$.
p
b p ()
a
p
= p
b
sol) f (1) = 1, f (2) = 2, f (3) =
f (A) = { 2, 1, 2}t‰.
p
⇠¥ p
¨⇠ ⇠îp, tî ®⌧t‰. 0|⌧ b = 0
t‰. t| µt a = 0у L ⇠ à‰.
1, f (4) = 2t¿\ XÌ@
2
5. ‰⇠ a–
proof )
XÏ a · 0 = 0ÑD ùÖX‹$.
=) 0 = a · 0
a·0=a·0+a·0
f (x)
a · 0 = a · (0 + 0) = a · 0 + a · 0
a·0=a·0
0
t¿\ a · 0 = 0t‰.
6. ‰⇠ a, b, c– XÏ ab = act‡ a 6= 0tt b = cÑD Ùt‹
2
$.
1
proof ) a 6= 0t¿\ ¯ Ì– t t¨X‡
a
0
1
2
3
4
5
✓
◆
x
1
1
b=1·b=
· a · b = · (ab)
a
a
2. h⇠
✓
◆
1
1
2
2
= (ac) =
·a ·c=1·c=c
f (x) = + 3x, g(x) =
+ 9x + 5
a
a
x
x
– XÏ (f + g)(x)@ t h⇠X XÌD lX‹$.
t¿\ ùÖt DÃ⇠»‰.
sol) f : X ! Y – t Dom(f ) = X| f X XÌt|‡
Xê. Dom(f + g) = Dom(f ) \ Dom(g)ÑD t©\‰. êÖà
1
1
7. ‰⇠ a, b– XÏ 0 < a < btt 0 < < ÑD Ùt‹$.
b
a
1 1
1
1
proof ) a > 0, b > 0t¿\ , > 0@ êÖX‰. Ã}
a b
b
a
t| Xt,
Dom(f ) = Dom(g) = R
t‰. 0|⌧ (f + g)(x) = 12x + 5t‡
Dom(f + g) = R
1
1
1
a· <b· =1
a
b
b
=) 1 < 1
1=a·
⇠¥ ®⌧t‰. 0|⌧ 0 <
8. ÑXX 8 ‰⇠ x, y, z–
$.
{0}
{0}
t‰.
3. h⇠
1
1
< t 1Ω\‰.
b
a
f (x) = x2
t ‰L ÄÒ›t 1ΩhD Ùt‹ –
XÏ f /gX
4x + 1,
g(x) = (x
2)(x + 3)
XÌD lX‡, (f /g)(x)X ›D ‹$.
sol) Dom(f /g) = Dom(f ) \ Dom(1/g)ÑD t©\‰.
Dom(f ) = R, Dom(1/g) = R { 3, 2}t¿\ h⇠ f /gX
XÌ@
|x + y + z|  |x| + |y| + |z|
proof )
Dom(f /g) = R
|x + y + z| = |(x + y) + z|
{ 3, 2}
t‡, ¯ ›@
 |(x + y)| + |z|
 |x| + |y| + |z|
x2 4x + 1
f
(x) =
g
(x 2)(x + 3)
t¿\ ùÖt DÃ⇠»‰.
8
SOLUTION
6. h⇠ f : R ! R, f (x) = x2
t‰.
4. h⇠
f (x) =
XÏ f X
4x + 9X XÌD lX‹$.
sol) h⇠ f X XÌ@ f (R) = [5, 1)t‰.
1
1 + 1+1 1
7. h⇠ f : R ! R, f (x) = 5x + |x|X Ìh⇠| lX‹$.
x
–
1 —i¸ h⇠
XÌD lX‡, f (1/2)X ✓D lX‹$.
sol) h⇠ f | ‰‹ t
sol) h⇠ g, h, i| ‰L¸ ⇡t
1
g(x) = 1 + ,
x
XXt
1
h(x) = ,
x
f (x) =
1
i(x) =
,
1+x
(
6x,
4x,
x 0
x<0
t¿\ ¯ Ìh⇠î
h⇠ gî x = 0–⌧ X⇠
(
1
x
, x 0
¿ Jî‰. ⇣, h⇠ h gî h(g(x)) =
t¿\ g(x) = 0x
1
f (x) = x6
g(x)
4, x < 0
x–⌧ X⇠¿ Jî‰. â, x = 1–⌧ X⇠¿ Jî‰. t
1
⌧ »¿…<\ h⇠ i h gî i(h(g(x))) =
t¿\ t‰.
1 + h(g(x))
1
h(g(x)) =
= 1 () g(x) = 1x x–⌧ X⇠¿ Jî
g(x)
8. h⇠ f : R ! R, f (x) = 3x 3 x X Ìh⇠| lX‹$.
1
‰. â, x =
–⌧ X⇠¿ Jî‰. 0|⌧, h⇠ f X XÌ@
2
sol) 3x = t, y = f (x)\ XXXt
1
R { 1,
, 0}
1
2
y=t
() t2 yt 1 = 0
t
p
t‰.
y ± y2 + 4
1
=) t =
t⌧ f î x = –⌧ ò X⇠‡, ¯ h+✓@
2
2
p
✓ ◆
y
y2 + 4
1
3
t‰. tL, 3x = t > 0 t¿\, t =
< 0x Ω∞î
=
f
2
2
4
⌧x⌧‰. 0|⌧, Ìh⇠| f 1 t| àD L,
p
p
t‰.
1
x + x2 + 4
x + x2 + 4
3f (x) =
=) f 1 (x) = log3
2
2
f (x) = (i
h
g)(x)t‰. <
5. h⇠
f (x) =
fX
5
⌧‰.
p
x2 9
16 x2
9. h⇠
XÌD lX‹$.
sol) h⇠ g, h| ‰L¸ ⇡t
p
g(x) = 5
x2
f (x) =
– XÏ i1h⇠ f
sol)
XXt
9,
h(x) = 16
x2 ,
1+x
,
3 x x2
1
1
3
2
3x2 x 1
x
x
(g f )(x) =
=
.
1
x(x + 1)
1+
x
(f
f (x) = (g/h)(x)t‰. ⇣, h⇠ f X
XÌD lX0 ⌅t
Dom(f ) = Dom(g/h) = Dom(g) \ Dom(1/h)ÑD t©\‰.
Dom(g) = R ( 3, 3), Dom(1/h) = R { 4, 4}t¿\ h⇠
f = g/hX XÌ@
Dom(f ) = R
1
3 x x2
, g(x) =
x
1+x
g, g f | lX‹$.
g)(x) =
({ 4, 4} [ ( 3, 3))
10. h⇠ f : R ! R, f (x) = x5 + 2î | | QÑD Ùt‹$.
t‰.
9
SOLUTION
1
proof ) h⇠ g : R ! R, g(x) = (x 2) 5 î XÌ–⌧ ò
⇠‡, f X Ìh⇠t¿\ h⇠ f î | | Qt‰.
11. h⇠ f : X ! Y @ A ⇢ B ⇢ X–
Ùt‹$.
f (A) ⇢ f (B),
X 0|⌧ h⇠ gƒ ⌅¨h⇠t‰.
XÏ ‰Lt 1ΩiD
f (A [ B) = f (A) [ f (B).
2 ˘\
2
˘\
µ8⌧ 2.1
1. ⇠Ù {xn }, {yn }, {zn } t
‰⇠ 2, 3, 0<\ ⇠4X‡,
{wn }@ 1\ ⌧∞\‰. ‰L ⇠ÙX ⇠4 ÏÄ| UxX‡, ⇠4
Xî Ω∞–î ⇠ÙX ˘\D lX‹$.
proof ) y 2 f (A)tt ¥§ x 2 A– t y = f (x)t‰. A ⇢ B
t¿\ x 2 Bt¥⌧ y = f (x) 2 f (B)t‰. 0|⌧ f (A) ⇢ f (B)
1Ω\‰.
a) {xn + 2yn }
t⌧ p 2 f (A [ B)tt ¥§ q 2 A [ B– t p = f (q)t‰. sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 8.
0|⌧
p = f (q) 2 f (A) or p = f (q) 2 f (B)
b) {2xn yn }
t¿\, p 2 f (A) [ f (B)t‰. ¯Ï¿\ f (A [ B) ⇢ f (A) [ f (B) sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 1.
t‰. \∏ A, B ⇢ A [ Bt¿\ f (A), f (B) ⇢ f (A [ B)t‡,
0|⌧
c) {xn yn }
f (A) [ f (B) ⇢ f (A [ B)
sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 6.
t‰. t| µt f (A [ B) = f (A) [ f (B)ÑD L ⇠ à‰.
d) {xn zn }
12. h⇠ f : X ! Y @ A ⇢ B ⇢ X– XÏ ‰Lt 1ΩiD sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 0.
Ùt‹$.
⇢
f (A \ B) ⇢ f (A) \ f (B).
xn
e)
¯¨‡ f (A \ B) 6= f (A) \ f (B)x | ⌧‹X‹$.
yn
2
sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: .
3
proof ) A \ B ⇢ A, Bt¿\ f (A \ B) ⇢ f (A), f (B)t‡, 0|⌧
⇢
f (A \ B) ⇢ f (A) \ f (B) 1Ω\‰.
1
f)
x n yn
h⇠ f : R ! R, f (x) = |x|– t A = ( 1, 0), B = (0, 1) sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 1 .
6
\ ì<t f (A) = f (B) = (0, 1)\ f (A) \ f (B) = (0, 1)t¿Ã
⇢
f (A \ B) = ;t¿\ f (A \ B) 6= f (A) \ f (B)t‰.
zn
g)
xn
13. P h⇠ f : A ! B, g : B ! CX i1h⇠ g f | |h sol) ¨ 2.1.2– XXÏ ⇠4. ˘\✓: 0.
⇠tt f ƒ | |h⇠ÑD Ùt‹$.
⇢
xn
h)
w
proof ) h⇠ f | |h⇠ D»|‡
Xê. â, f (x1 ) =
n
f (x2 ) t¿Ã x1 6= x2 | ÃqXî x1 , x2 2 A t¨\‰. X¿Ã sol) ¨ 2.1.5– XXÏ lim 1 = 0 t‰.
n!1 wn
h⇠ g f | |h⇠t‡ f (x1 ) = f (x2 )t¿\,
⇠4. ˘\✓ : 0.
(g f )(x1 ) = (g f )(x2 ) =) x1 = x2
⇠¥ ®⌧t‰. 0|⌧, h⇠ f î | |h⇠t‰.
¨ 2.1.5– XXÏ
2. ‰L ⇠Ùt ù ⇠Ùx¿ ⇣ X‹$.
2
5}
⌅¨h⇠ a) {n + 4n +
sol) an = n2 + 4n + 5|‡ Xê. ¯Ït ®‡ ê⇠ n
1–
t
h
i
2
proof ) h⇠ g f ⌅¨h⇠t¿\ (g f )(A) = g(f (A)) = C
an+1 an = (n + 1) + 4 (n + 1) + 5
n2 + 4n + 5
1Ω\‰. ⇣\ f (A) ⇢ Bt¿\, g(f (A)) ⇢ g(B) 1ΩX‡,
= 2n + 5 > 0
g(B) ⇢ Ct¿\ C = g(f (A)) ⇢ g(B) ⇢ Ct‡ g(B) = Ct‰.
14. P h⇠ f : A ! B, g : B ! CX i1h⇠ g f
tt gƒ ⌅¨h⇠ÑD Ùt‹$.
10
SOLUTION
t¿\ ¸¥ƒ ⇠Ù@ ù ⇠Ùt‰.
2 ˘\
t‰. t⌧ n t @⇠| L an+1 an î L⇠✓D ¿‡ n t
›⇠| L ë⇠✓D ¿¿\ ¸¥ƒ ⇠Ù@ ù ⇠Ùƒ, ⇣å⇠
Ùƒ D»‰.
b) {n2 n}
sol) an = n2
4n + 5 |‡ Xê. ¯Ït ®‡ ê⇠ n–
⇥
⇤
an+1 an = (n + 1)2 (n + 1)
(n2 n)
XÏ
= 2n > 0
t¿\ ¸¥ƒ ⇠Ù@ ù ⇠Ùt‰..
⇢
(n + 2)2
2n+2
(n + 2)2
sol) an =
| Xt ®‡ ê⇠ n–
2n+2
h)
an+1
c) {2n }
sol) an = 2n |‡ Xê. ¯Ït ®‡ ê⇠ n
1– t
an+1 an = 2n > 0 t‰. 0|⌧ ¸¥ƒ ⇠Ù@ ù ⇠Ùt‰.
an+1
an =
(
2,
2,
n t @⇠| L
n t ›⇠| L
t‰. 0|⌧
ÑD L ⇠ à‡ 0|⌧ ¸¥ƒ ⇠Ù@ ù ⇠Ùƒ, ⇣å⇠Ùƒ
i)
D»‰.
e)
⇢
2
(n + 3)2
(n + 2)2
2n+3
2n+2
2
(n + 3)
2(n + 2)2
n2 2n + 1
=
=
n+3
2
2n+3
2
n
n n 1
=
2n+3
n2 n
n2 n 1 + 1
<
=
<0
n+3
2
2n+3
an =
n
d) {( 1) }
n
sol) an = ( 1) | Xt
⇢
⇢
(n + 2)2
2n+2
@ ⇣ååÙt‰.
2n+2
(n + 2)2
2n+2
sol) an =
| Xê. ¯Ït ®‡ n–
(n + 2)2
1
n2
1
| Xt ®‡ ê⇠ n 1– t
n2
✓
◆ ✓
◆
1
1
an+1 an = 2
2
(n + 1)2
n2
1
1
= 2
n
(n + 1)2
2n + 1
= 4
>0
n + 2n3 + n2
an+1
sol) an = 2
XÏ
XÏ
2n+3
2n+2
(n + 3)2
(n + 2)2
2(n + 2)2 (n + 3)2
n2 + 2n 1
n+2
= 2n+2
=
2
(n + 3)2 (n + 2)2
(n + 3)2 (n + 2)2
2
2
n +n+n 1
n +n+1 1
= 2n+2
2n+2
(n + 3)2 (n + 2)2
(n + 3)2 (n + 2)2
n2 + n
= 2n+2
>0
(n + 3)2 (n + 2)2
an =
t‰. 0|⌧ ¸¥ƒ ⇠Ù@ ù ⇠Ùt‰.
⇢
✓ ◆n
1
f)
1
2 ✓ ◆
n
1
sol) an = 1
| Xt ®‡ ê⇠ n
2
✓ ◆n+1
1
an+1 an =
>0
2
t‰. 0|⌧
1–
t
t‰. 0|⌧ ¸¥ƒ ⇠Ù@ ù ⇠Ùt‰.
g)
⇢
2
✓
sol) an = 2
an+1
◆n
1
2✓
1
2
◆n
an =
✓
| Xt
1
2
◆n
✓
1
2
◆n+1
=
3
2
✓
1
2
◆n
⇢
2n+2
(n + 2)2
î ù ⇠Ùt‰.
3. {xn } t ⇣å⇠Ùt| Xê. ¥§ ¡⇠ m 2 R t t¨XÏ
®‡ n 2 N – t xn m tt {xn } @ ⇠4hD ùÖX‹$.
proof ) {xn } t ⇣å⇠Ùt¿\ { xn } @ ù ⇠Ùt‰. ¯¨‡
®‡ ê⇠ n– t xn  m t‰. 0|⌧ Ëp⇠4 ¨–
XXÏ { xn } @ ¥§ ‰⇠ x\ ⇠4\‰. ⇣ ¨ 2.1.2– XXÏ
{xn }@ x\ ⇠4\‰.
4. a1 = 1, an+1 = (2a2n + 1)1/5 ,(n 1) X ⇠4 ÏÄ| ⇣ X‹
$.
sol) ⇠Y ¿©ïD t©XÏ an  an+1 < 3 ÑD ùÖXê.
n = 1 | L a1 = 1  31/5 < 3 t¿\ 1Ω\‰. t⌧ n = k |
L 1Ω\‰‡
Xê. â, ak  ak+1 < 3 t| Xê. ¯Ït
ak+2 = 2a2k+1 + 1
11
1/5
< 191/5 < 2431/5 = 3
SOLUTION
t‡
ak+2 = 2a2k+1 + 1
1/5
> (2a2k + 1)1/5 = ak+1
2 ˘\
µ8⌧ 2.2 .
1. ¸¥ƒ h⇠X å˘\¸ ∞˘\D lX‹$. Dò–⌧ [a]î
‰⇠ aÙ‰ ëpò ⇡@ ⇠ ⌘– \ X ⇠t‰.
t‰. 0|⌧ n = k + 1 |Lƒ 1ΩX‡ ⇠Y ¿©ï– XXÏ
|x|
|x|
®‡ ê⇠ n– XÏ an  an+1 < 3 t‰. {an } @ ù ⇠
a) lim
, lim+
Ùt‡ ®‡ ê⇠ n– XÏ an  3 t¿\ Ëp⇠4 ¨–
x x!0 x
x!0
|x|
|x|
XXÏ {an } @ ⇠4\‰.
sol) x > 0| L
= 1, x < 0| L
=
x
x
1 t¿\
|x|
|x|
5. ⇠Ù {an }t a1 = 2t‡ ®‡ ê⇠ n– XÏ an+1 =
lim
= lim 1 = 1, lim+
= lim+ 1 = 1
an + 6
x
x
x!0
x!0
x!0
x!0
D Ãq` L {an }t ⇠4hD ùÖX‡ ˘\✓D lX‹
2
t‰.
$.
sol) ⇠Y ¿©ïD t©XÏ ®‡ ê⇠ n– XÏ an 


an+1 < 6 ÑD ùÖXê. n = 1 | L a1 = 2  4 = a2 < 6
1
1
b)
lim
,
lim
t¿\ 1Ω\‰. t⌧ n = k | L 1Ω\‰‡
Xê. â,
x x!0+ x
x!0
ak  ak+1 < 6 t| Xê. ¯Ït
sol) x =
6 0|L
ak+1 + 6
12

ak+2 =
<
=6
1
1
1
2
2
1

x
x
x
t‡
ÑD t©Xt
ak+1 + 6
ak + 6
ak+2 =
= ak+1


2
2
1
1
lim
= 1, lim+
=1
t‰. 0|⌧ n = k + 1 |Lƒ 1Ω\‰. ⇠Y ¿©ï– XXÏ
x
x
x!0
x!0
®‡ ê⇠ n– XÏ an  an+1 < 6 t‰. {an } t ù ⇠
Ùt‡ ®‡ ê⇠ n– XÏ an < 6 t¿\ Ëp⇠4 ¨– t‰.
an + 6
XXÏ {an }@ ¥§ ‰⇠ ↵\ ⇠4\‰. an+1 =
– nt
2
c) lim
[tan x], lim
[tan x]
↵+6
+
x! ⇡
x! ⇡
4\à ‰¿î ˘\D ËXt ↵ =
t¿\ ↵ = 6 t‰.
2
2
2
sol) b) @ ⇡t ∞§ h⇠X 1»D t©Xt
0|⌧ lim a = ↵ = 6 t‰.
n!1
n
6. a1 = 1 t‡ an+1 =
1
an + 4| L {an }X ⇠4 ÏÄ@ ⇠4X
2
t‰.
lim [tan x] = 1, lim
[tan x] =
⇡+
x! ⇡
2
x! 2
1
t ˘\✓D lX‹$.
sol) ⇠Y ¿©ïD t©XÏ ®‡ ê⇠n–
XÏ an 
p
p
1 cos2 x
1 cos2 x
9
, lim+
an+1 < 8 ÑD ùÖXê. n = 1| L a1 <
= a2 < 8 d) lim
x
x
x!0
x!0
2
p
t¿\ 1Ω\‰. t⌧ n = k | L 1Ω\‰‡
Xê. â, sol) 1 cos2 x = |sin x|ÑD t©Xt
ak  ak+1 < 8 t| Xê. ¯Ït
p
p
1 cos2 x
1 cos2 x
1
1
lim
= 1, lim
=1
ak+2 = ak+1 + 4 < · 8 + 4 = 8
x
x
x!0
x!0+
2
2
t‰.
t‡
ak+2 =
1
ak+1 + 4
2
1
ak + 4 = ak+1
2
2. ‰L ˘\t t¨Xî¿ ⇣ X‡, t¨` Ω∞ ¯ ˘\✓D
t‰. n = k + 1 |Lƒ 1ΩX¿\ ⇠Y ¿©ï– XXÏ ®‡ lX‹$.
ê⇠ n– XÏ an  an+1 < 8 t 1Ω\‰. {an }@ ù ⇠
✓
◆
Ùt‡ ®‡ ê⇠ n– XÏ an < 8 t¿\ Ëp⇠4 ¨–
|x|
x
a)
lim
1
x!0
x
|x|
XXÏ {an }@ ¥§ ‰⇠ ↵\ ⇠4\‰. an+1 = an + 4 – nt
2
✓
◆
1
x
4\t ‰¿î ˘\D ËXt ↵ = ↵ + 4 t¿\ ↵ = 8 t‰. sol) lim |x|
= lim+ (1 1) = 0 t‡
2
x ◆|x|
x!0+
x!0
✓
0|⌧ lim an = ↵ = 8 t‰.
n!1
|x|
x
lim
= lim [ 1 ( 1)] = 0 t¿\ ¨ 2.2.4–
x
|x|
x!0
x!0
12
SOLUTION
Xt ˘\✓t t¨X‡, ¯ ✓@ 0 t‰.
2 ˘\
i) lim x sin
x!1
b) lim
3x + 1
2
–
⇡
sol) x
x!1 x2 + 1
sol) lim (3x + 1) = 4 t‡ lim x2 + 1 = 2t¿\
x!1
1
x
t
¨ 2.2.1–
x!1
cos
Xt ˘\✓t t¨X‡, ¯ ✓@ 2 t‰.
x2 + x 2
x!1
x 1
c) lim
t‡ lim cos
x!1
2
x +x 2
(x + 2)(x
= lim
x!1
x!1
x 1
x 1
sol) lim
1)
j) lim
= lim (x + 2) = 3.
x!1
x!0
1 3x2
x!1 3x2 + x
✓ ◆
✓ ◆
1
1
 x sin
1
x
x
✓ ◆
1
1
= 1t¿\ lim x sin = 1t‰.
x!1
x
x
sin x
x
sin x
1
= lim tsin = 1.
t!1
x
t
sin x
k) lim
x!1 x
d) lim
sol) lim
x!0
1
3
1 3x2
3
x2
=
lim
=
=
1
2
x!1 3x + x
x!1 3 +
3
x
sol) lim
1.
sin x
1
1

t‡ lim
= 0t¿\
x!1 x
x
x
sin x
2.2.5– Xt lim
= 0t‰.
x!1 x
sol) x 6= 0–
2x 3
x! 1 x2 + 2
e) lim
2
3
2x 3
0
x
x2
=
lim
= = 0.
x! 1 x2 + 2
x! 1 1 + 22
1
x
p
3
x 1
f) lim
x!1 x
1
sol) lim
t
cos x
x!0 x
l) lim
cos x
cos x
= 1, lim
=
x!0
x
x
X¿ Jî‰.
sol) lim
x!0+
sol)
lim
x!1
p
3
x 1
x 1
= lim 2
p
1
x!1 (x 3 + x 3 + 1)( 3 x
x 1
1
1
= lim 2
= .
1
x!1 x 3 + x 3 + 1
3
p
3
x!0
sol) lim+ ln x =
x!0
1. ( ¨ 3.7.12 8p)
n) lim (ln (2x + 1)
x!1
x!1
1–
sin(2n⇡) = 0,
1t¿\ ˘\✓@ t¨
m) lim+ ln x
1)
g) lim sin x
sol) ®‡ ê⇠ n
¨
t
ln (x))
sol) x > 1| L
✓
1
sin((2n + )⇡) = 1
2
1
ln 2  ln 2 +
x
◆
 ln 2 +
t¿\, ¥§ ë⇠ M > 0D ‡tT|ƒ x = 2n⇡ > M | Ãq
1
Xî ê⇠ n@ 4\à Œ<¿\ sin xî x ©Ñà lT|ƒ t‡ lim ln 2 +
= ln 2t¿\
x!1
x
|\ ✓<\ ⇠4X¿ ª\‰. 0|⌧, ˘\✓@ t¨X¿ Jî lim (ln (2x + 1) ln (x)) = ln 2t‰.
x!1
‰.
1
x
¨
sin 4x
1
h) lim x sin
x!0
x
o) lim
1
sol) x 6= 0– t x sin
 |x|t‡ lim |x| = 0t¿\
x!0
x
1
2.2.3– Xt lim x sin = 0t‰.
x!0
x
sol) lim
x!0 sin 2x
¨
sin 4x
x!0 sin 2x
p) lim
x!4 5
13
= lim
x!0
4 x
p
x2 + 9
sin 4x
2x
4x
·
·
= 2.
4x
sin 2x 2x
2.2.3–
Xt
SOLUTION
sol) (p”)
⇠@: h⇠ f, g| ‰L¸ ⇡t XXt,
(
1, x > 0
f (x) =
, g(x) = x,
0, x  0
sol)
p
4 x
(4 x)(5 + x2 + 9)
p
= lim
2
16 x2
x + 9 x!4
p
5 + x2 + 9
5
= lim
= .
x!4
4+x
4
lim
x!4 5
q) lim
x!0
✓
1
2
+
x (x 1) x (x + 2)
f (g(x)) = f (x)t‡, a = 0| L lim g(x) = 0t¿Ã
x!0
◆
lim f (g(x)) = f (0+) = 1,
x!0+
lim
x!0
✓
1
2
+
x (x 1) x (x + 2)
3. ‰L Ö⌧
◆
3
=
1)(x + 2)
= lim
x!0 (x
3
.
2
e) h⇠ f : R ! R– t ¥§ ¡⇠ M > 0t t¨XÏ ®‡
‰⇠ x– t |f (x)|  M tt lim xf (x) = 0t‰.
x!0
8tt ùÖX‡, p”tt ⇠@| ⌧‹X‹$.
a) ®‡ ‰⇠ x–
proof ) (8)
¸¥ƒ pt– Xt
XÏ f (x) < g (x)tt lim f (x) < lim g (x)
x!a
t‰.
sol) (p”)
⇠@ : h⇠ f, g : R ! R |
(
0,
f (x) =
1,
x!a
|xf (x)|  M |x|,
x2R
t 1ΩX‡ lim M |x| = 0t¿\ DP ¨ 9@ Ë⌅X
x!0
¨–
Xt lim xf (x) = 0t‰.
x!0
x 6= 0
,
x=0
g(x) = |x|
4. ¡⇠ a 2 R – t h⇠ f l⌅ [a, 1)–⌧ ⇣åh⇠( X
3.3.1)| Xê. ¥§ ¡⇠ m 2 Rt t¨XÏ ®‡ x 2 [a, 1)–
XXt R–⌧ f (x) < g(x) t¿Ã lim f (x) = lim g(x) = 0
t f (x) mtt x ! 1 | L f (x) ⇠4hD Ùt‹$.
x!0
t‰.
lim f (g(x)) = f (0 ) = 0
x!0
t¿\ ˘\✓t t¨X¿ Jî‰.
sol)
\
2 ˘\
x!0
proof ) h⇠ f l⌅ [a, 1)–⌧ ⇣åh⇠t¿\ h⇠ f î l
b) lim f (x) = 1 t‡ lim g (x) = 0 tt lim f (x) g (x) = 0 ⌅ [a, 1)–⌧ ù h⇠t‰. ®‡ x 2 [a, 1)– t f (x) m
x!a
x!a
x!a
t¿\, ®‡ x 2 [a, 1)– t f (x)  mt‰. ¨ 2.2.7–
t‰.
XXÏ x ! 1|L f (x)î ¥§ ‰⇠ l\ ⇠4\‰. t⌧ ¨
2.2.1– XXÏ lim f (x) = l t‰.
x!1
sol) (p”)
1
⇠@: f (x) = 2 , g(x) = x2 , a = 0\ Pt lim f (x) = µ8⌧ 2.3 .
x!0
x
1. ¸¥ƒ h⇠ x = 0–⌧ çx¿ ⇣ X‹$.
1, lim g(x) = 0t¿Ã lim f (x) g (x) = 1 6= 0t‰.
x2
x!0
x!0
a) y =
x
sol) x = 0–⌧ h+✓t X⇠¿ J<¿\ x = 0–⌧ çt
f (x)
c) lim f (x) = 1t‡ lim g(x) = 0tt lim
= 1 tpò D»‰.
x!a
x!a
x!a g(x)
f (x)
lim
= 1t‰.
x!a g(x)
b) y = x [x]
sol) lim (x [x]) = lim x = 0 t‡
x!0+
x!0+
sol) (p”)
lim x [x] = lim (x+1) = 1 t¿\ x = 0–⌧ çt D»‰.
⇠@: f (x) = 1, g(x) = x, a = 0<\ Pt lim f (x) = 1 x!0
x!0
x!0
f (x)
1
t‡ lim g(x) = 0 t¿Ã x ! 0| L
= X ˘\@ t¨ c) y = [x [x]]
x!0
g(x)
x
lim [x
[x]] = lim [x] = 0 t‡ lim [x
[x]] =
X¿ J<p, ëX 4\ ⇣î LX 4\ \ƒ ⌧∞X¿ Jî‰. sol) x!0
+
x!0+
x!0
lim [x + 1] = 0 t‡ [0 [0]] = 0 t¿\ x = 0–⌧ ç
d) lim g(x) = btt, lim f (g(x)) = f (b)t‰.
x!a
x!a
x!0
t‰.
14
SOLUTION
d) y = [|x|]
sol) lim [|x|] = 0 t‡ [|0|] = 0 t¿\ x = 0–⌧ çt‰.
t¿\ x = 0–⌧ çt ⇠ƒ] h+✓D
2 ˘\
X` ⇠ ∆‰.
x!0
sin(x + 1)
x2 + x
e) y = |[x]|
sol)
x
=
1,
0 –⌧ h+✓t X⇠¿ J<¿\ t ⇣‰–⌧
sol) lim |[x]| = 0 t‡ lim |[x]| = 1t¿\ x = 0–⌧ çt
sin(x + 1)
x!0+
x!0
àçt‰. f ( 1) = lim
= 1 \ XXt ¸¥
D»‰.
x! 1 x2 + x
ƒ h⇠ x = 1–⌧ çt ⇠ƒ] X` ⇠ à‰. X¿Ã
sin(x + 1)
sin(x + 1)
= 1 t‡ lim
= 1 t¿\ x = 0
2. ¸¥ƒ h⇠‰X àç⇣D ®P >‡,
àç⇣–⌧ ¯ lim+
2
x +x
x2 + x
x!0
x!0
h⇠ çt ⇠ƒ] X` ⇠ àî¿ ıX‹$.
–⌧ çt ⇠ƒ] h+✓D X` ⇠ ∆‰.
g) f (x) =
x2
2x + 1
x 1
sol) x = 1–⌧ h⇠✓t X⇠¿ J<¿\ x = 1–⌧ àçt
x2 2x + 1
‰. f (1) = lim
= lim (x 1) = 0\ XXt »\
x!1
x!1
x 1
X⌧ h⇠î ‰⇠ ⌅¥–⌧ çt ⌧‰.
a) f (x) =
1
3x + 2
sol) x = 1, 2–⌧ h+✓t X⇠¿ J<¿\ t ⇣‰–⌧ à
x 1
1
çt‰. f (1) = lim 2
= lim
= 1 \ XXt
x!1 x
3x + 2 x!1 x 2
x = 1–⌧ çt ⌧‰. x = 2–⌧î ˘\t ⌧∞X0 L8–
x = 2–⌧ çt ⇠ƒ] h+✓D X` ⇠ ∆‰.
b) f (x) =
x
x2
cos x
x ⇡2
⇡
sol) x = ⇣
h+✓t X⇠¿ J<¿\ x = ⇡2 –⌧ à
2 –⌧
⌘
⇡
cos x
çt‰. f
= lim⇡
1\ XXt »\ X⌧
⇡ =
x! 2 x
2
2
h⇠î ‰⇠ ⌅¥–⌧ çt ⌧‰.
h) f (x) =
3. ‰L <L– ıX‹$.
a) ) › x3 x 1 = 0@ l⌅ [1, 2]–⌧ ¸D
–D Ùt‹$.
proof ) f = x3 x 1 t| XXê. f (1) = 1 t‡ f (2) = 5
t¿\ ¨t✓ ¨– XXÏ f (c) = 0 D ÃqXî c (1, 2)–
t¨\‰. 0|⌧ ) › x3 x 1 = 0 @ l⌅ [1, 2]–⌧ ¸D
ƒ‰.
ln|x + 1|
x
x
sol) x = 1, 0–⌧ h+✓t X⇠¿ J<¿\ t ⇣‰–⌧ b) ) › 3x = e @ l⌅ (0, 1)–⌧ ¥ƒ P ⌧X ¸D –D
Ùt‹$.
1
ln|x + 1|
àçt‰. f (0) = lim
= lim ln(x + 1) x = ln e = 1
x!0
x!0
x
\ XXt ¸¥ƒ h⇠| x = 0 –⌧ çt ⇠ƒ] X` ⇠
proof ) f (x) = ex
3x|
XXê. f (0) = 1 > 0 t‡
à‰.X¿Ã lim f (x) = 1t¿\ x = 1–⌧î çt ⇠ƒ]
f (1) = e 3 < 0 t¿\ ¨t✓ ¨– XXÏ ) › f (x) = 0
x! 1
h+✓D X` ⇠ ∆‰.
@ 0¸ 1¨t– ¥ƒ 1⌧X ¸D ƒ‰. ⇣\ f (1) = e 3 < 0
t‡ f (4) = e4 12 > 0 t¿\ ¨t✓ ¨– XXÏ ) ›
f (x) = 0@ 1¸ 4¨t–⌧ƒ ¥ƒ 1⌧X ¸D ƒ‰. 0|⌧
1
d) f (x) = x sin
3x = ex î (0, 1)–⌧ ¥ƒ P ⌧X ¸D ƒ‰.
x
sol) x = 0–⌧ h+✓t X⇠¿ J<¿\ x = 0–⌧ àç
1
t‰. f (0) = lim x sin = 0 <\ XXt »\ X⌧ h⇠î c) ) › x3 4x + 1 = 0 @ 8 ‰¸D –D Ùt‹$.
x!0
x
‰⇠ ⌅¥–⌧ çt ⌧‰.
c) f (x) =
1
e) f (x) = x2 cos
x
sol) x = 0–⌧ h+✓t X⇠¿ J<¿\ x = 0–⌧ àç
1
t‰. f (0) = lim x2 cos = 0<\ XXt »\ X⌧ h⇠î
x!0
x
‰⇠ ⌅¥–⌧ çt ⌧‰.
p
proof ) ∞
¸¥ƒ ) ›@ 3( ) › t¿\ \
3⌧X
‰¸D ƒ‰. h⇠ f : R ! R| f (x) = x3
4x + 1 \
XXê. f ( 3) = 14 < 0 t‡ f ( 1) = 4 > 0 t¿\ ¨
t✓ ¨– XXÏ ) › f (x) = 0@ ( 3, 1)–⌧ ¥ƒ 1
⌧X ‰¸D ƒ‰. ⇣, f ( 1) = 4 > 0 t‡ f (1) = 2 < 0
t¿\ ) › f (x) = 0@ ( 1, 1) –⌧ ¥ƒ 1⌧X ‰¸D
¿‡ f (1) = 2 < 0 t‡ f (3) = 16 > 0t¿\ ) ›
f (x) = 0@ (1, 3)–⌧ ¥ƒ 1⌧X ‰¸D ƒ‰. 0|⌧, )
› x3 4x + 1 = 0@ 8 ⌧X ‰¸D ƒ‰.
cos2 x
x
sol) x = 0–⌧ h+✓t
X⇠¿ J<¿\ x =p0–⌧ àçt
p
d) ) › ⇡ sin x = x î \å\ 8 ‰¸D
1 cos2 x
1 cos2 x
‰. X¿Ã lim
= 1t‡ lim+
=1
x
x
x!0
x!0
f) f (x) =
1
15
–D Ùt‹$.
SOLUTION
proof ) h⇠ f : R ! R| f (x) = ⇡ sin x x\ XXê.
⇣ ⇡⌘
⇡
f ( ⇡) = ⇡ > 0, f
=
< 0,
2
2
⇣⇡⌘ ⇡
f
= > 0, f (⇡) = ⇡ < 0
2
2
⇣
⇡⌘ ⇣ ⇡ ⇡⌘ ⇣⇡ ⌘
t¿\ ) › f (x) = 0@ l⌅
⇡,
,
,
, ,⇡
2
2 2
2
–⌧
¥ƒ 1⌧X ¸D ƒ‰. 0|⌧ ) › ⇡ sin x = x
î \å\ 8 ⌧X ‰¸D ƒ‰.
⌅ Ùp ¨\Ä0 ↵ 6= 0x ‰⇠– t xn ! ↵, yn ! ↵| Ã
qXî ¨⇠Ù {xn }¸ 4¨⇠Ù {yn }t t¨\‰. 0|⌧, f
çt|‡
Xt
f (↵) = f ( lim xn ) = lim f (xn ) = 0,
n!1
|c||  |x
n!1
f (↵) = f ( lim yn ) = lim f (yn ) = lim yn = ↵
n!1
n!1
n!1
⇠¥ ↵ = 0t¥| \‰. X¿Ã tî ↵ 6= 0t|î
®⌧t‰. 0|⌧ h⇠ f î x = 0–⌧Ã çt‰.
4. h⇠ y = |x| çh⇠ÑD Ùt‡, t| t©XÏ ÑXX
çh⇠ f – XÏ y = |f (x)| çÑD Ùt‹$.
proof ) ÑXX x, c 2 R – XÏ x 6= c t‡ |x c| ! 0tt
||x|
3 ¯Ñ
3
–
¯Ñ
µ8⌧ 3.1 .
1. ‰L h⇠X ƒh⇠| lX‹$.
c| ! 0
t¿\ y = |x| î çh⇠t‰.
t⌧ ÑXX h⇠ f ‰⇠ ⌅¥–⌧ çt|‡
Xê. c 2 R a) y = (x3 + 1)(x2 + x + 1)
D ‡ Xt f ‰⇠⌅¥–⌧ çt¿\ ÑXX x 2 R– X sol)
Ï x 6= ct‡ |x c| ! 0 tt |f (x) f (c)| ! 0t‰. º ÄÒ
›– XXÏ
y 0 = (x3 + 1)0 · (x2 + x + 1) + (x3 + 1) · (x2 + x + 1)0
||f (x)|
|f (c)||  |f (x)
ÑD L ⇠ à‰. 0|⌧ h⇠ f
= (3x2 ) · (x2 + x + 1) + (x3 + 1) · (2x + 1)
f (c)|
= (3x4 + 3x3 + 3x2 ) + (2x4 + x3 + 2x + 1)
çtt |f | ƒ çh⇠t‰.
= 5x4 + 4x3 + 3x2 + 2x + 1.
5. h⇠ f : R ! R
x
¨⇠| Lî f (x) = 0t‡, x 4
¨⇠| Lî f (x) = xt‰. t h⇠ çx ⇣D ®P ><‹$. b) y = (x + 1)(x2 + 1)(3x
sol)
sol) |f (x)|  |x| t‡ lim |x| = 0 t¿\
x!0
¨ 2.2.5– XXÏ
y 0 = (x + 1)0 (x2 + 1)(3x
lim f (x) = 0 = f (0) t‰. 0|⌧ f î x = 0–⌧ çt‰.
2
+ (x + 1)(x + 1)(3x
x!0
t⌧ x = 0D ⌧x\ ‰⇠‰–⌧ àçÑD
Xò| å⌧\‰:
lemma) ÑXX ‰⇠ ↵ 2 R–
¸ 4¨⇠Ù {yn }t t¨\‰.
proof ) ↵ 2 R – t
[n↵]
xn =
,
n
t ↵\ ⇠4Xî
yn =
àê. Ùp ¨
= (x + 1)(3x
= (3x3
p
2
+ xn
n
1)
1) + 2x(x + 1)(3x
1) + 3(x + 1)(x2 + 1)
1)
3
2x) + (3x3 + 3x2 + 3x + 3)
2
1)
0
= 12x3 + 6x2 + 4x + 2.
c) y =
sol)
x+2
x2 + 3x 1
(x + 2)0 (x2 + 3x 1) (x + 2)(x2 + 3x
(x2 + 3x 1)2
2
(x + 3x 1) (x + 2)(2x + 3)
=
(x2 + 3x 1)2
2
x
4x 7
= 2
.
(x + 3x 1)2
y0 =
1
 xn  ↵
n
t¿\ pÑ ¨ 9@ DP ¨– Xt n ! 1| L xn ! ↵,
p
2
lim yn = lim
+ lim xn = ↵
n!1
n!1 n
n!1
t¿\ yn ! ↵ t‰.
1) + (x + 1)(x2 + 1)0 (3x
x2 + 3x
+ (6x + 4x
¨⇠Ù {xn }
| Xt {xn }@ ¨⇠Ùt‡ {yn }@ 4¨⇠Ùt‰ (Ë, [x]î x
Ù‰ ë@ \ X ⇠). ⇣\,
↵
2
1)
d) y =
16
x
x2
+ 2
x+1 x +1
1)0
SOLUTION
f) y =
sol)
sol)
✓
◆0
✓
◆0
x
x2
+
x+1
x2 + 1
1
2x
=
+ 2
(x + 1)2
(x + 1)2
x4 + 2x3 + 6x2 + 2x + 1
=
.
(x + 1)2 (x2 + 1)2
y0 =
2. ‰L h⇠X ƒh⇠
a) y = (5x2
sol)
7)3
3x2
7)2 (10x) = 30x(5x2
p
7)2 .
=
sol)
1
2
2x + 1)
\ ¸¥ƒ‰. f 0 (x) = p
0
sol)
✓
(19(x
2
4)
18
· 2x ·
p
x
4)18 (75x2 + 4)
(x2
3
2x 2
1
x
t¿\ x = a = 0–⌧X |(¸
f (x) ⇡ f (0) + f 0 (0)(x
= 1 + ( 1)x = 1
0)
x
1
p
x)
(x
2 x
2
4)
b) f (x) = (1
x)3 ,
a = 0.
sol) f 0 (x) =
3(1
x)2 t¿\ x = a = 0–⌧X |(¸¨›@
19
.
f (x) ⇡ f (0) + f 0 (0)(x
= 1 + ( 3)x = 1
0)
3x
t‰.
1
x
2x
t‰.
y = q
·
x+1
2 2x
1
e) y =
1
1
a)
¨›@
x+1
2x 1
0
a = 0.
f (x) ⇡ f (a) + f 0 (a)(x
4)19
x
y0 =
d) y =
2x,
sol) h⇠ f X x = a–⌧X |(¸¨›@
sol)
r
1
2x + 1
y = p
· (3x
2 3x2 2x + 1
3x 1
=p
.
3x2 2x + 1
(x2
p
7)0
7)2 · (5x2
0
c) y =
x+1
3. ⇣ x = a ¸)–⌧ ¸¥ƒ h⇠X |(¸¨›D lX‹$.
a) f (x) =
= 3(5x2
p
pp
p
1
y0 = p p
· ( x + 1)0
2 ( x + 1)
1
1
= pp
· p
2
x
2 ( x + 1)
1
= p p
.
4 x x+x
dy
| lX‹$.
dx
y 0 = 3(5x2
b) y =
sol)
3 ¯Ñ
✓
x+1
2x 1
◆0
1
=
2
=
r
2x 1
3
·
x + 1 (2x 1)2
3
1
2
2(x + 1) (2x
◆7
3
1) 2
c) f (x) =
p
x2 + 3,
a = 1.
x
sol).
f 0 (x) = p
t¿\ x = a = 1–⌧X |(¸¨›@
2
x +3
.
f (x) ⇡ f (1) + f 0 (1)(x 1)
1
1
3
= 2 + (x 1) = x +
2
2
2
t‰.
0
✓
y =7 x
✓
=7 x
1
x
1
x
◆6 ✓
· x
1
x
◆0
◆6 ✓
◆
1
· 1+ 2 .
x
4. |(¸¨›D t©XÏ ‰L ✓X ¸ø✓D lX‹$.
a)
17
p
3
26
SOLUTION
sol) h⇠ f : R ! RD ‰L¸ ⇡t
f (x) =
p
3
27
sol) h⇠ f : R ! RD ‰L¸ ⇡t
XXt
x,
f (x) ⇡ f (0) + f 0 (0)(x
1
1
=
x
4 768
1
80
=
= 2.962
27
27
26 = f (1) ⇡ 3
t‰. 0|⌧ (65)
(65)
t‰.
1
3
sol) h⇠ f : R ! RD ‰L¸ ⇡t
XXt
5. ‰L h⇠–
f (x) = (x + 1)10 ,
a) y = x5 + x1/2
x = 0–⌧X f X |(¸¨›@
f (x) ⇡ f (0) + f 0 (0)(x
X ¸ø✓@
= f (1) ⇡
1
4
1
191
=
⇡ 0.2487
768
768
t‰.
1
c) (15) 4
XXt
b) y =
1
x) 4 ,
x = 0–⌧X f X |(¸¨›@
1
(2x + 3)2
sol) (ıÃ Ö‹\‰.)
0
f (x) ⇡ f (0) + f (0)(x
1
=2
x
32
0)
dy
4
=
,
dx
(2x + 3)3
d2 y
24
=
,
2
dx
(2x + 3)4
d3 y
192
=
.
3
dx
(2x + 3)5
1
t‰. 0|⌧ (15) 4 X ¸ø✓@
1
(15) 4 = f (1) ⇡ 2
1
x
dy
1 1
1
= 5x4 + x 2 + 2 ,
dx
2
x
d2 y
1 3
2
= 20x3
x 2
,
2
dx
4
x3
d3 y
3 5
6
= 60x2 + x 2 + 4 .
dx3
8
x
(1.01)10 = f (0.01) ⇡ 1 + 10 · 0.01 = 1.1
f (x) = (16
dy d2 y d3 y
,
,
| lX‹$.
dx dx2 dx3
sol) (ıÃ Ö‹\‰.)
t‰. 0|⌧ (1.01)10 X ¸ø✓@
sol) h⇠ f : ( 1, 16] ! RD ‰L¸ ⇡t
t
0)
= 1 + 10x
1
63
=
= 1.96875
32
32
t‰.
1
3
1
3
0)
t‰.
b) (1.01)10
d) (65)
,
0)
26X ¸ø✓@
p
3
1
3
x = 0–⌧X f X |(¸¨›@
f (x) ⇡ f (0) + f 0 (0)(x
1
=3
x
27
p
3
XXt
f (x) = (64 + x)
x = 0–⌧X f X |(¸¨›@
t‰. 0|⌧
3 ¯Ñ
p
c) y = x x + 1
18
SOLUTION
sol) (ıÃ Ö‹\‰.)
a) f (x) =
dy
3x + 2
= p
,
dx
2 x+1
d2 y
3x + 4
=
3 ,
dx2
4(x + 1) 2
3 ¯Ñ
1
(x
1)m
sol) h⇠ f X nƒƒh⇠
f (n) (x) = ( 1)n
d3 y
3x 6
=
5 .
3
dx
8(x + 1) 2
(m + n
(m 1)!(x
ÑD Ùtê.
n = 1| L
f 0 (x) = ( m)(x
d) y =
x
x2 + 1
1) m 1 ,
t¿\ ⌅ ›@ n = 1| L êÖà 1Ω\‰. t⌧ n = k| L f
X kƒƒh⇠
sol) (ıÃ Ö‹\‰.)
f (k) (x) = ( 1)k
2
dy
x
1
=
,
dx
(x2 + 1)2
d2 y
2x(x2 3)
=
,
2
dx
(x2 + 1)3
d3 y
6(x4 6x2 + 1)
=
.
3
dx
(x2 + 1)4
e) y =
1)!
1)m+n
<\ ¸¥ƒ‰‡
(m + k 1)!
(x
(m 1)!
1) m k
Xt
f (k+1) (x) = (f (k) (x))0
(m + k 1)!(m + k)
(x
(m 1)!
(m + k)!
= ( 1)k+1
(m 1)!(x 1)m+k+1
= ( 1)k ( 1)
sin x
x
1) m k 1
t¿\ f X (k + 1)ƒƒh⇠– t⌧ƒ Ë ⌅X ›D Ãq\‰.
0|⌧ ⇠Y ¿©ï– Xt h⇠ f X nƒƒh⇠î
sol) (ıÃ Ö‹\‰.)
dy
x cos x sin x
=
,
dx
x2
d2 y
x2 sin x 2 sin x + 2x cos x
=
,
2
dx
x3
d3 y
x3 cos x 3x2 sin x + 6 sin x
=
dx3
x4
f (n) (x) = ( 1)n
(m
b) f (x) =
6x cos x
.
x 3
x+1
sol) h⇠ f X nƒƒh⇠
f) y = ex cos x
dy
= ex (cos x sin x),
dx
d2 y
= 2ex sin x,
dx2
d3 y
= 2(ex sin x + ex cos x).
dx3
1)!
1)m+n
t‰.
f (n) (x) =
sol) (ıÃ Ö‹\‰.)
(m + n
1)!(x
ÑD Ùtê.
n = 1| L
f 0 (x) =
( 1)n+1 4n!
(x + 1)n+1
4
(x + 1)2
t¿\ ⌅ ›@ n = 1| L êÖà 1Ω\‰. n = k| L f X k
( 1)k+1 4k!
ƒƒh⇠ f (k) (x) =
\ ¸¥ƒ‰‡
Xt,
(x + 1)k+1
f (k+1) (x) = (f (k) (x))0
( 1)k+1 4(k + 1)(x + 1)k k!
(x + 1)2k+2
( 1)k+2 4(k + 1)!
=
(x + 1)k+2
=
6. ‰L h⇠X nƒƒh⇠| lX‹$.
19
SOLUTION
t¿\ f X (k + 1)ƒƒh⇠– t⌧ƒ Ë ⌅X ›D Ãq\‰.
0|⌧ ⇠Y ¿©ï– Xt h⇠ f X nƒƒh⇠î
f (n) (x) =
( 1)n+1 4n!
(x + 1)n+1
3 ¯Ñ
(f g)(n+1) (x)
n ✓ ◆⇣
⌘
X
n
=
f (k+1) (x)g (n k) (x) + f (k) (x)g (n+1 k) (x)
k
k=0
t‰.
=
+
k=0
n!
1)!(n + 1
k)!
f (k) (x)g (n+1 k) (x)
n!
f (k) (x)g (n+1 k) (x)
k!(n k)!
✓ ◆
n
=
f (x)g (n+1) (x)
0
◆
n ✓
X
n!
n!
+
+
f (k) (x)g (n+1 k) (x)
(k 1)!(n + 1 k)! k!(n k)!
k=1
✓ ◆
n (n+1)
+
f
(x)g(x)
n
✓
◆
n+1
=
f (x)g (n+1) (x)
0
◆ ✓ ◆◆
n ✓✓
X
n
n
+
+
f (k) (x)g (n+1 k) (x)
k 1
k
k=1
✓
◆
n + 1 (n+1)
+
f
(x)g(x)
n+1
n+1
X ✓n + 1 ◆
=
f (k) (x)g (n+1 k) (x)
k
x 6= 0–⌧ ¯Ñ •h@ êÖX‰. ⇣\, x = 0
f (x)
x
(k
k=1
n
X
8
<x2 sin 1 , x 6= 0
7. h⇠ f (x) =
X ƒh⇠ f 0 @ ‰⇠ ⌅¥–⌧
x
:0,
x=0
t¨X¿Ã 2ƒƒh⇠ f 00 î x = 0–⌧ t¨X¿ JLD Ùt‹
$.
proof ) f (x)
–⌧
n+1
X
x2 sin x1
f (0)
=
0
x
1
= x sin
 |x|
x
t¿\ DP ¨ 9@ pÑ ¨– Xt x = 0–⌧X f X ¯Ñƒ
⇠î 0t‰. 0|⌧ f 0 î ‰⇠ ⌅¥–⌧ t¨\‰. ⇣, f 0 | ›<\
ò¿¥t
8
<2x sin 1 cos 1 , x 6= 0
f 0 (x) =
x
x
:0,
x=0
k=0
t¿\ n + 1–
ï– Xt Ö⌧
t⌧ƒ Ö⌧
ùÖ⌧‰.
1Ω\‰. 0|⌧ ⇠Y
t‰. t⌧ x = 0–⌧X f X 2ƒƒh⇠| p¨Xê. ¯Ñƒ⇠X
X– Xt x = 0–⌧ f 00 t¨X$t x ! 0| L
µ8⌧ 3.2 .
1. ‰L h⇠X ˘◆✓¸ ˘ü✓D lX‹$.
2x sin x1 cos x1
f 0 (x) f 0 (0)
=
x 0
x
a) y = x3 3x + 1
1
1
1
= 2 sin
cos
sol) y 0 = 3x2 3 = 0 () x = 1, 1.
x x
x
˘ü✓@ f (1) = 1, ˘◆✓@ f ( 1) = 3t‰.
X ˘\✓t t¨t| \‰. X¿Ã ⌅ ›@ x ! 0| L ⌧∞\‰.
8. P h⇠ f @ g
ùÖX‹$.
(f g)(n) (x) =
nà ¯Ñ
n ✓ ◆
X
n
•` L ‰L |t⌅»
f (k) (x)g (n k) (x),
p
1
x+ p
x
ïYD sol) y 0 = x 1 = 0 () x = 1.
3
2x 2
˘ü✓@ f (1) = 2, ˘◆✓@ t¨X¿ Jî‰.
✓ ◆
n
n!
=
k
k!(n k)!
b) y =
2 + x x2
(x 1)2
x 5
sol) y 0 =
= 0 () x = 5.
(x 1)3
proof ) n = 1| L ⌅ ›t 1ΩXî É@ êÖX‰. t⌧ ¥§
9
ëX ⇠n– t Ö⌧ 1Ω\‰‡
Xt, n + 1à ¯Ñ
˘ü✓@ f (5) =
, ˘◆✓@ t¨X¿ Jî‰.
8
•\ h⇠ f, g– t
k=0
k
c) y =
20
¿©
SOLUTION
x2 3x
x2 + 3
3(x2 + 2x 3)
sol) y 0 =
= 0 () x = 3, 1.
(x2 + 3)2
1
3
˘ü✓@ f (1) =
, ˘◆✓@ f ( 3) = .
2
2
p
p
x 3x x
1 9x
1
sol) y 0 = p = 0 () x = .
9
2✓ x◆
1
2
1
˘◆✓@ f
= t‡, ⇣\ x < | L f (x)
0 = f (0)
9
9
3
t¿\ t h⇠î x = 0–⌧ ˘ü✓ f (0) = 0D ƒ‰.
2. ¸¥ƒ l⌅–⌧ ‰L h⇠X \◆✓¸ \ü✓D lX‹$.
f (1) = 0,
f) y = x1/2 x3/2 ; [0, 4]
1 3x
1
sol) y 0 = p = 0 () x = t‡,
3
2 x
✓ ◆
1
2
f (0) = 0, f
= p ,
3
3 3
3. ‰L h⇠ ¸¥ƒ l⌅–⌧ …‡✓
| ®P lX‹$.
f ( 2) = 3,
f ( 1) = 10,
f (3) =
f (4) =
sol) l⌅ [0, 2]–⌧ y 0 =
t‰. 0|⌧ …‡✓
t¿\ \ü✓@ 0, \◆✓@
f (2) =
p
e) y = x 1
1
sol) y 0 = p
1
2
p
x
f (b)
b
2
t‰.
3
f (a)
f (5) f (2)
=
a
5 2
2 1
1
=
=
5 2
3
t‡
f 0 (c) =
1
1
1
13
() p
=
() c =
3
3
4
2 c 1
t‰. 0|⌧ …‡✓
3
,
17
c) f (x) = (x
sol)
1
1
1
p , \◆✓@ + p t‰.
2
2
2
x2 ; [ 1, 1]
2x2
1
= 0 () x = ± p t‡,
2
2
1 x
✓
◆
1
1
f ( 1) = f (1) = 0, f ± p
=± ,
2
2
¨| ÃqXî cî c = 0t‰.
1; [2, 5]
2
,
3
1 x
d) y = 2
; [ 2, 4]
x +1
2
p
x
2x 1
sol) y 0 =
= 0 () x = 1 ± 2t‡,
(x2 + 1)2
p
3
1
1
f ( 2) = , f (1 ± 2) = ⌥ p , f (4) =
5
2
2
t¿\ \ü✓@
b) f (x) =
sol)
1
> 0t‡,
(x + 1)2
f (0) = 0,
f (a)
f (1) f ( 1)
=
a
1 ( 1)
1 ( 3)
=
=2
1 ( 1)
f 0 (c) = 2 () 2c + 2 = 2 () c = 0
15,
t¿\ \ü✓@ -22, \◆✓@ 10t‰.
x
c) y =
; [0, 2]
x+1
¨| ÃqXî ‰⇠ c
t‡
1, 3t‡,
22,
6,
2; [ 1, 1]
f (b)
b
f (3) = 4,
t¿\ \ü✓@ 0, \◆✓@ 4t‰.
b) y = x3 3x2 9x + 5; [ 2, 4]
sol) y 0 = 3x2 6x 9 = 0 () x =
f (4) =
2
t¿\ \ü✓@ -6, \◆✓@ p t‰.
3 3
a) f (x) = x2 + 2x
sol)
a) y = (x 1)2 ; [ 1, 3]
sol) y 0 = 2(x 1) = 0 () x = 1t‡,
f ( 1) = 4,
1
1
, \◆✓@ t‰.
2
2
t¿\ \ü✓@
d) y =
e) y =
3 ¯Ñ
¨| ÃqXî cî c =
13
t‰.
4
1)2/3 ; [1, 2]
f (b)
b
f (a)
f (2) f (1)
=
a
2 1
1 0
=
=1
2 1
t‡
f 0 (c) = 1 ()
21
2
3(c
1
1) 3
= 1 () c =
35
27
SOLUTION
t‰. 0|⌧ …‡✓
35
t‰.
27
¨| ÃqXî cî c =
3 ¯Ñ
6. h⇠ f (x) =
n
X
ai )2 X \ü✓D lX‹$.
(x
i=1
1
d) f (x) = x + ; [1, 4]
x
sol)
f (b)
b
sol) h⇠ f î Dò\ ¸]\ t(h⇠‰X
⇣–⌧ ˘ü✓tê \ü✓D ƒ‰. ⇣\
f (a)
f (4) f (1)
=
a
4 1
4 + 14 (1 + 1)
3
=
=
4 1
4
0
f (x) = 2
3
() 1
4
1
3
=
() c2
2
c
4
t‡ c 2 (1, 4)t¥| X¿\ …‡✓
t‰.
ai ) = 0 () x =
(x
i=1
t¿\ x =
t‡
f 0 (c) =
n
X
Pn
i=1 ai
n
–⌧ \ü✓D
n ✓ Pn
X
(
n
i=1
¨| ÃqXî cî c = 2
t‰.
(Remark: ∏(⌧ÒiD
Pn
i=1 ai
n
¿‡ ¯ ✓@
i=1 ai )
4=0
∞it¿\ у
ai
◆2
• ëå Ëî É@ \¯X …‡t‰.)
4. ‰LD ùÖX‹$.
a) h⇠ f Ù∞ l⌅ I–⌧ ¯Ñ •X‡ t l⌅–⌧ f = 0
tt f î ¡⇠h⇠t‰.
0
proof ) h⇠ f
¡⇠h⇠
D»|‡
Xê. â,
7. ‰⇠ ai
t¨\‰‡
n
X
i=1
X \ü✓D lX‹$.
|x
ai |
sol) h⇠ g(x)X ƒh⇠| lXt
f (x1 ) 6= f (x2 )
| ÃqXî ⇣ x1 , x2 2 I
¨– Xt
f (x2 )
x2
a1 < a2 < . . . < an D Ãq` L g(x) =
g 0 (x) =
Xê. t⌧ …‡✓
n
X
sgn(x
x 6= ai
ai ),
i=1
t‰ (Ë, sgn @ Ä8| ò¿¥î h⇠). t| ‰⇠ a1 , a2 , · · · , an
X ⌘Y✓t| Xt x < t| ÃqXî ⇣ t X ˘\ ¸)X
0
| ÃqXî ⇣ c 2 I
x1 ¸ x2 ¨t– t¨\‰. X¿Ã ⌅ ⇣ x –⌧ g (x) 0 0 t‡ x > t| ÃqXî ⇣ t X ˘\
›–⌧ f 0 (c) 6= 0t¿\, f 0 (x) = 0 |î
– ®⌧t‰. 0|⌧ ¸)X ⇣ x –⌧ g (x) 0 t¿\ (|ƒƒh⇠ ⇣ ï– Xt)
h⇠ gî t–⌧ ˘ü✓D ƒ‰. ⇣\, h⇠ g R–⌧ çt‡
f î ¡⇠h⇠t‰.
lim g(x) = +1t¿\ ò⇠ gî \ü✓D ƒ‰. \ü✓‰@
f (x1 )
= f 0 (c)
x1
x!±1
˘å⇣–⌧ ò¿ò¿\ 0|⌧ g(t) lX‡ê Xî \ü✓t‰.
b) h⇠ f @ g
Ù∞ l⌅ I–⌧ ¯Ñ •X‡ t l⌅–⌧
(Remark: ƒ \ X¡ ⌘Y✓@ |X¿ JD ⇠ à‰. ›⇠
f 0 = g 0 tt f (x) = g(x) + C (Cî ¡⇠)t‰.
⌧ \¯X Ω∞, ⇠¡ ¡ an/2 @ an/2+1 ¨tX ✓‰ ⌅Ä
⌘Y✓t
⇠ à‰.)
proof ) h⇠ h : I ! RD h(x) = f (x) g(x)\ XXt I
–⌧ h0 (x) = 0t¿\ ⌅ µ8⌧– Xt hî ¡⇠h⇠t‰. â,
8. ‰⇠ a > 0| L h⇠
f (x) = g(x) + C 1Ω\‰.
5. h⇠ f (x) = x x + 7x + 3x 11X ¯ò⌅î
⇠…⌘ D ƒ‰î ÉD ùÖX‹$.
4
3
2
g(x) =
¥ƒ XòX
1
1
+
1 + |x| 1 + |x
a|
X ˘ü✓D lX‹$.
proof ) ⇠…⌘ X t¨1@ f 0 (x) = 0D ÃqXî ⇣ x t
¨Xî ÉD Ùtt ©ÑX‰. f 0 (x) = 4x3 3x2 + 14x + 3@ sol) h⇠ g| ‰‹ ò¿¥t
8
‰⇠ ⌅¥–⌧ çt‡ f 0 ( 1) = 18 < 0, f 0 (1) = 18 > 0D
1
1
>
0
< 1 x + 1+a x ,
ÃqX¿\ ¨t✓ ¨– Xt f (x) = 0D ÃqXî ⇣ x
1
1
g(x) = 1+x + 1+a x ,
( 1, 1)– t¨\‰. 0|⌧ f (x)î ⇠…⌘ D ƒ‰.
>
: 1
1
1+x + 1+x a ,
22
x<0
0x<a
x a
SOLUTION
3 ¯Ñ
f 00 (x) = 6xt‰.
( 1, 3],[3, 1)X ¥Ä ( 1, 3), (3, 1) HX ®‡ ⇣ x–
x<0
XÏ f (x) > 0 t¿\ ( 1, 3],[3, 1)–⌧î ⌧ù X‡ [ 3, 3]
X ¥Ä ( 3, 3)HX ®‡ ⇣ x– XÏ f (x) < 0t¿\ [ 3, 3]
0<x<a
–⌧î ⌧⇣å\‰.
x>a
f 0 (3) = 0 t‡ f 00 (3) > 0 t¿\ 3@ f X ˘å⇣t‡ ˘ü✓@
f (3) = 54t‰. f 0 ( 3) = 0t‡ f 00 ( 3) < 0 t¿\ 3@ f X
1
1
a
t¿\
+
=
0
()
x
=
ÑD
L
⇠
˘ ⇣t‡ ˘◆✓@ f ( 3) = 54t‰.
(1 + x)2
(1 + a x)2
2
a
4
à‰. ⇣, x = –⌧ ˘ü✓D ¿p, ¯ ✓@
t‰.
2
2+a
b) f (x) = x3 6x2 + 9x
t‡ gX ƒh⇠î
8 1
1
>
< (1 x)2 + (1+a x)2 > 0,
1
1
g 0 (x) =
(1+x)2 + (1+a x)2 ,
>
:
1
1
(1+x)2 +
(1+x a)2 < 0,
9. ‰L h⇠X ˘ ⇣¸ ˘å⇣D ><‹$.
(
x, xî ¨⇠
f (x) =
0, xî 4¨⇠
sol) , 1 , 2 > 0 t| Xê.
i) x = 0x Ω∞: ¥†\ xX ¸) (x
1
°T|ƒ < nD ÃqXî ê⇠ n–
✓
◆
,x + ) = (
t
sol) f (x)| ¯ÑXt f 0 (x) = 3x2 12x + 9t‡ f 0 (x)| ¯ÑX
t f 00 (x) = 6x 12t‰.
[1, 3]X ¥Ä (1, 3) HX ®‡ ⇣ x– XÏ f 0 (x) < 0 t¿\ f
î [1, 3] –⌧ ⌧⇣åX‡ ( 1, 1], [3, 1)X ¥Ä ( 1, 1), (3, 1)
HX ®‡ ⇣ x– XÏ f (x) > 0 t¿\ f î ( 1, 1], [3, 1)
–⌧ ⌧ù \‰.
f 0 (1) = 0 t‡ f 00 (1) = 6 < 0 t¿\ x = 1 –⌧ ˘◆✓
0
00
, )| f (1) = 4 | ƒ‰. ⇣ f (3) = 0t‡ f (3) = 6 > 0t¿\
x = 3 –⌧ ˘ü✓ f (3) = 0D ƒ‰.
✓ ◆
1
n
c) f (x) = (1 + x)2 (x + 2)2
sol) f (x)| ¯ÑXt f 0 (x) = 2(x + 1)(x + 2)(2x + 3)t‡ f 0 (x)
| ¯ÑXtf 00 (x) = 12x2 + 36x + 26 t‰.✓
◆
3
3
t¿\ x = 0@ ˘⇣t D»‰.
( 1, 2] [
, 1 X ¥Ä ( 1, 2) [
, 1 HX ®‡
2
2
3
ii) x > 0x Ω∞: x
¨⇠tt xX ¥†\ ¸) (x 1 , x + 1 ) ⇣ x– XÏ f 0 (x) < 0 t¿\ f î ( 1, 2],
, 1 –⌧
– t q1 2 (x 1 , x), q2 2 (x, x + 1 )D ÃqXî ¨⇠ q1 , q2

✓
◆ 2
3
3
t¨XÏ
⌧⇣åX‡
2,
[ [ 1, 1)X ¥Ä
2,
[ ( 1, 1) H
2
f (q1 ) = q1 < x < q2 = f (q2 )
 2
3
0
2,
[ [ 1, 1)
t¿\ x 2 Q+ î ˘⇣t D»‰. Ã} x 4¨⇠tt, Ù∞l⌅ X ®‡ ⇣ x– XÏ f (x) > 0 t¿\ f î
2
x 3x
✓
◆
✓
◆
= I–⌧ y 2 Itt f (y)
0t¿\ x–⌧ ˘å⇣D
2, 2
3
3
–⌧ ⌧ù \‰. f 0
= 0 t‡ f 00
= 1 < 0 t
ƒ‰.
2
2
✓
◆
3
1
¿\ ˘◆✓@ f
=
tp f 0 ( 1) = f 0 ( 2) = 0t‡
iii) x < 0x Ω∞: x
¨⇠tt xX ¥†\ ¸) (x 2 , x+ 2 )
2
16
– t r1 2 (x 2 , x), r2 2 (x, x + 2 )D ÃqXî ¨⇠ r1 , r2 f 00 ( 1) = f 00 ( 2) = 2 > 0t¿\ ˘ü✓@ f ( 1) = 0 = f ( 2)
t‰.
t¨XÏ
f (r1 ) = r1 < x < r2 = f (r2 )
t¿\ x 2 Q î ˘⇣t D»‰. Ã} x 4¨⇠tt, Ù∞l⌅ d) f (x) = 3px xpx
3x x
= J–⌧ w 2 Jtt f (w)  0t¿\ x–⌧ ˘ ⇣D
3
3 1/2
2 , 2
sol) f (x)| ¯ÑXt f 0 (x) = x 1/2
x t‡ f 0 (x)| ¯Ñ
ƒ‰.
2
2
3 3/2 3 1/2
Xt f 00 (x) =
x
x
t‰.
4
4
0|⌧ h⇠ f î x < 0x ®‡ 4¨⇠⇣–⌧ ˘ ⇣t‡, x > 0 [0, 1] X ¥Ä (0, 1)X ®‡ ⇣ x– XÏ f 0 (x) > 0 t¿\ f
x ®‡ 4¨⇠⇣–⌧ ˘å⇣t‰.
î [0, 1] –⌧ ⌧ù X‡ [1, 1)X ¥Ä (1, 1) H– ®‡ ⇣ x–
XÏ f 0 (x) < 0 t¿\ f î [1, 1)–⌧ ⌧⇣å\‰. f î x = 1
3
µ8⌧ 3.3 .
–⌧ f 0 (1) = 0, f 00 (1) =
< 0t¿\ ˘◆✓ 2 = f (1) D ¿
1. ‰L h⇠X ù , ⇣å l⌅D
lX‡ ˘◆✓¸ ˘ü✓D
2
‡ 0  x  1 –⌧ f (0)  f (x) t¿\ ˘ü✓ f (0) = 0D ƒ‰.
lX‹$.
f
1
n
=
1
1
<0< =f
n
n
a) f (x) = x3 27x
sol) f (x)| ¯ÑXt f 0 (x) = 3x2
e) f (x) = 2x3 3x2
27t‡ f (x)| ¯ÑXt sol) f (x)| ¯ÑXt f 0 (x) = 6x2
0
23
6xt‡ f 0 (x)| ¯ÑXt
SOLUTION
16
> 0
3
t¿\ ˘ü✓ f (±1) = 3D ƒ‰. ⇣\, x 2 ( 1, 1) –⌧
f (0) = 0 f (x) t¿\ ˘◆✓ f (0) = 0D ƒ‰.
f 00 (x) = 12x 6 t‰.
( 1, 0] [ [1, 1)X ¥Ä ( 1, 0) [ (1, 1)X ®‡ ⇣ x– XÏ
f 0 (x) > 0t¿\ f î ( 1, 0] [ [1, 1)–⌧ ⌧ù X‡ [0, 1]X
¥Ä (0, 1)X ®‡ ⇣ x– XÏ f 0 (x) < 0t¿\ f î [0, 1]–⌧
⌧⇣å\‰. f î x = 0–⌧ f 0 (0) = 0, f 00 (0) = 6 < 0 t¿\
˘◆✓ f (0) = 0D ¿‡ x = 1–⌧ f 0 (1) = 0, f 00 (1) = 6 > 0
t¿\ ˘ü✓ f (1) = 1D ƒ‰.
⌧ù \‰. ⇣\ x = ±1–⌧ f 0 (±1) = 0, f 00 (±1) =
j) f (x) = x1/3 (1
2x1/3
(1 x)2/3
+
t‡
3(1 x)1/3
3x2/3
2
f 0 (x)| ¯ÑXt f 00 (x) = 5/3
t‰.
4/3
9x
(1
x)
✓
✓
◆
1
1
( 1, 0) [ 0, X ¥Ä ( 1, 0) [ 0,
HX ®‡ ⇣ x–
3
✓3
1
XÏ f 0 (x) > 0t¿\ f î ( 1, 0) [ 0, –⌧ ⌧ù \‰.
3
⇣\ ⌅ l⌅ H–⌧X x = 0X ˘\ ¸)–
tx<0<y
✓
1
tt f (x) < f (0) = 0 < f (y)t¿\ f î
1, –⌧ ⌧ù
3

✓
◆
1
1
\‰. ¯¨‡
, 1 X ¥Ä
, 1 HX ®‡ ⇣ x– XÏ
3 
3
1
f 0 (x) < 0 t¿\ f î
, 1 –⌧ ⌧⇣å\‰. î
<\ (1, 1)
3
–⌧ f 0 (x) > 0 ✓
t¿\
◆ f î [1,✓1)◆–⌧ ‰‹ ⌧ù \‰. t⌧
1
1
1
x =
–⌧ f 0
= 0, f 00
= 3 · 2 1/3 < 0 t¿\
3
3
3
p
✓
◆
3
1
4
1 3
f î x = –⌧ ˘◆✓
D ƒ‰. ⇣, x 2
,
–⌧
3
3
2 2
f (1) = 0  f (x) t¿\ ˘ü✓ f (1) = 0 D ƒ‰.
g) f (x) = 8x5 5x4 20x3
sol) f (x)| ¯ÑXt f 0 (x) = 40x4 20x3 60x2 t‡ f 0 (x)|
3
¯ÑXt f 00 (x)
60x2 120x t‰.
 = 160x
◆
✓
◆
3
3
( 1, 1] [
, 1 X ¥Ä ( 1, 1) [
, 1 X ®‡ ⇣ x
2
2 
◆
3
– XÏ f 0 (x) > 0 t¿\ f î ( 1, 1] [
, 1 –⌧ ⌧
2

✓
◆
3
3
ù X‡
1, X ¥Ä
1,
–⌧ f 0 (x)  0t¿\ f î
2
2

3
1, –⌧ ⇣å\‰. f î x = 1–⌧ f 0 ( 1) = 0, f 00 ( 1) =
2
3
100 < 0 t¿\ ˘◆✓ f ( 1) = 7D ¿‡ x = –⌧
✓ ◆
✓ ◆
✓ ◆ 2
3
3
3
513
f0
= 0, f 00
= 225 > 0 t¿\ ˘ü✓ f
=
2
2
2
16
D ƒ‰.
x1/3
x)2/3
sol) f (x)| ¯ÑXt f 0 (x) =
f) f (x) = 3x4 4x3 12x2 + 8
sol) f (x)| ¯ÑXt f 0 (x) = 12x3 12x2 24xt‡ f 0 (x)|
¯ÑXt f 00 (x) = 36x2 24x 24 t‰.
( 1, 1] [ [0, 2]X ¥Ä ( 1, 1) [ (0, 2)X ®‡ ⇣ x–
XÏ f 0 (x) < 0 t¿\ f î ( 1, 1] [ [0, 2]–⌧ ⌧⇣åX‡
[ 1, 0] [ [2, 1)X ¥Ä ( 1, 0) [ (2, 1)X ®‡ ⇣ x–
X
Ï f 0 (x) > 0t¿\ f î [ 1, 0] [ [2, 1)–⌧ ⌧ù \‰. f
î x = 1, 2–⌧ f 0 ( 1) = f 0 (2) = 0t‡ f 00 ( 1) = 36 >
0, f 00 (2) = 72 > 0 t¿\ ˘ü✓ f ( 1) = 3, f (2) = 24|
¿‡ x = 0–⌧ f 0 (0) = 0, f 00 (0) = 24 < 0 t¿\ ˘◆✓
f (0) = 8D ƒ‰.
h) f (x) = 1
3 ¯Ñ
2. tƒ ¯Ñƒ⇠ ⇣ ïD t©XÏ ‰L h⇠X ˘◆✓¸ ˘ü
✓D lX‹$.
a) f (x) = x2 4x + 5
sol) f (x)| ¯ÑXt f 0 (x) = 2x
4t‡ f 0 (x)| ¯ÑXt
00
f (x) = 2t‰.
f 0 (2) = 0 t‡ f 00 (2) = 2 > 0 t¿\ f î x = 2 –⌧ ˘ü✓
f (2) = 1D ƒ‰.
1 2/3
sol) f (x)| ¯ÑXt f 0 (x) =
x
t‡ f 0 (x)| ¯ÑXt
3
2
f 00 (x) = x 5/3 t‰.
9
f çt‡ R {0} X ®‡ ⇣ x– t f 0 (x)  0 t¿\
f î R–⌧ ⇣å\‰. ⇣\, f î ˘◆✓¸ ˘ü✓D ¿¿ Jî‰.
b) f (x) = 5 6x x2
sol) f (x)| ¯ÑXt f 0 (x) = 2x 6t‡ f 0 (x)| ¯ÑXt
f 00 (x) = 2t‰.
f 0 ( 3) = 0 t‡ f 00 ( 3) = 2 < 0 t¿\ f î x = 3–⌧
˘◆✓ f ( 3) = 14D ƒ‰.
c) f (x) = x3 3x2
sol) f (x)| ¯ÑXt f 0 (x) = 3x2 6xt‡ f 0 (x)| ¯ÑXt
8
00
sol) f (x)| ¯ÑXt f 0 (x) = p
(x2 1)t‡ f 0 (x)| ¯ÑX f (x) = 6x 6t‰.
33x
f 0 (0) = 0 t‡ f 00 (0) = 6 < 0t¿\ f î x = 0–⌧ ˘◆✓
2
8(5x
+
1)
f (0) = 0D ƒ‰. ¯¨‡ f 0 (2) = 0 t‡ f 00 (2) = 6 > 0t¿\
t f 00 (x) =
t‰.
4/3
9x
f î x = 2–⌧ ˘ü✓ f (2) = 4D ƒ‰.
( 1, 1] [ [0, 1] X ¥Ä ( 1, 1) [ (0, 1) –⌧ f 0 (x) < 0 t
¿\ f î ( 1, 1] [ [0, 1] –⌧ ⌧⇣åX‡ [ 1, 0] [ [1, 1) X
¥Ä ( 1, 0) [ (1, 1) –⌧ f 0 (x) > 0 t¿\ [ 1, 0] [ [1, 1) –⌧ d) f (x) = x5 + x
i) f (x) = x2/3 (x2
4)
24
SOLUTION
3 ¯Ñ
sol) f (x)| ¯ÑXt f 0 (x) = 5x4 + 1t‰.
¯Ï¿\ ®‡ x 2 R– XÏ f 0 (x) > 0 t¿\ f î ˘◆✓¸
˘ü✓D ¿¿ Jî‰.
e) f (x) = x4 4x2
sol) f (x)| ¯ÑXt f 0 (x) = 4x3 8xt‡ f 0 (x)| ¯ÑXt
2
f 00 (x)
p = 12x p8t‰.
p
p
2) = f ( 2)p= p
0 t‡ f 00 (
2) = p
f 00 ( 2) =
f(
p 16 > 0
t¿\ f î x =
2, 2 –⌧ ˘ü✓ f (
2) = f ( 2) = 4
D ƒ‰. f 0 (0) = 0t‡ f 00 (0) = 8 < 0t¿\ f î x = 0–⌧
˘◆✓ f (0) = 0D ƒ‰.
1
.
x
sol) ¸¥ƒ h⇠X ¯ò⌅î ‰L¸ ⇡‰.
b) y = x2 + 1 +
f) f (x) = x2 (x 1)2
sol) f (x)| ¯ÑXt f 0 (x) = 2x(2x 1)(x 1)t‡ f 0 (x)|
¯ÑXt f 00 (x) = 2(6x2 6x + 1)t‰.
f 0 (0) = 0t‡ f 00 (0) ✓
= ◆
2 > 0t¿\ f✓î ◆x = 0–⌧ ˘ü✓
1
1
f (0) = 0D ƒ‰. f 0
= 0 t‡ f 00
= 1 < 0 t¿\
2 ✓ ◆
2
1
1
1
fî x =
–⌧ ˘◆✓ f
=
D ƒ‰. f 0 (1) = 0t‡
2
2
16
f 00 (1) = 2 > 0t¿\ f î x = 1–⌧ ˘ü✓ f (1) = 0D ƒ‰.
g) f (x) = x3 (x + 2)2
sol) f (x)| ¯ÑXt f 0 (x) = x2 (5x + 6)(x + 2)t‡ f 0 (x)|
x2 x 2
¯ÑXt f 00 (x) = 4x(5x2 + 12x + 6)t‰.
4. h⇠ f (x) = 2
X ˘⇣, ¿·⇣, ⇣¸ D
0
00
x
2x + 1
f ( 2) = 0 t‡ f ( 2) =✓ 16 ◆
< 0t¿\ f î✓x =◆ 2–⌧ ˘◆
‡, ¯ò⌅X ⌧ D ¯¨‹$.
6
6
144
✓ f ( 2) = 0D ƒ‰. f 0
= 0t‡ f 00
=
> 0 sol)h⇠ f (x)X ƒh⇠@ tƒƒh⇠| ltÙê.
5 ✓
5
25
◆
6
6
3456
t¿\ f î x =
–⌧ ˘ü✓ f
=
D ƒ‰.
5
5
3125
5 x
f 0 (x) =
(x 1)3
h) f (x) = (x 1)2 (x 2)3
sol) f (x)| ¯ÑXt f 0 (x) = (x 2)2 (5x 7)(x 1)t‡ f 0 (x) t‡
| ¯ÑXt f 00 (x) = 20x3 96x2 + 150x 76t‰.
f 0 (1) = 0t‡ f 00 (1) ✓
= ◆2 < 0t¿\✓f î
◆ x = 1–⌧ ˘◆✓
7
7
18
0
00
f (1) = 0D ƒ‰. f
= 0t‡ f
=
> 0<¿\ f
5 ◆
5
25
✓
7
7
108
î x = –⌧ ˘ü✓ f
=
D ƒ‰.
5
5
3125
3. ‰L h⇠X ¯ò⌅| ¯¨‹$.
x2 + 3
a) y =
.
x 1
sol) ¸¥ƒ h⇠X ¯ò⌅î ‰L¸ ⇡‰.
f 00 (x) =
lX
2(x 7)
(x 1)4
✓
◆
9
t‰. 0|⌧ h⇠ f (x) î x = 5 –⌧ ˘⇣ 5,
D ¿‡
8
✓
◆
10
x = 7 –⌧ ¿·⇣ 7,
D ƒ‰. x ! ±1 | L f (x) ! 1
9
t‡ x ! 1 | L f (x) ! 1 t¿\ h⇠ f (x)X ⇣¸
@ x = 1, y = 1t‰. f (x)X ¯ò⌅ ⌧ @ Dò@ ⇡‰.
25
SOLUTION
3 ¯Ñ
d) x = t3 + t + 1, y = t3 + 2
sol)
dy
dy
3t2
dt
= dx
= 2
.
dx
3t + 1
dt
3. ¸¥ƒ ⇣–⌧ ‰L h⇠X ⌘ X ) ›D lX‹$.
a) y = x3
7x + 5, x = 1
µ8⌧ 3.4 .
1. h⇠ f : R ! R f (x) = x5 + 3x3 + x 7\ X⇠»‰. h⇠
sol) ⌅ h⇠| f (x)\ ì‡ f 0 (x) = 3x2
f X Ìh⇠ f 1 t¨hD Ùt‡ (f 1 )0 ( 2)X ✓D lX‹$.
–⌧X ⌘ X ) ›@
y = f 0 (1)(x
sol) f ¯Ñ •X‡ ‰⇠ ⌅¥–⌧ f 0 (x) = 5x4 + 9x2 + 1 > 0
t¿\ Ìh⇠ f 1 : R ! R
t¨X‡ ¯Ñ •X‰. ⇣,
f (1) = 2t‡ f 0 (1) = 15t¿\
1 0
(f
) ( 2) =
1
1
=
f 0 (1)
15
2. ‰L ›–⌧
a) x =
=
1
,
x
–⌧X ⌘ X ) ›@
t2 1
2t
,
y
=
t2 + 1
t2 + 1
=
2t
1
t
2t(t2 +1) 2t(t2 1)
(t2 +1)2
2(t2 +1) 4t2
(t2 +1)2
=
2
x
.
y
t‰.
c) y =
4
x2 ,
x=1
x = 1–⌧X ⌘ X ) ›@
sol)
y = f 0 (1)(x
dy
dy
dt
= dx
=
dx
dt
=
c) x = t
p
sol) ⌅ h⇠| f (x)\ ì‡ f 0 (x) =
b) x = sin t + 1, y = cos 2t
sin t, y = 1
1) + f (1) =
2 sin 2t
cos t
4 sin t = 4
4x.
1
ÑD t©Xt x = 2
x2
2) + f (2) =
sol)
dy
1
3
5
(x 2) +
4
2
3
= x+1
4
y = f 0 (2)(x
dy
dt
= dx
=
dx
dt
1)
4x + 3
x=2
sol) ⌅ h⇠| f (x)\ ì‡ f 0 (x) = 1
dy
| lX‹$.
dx
4(x
t‰.
b) y = x +
t‰.
1) + f (1) =
7ÑD t©Xt x = 1
=
p
x
4
x2
ÑD t©Xt
p
1
p (x 1) + 3
3
1
4
p x+ p
3
3
t‰.
cos t
d) y = (1 + x2 )3 (1 + x3 )2 ,
x=0
sol)
dy
dy
sin t
dt
= dx
=
.
dx
1 cos t
dt
sol) ⌅ h⇠| f (x)\ ì‡
f 0 (x) = 6x(1 + x2 )2 (1 + x3 )(1 + x + 2x3 )
26
SOLUTION
ÑD t©Xt x = 0–⌧X ⌘ X ) ›@
y = f 0 (0)(x
0) + f (0) = 0 · (x
t¿\ x =
0) + 1
=1
t‰.
e) y =
✓
x 1
x+1
◆3
3 ¯Ñ
3
3
,y =
() t = 1–⌧X ⌘ X ) ›@
2
2
3
3
y = 1(x
)+
2
2
= x+3
t‰.
,
x=2
h) x 3 + y 3 = 2,
x = 1,
y=1
6(x 1)2
ÑD t©Xt sol) ¸¥ƒ ⇣ ¸)–⌧ h⇠|
(x + 1)4
x = 2–⌧X ⌘ X ) ›@
1
1
1
3
x = ,y =
t
1 ,
3
t
2
(2 t ) 3 2
2
1
y = f 0 (2)(x 2) + f (2) =
· (x 2) +
27
27
<\ ‰⌧TXt
2
3
=
x
dy
dy
27
27
dt
= dx
= t4 (2 t3 ) 4/3
dx
dt
t‰.
t¿\ x = 1, y = 1 () t = 1–⌧X ⌘ X ) ›@
p
1
1
p
f) x + y = 1, x = , y =
y = 1(x 1) + 1
4
4
= x+2
2
2
sol) ¸¥ƒ ⇣ ¸)–⌧ h⇠| x = t , y = (1 t) , 0  t  1\ t‰.
‰⌧TXt
sol) ⌅ h⇠| f (x)\ ì‡ f 0 (x) =
dy
i) x = t2 + 1,
dy
t 1
dt
= dx
=
dx
t
dt
t¿\ x =
1
1
1
,y =
() t = –⌧X ⌘ X ) ›@
4
4
2
y=
1(x
=
x+
1
1
)+
4
4
y = t3 + 1,
sol)
dy
dy
3t
dt
= dx
=
dx
2
dt
t¿\ t = 1 () x = 2, y = 2–⌧X ⌘ X ) ›@
1
2
3
(x 2) + 2
2
3
= x 1
2
y=
t‰.
g) x3 + y 3 = 3xy,
x=
3
,
2
y=
3
2
t‰.
sol) y = xt\ P‡ Ät
3
j) x =
3
3
3
2
x + y = 3xy =) x + (xt) = 3x t
=) (t3 + 1)x3 = 3tx2
=) x =
3t
t3 + 1
t‰. ⇣\ ¸¥ƒ ⇣ ¸)–⌧X h⇠|
⌅ ›<\ ‰⌧TXt
dy
t=1
,y =
y=
t2
,
t3 + 1
t=1
sol)
3t2
dy
dy
2t t4
dt
= dx
=
dx
1 2t3
dt
t3 + 1
1
3
t x
2
2
dy
2t t4
dt
= dx
=
dx
1 2t3
dt
t
,
3
t +1
Ì–⌧ t¿\ t = 1 () x = 1 , y = 1 –⌧X ⌘ X ) ›@
2
2
1
1
y = 1(x
)+
2
2
= x+1
27
SOLUTION
t‰.
4. (Lh⇠ ¯Ñï) …t–⌧ ‰L
Ò›D ÃqXî ⇣ (x, y)
X —iD › Xê. xï–⌧ ¥§ Ù∞ l⌅ I t¨XÏ
Ò›D y = y(x)(x 2 I)\ ò¿»‰‡ ` L, i1h⇠ ¯ÑïD
dy
t©XÏ
(x 2 I)| lX‹$.
dx
5. Lh⇠ x2 xy + y 2 = 9 \ ¸¥ƒ h⇠ y = f (x) à‰. ·
⌅X ⇣ (3, 0)–⌧ f (x)X 2ƒ ¯Ñƒ⇠| lX‹$.
sol) Lh⇠ ¯ÑïD t©XÏ x2 xy + y 2 = 9X ë¿D x–
XÏ ¯ÑXt
2x
y
x
dy
dy
dy
+ 2y
= 0 =)
=
dx
dx
dx
dy
= 2 t‰. t⌧ 2x
dx (3,0)
X ë¿D x– XÏ ¯ÑXt
t‡ 0|⌧
a) x = 5y y 3
sol) Lh⇠ ¯Ñï– Xt
x = 5y
3 ¯Ñ
y 3 =) 1 = 5
=)
dy
dx
3y 2
dy
dx
2
dy
1
=
dx
5 3y 2
t‰.
b) x3 + y 3 = 1
sol) Lh⇠ ¯Ñï– Xt
y
2x + y
x + 2y
x
dy
dy
+ 2y
=0
dx
dx
d2 y
dy dy
d2 y
+
2
·
+
2y
=0
dx2
dx dx
dx2
⇣ ⌘2
dy
dy
2 dx
2 dx
2
d2 y
=)
=
2
dx
2y x
dy
dx
dy
dx
x
d2 y
= 2 t‰. ¯Ï¿\ (3, 0)–⌧ f (x)X 2ƒ
dx2 (0,3)
¯Ñƒ⇠î 2t‰.
t‡ 0|⌧
x3 + y 3 = 1 =) 3x2 + 3y 2
=)
dy
=
dx
dy
=0
dx
x2
y2
µ8⌧ 3.5 .
1. ‰L h⇠X ƒh⇠| lX‹$.
(ıÃ Ö‹\‰.)
t‰.
c) x2 y = xy 2 + 1
sol) Lh⇠ ¯Ñï– Xt
dy
dx
x2 y = xy 2 + 1 =) 2xy + x2
= y 2 + 2xy
dx
dy
dy
2xy y 2
=)
=
dx
2xy x2
t‰.
(x2
2 2
dy
dy
)=y+x
dx
dx
dy
4x3 4xy 2 y
=)
=
dx
4y 3 + 4yx2 + x
y 2 )2 = xy =) 2(x2
y 2 )(2x
2y
t‰.
e) x2/3 + y 2/3 = 1
sol) Lh⇠ ¯Ñï– Xt
x
t‰.
2/3
+y
2/3
b) y = cos(sin x2 )
dy
sol)
= 2x sin(sin(x2 )) cos(x2 ).
dx
c) y = sec2 x tan x
dy
sol)
= sec4 x + 2 tan2 x sec2 x.
dx
d) (x
y ) = xy
sol) Lh⇠ ¯Ñï– Xt
2
a) y = sin(2x cos 3x)
dy
sol)
= 2 cos(2x cos(3x))(cos(3x)
dx
2
2
dy
= 1 =) x 1/3 + y 1/3
=0
3
3
dx
dy
y 1/3
=)
=
dx
x1/3
p
d) y = (cot x)2 + 1
dy
cot x csc2 x
sol)
= p 2
.
dx
cot x + 1
p
e) y = arctan 2x
dy
1
sol)
=p
.
dx
2x(2x + 1)
f) y = (arcsin(3x2 )) 2
dy
12x
sol)
= p
.
4
dx
1 9x (arcsin(3x2 ))3
g) y = csc2 (arccos(x))
dy
2x
sol)
=
.
dx
(1 x2 )2
28
3x sin(3x)).
SOLUTION
p
dy
sinh x
p .
sol)
=
dx
2 x
2
h) y = e x /2
dy
2
sol)
= xe x /2 .
dx
p
2
u) y = sinh ex
dy
2
2
sol)
= 2xex cosh ex .
dx
i) y = e
p
dy
e x
sol)
= p .
dx
2 x
x
v) y = ln(cosh 3x)
dy
sol)
= 3 tanh 3x.
dx
ex 1
ex + 1
dy
2ex
sol)
= x
.
dx
(e + 1)2
j) y =
w) y = sin(sinh x)
dy
sol)
= cosh x cos(sinh x).
dx
k) y = earcsin x
dy
earcsin x
sol)
=p
.
dx
1 x2
2. ‰L ✓DplX‹$.
a) arcsin(
3/2)
sol) arcsin x X XÌ@ {x :
l) y = ln(ln x)
dy
1
sol)
=
.
dx
x ln x
⇡
⇡
 x  } t‡ arcsin @ sin
2p
2
p
⇡
3
¸ Ìh⇠ ƒtt⌧ sin
=
t¿\ arcsin(
3/2) =
3
2
⇡
t‰. b@ c ƒ t@ ⇡@ )ï<\ Ät ⇠¿\ tƒ Ät¸
3
@ ›µ\‰.
p
m) y = ln( x2 + 1 1)
dy
x
p
sol)
=
.
dx
x2
x2 + 1 + 1
n) y = log2 x + log3 x
dy
ln 6
sol)
=
.
dx
x ln 2 ln 3
b) arccos(1/2)
⇡
sol) arccos(1/2) = .
3
o) y = ln(ex + x2 )
dy
2x + ex
sol)
= x
.
dx
e + x2
p) y = x
3 ¯Ñ
c) arctan( 1)
sol) arctan( 1) =
x
⇡
.
4
y0
= ln x + 1 t‰.
y
3. ‰L Ò›D ùÖX‹$.
dy
a) sinh(x + y) = sinh x cosh y + cosh x sinh y
0|⌧
= y 0 = y(ln x + 1) = xx (ln x + 1)t‰.
dx
proof )
sol) ln y = x ln xt‡ Lh⇠ ¯Ñï– Xt
q) y = xsin x
sol) p)@ ⇡@ )ï<\ ƒh⇠| l` ⇠ à‰. l¥ x $Ö@
›µX‡ ıÃ ⌧‹\‰.
dy
sin x
= xsin x (cos x ln x +
).
dx
x
sinh x cosh y + cosh x sinh y
ey + e y
ex + e x ey e y
+
·
2
2
2
2
ex ey + ex e y e x ey e x e y
=
4
ex ey ex e y + e x ey e x e y
+
4
ex ey e x e y
ex+y e (x+y)
=
=
2
2
= sinh(x + y).
=
2
r) y = 3x
dy
2
sol)
= 3x 2x ln 3.
dx
s) y = sinh2 (3x 2)
dy
sol)
= 6 sinh(3x
dx
t) y = cosh
p
x
2) cosh(3x
2).
ex
e x
·
b) cosh(x + y) = cosh x cosh y + sinh x sinh y
29
SOLUTION
proof )
<\Ä0 L ⇠ à‰. t⌧
´X ¯ÑD \©Xt
cosh x cosh y + sinh x sinh y
x
x
y
y
x
x
e +e
e +e
e
e
e
=
·
+
·
2
2
2
ex ey + ex e y + e x ey + e x e y
=
4
ex ey ex e y e x ey + e x e y
+
4
ex ey + e x e y
ex+y + e (x+y)
=
=
2
2
= cosh(x + y).
y
e
2
sinh(2x) = sinh(x + x)
= sinh x cosh x + cosh x sinh x
= 2 sinh x cosh x
t‰. Ï0⌧ cosh2 x sinh2 x = 1ÑD \©X ‰.
d
t⌧
coth x = csch2 x ÑD ùÖXê. ⌅@ ⇡@ )ï<\
dx
✓
◆
d
d cosh x
coth x =
dx
dx sinh x
sinh x · sinh x cosh x · cosh x
=
sinh2 x
1
=
= csch2 x
sinh2 x
t‰.
t‰.
t¥⌧
cosh 2x = cosh(x + x)
= cosh x cosh x + sinh x sinh x
= cosh2 x + sinh2 x
t‰.
·h⇠ tanh, coth, sech, cschî ‰L¸ ⇡t
sinh x
,
cosh x
1
sechx =
,
cosh x
tanh x =
X⌧‰.
cosh x
,
sinh x
1
cschx =
.
sinh x
coth xcschxу Ùtê. tƒ ⇡@ )
✓
◆
d
d
1
cschx =
dx
dx sinh x
cosh x
=
sinh2 x
cosh x
1
=
·
sinh x sinh x
= coth xcschx
‰L Ò›D ùÖX‹$.
d
d
sinh x = cosh x t‡
cosh x = sinh x ÑD
dx
dx
ùÖXê. tî
✓
◆
d
d ex e x
ex + e x
sinh x =
=
= cosh x,
dx
dx
2
2
✓
◆
d
d ex + e x
ex e x
cosh x =
=
= sinh x
dx
dx
2
2
tanh xsechxÑD Ùtê. tî
–⌧ L ⇠ à‰.
d
»¿…<\
cschx =
dx
ï<\
coth x =
d
d
tanh x = sech2 x,
coth x = csch2 x,
dx
dx
d
d
sechx = tanh xsechx,
cschx = coth xcschx.
dx
dx
d
sechx =
dx
✓
◆
d
d
1
sechx =
dx
dx cosh x
sinh x
=
cosh2 x
sinh x
1
=
·
cosh x cosh x
= tanh xsechx
d) cosh 2x = cosh2 x + sinh2 x
proof ) b)– XXÏ
proof ) <
d
tanh x = sech2 x ÑD ùÖXê.
dx
✓
◆
d
d sinh x
tanh x =
dx
dx cosh x
cosh x · cosh x sinh x · sinh x
=
cosh2 x
1
=
= sech2 x
cosh2 x
y
c) sinh(2x) = 2 sinh x cosh x
proof ) a)– XXÏ
4.
3 ¯Ñ
t‰. 0|⌧ ùÖt ®P DÃ⌧‰.
5. ¸¥ƒ ⇣–⌧ ‰L h⇠X ⌘ X ) ›D lX‹$.
a) y = xe x , x = 0
sol) f (x) = xe x |‡ Xt f 0 (x) = (1 x)e x t‰. 0|⌧ l
Xî ⌘ X ) ›@
y = f 0 (0)(x
=x
30
0) + f (0)
SOLUTION
3 ¯Ñ
f) arcsin x + arcsin y = ⇡/2,
sol) ¸¥ƒ ›X ë¿D x–
t‰.
b) y = x ln x, x = e
sol) f (x) = x ln x|‡ Xt f 0 (x) = ln x + 1 t‰. 0|⌧ lXî
⌘ X ) ›@
y = f 0 (e)(x
= 2x
p
e
1
p
y=
x=8
sol) f (x) = log3 (log2 x) |‡ Xt f 0 (x) =
⌧ lXî ⌘ X ) ›@
1
t‰. 0|
x ln 3 ln x
=
1
1
✓
◆ p
1
1
3
p
x
+
2
2
3
p
1
2 3
p x+
3
3
arctan x + arctan y = arctan
f (x) = arctan
1
dy
+ 5x ln 5 + y3xy ln 3 + x3xy ln 3
=0
x
dx
0
f (x) =
1
(1 + 5 ln 5 + 3 ln 3)
3 ln 3
1+
1) + 1
t‰.
y = ⇡/6
dy
= sec2 x sin2 y
dx
ÑD L ⇠ à‰. 0|⌧ lXî ⌘ X ) ›@
sol) ¸¥ƒ ›X ë¿D x–
XÏ ¯ÑXt
dy
(x
dx x=⇡/3,y=⇡/6
⇡
⇡
= 1 · (x
)+
3
6
⇡
=x
6
y=
⇡
⇡
)+
3
6
x+y
1 xy
arctan x
arctan y
⇣
1
x+y
1 xy
⌘2 ·
✓
(1
xy) (x + y)( y)
(1 xy)2
◆
✓
◆
(1 xy)2
1 xy + xy + y 2
=
·
(1 xy)2 + (x + y)2
(1 xy)2
2
1+y
1
=
2
2
2
2
1 2xy + x y + x + 2xy + y
1 + x2
2
1+y
1
=
2
2
2
2
1+x +y +x y
1 + x2
2
1+y
1
=
2
2
2
(1 + y ) + x (1 + y ) 1 + x2
1
1
=
=0
1 + x2
1 + x2
t‡ lXî ⌘ X ) ›@
x = ⇡/3,
(|x| < 1, |y| < 1)
|‡ Xê. t⌧ f (x)| ¯ÑXt
D ªD ⇠ à‡ 0|⌧
1
(1 + 5 ln 5 + 3 ln 3)(x
3 ln 3
x+y
1 xy
proof ) y | ÑXX ¡⇠\ ‡ ‹§‡
d) ln x + 5x + 3xy = 8, x = 1, y = 1
sol) ⌅ Ò›X ë¿D x– XÏ ¯ÑXt
p
e) tan x + cot y = 2 3,
=0
6. ‰L Ò›D ùÖX‹$.
t‰.
y=
3/2
t‰.
y = f 0 (8)(x 8) + f (8)
1
1
=
x+1
24 ln 3 ln 2
3 ln 3 ln 2
dy
=
dx x=1,y=1
dy
y 2 dx
p
y2
t‰. 0|⌧ lXî ⌘ X )
x2
1
1
t‰.
c) y = log3 (log2 x),
+p
x2
p
dy
t‡ 0|⌧
=
dx
›@
e) + f (e)
1
x = 1/2, y =
XÏ ¯ÑXt
1
1 + x2
1
1 + x2
t‰. µ8⌧ 3.4 X 4àX a) – XXÏ f î ¡⇠h⇠t‡
f (0) = 0 t¿\ f (x) = 0 t‰. â,
arctan x + arctan y = arctan
x+y
1 xy
t‰. 0|⌧ ùÖ@ ]¨‰.
t‰.
(ƒt) ¯ÑD ¨©X¿ J‡ Ä ⇠ àî )ïƒ à‰.
31
SOLUTION
proof ) A = arctan x + arctan y|‡ Xê. ¯Ït
3 ¯Ñ
t‰. t⌧ (an , fn (an )) t h⇠ fn (an )X ¿·⇣t¿\
f 00 (an ) = (4n2
2n) sin2n 2 an
1
=) sin2 an = 1
2n
tan A = tan(arctan x + arctan y)
tan(arctan x) + tan(arctan y)
1 tan(arctan x) tan(arctan y)
x+y
=
1 xy
=
t‰. ¯Ï¿\
lim fn (an ) = lim sin2n an = lim (sin2 an )n
n!1
n!1
✓
◆n
1
1
= lim 1
=p
n!1
2n
e
n!1
t¿\
arctan x + arctan y = A
= arctan(tan A)
x+y
= arctan
1 xy
t‰.
10. h⇠ f (x) = sinh x X Ìh⇠ f
⌅¥–⌧ X⇠‡,
t‰. 0|⌧ ùÖ@ ]¨‰.
f
7. l⌅ (1, 5)–⌧ f (x) = x ln x+2x\ X⌧ h⇠ f X Ìh⇠|
f 1 | ` L, (f 1 )0 (3e)X ✓D lX‹$.
tp
sol) f (e) = 3e t‡ f 0 (x) = ln x + 3 t¿\ f 0 (e) = 4 t‰. ¨
3.4.1– XXÏ
(f
1 0
) (3e) =
d2 y
| lX‹$.
dx2 ( ⇡ , ⇡ )
3 6
sol) sin y + cos x = 1 X ë¿D x– XÏ ¯ÑXt
8. sin y + cos x = 1 | L, y 00 =
y 0 cos y
sin x = 0 =) y 0 =
dy
= 1 t‰. y 0 cos y
dx ( ⇡ , ⇡ )
3 6
XÏ ¯ÑXt
t‡ 0|⌧
y 00 cos y
(y 0 )2 sin y
t‡ 0|⌧
sin x
cos y
sin x = 0 X ë¿D x
6
n!1
h⇠ fn (x)X tƒƒh⇠| lXê.
–⌧
= 2n(2n
= (4n2
1) sin2n 2 x · cos2 x
1) sin2n 2 x · (1
2n) sin2n 2 x
1
x2 + 1
1
x
p
(1 + p
)
2
2
x+ x +1
x +1
p
1
x + x2 + 1
p
p
=
2
x+ x +1
x2 + 1
1
=p
2
x +1
D ªD ⇠ à‰.
11.
⇠n
( 1)2 ex
2n sin2n 1 x · sin x
sin2 x)
x2 + 1)
0–
XÏ h⇠ Hn (x) = ( 1)n ex
‡ ` L, ‰L Ò›D ùÖX‹$.
a) Hn+1 (x) 2xHn (x) + 2nHn 1 (x) = 0, (n
proof ) n = 1 | L,
fn0 (x) = 2n sin2n 1 x · cos x
fn00 (x) = 2n(2n
p
‰⇠
(x))0 =
(y 0 )2 sin y + cos x
cos y
9. ê⇠ n– XÏ fn (x) = sin2n x(0 < x < ⇡2 ) | L,
y = fn (x)X ¿·⇣D (an , fn (an ))t|‡ Xê. t L, ˘\
lim fn (an )D lX‹$.
sol) ∞
1
dy
2
= p t‰.
dx ( ⇡ , ⇡ )
3
3
(x) = sinh 1 x
(x) = sinh 1 x = ln(x +
(sinh 1 x)0 = p
(f
cos x = 0 =) y 00 =
1
1
ÑD ùÖX‹$.
proof ) ∞ sinh x X Ìh⇠ ‰⇠ ⌅¥–⌧ X(D àê.
sinh x
‰⇠ ⌅¥–⌧ çt‡ x ! 1 | L f (x) ! 1 t‡
x ! 1 | L f (x) ! 1t¿\ f X XÌ@ ‰⇠ ⌅¥t‡
f 0 (x) = cosh x > 0 t¿\ f î | | Qt‰. 0|⌧ f X Ì
h⇠î ‰⇠ ⌅¥–⌧ ò X⌧‰.
ex e x
y = sinh x =
t¿\ e2x 2yex 1 = 0 t‰. t⌧
2
p
x
¸X ı›D t©Xt ep
= y ± y 2 + 1 ÑD L ⇠ à‰.
p ¯p
x
x
2
e > 0 t¿\ e =py+ y + 1 t‡ 0|⌧ x = ln(y+ y 2 + 1)
2
t‰.. x = ln(y
p + y + 1) | y = x–1 XÏ mtŸ
p ܤt
2
y = ln(x + x + 1) t‡ p
0|⌧ f (x) = ln(x + x2 + 1)
t‰. t⌧ f 1 (x) = ln(x + x2 + 1) D x – XÏ ¯ÑXt
1
1
=
f 0 (e)
4
t‰.
–
4n2 sin2n an = 0
2n sin2n x
= (4x2
4n2 sin2n x
=0
32
2
2
dn
2
(e x )t|
dxn
1)
2
2 d
2
2
2
d2
(e x ) 2x( 1)ex
(e x ) + 2 · ex · e x
dx2
dx
2) 4x2 + 2
SOLUTION
3 ¯Ñ
t<\ ¸¥ƒ Ò›t 1Ω\‰. t⌧ n = k(k î 1 t¡X ê t‰. t⌧ a) – XXÏ 2xHn (x)
⇠) | L ¸¥ƒ Ò›t 1Ω\‰‡ Xê. â,
\ Hn0 (x) = 2nHn 1 (x) t‰.
Hk+1 (x)
2xHk (x) + 2kHk 1 (x)
k
2
dk+1
x2
k x2 d
= ( 1)k+1 ex
(e
)
2x(
1)
e
(e x )
dxk+1
dxk
k 1
2 d
2
+ 2k( 1)k 1 ex
(e x )
k
1
dx
=0
2
1Ω\‰‡
Xê. ⌅ Ò›X ë¿D x –
2
= 2x (Hk+1 (x)
(Hk+2 (x)
0
Hn+1
(x) = 2(n + 1)Hn (x)
t‰. ¯¨‡ a)X ›D x–
2xHk+1 (x) + 2(k + 1)Hk (x))
2xHk+1 (x) + 2(k + 1)Hk (x)) = 0
t ¯ÑXt
0
Hn+1
(x) = 2Hn (x) + 2xHn0 (x)
0
Hn+1
(x) = 2Hn (x) + 2xHn0 (x)
Hn00 (x)
ÑD L ⇠ à‡ 0|⌧
2(n + 1)Hn (x) = 2Hn (x) + 2xHn0 (x)
Hn00 (x)
t‡ ¯Ï¿\
Hn00 (x)
2xHn0 (x) + 2nHn (x) = 0
t‰. 0|⌧ ùÖ@ ]¨‰.
µ8⌧ 3.6 .
1. ‰L ˘\D lX‹$.
ex + x
x!0
x2
a) lim
1
sol) h⇠ f, g : ( 1, 1)
g(x) = x2 \ XXt
{0} ! R|
lim f (x) = 0,
t‰. 0|⌧
x!0+
Hk+2 (x)
2nHn0 1 (x)
t‡ ⇣ b)– XXÏ 2nHn0 1 (x) = Hn00 (x)t¿\
2xHk (x) + 2kHk 1 (x))
(Hk+2 (x)
=
c) Hn00 (x) 2xHn0 (x) + 2nHn (x) = 0
0
proof ) Hn+1
(x)| ƒ∞XÏ Ùê. b)– XXÏ
XÏ ¯ÑXt
k+2
2
2 d
2
dk+1
(e x ) + ( 1)k+1 ex
(e x )
k+1
dx
dxk+2
k
k
2 d
2
x2
2
k x2 d
2( 1)k ex
)
4x
(
1)
e
(e
(e x )
k
k
dx
dx
k+1
k 1
2
k x2 d
x2
k 1 x2 d
2x( 1) e
(e
) + 4xk( 1)
e
(e x )
k+1
dx
dxk 1
k
2 d
2
+ 2k( 1)k 1 ex
(e x )
k
dx
✓
k+1
k
2
2 d
2
k+1 x2 d
= 2x ( 1)
e
(e x ) 2x( 1)k ex
(e x )
k+1
dx
dxk
◆
k 1
k 1 x2 d
x2
+ 2k( 1)
e
(e
)
dxk 1
✓
k+2
k+1
2 d
2
x2
k+1 x2 d
( 1)k+2 ex
(e
)
2x(
1)
e
(e x )
dxk+2
dxk+1
◆
k
k x2 d
x2
+ 2(k + 1)( 1) e
(e
)
dxk
( 1)k+1 2xex
Hn+1 (x) = 2nHn 1 (x) t¿
f (x) = ex + x
1@
lim g(x) = 0
x!0+
2xHk+1 (x) + 2(k + 1)Hk (x) = 0
t‰. ⇣\ f @ gî ( 1, 1) {0}–⌧ ¯Ñ •X‡ f 0 (x) = ex +1
tp g 0 (x) = 2xt‰. ¯¨‡ ®‡ x 2 ( 1, 1) {0}– t
t‡ n = k + 1 | Lƒ 1ΩX¿\ ⇠Y ¿©ï– XXÏ
|g 0 (x)| > 0t‰. t⌧
¸¥ƒ Ò›@ ®‡ ëX ⇠ n – XÏ 1Ω\‰.
(Remark) Hn (x)î –t¯∏ ‰m›(Hermite polynomial)t|
f 0 (x)
lim f 0 (x) = 1,
lim g 0 (x) = 0,
lim 0
= +1
à¨p, ëêÌY–⌧ ¸\ Ò•\‰. a)X ›@ t n( –t¯∏
x!0+
x!0+
x!0+ g (x)
‰m›D `tå ƒ∞` ⇠ àå ƒ@¸î ⇣T›t‰.
t‰. 0|⌧, \<» ïY– Xt
b) Hn0 (x) = 2nHn 1 (x), (n
proof )
1)
f (x)
f 0 (x)
= lim 0
= +1
x!0+ g(x)
x!0+ g (x)
lim
n
n+1
2 d
2
2 d
2
d
Hn (x) = ( 1)n 2xex
(e x ) + ( 1)n ex
(e x )
n
dx
dx
dxn+1
= 2xHn (x) Hn+1 (x)
t‰. ⇡@ )ï<\
lim
x!0
t‰. â,
Hn0 (x) = 2xHn (x)
Hn+1 (x)
f (x)
f 0 (x)
= lim 0
=
g(x) x!0 g (x)
1
t‰. 0|⌧ lX‡ê Xî ˘\@ t¨X¿ Jî‰.
33
SOLUTION
sin x2
x!0
x
sol)
b) lim
sol)
x arctan x
1 cos x
= lim
(1+x2 )2
x!0
lim
x!0
x
e) lim
x!0 x
sol)
3x
2x
x
x)
(1
1
= lim { (1+x)
x!0 3
2
3
1
+ (1 x)
3
cos x
= 2.
= lim
x!0
3x ln(3)
2x ln(2)
1
✓ ◆
3
= ln
.
2
1 p1
x
2 · x
p = lim 1 1
lim+ p
p
1
x + sin( x) x!0+ 2 · px + cos( x) · 2p
x!0
x
= lim+
tan x
1 sec2 x
= lim
x!0 1
sin x
cos x
(cos x + 1)(cos x 1)
= lim
x!0
cos2 x(1 cos x)
1 + cos x
= lim
= 2.
x!0
cos2 x
x!0
m) lim (
x!0
sol)
1
1
p = .
2
1 + cos( x)
1
sin x
1
)
x
1
sin x
1
x sin x
1 cos x
) = lim
= lim
x!0 x sin x
x!0 sin x + x cos x
x
lim (
ex e x
x!0
sin x
sol)
f) lim
= lim
sin x
x!0 cos x + cos x
ex e x
ex + e x
= lim
= 2.
x!0
x!0
sin x
cos x
x sin x
n) lim
ex 1
x!0 ln(x + 1)
sol)
g) lim
o) lim x ln x
ex 1
ex
lim
= lim 1 = lim (x + 1)ex = 1.
x!0 ln(x + 1)
x!0
x!0
x+1
x!0+
sol)
⌧ 3.6.1. 8‡.
lim x ln x = 0.
x!0+
x!0
sol)
lim
x!0
cosh x
x2
p) lim xx
x!0+
1
cosh(x)
= lim
x!0
x2
sinh(x)
= lim
x!0
2x
cosh(x)
=
2
= 0.
x2 + 1
x!1 x ln x
sol)
x2 + 1
2x
2
lim
= lim
= lim 1 = 1.
x!1 x ln x
x!1 ln x + 1
x!1
x
lim
1
}=
p
x!0
h) lim
2
3
2x
tan x
sin x
x
lim
x!0 x
x
ln(1 + x2 )
1+x2
= lim
= 0.
x!0 ex
cos x x!0 ex + sin x
p
1 cos x
k) lim+
x
x!0
sol)
p
p
1
sin( x) · 2p
1 cos( x)
1
x
lim
= lim
= .
x
1
2
x!0+
x!0+
p
x
p
l) lim p
x + sin( x)
x!0+
sol)
x
x!0
p
3
lim
2x
sol)
p
3
(1 + x)
ln(1 + x2 )
x!0 ex
cos x
sol)
x arctan x
arctan x + x/(1 + x2 )
= lim
x!0 1
x!0
cos x
sin x
!
2
3x
x
j) lim
lim
d) lim
1
x
sol)
lim
p
3
1+x
x!0
x!0
x!0
p
3
i) lim
sin x2
cos x2 · 2x
lim
= lim
= 0.
x!0
x!0
x
1
c) lim
3 ¯Ñ
1
.
2
sol) h⇠ f (x) : (0, 1) ! R | f (x) = xx |‡ Xt
f (x) = xx = ex ln(x)
34
2
.
3
SOLUTION
3 ¯Ñ
|‡ ¯ ⇠ à‰. \∏ h⇠ ex î ‰⇠ ⌅¥–⌧ çt¿\ ‰L t) lim+ xsin x
x!0
¸ ⇡t ⌧⌧| ¿ ⇠ à‰. (∏X| ⌅t <Ät ∆î \ lim sol)
08X ⌘D ›µ\‰.)
lim xsin x = lim esin x ln x = elim(sin x·ln x)
lim+ f (x) = lim+ ex ln(x) = elim x ln(x)
x!0
x!0
x!0+
x!0+
t‰. \∏ o)à 8⌧| 8‡Xt
lim sin x · ln x = lim
0|⌧ ⌅– àî ˘\✓D xî 8⌧\ ⇣‰. 8⌧ o)– Xt
x!0+
x!0+
sin x
· x ln x = 0
x
t‰. 0|⌧
lim x ln(x) = 0
elim(sin x·ln x) = 1
x!0+
t‰.
t¿\ lim+ f (x) = e0 = 1t‰.
x!0
1
q) lim x x
u) lim x cot x
sol)
sol)
x!1
x!0
1
1
lim x x = lim e x ln(x)
x!1
x!1
t‰. \∏
1
ln(x)
= lim x = 0
x!1
x!1 1
x
x!0
v) lim x(e1/x
lim
x!1
1
lim ex ln(x) = elim x ln(x) = e0 = 1.
x!1
x!1
1
x
p
3
x3 + 2x + 5
x)
w) lim (
x!1
x!1
sol)
1
lim (ln x) x = lim e x ln(ln x)
x!1
x!1
t‰. \∏
1
ln(ln x)
= lim x = 0
x!1
x!1 ln x
x
lim
1
lim e x ln(ln x) = elim
ln(ln x)
x
eh
1
h
h!0
h!0
eh
= 1.
1
◆
(x3 + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2
· 3
(x + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2
x3 + 2x + 5 x3
= lim
x!1 (x3 + 2x + 5)2/3 + x(x3 + 2x + 5)1/3 + x2
2 + x5
= lim 1 3
=0
x!1 (x + 2x + 5)2/3 + (x3 + 2x + 5)1/3 + x
x
t‰.
t‰.
x!0
sol)
1
1
x
lim (ex + x) x = lim e x ln(e +x)
x!0
x!0
x) lim
x!0
t‰. \∏
sol)
x
ln(e + x)
lim
= lim
x!0
x!0
x
ex +1
ex +x
1
x
e +1
= lim x
=2
x!0 e + x
✓
1
x
lim
x!0
✓
1
ln(x + 1)
1
x
◆
1
ln(x + 1)
◆
ln(1 + x) x
x!0 x ln(x + 1)
= lim
= lim
t¿\, 0|⌧
1
x+1
1
x!0 ln(x + 1) +
1
x
lim e x ln(e +x) = elim
x!0
= lim+
x!1
=1
s) lim (ex + x)1/x
= lim+
sol) (a b)(a2 + ab + b2 ) = a3 b3 ÑD t©XÏ |‰. Ñ® Ñê
– (a2 + ab + b2 )4D Òt ‰. ¯Ït
p
3
lim ( x3 + 2x + 5 x)
x!1
✓ p
3
= lim ( x3 + 2x + 5 x)
t¿\, 0|⌧
x!1
1
1) = lim
x!1
1
x
ex
lim x(e1/x
t‰.
1
1)
sol)
t¿\
r) lim (ln x)
cos x
x
= lim
· cos x = 1.
x!0 sin x
sin x
lim x cot x = lim x ·
x!0
ln(ex +x)
x
= e2
= lim
1
x
x+1
(x + 1)
x!0 (x + 1)(ln(x + 1)) + x
t‰.
= lim
1
x!0 ln(x + 1) + 2
35
=
1
.
2
SOLUTION
a) lim (1 + hx)1/h = ex
h!0
proof ) x| ‡ X‡ ⌅ ›D ‰‹ t
h!0
ln(1+hx)
h
lim
t‰. \∏ \<» ïYD t©Xt
lim
f (x + h)
h)
f 0 (x + h) + f 0 (x
h!0
2
= lim
h)
= f 0 (x)
t‰.
ln(1 + hx)
x
= lim
=x
h!0 1 + hx
h
2f (x) + f (x h)
= f 00 (x)
h2
proof ) ÑXX x 2 R D ›Xê. Ñ®@ ÑêX h⇠î ( 1, 1)
–⌧ ¯Ñ •Xp, h 2 ( 1, 1) {0}–⌧ |(h2 )0 | = |2h| > 0
t‰. ⇣\ Ñ® Ñê ®P h ! 0 | L 0<\ ⇠4X¿\ \<»
¨| ¯ ⇠ à‰. 0|⌧
b) lim
f (x + h)
h!0
t¿\
elim
ln(1+hx)
h
= ex
t‰.
b) lim 1 +
n!1
proof )
x n
= ex
n
lim
1
x n
= lim 1 + hx) h
h!0
n
\ ∏¥ ¯ ⇠ à<¿\ a)à– Xt ˘\✓@ ex t‰.
lim 1 +
n!1
3. ‰L ˘\D lX‹$.
(ıÃ Ö‹\‰.)
f (x + h)
h!0
2f (x) + f (x
h2
h)
= lim
f 0 (x + h)
f 0 (x
h)
2h
h!0
t‰. h⇠ f Pà ¯Ñ •X¿\ a)à Ät¸ D
˘\✓ f 00 (x)| ªD ⇠ à‰.
©Xt
2
5. ˘\ lim (e 1/x )cos(x) 1 | lX‹$.
x!0
sol) ⌅ ›D ‰‹ t
a) lim (1 + 2x)1/x
2
lim (e 1/x )cos(x) 1 = elim
x!0
1
cos(x)
x2
x!0
sol)
lim (1 + 2x)1/x = lim (1 + 2x)2/2x = e2 .
x!0
x!0
t‰. \∏
1 n
n
b) lim 1
n!1
sol)
lim
1
x!0
cos(x)
sin(x)
1
= lim
=
x!0 2x
x2
2
t¿\
lim 1
n!1
1 n
1 ( n)( 1)
) = lim 1 +
)
= e 1.
n!1
n
( n)
1 n
c) lim 1 +
n!1
3n
sol)
lim 1 +
n!1
d) lim (1 +
n!1
sol)
f (x
2h
h!0
h!0
h!0
f (x + h)
h!0
2. \<» ïYD t©XÏ ‰LD ùÖX‹$.
1
Ñ
f (x h)
= f 0 (x)
2h
proof ) Ñ®@ ÑêX h⇠î (-1,1)–⌧ ¯Ñ •Xp, h 2
( 1, 1) {0}–⌧ |(2h)0 | = |2| > 0 t‰. ⇣\ Ñ®, Ñê ®
P h 0<\
L 0<\ ⇠4X¿\ \<» ¨| ¯ ⇠ à‰.
0|⌧
a) lim
lim (1 + hx)1/h = lim e h ln(1+hx) = elim
4
elim
1
cos(x)
x2
= e1/2
t‰.
1 n
1 3n· 1
= lim 1 +
) 3 = e1/3 .
n!1
3n
3n
3 2n
)
n
4
Ñ
µ8⌧ 4.1 .
1. ‰L
ÑX ✓D
3 2n
3 n3 ·6
lim 1 +
= lim 1 +
= e6 .
n!1
n!1
n
n
X| ¨©XÏ lX‹$.
Z 1
1
(x + 2)dx
4. h⇠ f
P à ¯Ñ •` L \<» ïYD t©XÏ ‰LD
ùÖX‹$.
sol) < Ñh⇠ l⌅ [ 1, 1]–⌧ çt¿\, ®‡ ¨Ãit
⇡@ ⇠\ ⇠4\‰. l⌅ [ 1, 1]| n ÒÑXÏ ª@ Ñ`⇣
36
SOLUTION
4
2k
(k = 0, 1, 2, . . . , n)|
n
›Xt
Ñ
X ål⌅–⌧X \¯ sol) ‰L¸ ⇡t ƒ∞\‰.
⇢
⇣<\
ln(n + 2) ln n ln(n + 4) ln n
ln(3n) ln n
lim
+
+ ···
n!1
n+2
n+4
3n
Z 1
n 
X
2k
2
n
X
(x + 2)dx = lim
1+
+2
ln(n + 2k) ln n
n!1
n
n
= lim
1
k=1
n!1
n + 2k
◆
k=1
n ✓
X
2k 2
n
X
= lim
1+
ln 1 + 2k
1
2
n
n!1
n n
=
lim
·
k=1
2k
n!1
2
n
1+ n
✓
◆
k=1
4 n(n + 1)
Z
Z
3
ln
3
= lim 2 + 2 ·
=4
1
ln x
1
1
2
n!1
n
2
=
dx =
tdt = (ln 3)
2 1 x
2 0
4
t‰.
µ8⌧ 4.2 .
1. ‰L Ä
ÑD lX‹$.
2.
Ñ<\ X⌧ ‰L h⇠‰X ƒh⇠| lX‹$.
(ıÃ Ö‹\‰.)
(ıÃ Ö‹\‰.)
Z
2x7 + 3x4 5x2 1
Z x
a)
dx
p
3
x3
a) f (x) =
t 1dt
sol)
1
sol)
Z
p
3
2x7 + 3x4 5x2 1
2
3
1
f 0 (x) = x 1.
dx = x5 + x2 5 ln |x| + 2 + C.
3
x
5
2
2x
Z sin(x) p
b) f (x) =
3 + 2t4 dt
(Ë, C î ¡⇠t‰. ^<\ C î t E–⌧ ‰x ∏ t ∆î \
0
Ñ¡⇠\ Ë \‰.)
sol)
q
Z
f 0 (x) = 3 + 2 sin4 (x) · cos(x).
p
3
b) (2x4 + x 1)dx
xk =
1+
c) g(x) =
Z x4
e
p
sol)
t
dt,
x>0
x2
sol)
g 0 (x) = 4x3 ex
3. h⇠ g(x) =
Z sin 1 x
p
1
2
2xex .
sin tdt (0 < x < 1)| L, g
0
X ✓D lX‹$.
sol) ¯ ÑYX 0¯ ¨– XXÏ
g 0 (x) =
=
q
p
1
1
0
✓
1
(
2
sin(sin 1 x) · (sin 1 x)0
x· p
1
1
◆
Z
p
c)
x
✓ ◆ r
1
2
t‰. 0|⌧ g
=
t‰.
2
3
lim
p
3
x
1)dx =
x
1dx =
2 5 3 4/3
x + x
5
4
x + C.
1dx
sol)
Z
3x 1
d)
dx
2x2
sol)
Z
p
Z
3x 1
3
1
dx = ln |x| +
+ C.
2
2x
2
2x
2
(x
3
1)3/2 + C.
Z
1 + cos3 x
e)
dx
cos2 x
sol)
Z
4. ‰L ˘\D lX‹$.
n!1
(2x4 +
1
=p
1+x
x2
0
⇢
Z
ln(n + 2) ln n ln(n + 4) ln n
ln(3n) ln n
+
+ ···
n+2
n+4
3n
37
Z
1 + cos3 x
dx = (sec2 x + cos x)dx
cos2 x
= tan x + sin x + C.
SOLUTION
Z
2
f) p
dx
2 2x2
sol)
p
Z
Z
2
2
p
p
dx =
dx
2
2 2x
1 x2
p
= 2 arcsin(x) + C.
Z
g)
t‰.
Z
c)
4
x
dx
2x2 (1 + ln x)
sol) u = 1 + ln x|‡ Pê. ¯Ït
Z
3
dx
4 + (2x)2
x
dx =
2x2 (1 + ln x)
3
dx =
4 + (2x)2
=
Z
h) 2e2x dx
sol)
Z
Z
i) 5 sin(4x)dx
sol)
Z
Z
1
ln |u| + C
2
1
= ln |1 + ln x| + C
2
3
dx
4(1 + x2 )
3
arctan(x) + C.
4
t‰.
Z
p
d) x2 x3
3dx
sol) u = x3
3|‡ Pê. ¯Ït
2e2x dx = e2x + C.
5 sin(4x)dx =
Z
x2
p
x3
5
cos(4x) + C.
4
du
= 3x2 t‡
dx
Z
1p
3dx =
udu
3
3
2
= (u) 2 + C
9
3
2
= (x3 3) 2 + C
9
t‰.
Z
2 sin x
e)
dx
cos3 x
µ8⌧ 4.3 .
1. ‰L
Ä
ÑD lX‹$.
Z
a)
1
du
2u
=
sol)
Z
Z
du
= 1/xt‡
dx
sin x cos3 x dx
du
sol) u = cos x|‡ Pê. ¯Ït
= sin xt‡
dx
Z
Z
sin x cos3 xdx =
u3 du
=
=
u4
+C
4
cos4 x
+C
4
t‰.
Z
2
b) xex dx
du
sol) u = x2 |‡ Pê. ¯Ït
= 2xt‡
dx
Z
Z u
e
x2
xe dx =
du
2
eu
=
+C
2
2
ex
=
+C
2
du
= sin xt‡
dx
Z
Z
2 sin x
2
dx
=
du
cos3 x
u3
= (u) 2 + C
sol) u = cos x|‡ Pê. ¯Ït
= sec2 x + C
t‰.
Z
f) etan x sec2 x dx
du
= sec2 xt‡
dx
Z
Z
etan x sec2 xdx = eu du
sol) u = tan x|‡ Pê. ¯Ït
= eu + C = etan x + C
t‰.
Z
g) 2(ln x)2 dx
38
Ñ
SOLUTION
sol) u = ln x|‡ Pê. ¯Ït
Z
Z
du
= 1/xt‡
dx
1
2(ln x)2 dx = 2x(ln x)2 · dx
x
Z
Z
u 2
= 2e (u )du = 2u2 eu du
t‡ D∑\ )ï<\
Z
e x cos x dx =
=
4ueu + 4eu + C
= 2eu (u2
2u + 2) + C
= 2x((ln x)2
x
e
cos x
e x cos x
Z
e x sin x dx
Z
e x sin xdx)
I
e x sin x + ( e x cos x
I=
e x sin x
=
=) 2I =
=) I =
e x cos x
I
x
x
e sin x e cos x
1 x
e (sin x + cos x) + C
2
t‰.
2 ln x + 2) + C
2. h⇠ F
‹$.
t‰.
Z
h) x2 cos 2x dx
h⇠ f X –‹h⇠| L ‰L Ò›t 1ΩhD Ùt
Z
f (↵x + ) =
1
F (↵x + ) + C,
↵
(↵ 6= 0)
✓
◆0
sol) f (x) = x2 , g 0 (x) = cos 2x|‡ Pê. ¯Ït ÄÑ Ñï–
1
proof )
F (↵x + ) = f (↵x + )t¿\ Ä
Xt
↵
Z
Z
1 2
0|
2
Z
x sin 2x dx
x cos 2x dx = x sin 2x
1
2
f (↵x + ) = F (↵x + ) + C
↵
t‡. ‰‹ f (x) = x, g 0 (x) = sin(2x)|‡ Pt,
Z
Z
1
1
x sin 2xdx =
x cos 2x +
cos 2x dx
2
2
1
1
=
x cos 2x + sin 2x + C ⇤
2
4
t¿\
Z
1
1
1
x2 cos(2x)dx = x2 sin 2x (
x cos 2x + sin 2x) + C
2
2
4
1 2
1
1
= x sin 2x + x cos 2x
sin 2x + C
2
2
4
1Ω\‰.
3. ‰L
a)
Z 2
1
ÑD lX‹$.
|x| dx
sol)
Z 2
1
t‰.
i)
Z
|x| dx =
=
e
x
sol) I =
=
e x sin x dx\ P‡, f (x) = sin x, g 0 (x) = e x \
Pt ÄÑ Ñï– Xt
Z
I = e x sin x dx =
e
x
sin x +
Z
e x cos x dx
Z 0
1
Z 0
Z 1/p2 p
b)
1
1
39
x2 dx

1
|x| dx +
Z 2
( x)dx +
0
|x| dx
Z 2
0
xdx
 2 2
0
x2
x
+
2
2 0
1
1 4
5
= + = .
2 2
2
sin x dx
Z
Ñ
t¿\
t‰. f (u) = u2 , g 0 (u) = 2eu |‡ Xê. ¯Ït ÄÑ Ñï–
Xt
Z
Z
Z
2 u
0
2u e du = f (u) · g (u)du = f (u)g(u)
f 0 (u)g(u)du
Z
= 2u2 eu
4ueu du
= 2u2 eu
4
ÑX
X–
SOLUTION
sol) x2 = t\ Pt XX Ñ– Xt
sol)
x = sin t\ P‡ XXXt ‰Lt 1Ω\‰.
Z 1/p2 p
1
x2 dx =
1
Z ⇡/4
⇡/2
=
=
Z 3
1
1
1
sin 2t + t
4
2
⇡/2
g)
Z 1
(x
1
 2
x
2x3 )dx =
2
✓
9
=
2
x
2
81
2
36.
◆
✓
1
2
1
2
a) y = e2x ,
3
Z 2
1
1 3
t dt
2
y = x2 ,
x = 0,
ÌX ◆t| lX‹$.
x=1
5
3 4
5 8
x3 + x5
4
8
✓
◆1
3 4
5 8
3 5
=
53 + 55
( + )
4
8
4 8
3 4
5 8
= (5 3 1) + (5 5 1).
4
8
(x 3 + x 5 )dx =
1
x3 dx +
µ8⌧ 4.4 .
1. ‰L– ¸¥ƒ · ‰\ Xϯx

x(x2 + 1)3 dx
 4 1  4 2
x
t
=
+
4 0
8 1
1
1
17
= + (2
)=
.
4
8
8
◆
3
1
Z 1
Z 1
0
0
1
sol)
Z ⇡2
4
1 t
e4 e
e
=
2 1
2
0
=
1
e)
(x3 + (x2 + 1)3 x)dx =
4 3
(x 3 + x 5 )dx
Z 5
x3 dx +
1
(x3 + (x2 + 1)3 x)dx
Z 1
=
Z 5
Z 1

sol)
sol)
1
1 t
e dt
2
0
0
Z 3
Z 4
t‰.
⇡/4
1
d)
xe dx =
cos 2t + 1
dt
2
⇡/2
2x3 )dx
(x
x2
=
1 3⇡
= +
.
4
8
c)
Z 2
|cos t| cos tdt
Z ⇡/4

4
sol)
10
y = e2x
y = x2
8
sin x cos2 xdx
6
0
sol) cos x = t\ Pt XX Ñ– Xt
Z ⇡2
0
sin x cos2 xdx =
Z 0
4
t2 dt
2
1
 3 1
t
1
2
=
t dt =
=
3 0
3
0
Z 1
1
0.5
0.5
1
1.5
t‰.
f)
Z 2
lX‡ê Xî
ÌX ◆tî
Z 1
2
xex dx
1
0
40
(e2x
x2 )dx
2
Ñ
SOLUTION
2
t‰. t| ƒ∞Xt
Z 1

y = 2 sin x
y = sin 2x
1
1 2x 1 3
e
x
2
3
✓ 2
◆ ✓0
e
1
1
=
2
3
2
2
e
5
=
2
6
(e2x
x2 )dx =
0
1
◆
0
1
2
3
4
5
1
t‰.
b) y =
4
p
1
x,
x = 0,
y=0
2
lX‡ê Xî
sol)
y=
2
p
1
ÌX ◆tî
Z ⇡
x
(2 sin x
sin 2x)dx
0
t‰. t| ƒ∞Xt
Z ⇡
1.5

⇡
1
2 cos x + cos 2x
2
0
✓
◆ ✓
◆
1
1
2+
=4
= 2+
2
2
(2 sin x
sin 2x)dx =
0
1
t‰.
0.5
4
3
lX‡ê Xî
2
1
1
2
x
x2 + 3
,
y=
1,
x = 0,
x=2
sol)
ÌX ◆tî
Z 1
d) y = p
2
p
1
p
y = x/ x2 + 3
y= 1
xdx
0
1
t‰. t| ƒ∞Xt
Z 1
p
1
xdx =
0

=0
2
(1
3
✓
3
1
x) 2
0
◆
2
2
=
3
3
1
1
2
3
1
t‰.
2
c) y = 2 sin x,
y = sin 2x,
x = 0,
x=⇡
lX‡ê Xî
ÌX ◆tî
Z 2✓
sol)
0
41
p
x
x2 + 3
◆
+ 1 dx
4
Ñ
SOLUTION
4
t‰. t| ƒ∞Xt
Z 2✓
0
p
x
x2 + 3
1
◆
+ 1 dx =
Z 2
Z 2
x
p
y = x3
y = x3
1
dx +
1dx
x2 + 3
0
Z 7
1
p dt + 2
=
3 2 t
hp i 7
p
p
=
t +2= 7
3+2
0
1
1
2
3
1
t‰.
e) y = x4 ,
x2
y = 2x
lX‡ê Xî
ÌX ◆tî
Z 1
sol)
1
y = x4
y = 2x x2
1.5
Z 1
1
|x3
1
x 3 |dx = 2
0.5
1
2
3

g) y = sin
⇡
x,
2
sol)
(2x
x2
x4 )dx
y = sin ⇡2 x
y=x
t‰. t| ƒ∞Xt
x2
0
1

x4 )dx = x2
=1
1
1 3 1 5
x
x
3
5
0
1 1
7
=
3 5
15
1
t‰.
f) y =
1
y=x
0
(2x
x3 )dx
0
ÌX ◆tî
Z 1
Z 1
1
(x 3
t‰.
0.5
lX‡ê Xî
Z 1
3 4
x4
x3
4
4
✓
◆ 0
3 1
=2
=1
4 4
=2
1
x 3 |dx
t‰. t| ƒ∞Xt
1
2
1
|x3
1
2
1
p
3
x,
y = x3
lX‡ê Xî
ÌX ◆tî
Z 1
sol)
1
42
|sin
⇡
x
2
x|dx
( m)
Ñ
SOLUTION
4
5
t‰. t| ƒ∞Xt
Z 1
⇡
|sin x
2
1
x|dx = 2
Z 1⇣

0
⌘
⇡
sin x
2
y=
( m)
x dx
cos x
4
1
2
⇡
x2
cos x
⇡
2
2 0
✓✓
◆ ✓
◆◆
1
2
4
=2
0
0
=
2
⇡
⇡
=2
p
Ñ
3
1
2
t‰.
1
2. ‰L · <\ Xϯx ÌD ¸¥ƒ ï ⇣î ¡ D ⌘Ï<
\ å⌅‹® å⌅¥X Ä<| lX‹$.
3
a) y =
p
x,
y = 0,
x = 4;
xï
2
1
1
⌅ ¯º–⌧ `tƒ
<î
⇡
y=
p
Z ⇡/2
p
( cos x)2 dx
0
x
t‰. t| ƒ∞Xt
4
⇡
Z ⇡/2
p
( cos x)2 dx = ⇡
0
3
2
t‰.
1
c) y = 1
1
1
2
⌅ ¯º–⌧ `tƒ
<î
3
4
5
x2 ,
y=
⇡
p
Z ⇡/2
cos xdx
0
⇡/2
= ⇡ [sin x]0
1;
=⇡
yï
sol)
y=1
y=
ÌD xï<\ å⌅‹0D L å⌅¥X Ä
Z 4
3
ÌD xï<\ å⌅‹0D L å⌅¥X Ä
sol)
5
2
0.5
x2
1
2
( x) dx
0
2
t‰. t| ƒ∞Xt
⇡
Z 4
p
2
( x) dx = ⇡
0
Z 4
1
1
2
0.5
xdx
0
 2 4
x
=⇡
= 8⇡
2 0
1
1.5
t‰.
b) y =
p
cos x,
y = 0,
x = 0,
x=
⇡
;
2
xï
⌅ ¯º–⌧ `tƒ
<î
sol)
ÌD yï<\ å⌅‹0D L å⌅¥X Ä
Z 1 p
⇡
( 1
1
43
y)2 dy
SOLUTION
4
p
y = x+1
y=2
t‰. t| ƒ∞Xt
Z 1 p
⇡
( 1
y)2 dy = ⇡
1
Z 1

=⇡ y
=⇡
4
(1
Ñ
y)dy
1
✓✓
1
y2
2
1
1
2
1
2
◆
✓
1
2
1
◆◆
= 2⇡
4
2
4
2
6
t‰.
d) x =
p
y,
2
sol)
3
x=y ;
yï
⌅ ¯º–⌧ `tƒ ÌD y = 3D 0 <\ å⌅‹0D L å
⌅¥X Ä<î `tƒ ÌD yï<\ 3Ã|, xï<\ 1Ã|
…âtŸ ‹® ƒ xï– t å⌅‹® å⌅¥X Ä<@ ⇡<¿\
sol)
3
4
p
x= y
x = y3
3
p
y= x 3
y= 1
2
1
2
4
1
2
2
2
4
6
2
Z 4
( 1)2 dx
Z 4
( 1)2 dx
1
1
1
2
2
1
lX‡ê Xî Ä<î
⌅ ¯º–⌧ `tƒ
<î
⇡
ÌD yï<\ å⌅‹0D L å⌅¥X Ä
⇡
Z 1
p
( y)2 dy
⇡
0
Z 1
⇡
p
2
( y) dy
0
(y 3 )2 dy
0
( x
3) dx
⇡
0
t‰. t| ƒ∞Xt
Z 4
( x
=⇡
(x
⇡
0
 2 1
 7 1
y
y
⇡
(y 3 )2 dy = ⇡
⇡
2
7 0
0
0
⇡ ⇡
5
=
=
⇡
2
7
14
Z 1
t‰.
e)y =
p
0
t‰. t| ƒ∞Xt
Z 1
Z 4
p
Z 4
2
3) dx
⇡
0
p
6 x + 9)dx
=⇡
 2
x
2
4
4x3/2 + 9x
4⇡
0
= ⇡ (8
32 + 36)
4⇡ = 8⇡
x = 1,
y = 1;
x=1
t‰.
p
x + 1,
x=
1,
y = 2;
y=3
4⇡
0
f) y = x4 + 1,
44
SOLUTION
4
y = x4 + 1
y=1
3
Ñ
y = x2
y=k
2
2
1
1
1
1
2
2
3
1
1
sol)
1
2
1
⌅ ¯º–⌧ `tƒ ÌD x = 1D 0 <\ å⌅‹0D L å sol) ⌅ ¯º–⌧ `tƒ ÌD y = kD 0 <\ å⌅‹0D
⌅¥X Ä<î `tƒ ÌD yï<\ 1Ã|, xï<\ 1Ã| L å⌅¥X Ä<î Dtƒ ÌD yï<\ kÃ| …âtŸƒ
…âtŸ ‹® ƒ yï– t å⌅‹® å⌅¥X Ä<@ ⇡<¿\ xï– t å⌅‹® å⌅¥X Ä<@ ⇡<¿\
y = (x + 1)4
y=0
2
2
1
1
2
2
y = x2 k
y=0
1
1
1
2
1
2
(k
x2 )2 dx
1
1
2
3
2
lX‡ê Xî Ä<î
lX‡ê Xî Ä<î
⇡
Z 1
⇡
(y
1/4
2
1) dy
0
t‰. t| ƒ∞Xt
⇡
Z 1
(y 1/4
1)2 dy = ⇡
0

p
k
t‰. t| ƒ∞Xt
Z pk
Z pk
2 2
⇡ p (k x ) dx = 2⇡
(k
k
(y 1/2
2y 1/4 + 1)dy
1
x2 )2 dx
0
= 2⇡
0
2 3/2 8 5/4
y
y
+y
3
5
0
✓
◆
2 8
⇡
=⇡
+1 =
3 5
15
=⇡
t‰.
Z 1
Z pk
Z pk

(k 2
0
= 2⇡ k 2 x
=
16 2 p
⇡k k
15
2kx2 + x4 )dx
p
2 3 1
kx +
3
5 0
k
y = x2 ¸ y = k\ Xϯx ÌD y = k| ⌘Ï<\
81⇡
16 2 p
81
å⌅‹® å⌅¥X Ä<
t¿\
⇡k k =
⇡ t‰.
10
15
10
2
9
3. · y = x ¸ y = k\ Xϯx ÌD y = k| ⌘Ï<\
0|⌧ k = t‰.
81⇡
4
å⌅‹® å⌅¥X Ä<
| L, ë⇠ kX ✓D lX‹$.
10
t‰. ·
45
SOLUTION
2
4. · y = e x + x@ x = 1, x = 2,xï<\ Xϯx
ïD ⌘Ï<\ å⌅‹® å⌅¥X Ä<| lX‹$.
ÌD y
Z x0
4
k sin x)dx = 1D Ãq\‰.
(2 cos x
0
Z x0
2
= 2 sin x0 + k cos x0
t‡, tan x0 =
2
1
2
t‰. 0|⌧
Z x0
2 cos x
2
1
Z 2
2xe x dx + ⇡
⇡
e t dt + ⇡
2
1
=
Z 4
1
⇥
⇤
t 4

p
Z 2
Z 2
2x2 dx
4
k2
k2 + 4
k=1
3
2
µ8⌧ 4.5 .
1. ‰L πt ÑX ✓D lX‹$.
a)
Z
1
2
xe x dx
sol)
Z 1
1
2
xe x dx =
1
=
b)
Z 1
sol)
M! 1
0
lim
M! 1
1
.
2e
Z
1
0
M
✓
2
◆t| · y = k sin x tÒÑ` L, ¡⇠ kX ✓D lX‹$.
⇡
sol) 0  x 
–⌧ y = cos x @ y = k sin x Ãòî ⇣
2
Z ⇡/2
X xå\| x0 | Xê.
2 cos xdx = 2 t¿\, lXî kî
lim
M! 1
◆
1 1 1 M2
e + e
2
2
1
dx = lim
M !1
(x + 1)2
= lim
ÌX
2
xe x dx =
⇥
1 x2 ⇤ 1
e
M
2
1
dx
(x + 1)2
Z 1
,xï,yï<\ Xϯx
lim
=
t‰.
y = 2 cos x 0  x 
k2
k2 + 4
2
a
5. ·
k2 + 4
+p
t‰.
1
2⇡xf (x)dx
⇡⌘
+p
k=
[8‡] –µ ï– X\ å⌅¥X Ä<
ëX l⌅ [a, b]–⌧ h+✓t ëx çh⇠ y = f (x)X ¯ò⌅
@ x ï ✏ P ⇠¡ x = a,x = b\ Xϯx ÌD y ïD
⌘Ï<\ å⌅‹® å⌅¥X Ä< V î
⇣
4
k2 + 4
1
t‰.
V =
4
k
k2 + 4
–⌧
2x2 dx
2
= ⇡ e 1 + ⇡ x3
3
1
14⇡
= ⇡( e 4 + e 1 ) +
3
14
= ⇡(e 1 e 4 + )
3
Z b
, cos x0 = p
t¿\
sol) –µ ï– XXÏ lX‡êXî å⌅¥X Ä<î
2⇡x(e x + x)dx = ⇡
2
k2 + 4
k sin xdx = p
0
3
k
2
t¿\
k
sin x0 = p
1
Z 2
x
k sin x)dx = [2 sin x + k cos x]0 0
(2 cos x
0
y =e x +x
1
Ñ
M !1
Z M
0
1
dx
(x + 1)2
1
1
dt
t2
Z M +1
1 ⇤M +1
t 1
1
= lim [
+ 1] = 1.
M !1
M +1
= lim
M !1
0
46
⇥
SOLUTION
c)
Z ln 3
sol)
0
ex
p x
e
1
Z ln 3
0
ex
p x
= lim
e
1 ✏!0
Z ln 3
ex
p x
dx
e
1
✏
Z 2
1
p dt
= lim
✏!0 e✏ 1
t
p 2
= lim [2 t]e✏ 1
✏!0
p
p
p
= lim []2 2 2 e✏ 1] = 2 2.
4
µ8⌧ 4.6 .
1. Ñ l⌅D 8ÒÑXÏ ‰L
ÑX ¸ø✓D ¨‰¨4 )
ï¸ Ï® )ï<\
lX‹$.
a)
Z 8
x3 dx
1
sol) f (x) = x3 <\ Pt
x
1
1.875
2.75
3.625
4.5
5.375
6.25
7.125
8
✏!0
d)
Z 1
sol)
xe x dx
0
Z 1
0
xe x dx = lim
M !1
Z M
⇥
= lim
M !1
= lim
M !1
= lim
M !1
2. p
xe x dx
0
✓
xe
⇤
x M
0
+
Z M
0
M
+
eM
Z M
M
eM
1
+1
eM
Z 1
1
dx
xp
ëX ¡⇠| L, πt Ñ
0
e x dx
0
◆
!
= 1.
✏!0
lim+ ln(✏) = 1
✏!0
t‰.
f (x)
1
6.5917968
20.796875
47.634765
91.125
155.2871093
244.140625
361.7050781
512
t‰. ¨‰¨4 )ï<\ ƒ∞\ )ï@
f (1) + 2f (1.875) + · · · + 2f (7.125) + f (8) 7
· = 1035.80859305,
2
8
t‡, Ï® )ï<\ ƒ∞\ )ï@
f (1) + 4f (1.875) + 2f (2.75) + · · · + 4f (7.125) + f (8) 7
· = 1023.7499
3
8
⇠4Xî pX î⌅|
lX‹$.
sol)
case 1) p = 1 x Ω∞,
Z 1
Z 1
1
1
dx = lim+
dx
x
x
✏!0
0
✏
= lim+ [ln x]1✏
=
e x dx
!
Ñ
t‰. ‰⌧
Ñ<\ ƒ∞\ ✓@
Z 8
4095
x3 dx =
= 1023.75
4
1
t‰.
Z 1
2
b)
e x dx
0
2
sol) f (x) = e x <\ Pt
x
0.000
0.125
0.250
0.375
0.500
0.625
0.750
0.875
1.000
case 2) ¯ xX Ω∞
Z 1
Z 1
1
1
dx = lim+
dx
p
p
x
x
✏!0
0
✏
1
= lim+ [
x1 p ]1✏
p
✏!0 1
1
=
lim (1 ✏1 p )
1 p ✏!0+
f (x)
1.0000000
0.9844964
0.9394131
0.8688151
0.7788008
0.6766338
0.5697828
0.4650432
0.3678794
t‰. 0 < p < 1| Ω∞ 0 < 1 p < 1t ⇠¿\ ¸¥ƒ ˘\@ t‰. ¨‰¨4 )ï<\ ƒ∞\ )ï@
1/(1 p)\ ⇠4Xp, p > 1| Ω∞ 1 p < 0t ⇠¥ ⌧∞\‰.
f (0) + 2f (0.125) + · · · + 2f (0.875) + f (1)
0|⌧ 0 < p < 1x Ω∞–Ã πt Ñt ⇠4\‰.
= 0.7458656,
16
47
SOLUTION
t‡, Ï® )ï<\ ƒ∞\ )ï@
d)
f (0) + 4f (0.125) + 2f (0.250) + · · · + 2f (0.750)
24
4f (0.875) + f (1)
+
24
= 0.7468261
5 4\ ⇠
1 ✓
X
5
n=1
⇠
sol)
1
X
1
n=1
Xt
1 ✓
X
5
n=1
t‰.
t‰.
◆
7
+ n
2n
3
2
2n
+
n
1
X
1
,
n=1
3n
@ ⇠4Xî
◆
7
ƒ ⇠4Xî
3n
⇠t¿\
⇠t‡ ¯ ˘\@ 5+
◆
1
1
5 4\ ⇠
n+1 n+2
n=1
sol) Sn D ⇠X ÄÑi<\ Pt
µ8⌧ 5.1 .
◆
X ✓ 1
1. ‰L ⇠X ⇠4, ⌧∞ ÏÄ| ⇣ X‡, ⇠4Xî Ω∞– ¯
1
1
S
=
=
n
˘\D lX‹$.
k+1 k+2
2
n
e)
1
n+2
3n
n=1
sol) Sn D
⇠X ÄÑi<\ Pt
Sn =
n
X
1
k=1
1 1
= ·
3k
3 1
1
3n
1
3
1
=
2
✓
1
1
3n
1
t‡ lim Sn = t¿\ Sn @ ⇠4\‰. 0|⌧
n!1
2
1
⇠4X‡, ¯ ˘\@ t‰.
2
b)
1
t‡ lim Sn
=
t¿\ Sn @ ⇠4\‰.
n!1
2
✓
◆
1
X
1
1
1
@ ⇠4X‡, ¯ ˘\@ t‰.
n+1 n+2
2
n=1
1
X
1
◆
⇠
1
X
1
3n
n=1
n
X
1
k=1
1 1
= ·
2k
2 1
1
2n
1
2
=
✓
1
1
2n
Sn =
=
◆
t‡ lim Sn = 1t¿\ Sn @ ⇠4\‰. 0|⌧
n!1
⇠4X‡, ¯ ˘\@ 1t‰.
⇠
n
X
(2k
k=1
n ✓
X
1
2
1
1)(2k + 1)
k=1
1
2k
1
1
2k + 1
◆
=
1
t‡ lim Sn
=
t¿\ Sn @
n!1
2
✓
◆
1
X
1
1
= t‰.
(2n 1)(2n + 1)
2
n=1
⇠X ÄÑi<\ Pt
Sn =
1
X
@
2n
sol) Sn D
0|⌧
1
(2n
1)(2n
+ 1)
n=1
sol) Sn D ⇠X ÄÑi<\ Pt
f)
1
X
1
n=1
7
17
=
2
2
1 ✓
X
k=1
a)
¨ 5.1.2–
g)
1
X
1
2n
n=1
1
X
p
( n+1
p
1
2
1
4n + 2
⇠4\‰.
0|⌧
n)
n=1
@
sol) Sn D
⇠X ÄÑi<\ Pt
Sn =
n
X
p
n+1
p
n =
p
n+1
1
k=1
c)
1 ✓
X
1
n=1
sol)
1
+ n
2n
3
⇠
1
X
1
n=1
Xt
1 ✓
X
1
n=1
t‰.
2
2n
+
n
◆
,
t‡
1
X
1
n=1
3n
@ ⇠4Xî
◆
1
ƒ ⇠4Xî
3n
⇠t¿\
¨ 5.1.2–
⇠t‡ ¯ ˘\@ 1 +
1
3
=
2
2
1
X
lim Sn
p
n+1
n=1
2. ‰L
a)
p
1
t¿\
Sn @
⌧∞\‰.
n = 1 t‰.
⇠X ⇠4, ⌧∞ ÏÄ| ⇣ X‹$.
1
X
2n2 + 1
n=1
48
=
n!1
n2 + 1
0|⌧
SOLUTION
2n2 + 1
<\ Pt lim an = 2 6= 0t¿\ ¸¥ƒ f î
n!1
n2 + 1
⇠î ⌧∞\‰.
5 4\ ⇠
XÌ–⌧ ⇣åX¿\
sol) an =
b)
3n
n+1
n=1
1
X
( 1)n sin
n=1
f (x)dx =
1
1
X
=
t¿\ an @ ⇠4Xî ⇠Ùt D»‰. 0|⌧ ¸¥ƒ
∞\‰.
⇠î
⇠¥ πt Ñ
1
X
⇠î ⌧
n=1
c)
µ8⌧ 5.2 .
1. ‰L ⇠X ⇠4, ⌧∞ ÏÄ| ⇣ X‹$.
1
a)
7n
+1
n=1
1
1
⇣ ⌘
5 px 2
5
dx
+1
Z 1
f (x)dxt ⇠4\‰. 0|⌧,
⇠
1
1
n2 + 5
ƒ ⇠4\‰.
1
X
1
(n + 1)3
n=1
t‡ f î
t L,
sol) f : [1, 1) ! R| f (x) =
XÌ–⌧ ⇣åX¿\
Z 1
Z 1
f (x)dx =
1
\ XXt f (x) > 0t‡
7x + 1
Ñ⇣ ïD ¨©` ⇠ à‰. t L,
1
\ XXt f (x) > 0
(x + 1)3
XÌ–⌧ ⇣åX¿\ Ñ⇣ ïD ¨©` ⇠ à‰.
Z 1
Z 1
⇠¥ πt Ñ
f (x)dxt ⌧∞\‰. 0|⌧,
⇠
1
1
ƒ ⌧∞\‰.
7n
+1
n=1
1
b)
2+5
n
n=1
sol) f : [1, 1) ! R| f (x) =
Z 1
f (x)dxt ⇠4\‰. 0|⌧,
⇠
1
1
X
1
ƒ ⇠4\‰.
(n
+
1)3
n=1
d)
1
X
Z 1
1
dx
(x
+
1)3
1

M
1
1
= lim
M !1
2 (x + 1)2 1
✓
◆
1
1
1
= lim
M !1
2 (M + 1)2
4
1
=
8
f (x)dx =
1
1
dx
7x
+1
1

M
1
= lim
ln(7x + 1)
M !1 7
1
✓
◆
1
1
= lim
ln(7M + 1)
ln 8
M !1 7
7
=1
⇠¥ πt Ñ
dx
sol) f : [1, 1) ! R| f (x) =
1
X
1
1
x2 + 5
1
1
p 2
=
dt
p
5t +1
1/ 5

M
1
= lim p arctan t p
M !1
5
1/ 5
✓
◆
1
1
= lim p
arctan M arctan p
M !1
5
5
✓
◆
1
⇡
1
=p
arctan p
5 2
5
n⇡
2
n⇡
<\ Pt
2
8
>
< 1, n = 4k + 1
an = 1,
n = 4k + 3
>
:
0,
txX Ω∞
1
X
Z1 1
Z 1
sol) an = ( 1)n sin
fî
Z 1
1
3n
sol) an =
<\ Pt lim an = 1t¿\ ¸¥ƒ
n!1
n+1
⌧∞\‰.
c)
Z 1
Ñ⇣ ïD ¨©` ⇠ à‰. t L,
1
X
n2 e n
n=2
1
\
2
x +5
XXt f (x) > 0t‡
sol) f : [2, 1) ! R| f (x) = x2 e x \ XXt f (x) > 0t‡
XÌ–⌧ f 0 = (x2 2x)e x  0t¿\ f î ⇣åh⇠t‰.
49
SOLUTION
0|⌧
Ñ⇣ ïD ©` ⇠ à‡,
Z 1
Z 1
f (x)dx =
x2 e x dx
2
2
= lim
M !1
= lim
⇥
M !1
=
⇠¥ πt Ñ
1
X
10
e2
Z 1
5 4\ ⇠
⇠¥ πt Ñ
1
1
X
⇤M
e x (x2 + 2x + 2) 2
✓ 2
◆
M + 2M + 2 10
eM
e2
g)
1
n
ln
n
n=2
f (x)dxt ⇠4\‰. 0|⌧,
î
⇠
XÌ–⌧ ⇣åX¿\
Z 1
3+8 n
n4
n=1
M
= lim [ln t]ln 2
p
M !1
3+8 x
\ XXt f (x) > 0
x4
XÌ–⌧ f î ⇣åX¿\ Ñ⇣ ïD ¨©` ⇠ à‰.
sol) f : [1, 1) ! R| f (x) =
= lim (ln M
M !1
⇠¥ πt Ñ
p
3+8 x
f (x)dx =
dx
x4
1
1

M
16 5/2
= lim
x 3
x
M !1
5
1
◆
✓
✓
◆
1
16 1
16
= lim
1
M !1
M3
5 M 5/2
5
21
=
5
Z 1
⇠¥ πt Ñ
f (x)dxt ⇠4\‰. 0|⌧,
⇠
1
p
1
X
3+8 n
ƒ ⇠4\‰.
n4
n=1
f)
Z 1
1
X
h)
1
X
n=1
n
sol) f : [1, 1) ! R | f (x) =
t‡ XÌ–⌧ f î ⇣åX¿\
t L,
Z 1
f (x)dx = lim
M !1
1
x2 + 1
\ XXt f (x) > 0
x3 + 3x
XÌ–⌧ f î ⇣åX¿\ Ñ) ïD ¨©` ⇠ à‰.
1
Z M

1
n=1
M
1
ln(x3 + 3x)
M !1 3
1
✓
1
3
= lim
ln(M + 3M )
M !1 3
= lim
⇠¥ πt Ñ
1
X
x2 + 1
dx
x3 + 3x
◆
1
ln 4 = 1
3
i)
Z M

1
\
Z 1
XXt f (x) > 0
Ñ⇣ ïD ¨©` ⇠ à‰.
x
dx
x2 + 1
M
◆
1
ln 2 = 1
2
f (x)dx @ ⌧∞\‰. 0|⌧,
⇠
1
n
n2 + 1
1
X
n=1
p
3
ƒ ⌧∞\‰.
1
n+1
sol) f : [1, 1) ! R| f (x) = p
3
50
x
x2 + 1
1
ln(x2 + 1)
M !1 2
1
✓
1
ln(M 2 + 1)
= lim
M !1 2
= lim
x2 + 1
dx = lim
M !1
x3 + 3x
⇠
n2 + 1
sol) f : [1, 1) ! R | f (x) =
Z 1
f (x)dxt ⌧∞\‰. 0|⌧,
1
ƒ ⌧∞\‰.
n ln n
n=2
n3 + 3n
t‡
t L,
Z 1
ln ln 2) = 1
2
1
X
n2 + 1
n=1
Z 1
1
dx
x
ln
x
2
Z M
1
= lim
dt
M !1 ln 2 t
f (x)dx =
2
p
Z 1
1
\ XXt f (x) > 0t‡ f
x ln x
Ñ⇣ ïD ¨©` ⇠ à‰. t L,
sol) f : [2, 1) ! R| f (x) =
n2 e n ƒ ⇠4\‰.
t‡
t L,
⇠
1
X
2
1
X
x2 + 1
dxî ⌧∞\‰. 0|⌧,
x3 + 3x
n2 + 1
ƒ ⌧∞\‰.
n3 + 3n
n=1
n=2
e)
Z 1
1
\
x+1
XXt f (x) > 0t‡
SOLUTION
XÌ–⌧ f î ⇣åX¿\ Ñ⇣ ïD ¨©` ⇠ à‰. t L,
Z 1
Z M
1
1
p
p
dx = lim
dx
3
3
M
!1
x
+
1
x
+1
1
1

M
3
= lim
(x + 1)2/3
M !1 2
1

3
3 2/3
2/3
= lim
(M + 1)
2
=1
M !1 2
2
Z 1
1
p
⇠¥ πt Ñ
dxî ⌧∞\‰. 0|⌧,
3
x
+1
1
1
X
1
p
ƒ ⌧∞\‰.
3
x+1
n=1
j)
1
X
ln n
n=2
b)
1
X
5 4\ ⇠
n(1 + 4n2 )p
n=1
sol)
⇠
4x2 )p dx
1
X
2 p
n(1 + 4n ) t ⇠4Xî É@ πt Ñ
Z 1
x(1 +
1
n=1
⇠4Xî ɸ ŸXt‰. t L,
Z 1
Z M p
t
x(1 + 4x2 )p dx = lim
dt
M !1 5
8
1
Z M
1
= lim
dt
M !1 5
8t p
p > 1, â p < 1| L ⇠4X‡ p  1, â
1
X
1| L ⌧∞X¿\ ⇠
n(1 + 4n2 )p î p < 1| L
t‡ ⌅ ˘\@
p
n=1
n2
⇠4X‡ p
1| L ⌧∞\‰.
ln n
ln x
= 0 t‡ f (x) = 2 @ x 2–⌧ ë⇠t‰.
2
n!1 n
x
3. ¨ 5.1.3X Ì@ 8x ? 8<t ÌD ùÖX‡, p”tt
x 2x ln x
0
⇣\ x
2–⌧ f (x) =
<
0
t¿\
⇣åh⇠t‰.
⇠@| >‡ $ÖX‹$.
x4
proof ) ¨ 5.1.3X Ì@ 8t D»‰.
⇣\
1
Z 1
Z 1
an =
t| Xê. ¯Ït lim an = 0 t‰. ¯Ïò Ù0 5.2.1
ln x
u
n!1
n
dx
=
du
(*
ln
x
=
u)
1
u
X
x2
1
2
ln 2 e
Z a
– XXÏ
@ ⌧∞\‰. 0|⌧ ¨ 5.1.3X Ì@ 1ΩX¿
u
n
n=1
= lim
du
a!1 ln 2 eu
Jî‰.
Z a
⇥
⇤
u a
u
= lim
ue
+ lim
e du
ln 2
a!1
a!1 ln 2
µ8⌧ 5.3 .
✓
◆
✓
◆
1. ‰L ⇠X ⇠4, ⌧∞ ÏÄ| ⇣ X‹$.
ln 2
1
u
1
= lim
+
+ lim
+
a
a
a!1
a!1
e
2
e
2
1
X
n 1
ln 2 + 1
a)
=
2n
2
n=1
n 1
1
X
sol) an =
t| Xê. ¯Ït
ln n
2n
t¿\ Ñ⇣ ï– XXÏ
t
⇠4\‰.
2
n
n=2
n
an+1
n+1
lim
= lim 2n 1
n!1 an
n!1
2. ‰L– ¸¥ƒ ⇠ ⇠4Xî p ✓D lX‹$.
22
n
1
= lim
= <1
1
n!1
2(n 1)
2
X
1
a)
p
1
n(ln n)
X
n=2
n 1
t¿\ D(⇣ ï– XX@ ⇠
@ ⇠4\‰.
1
2n
X
1
n=1
sol)
⇠
⇠4\‰î
É@
πt Ñ
1
n(ln n)p
X
n=2
1
Z 1
b)
1
2
n +n+4
dx ⇠4\‰î ɸ ŸXt‰. t L,
n=1
x(ln x)p
2
1
1
Z 1
Z M
sol) n
1– t 0 < 2
< 2 t‡ p ⇠ ⇣
1
1
n
+
n
+
4
n
dx = lim
dt
1
X
M !1 ln 2 tp
x(ln x)p
1
2
ï– Xt ⇠
⇠4X¿\ DP⇣ ï– Xt ⇠
n2
t‡ ⌅ ˘\@ p > 1| L ⇠4X‡ p  1| L ⌧∞X¿\ ⇠
n=1
1
1
X
X
1
1
@
p
>
1|
L
⇠4X‡
p

1|
L
⌧∞\‰.
@ ⇠4\‰.
p
2+n+4
n(ln
n)
n
n=2
n=1
sol) <
lim
51
SOLUTION
c)
1
X
n3 + 1
n4
n=4
p
1
X
1
5 4\ ⇠
p
n+5
n
5
n
n=10
@ ⇠4\‰.
1
1
X
X
4 + ( 1)n
1
n3 + 1
1
p
sol) n 4– t 0 < < 4
t‡ ⇠
t ⌧∞X¿ h)
(n + 1) n
n
n
1
n
n=1
n=4
1
X
n3 + 1
1
X
4 + ( 1)n
5
5
\ DP⇣ ï– Xt ⇠
@ ⌧∞\‰.
4
p  3/2 t‡ ⇠
sol) n 1– t 0 <
t
n
1
3/2
n=4
(n + 1) n
n
n
n=1
1
1
X
X
4 + ( 1)n
1 + cos2 n
p @ ⇠4\‰.
⇠4X¿\
DP⇣
ï–
Xt
⇠
d)
(n + 1) n
2n + 3
n=1
n=1
1
X
n
1 + cos2 n
2
1
p
i)
<
<
t‡
⇠
n
n
n
1
4
2 +3
2 +3
2
n +1
n=1
1
1
X
X
1
1 + cos2 n
t ⇠4X¿\ DP⇣ ï– Xt ⇠
sol) n 1– t
2n 1
2n + 3
n=1
n=1
@ ⇠4\‰.
sol) n
e)
1–
t0<
n
n4 + 1
1
X
sin4 n + cos2 n
n=1
n2 + 1
1
X
sin4 n + cos2 n
2
2

t‡
⇠
2
2
n +1
n
n2
n=1
1
X sin4 n + cos2 n
t ⇠4X¿\ DP⇣ ï– Xt ⇠
@
n2 + 1
n=1
⇠4\‰.
sol) n
f)
p
1–
t0<
1
X
n + 1 + ln n
n=1
t‡
1
X
n=1
j)
n2
p
⇠
n
1
X
n
n4 + n4
n
=p
2n2
1
=p
2n
p
1
p
t ⌧∞X¿\ DP⇣ ï– Xt
2n
n=1
n4 + 1
⇠
@ ⌧∞\‰.
1
X
2n
1 + 4n
n=1
1
X
2n
2n
1
1

=
t‡
⇠
t
1
n
n + 1 + ln n
n
1 + 4n
4n
2n
2
sol) n 1– t 0 < = 2 
t‡
⇠
n=1
2
n
n
n
n
1
n=1
X
2n
1
X
⇠4X¿\ DP⇣ ï– Xt ⇠
@ ⇠4\‰.
n + 1 + ln n
1 + 4n
t ⌧∞X¿\ DP⇣ ï– Xt ⇠
@ ⌧∞
n=1
2
n
n=1
1
X
\‰.
1 + 3n + n2
p
k)
1 + 3n2 + n6
n=1
p
p
1
X n+5
n 5
g)
sol) n 1– t
n
n=10
1 + 3n + n2
1 + 3n + n2
p
p
sol) n 10– t
1 + 3n2 + n6
n6 + 3n6 + n6
p
p
1 + 3n + n2
p
=
n+5
n 5
10
p
= p
5n3
n
n( n + 5 + n 5)
2
n
1
p
10
=p
3
 p
5n
5n
n n
1
X
1
10
p
t‡
⇠
t ⌧∞X¿\ DP⇣ ï– Xt
⇠
= 3/2
n
5n
n=1
1
X
1 + 3n + n2
1
X
10
p
@ ⌧∞\‰.
t‡
⇠
t ⇠4X¿\ DP⇣ ï– Xt
⇠
1 + 3n2 + n6
3/2
n=1
n
n=10
1
X
1
sol) n
52
1–
t0<
SOLUTION
l)
1
X
6 + 2n + ln n
n=1
µ8⌧ 5.4 .
1. ‰L– ¸¥ƒ ⇠
∞Xî¿ ⇣ X‹$.
(1 + 3n + n2 )2
sol) n
t
1–
6 + 2n + ln n
6 + 2n + ln n

2
2
(1 + 3n + n )
(n2 )2
6n + 2n + n

n4
9
= 3
n
t‡
1
X
9
⇠
n=1
1
X
6 + 2n + ln n
n=1
t ⇠4X¿\ DP⇣ ï– Xt
n3
(1 + 3n + n2 )2
@ ⇠4\‰.
1
X
sin ⇡
m)
4
⇠
1
X
1
n2
n=1
tan2
n=1
2. ‰L
=1
1
n2
t ⇠4X¿\ ˘\DP⇣ ï– Xt
✓ ◆
1
@ ⇠4\‰.
n
⇠
⇠4Xî p✓D lX‹$.
1 ✓
X
n=1
sol) p = 1tt
1 ✓
X
n=1
t¿\ ¸¥ƒ
1 ✓
X
n=1
t ⇠¥
( 1)n
2n2
5n3 + 4
f 0 (x) =
1
n
tan2
n!1
1
X
1
X
tan x
= 1ÑD t©Xt
x
lim
t‡
1
\ Pt an @ ëm Ëp⇣å⇠Ùt‡ ⇣\
n+1
1
X
( 1)n
p
lim an = 0t¿\ P
⇠⇣ ï– Xt ⇠
@
n!1
n+1
n=1
⇠
1
1
X
X
1
p
⇠4\‰. X¿Ã ⇠
|an | =
@ ⌧∞X¿\
n+1
n=1
n=1
⇠4X¿î Jî‰.
sol) an = p
2n2
\ Pê. an > 0 t‡ lim an = 0Ñ@ ˘X
n!1
5n3 + 4
2x2
‰. t⌧ h⇠ f : [2, 1) ! RD f (x) = 3
\ XXt f î
5x + 4
¯Ñ •X‡ XÌ–⌧
✓ ◆
1
n)
tan
n
n=1
x!0
⇠4Xî¿ D»t ⌧
sol) an =
2
sol) lim
⇠4Xî¿,
1
X
( 1)n
p
n+1
n=1
n=1
sol) êÖà ⌧∞\‰.
1
X
a)
b)
n
n=1
5 4\ ⇠
p
n+3
p
n+3
1
n+9
◆
◆
=
1
X
6
(n + 3)(n + 9)
n=1
⇠î ⇠4\‰. p 6= 1tt
p
n+3
1
n+9
⇠¥ f î Ëp⇣å\‰. 0|⌧ an @ n = 2 Ä0 Ë
1
X
2n2
p⇣åX¿\ P
⇠⇣ ï– Xt
⇠
( 1)n 3
5n + 4
⇠
n=2
@ ⇠4\‰. (0|⌧ n = 1| L–ƒ ⇠4) X¿Ã
⇠
1
1
2
X
X
2n
|an | =
@ ⌧∞X¿\
⇠4X¿î Jî‰.
3+4
5n
n=1
n=1
c)
1
X
( 1)n
n=1
sol) an =
⌧∞\‰.
1
n+9
◆
=
2x(8 5x3 )
0
(5x3 + 4)2
d)
(p 1)n + 9p 3
(n + 3)(n + 9)
n=1
3n 1
3
\ Pt lim an = 6= 0t¿\ ¸¥ƒ
n!1
2n + 1
2
1
X
2n + 1
( 1)n 1 n
3 +2
n=1
sol) n
1
X
3n 1
2n + 1
1–
t
2n + 1
2n + 1
( 1)n 1 n
= n
3 +2
3 +2
✓ ◆n ✓ ◆n
2n + 1
2
1

=
+
3n
3
3
⇠î ⌧∞\‰.
53
⇠î
SOLUTION
t‡
1 ✓ ◆n
X
2
¸
⇠4\‰.
DP⇣ ï–
3
n=1
ƒ
1
X
1
n=1
3
t ⇠4X¿\
n
1
X
2n + 1
( 1)n 1 n
3 +2
n=1
X‡ 0|⌧
1
X
2n + 1
( 1)n 1 n
î
3 +2
n=1
e)
XXÏ
1 ✓ ◆n
X
2
n=1
1
n
X
3
2 +1
3n + 2
n=1
+
✓ ◆n
1
3
t
sol) an = ( 1)n
t ⇠¥ D(⇣ ï– Xt
2. ‰L
⇠î
n3/4
=
1
X
( 1)n
n=1
a)
sin(0.1)
n3/4
n!
n=1
sol) n
n!1
n3/4
1
X
sin(0.1)
n=1
f)
n3/4
1
X
î ⇠4\‰. X¿Ã
( 1) sin
n=1
✓
1
p
n
1
X
n=1
@ ⌧∞X¿\
n
⇠
t
4–
cos(n⇡/3)
1

n!
n!
1
 2
n
|an | =
⇠4X¿î Jî‰.
t‡
⇠
1
X
1
n=1
⇠4X¿\
n2
⇣ ï– Xt
◆
⇠
=
sin
n!
n=4
✓
p1
n
n!1 sin p1
n
‰.
g)
1
X
n=1
=1
1
X
1
p t ⌧∞X¿\ ˘\DP⇣ ï– Xt ⌧∞\
⇠
n
n=1
( 1)n
nn
n!
1
X
1
n=4
î
n2
n!
lim an = lim
=
t¿\
c)
1
X
n!
n=2
54
⇠
nn
1
X
1
n!1 n sin ⇡
n
n!1
1
6= 0
⇡
1
î ⌧∞\‰.
n sin n⇡
n=8
⇣\ ⇠4X‡ DP
⇠4\‰. 0|⌧
1
X
cos(n⇡/3)
lim
t‡
⇠
1
X
cos(n⇡/3)
◆
1
n=4
p \ Pt an @ ëm Ëp⇣å⇠Ùt
n
⇣\ ⇠4\‰.
‡ ⇣\ lim an = 0t¿\ P
⇠⇣ ï– Xt
n!1
✓
◆
1
1
X
X
1
1
⇠
( 1)n sin p @ ⇠4\‰. X¿Ã
⇠
|an | =
X
1
n
b)
n=1
n=1
⇡
✓
◆
1
n
sin
X
n
n=8
1
sin p
@
n
1
n=1
sol) an =
\ Pt
n sin n⇡
sol) an
nn
@ ⌧∞\‰. ⇣\
n!
1
X
cos(n⇡/3)
sin(0.1)
\ ì<t an @ ëm ⇣
n3/4
å⇠Ùt‡ lim an = 0t‰. 0|⌧ P
⇠⇣ ï– Xt
n=1
( 1)n
⇠X ⇠4, ⌧∞D ⇣ X‹$.
\ ‰‹ ¯ ⇠ à‡ an =
1
X
sin(n⇡ + 0.1)
1
X
n=1
1
X
sin(n⇡ + 0.1)
⇠
⇠
⇠4X¿ƒ Jî‰.
sin(n⇡ + 0.1)
n3/4
n=1
n=1
t |an | > 0t‡
1–
lim
ƒ ⇠4\‰. ¯Ï¿\
⇠4\‰.
nn
<\ Pt n
n!
|an+1 |
(n + 1)n+1 /(n + 1)!
= lim
n!1 |an |
n!1
nn /n!
✓
◆n
1
= lim 1 +
=e>1
n!1
n
⇠4
1
X
sol) ⌅
5 4\ ⇠
SOLUTION
sol) an =
n!
<\ Pt
nn
g)
|an+1 |
(n + 1)!
nn
= lim
·
n!1 |an |
n!1 (n + 1)n+1
n!
✓
◆n
n
= lim
n!1 n + 1
✓
◆n
1
= lim 1
=e 1<1
n!1
n+1
lim
t¿\ D(⇣ ï– Xt
d)
1
X
( 5)n
n!
n=2
sol) an =
⇠
1
X
n!
@ ⇠4\‰.
nn
n=2
( 5)
<\ Pt
n!
e)
1
X
n!1
1
X
( 5)n
@ ⇠4\‰.
n!
n=2
sol) an =
1
n
n!1 31/n
t‡
1
X
1
⇠
n
n=2
1
X
p
n
( 3
1
=
⇠
t ⌧∞X¿\ ˘\DP⇣ ï– Xt
1
X
1
@ ⇠4\‰.
(2n)!
n=1
q ⇠@ L|Ï
¨
2 · 4 · 6 · · · (2n)
<\ Pt
n!
an+1
(3x)n+1
= lim
= lim 3|x| = 3|x|
n!1
n!1
n!1
an
(3x)n
–⌧ ⇢ < 1 x Ω∞
⇠
|an+1 |
2 · 4 · 6 · · · (2n) · (2n + 2)
n!
lim
= lim
·
n!1 |an |
n!1
(n + 1)!
2 · 4 · 6 · · · (2n)
2n + 2
= lim
=2>1
n!1 n + 1
n=1
⇠
1
X
⇠4X¿\
(3x)n
@ ⌧∞\‰.
1
1
<x<
3
3
⇠4\‰.
n=0
1
|L
3
⇣, x =
1
X
n
(3x) =
n=0
t¿\
t¿\
1
X
( 1)n
n=0
⇠î ƒŸX‡ ⇠4X¿ Jî‰.
1
|L
3
(3x)n =
n=0
n!
lX‹$.
⇢ = lim
1
X
1
X
2 · 4 · 6 · · · (2n)
⇠
1) @ ⌧∞\‰.
¯¨‡ x =
⇠
1
ln 3
n=2
n!
t¿\ D(⇣ ï– Xt
= ln 3ÑD t©Xt
lim
x î⌅–⌧
1
X
2 · 4 · 6 · · · (2n)
n=1
1
x
x!0
⇢ = 3|x| < 1 ()
|an+1 |
(2n)!
= lim
n!1 (2n + 2)!
|an |
1
= lim
=0<1
n!1 (2n + 2)(2n + 1)
t¿\ D(⇣ ï– Xt
3x
n=0
1
<\ Pt
(2n)!
lim
sol) lim
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an = (3x)n \ P‡
D(⇣ ïD t©Xt
1
(2n)!
n=1
sol) an =
f)
⇠
1)
µ8⌧ 6.1 .
1. ‰L q ⇠X ⇠4⇠¿Ñ¸ ⇠4l⌅D
1
X
a)
(3x)n
|an+1 |
5n+1
n!
lim
= lim
·
n!1 |an |
n!1 (n + 1)! 5n
5
= lim
=0<1
n!1 n + 1
t¿\ D(⇣ ï– Xt
1
X
p
n
( 3
¨
n=2
6
n
6 q ⇠@ L|Ï
1
X
n=0
(1)n =
1
X
1
n=0
⇠î ⌧∞\‰.
1
¯Ï¿\ t ⇠X ⇠4⇠¿Ñ@ R = t‡ ⇠4l⌅@
3
✓
◆
1 1
,
t‰.
3 3
b)
1
X
n=0
55
(2x
1)n
SOLUTION
1
1
| Lî êÖX‰. x 6= | L, an = (2x
2
2
P‡ D(⇣ ïD t©Xt
sol) x =
an+1
(2x 1)
= lim
n!1
an
(2x 1)n
n!1
1
X
(2x
–⌧ ⇢ < 1 x Ω∞
⇠
⇠
1
X
t¿\
1)n
⇠4\‰.
d)
n=0
⇣, x = 0| L
1
X
1)n =
(2x
n=0
t¿\
( 1)n
n
(2x
1) =
1
X
n
(1) =
n=0
1
X
(3x+1)n+1
7·2n+1
(3x+1)n
7·2n
–⌧ ⇢ < 1 x Ω∞
⇠4X¿\
|3x + 1|
()
2
⇢=
x î⌅–⌧
⇠
1
X
(3x + 1)n
n=1
⇣, x =
1
X
(2x
⇠
1
n=0
1
⇠X ⇠4⇠¿Ñ@ R = t‡ ⇠4l⌅@ (0, 1)
2
t‰.
1
(3x + 1)n
| L, an =
3
7 · 2n
an+1
⇢ = lim
= lim
n!1
n!1
an
⇠î ⌧∞\‰.
¯Ï¿\ t
n=1
7 · 2n
n=0
n=0
c)
⇠î ⌧∞\‰.
1
| Lî êÖX‰. x 6=
3
\ P‡ D(⇣ ïD t©Xt
¯¨‡ x = 1| L
t¿\
1
1
X
X
n
=
(1)
=
1
42n
n=1
n=1
n=1
sol) x =
⇠î ƒŸX‡ ⇠4X¿ Jî‰.
1
X
1
X
16n
1
X
(3x + 1)n
n=1
1
X
=
¯Ï¿\
t
⇠X ⇠4⇠¿Ñ@ R = 8t‡ ⇠4l⌅@
✓
◆
13 19
,
t‰.
2 2
1| < 1 () 0 < x < 1
(2x
3)n
42n
n=1
n!1
⇠4X¿\
⇢ = |2x
x î⌅–⌧
= lim |2x 1| = |2x 1|
¨
19
|L
2
1)n \ ¯¨‡ x =
n+1
⇢ = lim
6 q ⇠@ L|Ï
1| L
7 · 2n
1
X
(3x + 1)n
3)n
42n
n=1
7 · 2n
=
3x + 1
|3x + 1|
=
2
2
= lim
n!1
1<x<
1
3
⇠4\‰.
1
1
X
X
( 2)n
( 1)n
=
7 · 2n
7
n=1
n=1
3
3
(2x 3)n
t¿\ ⇠î ƒŸX‡ ⇠4X¿ Jî‰.
| Lî êÖX‰. x 6= | L, an =
\
2
2
42n
¯¨‡ x = 13 | L
P‡ D(⇣ ïD t©Xt
sol) x =
an+1
⇢ = lim
= lim
n!1
n!1
an
(2x 3)n+1
42n+2
(2x 3)n
42n
–⌧ ⇢ < 1 x Ω∞
⇠4X¿\
⇠
|2x 3|
⇢=
< 1 ()
16
x î⌅–⌧
⇠
1
X
(2x
n=1
⇣, x =
n=1
3)n
42n
=
n=1
t¿\
13
19
<x<
2
2
⇠4\‰.
13
|L
2
1
X
(2x
t¿\
3)n
42n
1
X
(3x + 1)n
2x 3
|2x 3|
= lim
=
2
n!1
4
16
1
1
X
X
( 16)n
=
( 1)n
2n
4
n=1
n=1
7 · 2n
=
1
1
X
X
2n
1
=
n
7
·
2
7
n=1
n=1
⇠î ⌧∞\‰.
2
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R =
t‡ ⇠4l⌅@
3
✓
◆
1
1,
t‰.
3
e)
1
X
xn
n+1
n=0
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
D(⇣ ïD t©Xt
an+1
⇢ = lim
= lim
n!1
n!1
an
⇠î ƒŸX‡ ⇠4X¿ Jî‰.
56
xn+1
n+2
xn
n+1
= lim
xn
\ P‡
n+1
n+1
n!1 n + 2
|x| = |x|
SOLUTION
–⌧ ⇢ < 1 x Ω∞
⇠
⇠4X¿\
⇢ = |x| < 1 ()
x î⌅–⌧
⇣, x =
1
X
xn
n+1
n=0
⇠
1| L
t¿\
1
X
(x
g)
n=1
1
X
( 1)n
xn
=
n + 1 n=0 n + 1
n=0
⇠î P
⇠t‡ ⇠Ù
1
= 0 t¿\ P
n!1 n + 1
⇢
Ùtp lim
5)n
n2
P‡ D(⇣ ïD t©Xt
⇠⇣ ï– Xt ⇠4\‰.
= |x
⇠
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ [ 1, 1)
n2
n2
|x
n!1 (n + 1)2
= lim
\
5|
⇠4X¿\
⇢ = |x
⇠î ⌧∞\‰.
¯Ï¿\ t
t‰.
5)n
(x
5|
–⌧ ⇢ < 1 x Ω∞
1
1
1
X
X
X
xn
1n
1
=
=
n + 1 n=0 n + 1 n=0 n + 1
n=0
(x 5)n+1
(n+1)2
(x 5)n
n2
an+1
⇢ = lim
= lim
n!1
n!1
an
¯¨‡ x = 1| L
t¿\
⇠X ⇠4⇠¿Ñ@ R = 2t‡ ⇠4l⌅@ [ 2, 2)
sol) x = 5 | Lî êÖX‰. x 6= 5 | L, an =
1
î ëm Ëp⇣å⇠
n+1
¨
⇠î ⌧∞\‰.
¯Ï¿\ t
t‰.
1<x<1
⇠4\‰.
1
X
t¿\
6 q ⇠@ L|Ï
x î⌅–⌧
⇠
5| < 1 () 4 < x < 6
1
X
(x
n=1
5)n
⇠4\‰.
n2
⇣, x = 4| L
f)
1
X
xn
n2n
n=1
n=1
n
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
D(⇣ ïD t©Xt
an+1
⇢ = lim
= lim
n!1
n!1
an
xn+1
(n+1)2n+1
xn
n2n
–⌧ ⇢ < 1 x Ω∞
⇠4X¿\
⇢=
x î⌅–⌧
⇠
⇠
|x|
< 1 ()
2
1
X
xn
n=1
⇣, x =
1
X
(x
2| L
1
X
xn
n2n
x
\ P‡
n2n
t¿\
n
|x|
= lim
|x| =
n!1 2(n + 1)
2
p lim
⇠î P
1
n!1 n2
=
1
X
( 1)n
n2
n=1
⇢
1
⇠t‡ ⇠Ù 2 î ëm Ëp⇣å⇠Ùt
n
⇠⇣ ï– Xt ⇠4\‰.
¯¨‡ x = 6| L
1
X
(x
n=1
t¿\
1
1
X
X
( 2)n
( 1)n
=
=
n2n
n2n
n
n=1
n=1
n=1
n2
= 0 t¿\ P
2<x<2
⇠4\‰.
5)n
1
X
(x
n=1
n2
=
1
X
1n
n=1
n
=
2
1
X
1
n=1
n2
⇠î ⇠4\‰.
¯Ï¿\ t
t‰.
h)
5)n
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ [4, 6]
1)n
n2 + 3
⇢
1
(x 1)n
t¿\ ⇠î P
⇠t‡ ⇠Ù
î ëm Ëp⇣å⇠Ùtp sol) x = 1 | Lî êÖX‰. x 6= 1 | L, an =
\
n
n2 + 3
1
P‡ D(⇣ ïD t©Xt
lim
= 0 t¿\ P
⇠⇣ ï– Xt ⇠4\‰.
n!1 n
(x 1)n+1
an+1
n2 + 3
(n+1)2 +3
¯¨‡ x = 2| L
⇢ = lim
= lim
=
lim
|x 1|
n
(x 1)
n!1
n!1
n!1 (n + 1)2 + 3
an
1
1
1
2
n
n
X x
X 2
X1
n +3
=
=
n
n
=
|x
1|
n2
n2
n
n=1
n=1
n=0
57
SOLUTION
–⌧ ⇢ < 1 x Ω∞
⇠
⇢ = |x
x î⌅–⌧
⇠
⇣, x = 0| L
1)n
n2 + 3
1
1
1
X
X
X
( 1)n+1 xn
( 1)n+1 (1)n
( 1)n+1
=
=
n(n + 1)
n(n + 1)
n(n + 1)
n=1
n=1
n=1
⇠4\‰.
t¿\
1
X
(x
1
X
1)n
( 1)n
=
2
n +3
n2 + 3
n=1
n=1
⇢
1
t¿\ ⇠î P
⇠t‡ ⇠Ù 2
î ëm Ëp⇣å⇠
n +3
1
Ùtp lim 2
= 0 t¿\ P
⇠⇣ ï– Xt ⇠4\
n!1 n + 3
‰. ¯¨‡ x = 2| L
1
X
(x
1
1
X
X
1)n
1
1
=

2+3
2
n2 + 3
n
n
n=1
n=1
n=1
t‡
⇠
1
X
1
¯Ï¿\ t
t‰.
i)
–⌧ ⇢ < 1 x Ω∞
⇠
( 1)n+2 xn+1
(n+1)(n+2)
( 1)n+1 xn
n(n+1)
n
= lim
|x| = |x|
n!1 n + 2
4\‰.
¯Ï¿\ t
t‰.
j)
1
X
( 1)n
t‡
1
X
( 1)n+1 xn
n(n + 1)
n=1
⇠
–⌧ ⇢ < 1 x Ω∞
⇢=
x î⌅–⌧
1
X
1
X
1
n2
n=1
(x 2)n
p
4n n
|x
1
X
⇠4X¿\
2|
4
< 1 ()
2<x<6
( 1)n
(x 2)n
p
4n n
⇠4\‰.
2| L
1
X
( 1)n
n=1
⇠4\‰.
t¿\
1
1
X
X
(x 2)n
( 1)n ( 4)n
1
p
p
p
=
=
n n
4n n
4
n
n=1
n=1
⇠î ⌧∞\‰.
¯¨‡ x = 6| L
1
1
1
X
X
X
( 1)n (x 2)n
( 1)n 4n
( 1)n
p
p
p
=
=
4n n
4n n
n
n=1
n=1
n=1
t¿\
⇠4X¿\
DP⇣ ï–
1
î ⇠4\‰. ¯Ï¿\
n(n
+ 1)
n=1
4\‰.
⇠
⇠
⇠
n=1
1<x<1
1
n(n + 1)
n=1
⇠
(x 2)n
p
4n n
p
an+1
4n+1 n+1
⇢ = lim
= lim
n
( 1) (x 2)n
n!1
n!1
an
p
4n n
p
n
|x 2|
= lim p
|x 2| =
n!1 4 n + 1
4
1
1
1
X
X
X
( 1)n+1 xn
( 1)n+1 ( 1)n
( 1)2n+1
=
=
=
n(n + 1)
(n + 1)
n(n + 1)
n=1
n=1
n=1
1
X
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ [ 1, 1]
\ P‡ D(⇣ ïD t©Xt
⇣, x =
⇠4X¿\
⇢ = |x| < 1 ()
=
⇠⇣ ï– Xt ⇠
( 1)n+1 (x 2)n+1
an+1
⇢ = lim
= lim
n!1
n!1
an
1| L
= 0 t¿\ P
n!1 n(n + 1)
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ [0, 2]
( 1)n+1 xn
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
\
n(n + 1)
P‡ D(⇣ ïD t©Xt
⇣, x =
1
⇠Ùtp lim
1
î ëm Ëp⇣å
n(n + 1)
sol) x = 2 | Lî êÖX‰. x 6= 2 | L, an = ( 1)n
1
X
( 1)n+1 xn
n(n + 1)
n=1
x î⌅–⌧
⇢
⇠t‡ ⇠Ù
⇠î P
n=1
⇠4X¿\ DP⇣ ï– Xt ⇠4\‰.
n2
n=1
¨
¯¨‡ x = 1| L
1| < 1 () 0 < x < 2
1
X
(x
n=1
⇠4X¿\
6 q ⇠@ L|Ï
⇠
Xt
⇠î P
1
⇠t‡ ⇠Ù p î ëm Ëp⇣å⇠Ùt
n
1
p lim p = 0 t¿\ P
n!1
n
1
X
1
î ⇠ ¯Ï¿\ t
n(n
+
1)
t‰.
n=1
58
⇢
⇠⇣ ï– Xt ⇠4\‰.
⇠X ⇠4⇠¿Ñ@ R = 4t‡ ⇠4l⌅@ ( 2, 6]
SOLUTION
k)
1
X
6 q ⇠@ L|Ï
¯¨‡ x = 6| L
nxn
1
X
n
n=1
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an = nxn \ P‡
t¿\
D(⇣ ïD t©Xt
an+1
(n + 1)xn+1
n+1
= lim
= lim
|x| = |x|
n!1
n!1
n!1
an
nxn
n
⇢ = lim
–⌧ ⇢ < 1 x Ω∞
⇠
⇠4X¿\
⇢ = |x| < 1 ()
x î⌅–⌧
⇠
1
X
nxn
n=1
1
X
n=1
1<x<1
x =
⇠î ⌧∞\‰.
¯Ï¿\ t
t‰.
m)
6n
n
1
X
⇠X ⇠4⇠¿Ñ@ R = 6t‡ ⇠4l⌅@ ( 6, 6)
2n
n2 + 1
xn
P‡ D(⇣ ïD t©Xt
n=1
⇣, x =
nxn =
n=1
t¿\
1
X
nxn =
n=1
1
X
–⌧ ⇢ < 1 x Ω∞
⇠
⇠4X¿\
⇢ = 2|x| < 1 ()
n
x î⌅–⌧
n=1
⇠
⇠î ⌧∞\‰.
¯Ï¿\ t
t‰.
2n+1
n+1
(n+1)2 +1 x
n
2
n
n2 +1 x
= lim
n=1
1
X
2n
xn \
n2 + 1
2(n2 + 1)
|x| = 2|x|
n!1 (n + 1)2 + 1
( 1)n n
⇠î ƒŸX‡ ⇠4X¿ Jî‰.
¯¨‡ x = 1| L
t¿\
an+1
⇢ = lim
= lim
n!1
n!1
an
1| L
1
X
n
n=1
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
⇠4\‰.
¨
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ ( 1, 1)
1
|L
2
⇣, x =
1
X
2n
xn
2+1
n
n=1
1
1
<x<
2
2
⇠4\‰.
✓
◆n X
1
1
X
2n
2n
1
( 1)n
n
x
=
=
1
n2 + 1
n2 + 1
2
n2 + 1
X
n n
n=1
n=1
n=1
x
l)
⇢
6n
n=1
1
t¿\ ⇠î P
⇠t‡ ⇠Ù 2
î ëm Ëp⇣å⇠
n n
n +1
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an = n x \ P‡
1
6
Ùtp lim 2
= 0 t¿\ P
⇠⇣ ï– Xt ⇠4
D(⇣ ïD t©Xt
n!1 n + 1
\‰.
n+1 n+1
an+1
n+1
|x|
6n+1 x
1
⇢ = lim
= lim
= lim
|x| =
n n
¯¨‡ x = | L
n!1
n!1
n!1 6n
an
6
6n x
2
1
X
–⌧ ⇢ < 1 x Ω∞
⇢=
x î⌅–⌧
⇠
⇠
|x|
< 1 ()
6
1
X
n
n=1
⇣, x =
x
n
6<x<6
t‡
6
n=1
xn =
n
⇠4\‰.
1
X
n
6
n=1
( 6)n =
n
⇠
1
X
1
n=1
\‰.
6| L
1
X
n
t¿\
6n
✓ ◆n X
1
1
X
2n
2n
1
1
n
x
=
=
2+1
2+1 2
2+1
n
n
n
n=1
n=1
n=1
1
X
⇠4X¿\
1
X
( 1)n n
⇠4X¿\ DP⇣ ï– Xt
⇠î ⇠4
1
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R =
t‡ ⇠4l⌅@
2

1 1
,
t‰.
2 2
n=1
⇠î ƒŸX‡ ⇠4X¿ Jî‰.
n2
n)
1
X
2n
n=1
59
2n
1 2(n 1)
x
SOLUTION
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
\ P‡ D(⇣ ïD t©Xt
an+1
= lim
n!1
n!1
an
⇢ = lim
2n+1 2n
2n+1 x
2n 1 2(n 1)
2n x
2n 1 2(n 1)
x
2n
⇠
x î⌅–⌧
p
⇣, x =
⇠
1
X
2n
n=1
2
p
1 2(n 1)
x
n
2<x<
p
⇠t‡ ⇠Ù
⇢
1
î ëm Ëp⇣å
n(ln n)2
2
¯¨‡ x = 1| L
1
X
1
X
1
1
n
x
=
2
n(ln
n)
n(ln
n)2
n=2
n=2
⇠4\‰.
t‡
1
1
1 2(n 1) X 2n 1 p 2(n 1) X 2n 1
x
=
(
2)
=
2n
2n
2
n=1
n=1
t‡
⇠î ⌧∞\‰.
p
¯¨‡ x = 2| L
1
X
1
X
2n 1 2(n 1)
2n 1 p 2(n 1)
2n 1
x
=
( 2)
=
n
n
2
2
2
n=1
n=1
n=1
⇠î ⌧∞\‰.
¯Ï¿\
⇠X ⇠4⇠¿Ñ@ R =
p p t
(
2, 2) t‰.
p
⇠⇣ ï– Xt ⇠4
\‰.
2| L
1
X
⇠î P
1
= 0 t¿\ P
n!1 n(ln n)2
1
X
2n
o)
1
X
1
( 1)n
n
x
=
n(ln n)2
n(ln n)2
n=2
n=2
⇠Ùtp lim
⇠4X¿\
x2
⇢=
< 1 ()
2
t‡
1
X
t¿\
–⌧ ⇢ < 1 x Ω∞
¨
1| L
2n + 1 2
x
n!1 2(2n
1)
x
2
n=1
⇣, x =
= lim
2
=
6 q ⇠@ L|Ï
2t‡ ⇠4l⌅@
Z 1
2
1
1
dx =
t¿\
x(ln x)2
ln 2
Ñ⇣ ï– Xt
⇠î
⇠4\‰.
¯Ï¿\ t
t‰.
p)
1 ✓
X
n=1
1+
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ [ 1, 1]
2
n
◆n
xn
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
P‡ D(⇣ ïD t©Xt
✓
2
1+
n
◆n
xn \
2
(1 + n+1
)n+1 xn+1
an+1
= lim
n!1
n!1
an
(1 + n2 )n xn
⇢ = lim
1
X
1
xn
2
n(ln
n)
n=2
= lim
2
1 + n+1
n+1
2
2
2
|x| =
x =
1 ✓
X
n
2
e2
|x| = |x|
e2
n!1
1 + n2
1
n
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
x \
n(ln n)2
–⌧ ⇢ < 1 x Ω∞ ⇠ ⇠4X¿\
P‡ D(⇣ ïD t©Xt
⇢ = |x| < 1 () 1 < x < 1
1
n+1
an+1
(n+1)(ln (n+1))2 x
◆n
⇢ = lim
= lim
1 ✓
X
1
n
n!1
n!1
2
an
n(ln n)2 x
x î⌅–⌧ ⇠
1+
xn ⇠4\‰.
n
2
n=1
n(ln n)
= lim
|x| = |x|
n!1 (n + 1)(ln (n + 1))2
⇣, x = 1| L
✓
2 ln n
2
(ln n)
(ln n)
◆n
◆n
1 ✓
1 ✓
n
X
X
* lim
= lim 2 ln (n+1)
= lim
2
2
n
n!1 ln (n + 1))2
n!1
n!1 ln (n + 1)
1
+
x
=
1
+
( 1)n
n+1
n
n
◆
n=1
n=1
1
✓
◆n
= lim n1 = 1
n!1
2
n+1
t‡ lim 1 +
( 1)n @ ¯ ˘\✓t t¨X¿ J<¿\
n!1
n
–⌧ ⇢ < 1 x Ω∞ ⇠ ⇠4X¿\
⇠î ⌧∞\‰.
⇢ = |x| < 1 ()
x î⌅–⌧
1
X
1
⇠
xn
2
n(ln
n)
n=2
1<x<1
¯¨‡ x = 1| L
1 ✓
X
⇠4\‰.
n=1
60
2
1+
n
◆n
n
n=1
2
1+
n
◆n
SOLUTION
✓
2
n!1
n
⇠î ⌧∞\‰.
t‡ lim
1+
¯Ï¿\ t
t‰.
q)
1
X
3n (x
n=1
◆n
= lim
n!1
✓✓
1+
2
n
◆ n2 ◆2
t‰. ⇢ < 1 x Ω∞
1
X
‰⇠ x–⌧ ⇠
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ ( 1, 1)
t)
n!
an+1
⇢ = lim
= lim
n!1
n!1
an
3n+1 (x 2)n+1
(n+1)!
3n (x 2)n
n!
⇠
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@
( 1, 1) t‰.
2)n
P‡ D(⇣ ïD t©Xt
¨
⇠4Xîp ⇢ = 0 < 1 t¿\ ®‡
xn
î ⇠4\‰.
2 · 4 · 6 · · · · ·(2n)
n=1
= e2 6= 0 t¿\
sol) x = 2 | Lî êÖX‰. x 6= 2 | L, an =
6 q ⇠@ L|Ï
1
X
( 1)n x2n
22n (n!)2
n=1
3n (x 2)n
( 1)n x2n
\
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an = 2n
\
n!
2 (n!)2
P‡ D(⇣ ïD t©Xt
3
|x 2| = 0
an+1
n!1 n + 1
⇢ = lim
= lim
n!1
n!1
an
= lim
( 1)n+1 x2n+2
22n+2 {(n+1)!}2
( 1)n x2n
22n (n!)2
= lim
x2
n!1 22 (n + 1)2
=0
t‰. ⇢ < 1 x Ω∞ ⇠ ⇠4Xîp ⇢ = 0 < 1 t¿\ ®‡
1
X
t‰. ⇢ < 1 x Ω∞ ⇠ ⇠4Xîp ⇢ = 0 < 1 t¿\ ®‡
3n (x 2)n
1
‰⇠ x–⌧ ⇠
î ⇠4\‰.
X
( 1)n x2n
n!
‰⇠ x–⌧ ⇠
î ⇠4\‰.
n=1
22n (n!)2
n=1
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@
( 1, 1) t‰.
( 1, 1) t‰.
1
X
x2n 1
r)
( 1)n 1
1
1
(2n 1)!
X
X
n=1
2.
an 4n t ⇠4X‡
an 5n t ⌧∞` L, ‰L ⇠X ⇠4,
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
n=0
n=0
2n 1
⌧∞ ÏÄ|
⇣ X‹$.
n 1 x
( 1)
\ P‡ D(⇣ ïD t©Xt
sol) ¸¥ƒ q ⇠î ⇠4⇠¿ÑD ¿î ⇠t‰. t⌧ ⇠
(2n 1)!
1
1
X
X
n
a
x
X
⇠4
⇠¿ÑD
Rt|
Xê.
an 4n t ⇠4X¿
n 2n+1
n
( 1) x
an+1
(2n+1)!
n=0
n=0
⇢ = lim
= lim ( 1)n 1 x2n 1
1
X
n!1
n!1
an
(2n 1)!
\ R
4 t‡
an 5n t ⌧∞X¿\ R  5 t‰. 0|⌧
2
n=0
x
= lim
=0
4  R  5 t‰.
n!1 (2n)(2n + 1)
1
X
a)
an ( 3)n
n=0
t‰. ⇢ < 1 x Ω∞ ⇠ ⇠4Xîp ⇢ = 0 < 1 t¿\ ®‡
1
2n 1
X
x
sol) | 3| = 3 < 4  R t¿\ ¸¥ƒ ⇠î ⇠4\‰.
‰⇠ x–⌧ ⇠
( 1)n 1
î ⇠4\‰.
(2n
1)!
✓ ◆n
n=1
1
X
11
an
¯Ï¿\ t
⇠X ⇠4⇠¿Ñ@ R = 1t‡ ⇠4l⌅@ b)
2
n=0
( 1, 1) t‰.
11
sol) R < 5 <
t¿\ ¸¥ƒ ⇠î ⌧∞\‰.
1
X
2
xn
s)
2 · 4 · 6 · · · · ·(2n)
µ8⌧ 6.2 .
n=1
sol) x = 0 | Lî êÖX‰. x 6= 0 | L, an =
\ P‡ D(⇣ ïD t©Xt
an+1
⇢ = lim
= lim
n!1
n!1
an
xn+1
2·4·6·····(2n)·(2n+2)
xn
2·4·6·····(2n)
1. ‰L h⇠X q ⇠ ⌅⌧›D lX‹$.
1
a)
1 4x
xn
2 · 4 · 6 · · · · ·(2n)
sol)
|x|
= lim
=0
n!1 2n + 2
61
⇠
1
1
x
=
1
X
n=0
xn
(|x| < 1)
SOLUTION
–⌧ x| 4x\
‡Xt
1
1
t‰. ¯Ï¿\
1
X
=
4x
1
(4x)n
(|4x| < 1)
4n xn
✓
n=0
1
X
=
n=0
|x| <
1
4
x
sol)
⇠
+
1
x
\
8
=
x
1
X
xn
(|x| < 1)
n=0
‡Xt
1
1
1 ⇣ ⌘n
X
x
x =
8
n=0
1 ✓
X
⇣ x
8
n=0
1
8
2x
(1 + x2 )2
sol) ⇠
1
8
◆n
8
xn
⌘
<1
–⌧ x|
sol)
⇠
n=0
x2 \
x
=
+
1 ✓ ◆n+1
X
1
8
t‡ mƒ¯Ñ– Xt
x
n
(|x| < 8)
✓
1
1 + x2
◆0
(|x| < 1)
1
X
( x)2 < 1
( 1)n x2n
(|x| < 1)
1
X
2x
=
2n( 1)n x2n 1
(1 + x2 )2
n=1
=
=
1
X
2(n + 1)( 1)n+1 x2n+1
(|x| < 1)
n=0
x2
t‰.
1
x
=
1
X
xn
(|x| < 1)
n=0
‡Xt
1
1
X
x2n
x2 < 1
n=0
t‡ ë¿– x| ÒXt
x
x2
1
(1 x)3
sol) ⇠
e)
1
=
x2
1
1
xn
n=0
n=0
1
–⌧ x| x2 \
1
X
n=0
‡Xt
=
x
1
x
=
1
X
1
=
( x2 ) n
1 + x2
n=0
(|x| < 8)
t‰.
x
x2n+1
d)
1
t‡ ë¿– D ÒXt
8
1
n=0
1
=
c)
x +
1
X
= 1 + 2x + x2 + 2x3 + x4 + · · ·
1
X
=
an xn (|x| < 1)
1
1
=
x2
1
n
t‰. Ï0⌧ an @ ‰L¸ ⇡‰.
(
2, (n = 2k + 1, k 2 Z+ )
an =
1. (n = 0 or 2k, k 2 Z+ )
1
–⌧ x|
x
1
X
n=0
1
8
1
x
◆
t‰.
b)
6 q ⇠@ L|Ï
=
=
1
X
n=0
1
X
x2n+1
(|x2 | < 1)
x2n+1
(|x| < 1)
1
x
=
1
X
xn
(|x| < 1)
n=0
| mƒ¯ÑXt
✓
◆0
1
X
1
1
=
=
nxn 1
1 x
(1 x)2
n=1
1ΩX‡, ‰‹ mƒ¯ÑD Xt
✓
◆00
1
X
1
2
=
=
n(n
1 x
(1 x)3
n=2
n=0
62
(|x| < 1)
1)xn 2
(|x| < 1)
¨
SOLUTION
t‰. ë¿D 2\ ò⌅t
1
(1
x)
1
n(n
2
n=2
1
X
1
=
2
n=0
1)xn 2
ln (4
x) = C +
1 ✓ ◆n+1 n+1
X
1
x
4
n=0
(n + 1)(n + 2)xn
(|x| < 1)
t‰. ë¿–
t‰.
ln (4
n+1
1
2
1
x
sol)
⇠
x
1 ✓ ◆n+1 n+1
X
1
x
x) = ln 4
ò(‰.
h) sinh x
î
1
1
1
X
=
x
sol) Ù0 6.2.4–⌧ ex =
xn
(|x| < 1)
4
1
x\
n=0
=
x)2
(1
‡Xt
1
X
.
e
nx
n 1
(|x| < 1)
x)2
=
1
X
2nx
n 1
1
X
=
n=1
ex
1
x
(1
x)
1
X
=
2
2(n + 1)x
=
1
X
xn
n=0
1
X
2(n + 1)xn
n=0
(2n + 1)xn
n!
1
X
( x)n
n!
n=0
= 2x +
tp ë¿D 2\ ò⌅t
ex
(|x| < 1)
e x
2
1Ω\‰.
= sinh x =
1
X
x2n+1
(2n + 1)!
n=0
(x 2 R)
ò(‰.
g) ln(4 x)
sol) ⇠
1
1
x
\
4
=
x
1
X
i) cosh x
xn
(|x| < 1)
sol) Ù0 6.2.4–⌧ ex =
n=0
1
x
4
=
x\ ∏t
=
1 ⇣ ⌘n
X
x
n=0
1 ✓
X
⇣ x
4
n=0
1
4
◆n
4
xn
⌘
<1
e x=
t‰. ¯Ï¿\
(|x| < 4)
4
x
=
n=0
(x 2 R)
1
X
xn
1
X
( x)n
n!
n=0
n=0
4
xn
n!
+
2x2
2x4
+
+ ···
2!
4!
1
X
x2n
=2
(x 2 R)
(2n)!
n=0
=2+
1 ✓ ◆n+1
X
1
(x 2 R)ÑD t©Xê. x|
n!
1
X
( x)n
n!
n=0
ex + e x =
1
D ÒXt
4
1
1
X
xn
n=0
‡Xt
1
t‡ ë¿–
(x 2 R)
2x3
2x5
+
+ ···
3!
5!
1
X
x2n+1
=2
(x 2 R)
(2n + 1)!
n=0
(|x| < 1)
n=0
–⌧ x|
1
X
xn
n=0
n
n=0
2
1
X
( x)n
=
n!
n=0
e x=
t¿\
1
x
(x 2 R)ÑD t©Xê. x|
n!
t‰. 0|⌧
n=1
ÑD t©Xê. ⌅ ›X ë¿– 2| ÒXt
(1
(|x| < 4)
n+1
1
X
xn
n=0
t‡ 8⌧ e)–⌧
2
(C 2 R)
x)2
(1
1
1
(|x| < 4)
1D ÒX‡, x = 0D #¥ ¡⇠| lXt
n=0
f)
¨
t‰. mƒ Ñ– Xt
1
X
=
3
6 q ⇠@ L|Ï
(|x| < 4)
63
SOLUTION
t‡ ë¿D 2\ ò⌅t
6 q ⇠@ L|Ï
t‰. t⌧ 1)¸ 2)| \©Xê. ¯Ït
1
X
ex + e x
x2n
= cosh x =
2
(2n)!
n=0
1
1
1 X 2n 1
1 X n
+
= +
2 n=2 2n
2 n=2 2n 1
(x 2 R)
t‰.
=
1
2. |x| < 1| L
1
X
n2
n=1
2n
1
X ✓D
1
X
=
x
n=0
sol)
1
2=
1
X
n
2n
n=1
1
↵=↵
2
x
1
1
1
n
,
,
n
2 n=1 2n
n=0
1 =
2
1
X
1
n=0
···
2n
=
=
n
n
2
n=1
1)
1
X
1
X
1
2n
1
2n
1
2
=
=
n=1
1
X
2n
=
1
1 X 1
+
2 n=2 2n
p
n+1
p
n
⇢ = lim
n!1
n2
2n
n=1
n=1
1
= +
2 n=2 2n
=
xn
= lim
p(p 1)···(p n+1)(p n)
(n+1)!
p(p 1)···(p n+1)
n!
= lim
p n
=1
n+1
n!1
t‡
f 0 (x) =
2 n=1 2n
1
X
n2
n
1
= R = 1 t¿\ ⇠4⇠¿Ñ@ R = 1 t‰.
⇢
pf (x)
b) R < x < R–⌧ f 0 (x) =
ÑD Ùt‹$.
1+x
proof ) |R| < 1t¿\ 1 + x 6= 0t‡ 0|⌧
1=1
1
1 X n2
1
X
2n
a) f (x)X ⇠4⇠¿Ñ RD lX‹$.
sol) ¨6.1.3– Xt
2)
1
1
X
X
1
1
1
↵=
=
n
n
2
2
2
n=1
n=0
1
X
n2
n=2
=3
1 ✓ ◆
X
p
n!1
t¿\ ↵ = 2t‰.
1
X
n2
t⌧ =
|‡ Xê. ¯Ït
2n
n=1
1
X
1
|‡ Xê.
1
X
n
···
2n
2n
= 6t‰.
n=0
n
n+1
2
n=1
n=2
1
X
n
n=1
1
1X n
2 n=1 2n
1
1 X n
+
2 n=2 2n
n=1
=
=1+
f (x) =
1
X
n=2
1
X
3. p 0t DÃ ‰⇠| Xê. nt ëX ⇠| L, U•⌧ tmƒ
⇠|
✓ ◆
p
p(p 1) · · · (p n + 1)
=
n
n!
✓ ◆
p
\ XX‡,
= 1\ X\‰. tL,
0
(|x| < 1)
n=0
1
X
1
n
↵=
2
2n
n=1
t‰. 1)– Xt
xn
1
X
1
1
1 X n
1
+
+
n
2 n=1 2
2n
n=1
|‡ Xê. ¯Ït
=
1
2
1
X
t‰. 0|⌧
1
X
=
t¿\
t‰. ↵ =
xn ÑD t©XÏ
1
X
lX‹$.
1
¨
t‰. ⇣\
⇠
n2
2n+1
1
X
(n
n=2
f (x) =
1 ✓ ◆
X
p
n=0
1)2
2n
p
f (x) () (1 + x)f 0 (x) = pf (x)
1+x
n
xn = 1 +
✓ ◆
✓ ◆
p
p 2
x+
x + ···
1
2
| mƒ¯ÑXt
1
1 X 2n 1
+
2 n=2 2n
✓ ◆
✓ ◆
✓ ◆
✓ ◆
1
X
p n 1
p
p
p 2
f (x) =
n
x
=
+2
x+3
x + ···
n
1
2
3
n=1
0
64
SOLUTION
t‰. t⌧ (1+x)f 0 (x) = pf (x)ÑD Ùt0 ⌅t n(mX ƒ⇠|
DPXê. (1 + x)f 0 (x)X n(mX ƒ⇠î
✓
◆
✓ ◆
p
p
(n + 1)
+n
n+1
n
✓ ◆
p
t‡ pf (x)X n(mX ƒ⇠î p
t‰. \∏
n
✓
◆
✓ ◆
p
p
(n + 1)
+n
n+1
n
p(p 1) · · · (p n)
p(p 1) · (p n + 1)
= (n + 1) ·
+n·
(n + 1)!
n!
p(p 1) · · · (p n)
p(p 1) · · · (p n + 1)
=
+n·
n!
n!
p(p 1) · · · (p n + 1)
=
· {(p n) + n}
n!
✓ ◆
p
=
·p
n
6 q ⇠@ L|Ï
1
1
1
t¿\ a0 = 1, a1 = , a2 =
, a3 =
t‰.
2
8
16
1
b) p
1 x
1 ✓ ◆
X
p n
sol)
x = (1 + x)p –⌧ x| x\ ‡X‡ p =
n
n=0
ÖXt
p
1 ✓ 1◆
X
1
2 ( x)n = 1 + 1 x + 3 x2 + 5 x3 + · · ·
=
2
8
16
1 x n=0 n
t¿\ a0 = 1, a1 =
p
p(1 + x) p 1 f (x) + (1 + x) p f 0 (x)
t‡ ë¿D
=
‡X‡ p =
1
2
1
x2
1
=
1 ✓
X
n=0
1◆
2 (
n
x2 ) n
ÑXt
1
3
5
35 8
= 1 + x2 + x4 + x6 +
x + ···
2
8
16
128
1
3
5 7
35 9
=) arcsin(x) = x + x3 + x5 +
x +
x + ···
6
40
112
1152
p
p
f (x) =
f (x)
1+x
0
h (x) =
x2 \
1
3
5
35 8
= 1 + x2 + x4 + x6 +
x + ···
2
8
16
128
t‰. \∏ 8⌧ b)– Xt⌧
0
1
3
5
, a2 = , a3 =
t‰.
2
8
16
c) arcsin(x)
1 ✓ ◆
X
p n
sol)
x = (1 + x)p –⌧ x|
n
n=0
| ÖXt
c) h(x) = (1 + x) p f (x)X ¯ÑD t©XÏ f (x) = (1 + x)p ÑD
Ùt‹$.
proof ) 8⌧–⌧ ¸¥ƒ h⇠ h(x)| ¯ÑXt
t¿\ h0 (x)–
1
|
2
⇡<¿\ (1 + x)f 0 (x) = pf (x)t‡ 0|⌧
t‰. n(mX ƒ⇠
ùÖt DÃ⌧‰.
h0 (x) =
¨
ÖXt
p(1 + x) p 1 f (x) + (1 + x) p f 0 (x)
p
p(1 + x) p 1 f (x) + (1 + x) p ·
f (x) = 0
1+x
1
x2
1
t¿\ a0 = 0, a1 = 1, a2 = 0, a3 =
1
t‰.
6
1
t‰. 0|⌧ h(x) = ct‰. (Ë cî ¡⇠)
X
5.
‰L
h⇠X
q
⇠
⌅⌧›t
an xn t| `L, a0 , a1 , a2 ,
cX ✓D lX0 ⌅t x = 0D ÖXt h(x) = 1t (D L ⇠
n=0
à‡ 0|⌧ f (x) = (1 + x)p ÑD L ⇠ à‰.
a3 D lX‹$.
4. ‰L h⇠X q ⇠ ⌅⌧›t
1
X
an xn t| ` L, ⌅X ∞¸| a) sec x
n=0
t©XÏ a0 , a1 , a2 , a3 D lX‹$.
sol) sec x =
1
X
an xn \ Pê.
n=0
a)
p
1+x
sol) 8⌧ 3–⌧
1
p= |L
2
p
1 ✓ ◆
X
p
n=0
1+x=
n
n
cos x =
p
x = (1 + x) ÑD t©Xt
1
X
( 1)n 2n
x
(2n)!
n=0
(x 2 R)
t¿\
1 ✓1◆
X
2
n=0
n
xn = 1 +
x
2
x2
x3
+
+ ···
8
16
1 = cos x · sec x =
65
1
X
( 1)n 2n
x
(2n)!
n=0
!
·
1
X
n=0
an x
n
!
SOLUTION
d) tanh x
ÑD L ⇠ à‰.
1
a0 )x2 + (a3
2
a0 + a1 x + (a2
@ ⇡t
6 q ⇠@ L|Ï
1
a1 )x3 + · · · = 1
2
sol) tanh x =
1
X
an xn \ Pê.
n=0
¨| ` ⇠ à‡ ƒ⇠DP| µt
a0 = 1,
a1 = 0,
a2 =
cosh x =
1
,
2
a3 = 0
sol) tan x =
1
X
1
X
x2n
(2n)!
n=0
cosh x · tanh x =
an xn \ Pê.
1
X
( 1)n 2n
x ,
(2n)!
n=0
sin x =
1
X
( 1)n+1 2n+1
x
(2n + 1)!
n=0
t¿\
cos x · tan x =
@ ⇡t
1
X
x2n+1
(2n + 1)!
n=0
1
X
( 1)n 2n
x
(2n)!
n=0
1
a0 )x2 +(a3
2
!
1
X
an x
n
n=0
!
= sin x
n=0
(2n)!
1
a1 )x3 +· · · = x
2
x3
x5
+
+ ···
3!
5!
a1 = 1,
a2 = 0,
a3 =
!
1
X
an x
n
n=0
!
1
X
x2n+1
(2n + 1)!
n=0
1
1
x3
x5
a0 +a1 x+(a2 + a0 )x2 +(a3 + a1 )x3 +· · · = x +
+
+ ···
2
2
3!
5!
t‡ ƒ⇠DP| µt
a0 = 0,
a1 = 1,
a2 = 0,
a3 =
1
3
t‰.
ln(1 + x)
1 x
1
ln(1 + x) X
sol)
=
an xn |‡ Xê.
1 x
n=0
1
= 1 + x + x2 + x3 + · · ·
1 x
e)
¨| ` ⇠ à‰. ƒ⇠DP| µt
a0 = 0,
1
X
x2n
ÑD L ⇠ à‰.
ÑD L ⇠ à‡,
a0 +a1 x+(a2
sinh x =
= sinh x =
n=0
cos x =
,
t¿\
ÑD L ⇠ à‰.
b) tan x
¨
1
3
ÑD L ⇠ à‰.
(|x| < 1)
t¿\
c) ln|sec x + tan x|
sol) ln|sec x + tan x| =
1
X
1
=1
1+x
an xn \ Pê. a)à 8⌧| µt sec x
n=0
X q ⇠ ⌅⌧›D L ⇠ à<p
0
(ln|sec x + tan x|) = sec x
t¿\ mƒ
Ñ– Xt
◆
Z x
Z x✓
1
ln|sec x + tan x| =
sec tdt =
1 + t2 + · · · dt
2
0
0
1 3
= x + x + ···
6
x + x2
t‰. mƒ Ñ– Xt
Z x
ln(1 + x) =
0
ÑD L ⇠ à‡
✓
ln(1 + x)
= x
1 x
x3 + · · ·
1
dt = x
1+t
1 2 1 3
x + x
2
3
(|x| < 1)
1 2 1 3
x + x
2
3
···
◆
· · · · 1 + x + x2 + x3 + · · ·
D µt
ÑD L ⇠ à‰. 0|⌧
a0 = 0,
a1 = 1,
a2 = 0,
a3 =
1
6
1
5
x + x2 + x3 + · · · = a0 + a1 x + a2 x2 + a3 x3 + · · ·
2
6
ÑD L ⇠ à‰. 0|⌧
t‰.
a0 = 0,
66
a1 = 1,
a2 =
1
,
2
a3 =
5
6
SOLUTION
t‰.
6 q ⇠@ L|Ï
¨
t‡ mƒ¯ÑD \à T Xt
f) ex ln(1 + x)
sol) ex ln(1 + x)
1
X
f 00 (x) =
an xn \ Pê.
1
X
(2n + 1)(2n + 2)( 1)n+1 x2n
22n+2 {(n + 1)!}2
n=0
n=0
ex =
1
X
xn
n=0
t¿\
n!
,
x
e ln(1 + x) =
1 2 1 3
x + x
2
3
ln(1 + x) = x
1
X
xn
n=0
n!
! ✓
· x
···
1 2 1 3
x + x
2
3
···
ÑD L ⇠ à‰. x2 f 00 (x),xf 0 (x),x2 f (x)î ‰L¸ ⇡‰.
1
X
(2n + 1)(2n + 2)( 1)n+1 x2n
,
x f (x) = x
22n+2 {(n + 1)!}2
n=0
2 00
◆
xf 0 (x) = x
= a 0 + a 1 x + a 2 x2 + a 3 x3 + · · ·
ÑD L ⇠ à‡
1
X
(2n + 2)( 1)n+1 x2n+1
,
22n+2 {(n + 1)!}2
n=0
x2 f (x) = x2
1
1
x + x2 + x3 + · · · = a0 + a1 x + a2 x2 + a3 x3 + · · ·
2
3
t‰. 0|⌧
a0 = 0,
a1 = 1,
a2 =
1
,
2
a3 =
1
3
arctan(x)
1+x
1
arctan(x) X
sol)
=
an xn |‡ Xê.
1+x
n=0
1
X
( 1)n x2n+1
1
arctan(x) =
,
=1
2n
+
1
1
+
x
n=0
( 1)n
(2n + 2)( 1)n+1
(2n + 1)(2n + 2)( 1)n+1
+ n+1
+
n
2
2
4 (n!)
4
{(n + 1)!}
4n+1 {(n + 1)!}2
( 1)n
(2n + 2)2 ( 1)n+1
= n
+ n+1
2
4 (n!)
4
{(n + 1)!}2
( 1)n
4(n + 1)2 ( 1)n+1
= n
+ n+1
2
4 (n!)
4
{(n + 1)!}2
( 1)n
( 1)n+1
= n
+ n
=0
2
4 (n!)
4 (n!)2
g)
t¿\
1
X
( 1)n x2n+1
2n + 1
n=0
!
· 1
x + x2
x + x2
x3 + · · ·
x3 + · · ·
= a0 + a1 x + a2 x2 + a3 x3 + · · ·
ÑD L ⇠ à‰.
t¿\ ùÖt DÃ⌧‰.
7. q ⇠\ \⌅⌧ h⇠ f (x) =
1
X
an x3n î f 00 (x)
n=0
2
x2 + x3 + · · · = a0 + a1 x + a2 x2 + a3 x3 + · · ·
3
t¿\ ƒ⇠DP| µt
x
a0 = 0,
1
X
( 1)n x2n
.
22n (n!)2
n=0
t⌧ ÑXX n( mX ƒ⇠‰X it 0t (D Ùtê. ƒ∞–
Xt
t‰.
arctan(x)
=
1+x
2
a 1 = 1 , a2 =
1,
D Ãq\‰. a0 = 1| L, a1 , a2 , a3 | lX‹$.
sol) f (x)| mƒ¯ÑXt
2
a3 =
3
f 0 (x) =
1
X
(3n)an x3n 1
n=1
ÑD L ⇠ à‰.
t‡ ‰‹ mƒ¯ÑXt
1
X
n 2n
( 1) x
t|‡ `L, x2 f 00 (x) + xf 0 (x) +
2n (n!)2
2
n=0
x2 f (x) = 0ÑD Ùt‹$.
proof ) f (x)| mƒ¯ÑXt
6. f (x) =
f 0 (x) =
f 00 (x) =
1
1
X
X
(2n)( 1)n x2n 1
(2n + 2)( 1)n+1 x2n+1
=
22n (n!)2
22n+2 {(n + 1)!}2
n=1
n=0
=
1
X
n=1
1
X
n=0
67
(3n)(3n
1)an x3n 2
(3n + 3)(3n + 2)an+1 x3n+1
xf (x) = 0
SOLUTION
t‰. h⇠ f (x) f 00 (x)
tÙt
f 00 (x)
xf (x) =
=
1
X
n=0
1
X
⇠
x
1
X
an x3n
n=0
x3n+1 {(3n + 3)(3n + 2)an+1
⇠4⇠Ω¥X ®‡ x–
an+1 =
n=4
f 0 (x) =
sin x
f 00 (x) =
cos x,
t 0 t¥| X¿\
1
2
ÖXt
1
1
1
, a2 =
, a3 =
6
180
12960
⇡
–⌧ f X 4( L|Ï ‰m›@
3
p
p
3
⇡
1
⇡ 2
3
⇡ 3
1
(x
)
(x
) +
(x
) + (x
2
3
4
3
12
3
48
t¿\ x =
µ8⌧ 6.3 .
¿‹⌧ ⇣ x = a –⌧ ¸¥ƒ h⇠ f (x)X n( L|Ï ‰m›D
lX‹$.
1. f (x) = sin2 x;
a = 0, n = 5
sol)
sol)
⇡
,
6
a=
n=6
f 0 (x) = cos x,
f 00 (x) =
f
(3)
sin x,
(x) =
cos x,
f (4) = sin x,
f 0 (x) = sin 2x,
f 00 (x) = 2 cos 2x
f (5) = cos x,
f (3) (x) =
f (6) =
f (4) =
4 sin 2x,
8 cos 2x,
sin x
⇡
–⌧ f X 6( L|Ï ‰m›@
6
p
p
1
3
⇡
1
⇡ 2
3
⇡ 3
+
(x
)
(x
)
(x
)
2
2
6
4 p 6
12
6
1
⇡ 4
3
⇡ 5
1
+ (x
) +
(x
)
(x
48
6
240
6
1440
t¿\ x =
f (5) = 16 sin 2x
t¿\, x = 0–⌧ f X 5( L|Ï ‰m›@
x2
⇡ 4
)
3
t‰.
4. f (x) = sin x;
1 4
x
3
⇡ 6
)
6
t‰.
t‰.
a = 0,
n=4
5. f (x) = tan x;
a=
sol)
⇡
,
4
n=4
f 0 (x) =
sin 2x,
f 0 (x) = sec2 x,
f 00 (x) =
2 cos 2x,
f 00 (x) = 2 tan x sec2 x,
f (3) (x) = 4 sin 2x,
f (3) (x) = 2 sec4 x + 4 tan2 x sec2 x,
f (4) (x) = 8 cos 2x
f (4) (x) = 16 tan x sec4 x + 8 tan3 x sec2 x
t¿\ x = 0–⌧ f X 4( L|Ï ‰m›@
1
t‰.
⇡
,
3
f (4) = cos x,
ÑD L ⇠ à‰.
2. f (x) = cos2 x;
sol)
a=
¨
f (3) = sin x,
an }
an
(3n + 3)(3n + 2)
t¥| \‰. a0 = 1 t¿\ ›–
a1 =
Ö 3. f (x) = cos x;
sol)
(3n + 3)(3n + 2)an+1 x3n+1
n=0
t‰. ⌅
xf (x) = 0| ÃqX¿\ ⌅ ›D
6 q ⇠@ L|Ï
t¿\ x =
1
x2 + x4
3
1 + 2(x
t‰.
68
⇡
–⌧ f X 4( L|Ï ‰m›@
4
⇡
) + 2(x
4
⇡ 2 8
) + (x
4
3
⇡ 3 10
) + (x
4
3
⇡ 4
)
4
SOLUTION
6. f (x) = sec x;
a=
sol)
⇡
,
4
6 q ⇠@ L|Ï
¨
t¿\ x = 0–⌧ f X 6( L|Ï ‰m›@
n=4
1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + 7x6
f 0 (x) = sec x tan x,
t‰.
f 00 (x) = sec x tan2 x + sec3 x,
f (3) = sec x tan3 x + 2 sec3 tan x + 3 sec3 x tan x,
f
(4)
4
3
2
3
2
= sec x tan x + 3 sec tan x + 6 sec x tan x
9. f (x) = ex ;
a = 1, n = 5
sol) f (n) (x) = ex t¿\ x = 1–⌧ f X 5( L|Ï ‰m›@
+ 2 sec5 x + 9 sec3 x tan2 x + 3 sec5 x
t¿\ x =
e + e(x
⇡
–⌧ f X 4( L|Ï ‰m›@
4
p
⇡
3p
⇡ 2
)+
2(x
)
4
2
4
11 p
⇡ 3 19 p
⇡ 4
+
2(x
) +
2(x
)
6
4
8
4
2+
p
2(x
t‰.
e
e
1) + (x 1)2 + (x 1)3
2
6
e
e
+ (x 1)4 +
(x 1)5
24
120
t‰.
10. f (x) = 2x ;
a = 1, n = 4
sol) f (n) = 2x (ln 2)n t¿\ x = 1 –⌧ f X 4( L|Ï ‰m›
@
1) + (ln 2)2 (x
2 + 2 ln 2(x
1
7. f (x) = ;
x
sol)
a = 2,
n=6
f 0 (x) =
+
(ln 2)3
(x
3
1)3 +
(ln 2)4
(x
12
⇡
,
6
n=2
f 00 (x) = 2x 3 ,
f
(4)
f (5) (x) =
11. f (x) = ln(cos x);
6x 4 ,
5
(x) = 24x
a=
sol)
,
f 0 (x) =
120x 6 ,
f 00 (x) =
f (6) (x) = 720x 7
t¿\ x = 2 –⌧ f X 6( L|Ï ‰m›@
1
2
1
1
1
(x 2) + (x 2)2
(x 2)3
4
8
16
1
1
1
+ (x 2)4
(x 2)5 +
(x 2)6
32
64
128
t¿\ x =
(1
x)2
;
a = 0,
t‰.
sol)
n=6
f 00 (x) =
x)
3
f (x) = 3!(1
x)
4
f (3) (x) = 4!(1
x) 5 ,
f (x) = 2!(1
00
f (4) = 5!(1
x) 6 ,
f (5) = 6!(1
x) 7 ,
f
(6)
= 7!(1
x)
a=
⇡
,
3
n=2
f 0 (x) = cos x · esin x
sol)
0
sin x sec x tan x
⇡
–⌧ f X 2( L|Ï ‰m›@
6
p
p
3
3
⇡
2
⇡ 2
ln(
)
(x
)
(x
)
2
3
6
3
6
12. f (x) = esin x ;
8. f (x) =
sin x sec x,
cos x sec x
t‰.
1
1)4
t‰.
x 2,
f (3) (x) =
1)2
,
,
sin x · esin x + cos2 x · esin x
⇡
–⌧ f X L|Ï 2( ‰m›@
3
p
p
3
1 p3
⇡
1
3 p3
2
2
e + e (x
)+(
)e 2 (x
2
3
8
4
t¿\ x =
t‰.
8
69
⇡ 2
)
3
SOLUTION
µ8⌧ 6.4 .
1. ÑXX ‰⇠ x–
XÏ cos x =
1
X
n=0
6 q ⇠@ L|Ï
¨
p
p
3. x = 0–⌧ x + 1X 2( L|Ï ‰m›D t©XÏ 1.1X
¸ø✓D l` L, $(X l0– XÏ $ÖX‹$.
( 1)n 2n
x ÑD L|Ï sol) x = 0–⌧ px + 1 X 2( L|Ï ‰m›@
(2n)!
¨| t©XÏ Ùt‹$.
1
1 2
1+ x
x
proof ) 176p Ù0 6.4.1. Ät@ D∑X‰. R–⌧ X⌧ T¨x
2
8
p
h⇠– L|Ï ¨| ©\‰. ∏X¡ f (x) = cos x| Pê.
t‡ t| t©XÏ l\ 1.1X ¸ø✓@ 1.04875 t‰. ¯¨‡
ÑXX x 2 R– t
L|Ï ¨– XXÏ
2n
✓
◆
(k)
(2n+1)
X
p
f (0) k f
(c) 2n+1
5
1
1
1
2
cos x =
x +
x
| 1.1
1 + (0.1)
(0.1) | =
(c + 1) 2 (0.1)3
k!
(2n + 1)!
2
8
16
k=0
=
n
X
( 1)k 2k ( 1)n+1 sin c 2n+1
x +
x
(2k)!
(2n + 1)!
D Ãq‹§î c 2 (0, 0.1) t t¨\‰. c 2 (0, 0.1) t¿\
1
(c + 1)
16
k=0
0¸ x¨t– t¨\‰. ¯Ï¿\ ÑXX x 2 R
t ‰L ÄÒ›t 1Ω\‰.
| ÃqXî c
@ ê⇠ n–
|cos x
n
X
( 1)k 2k
|sin c|
|x|2n+1
x |=
|x|2n+1 
(2k)!
(2n + 1)!
(2n + 1)!
5
2
(0.1)3 <
1
(0.1)3 = 6.25 ⇥ 10 5
16
p
t‰. ¯Ï¿\
x = 0–⌧ x + 1X 2( L|Ï ‰m›D t©X
p
Ï l\ 1.1X ¸ø✓X $(X l0î 6.25 ⇥ 10 5 Ù‰ ë‰.
4. x = 0–⌧ ex X 3( L|Ï ‰m›D t©XÏ e0.1 X ˘ø✓D
l` L, $(X l0– XÏ $ÖX‹$.
X |x|2n+1
x
D(⇣ ï– 0tt ÑXX x 2 R– t
t ⇠4 sol) x = 0 –⌧ e X 3( L|Ï ‰m›@
(2n + 1)!
1
1
|x|2n+1
1 + x + x2 + x3
\‰. ¯Ï¿\ ÑXX x 2 R– t lim
= 0t‰.
2
6
n!1 (2n + 1)!
0|⌧ ë¿– n ! 1x ˘\D ËXt R–⌧ T¨x h⇠X t‡ t| t©XÏ l\ e0.1 X ¸ø✓@ 1.105166 . . . t‰. L
q ⇠ ⌅⌧›D ªî‰.
|Ï ¨– XXÏ
✓
◆
1
1
ec
x
2
3
|e
1
+
(0.1)
+
(0.1)
+
(0.1)
|
=
(0.1)4
1
n
Xx
2
6
24
x
2. ÑXX ‰⇠ x– XÏ e =
ÑD L|Ï ¨| t
n!
n=0
D ÃqXî c 2 (0, 0.1)
t¨\‰. c 2 (0, 0.1) t¿\
©XÏ Ùt‹$.
ec
e0.1
proof ) ∞ ∏X¡ f (x) = ex |‡ Xê. L|Ï ¨– XXÏ
(0.1)4 <
(0.1)4 ⇡ 4.6 ⇥ 10 6
24
24
ÑXX x 2 R– XÏ
k=0
x
e =
=
n
X
f k (0)
k=0
n
X
k=0
D ÃqXî c
f n+1 (c) n+1
x +
x
k!
(n + 1)!
k
1 k
ec
x +
xn+1
k!
(n + 1)!
0¸ x¨t– t¨\‰. 0|⌧
ex
n
X
1
k=0
n!
xk 
|ec |
|x|n+1
(n + 1)!
t‰. 0|⌧ x = 0 –⌧ ex X 3( L|Ï ‰m›D t©XÏ
l\ e0.1 X ¸ø✓X $(î 4.6 ⇥ 10 6 Ù‰ ë‰.
5. L|Ï ‰m›D t©XÏ $( 0.00005 Ù‰ ëå ⇠ƒ]
‰L
p ✓X ˘ø✓D lX‹$.
a) 10
sol) L|Ï ¨– Xt ⇣ x = 0 –⌧
n ✓1◆
X
p
2 31 2k xk + R (x),
9+x=
n
k
k=0
f (n+1) (c) n+1
Rn (x) =
x
=
(n + 1)!
✓
1
2
◆
2n+1
(9 + c) 2 xn+1
D(⇣ ï– XXÏ ÑXX x 2 R–
XÏ
n+1
c
e
|x|n+1 t ⇠4X¿\ ÑXX x 2 R –
XÏ
D ÃqXî c
0 ¸ x ¨t– t¨\‰. ⇣\ x 2 (0, 1)–
(n + 1)!
c
e
✓
◆
✓ 1 ◆
1
lim
|x|n+1 = 0 t‰. 0|⌧ n ! 1x ˘\D ËX
2n+1
1
2
2
n!1 (n + 1)!
(9 + c) 2 xn+1 
2n+1
n
+
1
n
+
1
3
1
X xn
t ex =
ÑD L ⇠ à‰.
1 1
n!

n=0
2 32n+1
t‡
X
70
t
SOLUTION
6 q ⇠@ L|Ï
t‰. t⌧
Rn (x) = ( 1)k
1 1
< 0.00005 () n
2 32n+1
t¿\ x = 0 –⌧ ⌅⌧\ 4( L|Ï ‰m›¸ f (x)@X $(î
0.00005 Ù‰ ë‰. 0|⌧, ¸ø✓
p
10 =
p
9+1⇡
4 ✓1◆
X
2
k
k=0
( 1)
1
1
12
13
+
6 216 3888
= 3.162276...
5
885239
=
279936
279936
=3+
0 ¸ x ¨t– t¨\‰. ⇣\ x 2
t
1 2k k
3
1
n
(1 + c)n+1 (n + 1)
✓
1
3
◆n+1

✓
◆
1
,0 –
3
1
2n+1 (n + 1)
t‡
1
< 0.00005 () n
2n+1 (n + 1)
D ªî‰.
10
t¿\ x = 0 –⌧ ⌅⌧\ 10( L|Ï ‰m›¸ f (x)@X $(î
0.00005 Ù‰ ë‰. 0|⌧, ¸ø✓
p
3
b) 26
sol) L|Ï
¨– Xt ⇣ x = 0 –⌧
✓1◆
n
X
p
1 3k
3
k 3
27 x =
( 1)
(27) 3 xk + Rn (x),
k
ln
✓
2
= ln 1
3
1
3
◆
k=0
Rn (x) = ( 1)
n+1
✓
1
3
n+1
◆
(27
c)
3n+2
3
⇡
=
✓
◆k
10
X
( 1)k+1
1
k
3
k=1
20111503
=
49601160
0.4054643...
xn+1
D ÃqXî c
0 ¸ x ¨t– t¨\‰. ⇣\ x 2 (0, 1)–
✓ 1 ◆
✓ 1 ◆
3n+2
3n+2
3
3
( 1)n+1
(27 c) 3 xn+1 
8 3
n+1
n+1
1
1
1

·
· 3n+2
3 n+1 2
t‡
1
2
= ln
ln 2 t‡ ¯8–⌧ t¯ l\
3
3
t
ln 2 ⇡ 0.6931471... D å⇠⇣ Ï/¯ ê¨L¿
à J¥⌧
\©Xt
D ªî‰. t⌧ ln
ln
1
2
⇡ ln
3
3
20111503
0.693147
49601160
= 1.098612...
ln 2 =
D ªî‰.
1
1
1
·
· 3n+2 < 0.00005 () n
3 n+1 2
3
d) sin 1
t¿\ x = 0 –⌧ ⌅⌧\ 3( L|Ï ‰m›¸ f (x)@X $(î sol) L|Ï
0.00005 Ù‰ ë‰. 0|⌧, ¸ø✓
p
3
1
xn+1
(1 + c)n+1 (n + 1)
4
D ÃqXî c
¨
p
26 = 3 27
3 ✓1◆
X
3 31 3k (
1⇡
k=0
k
1
12
27 2187
= 2.962496...
=3
sin x =
1)k
n
X
( 1)k 2k+1
x
+ Rn (x),
(2k + 1)!
k=0
5
1574392
=
531441
531441
Rn (x) =
D ÃqXî c
D ªî‰.
1
c) ln
3
sol) L|Ï
¨– Xt ⇣ x = 0 –⌧
( 1)n+1 sin c 2n+2
x
(2n + 2)!
0 ¸ x ¨t– t¨\‰. ⇣\ x 2 (0, 1)–
( 1)n+1 sin c 2n+2
1
x

(2n + 2)!
(2n + 2)!
¨– Xt ⇣ x = 0 –⌧
t‡
n
X
( 1)k+1 k
ln(1 + x) =
x + Rn (x),
k
1
< 0.00005 () n
(2n + 2)!
k=1
71
3
t
SOLUTION
t¿\ x = 0 –⌧ ⌅⌧\ 2 · 3 + 1 = 7( L|Ï ‰m›¸ f (x)
@X $(î 0.00005 Ù‰ ë‰. 0|⌧, ¸ø✓
3
X
( 1)k 2k+1
sin 1 ⇡
1
(2k + 1)!
k=0
13
15
=1
+
6
120
= 0.841468...
1+x
;
1 x
a=0
f (x) = ln
1+x
= ln(1 + x)
1 x
e) f (x) = ln
sol)
t‡
17
4241
=
5040
5040
f 0 (x) =
µ8⌧ 6.5 .
1. ¿‹⌧ ⇣ x = a–⌧ ¸¥ƒ h⇠ f (x)X L|Ï
‹$.
1
X
1
=
( 1)n xn ,
1 + x n=0
⇠| lX t¿\
f 0 (x) = 2
t‡ 0|⌧
(x 2 R)
f (x) = 2
a=1
✓ ◆✓
◆n
3
X
3
x 1
=
8
n
2
n=0
1
1
X
1) + 6(x
✓
x
1
2
◆
d) f (x) =
sol)
3
x
;
1)2 + (x
sol) cos x = cos x·cos x =
1)2 · · · .
=
!
!
1
1
X
X
( 1)n 2n
( 1)n 2n
x
·
x
(2n)!
(2n)!
n=0
n=0
x2
x4
+
+ · · · ) · (1
2!
4!
f (x) = (1
x2
x4
+
+ ···)
2!
4!
t¿\ ƒ⇠DPïD µt
a0 = 1,
(|x
1| < 2)
a1 = a3 = 0,
1
1
1
+ (x + 1) + 3 (x + 1)2 + · · · .
4 16
4
1
1) cos x
1) cos x = (x2
1
x+1
4
a2 =
ÑD L ⇠ à‰.
b) f (x) = (x2
sol)
1
1
1
=
x
4 (x + 1)
41
✓
◆n
1
X
1 x+1
=
4
4
n=0
=
an xn t| ` L,
1)3 .
t‡
a=
1
X
⇠
n=0
(x2
1
3
1
(|x| < 1)
2n + 1
a0 , a1 , a2 , a3 D lX‹$.
2
1
1) + (x
8
(|x| < 1)
(|x| < 1)
2. Dò– ¸¥ƒ f (x)X ‰t\∞
a=1
1
(x
4
xn ,
n=0
)3
1
1
=
=
1+x
2 + (x 1)
1 ( x2 1 )
✓
◆n
1
1X
x 1
=
( 1)n
2 n=0
2
1
2
x2n
1
X
t‰.
1
2
=
x
=
1
X
x2n+1
a) f (x) = cos2 x
1
c) f (x) =
;
1+x
sol)
1
1
+
1+x 1 x
1
n=0
1 + 2)3 = 8(1 +
= 8 + 12(x
x)
n=0
a) f (x) = (x + 1)3 ; a = 0
sol)
(x + 1)3 = 1 + 3x + 3x2 + x3 .
(x + 1)3 = (x
ln(1
¨
t‰. ⇣\
D ªî‰.
b) f (x) = (x + 1)3 ;
sol)
6 q ⇠@ L|Ï
1) · (1
x2
x4
+
+ ···)
2!
4!
t¿\ ƒ⇠DPïD µt
a0 =
(|x + 1| < 4)
ÑD L ⇠ à‰.
72
1 a1 = 0,
a2 =
3
,
2
a3 = 0
SOLUTION
c) f (x) = ex sin x
sol)
6 q ⇠@ L|Ï
t‰.
ex · sin x = (1 + x +
x2
+ · · · ) · (x
2!
x3
+ ···)
3!
t¿\ ƒ⇠DPïD µt
a0 = 0,
a1 = 1,
a2 = 1,
1
a3 =
3
3. Dò– ¸¥ƒ F (x)X ‰t\∞ ⇠| t©XÏ $(
Ù‰ ëå ⇠ƒ] F (1)X ¸ø✓D lX‹$.
e
1
sol)
ex ·
0.001
[Ùp ¨] t Ùp ¨| 8‡Xê.
{an } t Ëp⇠4tt⌧ 0 <\ ⇠4Xt
1
X
ÑD L ⇠ à‰.
d) f (x) =
¨
x
m
X
( 1)k ak
k=0
x
k=0
( 1)k ak  |am+1 |
t‰.
2
1
1
x
= (1 + x +
3
x
x
+
+ · · · )(1 + x + x2 + x3 + · · · )
2!
3!
t¿\ ƒ⇠DPïD µt
a0 = 1,
a) F (x) =
sin t2 dt
0
a1 = 2,
5
a2 = ,
2
8
a3 =
3
sol) ‰⇠ ⌅¥–⌧
sin x =
ÑD L ⇠ à‰.
cos x
e) f (x) = p
1 x
sol)
1
X
( 1)n 2n+1
x
(2n + 1)!
n=0
t¿\
1 2 x4
cos x = 1
x +
+ ·,
2
24
1
1
3
5
p
= 1 + x + x2 + x3 + · · ·
2
8
16
1 x
t¿\
cos x
1
p
=1+ x
2
1 x
t‰. ¯Ï¿\
a0 = 1,
Z x
a1 =
1
,
2
1 2
1
x + x3 + · · ·
8
16
a2 =
1
,
8
a3 =
sin x2 =
t‰. ⌅X ⇠ƒ ‰⇠ ⌅¥–⌧ ⇠4Xî q ⇠ t‰. 0|⌧,
‰L¸ ⇡@ mƒ Ñt 1Ω\‰.(¿ò ƒ ã@ ƒ¿–: l¥
1
X
( 1)n 4n+2
<\, 1Ù‰ p ë⇠ – t [ , ]–⌧
x
(2n + 1)!
n=0
‡Ò⇠4(uniformly convergent)X¿\, 0Ä0 1L¿X mƒ Ñ
t 1Ω\‰.)
1
16
1
X
( 1)n 4n+2
t
(2n + 1)!
n=0
Z x
=) F (x) =
sin t2 dt
sin t2 =
t‰.
p
f) f (x) = cos( 1 + x)
sol) İРXt
f (0) = cos 1,
sin 1
f 0 (0) =
,
2
sin 1 cos 1
f 00 (0) =
,
4
3 cos 1 2 sin 1
f (3) (0) =
8
1
X
( 1)n 4n+2
x
(2n + 1)!
n=0
0
=) F (x) =
1
X
( 1)n
x4n+3
(2n
+
1)!(4n
+
3)
n=0
t‰. mƒ Ñ⌧ q ⇠X ⇠4⇠¿Ñƒ < Ñ h⇠ X q
⇠X ⇠4 ⇠¿Ñ¸ ⇡<¿\ F (1) @ t¨\‰. an =
( 1)n
t| Xê. ¯Ït {an }@ 0<\ ⇠4Xî Ë
(2n + 1)!(4n + 3)
p⇣å⇠Ùt¿\
t¿\
1
X
sin 1
a0 = cos 1,
a1 =
,
2
sin 1 cos 1
3 cos 1 2 sin 1
a2 =
, a3 =
8
48
k=0
73
( 1)k
(2k + 1)!(4k + 3)
m
X
k=0
1
=
(2m + 3)!(4m + 3)
( 1)k
 |am+1 |
(2k + 1)!(4k + 3)
SOLUTION
t‰. m
6 q ⇠@ L|Ï
1
< 0.001 t¿\ F (1)X t¿\
(2m + 3)!(4m + 3)
1|L
¸ø✓@
2
e x =
1
X
k=0
t‰.
b) F (x) =
Z x
( 1)k
= 0.309524
(2k + 1)!(4k + 3)
p
cos
p
x=
=) F (x) =
n
( 1) n
x
(2n)!
n=0
1
X
t=
( 1)k
(2k + 1)(k!)
k=0
m
X
( 1)
(2k)!(k + 1)
k=0
t‰. m
0
1
X
( 1)n
xn+1
(2n)!(n
+
1)
n=0
k
( 1)
 |am+1 |
(2k)!(k + 1)
=
t‰. m
2 | L
k=0
t‰.
c) F (x) =
Z x
m
X
1
(2m + 2)!(m + 2)
k=0
4 | L
1
(m + 1)!(2m + 3)
1
< 0.001 t¿\ F (1)X
(m + 1)!(2m + 3)
4
X
k=0
( 1)k
= 0.7474868
(2k + 1)(k!)
Z x
1
1
x
=
1
X
xn
n=0
1
X
1
=
( 1)n t7n
1 + t7
n=0
k
( 1)
= 0.7635417
(2k)!(k + 1)
t‰. ⌅X
⇠ƒ ⇠4⇠Ω ¥–⌧ mƒ Ñt
1
X
1
=
( 1)n t7n
1 + t7
n=0
Z x
1
=) F (x) =
dt
1
+
t7
0
1
X
x7n+1
=) F (x) =
( 1)n
7n + 1
n=0
2
e t dx
0
sol) ‰⇠ ⌅¥–⌧
1
X
xn
n=0
x2n+1
(2n + 1)(n!)
( 1)k
 |am+1 |
(2k + 1)(k!)
1
dt
7
0 1+t
sol) |x| < 1 –⌧
d) F (x) =
1
< 0.001 t¿\ F (1)X
(2m + 2)!(m + 2)
t¿\
ex =
( 1)n
¸ø✓@
¸ø✓@
2
X
1
X
=
t‰. mƒ Ñ⌧ q ⇠X ⇠4⇠¿Ñƒ < Ñ h⇠ X q ⇠
( 1)n
X ⇠4 ⇠¿Ñ¸ ⇡<¿\ F (1) @ t¨\‰. an =
(2n)!(n + 1) t‰.
t| Xê. ¯Ït {an }@ 0<\ ⇠4Xî Ëp⇣å⇠Ùt¿\
k
2
e t dt
t‰. mƒ Ñ⌧ q ⇠X ⇠4⇠¿Ñƒ < Ñ h⇠ X q ⇠
1
X ⇠4 ⇠¿Ñ¸ ⇡<¿\ F (1) @ t¨\‰. an =
(2n + 1)(n!)
t| Xê. ¯Ït {an } @ 0<\ ⇠4Xî Ëp⇣å⇠Ùt¿\
k=0
1
X
( 1)n n
t
(2n)!
n=0
Z x
p
=) F (x) =
cos tdt
1
X
Z x
n=0
t‰. ⌅X ⇠ƒ ‰⇠ ⌅¥–⌧ ⇠4Xî q ⇠ t‰. 0|⌧,
‰L¸ ⇡@ mƒ Ñt 1Ω\‰.
=) F (x) =
t2n
n!
0
1
X
( 1)n 2n
cos x =
x
(2n)!
n=0
p
( 1)n
=) F (x) =
sol) ‰⇠ ⌅¥–⌧
cos
1
X
n=0
tdt
0
1
X
x2n
n!
t‰. ⌅X ⇠ƒ ‰⇠ ⌅¥–⌧ ⇠4Xî q ⇠ t‰. 0|⌧,
‰L¸ ⇡@ mƒ Ñt 1Ω\‰.
e t =
t¿\
( 1)n
n=0
2
cos
1
X
¨
n!
74
•X¿\
SOLUTION
1
X
( 1)n
7n + 1
n=0
t‰.
1\
⇠4X¿\ D®
î å˘\D ËXt F (1) =
6 q ⇠@ L|Ï
¨
32
¨– XXÏ F (x)– t¿\ ˘\✓@ 3 t‰.
1
X
( 1)n
7n + 1
n=0
1Ω\‰. b) lim e
2x
e 2x
x!0
x
sol) x = 0 –⌧ e2x , e 2x X L|Ï ⇠î
1
an =
t| Xê. {an }@ 0<\ ⇠4Xî Ëp⇣å⇠Ù
7n + 1
4
t¿\
e2x = 1 + 2x + 2x2 + x3 + · · ·
3
1
m
X
X
( 1)k
( 1)k
 |am+1 |
4 3
7k + 1
7k + 1
k=0
k=0
e 2x = 1 2x + 2x2
x + ···
3
1
=
7m + 8
t‡
t‰. m
142 | L
1
< 0.001 t¿\ F (1)X ¸ø✓@
7m + 8
142
X
( 1)k
= 0.9159802
7k + 1
e2x
e 2x
x
=
4x + 83 x3 + · · ·
8
= 4 + x2 + · · ·
x
3
t¿\ ˘\✓@ 4 t‰.
k=0
t‰.
[D®] l⌅ ( R, R) ⌅–⌧ f (x) =
1
X
an xn t 1Ω\‰‡
2
cos x x2
x!0
x4
sol) x = 0–⌧ cos xX L|Ï ⇠î
c) lim
1
n=0
Xê. Ã| l⌅ ( R, R) X ]⇣ x = R –⌧
1
X
cos x = 1
⇠
t‡
an R n
n=0
t ⇠4Xt
1
ƒ›
lim f (x) =
x!R
t 1Ω\‰.
1
X
an R n
x2
2
=
x4
x6
4! + 6! + · · · +
=
x4
1
x2
+
+ ···
4!
6!
1
t‰.
24
t¿\ ˘\✓@
n=0
tan 1 x sin x
x!0
x3 cos(2x)
1
⇠| t©XÏ ‰LX ˘\✓D lX‹$. sol) x = 0–⌧ tan x, sin x,
l\ ƒ 8⌧X ›– ÖXt
Remark) ⇠4⇠Ωt t¨Xî q ⇠– t, ⇠ ⇠4⇠Ω
¥X ÑXX ƒ–l⌅–⌧ t
t‡ ‡Ò⇠4X¿\, ˘\¸
i 08X ⌧⌧| ∏¥ ƒ∞` ⇠ à<¿\ ⇠X
m–
˘\D ËX‡ TXσ 4)X‰.
(x
tan 1 x sin x
=
3
x cos(2x)
=
(1 + 4x + 8x2 )
x!0
x3
sol) x = 0 –⌧ e4x X L|Ï ⇠î
e4x
e4x = 1 + 4x + 8x2 +
=
32 3
x + ···
3
t‡
e4x
cos x
x4
d) lim
4. x = 0–⌧ L|Ï
a) lim
x2
x4
+
+ ···
2!
4!
4
( 32 x3 + 32
32 32
(1 + 4x + 8x2 )
3 x + ···)
= 3
=
+ x + ···
3
3
x
x
3
3
t‰. 0|⌧ ˘\✓@
cos(2x)X L|Ï
⇠|
3
5
x3
x5
(x x3! + x5! + · · · )
3 + 5 + ···)
2
16x4
x3 (1 4x
2! + 4! + · · · )
1
1
1
1
3
5
3!
3 x + 5
5! x + + · · ·
x3 (1 2x2 + 23 x4 + · · · )
1
23 2
1
1
4
6 + 120 x + 7!
7 x
(1 2x2 + 23 x4 + · · · )
1
t‰.
6
sin(3x2 )
x!0 1
cos(2x)
sol) x = 0 –⌧ sin(3x2 ),
e) lim
75
cos(2x)X L|Ï
⇠|
l\
SOLUTION
ƒ 8⌧X ›–
ÖXt
2 3
t‰.
r¸ ✓X î⌅ H– àî É<\ ‡Xt ı@
p t| ¸¥ƒ
5, arctan( 12 ) t‰.
2 5
3x2 (3x3! ) + (3x5! ) + · · ·
sin(3x2 )
=
2
(2x)4
1 cos(2x)
1 1 + (2x)
+ ···
=
=
t¿\ ˘\✓@
7
3x
3
2
p
b) ( 3, 1)
2!
4!
9 6
35 1
2 x + 5! x 0 + · · ·
2x2 23 x4 + · · ·
9 4
35 8
2 x + 5! x + · · ·
2 23 x2 + · · ·
sol) ˘å\\ ò¿¥t
3
t‰.
2
d) (
⇡⌘
sol) ˘å\\ ò¿¥t
⇡
⇡ p
= 1, y = 2 · sin = 3
3
3
p
t‰. 0|⌧ ı@ (1, 3)t‰.
x = 2 · cos
⇣
1,
⇡⌘
3
sol) ¡På\\ ò¿¥t
c)
✓
2,
3⇡
2
11⇡
6
◆
t‰.
p p
2, 2)
3
sol) x = r cos ✓, y = r sin ✓t¿\
b)
2,
p
sol) ˘å\\ ò¿¥t ( 5, ⇡ + arctan(2)) t‰.
µ8⌧ 7.1 .
1. ‰L¸ ⇡t ˘å\\ \⌅⌧ ⇣‰D ¡På\\ ò¿¥‹$.
a) 2,
✓
c) ( 1, 2)
å\ƒ
⇣
7 å\ƒ
1
,
2
✓
2,
3⇡
4
◆
t‰.
3. ‰L¸ ⇡t ˘å\\ ¸¥ƒ · X ⌧ D ¯¨‹$.
a) (Cardioid) r = 1 + cos ✓
sol) ¸¥ƒ · X ⌧ @ ‰L¸ ⇡‰.
90
120
60
p !
3
t‰.
2
◆
150
30
sol) ¡På\\ ò¿¥t (0, 2) t‰.
d)
⇣
1,
⇡⌘
6
sol) ¡På\\ ò¿¥t
0
180
p
3
,
2
1
2
!
0.5
1
1.5
0
t‰.
210
330
2. ‰L¸ ⇡t ¡På\\ \⌅⌧ ⇣‰D ˘å\\ ò¿¥‹$.
240
a) (2, 1)
sol) x = r cos ✓, y = r sin ✓, r
2
270
300
0, 0  ✓  2⇡ t¿\
2 = r cos ✓, 1 = r sin ✓ =) r2 = 5, tan ✓ =
1
2
b) r = 2a sin ✓ (aî ëX ¡⇠)
sol) ¸¥ƒ · X ⌧ @ ‰L¸ ⇡‰.
76
SOLUTION
90
120
150
0.5
1
1.5
2
210
240
0
0.5
210
1
0
330
240
300
270
30
180
0
330
a>0
60
150
30
0
90
120
60
180
7 å\ƒ
300
270
4. ¡På\\ \⌅⌧ · D ˘å\\ ò¿¥‹$.
a) y = 3x
sol) x = r cos ✓,
p
c) r = 3 + 2 sin ✓
sol) ¸¥ƒ · X ⌧ @ ‰L¸ ⇡‰.
90
120
60
y = r sin ✓|
ÖXt
tan ✓ = 3
D ªD ⇠ à‰. DòX 8⌧‰ƒ ⇡@ )ï<\ Ä ⇠ à‰.
150
b) x2 + y 2 = 9
sol) ¸¥ƒ · D ˘å\\ ò¿¥t r = 3 t‰.
30
c) x2 y 2 = 1
sol) ¸¥ƒ · D ˘å\\ ò¿¥t r2 cos(2✓) = 1 t‰.
0
180
1
2
210
3
4
0
330
240
270
300
d) x2 + (y 2)2 = 4
sol) ¸¥ƒ · D ˘å\\ ò¿¥t r = 4 sin ✓ t‰.
5. ¡På\\ \⌅⌧
ÌD ˘å\\ ò¿¥‹$.
p
a) 3  x  3, 0  y  9 x2
sol) ¸¥ƒ Ì@ –⇣t (0, 0)t‡ ⇠¿Ñt 3x –⇣X xï
⌅– ìx ÄÑt‰. 0|⌧ t| ‰‹ ˘å\\ \⌅Xt
0  r  3,
0✓⇡
t‰.
d) r = sin(2✓)
sol) ¸¥ƒ · X ⌧ @ ‰L¸ ⇡‰.
b) x2 + y 2  1, x 0, y 0
sol) ¸¥ƒ ÌD ˘å\\ \⌅Xt
0  r  1,
77
0  ✓  ⇡/2
SOLUTION
7 å\ƒ
t‰.
120
p
c) {(x, y) 2 R2 |0  x  1, 0  y p 3x}
[ {(x, y) 2 R2 |1  x  2, 0  y  4 x2 }
sol) ¸¥ƒ ÌD ˘å\\ \⌅Xt
0  r  2,
60
150
30
0  ✓  ⇡/3
0
180
t‰.
6. ˘å\\ \⌅⌧
90
0.5
1
1.5
2
0
ÌD ¡På\\ ò¿¥‹$.
210
a) 0  r  3, ⇡/2  ✓  ⇡
sol) ¸¥ƒ Ì@ –⇣t (0, 0)t‡ ⇠¿Ñ 3x –⇣X ⌧ 2¨Ñ
t– t˘Xî ÄÑt‰. 0|⌧ t| ‰‹ ¡På\\ \⌅Xt
3  x  0,
0y
t‰.
p
9
b) 1  r  2, 0  ✓  2⇡
sol) ¸¥ƒ ÌD ¡På\\ \⌅Xt
1  x2 + y 2  4
t‰.
c) ⇡/2  ✓  ⇡/2, 0  r  2 cos ✓
sol) ¸¥ƒ ÌD ¡På\\ \⌅Xt
0  (x
x2
330
240
270
300
¸¥ƒ pt‰ ⌘
p
r2 = 4 cos ✓ =) r = ± 4 cos ✓
⇠¥ $xΩ › r = 1 cos ✓| ΩXt r 0, 0  ✓ < 2⇡x
î⌅–⌧ \⌅⌧ 3⌧X P⇣
p
p
p
p
(2 2 2, cos 1 (3 2 2)), (2 2 2, 2⇡ cos 1 (3 2 2)), (2, ⇡)
| ªD ⇠ à‰. ⇣, r = 0x Ω∞| › Xt P⇣ (0, 0) ƒ ªD
⇠ à‰. ª@ P⇣‰D ⌅X ¯º¸ DPXt
4⌧X P⇣D
–D Ux` ⇠ à‰.
b) r = sin ✓,
sol)
1)2 + y 2  1
r = cos 2✓
120
90
60
t‰.
150
d) 0  ✓  ⇡, 0  r  2 sin ✓
sol) ¸¥ƒ ÌD ¡På\\ \⌅Xt
0  x2 + (y
1)2  1
30
0
180
0.5
1
0
t‰.
7. P · ‰X P⇣‰D ®P lX‹$.
a) r2 = 4 cos ✓,
r=1
210
cos ✓
330
240
sol)
78
270
300
SOLUTION
7 å\ƒ
90
¸¥ƒ P · X ›D ΩXt
120
2 sin2 ✓
sin ✓ = cos(2✓) =) sin ✓ = 1
=) 2 sin2 ✓ + sin ✓ 1 = 0
1
=) sin ✓ = , sin ✓ = 1
2
t¿\ r
150
0, 0  ✓ < 2⇡x î⌅–⌧ \⌅⌧ P⇣ 3⌧
✓
◆ ✓
◆ ⇣
1 ⇡
1 5⇡
⇡⌘
,
,
,
, 1,
2 6
2 6
2
sin ✓,
120
0
1.5
0
300
270
90
60
0
¸¥ƒ P · X ›D r
0.5
1
1.5
210
r = sin ✓,
(p
3 cos ✓,
p
r=
3 cos ✓,
2
0
330
270
0  ✓  ⇡/2, 3⇡/2  ✓ < 2⇡
⇡/2  ✓  3⇡/2
t¿\ P⇣ 2⌧
3 ⇡
,
2 3
300
!
,
p
3 2⇡
,
2 3
!
| ªD ⇠ à‰. ⇣, r = 0x Ω∞| › Xt P⇣ (0, 0) ƒ ªD
⇠ à‰. ª@ P⇣‰D ⌅X ¯º¸ DPXt
3⌧X P⇣D
–D Ux` ⇠ à‰.
¸¥ƒ P · X ›D ΩXt
sin ✓ = 1 + cos ✓ =) tan ✓ =
0, 0  ✓ < 2⇡x î⌅–⌧
\ ‰‹ ¯ ⇠ à‡, t XD ΩXt
(p
3,
0  ✓  ⇡/2
p
tan ✓ =
3, ⇡/2  ✓  3⇡/2
p
1
1
330
240
30
240
0.5
210
r = 1 + cos ✓
150
180
30
180
D ªD ⇠ à‰. ⇣, r = 0x Ω∞| › Xt P⇣ (0, 0) ƒ ªD
⇠ à‰. ª@ P⇣‰D ⌅X ¯º¸ DPXt
4⌧X P⇣D
–D Ux` ⇠ à‰.
c) r = 1
sol)
60
1
(Remark) T ¸¥ƒ P · X ›D Ë⌧à ΩXî É Ã
<\î ®‡ P⇣‰D l` ⇠ ∆î Ω∞, 9@ ‰⌧\ t¨X¿
t¿\ r 0, 0  ✓ < 2⇡x î⌅–⌧ \⌅⌧ P⇣ 2⌧
Jî P⇣‰D lXå ⇠î Ω∞ ›4‰. • p t \î
✓
◆ ✓
◆
· r = f (✓)»‰ ) ›D ÃqXî ®‡ ⇣‰D \⌅X0 ⌅
1 3⇡
1 7⇡
1 p ,
, 1+ p ,
t Dî\ ✓ X î⌅ ‰tp, ⇡@ ✓ ✓t|ƒ êË ¯$¿î
2 4
2 4
Ìt ‰t0 L8t‰. 0|⌧,
· D ¡⌘ ¯$Ù‡, µ
<\ò» P⇣X ⌧⇠| EX‡ xî Ét ⌧| å¡X‰.
| ªD ⇠ à‰. ⇣, r = 0x Ω∞| › Xt P⇣ (0, 0) ƒ ªD
⇠ à‰. ª@ P⇣‰D ⌅X ¯º¸ DPXt
3⌧X P⇣D
–D Ux` ⇠ à‰.
µ8⌧ 7.2 .
1. ‰L¸ ⇡t ¡På\\ \⌅⌧ ⇣‰D –0eå\\ ò¿¥
d) r = sin ✓, r2 = 3 cos2 ✓
‹$.
sol)
79
SOLUTION
a) (0, 2, 1)
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©Xt
0 = r cos ✓, 2 = r sin ✓, 1 = z
b) z =
7 å\ƒ
p
3r
(r
0)
sol) ¸¥ƒ ·tX ⌧ @ ‰L¸ ⇡‰.
2
=) r = 4, sin ✓ = 1, z = 1
t¿\ t| r
t‰.
0, 0  ✓ < 2⇡, z 2 Rx î⌅–⌧ \⌅Xt
⇣ ⇡ ⌘
2, , 1
2
raawi.EE
2J
b) (2, 1, 0)
sol) ¸¥ƒ å\| –0eå\\ \⌅Xt
✓
✓ ◆ ◆
p
1
5, 2⇡ arctan
,0
2
c) z = r2
t‰.
sol) ¸¥ƒ ·tX ⌧ @ ‰L¸ ⇡‰.
c) (1, 1, 1)
sol) ¸¥ƒ å\| –0eå\\ \⌅Xt
⇣p ⇡ ⌘
2, , 1
4
t‰.
p p
d) (2, 2, 2)
sol) ¸¥ƒ å\| –0eå\\ \⌅Xt
!
p !
p
p
2
6, arctan
, 2
2
一一
d) ✓ = z
t‰.
2. ‰L¸ ⇡t –0eå\\ ¸¥ƒ ·tX ⌧ D ¯¨‹$.
sol) ¸¥ƒ ·tX ⌧ @ ‰L¸ ⇡‰.
a) ✓ = ⇡2
sol) ¸¥ƒ ·tX ⌧ @ ‰L¸ ⇡‰.
3. ¡På\\ \⌅⌧ ) ›D –0eå\\ ò¿¥‹$.
a) x2 + y 2 + z 2 = 9
80
SOLUTION
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©Xt
r2 = 9
z2
7 å\ƒ
b) 0  r  1, 0  ✓  2⇡, 0  z  1
sol) ¸¥ƒ ÌD ¡På\\ \⌅Xt
x2 + y 2  1,
D ªD ⇠ à‰.
0z1
t‰.
b) x + y + z = 1
sol) ¸¥ƒ ) ›D –0eå\\ ò¿¥t
r(cos ✓ + sin ✓) + z = 1
p
c) 0  r  1, 0  ✓  2⇡, 0  z  3r
sol) ¸¥ƒ ÌD ¡På\\ \⌅Xt
t‰.
0z
c) x2 + y 2 + z 2 = x + y + z
sol) ¸¥ƒ ) ›D –0eå\\ ò¿¥t
r2 + z 2 = r(cos ✓ + sin ✓) + z
t‰.
4. –0eå\\ \⌅⌧ ) ›D ¡På\\ ò¿¥‹$.
a) r = 2 sin ✓
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©Xt
t‰.
6. ¡På\\ \⌅⌧ ÌD –0eå\\ ò¿¥‹$. r, ✓, z
ÃqXî ptD ®P ‹$.
a) x2 + y 2  4, x2 + y 2  z  16 3x2 3y 2
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©XÏ r
2⇡, z 2 R x î⌅–⌧ \⌅Xt
r2  4,
=) r2 = 2r sin ✓
=) x2 + (y
3r2
0  r2  z  16
3r2 ,
0  ✓ < 2⇡
t‰.
1)2 = 1
D ªD ⇠ à‰.
b) r2 cos(2✓) = z
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©Xt
r2 cos(2✓) = z
=) r2 (cos2 ✓
=) x
0, 0  ✓ <
2
=) x + y = 2y
2
r2  z  16
=) 0  r  2,
r = 2 sin ✓
2
p
p
3(x2 + y 2 )  3
sin2 ✓) = z
2
y =z
p
b) x2 + y 2  1,
x2 + y 2  z  1.
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©XÏ r
2⇡, z 2 R x î⌅–⌧ \⌅Xt
p
r2  1,
r2  z  1
=) 0  r  1,
0  ✓ < 2⇡
t‰.
µ8⌧ 7.3 .
1. ‰L¸ ⇡t ¡På\\ \⌅⌧ ⇣‰D ltå\\ ò¿¥‹$.
(0, 2, 1),
D ªD ⇠ à‰.
0  r  z  1,
0, 0  ✓ <
(2, 1, 0),
(1, 1, 1),
(2,
p p
2, 2)
sol) (0, 2, 1)D x = ⇢ sin cos ✓, y = ⇢ sin sin ✓, z = ⇢ cos ÑD
t©t⌧ ltå\\ ò¿¥t
5. –0eå\\ \⌅⌧
ÌD ¡På\\ \⌅X‹$.
0 = ⇢ sin cos ✓,
a) r2  z  5 r2
sol) x = r cos ✓, y = r sin ✓, z = zÑD t©Xt
x2 + y 2  z  5
x2
y2
=) ⇢2 = 5,
t¿\ t| ⇢
D ªD ⇠ à‰.
⇢=
81
0, 0 
p
5,
2 = ⇢ sin sin ✓, 1 = ⇢ cos
1
cos = p , sin ✓ = 1
5
 ⇡, 0  ✓ < 2⇡x î⌅–⌧ ò¿¥t
= cos
1
✓
1
p
5
◆
,
✓=
⇡
2
SOLUTION
⌘
2
2
2
, ⇡2 \ ò¿º ⇠ a) x + y + z = 9
sol) x = ⇢ sin cos ✓, y = ⇢ sin sin ✓, z = ⇢ cos ÑD t©XÏ
à‰. ò8¿ 8 ⇣‰ƒ
⇢ 0, 0   ⇡, 0  ✓ < 2⇡x î⌅–⌧ ò¿¥t
✓
✓ ◆◆ ✓
✓
◆ ◆
p ⇡
p
1
1
⇡
5, , 2⇡ tan 1
,
3, cos 1 p
,
,
x2 + y 2 + z 2 = 9 () ⇢2 = 9
2
2
4
3
✓
✓
◆◆
=) ⇢ = 3,
p ⇡
1
1
p
2 2, , tan
3
2
t¿\ (0, 2, 1)@ ltå\
⇣p
5, cos 1
⇣
7 å\ƒ
p1
5
⌘
=) ⇢ = 3,
¸ ⇡t ltå\\ ò¿º ⇠ à‰.
0
 ⇡,
0  ✓ < 2⇡
t‰.
2. ‰L¸ ⇡t ltå\\ \⌅⌧ ·tD ¡På\\ ò¿¥‹$.
b) x + y + z = 1
sol) ¸¥ƒ ) ›D ltå\\ ò¿¥t
a) ⇢ = 2
sol) x = ⇢ sin cos ✓, y = ⇢ sin sin ✓, z = ⇢ cos ÑD t©t⌧
⇢(cos + sin (cos ✓ + sin ✓)) = 1,
¡På\\ ò¿¥t
(0   ⇡, 0  ✓ < 2⇡)
p
⇢ = 2 ()
x2 + y 2 + z 2 = 2
t‰.
() x2 + y 2 + z 2 = 4
t¿\ ¸¥ƒ ·tD ¡På\\ ò¿¥t x2 +y 2 +z 2 = 4t‰. c) x2 + y 2 + z 2 = x + y + z
sol) ¸¥ƒ ) ›D ltå\\ ò¿¥t
b) = ⇡/6
⇢2 = ⇢(cos + sin (cos ✓ + sin ✓)),
sol) ¸¥ƒ ·tD ¡På\\ ò¿¥t
(0   ⇡, 0  ✓ < 2⇡)
p
z = 3x2 + 3y 2
t‰.
t‰.
c) ⇢ cos = 2
sol) ¸¥ƒ ·tD ¡På\\ ò¿¥t
z=2
4. ltå\\ \⌅⌧ ÄÒ›D ¡På\\ ò¿¥‹$.
a) 1  ⇢  2
sol) x = ⇢ sin cos ✓, y = ⇢ sin sin ✓, z = ⇢ cos ÑD t©Xt
1  ⇢  2 () 1  x2 + y 2 + z 2  4
t‰.
t‰.
d) ⇢ sin = 1
sol) ¸¥ƒ ·tD ¡På\\ ò¿¥t
2
2
x +y =1
t‰.
e) cos = ⇢ sin2
sol) ¸¥ƒ ·tD ¡På\\ ò¿¥t
z = x2 + y 2
b) 0  ⇢  3, 0   ⇡/4
sol) ¸¥ƒ ÄÒ›D ¡På\\ ò¿¥t
p
p
x2 + y 2  z  9 x2
t‰.
c) 0  ⇢  1, 0   ⇡/2, 0  ✓  ⇡
sol) ¸¥ƒ ÄÒ›D ¡På\\ ò¿¥t
1  x  1,
t‰.
3. ¡På\\ \⌅⌧ ) ›D ltå\\ ò¿¥‹$.
y2
t‰.
82
y
0,
z
0,
x2 + y 2 + z 2  1
SOLUTION
8
°0@ â,
t‡
µ8⌧ 8.1 .
1. ‰L¸ ⇡t ¸¥ƒ P °0–
|
sol)
1
p (2, 3, 5),
38
1
p (3, 1, 4),
26
c) a = i
sol)
j,
kbk = k bk = ka
=) kbk
bk
1Ω\‰.
1
p (4, 1, 3),
26
b=i
1
p (1, 1, 0),
2
kak  ka
ak  ka
bk
bk + kak
kak
kbk  ka
bk
p
2 6.
2. a = (2, 3, 1)¸ b = (2, 1, 2)– t °0 b– …âx °0
c@ b– ⇠¡x °0 dX it a@ ⇡D L, °0 c@ °0 d|
lX‹$.
30.
sol) °0 c °0 b– …âX‡ °0 dî °0 b– ⇠¡t¿\
c = k(2, 1, 2)\ ¯ ⇠ à‡, (Ë k 6= 0) dî
k
1
p (1, 0, 1),
2
b=i+j
(10, 1, 11),
b
t¿\
b = (4, 1, 3)
b) a = 3i + j + 4k,
sol)
a
b
,
, a + 2b, ka
kak kbk
t
lX‹$.
a) a = (2, 3, 5),
8 °0@ â,
(5, 1, 2),
p
d = (x, y, z)
3k
1
p (1, 1, 3),
11
(3, 1, 6),
p
13.
(2x + y
2z = 0)
\ ¯ ⇠ à‰. c + d = at¿\ ‰Lt 1Ω\‰.
8
8
>
>
<2k + x = 2
<4k + 2x = 4
=) k + y = 3
k+y = 3
>
>
:
:
2k + z = 1
4k 2z = 2
2. °0 (5, 3)D P °0 (2, 1)¸ (1, 3)X |(∞i<\ \⌅
$xΩX ) ›D ®P TXt x, y, zî ‰ åp⇠¥ 9k = 1D
X‹$.
ªD ⇠ à‡, l\ k| ò8¿ ) ›– ÖXt x, y, zƒ l`
sol)
⇠ à‰. 0|⌧
12
11
(5, 3) =
(2, 1) + (1, 3)
7
7
1
1
c=
(2, 1, 2), d = (20, 26, 7)
9
9
3. °0 (3, 2, 7)D 8 °0 (1, 1, 1), ( 2, 0, 3)¸ (3, 1, 2)X t‰.
|(∞i<\ \⌅X‹$.
sol)
3. °0 a = (4, 4, 1)| ¥§ °0 c@ dX i<\ ò¿¥»‰.
13
41
33
°0 cî b = (1, 2, 2)@ …â\ °0t‡, dî b– ⇠¡x °0
(3, 2, 7) =
(1, 1, 1) + ( 2, 0, 3) + (3, 1, 2)
10
10
10
| L, 4c + d| lX‹$.
4. °0 (4, 6, 7)¸ ⇡@ )•tt⌧ l0 3x °0| lX‹ sol) µ8⌧ 2ภ⇡@ )ï<\ Ä ⇠ à‰. 0|⌧ ı@
$.
2
1
(4, 6, 7)
(4, 6, 7)
c=
(1, 2, 2), d = (14, 8, 1)
sol) °0
= p
@ °0 (4, 6, 7) ¸ )•t
3
3
|(4, 6, 7)|
101
=)
4c
+
d
=
(2,
8,
5)
⇡<t⌧ l0 1x °0t‰. 0|⌧ °0 (4, 6, 7)¸ ⇡@
3
t‰.
)•tt⌧ l0 3x °0î p
(4, 6, 7)t‰.
101
µ8⌧ 8.2 .
1. ÑXX °0 a@ b–
4.
XÏ
kak
kbk  ka
bk
b + bk  ka
bk + kbk
proof ) °0 cX Ñ®î ëX ‰⇠t¿\ °0X l0–à •D
¸¥ ƒ@î  t ∆‰. 0|⌧, cX Ñê a + ↵bà ‡$tƒ
©ÑX‰. â, a@ cX ¨á ¸ a@ d = a+↵bX ¨á @ ⇡‰.
t⌧ a@ bX ¨á D 0  ✓  ⇡\ ìê. ƒ∞– Xt
ÑD Ùt‹$.
proof ) º ÄÒ›– Xt
kak = ka
=) kak
kbk  ka
DÃ P °0 a@ b– XÏ ↵ = kak, = kbk| X
a + ↵b
ê. °0 c =
î a@ bX ¨á D tÒÑhD Ùt‹$.
↵+
°0
a · d = a · ( a + ↵b)
= ↵2 + ↵2 cos ✓,
bk
83
SOLUTION
p
(d · d) = ( a + ↵b) · ( a + ↵b)
p
p
= 2↵ 1 + cos ✓
kdk =
p
t‡, °0 a@ °0 dX ¨á D 0 
cos
! !
!
!
!
AC · BD = (BC BA) · BD
! !
! !
= BC · BD BA · BD = 0
2
a·d
↵ + ↵ cos ✓
p
= p
kakkdk
↵ 2↵ 1 + cos ✓
r
1 + cos ✓
✓
=
= cos
2
2
=
t ⇠¥ »Ñ®X
°0 DÃ P °0 a, bX l0 ⇡‡ k2a + bk = ka
1Ω\‰. t L P °0 a, bX ¨á ✓| lX‹$.
3bk
|\ ¡¨
@
t‡ ⌅ ›D ƒ∞Xt
=) ((2a + b) · (2a + b)) = ((a
2
3b) · (a
3b))
a 2 = b2
2
=) 4kak + 4kakkbk cos ✓ + kbk
= kak2
‰t ¡PXî
proof ) -”⇣t A(0, 0), B(a, 0), C(a, b), D(0, b)x (Ë, a, b > 0)
⌘¿X 8t a t‡ ít bx ¡¨
ABCD – t⌧Ã ›
tƒ ©ÑX‰. ¡¨
ABCDX
‰t ¡PX0 ⌅\
Dî©Ñpt@
! !
AC · DB = 0
3bk
sol) İРXt
k2a + bk = ka
‰@ ¡P\‰.
)
⇠t‰. 0|⌧ °0 8. (¡PXî
¨
ÑD Ùt‹$.
t¿\ c@ aX ¨á @ a@ bX ¨á X
cî a@ bX ¨á D tÒÑ\‰.
5.
proof ) »Ñ®X -”⇣D A, B, C, D\ P‡ P
D
! !
!
!
!
AC, BD\ Pê.
BD
°0 BA@ BCX ¨á D t
ÒÑhD t©Xt
 ⇡\ ì<t,
2
8 °0@ â,
t¿\ a = bt¥| \‰. 0|⌧ ¡¨
t‰.
6kakkbk cos ✓ + 9kbk2
ABCDî
¨
=) 10 cos ✓ = 5
t‰. 0  ✓  ⇡ ⌘ 10 cos ✓ = 5| ÃqXî ✓X ✓@
9. ⇣ A(1, 2, 3)–⌧ ⇣ B(2, 1, 1))•<\ ‹ [t xy…t– ⇠¨
⇠¥ ò‰. ⇠¨⌧ [X °0| lX‹$.
⇡
t‰.
3
6. (– ⌅X ¡P1) AB ⌘Ït Ox –X ¿Ñt| Xê. t – sol) A–⌧ B )•<\ ‹ [X °0î v = B A = (1, 1, 2)
!
!
⌅X ⇣ C C 6= At‡ C 6= B| ÃqXt P °0 CA @ CB t‰. t⌧ °0 v xy…t– ⇠¨⇠»<¿\ z1ÑX Ä8
¡PhD Ùt‹$.
⇠
⌧‰. 0|⌧ ⇠¨⌧ [X °0î (1, 1, 2)t‰.
!
!
proof ) °0 CA @ CB|
!
!
CA = OA
!
!
CB = OB
!
\ P‡, OB =
10. ¸¥ƒ °0– t x
¨á D lX‹$.
!
OC,
!
OC
a ⇥ b| lX‡ x D t©XÏ ¯
a) a = (2, 1, 5), b = (3, 1, 2)
! !
!
!
OA, kOAk = kOBk = kOCkÑD t©Xt
sol)
Xt
! !
!
!
!
!
CA · CB = (OA OC) · (OB OC)
! !
! !
! !
! !
= OB · OB OA · OC OB · OC + OC · OC
!
!
! !
! !
!
= kOAkkOBk OA · OC + OA · OC + kOCk2
X– 0| ƒ∞Xt a ⇥ b = (7, 11, 5)t‡, ¨á D ✓|
p
ka ⇥ bk
195
sin ✓ =
=p p
kakkbk
30 14
r !
13
=) ✓ = arcsin
28
=0
!
!
t¿\ CA @ CBî ¡P\‰.
t‰.
7. (»Ñ®X
) »Ñ®($ ¿X 8t
X
‰@ ¡PhD Ùt‹$.
®P ⇡@ ¨
)
b) a = ( 7, 3, 2), b = (0, 1, 2)
84
SOLUTION
X– 0| ƒ∞Xt a ⇥ b = (8, 14, 7)t‡, ¨á D ✓| t‰.
sol)
Xt
p
ka ⇥ bk
309
sin ✓ =
=p p
kakkbk
62 5
!
r
309
=) ✓ = arcsin
310
12. 8 °0 a, b, c
⌧\ ¡PXt a ⇥ (b ⇥ c) = 0ÑD Ùt‹$.
sol) Ë⌧ ƒ∞– Xt
a ⇥ (b ⇥ c) = b(a · c)
c(a · b)
1Ω\‰. a, b, c ⌧\ ¡PX¿\, a · c = 0, a · b = 0t‰.
0|⌧ a ⇥ (b ⇥ c) = 0t‰.
t‰.
c) a = 3i + 5j
2k, b =
2i
3k
sol) X– 0| ƒ∞Xt a ⇥ b = ( 15, 13, 10)t‡, ¨á D ✓
| Xt
p
ka ⇥ bk
494
sin ✓ =
= p p =1
kakkbk
38 13
⇡
=) ✓ = arcsin (1) =
2
t‰.
11.
8 °0@ â,
13. °0 a, b, c–
Ùt‹$.
sol) °0 a, b, c| ‰L¸ ⇡t ì<t
a = (a1 , a2 , a3 )
b = (b1 , b2 , b3 )
c = (c1 , c2 , c3 )
x X
a 3 b2 , a 1 b3 + a 3 b1 , a 1 b2
= c 1 a 2 b3
X–
a 2 b1 )
a3 b1 )2 + (a1 b2
a 2 b1 ) 2
= a 1 b2 c 3
(c ⇥ a) · b = (c2 a3
= b1 c 2 a 3
a 3 b2 c 1
c3 a2 , (c1 a3
c3 a1 ), c1 a2
(a · b)2 = (a21 + a22 + a23 )(b21 + b22 + b23 )
2
= a22 b23 + a23 b22 + a21 b21 + a21 b22 + a22 b21
b1 c 3 a 2 + b2 c 3 a 1
b3 c 2 a 1
2a1 a2 b1 b2
14. 8 ⇣ A(2, 1, 0), B(3, 1, 1), C(1, 1, 1)D -”⇣<\
º
X ◆t| lX‹$.
sol) 8⌧$X. ⇣ A(2, 1, 0)D 0 <\ º
!
!
î °0| AB = x, AC = y\ Pt
t‰. ¯Ï¿\
x = (1, 0, 1)
2
2
|a ⇥ b| = |a| |b|
2
(a · b)
c2 a1 )
t¿\ (a ⇥ b) · c = (b ⇥ c) · a = (c ⇥ a) · b t‰.
t¿\
2a1 a3 b1 b3
a 1 b3 c 2 + a 2 b3 c 1
b 2 c 1 a 3 + b3 c 1 a 2
a · b = a 1 b1 + a 2 b2 + a 3 b3
2a2 a3 b2 b3
b2 c 1 )
· (b1 , b2 , b3 )
|a| = a21 + a22 + a23 , |b| = b21 + b22 + b23
(a1 b1 + a2 b2 + a3 b3 )
b3 c1 ), b1 c2
t‡,
t‰.
|a|2 |b|2
b3 c2 , (b1 c3
a 2 b 1 c 3 + a 3 b1 c 2
2a1 a2 b1 b2
c 3 a 2 b1
· (a1 , a2 , a3 )
= a22 b23 + a23 b22 + a21 b21 + a21 b22 + a22 b21
2a1 a3 b1 b3
a 2 b1 )
t‡,
(b ⇥ c) · a = (b2 c3
2a2 a3 b2 b3
a3 b1 ), a1 b2
c 1 a 3 b2 + c 2 a 3 b1
c 2 a 1 b3 + c 3 a 1 b2
t‡ 0|⌧
a3 b2 )2 + (a1 b3
a3 b2 , (a1 b3
· (c1 , c2 , c3 )
proof ) a = (a1 , a2 , a3 ), b = (b1 , b2 , b3 )| Xê. x X
Xt
|a ⇥ b| = (a2 b3
X– Xt
(a ⇥ b) · c = (a2 b3
¨ 8.2.8 X (v)| ùÖX‹$.
a ⇥ b = (a2 b3
XÏ, (a ⇥ b) · c = (b ⇥ c) · a = (c ⇥ a) · bÑD
2
y = ( 1, 0, 1)
85
¿î
X P ¿D ò¿¥
SOLUTION
8 °0@ â,
t¿\ x = yt‰. 0|⌧ 8 ⇣@ ⇡@ ¡ ¡– à<¿\ t‡ d = (p, q, r) 2 R3 \ ì<t
º
t
⇠ ∆‰.
( 1, 3, 7) = a = c + d = (k, 2k, 3k) + (p, q, r)
8
>
<k + p = 1,
15. $ ⇣ A(1, 2, 3), B(2, 0, 1), C( 2, 5, 1), D(2, 1, 4)| -”
=)
2k + q = 3,
>
⇣<\ ¿î ¨t¥X Ä<| lX‹$.
:
3k + r = 7
sol) ¨t¥X ÑXX 8 ¿D ò¿¥î °0 a, b, c–
<î
1
|(a ⇥ b) · c|
6
t ¯ Ä t‰. ⇣\ °0 c@ d ⇠¡t¿\ °0 b@ cƒ ⇠¡t‡
(1, 2, 3) · (p, q, r) = p 2q + 3r = 0
<\ ¸¥ƒ‰. (¨t¥X Ä<î …â!t¥X 16 0 t‰) a =
!
!
!
AB, b = AC, c = AD\ Pt
a = (1, 2, 2)
t 1Ω\‰. t⌧ òL ›D ˘à p Xt
8
8
>
>
<k + p = 1,
<k + p = 1,
() 4k 2q = 6,
2k + q = 3,
>
>
:
:
3k + r = 7
9k + 3r = 21
=) 14k =
b = ( 3, 7, 2)
2
| ªD ⇠ à‡, p = 1, q = 1, r = 1 ⇠¥ d = (1, 1, 1)t
‰. ⇣, °0 a@ d| P ¿<\ Xî …â¨¿ X ◆tî ka ⇥ dk
\ ¸¥¿‡, t| ƒ∞Xt
c = (1, 3, 7)
t‰. |(a ⇥ b) · c|| ƒ∞Xt
ka ⇥ dk = k( 1, 3, 7) ⇥ (1, 1, 1)k
p
p
= k( 10, 8, 2)k = 168 = 2 42
|(a ⇥ b) · c| = |((1, 2, 2) ⇥ ( 3, 7, 2)) · (1, 3, 7)|
= |(10, 8, 13) · (1, 3, 7)|
28 () k =
t‰.
= 57
t¿\ lX‡ê Xî Ä<î
57
19
=
t‰.
6
2
18. 8 ⇣ P (0, 2, 3), Q(2, 1, 5), R( 3, 2, 2)D -”⇣<\ X
! ! !
îº
X ◆t| lX‡, x⌘\ 8 ¿ OP , OQ, ORD ¿î
…â!t¥X Ä<| lX‹$.
16. 8 ⇣ P (1, 1, 2), Q(0, 2, 1), R(3, 1, 4)@ –⇣ O– XÏ
! ! !
8 °0 OP , OQ, OR\ É¥¿î …â!t¥X Ä<| lX‹ sol) P °0 a, b|
$.
! ! !
sol) OP , OQ, OR\ É¥¿î …â!t¥X Ä<î
!
!
!
(OP ⇥ OQ) · OR
\ ¸¥ƒ‰. ƒ∞– Xt Ä<
!
!
!
(OP ⇥ OQ) · OR = |((1, 1, 2) ⇥ (0, 2, 1)) · (3, 1, 4)|
= |( 5, 1, 2) · (3, 1, 4)| = 6
ÑD L ⇠ à‰.
\ ì<t lX‡ê Xî º
= |(7, 6, 4) · ( 3, 2, 2)| = 17
19. $ ⇣ P (1, 1, 2)Q(4, 1, 1), R( 1, 4, 3), S(3, 0, 7)– t 8
! ! !
°0 P Q, P R, P S– Xt É¥¿î …â!t¥X Ä<| lX
‹$.
b– …âX¿\
c = k(1, 2, 3),
X ◆tî
1
1
ka ⇥ bk = k(2, 3, 8) ⇥ ( 3, 4, 5)k
2
2
1
= k(17, 34, 17)k
2 p
17 6
=
2
t‰. ⇣, lX‡ê Xî …â!t¥X Ä<î
!
!
!
(OP ⇥ OQ) · OR = |((0, 2, 3) ⇥ (2, 1, 5)) · ( 3, 2, 2)|
17. °0 a = ( 1, 3, 7) b = (1, 2, 3)– …â\ °0 c@ °0
c– ⇠¡x °0 dX i<\ \⌅⌧‰‡ Xê. t L °0 a@ d t‰.
| P ¿<\ Xî …â¨¿ X ◆t| lX‹$.
sol) °0 c
!
a = P Q = (2, 3, 8)
!
b = P R = ( 3, 4, 5)
k 6= 0
86
SOLUTION
sol) İРXt
8 °0@ â,
sol) lX‡ê Xî ¡ X ‰⌧¿⇠) ›@
!
P Q = (3, 2, 1)
!
P R = ( 2, 5, 1)
!
P S = (2, 1, 5)
x=
t‡
2 + 5t,
y = 5t,
z=3
5t,
t2R
m) ›@
x+2
y
z 3
= =
1
1
1
t¿\ lX‡ê Xî …â!t¥X Ä<î
!
!
!
(P Q ⇥ P R) · P S = |((3, 2, 1) ⇥ ( 2, 5, 1)) · (2, 1, 5)|
= |(7, 1, 19) · (2, 1, 5)| = 108
t‰.
<\ ò¿º ⇠ à‰.
c) P (1, 3, 4)| ¿ò‡, v = (1, 2, 3) – …â\ ¡
sol) lX‡ê Xî ¡ X ‰⌧¿⇠) ›@
20. °0 a = (0, 2, 3), b = (2, 1, 5), c = ( 3, 2, 2)– Xt ∞
⇠î …â!t¥X Ä<| lX‹$.
x = 1 + t,
t‡
y=3
x
1
1
= |(13, 6, 4) · ( 3, 2, 2)| = 59
t‰.
z=
4
y
3
z+4
=
2
3
3t,
t2R
m) ›@
sol) lX‡ê Xî …â!t¥X Ä<î
|(a ⇥ b) · c| = |((0, 2, 3) ⇥ (2, 1, 5)) · ( 3, 2, 2)|
2t,
=
<\ ò¿º ⇠ à‰.
d) ⇣ P (1, 1, 2)| ¿ò‡ P …t 2x
3x + y + 4z = 7–
…âx ¡
3y + z = 10 ¸
21. °0 a = (2, 1, 0), b = (2, 1, 1), c = (1, 0, 2)| L, 8 °0 sol) P …t 2x 3y + z = 10 ¸ 3x + y + 4z = 7–
…âx
a, b, c\ É¥¿î …â!t¥X Ä<| lX‹$.
°0 lX‡ê Xî ¡ X )•°0 ⇠¿\ t )•°0|
u\ ì<t
sol) lX‡ê Xî …â!t¥X Ä<î
u = (2, 3, 1) ⇥ ( 3, 1, 4) = ( 13, 11, 7)
|(a ⇥ b) · c| = |((2, 1, 0) ⇥ (2, 1, 1)) · (1, 0, 2)|
t‰. 0|⌧, ¡ X ‰⌧¿⇠) ›@
= |(1, 2, 4) · (1, 0, 2)| = 7
x = 1 13t, y = 1 11t, z = 2 7t, t 2 R
t‰.
t‡ m) ›@
µ8⌧ 8.3 .
1. ‰L ¡ X ‰⌧¿⇠) ›¸
e) P (1, 3, 4)| ¿ò‡ …t 2x
sol) lX‡ê Xî ¡ X ‰⌧¿⇠) ›@
t‡
y=2
t,
z=
t,
y
2
z
=
1
1
m) ›@
x = 1,
1
y 1
z 2
=
=
13
11
7
<\ ò¿º ⇠ à‰.
a) P (1, 2, 0)¸ Q(1, 1, 1)D ¿òî ¡
x = 1,
x
m) ›D lX‹$.
t2R
sol) lX‡ê Xî ¡ t )•°0î ¸¥ƒ …tX ï °0@
⇡<¿\ ¡ X ‰⌧¿⇠) ›@
x = 1 + 2t,
t‡
y=
1
2
<\ ò¿º ⇠ à‰.
87
3
t,
z = 4 + 3t,
m) ›@
x
<\ ò¿º ⇠ à‰.
b) P ( 2, 0, 3)¸ Q(3, 5, 2)D ¿òî ¡
y + 3z = 5– ⇠¡x ¡
=
y+3
z 4
=
1
3
t2R
SOLUTION
2.
ptD ÃqXî …tX ) ›D lX‹$.
8 °0@ â,
t¿\ lX‡ê Xî …tX ) ›
8(x
!
a) ⇣ A(1, 2, 1)D ¿ò‡ °0 OA– ⇠¡x …t
1(x
1)
1) + 2(y
1) + 4(z
2) = 0
6y + 5z = 18
4. P …t x + y = z@ 2x z = 10– ⇠¡t‡, ⇣ P (3, 3, 2)|
¿òî …tX ) ›D lX‹$.
b) ⇣ P (1, 1, 1)| ¿ò‡ °0 n = (1, 2, 4)– ⇠¡x …t
1(x
6y + 5(z
2) = 0
2y + z = 6
t‰.
sol) lX‡ê Xî …tX ï °0
) ›@
() 8x
1)
5(z
D ªD ⇠ à‰.
1) = 0 () x
2(y + 2) + 1(z
() 8(x
!
OA = (1, 2, 1)t¿\
sol) lX‡ê Xî …tX ï °0
…tX ) ›@
1) + 6(y)
n = (1, 2, 4)t¿\ …tX
sol) lX‡ê Xî …tX ï °0î …t x + y = z@ 2x
10– Ÿ‹– ⇠¡t¿\ ï °0î
z=
(1, 1, 1) ⇥ (2, 0, 1) = ( 1, 1, 2)
1) = 0 () x + 2y + 4z = 7
t‰. 0|⌧, lX‡ê Xî …tX ) ›@
t‰.
1(x
3)
1(y
3)
2(z
2) = 0 () x + y + 2z = 10
c) –⇣, ⇣ P (1, 1, 2) @ Q(1, 2, 3)D ¿òî …t
t‰.
sol) lX‡ê Xî …tX ï °0 nD
!
!
n = OP ⇥ OQ
5. …t x + 2y + 3z = 3 ⌅X ⇣ P (2, 1, 1)| ¿ò‡ t …t–
⇠¡x ¡ D lt| ` L, ⇣ Q(3, 1, 3)¸ ¡ l¨tX p¨|
lX‹$.
\ ì<t ƒ∞– Xt n = (7, 1, 3)t¿\ lX‡ê Xî …
tX ) ›@
sol) ¡
7x y 3z = 0
lX ‰⌧¿⇠) ›@ ƒ∞– Xt
t‰.
x = 2 + t,
d) –⇣D ¿ò‡ …t 2x + 4y
1 + 2t,
z = 1 + 3t, t 2 R
\ ¸¥ƒ‰. t⌧ ⇣ Q–⌧ ¡ l– \Ëp¨\ ¯@ ¡ ¸
!
lX P⇣D R = (2 + k, 1 + 2k, 1 + 3k)\ Pt °0 RP =
!
z = 10X ( k, 2k, 3k)@ RQ = (1 k, 2 2k, 2 3k)î ⇠¡t¿\
z = 10– …âx …t
sol) lX‡ê Xî …tX ï °0î …t 2x + 4y
ï °0@ ⇡<¿\ ¯ ) ›@
2x + 4y
y=
! !
RP · RQ = 0 =) ( k, 2k, 3k) · (1
z=0
k, 2
2k, 2
3k) = 0
2
=) 14k = 11k
t‰.
t‰. k = 0tt ⇣ Q(3, 1, 3)t ⇡@ …t x + 2y + 3z = 3⌅–
11
t‡
3. P …t x
y + z = 5@ 3x
y = 4X P – … à¥| X¿Ã tî ®⌧t‰. 0|⌧, •\ ✓@ k =
14
âX‡ P ⇣ P (1, 0, 2), Q(3, 1, 0)| ¿òî …tX ) ›D lX‡ê Xî p¨î
ax + by + cz = d(a > 0)X ‹\ ıX‹$.
✓
◆
!
3 6
5
kRQk =
, ,
14 14 14
sol) x y + z = 5@ 3x y = 4| ΩXt 2x z = 1D ªD
p
70
⇠ à‡ x = t, t 2 R\ ì<t ‰⌧¿⇠›t
=
14
x = t, y = 4 + 3t, z = 1 + 2t
t‰.
x P X ) ›D ªD ⇠ à‰. lX‡ê Xî …tX ï °0
!
î P X )•°0@ °0 P Q = (2, 1, 2)– Ÿ‹– ⇠¡t¿\
6. ⇣ P ( 5, 3, 5)–⌧ …t 4x + 8y z = 17– ttî p¨|
…tX ï °0î
lX‹$.
(1, 3, 2) ⇥ (2, 1, 2) = ( 8, 6, 5)
88
SOLUTION
sol) ÑXX ⇣ (p, q, r)¸ …t ax + by + cz + d = 0 ¨tX p¨î L, º
|ap + bq + cr + d|
p
a 2 + b2 + c 2
AP QX ◆t| lX‹$.
sol) …t 3x 2y 2z = 19X ï °0î (3, 2, 2) t¿\ ⇣
P | ¿ò‡ …t 3x 2y 2z = 19– ⇠¡x ¡ X ‰⌧¿⇠)
›@
\ ¸¥ƒ‰. 0|⌧, lX‡ê Xî p¨î
|4( 5) + 8(3) 1(5) 17|
18
p
=
=2
9
42 + 82 + ( 1)2
t‰.
8 °0@ â,
x = 2 + 3t,
t‰. t| 3x
y=
1 + 2t,
2y
2z = 19–
6 + 9t
2 + 4t
z=1
2t,
t2R
ÖtÙt
2 + 4t =) t = 1
t‡ 0|⌧ A(5, 1, 1)t‰. P (2, 1, 1), A(5, 1, 1), Q(3, 1, 2)
!
!
7. ‰⌧¿⇠\ \⌅⌧ ¡ x = 1 + t, y = 1 t, z = 2tD Lt| t¿\ P A = (3, 2, 2), P Q = (1, 0, 3) t‰. º
AP QX
‡ Xê. ⇣ P (0, 1, 2)@ LD ®P ÏhXî …tX ) ›¸ ⇣ ◆tî
P (0, 1, 2)| ¿ò‡ L¸ ⇠¡<\ Ãòî ¡ X ) ›D
p
!
1 !
89
lX‹$,
|P A ⇥ P Q| =
2
2
sol) ¡ L ⌅X \ ⇣ Q(1, 1, 0)D ‡tê. ⇣P (0, 1, 2) L⌅–
!
t‰.
à¿ J<¿\ °0 P Q = (1, 0, 2)@ LX )•°0 (1, 1, 2)
î …ât D»‡, ⇣ P (0, 1, 2)@ LD ®P ÏhXî …tX ï
°0 ⌅ P °0– ⇠¡ÑD t©Xt ï °0î
10. …â\ P …t ax + by + cz d1 = 0¸ ax + by + cz d2 = 0
¨tX ⇠¡p¨ dî
(1, 0, 2) ⇥ (1, 1, 2) = ( 2, 4, 1)
|d1 d2 |
d= p
\ ¸¥¿‡ 0|⌧ lX‡ê Xî …tX ) ›@
a 2 + b2 + c 2
2(x
0)
4(y
1)
2) = 0 () 2x + 4y + z = 6
1(z
t‰. t⌧ lX‡ê Xî ¡ X )•°0| u\ P‡, ⇣ P –⌧
¡ L– ¯@ ⇠ X ⌧D R = (1 + k, 1 k, 2k)t| Xt °
!
0 P R = (1 + k, k, 2k 2)¸ LX )•°0 (1, 1, 2)
⌧\
⇠¡t¿\
ÑD Ùt‹$.
proof ) ‰⇠ a, b, c⌘ ¥ƒ Xòî 0t D»‰. |⇠1D É¿ J
‡ a 6= 0t| Xt, ⇣ P (d1 /a, 0, 0)@ …t ax + by + cz d1 = 0
⌅X ⇣t‡, P –⌧ …t ax + by + cz d2 = 0¨tX p¨î P
…t⌅X ⇠¡p¨ d¸ ⇡<¿\
!
d1
a + 0 · b + 0 · c d2
P R · (1, 1, 2) = 0 () (1 + k, k, 2k 2) · (1, 1, 2) = 0
d= a p
1
a 2 + b2 + c 2
=) k =
2
|d1 d2 |
=p
!
a 2 + b2 + c 2
t¥⌧ P R = 32 , 12 , 1 t¿\ u = (3, 1, 2)\ ì<t l
t¥⌧ ùÖt ]ú‰.
X‡ê Xî ¡ X ) ›@
x = 3t,
y=1
t,
z=2
2t, t 2 R
µ8⌧ 8.4 .
1. ‰L¸ ⇡t ¸¥ƒ P â,– t A+B, A B, cA+B, AT +
t‰.
B T , (A + B)T |
lX‹$.
(Remark) : AT + B T = (A + B)T t¿\ Ò8 |ΩX ∞¸Ã
8. ⇣ A(1, 2, 3)–⌧ ⇣ B(2, 1, 1) )•<\ ‹ [t xy …t– ⇠ Ö‹\‰.(8⌧ 6à 8‡)
¨⇠¥ ò‰. ⇠¨⌧ [X )•°0| lX‹$. (Ë, Ö¨ ¸
✓
◆
✓
◆
⇠¨ @ ⇡‰.)
6 3
1 2
a) A =
,B =
,c = 2
1
4
0
3
sol) ⇣ A@ ⇣ BD xy…t– t⌧ mt‡ ‹® ⇣D
A0 , B 0 |‡ Xê. ¯Ït A0 = (1, 2, 3), B 0 = (2, 1, 1)t‰. sol)
!
A0 B 0 = (1, 1, 2) t¿\ ⇠¨⌧ [X )•°0î (1, 1, 2) t‰.
9. ⇣ P (2, 1, 1)| ¿ò‡ …t 3x 2y 2z = 19– ⇠¡x …t
3x + 2y 2z = 19 ¸ Ãòî ⇣D A| Xê. ⇣ Q(3, 1, 2) |
89
◆
✓
◆
7 5
5 1
,A B =
,
1
7
1
1
✓
◆
✓
◆
13
8
7 1
cA + B =
, AT + B T =
.
2
11
5
7
A+B =
✓
SOLUTION
0
5
b) A = @0
3
1
0
0 3
1
1 2A , B = @ 7
2 4
5
sol)
0
4
A+B =@ 7
2
0
14
cA + B = @ 7
4
0
2
B0
c) A = B
@3
0
3
1
1
7
1
2
2
3
sol)
0
5
B5
B
A+B =@
1
3
0
13
B5
B
cA + B = @
11
3
0
4
7
2
7
2
3
2
1
3
2A , c = 3
1
sol)
1
0
1
6
6
2 0
4A , A B = @ 7 4 0A ,
5
8
0 3
1
0
1
2 12
4 7
2
0
8 A , AT + B T = @2
2
4A .
8 13
6 4
5
1
0
1
3
B5
1C
C,B = B
@ 4
4A
4
3
1
1
6
4
3
1
6
1
0
0
2
8
2
0
2
5
0
7
✓
3
10
0
3
=
✓
1
3
◆
5
@ B
0
2
1
BA, B T AT , (BA)T |
sol)
1
1
4C
C,
3A
5
=
lX‹$.
0
15
BA = @14
11
0
0
3
@5
2
1
✓
15
3
T T
A
25 , B A =
13
10
1
14 11
7
1A .
25 10
2
7
1
15
(BA)T = @ 2
15
4.
1
4
3A –
3
i) (tA)B = t(AB) = A(tB)
◆
14
,
15
(tA)B(i, j) =
=
n
X
t ‰LX Ò
(t 2 R)
proof ) ®‡ 1  i  m, 1  j  k–
XÏ
(tA)(i, l)B(l, j)
l=1
n
X
tA(i, l)B(l, j)
l=1
n
X
=t
A(i, l)B(l, j) = t(AB)(i, j),
l=1
1
37
4 A.
9
t(AB)(i, j) = t
=
=
n
X
(A)(i, l)B(l, j)
l=1
n
X
(A)(i, l)tB(l, j)
l=1
n
X
l=1
90
t
¨ 8.4.4| ùÖX‹$.
m ⇥ nâ, A, A1 , A2 @ n ⇥ k â, B, B1 , B2 –
›t 1Ω\‰.
(Remark: i) @ iii)Ã ùÖX0\ \‰.)
◆
6 3
1 2
,B =
1
4
0
3
✓
◆
✓
◆
6
21
6
1
T
sol) AB =
, (AB) =
.
1 14
21 14
0
1
0
1
5 0 3
1 2 3
3 2A
b) A = @0 1 2A , B = @ 7
3
2 4
5
2 1
0
1
0
20 4 18
20
3
7 4 A , (AB)T = @ 4
7
sol) AB = @ 3
37 4
9
18
4
0
1
0
1
2 3 1
1
3 1
2 2
B0 1 2
B
1C
8 3C
C,B = B 5 6
C
c) A = B
@3 1
A
@
2 4
4 1
2 1A
0 7 3
4
3 0 0 1
a) A =
1
20 21
30 15
B 6 8
12 4 C
C,
AB = B
@ 34 7
10 11A
11 45
62 20
0
1
20
6
34
11
B 21
8
7
45 C
C.
(AB)T B
@ 30
12
10
62A
15
4
11
20
3. â, A
1
2
3C
C,c = 5
1A
1
2. 1à–⌧ ¸¥ƒ â, A@ BX Ò AB@ ¯ÉX ⌅Xâ,
(AB)T | lX‹$.
◆
0
2
2
4
3
1
B
2C
C,A B = B 5
@7
5A
3
3
1
16
3
7
11
2
2C
C,
6
12 21 A
35 15
19
1
5
5
1 3
B4
7
2
7C
C.
AT + B T = B
@ 1
6
4 3A
3
2
5
3
✓
8 °0@ â,
(A)(i, l)(tB)(l, j) = A(tB)(i, j)
SOLUTION
t¿\ (tA)B = t(AB) = A(tB)
(t 2 R)
1Ω\‰.
iii) (Ñ0ïY) A(B1 + B2 ) = AB1 + AB2 t‡ (A1 + A2 )B =
A1 B + A2 B
proof )®‡ 1  i  m, 1  j  k–
A(B1 + B2 )(i, j) =
=
=
=
n
X
l=1
n
X
l=1
n
X
l=1
n
X
XÏ
A(i, l)(B1 + B2 )(l, j)
A(i, l)(B1 (l, j) + B2 (l, j))
(A(i, l)B1 (l, j) + A(i, l)B2 (l, j))
A(i, l)B1 (l, j) +
l=1
n
X
A(i, l)B2 (l, j)
t¿\ A(B1 + B2 ) = AB1 + AB2
(A1 + A2 )B(i, j) =
=
=
=
l=1
n
X
sol) â,Òt
0
1
c) @ 2
3
X⇠¿ Jî‰.
10
2
2
0A @2
2
3
3
1
4
1
3
1A
3
1
1
6
sol) â,Ò@ ò X⇠‡, ¯ Ò@
0
10
1 0
1
3 2
2 1
3
2
@ 2 1 0A @2
1 1 A=@ 2
3 4 2
3 6
3
8
16
3
5
\ ¸¥ƒ‰.
d)
✓
1
3
◆✓
5
3
0
4
2
1
sol) â,Òt
5
3
1
0
7A
19
◆
2
3
X⇠¿ Jî‰.
l=1
= AB1 (i, j) + AB2 (i, j)
n
X
8 °0@ â,
1ΩX‡, ⇣\
6. ¨ 8.4.8D ùÖX‹$.
proof )
(A + B)T (i, j) = (A + B)(j, i)
(A1 + A2 )(i, l)B(l, j)
= A(j, i) + B(j, i)
= AT (i, j) + B T (i, j)
= (AT + B T )(i, j)
(A1 (i, l) + A2 (i, l))B(l, j)
l=1
n
X
l=1
n
X
t¿\ (A + B)T = AT + B T t‰. ⇣\
(A1 (i, l)B(l, j) + A2 (i, l)B(l, j))
A1 (i, l)B(l, j) +
l=1
n
X
(AC)T (i, j) = (AC)(j, i)
n
X
=
A(j, l)C(l, i)
A2 (i, l)B(l, j)
l=1
l=1
= A1 B(i, j) + A2 B(i, j)
=
t¿\ (A1 + A2 )B = A1 B + A2 Bƒ 1Ω\‰.
n
X
C T (i, l)AT (l, j)
l=1
T
= (C AT )(i, j)
5. ‰L â,X Òt
lX‹$.
a)
✓
2 1
3 2
◆✓
1
6
X⇠î¿ ⇣ËX‡,
3
4
\ ¸¥ƒ‰.
b)
1
3
0
◆ 5
3 @
4
1
3
µ8⌧ 8.5 .
◆
2
3
sol) â,Ò@ ò X⇠‡, ¯ Ò@
✓
◆✓
◆ ✓
2 1
1
3 2
4
=
3 2
6 4 3
9
✓
X⇠î Ω∞– ÒD t¿\ (AC)T = C T AT t‰.
1
4
6A
3
1. ¸¥ƒ â,X â,›D lX‹$.
10
17
◆
1
0
a)
✓
5
1
◆
2
4
sol) ¸¥ƒ â,D A\ Pt det(A) = 5 · 4
b)
✓
cos ✓
sin ✓
sin ✓
cos ✓
( 1) · 2 = 22t‰.
◆
sol) ¸¥ƒ â,D A\ Pt det(A) = cos2 ✓ + sin2 ✓ = 1t‰.
91
SOLUTION
✓
cosh ✓
sinh ✓
sinh ✓
cosh ✓
8 °0@ â,
◆
sol) ¸¥ƒ â,D A\ Pt det(A) = 2 6= 0t¿\ Ìâ,t
t¨X‡, Ìâ,D A 1 \ Pt
✓
◆
1 1 7
sol) ¸¥ƒ â,D A\ Pt det(A) = cosh2 ✓ sinh2 ✓ = 1t‰.
1
A =
2 0 2
0
1
1 1 1
t‰.
d)@2 3 4 A
✓
◆
4 9 16
1 0
c)
9 3
sol) ¸¥ƒ â,D A\ Pt
c)
det(A) = 1(3 · 16
=2
4 · 9)
1(2 · 16
4 · 4) + 1(2 · 9
3 · 4)
t‰.
0
0 1
e) @1 1
2 3
t‰.
1
2
3A
0
d)
sol) ¸¥ƒ â,D A\ Pt
det(A) = 0(1 · 0
=8
3 · 3)
1(1 · 0
3 · 2) + 2(1 · 3
2 · 1)
t‰.
2
f) @ 0
0
23
p
2
0
1
14
3A
10
sol) ¸¥ƒ â,D A\ Pt
p p
det(A) = 2( 2 · 10
+ 14(0 · 0
= 20
3 · 0)
p
2 · 0)
23(0 · 10
3 · 0)
cos ✓
sin ✓
◆
✓
x+1
x
x+2
x+1
◆
13 6= 0t¿\ Ìâ,t t‰.
◆
2
1
3. 3(
t‰.
◆
7
1
sin ✓
cos ✓
sol) ¸¥ƒ â,D A\ Pt ®‡ x 2 R– tdet(A) = 1 6= 0
t¿\ Ìâ,t t¨X‡, Ìâ,D A 1 \ Pt
✓
◆
x+1
(x + 2)
1
A =
x
x+1
◆
1 2
5 3
2
0
✓
sol) ¸¥ƒ â,D A\ Pt det(A) = 1 6= 0t¿\ Ìâ,t
t¨X‡, Ìâ,D A 1 \ Pt
✓
◆
1
sin ✓
cos ✓
1
A =
=A
cos ✓
sin ✓
1
f)
sol) ¸¥ƒ â,D A\ Pt det(A) =
t¨X‡, Ìâ,D A 1 \ Pt
✓
1
3
1
A =
5
13
b)
◆
5
4
t‰.
2. ¸¥ƒ â,X Ìâ,D lX‹$.
✓
1
2
sol) ¸¥ƒ â,D A\ Pt det(A) = 6 6= 0t¿\ Ìâ,t
t¨X‡, Ìâ,D A 1 \ Pt
✓
◆
1
4
5
A 1=
2 1
6
e)
t‰.
a)
✓
t‰.
0p
✓
sol) ¸¥ƒ â,D A\ Pt det(A) = 3 6= 0t¿\ Ìâ,t
t¨X‡, Ìâ,D A 1 \ Pt
✓
◆
1 3 0
A 1=
9 1
3
–
92
¨ â,
0
2
A = @1
0
0
1
0
1
2
2A
1
t det(A xI3 ) = 0t ⇠ƒ] ‰⇠ xX ✓D ®P lX‹$.
SOLUTION
sol)
0
2 x
xI3 ) = det @ 1
0
det(A
= (2
t¿\ det(A
x)(1
a)
x+2
3
1
x
0
x)2
sol) ⌅ â,D A\ Pt A
det(A) 6= 0t‰. ⇣,
3
x+2
¿ƒ] Xî xX ptD lX‹$.
Ìâ,D
» Dî©Ñpt@
✓
◆
0
5. 3(
() 4x2
1, 1| L A
–
0
x+2
3
4 6= 0
Ìâ,D
» Dî©Ñpt@
1
x
0 A 6= 0
x 2
ƒ‰.
¨ â,
0
◆
sol) ⌅ â,D A\ Pt A
det(A) 6= 0t‰. ⇣,
Ìâ,D
1
det A 6= 0 () det @x
0
xI3 ) = 0D ÃqXî xî x = 1 ⇣î x = 2t‰. t¿\ x 6=
4. ¸¥ƒ â,t Ìâ,D
✓
1
2
2 A
1 x
0
8 °0@ â,
t det A =
1
1
1
1
A = @x 1 x 2 x 3 A
x21 x22 x23
Y
(xj xi )ÑD Ùt‹$.
1i<j3
x+2
3
3
x+2
sol) â,›X 1»– Xt
0
1
1
1
1
() (x + 5)(x 1) 6= 0
det A = det @x1 x2 x3 A
x21 x22 x23
t¿\ x 6= 5, 1| L A Ìâ,D ƒ‰.
0
1
1
1
1
✓
◆
= det @ 0 x2 x1 x3 x1 A
2x 1
1
b)
x21
x22
x23
2
x+1
0
1
1
1
1
sol) ⌅ â,D A\ Pt A Ìâ,D » Dî©Ñpt@
= det @0 x2 x1 x3 x1 A
det(A) 6= 0t‰. ⇣,
0 x22 x21 x23 x21
✓
◆
= (x2 x1 )(x23 x21 ) (x22 x21 )(x3 x1 )
2x 1
1
Y
det A 6= 0 () det
6= 0
2
x+1
= (x2 x1 )(x3 x1 )(x3 x2 ) =
(xj
1i<j3
() (2x + 3)(x 1) 6= 0
det A 6= 0 () det
3
, 1| L A
2
1
2 0
1 1A
2 1
t¿\ x 6=
0
x
c) @x
0
()
1
d) @x
0
0
x+2
3
1
x
0 A
x 2
ƒ‰.
6. ‰L Ω) ›D â,X
t|
lX‹$.
Ìâ,D
0
x
det A 6= 0 () det @x
0
0
xi )
1Ω\‰.
Ìâ,D
sol) ⌅ â,D A\ Pt A
det(A) 6= 0t‰. ⇣,
t¿\ x 6= 0| L A
6= 0
3x 6= 0
Ìâ,D
» Dî©Ñpt@
1
2 0
1 1A 6= 0
2 1
a)
(
‹ Ax = b\ ò¿¥‡, A 1 @
x + 3y = 4
2x + 5y = 3
sol) ⌅ Ω) ›D Ax = b4\ ò¿¥t
✓
◆✓ ◆ ✓ ◆
1 3
x
4
=
2 5
y
3
t‰. ⇣\, ƒ∞– Xt
A 1=
ƒ‰.
✓
t¿\, Ω) ›X tî
✓ ◆
✓
x
5
1
=A b=
y
2
93
5
2
◆
3
1
◆✓ ◆ ✓
◆
3
4
11
=
1
3
5
SOLUTION
8. @⇠ n– t n( ¨ â, A
det A = 0ÑD Ùt‹$.
\ ¸¥ƒ‰.
b)
(
8 °0@ â,
5x 4y = 2
3x + 2y = 1
proof ) ÑXX ëX @⇠ n–
A + AT = O| ÃqXt
t
det(A) = det( AT ) = ( 1)n det(A)
sol) ⌅ Ω) ›D Ax = b4\ ò¿¥t
✓
◆✓ ◆ ✓ ◆
5
4
x
2
=
3 2
y
1
=
det(A)
=) det(A) = 0
t¿\ ùÖt DÃ⌧‰.
t‰. ⇣\, ƒ∞– Xt
A
1
1
=
22
✓
t¿\, Ω) ›X tî
✓ ◆
✓
1
x
2
= A 1b =
y
3
22
2
3
◆
4
5
9. ÑXX n( ¨ â, A
ÑD Ùt‹$.
c)
¿t det A = 0
proof ) â, AX ⌘⇡@ P âD fi∏¥ ª@ â,D B|‡
Xt â,›X 1»– Xt
◆✓ ◆ ✓ ◆
0
4
2
=
1
5
1
2
det(A) =
det(B)
t‰. t L, A = Bt¿\
\ ¸¥ƒ‰.
(
⌘⇡@ P âD
det(A) =
3x + 7y = 17
4x y = 10
det(A) =) det(A) = 0
t 1Ω\‰.
10. xy…t–⌧ 8 ⇣ (x1 , y1 , 0), (x2 , y2 , 0), (x3 , y3 , 0)D -”⇣
<\ ¿î º
X ◆tî
0
1
1
1
1
1
det @x1 x2 x3 A
2
y1 y2 y3
sol) ⌅ Ω) ›D Ax = b4\ ò¿¥t
✓
◆✓ ◆ ✓ ◆
3 7
x
17
=
4
1
y
10
t‰. ⇣\, ƒ∞– Xt
A
1
=
1
31
✓
t¿\, Ω) ›X tî
✓ ◆
✓
1
x
1
1
=A b=
y
4
31
1
4
◆
7
3
ÑD Ùt‹$.
sol) º
◆tî
!
a, bx ¿ P⌧!
a , b @ |x
✓–
t¯
!
1
1
ab sin ✓ = k!
a ⇥ bk
2
2
\ ¸¥ƒ‰. ⇣ (x1 , y1 , 0)D 0 <\ º
X P ¿D ò¿¥î
°0| A, B\ Pt
◆ ✓ ◆ ✓ 87 ◆
7
17
= 31
38
3
10
31
\ ¸¥ƒ‰.
7. n( ¨ â, AX â,›t a| L, ê⇠ k–
Ak X â,›D lX‹$.
X 8t
t â,
A = (x2
x1 , y2
y1 , 0)
B = (x3
x1 , y3
y1 , 0)
t‡, A ⇥ B| ƒ∞Xt
A ⇥ B = (0, 0, (x2 x1 )(y3 y1 ) (x3 x1 )(y2 y1 ))
sol) ê⇠ k– t det Ak = ak ÑD ùÖXê. k = 1| Lî
t‰. 0|⌧,
êÖX‰. t⌧ k = n| L det An = an t|‡
Xt,
1
1
ka ⇥ bk = |(x2 x1 )(y3 y1 ) (x3 x1 )(y2 y1 )|
n+1
n
det(A
) = det(A) · det(A )
2
2
1
n
n+1
=a·a =a
= |(x2 y3 x3 y2 ) (x1 y3 x3 y1 ) + (x1 y2 x2 y1 )|
2
k
k
1
1
1
t¿\ ⇠Y ¿©ï– Xt ®‡ ê⇠ k– t det A = a
1
x
x
x
=
1
2
3
t‰.
2
y1 y2 y3
94
SOLUTION
t 1Ω\‰.
8 °0@ â,
Q(2, 2, 2)| L, º
AP QX ◆t| lX‹$.
!
!
!
11. OA = (2, 1), CB = (4, 4), OB = (3, 8)| L, 8 ⇣ sol) 8⌧X ptD ÃqXî ¡ D lt| Xt lX ‰⌧¿⇠›@
A, B, C| -”⇣<\ ¿î º
X ◆t| lX‹$,
x = 1 + 2t, y = 1 + 5t, z = 1 7t, t 2 R
!
sol) BA = (2, 1)
lX‡ê Xî º
¸ ⇡‰. t| 2x + 5y
7z = 68– ÖXt t = 1t¿\
!
(3, 8) = ( 1, 7), BC = ( 4, 4)\ Pt A(3, 4, 6)t‰. t⌧
X ◆tî
!
a = AP = (1, 1, 1) (3, 4, 6) = ( 2, 5, 7),
! !
1
!
det(BA; BC)
b = AQ = (2, 2, 2) (3, 4, 6) = ( 1, 2, 4)
2
\ Pt lX‡ê Xî º
\ ¸¥¿¿\ ¯ ✓@
! !
1
1
det(BA; BC) =
det
2
2
✓
◆
4
= 12
4
1
7
1
1
ka ⇥ bk = k( 2, 5, 7) ⇥ ( 1, 2, 4)k
2
2
1
= k( 6, 1, 1)k
2p
38
=
2
t‰.
12. ¸¥ƒ P ¡
z
2
x
1
= y+1 =
2
z
3
3
¸
x+1
y
=
=
4
2
X ◆tî
t‰.
X P⇣D P |‡ Xê. t L ⇣ P, Q(1, 1, 1), R(2, 3, 1)D
2
14. °0 a = (2, 2, 7) °0 b = (4, 4, 8)– …â\ °0
-”⇣<\ ¿î º
X ◆t| lX‹$.
c(c 6= 0)@ c– ⇠¡x °0 dX i<\ \‹⌧‰‡ Xê. t L,
°0 a@ d| P ¿<\ Xî …â¨¿ X ◆t| lX‹$.
x 1
z 3
sol) < P⇣ P | lXê. ¡
=y+1=
D ‰⌧
2
3
¿⇠›<\ ò¿¥t
sol) °0 c b– …âX¿\
x = 1 + 2t, y =
1 + t, z = 3 + 3t,
x+1
y
z 2
=
=
–
4
2
2
¿\ P (1, 1, 3)t‰. t⌧
t‰. t| ¡
t2R
c = k(4, 4, 8),
ÖXt t = 0D ª<
t‡ d = (p, q, r), p, q, r 2 R\ ì<t
(2, 2, 7) = a = c + d = (4k, 4k, 8k) + (p, q, r)
8
>
<4k + p = 2,
=)
4k + q = 2,
>
:
8k + r = 7
!
a = (a1 , a2 , a3 ) = P Q = (1, 1, 1) (1, 1, 3) = (0, 2, 2),
!
b = (b1 , b2 , b3 ) = P R = (2, 3, 1) (1, 1, 3) = (1, 4, 4)
\ Pt lX‡ê Xî º
1
ka ⇥ bk
2
✓
✓
1
a
=
det 2
b2
2
t‰. ⇣\ °0 c@ d
X ◆tî
k 6= 0
⇠¡t¿\ °0 b@ dƒ ⇠¡t‡
(4, 4, 8) · (p, q, r) = 0 =) p
◆
a3
,
b3
✓
a
det 1
b1
◆
✓
a3
a
, det 1
b3
b1
a2
b2
◆◆
\ ¸¥¿¿\ ƒ∞– Xt
1
k(0, 2, 2)k
2
p
= 2
Area =
| ªD ⇠ à‰.
q + 2r = 0
t 1Ω\‰. t⌧ òL ›D ˘à p Xt
8
8
>
>
<4k + p = 2,
<4k + p = 2,
() 4k q = 2,
4k + q = 2,
>
>
:
:
8k + r = 7
16k + 2r = 14
3
=) 24k = 18 () k =
4
| ªD ⇠ à‡, p = 1, q = 1, r = 1 ⇠¥ d = ( 1, 1, 1)t‰.
⇣, °0 a@ d| P ¿<\ Xî …â¨¿ X ◆tî ka ⇥ dk\
¸¥¿‡, t| ƒ∞Xt
ka ⇥ dk = k(2, 2, 7) ⇥ ( 1, 1, 1)k
p
= k( 9, 9, 0)k = 9 2
13. ⇣ P (1, 1, 1)| ¿ò‡ …t 2x + 5y 7z = 68– ⇠¡x
¡ t …t 2x + 5y 7z = 68¸ Ãòî ⇣D A| Xê. ⇣
95
SOLUTION
t‰.
9 ıå⇠
t‰. 0|⌧ x = 5t‰.
p
p
15. 8 °0 a = (1, 3,
2), b = (3, 2, 1), c = (0, 2, 6)\ É¥ 18. 8 °0 a = ( 1, 2, 3), b = (1, 1, 0), c = (6, 6, 1)\ É¥¿
î ¨t¥X Ä<| lX‹$.
¿î …â!t¥X Ä<| lX‹$.
sol) lX‡ê Xî …â!t¥X Ä<
|det(a; b; c)|
\ ¸¥¿¿\ ¯ ✓@
0
1
3
|det(a; b; c)| = det @ p
= 56
2
1
p3 0
2 2A
1 6
sol) lX‡ê Xî ¨t¥X Ä<î °0 a, b, c\ É¥¿î …
1
â!t¥X 0 t¿\ ¯ Ä<î
6
0
1
1 1 6
1
1
|det(a; b; c)| =
det @ 2 1 6A
6
6
3 0 1
1
=
2
t‰.
t‰.
19. ⇣ O| –⇣<\ Xî å\ı⌅–⌧ ¨t¥ OABC
à
‰.
º
ACD,
ABC,
BCD,
ABD
$
…t
z
=
16. $ ⇣ P (2, 0, 3), Q(4, 1, 1), R( 1, 4, 3), S(3, 0, 7)–⌧ 8
! ! !
0, 3x + 2y = 6, x = 0, x 2y + z = 2⌅– àD L, ¨t¥
°0 P Q, P R, P S– Xt É¥¿î …â!t¥X Ä<| lX
OABCX Ä<| lX‹$.
‹$.
sol) ¸¥ƒ
sol)
!
a = P Q = (4, 1, 1) (2, 0, 3) = (2, 1, 4)
!
b = P R = ( 1, 4, 3) (2, 0, 3) = ( 3, 4, 6)
!
c = P S = (3, 0, 7) (2, 0, 3) = (1, 0, 10)
<\ Pt lX‡ê Xî …â!t¥X Ä<î
0
1
2
3 1
0A
|det(a; b; c)| = det @1 4
4 6 10
Ù‰\Ä0 ›‰D ΩXÏ Ät
A(2, 0, 0), B(0, 3, 8), C(0, 3, 0), D(0, 1, 0)
ÑD L ⇠ à‰. 0|⌧ ¨t¥ OABCX Ä<î
0
1
2 0 0
! ! !
1
1
det(OA; OB; OC) =
det @0 3 3A
6
6
0 8 0
=8
t‰.
= 100
9
t‰.
ıå⇠
17. 8 °0 a = (1, 2, x), b = (x, x 1, 6), c = (5, x, 9)\ É µ8⌧ 9.1 .
¥¿î …â!t¥X Ä< 1 t ⇠ƒ] Xî ê⇠ xX ✓D
lX‹$.
1. ‰LD ƒ∞XÏ a + bi ‹\ ò¿¥‹$.
sol) °0 a, b, c\ É¥¿î …â!t¥X Ä<î
0
1
1
x
5
|det(a; b; c)| = det @ 2 x 1 xA
x
6
9
= x3
t‰. t Ä<
x3
5x2
10x + 51
1t ⇠0 ⌅t⌧î
5x2
10x + 51 = 1,
a) (3 + 4i) + (5
sol) 8
2i.
b) (2 + 3i)
sol) 6
x2N
() x = 5
96
6i)
2i.
( 4 + 5i)
SOLUTION
c) (2 + 3i)(4
5i)
sol) 23 + 2i.
d)
1 + 3i
2 + 4i
sol)
9 ıå⇠
b) (x + y) + 2i =
1 + yi
sol)
2 = y =) x =
x+y =
1,
3,
5. x = 1 + i| L, x2 + 2x + 2@ x3
lX‹$.
sol)
1
(7 + i).
10
a
1
(a
2
a + b2
bi,
x2 + 3x + 8X ✓D
1 + i =) (x + 1)2 =
x=
t¿\
2.‰L ıå⇠X gH, ÒH– \ Ì–D
lX‹$.
(Remark) : Ã} ıå⇠ a + bi à<t gH, ÒH– \ Ì–
(a2 + b2 6= 0)@
t‡
y = 2.
1
x2 + 2x + 2 = (x + 1)2 + 1 = 0
x3
bi)
x2 + 3x + 8 = (x + 1)3
=
i
4x2 + 7
4( 1 + i)2 + 7
= 7(1 + i)
\ ƒ∞⌧‰.
t‰.
a) 1 + 2i
sol)
1
2i,
p
2i
1
p
b) 1 +
sol)
1
5 (1
6. (3 + 2i)z = 1 + 2i| ÃqXî ıå⇠ z| lX‹$.
sol)
2i).
1 + 2i
3 + 2i
1
=) z =
(7 + 4i).
13
(3 + 2i)z = 1 + 2i =) z =
1
3 (1
2i,
p
2i).
c) 1 + ⇡i
sol)
1
⇡i,
1
(1
1 + ⇡2
7. ‰LD ƒ∞X‹$.
⇡i).
3. ‰L ıå⇠ zX $ ıå⇠ z̄| lX‹$.
a) z = 1
3i
a) |(2 + 3i)(5
sol) |19 + 9i| =
b)
sol) z̄ = 1 + 3i.
3i)|
p
442.
5 2i
2+i
p
b) z = 5 + 2i
1
sol) (8
5
sol) z̄ = 5
c) 1 + i + i2 + · · · + i20 .
2i.
4. ‰L Ò›t ÃqXî ‰⇠ x, yX ✓D lX‹$.
a) (x
2) + 3yi = 0
sol) x = 2, y = 0.
9i) =
145
.
5
sol) i4 = 1ÑD t©Xt
1+i+i2 +· · ·+i20 = 1+5(i+i2 +i3 +i4 ) = 1+5(i 1 i+1) = 1
t‰.
d) 1 +
97
1
1
1
+ 2 + · · · + 20
i
i
i
SOLUTION
9 ıå⇠
t‰.
sol)
1
1
1
+ 2 + · · · + 20
i
i
i
1
1
1
1
= 1 + 5( + 2 + 3 + 4 )
i
i
i
i
= 1 + 5( i 1 + i + 1) = 1.
1+
8. ıå⇠ z = 2 + i–
t
2. ‰L ›D ˘ ›D t©XÏ ⌅Ëà ò¿¥‹$.
a) i77
sol)
z z̄
X ✓D lX‹$.
z + z̄
b)
sol)
z z̄
(2 + i)(2 i)
5
5
=
=
= i.
z + z̄
2 i+2 i
2i
2
9. f (x) = x20 +
1
| L, f
x20
✓
✓
1+i
1 i
◆77
✓
1+i
1 i
sol)
◆
1+i
X ✓D lX‹$.
1 i
i77 = (i sin
⇡ 77
77⇡
) = i sin
= i.
2
2
◆77
p
.
t¿\
✓
◆
⇡
⇡
2 cos
4 + i sin
4
p 77
77⇡
77⇡
2 cos 4 + i sin 4
= p 77
2 cos
1 i
=
= i.
1+i
sol)
1+i
1+i 1
=
·
1 i
1 i 1
p
=
2 cos ⇡4 + i sin ⇡4
i
=i
i
77⇡
4
!77
77⇡
4
+ i sin
µ8⌧ 9.2 .
p
c) (1 + i)(1 + 3)
sol) ˘ ›<\ ò¿¥t
p
p ⇣
⇡
⇡⌘
( 2 + 6) cos + i sin
4
4
t‰.
1. ‰L ıå⇠| ˘ ›<\ ò¿¥‹$.
3. ‰L Ò›D ÃqXî zX —iD ıå…t– ¯¨‹$.
f
t‰.
a) 2020i
sol)
1+i
1 i
= f (i) = 2
⇣
2020(0 + i) = 2020 cos
⇡⌘
⇡
+ i sin
2
2
b) 2020 + 2020i
sol)
p
p !
✓
◆
p
p
2
2
3⇡
3⇡
2020 2
+i
= 2020 2 cos
+ i sin
2
2
4
4
a) Re(z + 2) = 0
sol) z = a + bi\ Pt
Re(z + 2) = 0 () a =
t¿\ —i { 2 + bi|b 2 R}D ıå…t⌅– ò¿¥t
10
Imz
5
c) (10 + 10i)3
sol)
p
10 + 10i = 10 2
Rez
p !
p ⇣
2
2
⇡
⇡⌘
+i
= 10 2 cos + i sin
2
2
4
4
p
10
5
5
5
t¿\
p ⇣
⇡
⇡ ⌘3
(10 + 10i)3 = 2000 2 cos + i sin
4
4 ◆
✓
p
3⇡
3⇡
= 2000 2 cos
+ i sin
4
4
2
10
¸ ⇡‰.
98
10
SOLUTION
10 ·
b) |z 3 + 2i| = 3
1. ‰L
· ⇣î § –
sol) ⌅ ›D ÃqXî zî ıå…t–⌧ ⌘Ït 3-2i t‡ ⇠¿Ñ lX‡ ¯ ⌧ D ¯¨‹$.
t 3x –X êËt¿\ t| ¯¨t
a) {(x, y) 2 R2 |y = 2ex }
sol) ‰⌧¿⇠ · C(t)î
5
C(t) = (t, 2et ),
Imz
QXî ‰⌧¿⇠ ·
0
( 1 < t < 1)
t‡ ⌧ @ ‰L¸ ⇡‰.
(3
2i)
5
5
0
Rez
5
¸ ⇡‰.
4. …tX ⇣ P (a, b) (P 6= O)| –⇣ O| ⌘Ï<\
✓Ã|
å⌅XÏ ª@ ⇣@ (a cos ✓ b sin ✓, a sin ✓ + b cos ✓)ÑD ıå⇠
X ˘ ›D t©XÏ ùÖX‹$.
(¯ò⌧
å⌅tŸ<\
‹\ \⌅\‰t b) {(x, y) 2 R2 |x2 + 4y 2 = 1}
✓
◆ ✓ ◆ ª@ ⇣D â,X
cos ✓
sin ✓
a
sol) ‰⌧¿⇠ · C(t)î
\ ò¿º ⇠ à‰)
sin ✓
cos ✓
b
✓
◆
1
C(t) = cos t, sin t ,
proof ) (a, b)| ıå…t– ò¿¥t a + bi\ tt` ⇠ à‰.
2
®‡ ıå⇠‰@ ˘ ›<\ \⌅` ⇠ à<¿\
a + bi = r(cos ↵ + i sin ↵)
1ΩXî r, ↵
|‡ Xt
(r > 0, ⇡ < ↵  ⇡)
(0  t < 2⇡)
t‡ ⌧ @ ‰L¸ ⇡‰.
t¨\‰. ✓Ã| å⌅tŸ\ ✓D c + di
c + di = r(cos(↵ + ✓) + i sin(↵ + ✓))
t 1ΩXp gHı›D µt
cos(↵ + ✓) = cos ↵ cos ✓
sin ↵ sin ✓
sin(↵ + ✓) = sin ↵ cos ✓ + cos ↵ sin ✓
ÑD L ⇠ à‰. ⌅ ›¸ a = r cos ↵, b = r sin ↵|
c + di = (a cos ✓
ÑD L ⇠ à‰.
10
ÖXt
b sin ✓) + i(a sin ✓ + b cos ✓)
c) R3 –⌧ (0, 0, 0)¸ ⇣ (1, 2, 3)D ¿òî ¡
sol) ‰⌧¿⇠ · C(t)î
·
C(t) = (t, 2t, 3t),
µ8⌧ 10.1 .
t‡ ⌧ @ ‰L¸ ⇡‰.
99
(t 2 R)
C(t)|
SOLUTION
4. ‰⌧¿⇠· C(t) = (t, t2 , t3 ), 1  t  1X ⌧ D ¯¨‹
$.
sol) ⌧ @ ‰L¸ ⇡‰.
d) {(x, y) 2 R2 |16x2 + 9y 2 = 4}
sol) ‰⌧¿⇠ · C(t)î
C(t) =
✓
10 ·
◆
1
2
cos t, sin t ,
2
3
(0  t < 2⇡)
t‡ ⌧ @ ‰L¸ ⇡‰.
5. ‰⌧¿⇠·
C(t) = (cos2 t, cos t sin t, sin 2t),
⌧ D ¯¨‹$.
sol) ⌧ @ ‰L¸ ⇡‰.
⇡
⇡
t X
2
2
2. ‰⌧¿⇠· C(t) = (cosh t, sinh t), t > 0X ⌧ D ¯¨‹$.
sol) ⌧ @ ‰L¸ ⇡‰.
µ8⌧ 10.2 .
1. C(0) = (0, 5, 1) t‡ C 0 (t) = (4t, et , 4t3 ) | L C(t) X ›D
lX‹$.
sol) C(t) = (x(t), y(t), z(t)) |‡ Xê. ¯Ït
x0 (t) = 4t,
3. ‰⌧¿⇠· C(t) = (t2 , t3
$.
sol) ⌧ @ ‰L¸ ⇡‰.
y 0 (t) = et ,
z 0 (t) = 4t3
9t), 0  t  3X ⌧ D ¯¨‹ t‡ 0|⌧
x(t) = 2t2 + C1 ,
100
y 0 (t) = et + C2 ,
z 0 (t) = t4 + C3
SOLUTION
t‰. (Ë, C1 , C2 , C3 î ¡⇠)
C(0) = (0, 5, 1) t¿\ C1 = 0, C2 =
\
C(t) = (2t2 , et
10 ·
t‰. t = 0 –⌧X ⌘ X ) ›@
6, C3 = 1 t‰. ¯Ï¿
C 0 (0) · t + C(0) = (1, 0, 0) · t + (0, 0, 0)
= (t, 0, 0)
6, t4 + 1)
t‰.
t‰.
2. ‰LX
Ω∞– t çƒ C 0 (t) @ C 0 (0) D lX‹$.
a) C(t) = (sin ⇡t, cos ⇡t, t t2 )
sol)
b) C(t) = (sin ⇡t, cos ⇡t, 2t3/2 ), t = 1
sol) 烰0 C 0 (t) @ çƒ °0 C 00 (t) î
C 0 (t) = (⇡ cos ⇡t, ⇡ sin ⇡t, 3t1/2 ),
3
C 00 (t) = ( ⇡ 2 sin ⇡t, ⇡ 2 cos ⇡t, t 1/2 )
2
0
(sin ⇡t) = ⇡ cos ⇡t,
(cos ⇡t)0 =
⇡ sin ⇡t,
2 0
(t
t ) =1
2t
t‰. t = 1 –⌧ ⌘ X ) ›@
t¿\
C 0 (t) = (⇡ cos ⇡t, ⇡ sin ⇡t, 1
C 0 (1) · (t
2t)
1) + C(1) = ( ⇡, 0, 3) · (t
1) + (0, 1, 2)
= ( ⇡t + ⇡, 1, 3t
1)
t‡ 0|⌧ C 0 (0) = (⇡, 0, 1) t‰.
t‰.
b) C(t) = (et , sin t, cos t)
sol)
c) C(t) = (cos2 t, t t3 , t), t = 0
sol) 烰0 C 0 (t) @ çƒ °0 C 00 (t) î
(et )0 = et ,
(sin t)0 = cos t,
C 0 (t) = ( sin 2t, 1
(cos t)0 =
C 00 (t) = ( 2 cos 2t, 6t, 0)
sin t
t¿\
3t2 , 1),
t‰. t = 0 –⌧X ⌘ X ) ›@
C 0 (t) = (et , cos t,
sin t)
C 0 (0) · t + C(0) = (0, 1, 1) · t + (1, 0, 0)
t‡ 0|⌧ C 0 (0) = (1, 1, 0) t‰.
c) C(t) = (t2 , t3
sol)
= (1, t, t)
t‰.
4t + 1, 0)
d) C(t) = (0, 1, t), t = 1
sol) 烰0 C 0 (t) @ çƒ °0 C 00 (t) î
(t2 )0 = 2t,
(t
0
3
4t + 1) = 3t
2
4,
0
C 0 (t) = (0, 0, 1),
(0) = 0
C 00 (t) = (0, 0, 0)
t¿\
C 0 (t) = (2t, 3t2
4, 0)
t‰. t = 1 –⌧X ⌘ X ) ›@
0
t‡ 0|⌧ C (0) = (0, 4, 0) t‰.
C 0 (1) · (t
3. ‰L
· – t çƒ@ çƒ °0| lX‡ ¸¥ƒ t
✓–⌧ ⌘ X ) ›D lX‹$.
a) C(t) = (t, 3t2 , t3 ), t = 0
sol) 烰0 C 0 (t) @ çƒ °0 C 00 (t) î
C 0 (t) = (1, 6t, 3t2 ),
1) + C(1) = (0, 0, 1) · (t
1) + (0, 1, 1)
= (0, 1, t)
t‰.
4. R3 ¡– ìx · CX
(point)ÑD Ùt‹$.
proof ) ÑXX ⇣ t–⌧
C 00 (t) = (0, 6, 6t)
çƒ
0tt t · @ ¡
C 00 (t) = (0, 0, 0)
101
⇣î ⇣
SOLUTION
t|‡ Xê. ¯Ït
10 ·
t‰.
0
(ai 2 R),
C (t) = (a1 , a2 , a3 )
6. ¸¥ƒ l⌅–⌧ ¸¥ƒp· X 8t| lX‹$.
2 2 3/2
t ), [0, 1]
t‰. t · @ a1 = a2 = a3 = 0tt ⇣D ò¿¥‡, ¯ xX a) C(t) = (t sin t, t cos t,
3
Ω∞–⌧î ¡ D ò¿∏‰. 0|⌧ · CX çƒ 0tt sol) C 0 (t) = (t cos t + sin t, t sin t + cos t, p2t1/2 )t¿\
Cî ¡ ⇣î ⇣t‰.
p
kC 0 (t)k = k(t cos t + sin t, t sin t + cos t, 2t1/2 )k
p
5. ¸¥ƒ l⌅–⌧ ¸¥ƒ · X 8t| lX‹$.
= t2 + 1 + 2t
a) C(t) = (6t, 3t2 , t3 ), [0, 2]
= |t + 1|
sol) C 0 (t) = (6, 6t, 3t2 )t¿\
(ci 2 R)
C(t) = (a1 t + c1 , a2 t + c2 , a3 t + c3 )
kC 0 (t)k = k(6, 6t, 3t2 )k
p
= 36 + 36t2 + 9t4
t‰. 0|⌧ ·
Z 1
= 3 t2 + 2
t‰. 0|⌧ ·
CX 8tî
0
kC 0 (t)kdt =
0
kC 0 (t)kdt =
Z 2
3 t2 + 2 dt
0
⇥
⇤2
= t3 + 6t 0 = 20
t‰.
b) C(t) = (sin 4t, cos 4t, 2t3/2 ), [0, 1]
sol) C 0 (t) = (4 cos 4t, 4 sin 4t, 3t1/2 )t¿\
kC 0 (t)k = k(4 cos 4t, 4 sin 4t, 3t1/2 )k
p
= 16 cos2 4t + 16 sin2 4t + 9t
p
= 16 + 9t
t‰. 0|⌧ · CX 8tî
Z 1
Z 1
p
kC 0 (t)kdt =
16 + 9tdt
0
=

p
t
t
b) C(t) = (2 2t,
p 2e ,t2e ), t [0, 1]
0
sol) C (t) = (2 2, 2e , 2e )t¿\
p
kC 0 (t)k = k(2 2, 2et , 2e t )k
p
= 8 + 4e2t + 4e 2t
= 2(et + e t )
t‰. 0|⌧ ·
Z 1
CX 8tî
kC 0 (t)kdt =
2(et + e t )dt
0
✓
⇤1
e t) 0 = 2 e
⇥
0
c) C(t) = (t2 , sin t t cos t, cos t + t sin t),
sol) C 0 (t) = (2t, t sin t, t cos t)t¿\
Z 1
= 2(et
1
t‰.
[0, ⇡]
t‰.
2
c) C(t) = (t, 2t, t3/2 ), [0, 1]
3
sol) C 0 (t) = (1, 2, t1/2 )t¿\
kC 0 (t)k = k(1, 2, t1/2 )k
p
= 5+t
0
kC (t)k = k(2t, t sin t, t cos t)k
p
= 4t2 + t2 sin2 t + t2 cos2 t
p
= 5 |t|
|t + 1| dt
t‰.
0
2
122
(9t + 16)3/2 =
27
27
0
0
 2
1
t
3
=
+t =
2
2
0
CX 8tî
Z 2
Z 1
t‰. 0|⌧ ·
CX 8tî
Z 1
t‰. 0|⌧ · CX 8tî
Z ⇡
Z ⇡p
0
kC (t)kdt =
5 |t| dt
0
"0p #⇡
p
5 2
5 2
=
t
=
⇡
2
2
0
kC 0 (t)kdt =
102

p
5 + tdt
0
1
2
(5 + t)3/2
3
0
⌘
2 ⇣ 3/2
3/2
=
6
5
3
=
0
Z 1
1
e
◆
SOLUTION
t‰.
7. ¸¥ƒ ·
C(t) –
t‰. 0|⌧ lX‡ê Xî · X 8tî
Z ⇡
Z ⇡p
0
kC (t)kdt =
t2 + 2dt
t · X 8th⇠
s(t) =
Z t
a
10 ·
0
0
=
kC 0 (⌧ )d⌧ k
0
sinh
k↵0 (t)k = k(cosh t, sinh t, 1)k
p
= cosh2 t + sinh2 t + 1
p
p
= 2 cosh2 t = 2 cosh t
Z t
0
↵X 8th⇠î
k↵0 (⌧ )kd⌧ =
=
t‰. ⇣,
0
(t) = (cos t,
Z tp
2 sinh ⌧
it
0
=
p
9. t > 0 –⌧ X⌧ · C(t) = (2t, t2 , ln t) – t P ⇣
(2, 1, 0) ¸ (6, 9, ln 3) ¨tX · X 8t| lX‹$.
sol) C 0 (t) = 2, 2t, 1t t¿\
kC 0 (t)k =
2 sinh t
sin t, 1) t¿\
Z t
0
0
k (⌧ )kd⌧ =
=
Z tp
2d⌧
0
h p it p
2 = 2t
0
Z 3
1
t
◆
Z 3
C(t) = (0, 0, t),
f ds = 0ÑD ùÖX‹$.
C
proof )
f (C(t)) = f (0, 0, t) = 0 (0  t  1)
t¿\
8. ·
C(t) = (t, t cos t, t sin t) –
t P⇣ (0, 0, 0) ¸
(⇡, ⇡, 0) ¨tX · X 8t| lX‹$.
t‰.
sol) C 0 (t) = (1, cos t t sin t, sin t + t cos t)t¿\
t sin t, sin t + t cos t)k
1
t
1
t
1
2t + dt
t
1
⇥2
⇤3
= t + ln t 1 = 8 + ln 3
kC 0 (t)kdt =
µ8⌧ 10.3 .
1. ‰h⇠
f (x, y, z) = x@ ·
Z
t‰.
kC 0 (t)k = k(1, cos t
p
= t2 + 2
2, 2t,
t‰. 0|⌧ lX‡ê Xî · X 8tî
t‰.
X 8th⇠î
✓
= 2t +
k 0 (t)k = k(cos t, sin t, 1)k
p
= cos2 t + sin2 t + 1
p
= 2
t‰. 0|⌧, ·
p
(⇡/ 2)
t‰.
2 cosh ⌧ d⌧
0
hp
1
= [u + sinh u cosh u]0
h
isinh 1 (⇡/p2)
p
2
= u + sinh u 1 + sinh u
0
✓
◆
⇡
⇡p
1
p
= sinh
+
2 + ⇡2
2
2
(t) = (sin t, cos t, t)
X 8th⇠|
lX‹$.
sol) ↵0 (t) = (cosh t, sinh t, 1) t¿\
t‰. 0|⌧, ·
2 cosh2 udu
◆
Z sinh 1 (⇡/p2) ✓
1 + cosh 2u
=
2
du
2
0
p

sinh 1 (⇡/ 2)
1
= u + sinh 2u
2
0
î Öê ‹ a | L ú⌧XÏ ‹ t L¿ C(t) X § D 0|
¿¡x p¨| ò¿∏‰. â, s(t) î C(a) @ C(t) ¨tX · C
X 8tt‰. a = 0 | L P ·
↵(t) = (sinh t, cosh t, t),
Z sinh 1 (⇡/p2)
2. ‰L
103
Z
f ds =
C
Ω∞–
Z 1
f (C(t))kC 0 (t)kdt = 0
0
t
Z
f (x, y, z)ds| lX‹$.
C
0  t  1–
SOLUTION
a) f (x, y, z) = x + y + z,
sol) ·
t 2 [0, 2⇡]
C(t) = (cos t, sin t, t),
t¿\
Z
C(t)| ¯ÑXt
0
t 2 [0, 2⇡]
C (t) = ( sin t, cos t, 1),
10 ·
f (x, y, z)ds =
C
=
=
Z 2⇡ p

0
p
=
1
cos t + t2
2
2 sin t
p
= 2 2⇡ 2
sol) ·
◆ 2⇡
=2
t‰.
b) f (x, y, z) = yz, C(t) = (t, 2t, 3t),
sol) C(t)| ¯ÑXt
C 0 (t) = (1, 2, 3),
t 2 [0, 2⇡]
C(t) = (cos t, sin t, t),
f (x, y, z)ds =
C
Z 2⇡
Z
t 2 [0, 2⇡]
f (x, y, z)ds =
C
=
0
f (C(t))kC (t)kdt
Z 2⇡ p
= [ 2 sin t]2⇡
o
Z 3
Z 3p
t‰.
c) f (x, y, z) = x sin z, C(t) = (t, t2 , ⇡), t 2 [0, 1]
sol)
f (C(t)) = 0 , t 2 [0, 1]
t¿\
f (x, y, z)ds =
C
x+y
c) f (x, y, z) =
,
y+z
sol) C(t)| ¯ÑXt
=
C(t) =
C 0 (t) = (1,
Z
f (C(t))kC 0 (t)kdt
0dt = 0
p
f (x, y, z)ds =
C
t‰.
p
a) f (x, y, z) = exp( z),
sol) C(t)| ¯ÑXt
f (x, y, z)ds| lX‹$.
C
C(t) = (1, 3, t2 ),
C 0 (t) = (0, 0, 2t),
◆
2 3/2
t, t , t ,
3
t, 1),
t 2 [1, 2]
t 2 [1, 2]
Z 2
Z 2
f (C(t))kC(t)kdt

p
2 + tdt
1
2
2
(2 + t)3/2
3
1
2
2 p
= ·8
·3 3
3
3
p
16
=
2 3
3
=
Z
✓
1
=
0
t
14 · 6t2 dt
t¿\
0
Z 1
1
t‰.
=0
Z 1
f (C(t))kC(t)kdt
p
= [2 14t3 ]31
p
= 52 14
2 cos tdt
0
p
Ω∞–
t 2 [1, 3]
1
0
=
3. ‰L
t 2 [1, 3]
t¿\
C 0 (t) = ( sin t, cos t, 1),
Z
2et ]10
0
C(t)| ¯ÑXt
t‰. 0|⌧
Z
2tet dt
= [2tet
t‰.
b) f (x, y, z) = cos z,
Z 1
|2t| · e|t| dt
0
2(cos t + sin t + t)dt
✓
Z 1
0
0
=
f (C(t))kC(t)kdt
0
t‰. 0|⌧
Z
Z 2⇡
f (x, y, z)ds =
f (C(t))kC 0 (t)kdt
C
Z 1
t 2 [0, 1]
t‰.
t 2 [0, 1]
104
SOLUTION
4. ‰h⇠ f (x, y)–
t ˘å\\ ¸¥ƒ ·
a  ✓  b| 0x
Ñt
Z b
f (r cos ✓, r sin ✓)
a
r = r(✓), t| \©Xt ‰L¸ ⇡‰.
p Z 2⇡ p
p Z 2⇡
✓
✓
2
1 + sin ✓d✓ = 2
cos + sin d✓
2
2
0
0
✓
◆
✓
◆
p Z 3⇡/2
p Z 2⇡
✓
✓
✓
✓
= 2
cos + sin
d✓
2
cos + sin
d✓
2
2
2
2
0
3⇡/2

✓
◆ 3⇡

✓
◆ 2⇡
2
p
p
✓
✓
✓
✓
= 2 2 sin
cos
2 2 sin
cos
2
2 0
2
2
3⇡
p
r2 + (r0 )2 d✓
ÑD Ùt‹$.
proof ) · r = r(✓)| ‰⌧T Xt
C(✓) = (r(✓) cos ✓, r(✓) sin ✓)
(a  ✓  b)
2
= 8.
⌧‰. ⇣\ C(✓)| ¯ÑXt
C 0 (✓) = (r0 (✓) cos ✓
r(✓) sin ✓, r0 (✓) sin ✓ + r(✓)cos✓)
t‰. \∏
0
(r (✓) cos ✓
0 2
0
2
r(✓) sin ✓) + (r (✓) sin ✓ + r(✓)cos✓)
2
2rr0 cos(✓) sin(✓)
= (r ) cos (✓) + r2 sin2 (✓)
2
10 ·
+ (r0 )2 sin2 (✓) + r2 cos2 (✓) + 2rr0 cos(✓) sin(✓)
0|⌧ ¸¥ƒ · X 8tî 8t‰.
6. {(x, y, z) 2 R3Z
|y 6= 0}–⌧
X⌧ ‰h⇠ f (x, y, z) =
1
– t
Ñ
f (x, y, z)ds| lX‹$. Ï0⌧ C(t) =
y3
C
(ln t, t, 2), t 2 [1/2, 1]t‰.
sol) · C(t)| ¯ÑXt
= r2 + (r0 )2
C 0 (t) = (1/t, 1, 0)
t¿\
kC 0 (✓)k =
t‰. 0|⌧
Z
f (x, y)ds =
r
Z b
p
r2 + (r0 )2
t‰. ¯Ït ‰Lt 1Ω\‰.
Z
f (C(✓))kC 0 (✓)kd✓
f (x, y, z)ds =
C
=
a
=
Z b
f (r cos ✓, r sin ✓)
a
p
r2 + (r0 )2 d✓
5. · r = 1 + sin ✓, 0  ✓  2⇡X 8t| lX‹$.
sol) 4. | \©Xê. · X 8tî
ds =
r
Z 2⇡ p
r2 + (r0 )2 d✓
0
t‰. \∏
r2 + (r0 )2 = 2 + 2 sin ✓
ds =
r
Z 2⇡ p
Z 2⇡ p
1p
1 + 1/t2 dt
3
1/2 t
Z 5
1 p
(5 5
3
p
2 2)t‰.
7. h⇠ f (x, y) = 2x y@ ‰⌧¿⇠·
t  1)– t ‰L <L– ıX‹$.
r2 + (r0 )2 d✓
Z t p
s(t) =
2 2 |⌧ | d⌧
p
2 1 + sin ✓d✓
0
t‰.
0|⌧ ¸¥ƒ · X 8tî
0
=
f (C(t))kC 0 (t)kdt
1/2
Z 1
1
t‰.
Ñ✓D lX0 ⌅t ‰L¸ ⇡@ mÒ›D t©Xê.
⇣
(1 + sin x) = cos
⇣x⌘
2
+ sin
C(t) = (t2 , t2 )( 1 
a) CX 8th⇠ s(t)| lX‹$.(10.2 X 7à 8⌧| 8pX‹
$.)
sol) ¸¥ƒ · X 8th⇠î
t¿\
Z
Z 1
1p
udu
2 2
p
1 p
= (5 5 2 2).
3
=
1Ω\‰.
Z
t 2 [1/2, 1]
⇣ x ⌘⌘2
2
b) f X
105
Ñ
Z
f (x, y)ds | lX‹$.
C
SOLUTION
sol) lX‡ê Xî
Z
Ñ@
f (x, y, z)ds =
C
=
=
11 ‰¿⇠h⇠@ ¯Ñ
sol) ¸¥ƒ ¯ò⌅î …tt‰.
Z 1
1
Z 1
p
1
f (C(t))kC 0 (t)kdt
p
t2 2 2 |t| dt
2
t‰.
11
‰¿⇠h⇠@ ¯Ñ
µ8⌧ 11.1 .
1. h⇠ f X p•\ p• p
a) f (x, y) = 2x + 2y
sol) h⇠ f X XÌ@
p
1
b) f (x, y) = 4
XÌD ⌧ X‹$.
D = {(x, y) 2 R2 |x
x2
y2
sol) ¸¥ƒ ¯ò⌅î Ï<tt‰.
0, y
0}
D = {(x, y) 2 R2 |2x + y
0}
t‰.
p
b) f (x, y) = 2x + y
sol) h⇠ f X XÌ@
t‰.
c) f (x, y) = sin
sol) h⇠ f X
c) f (x, y) = 2x
3
xy
XÌ@
D = {(x, y) 2 R2 |x 6= 0, y 6= 0}
sol) ¸¥ƒ ¯ò⌅î …tt‰.
t‰.
x2 + y 2
(x2 y 2 )2
sol) h⇠ f X XÌ@
d) f (x, y) = ln
D = {(x, y) 2 R2 |x 6= y, x 6=
y}
t‰.
2. h⇠ f X ¯ò⌅
a) f (x, y) =
¥§ ·t(· )x¿ ⌧ X‹$.
3y + 5
1
d) f (x, y) = x 3
5
sol)
106
SOLUTION
b) f (x, y) = x2
e) f (x, y) =
p
11 ‰¿⇠h⇠@ ¯Ñ
y2 ;
c=
2, 1, 0, 1, 2
sol) ¸¥ƒ h⇠X ¯ò⌅@ c–⌧X Ò⌅ @ ‰L¸ ⇡‰.
1
x2
y2
sol) ¸¥ƒ ¯ò⌅î Ï<tt‰.
c) f (x, y) = x2
y;
c=
4, 4
sol) ¸¥ƒ h⇠X ¯ò⌅@ c–⌧X Ò⌅ @ ‰L¸ ⇡‰.
f) f (x, y) = 4x2 + 9y 2
sol) ¸¥ƒ ¯ò⌅î ¿–Ï<tt‰.
d) f (x, y) = 2y
cos x;
c = 0, 2, 4
sol) ¸¥ƒ h⇠X ¯ò⌅@ c–⌧X Ò⌅ @ ‰L¸ ⇡‰.
3. ¿ ⌧ ✓ c–
t h⇠ f X Ò⌅ D ¯¨‹$.
a) f (x, y) = 3x
y;
c = 4, 5
µ8⌧ 11.2 .
1. h⇠ f
çx ÛX ÌD 0 X‹$.
a) f (x, y) = 2y ln(1 + x)
sol) h⇠ (x, y) 7! y î R2 –⌧ çt‡ h⇠ (x, y) 7! ln(1+x)
î {(x, y) 2 R2 : x > 1} –⌧ çt¿\ h⇠ f î
{(x, y) 2 R2 : x >
sol) ¸¥ƒ h⇠X ¯ò⌅@ c–⌧X Ò⌅ @ ‰L¸ ⇡‰.
107
1}
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
✓ 2
◆
x
xy + 1
b)
lim
x2 + y 2
(x,y)!(2,0)
x2 xy + 1
x4 y
sol) h⇠ (x, y) 7!
î R2 {(0, 0)} –⌧ çt¿\
2 + y2
b) f (x, y) =
x
2
2
25 x
y
✓ 2
◆
sol) h⇠ (x, y) 7! x4 y î R2 –⌧ çt‡ h⇠ (x, y) 7!
x
xy + 1
5
2
2
lim
=
25 x
y @ Ñ® 0t DÃ Û–⌧ çt¿\ h⇠ f î
x2 + y 2
4
(x,y)!(2,0)
–⌧ çt‰.
{(x, y) 2 R2 : x2 + y 2 6= 25}
–⌧ çt‰.
t‰.
c)
lim
⇣y⌘
(x,y)!(0,0) x
◆
xy
sol) x 6= 0 | L, f (x, 0) = 0 t¿Ã f (x, x) = 1 t‰. ¯Ï¿\
c) f (x, y) = cos
xïD 0|⌧ ‰
t h⇠✓t 0<\, ¡ y = x | 0| ‰
1 + x2 + y 2
1
t h⇠✓t 1– LÉ. ¯Ï¿\ ˘\@ t¨X¿ Jî‰.
2
sol) h⇠ (x, y) 7! xy,
î R –⌧ çt‡ h⇠ f
1 + x2 + y 2
xy
✓ 3
◆
î h⇠ (x, y) 7!
@ z 7! cos z X i1t¿\ h⇠
x + 2x2 y xy 2y 2
1 + x2 + y 2
d)
lim
x + 2y
(x,y)!(2,2)
f î R2 –⌧ çt‰.
x3 + 2x2 y xy 2y 2
sol) h⇠ (x, y) 7!
î
x + 2y
d) f (x, y) = arcsin(xy)
sol) h⇠ (x, y) 7! xy î R2 –⌧ çt¿Ã h⇠ z 7! arcsin z
{(x, y) 2 R2 : x 6= 2y}
X XÌ@ {z : 1  z  1} t¿\ h⇠ f î
–⌧ çt¿\
✓ 3
◆
{(x, y) 2 R2 : 1  xy  1}
x + 2x2 y xy 2y 2
lim
=2
x + 2y
(x,y)!(2,2)
–⌧ çt‰.
✓
t‰.
e) f (x, y, z) = 3x e cos(xyz)
sol) h⇠ (x, y, z) 7! x4 , eyz , cos(xyz) î R3 –⌧ çt¿\ e)
lim ⇡ cos(x + y + z)
(x,y,z)!( ⇡
h⇠ f î R3 –⌧ çt‰.
2 , 2 ,0)
sol) T¨x h⇠î R3 ⌅¥–⌧ çt¿\
z+1
f) f (x, y, z) = 2
lim ⇡ cos(x, y, z) = 1
x + z2 1
(x,y,z)!( ⇡
2 , 2 ,0)
3
sol) h⇠ (x, y, z) 7! z + 1 @ R –⌧ çt‡ h⇠ (x, y, z) 7!
1
t‰.
î {(x, y, z) 2 R3 : x2 + z 2 6= 1} –⌧ çt¿\
x2 + z 2 1
✓
◆
h⇠ f î
sin(x2 + y 2 )
3
2
2
f)
lim
{(x, y, z) 2 R : x + z 6= 1}
x2 + y 2
(x,y)!(0,0)
2
2
sol) t = x + y <\ XXXt
–⌧ çt‰.
✓
◆
sin(x2 + y 2 )
sin t
lim
= lim
=1
2
2
t!0 t
x +y
2. ‰L ˘\t t¨Xî¿ ¥¥Ù‡, t¨\‰t ˘\✓D lX
(x,y)!(0,0)
‹$. ˘\t ✓
t¨X¿
◆ JD Ω∞, ¯ t | $ÖX‹$.
ÑD L ⇠ à‰.
1
a)
lim
x+
2
(x,y)!(2,5)
✓
◆
1
xy
sol) h⇠ (x, y) 7! x + î R2 –⌧ çt¿\
g)
lim
2
(x,y)!(0,0) 3x2 + 2y 2
sol) x 6= 0 | L, f (x, 0) = 0 t¿Ã f (x, x) = 1 t‰. ¯Ï¿\
✓
◆
1
5
xïD 0|⌧ ‰
t h⇠✓t 0<\, ¡ y = x | 0|
lim
x+
=
2
2
(x,y)!(2,5)
1
‰
t h⇠✓t – LÉ. ¯Ï¿\ ˘\@ t¨X¿
5
t‰.
Jî‰.
4 yz
108
SOLUTION
h)
lim
(x,y)!(0,0)
✓
3x3 y 3xy 3
x2 + y 2
◆
l)
1
sol) ∞ -0X …‡ ÄÒ›– XXÏ |xy|  (x2 +y 2 )  x2 +y 2
2
t¿\
3x3 y 3xy 3
 3 x2 y 2
x2 + y 2
3x3 y 3xy 3
 3 x2
x2 + y 2
t¿\
lim
(x,y)!(0,0)
✓
3x3 y 3xy 3
x2 + y 2
|y|
x
sol) < x 6= 0 | L f (x, 0) = 0x = 0t‰. 0|⌧ yïD
0|⌧ ‰
t h⇠✓t 0– LÉ. X¿Ã y 6= 0| L
0
f (0, y) = |y| = 1t¿\ xïD 0|⌧ ‰
t h⇠✓t 1–
x
LÉ. 0|⌧ ˘\
lim
|y| @ t¨X¿ Jî‰.
(x,y)!(0,0)
m)
y2 ! 0
◆
lim
(x,y)!(0,0)
✓
◆
2x2
(x,y)!(0,0) x2 + y 2
sol) y 6= 0 | L, f (0, y) = 0 t¿Ã f (y, y) = 1 t‰. ¯Ï¿\
yïD 0|⌧ ‰
t h⇠✓t 0 <\ ¿Ã, y = x | 0|
‰
t h⇠✓t 1– LÉ. ¯Ï¿\ ˘\@ t¨X¿
Jî‰.
ÑD L ⇠ à‰. 0|⌧ (x, y) ! (0, 0) | L,
0
11 ‰¿⇠h⇠@ ¯Ñ
lim
=0
n)
t‰.
lim
(x,y,z)!(0,0,0)
sin(x2 + y 2 + z 2 )
p
x2 + y 2 + z 2
!
sol)
✓ 3
◆
x + y3 + z3
!
(x,y,z)!(0,0,0) x2 + y 2 + z 2
sin(x2 + y 2 + z 2 )
p
lim
sol) x = ⇢ sin cos ✓, y = ⇢ sin sin ✓, z = ⇢ cos \ XXXt,
(x,y,z)!(0,0,0)
x2 + y 2 + z 2
p!0|L
✓
◆
sin(x2 + y 2 + z 2 ) p 2
✓ 3
◆
2 + z2
3
3
=
lim
·
x
+
y
x +y +z
x2 + y 2 + z 2
(x,y,z)!(0,0,0)
x2 + y 2 + z 2
=0
(⇢ sin cos ✓)3 + (⇢ sin sin ✓)3 + (⇢ cos )3
=
t‰.
⇢2
3
3
3
3
3
= ⇢(sin cos ✓ + sin sin ✓ + cos )
✓
◆
x2 1
1
 3⇢ ! 0
o)
lim
tan
x2 + (y 1)2
(x,y)!(0,1)
ÑD L ⇠ à‰. 0|⌧ Ë⌅X 9@ pÑ ¨– Xt
x2 1
sol) (x, y) ! (0, 1) | L, 2
! 1 t‰. 0|⌧
✓ 3
◆
x + (y 1)2
x + y3 + z3
lim
=0
✓
◆
(x,y,z)!(0,0,0) x2 + y 2 + z 2
x2 1
⇡
lim
tan 1
=
2
2
x + (y 1)
2
(x,y)!(0,1)
t‰.
i)
lim
◆
2x2 y xz 2
j)
lim
y 2 xz
(x,y,z)!(2, 1,1)
2x2 y xz 2
sol) h⇠ (x, y, z) 7!
î {(x, y, z) 2 R3 : y 2
y 2 xz
0} –⌧ çt¿\
✓ 2
◆
2x y xz 2
lim
= 10
y 2 xz
(x,y,z)!(2, 1,1)
t‰.
✓
xz 6=
t‰.
k)
lim
(x,y)!(0,0)
✓
x y
x2 + y 2
◆
1
sol) x 6= 0 | L, f (x, x) = 0 t¿Ã f (x, x) = t‰. x ! 0
x
| L, f (x, x) ! 0 t¿Ã f (x, x) î ⌧∞\‰. 0|⌧ ˘\@
t¨X¿ Jî‰.
✓
◆
x2 y
p)
lim
(x,y)!(0,0) x4 + y 2
1
2
sol) f (x, x2 ) = t‡ f (x, 2x2 ) = t‰. â, y = x2 D 0|⌧
2
5
1
‰
t h+✓t
– LÃ ¿¿Ã, y = 2x2 D 0|⌧ ‰
2
2
t h+✓t – LÉ. ¯Ï¿\ ˘\✓@ t¨X¿
5
Jî‰.
µ8⌧ 11.3 .
1. ¸¥ƒ h⇠ f –
X‹$.
(ıÃ Ö‹\‰.)
a) f (x, y) =
109
3 3/4
x
4
t 0∏0°0(¯ò∏∏) rf (x, y)| l
SOLUTION
✓
sol) rf (x, y) =
◆
9
1/4
x
,0 .
16
a) f (x, y) = 4x2
sol)
c) f (x, y) = x4 e3y
sol) rf (x, y) = 4x3 e3y , 3x4 e3y .
b) f (x, y) = ln (7x
sol)
d) f (x, y) = 3xy
sol) rf (x, y) = 3xy 1 y, 3xy ln x .
x+y+z
xy + yz + zx
@2f
@2f
(x, y) = 3ex cos y,
(x, y) = 3ex sin y,
2
@x
@x@y
@2f
@2f
(x, y) = 3ex sin y,
(x, y) = 3ex cos y.
@y@x
@y 2
y 2 + yz + z 2 , x2 + xz + z 2 , x2 + xy + y 2
.
(xy + yz + zx)2
b)f (x, y, z) = 3xez cos y
sol)
rf (x, y, z) = (3ez cos y, 3xez sin y, 3xez cos y).
d) f (x, y) =
@2f
40y 2 (y 2 3x2 ) @ 2 f
80yx(x2 y 2 )
(x,
y)
=
,
(x,
y)
=
,
@x2
(x2 + y 2 )3
@x@y
(x2 + y 2 )3
@2f
80xy(x2 y 2 ) @ 2 f
40x2 (x2 3y 2 )
(x, y) =
,
(x,
y)
=
.
@y@x
(x2 + y 2 )3
@y 2
(x2 + y 2 )3
rf (x, y, z)
!
✓ ◆ ✓ ◆z 1
✓ ◆ ✓ ◆ z 1 ✓ ◆z
z
x
xz
x
x
x
= 4
, 4
,4
ln
.
y
y
y2
y
y
y
sol)
2
1 + xyz 2
rf (x, y, z) =
3. h⇠ f –
t
| lX‹$.
(ıÃ Ö‹\‰.)
10(x2 y 2 )
x2 + y 2
sol)
✓ ◆z
x
c)f (x, y, z) = 4
y
sol)
✓
12y)
c) f (x, y) = 3ex cos y
sol)
sol)
d)f (x, y, z) = arcsin
7
@2f
49
@2f
84
(x, y) =
,
(x, y) =
,
2
2
@x
(7x 12y)
@x@y
(7x 12y)2
@2f
84
@2f
144
(x, y) =
,
(x, y) =
.
2
@y@x
(7x 12y)
@y 2
(7x 12y)2
2. h⇠ f – t 0∏0°0 rf (x, y, z)| lX‹$.
(ıÃ Ö‹\‰.)
rf (x, y, z) =
5xy 2 + 8y 5
@2f
@2f
(x, y) = 8,
(x, y) = 10y,
2
@x
@x@y
@2f
@2f
(x, y) = 10y,
(x, y) = 10x + 160y 3 .
@y@x
@y 2
b) f (x, y) = 3x + 3x2 y 4
sol) rf (x, y) = 3 + 6xy 4 , 12x2 y 3 .
a)f (x, y, z) =
11 ‰¿⇠h⇠@ ¯Ñ
4. h⇠ f (x, y) = x3 y 5 4x2 y + x– t
@3f
@3f
@3f
@3f
,
,
, 3 X ›D lX‹$.
3
@x @x@y@x @y@x@y @y
(ıÃ Ö‹\‰.)
sol)
◆
@3f
@3f
= 6y 5 ,
= 30xy 4 8,
3
@x
@x@y@x
@3f
@3f
= 60x2 y 3 ,
= 60x3 y 2
@y@x@y
@y 3
( 2yz 2 , 2xz 2 , 4xyz)
p
.
(1 + xyz 2 ) (1 + xyz 2 )2 4
@2f
@2f
@2f
@2f
(x, y),
(x, y),
(x, y), 2 (x, y)
2
@x
@x@y
@y@x
@y
5. g(x, y) = 3 sin xy| L,
@2g
@2g
=
t‡
@x@y
@y@x
@3g
@3g
=
ÑD Ùt‹$.
@x@x@y
@y@x2
110
SOLUTION
proof ) gD y–
t⌧ ∏¯ÑXt
proof ) zD x–
t‰. ¯¨‡ t| x– t⌧ ∏¯ÑXt
✓ ◆
@ @g
@2g
=
= 3 cos (xy) 3xy sin (xy)
@x @y
@x@y
t‰. t⌧ g| x–
@z
=
@x
t‡ t| x–
t⌧ ∏¯ÑXt
@g
= 3y cos (xy)
@x
t‰. â, a
3y sin (xy)
(3y sin (xy) + 3xy 2 cos (xy))
=
6y sin (xy)
3xy 2 cos (xy)
a2 e ay cos axt¿\
@z
@z
=a
t‰.
@y
@y
(x > 0)| L,
@z
@2z
=y 2
@x
@y
c
@z
@y
t⌧ ∏¯ÑXt
sol) zD x–
@z
y
= cxc 1 e y/x + 2 xc e y/x = cxc 1 e y/x + yxc 2 e y/x
@x
x
t‰. ¯¨‡ zD y–
t‡ t| y–
3xy 2 cos (xy)
1 c y/x
x e
x
t⌧ ⇣ ∏¯ÑXt
✓ ◆
@ @z
@2z
1
=
= 2 xc e y/x
2
@y @y
@y
x
t‰. â,
@2z
@y 2
@z
1
1
= y 2 xc e y/x + c xc e y/x
@y
x
x
@z
= yxc 2 e y/x + cxc 1 e y/x =
@x
y
@f
(x, y)| lX‹$. ⇣(a, b)–⌧
@y
@f
(a, b) t¨X¿ JD Ω∞, ¯ ⇣D ®P lX‹$.
@y
sol) f | y– t⌧ ∏¯ÑXt
t⌧ ∏¯ÑXt
@z
=
@y
@3g
@3g
=
t‰.
@x@x@y
@y@x2
@f
y2
(x, y) = 3
@y
(x + y 3 )2/3
ae ay cos ax
ÑD Ùt‹$.
=
6. f (x, y) = (x3 +y 3 )1/3 | L,
@z
=
@y
8. cî ¡⇠t‡ z = xc e y/x ,
@g
| x– t⌧ ∏¯ÑXt
@x
✓ ◆
@
@g
@2g
=
= 3y 2 sin (xy)
@x @x
@x2
t‰. ¯Ï¿\
t⌧ ∏¯ÑXt
@z
=
@y
@2g
@2g
t¿\
=
t‰.
@x@y
@y@x
@2g
¯¨‡
| x– t⌧ ∏¯ÑXt
@x@y
✓ 2 ◆
@
@ g
@3g
=
@x @x@y
@x@x@y
ae ay sin ax
t⌧ ⇣ ∏¯ÑXt
✓ ◆
@ @z
@2z
=
= a2 e ay cos ax
@z @x
@x2
t‰. \∏, zD y–
t‰. ¯¨‡ t| y– t⌧ ∏¯ÑXt
✓ ◆
@ @g
@2g
=
= 3 cos (xy) 3xy sin (xy)
@y @x
@y@x
t‰. t| y– t⌧ ∏¯ÑXt
✓
◆
@ @2g
@3g
=
= 6y sin (xy)
@y @x2
@y@x2
@2z
@z
= a ÑD Ùt‹$.
2
@x
@y
t⌧ ∏¯ÑXt
7. z = e ay cos ax| L,
@g
= 3x cos (xy)
@y
t‰. t⌧
11 ‰¿⇠h⇠@ ¯Ñ
c
t‰.
µ8⌧ 11.4 .
1. ¸¥ƒ ·t z = f (x, y)⌅– ¿‹⌧ ⇣ P –⌧ ⌘…tX )
›D lX‹$.
t‰. ⌅ ›–⌧ (x3 + y 3 )2/3 6= 0 t¥| X¿\ a3 + b3 = 0
@f
2 3
2
x ⇣(a, b)–⌧
(a, b) t¨X¿ Jî‰. ¯Ï¿\ a = b| a) z = x y + xy ; P ( 2, 1, 2)
@y
sol) h⇠ f î ⇣ p( 2, 1)–⌧ ¯Ñ •X‡
@f
ÃqXî ®‡ ⇣(a, b)–⌧
(a, b) t¨X¿ Jî‰.
@y
rf (x, y) = (2xy 3 + y 2 , 3x2 y 2 + 2xy)
111
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
t¿\ rf (p) = ( 3, 8)t‰. 0|⌧ lX‡ê Xî ⌘…tX ) e) z = xy ; P (2, 1, 2)
›@
sol) h⇠ f î ⇣ p(2, 1)–⌧ ¯Ñ •X‡
z = f (p) + rf (p) · (x
p)
= 2 + ( 3, 8) · (x + 2, y
=
â 3x
3x + 8y
rf (x, y) = (yxy 1 , xy ln x)
1)
t¿\ rf (p) = (1, 2 ln 2)t‰. 0|⌧ lX‡ê Xî ⌘…tX
) ›@
12,
12 t‰.
8y + z =
z = f (p) + rf (p) · (x
= 2 + (1, 2 ln 2) · (x
b) z = x2 + y 3 + xy 2 ; P ( 2, 1, 3)
sol) h⇠ f î ⇣ p( 2, 1)–⌧ ¯Ñ •X‡
= x + 2 ln 2y
rf (x, y) = (2x + y 2 , 3y 2 + 2xy)
â x + 2 ln 2y
p)
2, y
1)
2 ln 2,
z = 2 ln 2 t‰.
⇣ ⇡
⌘
sin y); P 0, , 1
2
⇣ ⇡⌘
sol) h⇠ f î ⇣ p 0,
–⌧ ¯Ñ •X‡
2
t¿\ rf (p) = ( 3, 1)t‰. 0|⌧ lX‡ê Xî ⌘…tX
f) z = ex (cos y
) ›@
z = f (p) + rf (p) · (x
p)
= 3 + ( 3, 1) · (x + 2, y
=
â 3x + y + z =
3x
y
1)
rf (x, y) = (ex (cos y
2,
t¿\ rf (p) = ( 1, 1)t‰. 0|⌧ lX‡ê Xî ⌘…tX
) ›@
2 t‰.
c) z = 3x4 y 7x3 y x2 + y + 2; P (1, 2, 5)
sol) h⇠ f î ⇣ p(1, 2)–⌧ ¯Ñ •X‡
rf (x, y) = (12x3 y
21x2 y
2x, 3x4
z = f (p) + rf (p) · (x
7x3 + 1)
t¿\ rf (p) = ( 20, 3)t‰. 0|⌧ lX‡ê Xî ⌘…tX
) ›@
z = f (p) + rf (p) · (x
=
=
1, y
d) z = x
⇡2
4
⇡
1, , 2
2
◆
y + sin y; P
⇣ ⇡⌘
sol) h⇠ f î ⇣ p 1,
–⌧ ¯Ñ •X‡
2
⇡
+ (2, ⇡) · (x
4
⇡2
= 2x ⇡y +
,
4
⇡y
z=
⇡2
t‰.
4
x
⇡
2
y+
⇡
2
⇡
)
2
1,
1 t‰.
= 3 + ( 2, 4) · (x
=
p)
2, y
1)
2x + 4y + 3
p)
2
â 2x
=
z = f (p) + rf (p) · (x
t¿\ rf (p) = (2, ⇡)t‰. 0|⌧ lX‡ê Xî ⌘…tX
) ›@
=2
1 + ( 1, 1) · (x, y
t¿\ rf (p) = ( 2, 4)t‰. 0|⌧ lX‡ê Xî ⌘…tX )
›@
rf (x, y) = (2x, 2y + cos y)
z = f (p) + rf (p) · (x
=
g) z =
3y + 21,
✓
2
p)
x+y
; P (2, 1, 3)
x y
sol) h⇠ f î ⇣ p(2, 1)–⌧ ¯Ñ •X‡ x 6= y| L
✓
◆
2y
2x
rf (x, y) =
,
(x y)2 (x y)2
2)
â 20x + 3y + z = 21 t‰.
2
â x+y+z =
p)
5 + ( 20, 3) · (x
20x
sin y), ex (sin y + cos y))
1, y
⇡
)
2
â 2x
4y + z = 3 t‰.
h) z = ln(x2 + y 2 ); P (2, 2, 3 ln 2)
sol) h⇠ f î ⇣ p(2, 2)–⌧ ¯Ñ •X‡ x2 + y 2 6= 0| L
✓
◆
2x
2y
rf (x, y) =
,
x2 + y 2 x2 + y 2
112
SOLUTION
t¿\ rf (p) =
) ›@
✓
◆
1 1
,
t‰. 0|⌧ lX‡ê Xî ⌘…tX t‰. ¸¥ƒ …t@ –⇣D ¿®t êÖX‰.
2 2
3. |(¸¨| ¨©XÏ 4f = f (Q) f (P )X ¸ø✓D lX‹$.
z = f (p) + rf (p) · (x p)
✓
◆
1 1
= 3 ln 2 +
,
· (x 2, y
2 2
1
1
= x + y + 3 ln 2 2
2
2
â x+y
2z = 4
11 ‰¿⇠h⇠@ ¯Ñ
p
a) f (x, y) = x2 + y 2 + 2; P (3, 5),
sol) ⇣ P –⌧X |(¸¨›
2)
Q(3.01, 4.96)
f (x) ⇡ f (P ) + rf (P ) · (x
–⌧ 4f = f (Q)
à‰. t⌧
6 ln 2 t‰.
f (P )X ¸ø✓@ rf (P ) · (Q
rf (x, y) =
2. ‰L <L– ıX‹$.
P)
p
x
x2 + y 2 + 2
,p
P )\ l` ⇠
y
x2 + y 2 + 2
!
x2
y2
t¿\ 4f X ¸ø✓@
+ 2 ⌅X ⇣
2
a
b
(x0 , y0 , z0 )–⌧ ⌘…tX ) ›D lX‹$.
4f ⇡ rf (P ) · (Q P )
✓
◆
3 5
=
,
· (3.01 3, 4.96 5)
6 6
sol) z = f (x, y)\ ì<t h⇠ f î ⇣ p(x0 , y0 )–⌧ ¯Ñ •X
‡
17
=
= 0.0283
✓
◆
600
2
2
rf (x, y) =
x,
y
ca2 cb2
t‰.
✓
◆
2
2
p
x0 , 2 y0 t‰. 0|⌧, lX‡ê
t¿\ rf (x0 , y0 ) =
b) f (x, y) = 2x3 + y 2 ; P (2, 3), Q(3.02, 2.97)
ca2
cb
Xî ⌘…tX ) ›@
sol) f ¯Ñ •\ ⇣–⌧
!
z = f (p) + rf (p) · (x p)
2
3x
y
✓
◆
rf (x, y) = p
,p
2
2
3 + y2
2x
2x3 + y 2
= z0 +
x
,
y
·
(x
x
,
y
y
),
0
0
0
0
ca2
cb2
t¿\ 4f X ¸ø✓@
2x0
2y0
â z z0 = 2 (x x0 ) + 2 (y y0 ) t‰.
ca
cb
4f ⇡ rf (P ) · (Q P )
✓
◆
⇣y⌘
12 3
b) R–⌧ ¯Ñ •\ |¿⇠ h⇠ g – t f (x, y) = xg
=
,
· (2.02 2, 2.97 3)
x
5 5
|‡ Xê (x 6= 0). f X ¯ò⌅– ⌘Xî …t@ m¡ –⇣D
3
¿®D Ùt‹$.
=
= 0.03
100
a) c
0t DÃ ¡⇠| L, Ï<t cz =
proof ) x0 6= 0, y0 2 R– t h⇠ f ⇣ p(x0 , y0 )–⌧ ¯Ñ
•X¿\
✓ ✓ ◆
✓ ◆
✓ ◆◆
y0
y0 0 y 0
y0
0
rf (x0 , y0 ) = g
g
,g
x0
x0
x0
x0
t‰.
sol) f
t‡, lX‡ê Xî ⌘…tX ) ›@
z = f (p) + rf (p) · (x p)
✓ ◆ ✓ ✓ ◆
y0
y0
= x0 g
+ g
x0
x0
y0 0
g
x0
✓
y0
x0
◆
, g0
· (x x0 , y y0 )
✓ ✓ ◆
✓ ◆◆
✓ ◆
y0
y0 0 y0
y0
= g
g
x + g0
y
x0
x0
x0
x0
✓
y0
x0
◆◆
1
; P (3, 6), Q(3.02, 6.05)
1+x+y
¯Ñ •\ ⇣–⌧
✓
◆
1
1
rf (x, y) =
,
(1 + x + y)2
(1 + x + y)2
c) f (x, y) =
t¿\ 4f X ¸ø✓@
4f ⇡ rf (P ) · (Q
P)
= ( 0.01, 0.01) · (3.02
=
113
0.0007
3, 6.05
6)
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
9⇡
9⇡
cos
10
20
sol) h⇠ f |
t‰.
b) sin
d) f (x, y) = x ln y y ln x; P (1, 1), Q(1.01, 1.02)
sol) f ¯Ñ •\ ⇣–⌧
✓
◆
y x
rf (x, y) = ln y
,
ln x
x y
f (x, y) = sin x cos y
<\
rf (x, y) = (cos x cos y, sin x sin y)
⇡⌘
t¿\ ⇣ p ⇡,
–⌧X f X |(¸¨›D t©Xt
2
⇣ ⇡⌘
9⇡
9⇡
sin
cos
= f ⇡,
10
20
2
✓✓
◆
◆
9⇡ 9⇡
⇡ f (p) + rf (p) ·
,
p
10 20
⇣ ⇡
⇡⌘
= 0 + (0, 0) ·
,
10 20
=0
P)
= ( 1, 1) · (1.01
1, 1.02
1)
= 0.01
t‰.
e) f (x, y) = tan 1 (1 + x + y); P (0, 0), Q(0.1, 0.2)
sol) f ¯Ñ •\ ⇣–⌧
✓
◆
1
1
rf (x, y) =
,
1 + (1 + x + y)2 1 + (1 + x + y)2
t‰.
1
2
c) (16.05) 4 (7.95) 3
sol) h⇠ f |
f (x, y) = x1/4 y 2/3
t¿\ 4f X ¸ø✓@
<\
4f ⇡ rf (P ) · (Q P )
✓
◆
1 1
=
,
· (0.1 0, 0.2
2 2
¯Ñ •\ ⇣–⌧
⇣
t¿\ 4f X ¸ø✓@
4f ⇡ rf (P ) · (Q
XXt f
0)
XXt f
¯Ñ •\ ⇣–⌧
✓
◆
1 3/4 2/3 2 1/4 1/3
rf (x, y) =
x
y , x y
4
3
t¿\ ⇣ p(16, 8)–⌧X f X |(¸¨›D t©Xt
= 0.15
1
2
(16.05) 4 (7.95) 3 = f (16.05, 7.95)
t‰.
⇡ f (p) + rf (p) · ((16.05, 7.95)
✓
◆
1 2
=8+
,
· (0.05, 0.05)
8 3
4. ‰L ¸ø✓D lX‹$.
⇡8
p
a) (3.012)2 + (3.997)2
sol) h⇠ f |
p
f (x, y) = x2 + y 2
<\
XXt f
t‰.
d) e0.1 ln 0.9
sol) h⇠ f |
¯Ñ •\ ⇣–⌧
rf (x, y) =
p
x
x2 + y 2
,p
y
x2 + y 2
!
t¿\ ⇣ p(3, 4)–⌧X f X |(¸¨›D t©Xt
p
(3.012)2 + (3.997)2 = f (3.012, 3.997)
⇡ f (p) + rf (p) · ((3.012, 3.997)
✓
◆
3 4
=5+
,
· (0.012, 0.003)
5 5
f (x, y) = ex ln y
<\
XXt f
¯Ñ •\ ⇣–⌧
✓
◆
ex
x
rf (x, y) = e ln y,
y
t¿\ ⇣ p(0, 1)–⌧X f X |(¸¨›D t©Xt
e0.1 ln 0.9 = f (0.1, 0.9)
p)
⇡ f (p) + rf (p) · ((0.1, 0.9)
= 0 + (0, 1) · (0.1, 0.1)
= 5 + 0.0048 = 5.0048
t‰.
0.0271 = 7.9729
=
t‰.
114
0.1
p)
p)
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
5. –‘(cone) ®ëX ®òT¯ à‰. t ®òT¯X ít
â 8x + 3y 5z = 0 t‰.
5cm–⌧ 5.01cm\, ⇠¿Ñt 4cm–⌧ 3.98cm\
¿àD L,
t ®òT¯X Ä<X ¿T…D |(¸¨| t©XÏ lX‹$.
c) f (x, y, z) = xy 2 cos z; ( 1, 2, 0)
sol) h⇠ f î ⇣ P ( 1, 2, 0)–⌧ ¯Ñ •X‡
⇡
sol) ít ht‡ ⇠¿Ñt rx –‘X Ä<î V (r, h) = r2 h
3
rf (x, y, z) = y 2 cos z, 2xy cos z, xy 2 sin z
t‰. p = (4, 5), q = (3.98, 5.01)|‡ Xt lX‡ê Xî Ä<X
¿T…@ 4V = V (q) V (p) \ ò¿º ⇠ à‰. ⇣\
t¿\ rf (P ) = (4, 4, 0)t‰. 0|⌧ lX‡ê Xî ⌘…tX
✓
◆
) ›@
2⇡
⇡ 2
rV (r, h) =
rh, r
3
3
w = f (P ) + rf (P ) · (x P )
4 + (4, 4, 0) · (x + 1, y
=
t‰. 0|⌧, ⇣ p–⌧X |(¸¨– Xt
= 4x
4V ⇡ rV (p) · (q p)
✓
◆
40⇡ 16⇡
=
,
· ( 0.02, 0.01)
3
3
64⇡
=
(0.01) = 0.670206
3
â 4x
4y
2, z)
4y + 8,
8 t‰.
w=
d) f (x, y, z) = e2x 4y z ; (0, 1, 1)
sol) h⇠ f î ⇣ P (0, 1, 1)–⌧ ¯Ñ •X‡
rf (x, y, z) = 2e2x 4y z , 4e2x 4y z , e2x 4y z
t‰.
t¿\ rf (P ) = 2e5 , 4e5 , e5 t‰. 0|⌧ lX‡ê Xî
6. ¸¥ƒ h⇠X ¯ò⌅ ⌅X ⇣ P –⌧ ⌘…tX ) ›D lX ⌘…tX ) ›@
‹$.
w = f (P ) + rf (P ) · (x P )
= e5 + 2e5 , 4e5 , e5 · (x, y + 1, z + 1)
a) f (x, y) = x4 6x2 8xy 2 + 12; (1, 2)
sol) h⇠ f î ⇣ P (1, 2)–⌧ ¯Ñ •X‡
rf (x, y) = 4x3
12x
= 2e5 x
8y 2 , 16xy
â 2e5 x
4e5 y
4e5 y
e5 z
e5 z
4e5 ,
w = 4e5 t‰.
t¿\ rf (P ) = ( 40, 32)t‰. 0|⌧ lX‡ê Xî ⌘…tX
7. ·t z = x2 + 3xy + 2y 2 6x + 8y ⌅X ⇣ P –⌧ ⌘…tt z
) ›@
ï ⌅– ⇠¡| Ñ, ⇣ P | ®P lX‹$.
z = f (P ) + rf (P ) · (x P )
sol) ·t z = f (x, y) X ⇣ P (x0 , y0 , f (x0 , y0 )) –⌧X zï–
⇠¡x ⌘…tD ¿$t rf (x0 , y0 ) = (0, 0) t¥| X¿\
= 25 + ( 40, 32) · (x 1, y + 2)
=
â 40x
40x + 32y + 79,
rf (x0 , y0 ) = (0, 0) () (2x0 + 3y0
() x0 =
32y + z = 79 t‰.
t¿\ rf (P ) =
) ›@
✓
p
4x
4x2 + y 2
,p
48, y0 = 34,
â ⇣ P ( 48, 34, 280)–⌧ zï– ⇠¡x ⌘…tD
p
b) f (x, y) = 4x2 + y 2 ; (2, 3)
sol) h⇠ f î ⇣ P (2, 3)–⌧ ¯Ñ •X‡
rf (x, y) =
6, 3x0 + 4y0 + 8) = (0, 0)
y
4x2 + y 2
!
ƒ‰.
8. ëX ¡⇠ c– t ·t z = c(x2 + y 2 )⌅X ⇣ P (a, b, c(a2 +
b2 ))– t ‰L <L– ıX‹$.
a) ⇣ P –⌧ ⌘…tX ) ›D lX‹$.
◆
8 3
,
t‰. 0|⌧ lX‡ê Xî ⌘…tX sol) ·t z = f (x, y)X ¯ò∏∏
5 5
rf (x, y) = (2xc, 2yc)
z = f (P ) + rf (P ) · (x P )
✓
◆
8 3
=5+
,
· (x 2, y 3)
5 5
8
3
= x + y,
5
5
t¿\ ⇣ P –⌧X ⌘…tX ) ›@
z = f (P ) + rf (P ) · (x
P)
2
2
2
2
= c(a + b ) + (2ac, 2bc) · (x
= 2acx + 2bcy
115
c(a + b )
a, y
b)
SOLUTION
t‰.
11 ‰¿⇠h⇠@ ¯Ñ
b) w =
b) ⇣ P –⌧ ⌘…t– ⇠¡\ ®‡ Ë⌅)•°0| lX‹$.
sol) ⌘…tX ï °0î (2ac, 2bc, 1)t¿\ ⌘…t– ⇠¡x
Ë⌅)•°0î
!
2ac
2bc
1
± p
,p
,p
4c2 (a2 + b2 ) + 1
4c2 (a2 + b2 ) + 1
4c2 (a2 + b2 ) + 1
t + c(a2 + b2 ), t 2 R
\ ò¿º ⇠ à<¿\ z = 0 () t = c(a2 + b2 )ÑD t©Xt
⇣ P (x0 , y0 , 0)
y0 = 2c2 b(a2 + b2 ) + b,
c) w = sin(xy 2 z 3 );
sol)
ï ¸ xy…tX P⇣t ⌧‰.
2 3
5t, z =
p
t
1
1
x = 3t, y = t 2 , z = t 3
2 3
3
2 3
= y z cos(xy z ) · 3 + 2xyz cos(xy z ) ·
✓
◆
1
2 2
2 3
+ 3xy z cos(xy z ) ·
2
3t 3
⇣ 5
⌘ ✓ 1 ◆
2
3
3
2
p
= t cos(3t ) · 3 + 6t cos(3t ) ·
2 t
✓
◆
⇣ 8
⌘
1
+ 9t 3 cos(3t3 ) ·
2
3t 3
= 9t2 cos(3t3 ).
1
+ c(a2 + b2 ),
2c
ï ¸ zïX P⇣t ⌧‰. Ã} a = b = 0tt ⇣
Q(0, 0, t),
1
,y=
t2
dw
@w dx @w dy @w dz
=
+
+
dt
@x dt
@y dt
@z dt
d) ⇣ P –⌧ ⌘…t– ⇠¡x ï ¸ zïX P⇣D lX‹$.
sol) a2 + b2 6= 0x Ω∞ ï X ) ›D c)–⌧@ ⇡t Pt
1
x = y = 0 () t =
t¿\ ⇣ P (0, 0, z0 ),
2c
z0 =
x=
dw
@w dx @w dy @w dz
=
+
+
dt
@x dt
@y dt
@z dt
✓
◆ ✓
◆ ✓
◆
z
2
2z
=
·
+
· ( 5)
x2 y 2
t3
xy 3
✓
◆ ✓
◆
1
1
p
+
·
xy 2
2 t
✓
◆ ✓
◆ ✓
◆
1 5/2
2
2
p
=
t
·
+
· ( 5)
25
t3
125 t
✓ ◆ ✓
◆
1
1
p
+
·
25
2 t
1
= p .
50 t
c) ⇣ P –⌧ ⌘…t– ⇠¡\ ï ¸ xy…tX P⇣D lX‹$.
sol) ⌘…t– ⇠¡x ï @
x0 = 2c2 a(a2 + b2 ) + a,
3;
sol)
t‰.
x = 2act + a, y = 2bct + b, z =
z
xy 2
t2R
t ï ¸ zïX P⇣t ⌧‰.
✓
2 t
µ8⌧ 11.5 .
dw
1. ƒïYD t©XÏ
(t)| lX‹$.
dt
x z
a) w =
; x = sin t, y = cos t, z = tan t
y
x
sol)
d) w = cos(xyz);
sol)
=
tan2 t + 1
sin2 t
sin2 t + cos t tan t
+
+
.
sin t
cos2 t
sin2 t
x = t, y = t2 , z = t3
dw
@w dx @w dy @w dz
=
+
+
dt
@x dt
@y dt
@z dt
= ( yz sin(xyz)) · 1 + ( xz sin(xyz)) · (2t)
dw
@w dx @w dy @w dz
=
+
+
dt
@x dt
@y dt
@z dt
✓ 2
◆
✓
◆
✓
◆
x + yz
x
1
=
·
cos
t
+
·
(
sin
t)
+
· sec2 t
x2 y
y2
x
✓ 2
◆
✓
◆
sin t + cos t tan t
sin t
=
· cos t +
· ( sin t)
cos2 t
cos t sin2 t
✓
◆
1
+
· (sec2 t)
sin t
=
+ ( xy sin(xyz)) · 3t2
t5 sin(t6 ) · 1 +
+
=
e) w = x2 ey ;
116
t4 sin(t6 ) · (2t)
t3 sin(t6 ) · 3t2
6t5 sin(t6 ).
(x, y) = (sin t, t3 )
1
p
◆
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
sol)
dw
@w dx @w dy
=
+
dt
@x dt
@y dt
y
= (2xe ) · cos t + x2 ey · 3t2
⇣ 3
⌘
⇣ 3
⌘
= 2et sin t · cos t + et sin2 t · 3t2
3
@w
@w @x @w @y @w @z
=
+
+
@v
@x @v
@y @v
@z @v
1
= p · (2v) + 2yz 3 · u + 3y 2 z 2 · 0
2 x
2v
= p
+ 2 · (uv) · 27u3 · u + 0
2 1 + u2 + v 2
v
=p
+ 54u5 v.
1 + u2 + v 2
3
= 3t2 et sin2 t + 2et cos t sin t.
f) w = x cos(yz 2 );
sol)
(x, y, z) = (sin t, t2 , et )
dw
@w dx @w dy @w dz
=
+
+
dt
@x dt
@y dt
@z dt
= cos(yz 2 ) · cos t + xz 2 sin(yz 2 ) · (2t)
+
=
e2t sin(t2 e2t ) sin t · (2t)
2t2 et sin(t2 e2t ) sin(t) · et
2t2 e2t sin(t2 e2t ) sin t
2te2t sin(t2 e2t ) sin t
+ cos(t2 e2t ) cos t.
@w
@w
(u, v) @
(u, v) | lX‹$.
@u
@v
2
2
2
2
x = u + v ,y = u
v
2. ƒïYD t©XÏ
a) w = x ln y;
sol)
@w
@w @x @w @y
=
+
@u
@x @u
@y @u
x
= (ln y) · (2u) + · (2u)
y
✓
◆
u2 + v 2
2
2
= 2u ln(u
v )+ 2
.
u
v2
b) w =
sol)
p
@w
@w @x @w @y
=
+
@v
@x @v
@y @v
x
= (ln y) · (2v) + · ( 2v)
y
✓
◆
u2 + v 2
= 2v ln(u2 v 2 )
.
u2 v 2
x + y2 z3 ;
(x, y, z) = (veu , u2 v 4 , uev )
@w
@w @x @w @y
@w @z
=
+
+
@u
@x @u
@y @u
@z @u
y
y
u
= · (ve ) + ln(xz) · 2uv 4 + · ev
x
z
u2 v 4
u2 v 4 v
u
u+v
4
=
·
(ve
)
+
ln(uve
)
·
2uv
+
·e
veu
uev
= u2 v 4 + 2uv 4 ln(uveu+v ) + uv 4 .
2xyz sin(yz 2 ) · et
= cos(t2 e2t ) · cos t +
+
c) w = y ln(xz);
sol)
@w
@w @x @w @y @w @z
=
+
+
@v
@x @v
@y @v
@z @v
y u
y
= · e + ln(xz) · 4u2 v 3 + · (uev )
x
z
u2 v 4 u
u2 v 4
u+v
2 3
=
·
e
+
ln(uve
)
·
4u
v
+
· (uev )
veu
uev
= u2 v 3 + 4u2 v 3 ln(uveu+v ) + u2 v 4 .
3. ƒïYD t©XÏ
@s
@s
@s
(x, y, z),
(x, y, z),
(x, y, z) |
@x
@y
@z
lX‹$.
a) s = eu+v+w ;
sol)
u = yz, v = xz, z = xy
@s
@s @u @s @v
@s @w
=
+
+
@x
@u @x @v @x @w @x
= eu+v+w · 0 + eu+v+w · z + eu+v+w · y
x = 1 + u2 + v 2 , y = uv, z = 3u
= (y + z) eu+v+w
= (y + z) exy+yz+zx .
@w
@w @x @w @y
@w @z
=
+
+
@u
@x @u
@y @u
@z @u
1
= p · 2u + 2yz 3 · v + 3y 2 z 2 · 3
2 x
2u
= p
+ 2 · (uv) · 27u3 · v
2 1 + u2 + v 2
+ 3 · u2 v 2 · 9u2 · 3
u
=p
+ 135u4 v 2 .
1 + u2 + v 2
@s
@s @u @s @v
@s @w
=
+
+
@y
@u @y
@v @y @w @y
= eu+v+w · z + eu+v+w · 0 + eu+v+w · x
= (z + x) eu+v+w
= (z + x) exy+yz+zx .
117
SOLUTION
@s
@s @u @s @v
@s @w
=
+
+
@z
@u @z
@v @z
@w @y
= eu+v+w · y + eu+v+w · x + eu+v+w · 0
b) w =
sol)
11 ‰¿⇠h⇠@ ¯Ñ
p
uvxy;
p
u=
w=
x
y, v =
p
x+y
p
uvxy
q
p
p
=
x y x + yxy
= (x + y) eu+v+w
1
= (x4 y 2
= (x + y) exy+yz+zx .
x2 y 4 ) 4
t‰. 0|⌧
b) s = uvw
sol)
u2
v2
w2 ;
@s
@s @u @s @v
@s @w
=
+
+
@x
@u @x @v @x @w @x
= (vw 2u) · 0 + (uw 2v) · 1 + (uv
= (u
2) (v + w)
= (y + z
@w
1
=
4x3 y 2
@x
4
@w
1
=
2x4 y
@y
4
(u, v, w) = (y + z, x + z, x + y)
2w) · 1
x4 y 2
x2 y 4
4x2 y 3
x4 y 2
x2 y 4
3
4
,
3
4
t‰.
c) w = uv
2) (2x + y + z) .
2xy 4
xy;
u=
x
x2 + y 2
y
,v =
x2 + y 2
sol)
@s
@s @u @s @v
@s @w
=
+
+
@y
@u @y
@v @y @w @y
= (vw 2u) · 1 + (uw 2v) · 0 + (uv
= (v
2) (w + u)
= (z + x
w = uv xy
✓
◆✓
◆
x
y
=
x2 + y 2
x2 + y 2
xy
=
xy
2
2
(x + y 2 )
!
1
= xy
1
2
(x2 + y 2 )
2w) · 1
2) (x + 2y + z) .
xy
t‰. 0|⌧
@s
@s @u @s @v
@s @w
=
+
+
@z
@u @z
@v @z
@w @y
= (vw 2u) · 1 + (uw 2v) · 1 + (uv
= (w
2) · (u + v)
= (x + y
@w
=y
@x
2w) · 0
@w
=x
@y
2) (x + y + 2z) .
1
2 2
(x2 + y )
1
(x2 + y 2 )
2
!
1
1
!
4x2 y
(x2 + y 2 )
3,
4xy 2
(x2 + y 2 )
3
t‰.
4.
@w
@w
(u, v, x, y) @
(u, v, x, y) | lX‹$.
@x
@y
a) w = u2 + v 2 + x2 + y 2 ;
sol)
u=x
y, v = x + y
w = u2 + v 2 + x2 + y 2
= (x
2
2
y) + (x + y) + x2 + y 2
= 3x2 + 3y 2
5. ‰L
Ò›D ÃqXî ⇣ (x, y, z) 2 R3 X —i@ ¿‹⌧ ⇣
P (x0 , y0 , z0 ) ¸)–⌧ ¯Ñ •\ h⇠ z = z(x, y) \ ò¿ú‰.
@z
@z
(x0 , y0 ) @
(x0 , y0 ) | lX‹$.
@x
@y
a) x3 + xz y 2 = 1; P (1, 2, 2)
sol) F (x, y, z) = x3 + xz y 2 + 1 t| Xê. ¯Ït ‰mh⇠
F î ⇣ (1, 2, 2) –⌧ ¯Ñ •X‡ F (1, 2, 2) = 0 t‰. 0|⌧
Lh⇠X ¯Ñï– Xt
@z
Fx (1, 2, 2)
(1, 2) =
= 5,
@x
Fz (1, 2, 2)
@z
Fy (1, 2, 2)
(1, 2) =
=4
@y
Fz (1, 2, 2)
t‰. 0|⌧
@w
= 6x,
@x
t‰.
@w
= 6y
@y
t‰.
118
SOLUTION
11 ‰¿⇠h⇠@ ¯Ñ
@w
@f @z
@f @b
@f @c
=
+
+
@z
@a @y
@b @z
@c @z
@f
@f
=
@c
@b
b) x2 + y 2 + z 2 = 14; P (1, 2, 3)
sol) F (x, y, z) = x2 + y 2 + z 2 14 t| Xê. ¯Ït ‰mh⇠
F î ⇣ (1, 2, 3) –⌧ ¯Ñ •X‡ F (1, 2, 3) = 0 t‰. 0|⌧
Lh⇠X ¯Ñï– Xt
t‰. 0|⌧
@z
Fx (1, 2, 3)
1
(1, 2) =
=
,
@x
Fz (1, 2, 3)
3
@z
Fy (1, 2, 3)
2
(1, 2) =
=
@y
Fz (1, 2, 3)
3
@w @w @w
+
+
=0
@x
@y
@z
ÑD L ⇠ à‰.
t‰.
8. x = r cos ✓, y = r sin ✓ | L, ‰L â,D r ¸ ✓ X ›<\
lX‹$.
c) x3 + y 3 + z 3 = 5xyz; P (2, 1, 1)
0
1
@r @r
sol) F (x, y, z) = x3 + y 3 + z 3 5xyz t| Xê. ¯Ït ‰mh⇠
B @x @y C
F î ⇣ (2, 1, 1) –⌧ ¯Ñ •X‡ F (2, 1, 1) = 0 t‰. 0|⌧
@ @✓ @✓ A
Lh⇠X ¯Ñï– Xt
@x @y
@z
Fx (2, 1, 1)
(2, 1) =
= 1,
sol) r¸ ✓| x, y– t \⌅Xt ‰L¸ ⇡‰.
@x
Fz (2, 1, 1)
p
y
@z
Fy (2, 1, 1)
r(x, y) = x2 + y 2 , ✓(x, y) = arctan( )
(2, 1) =
= 1
x
@y
Fz (2, 1, 1)
ƒïYD ¨©Xt
t‰.
6. f
R –⌧ ¯Ñ •\ |¿⇠ h⇠t‡ z = f (x
dz
dz
=
ÑD Ùt‹$.
dx
dy
proof ) t = x y |‡ Xê. ¯Ït
@z
@f @t
@f
@f
=
=
·1=
@x
@t @x
@t
@t
@z
@f @t
@f
=
=
· ( 1) =
@y
@t @y
@t
t‡ 0|⌧
7. f
y, y
dz
=
dx
y) | L,
dz
t‰.
dy
R –⌧ ¯Ñ •\ º¿⇠ h⇠t‡ w(x, y, z) = f (x
z, z x) | L, ‰L Ò›D ùÖX‹$.
@w @w @w
+
+
=0
@x
@y
@z
y, b = y
z, c = z
@r
y
=p
= sin ✓,
2
@y
x + y2
⇣ y ⌘
@✓
1
y
sin ✓
=
=
=
,
y 2
@x
1 + (x)
x2
x2 + y 2
r
✓ ◆
@✓
1
1
x
cos ✓
=
= 2
=
2
2
y
@y
x
x +y
r
1+ x
@f
@t
3
proof ) a = x
@r
x
=p
= cos ✓,
2
@x
x + y2
x |‡ Xt
@w
@f @a @f @b
@f @c
=
+
+
@x
@a @x
@b @x
@c @x
@f
@f
=
@a
@c
@w
@f @a @f @b
@f @c
=
+
+
@y
@a @y
@b @y
@c @y
@f
@f
=
@b
@a
ÑD L ⇠ à‰. 0|⌧
0
1
@r @r
B @x @y C
@ @✓ @✓ A =
@x @y
cos ✓
sin ✓
r
sin ✓
cos ✓
r
!
t‰.
9.(8⌧$X-¿Ω) x = r cos cos ✓, y = r cos sin ✓, z = r sin
| L, ‰L â,D r, , ✓ X ›<\ lX‹$.
0
1
@r @r @r
B @x @y @z C
B@
@
@ C
B
C
B
C
B @x @y @z C
@ @✓ @✓ @✓ A
@x @y @z
t \⌅Xt ‰L¸ ⇡‰.
p
r(x, y, z) = x2 + y 2 + z 2 ,
sol) r, , ✓| x, y, z–
119
SOLUTION
z
(x, y, z) = arcsin p
,
2
x + y2 + z2
⇣y⌘
✓(x, y, z) = arctan
.
x
ƒïYD ¨©Xt
ÑD L ⇠ à‰. 0|⌧
0
1
@r @r @r
0
cos cos ✓
B @x @y @z C
B@
C B sin cos ✓
@
@
B
C B
B
C=B
r
B @x @y @z C @
sin ✓
@ @✓ @✓ @✓ A
r cos
@x @y @z
@r
x
=p
= cos cos ✓,
@x
x2 + y 2 + z 2
@r
y
=p
= cos sin ✓,
2
@y
x + y2 + z2
@
=s
@x
1
=p
2
1 sin
sin sin ✓
,
r
1
z2
x2 + y 2 + z 2
1
1
1 sin2
cos
=
.
r
@w
@w @x @w @y
=
+
@v
@x @v
@y @v
@u
@w
u
= e sin v
+ eu cos v
@x
@y
x +y +z 2
r2
cos sin
( 1)
sin ✓
r
p
x2 + y 2 + z 2
C
C
C
A
@w
@w @x @w @y
=
+
@u
@x @u
@y @u
@w
@w
u
= e cos v
+ eu sin v
@x
@y
( 1)
cos sin
cos ✓
r
zp 2 y 2
1
1
proof )
x2 + y 2 + z 2
z2
x2 + y 2 + z 2
1
=p
@
=s
@z
x2 + y 2 + z 2
1
sin
cos
r
0
10. f
R2 –⌧ ¯Ñ •\ t¿⇠ h⇠t‡ w = f (x, y) tp,
u
x = e cos v, y = eu sin v | L, ‰L Ò›D ùÖX‹$.
"✓
✓
◆2 ✓
◆2
◆2 ✓
◆2 #
@w
@w
@w
@w
2u
+
=e
+
@x
@y
@u
@v
x2 + y 2 + z 2
1 sin2
sin cos ✓
,
r
@
=s
@y
=
x +y +z 2
z2
1
=p
=
zp 2 x 2
1
cos sin ✓
sin sin ✓
r
cos ✓
r cos
t‰.
@r
z
=p
= sin
2
@z
x + y2 + z2
t‰. ⇣\
11 ‰¿⇠h⇠@ ¯Ñ
zp 2 z 2
x +y +z 2
x2 + y 2 + z 2
z2
r3
t‡ 0|⌧
"✓
"
◆2 ✓
◆2 #
✓
◆2
✓
◆2 #
@w
@w
@w
@w
2u
2u
2u
2u
e
+
=e
e
+e
@u
@v
@x
@y
✓
◆2 ✓
◆2
@w
@w
=
+
@x
@y
t‰. t\h ùÖ@ ]¨‰.
1 ⇣
11. h⇠ f
R –⌧ ¯Ñ •X‡ w(t, r) = f t
r
⇣
⌘
p
r = x2 + y 2 + z 2 | L, ‰L Ò›D ùÖX‹$.
@2w
= a2
@t2
tp
⇣ y ⌘
@✓
1
y
sin ✓
=
=
=
,
y 2
@x
1 + (x)
x2
x2 + y 2
r cos
✓ ◆
@✓
1
1
x
cos ✓
=
= 2
=
2
y 2
@y
x
x
+
y
r
cos
1+ x
@✓
= 0.
@z
✓
@2w @2w @2w
+
+
@x2
@y 2
@z 2
r⌘
a
◆
proof ) ƒïY– Xt
@w
@w @r
=
@x
@r @x ⇣
1
=
f t
r2
120
r⌘
a
1 0⇣
f t
ar
r⌘
x
·p
,
2
a
x + y2 + z2
SOLUTION

@2w
2 ⇣
=
f t
@x2
r3
⇣
r⌘
2
+ 2 f0 t
a
ar
⇣
r⌘
1
+ 2 f 00 t
a
a r
x2
· 2
x + y2 + z2

1 ⇣
r⌘
1 0⇣
r⌘
f
t
+
f
t
r2
a
ar
a
h
1
2
2
2
2
2
· x +y +z
x x2 + y 2 + z 2
r⌘
a
3
2
proof ) ƒïY– Xt
@w
@w @r
=
@x
@r @x
i
⇣
r⌘
2
+ 2 f0 t
a
ar
⇣
r⌘
1
+ 2 f 00 t
a
a r
z2
· 2
x + y2 + z2

1 ⇣
r⌘
1 0⇣
r⌘
f
t
+
f
t
r2
a
ar
a
h
1
2
2
2
2
2
· x +y +z
z x2 + y 2 + z 2
ÑD L ⇠ à‰. ¯Ï¿\
@2w @2w @2w
+
+
@x2
@y 2
@z 2

⇣
⇣
⇣
2
r⌘
2
r⌘
1
= 3f t
+ 2 f0 t
+ 2 f 00 t
r
a
ar
a
a r

2 1 ⇣
r⌘
1 0⇣
r⌘
f t
+ f t
r r2
a
ar
a
1 00 ⇣
r⌘
= 2 f t
a r
a
2
1 @ w
= 2 2
a @t
x2 + y 2 + z 2
= f 0 (r) · p
x2 + y 2 + z 2
y
@w
@w @r
=
@z
@r @z
t‡ ⇣\
= f 0 (r) · p
,
z
@2w
y2
00
=
f
(r)
·
@y 2
x2 + y 2 + z 2
h
+ f 0 (r) · x2 + y 2 + z 2
i
@2w
z2
00
=
f
(r)
·
@z 2
x2 + y 2 + z 2
h
+ f 0 (r) · x2 + y 2 + z 2
r⌘
a
,
x2 + y 2 + z 2
@2w
x2
00
=
f
(r)
·
@x2
x2 + y 2 + z 2
h
+ f 0 (r) · x2 + y 2 + z 2
r⌘
a
3
2
x
= f 0 (r) · p
@w
@w @r
=
@y
@r @y
t‰. D∑\ )›<\

⇣
⇣
@2w
2 ⇣
r⌘
2
r⌘
1
r⌘
=
f t
+ 2 f0 t
+ 2 f 00 t
2
3
@y
r
a
ar
a
a r
a
2
y
· 2
x + y2 + z2

1 ⇣
r⌘
1 0⇣
r⌘
f
t
+
f
t
r2
a
ar
a
h
1
3i
2
2
2
2
2
· x +y +z
y x2 + y 2 + z 2 2 ,

@2w
2 ⇣
=
f t
@z 2
r3
11 ‰¿⇠h⇠@ ¯Ñ
1
2
1
2
1
2
x2 x2 + y 2 + z 2
y 2 x2 + y 2 + z 2
z 2 x2 + y 2 + z 2
t‰. 0|⌧,
3
2
3
2
i
,
i
,
3
2
i
@2w @2w @2w
+
+
@x2
@y 2
@z 2
1
= f 00 (r) + 3f 0 (r) p
2
x + y2 + z2
2
@ w 2 @w
=
+
@r2
r @r
f 0 (r) p
1
x2 + y 2 + z 2
t‰.
t‡ 0|⌧
@2w
= a2
@t2
✓
@2w @2w @2w
+
+
@x2
@y 2
@z 2
◆
t 1Ω\‰.
12.
R –⌧ ¯Ñ •X‡ w(x, y, z) = f (r) (r =
p h⇠ f
x2 + y 2 + z 2 ) | L, ‰L Ò›D ùÖX‹$.
13. x = r cos ✓ t‡ y = r cos ✓ | L, h⇠ f (x, y) = ln r +
@f
i✓ ((x, y) 6= (0, 0)) – t
(x, y) | lX‹$. (i î i2 = 1
@x
D ÃqXî ıå⇠)
sol) w = ln r + i✓ |‡ Xê. ¯Ït ,
@2w @2w @2w
d2 w 2 dw
+
+
=
+
@x2
@y 2
@z 2
dr2
r dr
121
@f
@w @r
@w @✓
=
+
@x
@r @x
@✓ @x
1 @r
@✓
=
+i
r @x
@x
SOLUTION
t‰. \∏ r(x, y) =
\
@r
x
=p
,
2
@x
x + y2
p
x2 + y 2 , ✓(x, y) = arctan(y/x) t¿ ⇣\ rf (P ) = (14, 13) t¿\
⇣ y ⌘
@✓
1
=
=
y 2
@x
1 + (x)
x2
Du f (P ) = u · rf (P )
✓
◆
1
1
p
p
=
,
· (14, 13)
2
2
p
27 2
=
2
y
x2 + y 2
t‰. 0|⌧
t‰.
12
12 ∏¯ÑX \©
@f
1
x
y
1
= p
+i 2
= 2 (x
2
2
2
@x
r x +y
x +y
r
t‰.
iy)
d) f (x, y, z) = 10✓+ xy + xz + yz,◆ P (1, 2, 1) , v = (1, 2, 1)
v
1
2
1
sol) u =
= p , p , p t‡ f ⇣ P –⌧ ¯Ñ •
kvk
6
6
6
Xp ⇣\ rf (P ) = (3, 2, 3) t¿\
∏¯ÑX \©
Du f (P ) = u · rf (P )
✓
1
2
= p ,p ,
6
6
2p
=
6
3
µ8⌧ 12.1 .
1. ¿ ⌧ ✓⇣ P –⌧
◆ °0 v )•<\ f X )•¯Ñƒ⇠
v
Du f (P )
u=
| lX‹$.
kvk
1
p
6
◆
· (3, 2, 3)
a) f (x, y) = ex cos
t‰.
✓ y, P (0, 0),
◆ v = (1, 1)
v
1
1
sol) u =
= p , p t‡ f ⇣ P –⌧ ¯Ñ •Xp
kvk
p
2
2
e) f (x, y, z) = x ✓yz, P (1, 4, 9)◆
, v =i+j k
⇣\ rf (P ) = (1, 0) t¿\
v
1
1
1
sol) u =
= p , p , p t‡ f ⇣ P –⌧ ¯Ñ •
kvk
3
3 ◆ 3
Du f (P ) = u · rf (P )
✓
✓
◆
3 1
Xp ⇣\ rf (P ) = 6, ,
t¿\
1
1
4 3
= p , p
· (1, 0)
2
2
1
Du f (P ) = u · rf (P )
=p
✓
◆ ✓
◆
2
1
1
1
3 1
= p ,p , p
· 6, ,
4 3
3
3
3
t‰.
p
77 3
=
⇣⇡ ⇡⌘
36
b) f (x, y) = cos x sin y, P
,
, v = (3, 4)
4 2
t‰.
✓
◆
v
3 4
sol) u =
=
,
t‡ f ⇣ P –⌧ ¯Ñ •Xp ⇣\
kvk
5◆ 5
✓
f) f (x, y, z) = xy✓+ exyz , P (1,
1
◆ 1, 0) , v = i k
p , 0 t¿\
rf (P ) =
v
1
1
2
sol) u =
= p , 0, p t‡ f ⇣ P –⌧ ¯Ñ •X
kvk
2
2
p ⇣\ rf (P ) = ( 1, 1, 1) t¿\
Du f (P ) = u · rf (P )
✓
◆ ✓
◆
3 4
1
Du f (P ) = u · rf (P )
p ,0
=
,
·
✓
◆
5 5
2
1
1
p
= p , 0, p
· ( 1, 1, 1)
3 2
2
2
=
10
=0
t‰.
t‰.
3
2
c) f (x, y) = x3 + 2xy
g) f (x, y, z) = xy✓
+ yz 2 + zx2◆
, P (2, 3, 4) , v = (0, 1, 1)
✓ + 3y ,◆ P (2, 1) , v = (1, 1)
v
1
1
v
1
1
sol) u =
= p , p t‡ f ⇣ P –⌧ ¯Ñ •Xp sol) u =
= 0, p , p t‡ f ⇣ P –⌧ ¯Ñ •X
kvk
kvk
2
2
2
2
122
SOLUTION
p
t¿\ \ )•¯Ñƒ⇠î krf (P )k = 4 70t‡, ¯ )•@
(1, 3, 5)t‰.
p ⇣\ rf (P ) = (25, 28, 28) t¿\
Du f (P ) = u · rf (P )
✓
◆
1
1
= 0, p , p
· (25, 28, 28)
2
2
=0
f) f (x, y, z) = xyz 2 , P (1, 1, 2)
sol) ⇣ P –⌧X \ )•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
rf (P ) = ( 4, 4, 4)
p
t¿\ \ )•¯Ñƒ⇠î krf (P )k = 4 3t‡, ¯ )•@
( 1, 1, 1)t‰.
t‰.
2. ¸¥ƒ ⇣ P –⌧ f X \
‹$.
)•¯Ñƒ⇠@ ¯X )•D lX
sol) ⇣ P –⌧X \
rf (P ) = (11, 33)
p
)•¯Ñƒ⇠î krf (P )k = 11 10t‡, ¯ )•@
b) f (x, y) = arctan
sol) ⇣ P –⌧X \
t¿\ \
(2, 1)t‰.
⇣
⇡⌘
P 1, ⇡,
2
)•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
g) f (x, y, z) = x2 z cos(xy),
a) f (x, y) = 2x + 3xy + 5y 2 , P (1, 3)
sol) ⇣ P –⌧X \ )•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
t¿\ \
(1, 3)t‰.
12 ∏¯ÑX \©
⇣y⌘
, P (1, 2)
x
)•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
✓
◆
2 1
rf (P ) =
,
5 5
rf (P ) = ( ⇡, 0, 1)
p
t¿\ \ )•¯Ñƒ⇠î krf (P )k = 1 + ⇡ 2 t‡, ¯ )•
@ ( ⇡, 0, 1)t‰.
3. h⇠ f (x, y) = 3x2 y 2 + 2x 6 i + jX )•<\ \
)•¯Ñƒ⇠| ƒ] Xî ®‡ ⇣D lX‹$.
sol) ⇣ p– t rf (p) i + j¸ ⇡@ )•| Ñ f \ )•
¯Ñƒ⇠| ƒ‰. f X ¯ò∏∏î
rf (x, y) = (6x + 2, 2y)
t¿\ 6x + 2 = k =
p
5
t‡, ¯ )•@
5
)•¯Ñƒ⇠î krf (P )k =
2y, (k > 0), â
x=
x ⇣ p(x, y)–⌧ \
k
2
6
,
y=
)•¯Ñƒ⇠|
k
2
ƒ‰.
4. Ò∞© ∞廧l
»‡ àî ⌘t‰. Ã} ∞X \tt
z✓ = 5
x◆2
2y 2 X ¯ò⌅@ ⇡@ ®ët‡, Ò∞ t ⇣
1 1 17
,
,
– à‰t, • h¨ ¥$ 0 ⌅\ )•@ ¥
2 2 4
x ?
✓
◆
1 1
e2 t‡, ¯ )•@
2
2
sol) z = f (x, y) = 5 x
2y , p =
,
\ Pt Ò∞ @
2 2
rf (p) )•<\ | • `tå ¥$⌅‰. f X ¯ò∏∏î
c) f (x, y) = x2 y + y 3 xy , P (e, 2)
sol) ⇣ P –⌧X \ )•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
rf (P ) = 0, 11
e2
t¿\ \ )•¯Ñƒ⇠î krf (P )k = 11
(0, 1)t‰.
d) f (x, y, z) = ex+y z , P (1, 1, 1)
sol) ⇣ P –⌧X \ )•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
rf (P ) = (e, e, e)
t¿\ \ )•¯Ñƒ⇠î krf (P )k =
(1, 1, 1)t‰.
p
rf (x, y) = ( 2x, 4y)
t¿\, Ò∞ t ∞–⌧
(1, 2)t‰.
• `tå ¥$ î )•@
rf (p) =
3et‡, ¯ )•@ 5. 2(– …tX \ ⇣ (x, y)–⌧ (ƒ
p
e) f (x, y, z) = xy 3 z 5 , P (2, 2, 2)
sol) ⇣ P –⌧X \ )•¯Ñƒ⇠î krf (P )k<\ ¸¥ƒ‰.
⇣ p
p
p ⌘
rf (P ) = 4 2, 12 2, 20 2
T (x, y) = 120e x
2
2y 2 x2 y 2
\ ¸¥L‰. ‰L <L– ıX‹$.
a) ⇣ P (1, 2)–⌧ ⇣ Q(3, 4)\X )•–
ƒ⇠| lX‹$.
123
\ (ƒ T X )•¯Ñ
SOLUTION
sol) Q
P = (2, 2)t‡ u =
)• u \X )•¯Ñƒ⇠î
✓
◆
1
1
p ,p
2
2
12 ∏¯ÑX \©
• `tå ¿Xî )•D lX‹$.
t¿\ T X ⇣ P –⌧ b) ⇣ P –⌧ ⌅⌅ V
sol) rf (P ) = (100, 0, 40) t¿\ (ƒ T
• `tå ¿Xî
)•@ ±(5, 0, 2)t‰.
Du f (P ) = u · rf (P )
✓
◆ ✓
1
1
1200
= p ,p
·
,
e13
2
2
p
1320 2
=
e13
1440
e13
◆
c) ⇣ P –⌧ ⌅⌅ V X \
sol)
)•¯Ñƒ⇠| lX‹$.
krf (P )k = k(100, 0, 40)k
p
= 1002 + 402
p
= 20 29
t‰.
p
)•¯Ñƒ⇠î 20 29t‰.
t¿\ ⇣ P –⌧X T X \
b) ⇣ P –⌧ (ƒ
• `tå
✓ T
◆ ù Xî )•D lX‹$.
1200 1440
sol) rf (P ) =
,
t¿\ (ƒ T
• `tå µ8⌧ 12.2 .
e13
e13
1. ¸¥ƒ h⇠X у⇣D ®P lX‹$.
ù Xî )•@ ( 5, 6)t‰.
c) ⇣ P –⌧ (ƒ T X \
sol)
krf (P )k =
)•¯Ñƒ⇠| lX‹$.
✓
r
1200
,
e13
1440
e13
◆
12002
14402
+ 26
26
e
e
p
240 p 2
240 61
2
= 13 5 + 6 =
e
e13
p
240 61
t¿\ ⇣ P –⌧X T X \ )•¯Ñƒ⇠î
t‰.
e13
=
5
sol) Ω) ›
rf (x, y) = (2x + 1, 2y) = (0, 0)
✓
◆
1
<\Ä0 f X у⇣@
, 0 t‰.
2
b) f (x, y) = 2x2 + 2xy + y 2
2x + 2y + 4
sol) Ω) ›
rf (x, y) = (4x + 2y
6. 3(– ı⌅X \ ⇣ (x, y, z)–⌧ ⌅⌅
V (x, y, z) = 10x2
a) f (x, y) = x2 + x + y 2
2, 2x + 2y + 2) = (0, 0)
<\Ä0 f X у⇣@ (2, 3)t‰.
6xy + 2xyz
4
c) f (x, y) = x4 + 4y 2 + x3
3
\ ¸¥L‰. ‰L <L– ıX‹$.
a) ⇣ P (5, 4, 3) –⌧ °0 v = i j k )•<\ ⌅⌅ V X )• sol) Ω) ›
¯Ñƒ⇠| lX‹$.
✓
◆
rf (x, y) = (4x3 + 4x2 , 8y) = (0, 0)
v
1
1
1
sol) u =
= p , p , p
t‡
kvk
3
3
3
<\Ä0 f X у⇣@ (0, 0), ( 1, 0)t‰.
rf (x, y, z) = (20x
6y + 2yz, 6x + 2xz, 2xy)
d) f (x, y) = e x
2
y 2 +4x
t¿\ ⇣ P –⌧ u )•<\X V X )•¯Ñƒ⇠î
Du f (P ) = u · rf (P )
✓
1
1
= p , p ,
3
3
p
= 20 3
t‰.
sol) Ω) ›
1
p
3
◆
· (100, 0, 40)
rf (x, y) = (( 2x + 4)e x
2
y 2 +4x
<\Ä0 f X у⇣@ (2, 0)t‰.
e) f (x, y) = (2x2 + 2y 2 )ex+y
124
, 2ye x
2
y 2 +4x
) = (0, 0)
SOLUTION
b) f (x, y) = 2x3
sol) Ω) ›
rf (x, y) = ((2x2 + 2y 2 + 4x)ex+y , (2x2 + 2y 2 + 4y)ex+y )
= (0, 0)
f) f (x, y, z) = x2 + xy + yz
12x2
3y 2
6xy
24x, 3x2
sol) Ω) ›
z3
t‡ ⇣ (0, 0)–⌧î
<\Ä0 f X у⇣@ (0, 0, 0),
✓
◆
2 4 2
, ,
t‰.
3 3 3
2. ¸¥ƒ h⇠X ®‡ у⇣D lX‡ ÑXX‹$.
x2 y
2xy
y 2 , 3x
x2
2xy) = (0, 0)
<\Ä0 f X у⇣@ (0, 0), (0, 3), (3, 0), (1, 1)t‰. \∏,
✓
◆
2y
3 2x 2y
Hf (x, y) =
3 2x 2y
2x
24
0
✓
0
det Hf (0, 0) = det
3
t¿\ ⇣ (0, 0)@ H•⇣t‰.
¯¨‡ ⇣ (0, 3)–⌧î
✓
6
det Hf (0, 3) = det
3
t¿\ ⇣ (0, 3)@ H•⇣t‰.
¯¨‡ ⇣ (3, 0)–⌧î
✓
0
det Hf (3, 0) = det
3
◆
3
=
0
9<0
◆
c) f (x, y) = e x
9<0
3
6
◆
=
9<0
2
1
◆
1
=3>0
2
2 < 0t¿\ ⇣ (1, 1)@ ˘ ⇣t‰.
◆
=
432 < 0
2
y2
) = (0, 0)
y2
sol) Ω) ›
rf (x, y) = ( 2xe x
2
y2
, 2ye x
<\Ä0 f X у⇣@ (0, 0)t‰. \∏,
✓ 2
2
x2 y 2 4x
Hf (x, y) = e
4xy
✓
2
0
◆
4xy
4y 2 2
◆
0
=4>0
2
t‡ fxx (0, 0) < 0t¿\ ⇣ (0, 0)@ ˘ ⇣t‰.
d) f (x, y) = xye x
t¿\ ⇣ (3, 0)@ H•⇣t‰.
⇣, ⇣ (1, 1)–⌧î
✓
2
det Hf (0, 0) = det
=
24
6
t¿\ ⇣ ( 4, 8)@ H•⇣t‰.
t¿\ ⇣ (0, 0)–⌧î
3
0
◆
0
= 144 > 0
6
t‡ fxx (0, 0) < 0t¿\ ⇣ (0, 0)@ ˘ ⇣t‰.
¯¨‡ ⇣ (2, 2)–⌧î
✓
◆
12
12
det Hf (2, 2) = det
= 216 < 0
12
6
t‡ ⇣ (0, 0)–⌧î
det Hf (1, 1) = det
✓
t¿\ ⇣ (2, 2)@ H•⇣t‰.
¯¨‡ ⇣ ( 4, 8)–⌧î
✓
24
det Hf ( 4, 8) = det
24
xy 2
sol) Ω) ›
rf (x, y) = (3y
det Hf (0, 0) = det
3z 2 ) = (0, 0, 0)
rf (x, y, z) = (2x + y, x + z, y
6y) = (0, 0)
<\Ä0 f X у⇣@ (0, 0), (2, 2), ( 4, 8)t‰. \∏,
✓
◆
12x 6y 24
6x
Hf (x, y) =
6x
6
sol) Ω) ›
t‡ fxx (1, 1) =
3x2 y
rf (x, y) = (6x2
<\Ä0 f X у⇣@ (0, 0), ( 1, 1)t‰.
a) f (x, y) = 3xy
12 ∏¯ÑX \©
2
y2
sol) Ω) ›
125
2x2 y)e x
2
y2
2
2
2xy 2 )e x y ) = (0, 0)
✓
◆ ✓
◆
1
1
1
1
p
p
p
p
<\Ä0 f X у⇣@ (0, 0),
,
,
,
,
2
2
2
✓
◆ ✓
◆ 2
1
1
1
1
p , p
p , p t‰. \∏,
,
2
2
2
2
✓
◆
2
2
2xy(2x2 3)
(2x2 1)(2y 2 1)
Hf (x, y) = e x y
(2x2 1)(2y 2 1)
2xy(2y 2 3)
rf (x, y) = ((y
, (x
SOLUTION
t‡ ⇣ (0, 0)–⌧î
t¿\ ⇣ (0, 1, 0)–⌧î
✓
0
det Hf (0, 0) = det
1
◆
1
=
0
t‰.
✓
1
p ,
2
1
p
2
◆
e) f (x, y, z) = x2 + yez
2
< 0t¿\ ⇣
y
✓
1
p ,
2
u1 = (1, 0, 0),
y)e
, 2yze
z2 y
0
2
(y 2)ez y
2
2z(1 y)ez y
1
0
0 A
2e 1
u2 = (0, 1, 0)
<\ Pt
uT1 Hf u1 = 2 > 0,
uT2 Hf u2 =
t¿\ ⇣ (0, 1, 0)@ H•⇣t‰.
ƒt) Ä] 12.4X ¥©D ©Xt
0
2
0
e 1
det Hf (0, 1, 0) = det @0
0
0
det
✓
2
0
0
e 1
◆
=
e 1<0
1
0
0 A=
2e 1
4e 2 < 0,
2e 1 < 0
t‡ fxx (0, 1, 0) > 0t¿\ ⇣ (0, 1, 0)@ H•⇣t‰.
3. h⇠ f (x, y) = 2x3 + xy 2 + 5x2 + y 2 X у⇣‰ ⌘ H•⇣D
lX‹$.
sol) Ω) ›
rf (x, y) = (6x2 + 10x + y 2 , 2y(x + 1)) = (0, 0)
✓
◆
5
<\Ä0 f X у⇣@ ( 1, 2), ( 1, 2), (0, 0),
, 0 t‰.
3
\∏,
✓
◆
12x + 10
2y
Hf (x, y) =
2y
2x + 2
det Hf ( 1, 2) = det
) = (0, 0, 0)
<\Ä0 f X у⇣@ (0, 1, 0)t‰. \∏,
0
2
@
Hf (x, y, z) = 0
0
t‰. t⌧ P Ë⌅°0 u1 , u2 |
(12.4 )
z2 y
0
e 1
0
◆
1
p @˘ ⇣
2
t‡ ⇣ ( 1, 2)–⌧î
sol) Ω) ›
rf (x, y, z) = (2x, (1
0
2
Hf (0, 1, 0) = @0
0
1<0
t¿\ ⇣ (0,
✓ 0)@ H•⇣t‰.
◆
1
1
¯¨‡ ⇣ p , p –⌧î
2
2
✓
◆
✓
◆
1
1
2 0
2
det Hf p , p
= e · det
= 4e 2 > 0
0
2
2
2
✓
◆
✓
◆
1
1
1
1
t‡ fxx p , p
< 0t¿\ ⇣ p , p @ ˘ ⇣t‰.
2
2
2
✓2
◆
1
1
p , p –⌧î
¯¨‡ ⇣
2
2
✓
◆
✓
◆
1
1
2 0
2
p ,p
det Hf
= e · det
= 4e 2 > 0
0 2
2
2
✓
◆
✓
◆
1
1
1
1
p ,p
p , p @ ˘å⇣t
t‡ fxx
> 0t¿\ ⇣
2
2
2
2
‰.
✓
◆
1
1
¯¨‡ ⇣ p , p –⌧î
2
2
✓
◆
✓
◆
1
1
2 0
2
det Hf p , p
= e · det
= 4e 2 > 0
0 2
2
2
✓
◆
✓
◆
1
1
1
1
t‡ fxx p , p
> 0t¿\ ⇣ p , p @ ˘å⇣t
2
2
2
2
‰.
✓
◆
1
1
p , p –⌧î
¯¨‡ ⇣
2
2
✓
◆
✓
◆
1
1
2 0
p , p
det Hf
= e 2 · det
= 4e 2 > 0
0
2
2
2
t‡ fxx
12 ∏¯ÑX \©
1
0
2
2z(1 y)ez y A
2
2y(1 + 2z 2 )ez y
✓
2
4
t¿\ ⇣ ( 1, 2)@ H•⇣t‰.
¯¨‡ ⇣ ( 1, 2)–⌧î
✓
2
det Hf ( 1, 2) = det
4
◆
4
=
0
4
0
◆
16 < 0
=
16 < 0
t¿\ ⇣ ( 1, 2)@ H•⇣t‰.
¯¨‡ ⇣ (0, 0)–⌧î
✓
◆
10 0
det Hf (0, 0) = det
= 20 > 0
0 2
126
SOLUTION
t¿\ ⇣ (0,
✓ 0)@◆H•⇣t D»‰.
5
¯¨‡ ⇣
, 0 –⌧î
3
det Hf
✓
5
,0
3
◆
t‡ y =
10
= det
0
4
3
0
!
=
40
>0
3
◆
5
t¿\ ⇣
, 0 @ H•⇣t D»‰.
3
¯Ï¿\ H•⇣@ ( 1, 2), ( 1, 2) t‰.
0D ®P ÃqXî
8⌧X✓ Ì@
X
-”⇣@
◆ ✓ ¨ ◆ t‡ ¨
1
14
(0, 0), (1, 0) ,
, 4 , 0,
t‰. f (x, y) = 4x + y|‡ Pt
2
3
✓
◆
✓
◆
1
14
14
f (0, 0) = 0, f (1, 0) = 4, f
, 4 = 6, f 0,
=
,
2
3
3
✓
◆
1
t‰. ¯Ï¿\ • p ✓x f
, 4 = 6î ˘◆✓t‡, • ë
2
@ ✓x f (0, 0) = 0î ˘ü✓t‰. t⌧ ò8¿ ⇣‰t ˘⇣x¿
⇣ƒtÙê.
sol)
y), 3(y 2
x)) = (0, 0)
<\Ä0 f X у⇣@ (0, 0), (1, 1)t‰. \∏,
✓
◆
6x
3
Hf (x, y) =
3 6y
t‡ ⇣ (0, 0)–⌧î
✓
0
3
3
0
◆
(i) ⇣ (1, 0) : y = 0,
=
f (x, 0) = 4x
t‡
y=8
✓
6
3
0x1|L
9<0
t¿\ ⇣(0, 0)@ H•⇣t‰.
¯¨‡ ⇣ (1, 1)–⌧î
det Hf (1, 1) = det
⇣ x
⌘
f x, + 1 = 4x + 4
2
ƒ‰. ¯
sol) Ω) ›
det Hf (0, 0) = det
2  x  0| L
b) 4x + 3y  14, 8x + y  8, x
0, y
Ì–⌧ 4x + yX ˘✓D lX‹$.
3xy + y 3 + 5î H•⇣D Xò
rf (x, y) = (3(x2
x
+ 1,
2
t¿\ ¨
X Ωƒ ⌅–⌧ ⇣ (0, 1) ¸)–⌧X f X ù⇣
t ¿X¿ Jî‰. â, (0, 1)@ ˘⇣t D»‰. 0|⌧ ˘⇣@
(1, 0), ( 2, 0), (0, 3)<\
8 ⌧t‡, ˘ ✓@ 5, ˘ü✓@ -4
t‰.
✓
4. h⇠ f (x, y) = x3
⇣D lX‹$.
12 ∏¯ÑX \©
◆
3
= 27 > 0
6
1
x1|L
2
8x,
f (x, 8
8x) = 8
4x
t¿\ ¨
X Ωƒ ⌅–⌧ ⇣ (1, 0) ¸)–⌧X f X ù
⇣t ¿X¿ Jî‰. â, (1, 0)@ ˘⇣t D»‰.
✓
◆
14
14
(ii) ⇣ 0,
: x = 0, 0  y 
|L
3
3
t¿\ ⇣ (1, 1)@ H•⇣t D»‰.
¯Ï¿\ H•⇣@ (0, 0) t‰.
5. ‰
®ëX —i–⌧ f (x, y) = ax + by + cX ˘◆✓¸ ˘ü
✓@ ò -”⇣–⌧à ª¥ƒ‰î ¨‰D t©XÏ ‰L <L–
ıX‹$.
f (0, y) = y
t‡
x=
a) $ ⇣ (0, 1), (1, 0), ( 2, 0), (0, 3)D -”⇣<\ Xî ¨
–⌧ 3x + 2y + 2X ˘✓D lX‹$.
14
3y
4
,
4y
f
✓
14
4
14
|L
3
◆
3y
, y = 14
◆
14
sol) f (x, y) = 3x + 2y + 2|‡ Pê. ¯Ït
t¿\ ¨
X Ωƒ ⌅–⌧ ⇣ 0,
¸)–⌧X f X
✓
◆ 3
14
f (0, 1) = 4, f (1, 0) = 5, f ( 2, 0) = 4, f (0, 3) = 4
ù⇣t ¿X¿ Jî‰. â, 0,
@ ˘⇣t D»‰.
3
✓
◆
t¿\ • p ✓x f (1, 0) = 5î ˘◆✓t‡, • ë@✓x
f ( 2, 0) = f (0, 3) = 4î ˘ü✓t‰. t⌧ ò8¿ ⇣‰t 0|⌧ ˘⇣@ 1 , 4 , (0, 0)<\
P ⌧t‡, ˘◆✓@ 6, ˘
2
˘⇣x¿ ⇣ƒtÙê. y = 1 x, 0  x  1 | L
ü✓@ 0 t‰.
f (x, 1 x) = x + 4
127
✓
2y
SOLUTION
µ8⌧ 12.3 .
1. ¸¥ƒ ⌧}pt–
12 ∏¯ÑX \©
0|⌧
t f X \◆✓ ⇣î \ü✓D lX‹$.
p
⇣ p
p ⌘ 18 3/5
f ± 2/5, 3/5 =
,
f (0, ±1) = 0,
25 p
a) x2 + y 2 = 1, f (x, y) = x2 + y 3 X \◆✓¸ \ü✓
⇣ p
⌘
p
18 3/5
sol) g(x, y) = x2 + y 2 1\ Pê. pt g(x, y) = 0D ÃqXî ⇣
f ± 2/5,
3/5 =
, f (±1, 0) = 0
25
‰X —iD D\ Pt Dî ƒ Îå —it‡ f t ⌅–⌧ 
p
çt¿\ f î D–⌧ \◆✓¸ \ü✓D ƒ‰. ⇣\, (x, y) 2 D
18 3/5
– t
t¿\ \◆✓@
t‰.
25
rg(x, y) = (2x, 2y) 6= (0, 0)
x2
y2
c)
+ y 2 = 1, f (x, y) = 2x x3
X \ü✓
4
3
t¿\ |¯ë¸ π⇠ï– Xt
x2
sol) g(x, y) =
+ y 2 1\ Pê. pt g(x, y) = 0D Ãq
4
rf (x, y) = rg(x, y) =) (2x, 3y 2 ) = (2x, 2y)
Xî ⇣‰X —iD D\ Pt Dî ƒ Îå —it‡ f t
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä ⌅–⌧ çt¿\ f î D–⌧ \◆✓¸ \ü✓D ƒ‰. ⇣\,
(x, y) 2 D– t
0
p
⇣x
⌘
(i) x, y 6= 0| L = 1 =) y = 2/3, x = ± 5/3,
rg(x, y) =
, 2y 6= (0, 0)
2
(ii) x = 0tt y = ±1,
t¿\ |¯ë¸ π⇠ï– Xt
(iii) y = 0tt x = ±1t‰.
✓
◆
⇣x
⌘
2
0|⌧
rf (x, y) = rg(x, y) =) 2 3x2 ,
y =
, 2y
3
2
⇣ p
⌘ 23
f ± 5/3, 2/3 =
, f (0, ±1) = ±1, f (±1, 0) = 1
27
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0
t¿\ \◆✓@ 1, \ü✓@ -1 t‰.
1
(i) =
tt
3
2
2
2 3
b) x + y = 1, f (x, y) = 3x y X \◆✓
sol) g(x, y) = x2 + y 2 1\ Pê. pt g(x, y) = 0D ÃqXî ⇣
x
1
x
= (2 3x2 ) =)
= 3(2 3x2 )
‰X —iD D\ Pt Dî ƒ Îå —it‡ f t ⌅–⌧ 
2
2
8
r
p
p
çt¿\ f î D–⌧ \◆✓¸ \ü✓D ƒ‰. ⇣\, (x, y) 2 D
>
1
865
1
2159
865
>
<x =
– t
+
=) y = ±
,
36
36
36
2
r
p
p2
=)
>
865
1 2159
865
>
:x = 1
rg(x, y) = (2x, 2y) 6= (0, 0)
=) y = ±
+
,
36
36
36
2
2
t¿\ |¯ë¸ π⇠ï– Xt
(ii) y = 0tt x = ±2 t‰.
rf (x, y) = rg(x, y) =) (6xy 3 , 9x2 y 2 ) = (2x, 2y)
0|⌧
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0
1
s
p
p
p
0
865 A 865 865 6479
@ 1 + 865 , ± 1 2159
f
=
36
36
36
2
2
23328
(i) x, y 6= 0 | L 3y 3 = , 9x2 y = 2 t¿\
⇡ 0.8218,
0
1
s
p
p
1
865
1 2159
865 A
f@
,±
+
=
36
36
36
2
2
2 = 9x2 y =) 9x2 y = 6y 3
=) 3x2 = 2y 2
8
q
q
<x = 2 =) y = ± 3 ,
5
5
q
q
=)
2
3
:x =
=)
y
=
±
5
5,
(ii) x = 0tt y = ±1,
(iii) y = 0tt x = ±1t‰.
⇡
1.36828,
f (±2, 0) = ⌥4
t¿\ \ü✓@ -4 t‰.
128
p
865 865 6479
23328
SOLUTION
✓
◆
x2
y2
d) xy = 8, f (x, y) = xy exp
X \◆✓
4
4
sol) g(x, y) = xy 8\ Pê. pt g(x, y) = 0D ÃqXî ⇣‰X
—iD D\ Pt (x, y) 2 D– t
(ii) x 6= 0 tt
y
e |y| cos(x2 ) = 2y
|y|
cos(x2 )
=) = 9 sin(x2 )e |y| = e |y|
2|y|
1
=) |y| =
18 tan(x2 )
x2
1
=)
+ 2
=1
9
18 tan2 (x2 )
2x sin(x2 )e |y| =
rg(x, y) = (y, x) 6= (0, 0)
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y) = rg(x, y)
✓ ✓
◆
✓
x2 +y 2
x2
4
=) y 1
e
,x 1
2
y2
2
◆
e
x2 +y 2
4
◆
12 ∏¯ÑX \©
2x
,
9
t¿\ t| ƒ∞Xt ¸¨ <\
= (y, x)
x = ± 2, 9853, ±2.5270, ±2.4868,
± 1.7919, ±1.7531, ±0.2359
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0 x = yt‡
p
p
(i) x = 2 2 =) y = 2 2,
p
p
(ii) x = 2 2 =) y = 2 2
t‡
1
> 0t¥| X¿\
18 tan(x2 )
•\ ⇣‰@
(±2.5270, ±0.5399), (⌥2.5270, ±0.5399),
(±1.7919, ±0.8002), (⌥1.7919, ±0.8002),
(±0.2359, ±0.9973), (⌥0.2359, ±0.9973)
t‰. 0|⌧,
t‰.
p
p
8
f (±2 2, ±2 2) = 4
e
⌅–⌧ l\ ⇣‰–⌧X h+✓‰@ ¸¨ <\
f (±2.5270, ±0.5399) = f (⌥2.5270, ±0.5399) = 0.5798,
8
t¿\ \◆✓@ 4 t‰.
e
f (±1.7919, ±0.8002) = f (⌥1.7919, ±0.8002) =
f (±0.2359, ±0.9973) = f (⌥0.2359, ±0.9973) = 0.3684,
e) ƒ∞ ⌅\¯® ¨© å• ✏ 2021Dƒ ⌧ ⇣X Ω∞ 8⌧–
$X à<¿\ (\ , \ü✓ ⌅Ä ⌧}pt ¥–⌧ t¨X¿
JL) ⌧ ⌅X 8⌧| ¯ \ ⇠]i»‰.
x2
+ y 2 = 1, f (x, y) = e |y| cos(x2 )X \ü✓
9
x2
sol) g(x, y) =
+ y 2 1\ Pê. pt g(x, y) = 0D Ãq
9
Xî ⇣‰X —iD D\ Pt Dî ƒ Îå —it‡ f t
⌅–⌧ çt¿\ f î D–⌧ \◆✓¸ \ü✓D ƒ‰. ⇣\,
(x, y) 2 D– t
rg(x, y) =
0.4473,
✓
2x
, 2y
9
◆
f (0, ±1) = 1/e ⇡ 0.3679
t‰. »¿…<\ y = 0| L, f (±3, 0) = cos 9 ⇡
\ t| ÖiXt \ü✓@ cos 9 t‰.
0.9111 t¿
f) x2 + 2y 2 + 2z 2 = 5, f (x, y, z) = xyzX \◆✓¸ \ü✓
sol) g(x, y) = x2 + 2y 2 + 2z 2 5\ Pê. pt g(x, y) = 0D
ÃqXî ⇣‰X —iD D\ Pt (x, y) 2 D– t
rg(x, y) = (2x, 4y, 4z) 6= (0, 0)
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y) = rg(x, y) =) (yz, xz, xy) =
6= (0, 0)
(2x, 4y, 4z)
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0 x, y, z 6= 0 | L 6= 0 t‡
t¿\ |¯ë¸ π⇠ï– Xt y 6= 0 tt
8
1
>
<x = 2 yz,
1 2 2 2
1
=) xyz =
x y z =) xyz = 32 3 ,
y = 4 xz,
rf (x, y) = g(x, y)
3
>
32
✓
◆
✓
◆
:
z = 41 xy
y
2x
=)
2x sin(x2 )e |y| ,
e |y| cos(x2 ) =
, 2y
8
1
2
|y|
9
>
<x = 2 xyz,
3
=) x2 + 2y 2 + 2z 2 =
xyz = 5
2y 2 = 21 xyz,
>
2
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
: 2
1
2z = 2 xyz
0
10
=) xyz =
(i) x = 0tt y = ±1,
3
129
SOLUTION
t¿\
1
=±
4
r
5
t‰. ⇣\,
3
y = 0 =) z = ±
x = 0,
z = 0 =) y = ±
y = 0,
(iii) f (x, y)
= 0| L
x = 0,
p
p
p
(iv) f (x, y)
x=
1 = (x
X⌧ h⇠ f X \◆✓¸ \ü✓D
2y
1  y  1x î⌅–⌧î
1 = (y
rf (x, y) = (6x, 2y) = (0, 0)
<\Ä0 f X у⇣@ (0, 0)t‡ f (0, 0) = 0t‰. t⌧ RX
Ωƒ· – t \ , \å| p¨tÙt
(i) f (x, y)
x = 0t‡ 0  y  2x î⌅–⌧î
11
(0  y  2)
t¿\ y = 0, y = 2–⌧ ˘✓D ƒ‰. ¯ ✓‰D DP
tÙt f (0, 0) = 0, f (0, 2) = 4 t¿\ ¸¥ƒ î⌅–⌧
f (0, y)X \◆✓@ 4, \ü✓@ 0t‰.
(ii) f (x, y)
0  x  2t‡ y = 0x î⌅–⌧î
f (x, 0) = 3x2
( 1  x  1)
1  y  1x î⌅–⌧î
f (1, y) = y + 8y + 5 = (y + 4)
0D
sol) Ω) ›
t¿\ x = 1, x = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D DPt
Ùt f ( 1, 1) = 2, f (1, 1) = 14 t¿\ ¸¥ƒ î⌅–⌧
f (x, 1)X \◆✓@ 14, \ü✓@ 2t‰.
2
( 1  y  1)
0, y
1  x  1t‡ y = 1x î⌅–⌧î
2
2
b) f (x, y) = 3x2 + y 2 , R@ x2 + y 2  4@ x
ÃqXî Ì
rf (x, y) = (2x + 5y + 3, 5x + 2y + 3) = (0, 0)
✓
◆
✓
◆
3 3
3 3
2
<\Ä0 f X у⇣@
,
t‡ f
,
=
7 7
7 7
7
t‰. t⌧ RX Ωƒ· – t \ , \å| p¨tÙt
x = 1t‡
1)2
Ì R–⌧ f X
f (0, y) = y 2
11
( 1  x  1)
2
»¿…<\ ⌅–⌧ l\ f X ✓‰D DPtÙt
\◆✓@ 14t‡ \ü✓@ 2t‰.
sol) Ω) ›
f (x, 1) = x2 + 8x + 5 = (x + 4)2
1)2
t¿\ y = 1, y = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D D
PtÙt f ( 1, 1) = 2, f ( 1, 1) = 2 t¿\ ¸¥ƒ
î⌅–⌧ f (1, y)X \◆✓@ 2, \ü✓@ 2t‰.
a) f (x, y) = (x + y)2 + 3(x + y + xy) + 1, R@ |x|  1¸ |y|  1
D ÃqXî Ì
(ii) f (x, y)
1t‡
f ( 1, y) = y 2
\
14⌧t‰. Dò 6⌧ ⇣–⌧î h+✓t ®P 0 t‰. t⌧
ò8¿ 8⌧ ⇣–⌧X h+✓D p¨Xê. ∏X| Xt ⇣‰X
1ÑX Ä8Ã ò¿¥t
p
5 15
f ( , , +) = f ( , +, ) = f (+, , ) = f (+, +, +) =
,
18
p
5 15
f (+, +, 1) = f (+, , +) = f ( , +, +) = f ( , , ) =
18
p
p
5 15
5 15
t¿\ \◆✓@
, \ü✓@
t‰.
18
18
(i) f (x, y)
2x
1x î⌅–⌧î
t¿\ x = 1, x = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D D
PtÙt f ( 1, 1) = 2, f (1, 1) = 2 t¿\ ¸¥ƒ
î⌅–⌧ f (x, 1)X \◆✓@ 2, \ü✓@ 2t‰.
5/2,
z = 0 =) x = ± 5,
Ì R–⌧
1  x  1t‡ y =
f (x, 1) = x2
5/2,
t¿\ t‰D ¨©Xt ¥¥¸ ⇣‰@
⇣ p
⌘ ⇣ p
⌘
p
p
p
p
± 5/3, ± 5/6, ± 5/6 , ⌥ 5/3, ± 5/6, ± 5/6 ,
⇣ p
⌘ ⇣ p
⌘
p
p
p
p
± 5/3, ⌥ 5/6, ± 5/6 , ± 5/3, ± 5/6, ⌥ 5/6 .
⇣
⌘ ⇣
⌘ ⇣ p
⌘
p
p
0, 0, ± 5/2 , 0, ± 5/2, 0 , ± 5, 0, 0
2. ¿ ⌧ …t
lX‹$.
12 ∏¯ÑX \©
t¿\ x = 0, x = 2–⌧ ˘✓D ƒ‰. ¯ ✓‰D DP
tÙt f (0, 0) = 0, f (2, 0) = 12 t¿\ ¸¥ƒ î⌅–⌧
f (x, 0)X \◆✓@ 12, \ü✓@ 0t‰.
(iii) f (x,p
y)
y= 4
x2 + y 2 = 4t‡ x
0, y
x2 (0  x  2) t¿\
f (x,
( 1  y  1)
t¿\ y = 1, y = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D DPt
Ùt f (1, 1) = 2, f (1, 1) = 14 t¿\ ¸¥ƒ î⌅–⌧
f (1, y)X \◆✓@ 14, \ü✓@ 2t‰.
130
(0  x  2)
p
4
x2 ) = 4 + 2x2
0x î⌅–⌧î
(0  x  2)
t‡ x = 0, x = 2–⌧ ˘✓D ƒ‰. ¯ ✓‰D DPt
Ùt pf (0, 2) = 4, f (2, 0) = 12 t¿\ ¸¥ƒ î⌅–⌧
f (x, 4 x2 )X \◆✓@ 12, \ü✓@ 4t‰.
SOLUTION
»¿…<\ ⌅–⌧ l\ f X ✓‰D DPtÙt
\◆✓@ 12t‡ \ü✓@ 0t‰.
c) f (x, y) =
Ì
x2
+ y2
2
2x
Ì R–⌧ f X
12 ∏¯ÑX \©
(i) f (x, y)
sol) Ω) ›
2) = (0, 0)
( 1  y  1)
1t‡ 0  x  1x î⌅–⌧î
y=x
f (x, x
<\Ä0 f X у⇣@ (2, 1)t¿Ã tî Ì R– çX¿ Jî
‰. t⌧ RX Ωƒ· – t \ , \å| p¨tÙt
y2
t¿\ y = 1, y = 0, y = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D
DPtÙt f (0, 1) = 1, f (0, 0) = 0, f (0, 1) = 1 t
¿\ ¸¥ƒ î⌅–⌧ f (0, y)X \◆✓@ 0, \ü✓@ 1
t‰.
(ii) f (x, y)
2, 2y
1  y  1x î⌅–⌧î
f (0, y) =
2y + 4, R@ x2 + 2y 2  4| ÃqXî
rf (x, y) = (x
x = 0t‡
1) = 2x
1
(0  x  1)
t¿\ x = 0, x = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D DPt
Ùt f (0, 1) = 1, f (1, 0) = 1 t¿\ ¸¥ƒ î⌅–⌧
f (x, x 1)X \◆✓@ 1, \ü✓@ 1t‰.
2
(i) f (x, p
y) x2 + 2yp
= 4t‡
p 0  x  2x î⌅–⌧î (iii) f (x, y) y = x + 1t‡ 0  x  1x î⌅–⌧î
2
x = 4 2y (
2  y  2) t¿\
⇣ p
p
p
p ⌘
f (x, x + 1) = 2x 1 (0  x  1)
f ( 4 2y 2 , y) = 2 4 2y 2 2y + 6
2y 2
t¿\ x = 0, x = 1–⌧ ˘✓D ƒ‰. ¯ ✓‰D DP
t‡
tÙt f (0, 1) = 1, f (1, 0) = 1 t¿\ ¸¥ƒ î⌅–⌧
p
f (x, x + 1)X \◆✓@ 1, \ü✓@ 1t‰.
4y
f 0 ( 4 2y 2 , y) = p
2=0
4 2y 2
»¿…<\ ⌅–⌧ l\ f X ✓‰D DPtÙt Ì R–⌧ f X
r
r
\◆✓@ 1t‡ \ü✓@ 1t‰.
p
2
2
–⌧ y =
t¿\ f ( 4 2y 2 , y)î y =
–⌧ ˘✓
3
3
⇣ p
⌘
p
p
2
x2 y 2 xy 3 , R@ 8 ⇣ (0, 0), (6, 0), (0, 6)D
D ƒ‰. ¯ ✓@ f 2 2/3, 2/3 = 6 2 6 t‰. ⇣, e) f (x, y) = 4xy
p
p
p
-”⇣<\ Xî º
Ì
]⇣–⌧î f 0,
2 = f 0, 2 = 6 + 2 2t‰.
(ii) f (x, y)p x2 + 2y 2 p
= 4t‡ p2  x  0x î⌅–⌧î sol) Ω) ›
x=
4 2y 2 (
2  y  2) t¿\
p
p
p
p
rf (x, y) = ( y 2 (2x + y 4), xy( 2x 3y + 8)) = (0, 0)
f(
4 2y 2 , y) = 2 4 2y 2 2y + 6 (
2  y  2)
<\Ä0 f X у⇣@ (x, 0), (0, 4), (1, 2)t‡ f (x, 0) = 0,
t‡
f (0, 4) = 0, f (1, 2) = 4t‰. t⌧ RX Ωƒ· – t \
p
4y
, \å| p¨tÙt
0
2
f(
4 2y , y) = p
2=0
4 2y 2
r
r
(i) f (x, y) x = 0t‡ 0  y  6x î⌅–⌧î
p
2
2
–⌧ y =
t¿\ f (
4 2y 2 , y)î y =
–⌧
3
3
⇣ p
⌘
f (0, y) = 0 (0  y  6)
p
p
˘✓D ƒ‰. ¯ ✓@ f
2 2/3,
2/3 = 6 + 2 6
t‰.
t‰.
(ii) f (x, y) 0  x  6t‡ y = 0x î⌅–⌧î
»¿…<\ ⌅–⌧
Ì R–⌧ f X
p l\ f X ✓‰D DPtÙt
p
\◆✓@ 6 + 2 6t‡ \ü✓@ 6 2 6t‰.
f (x, 0) = 0 (0  x  6)
d) f (x, y) = x2
\ Xî º
y 2 , R@ 8 ⇣ (1, 0), (0, 1), (0, 1)D -”⇣<
Ì
t‰.
(iii) f (x, y)
y=
x + 6t‡ 0  x  6x î⌅–⌧î
f (x, x + 6) =
sol) Ω) ›
2x3 + 24x2
72x
(0  x  6)
t‡
rf (x, y) = (2x, 2y) = (0, 0)
<\Ä0 f X у⇣@ (0, 0)t‡ f (0, 0) = 0t‰. t⌧ RX
Ωƒ· – t \ , \å| p¨tÙt
131
f 0 (x, x + 6) =
6x2 + 48x
72 =
6(x
2)(x
6) = 0
–⌧ x = 2, x = 6t¿\ f (x, x + 6)î x = 2, x = 6–⌧
˘✓D ƒ‰. ¯ ✓‰@ f (2, 4) = 64, f (6, 0) = 0t‰.
SOLUTION
»¿…<\ ⌅–⌧ l\ f X ✓‰D DPtÙt
\◆✓@ 4t‡ \ü✓@ 64t‰.
12 ∏¯ÑX \©
Ì R–⌧ f X t¿\ |¯ë¸ π⇠ï– Xt
✓
rf (x, y, z) = g(x, y, z) () (2x, 2y, 2z) =
f) f (x, y) = arctan (xy), R@ 8 ⇣ (0, 0), (4, 0), (0, 1)D -”⇣
| ÃqXî
<\ Xî º
Ì
0
(i)
sol) Ω) ›
rf (x, y) =
✓
y
x
,
1 + x2 y 2 1 + x2 y 2
◆
(ii)
= (0, 0)
x = 0t‡ 0  y  1x î⌅–⌧î
f (0, y) = 0
2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
p
= 2 =) x = ± 2,
p
= 3 =) z = ± 3,
p
f (0, 0, ± 3) = 3
p
t¿\ ·t¸ –⇣⌅X \åp¨î 2t‰.
(0  y  1)
f (x, y, z) = x2 + y 2 + z 2
– ·tX ) ›t ⌧}pt<\ ¸¥ƒ Ω∞X \ü✓X ë
X ⌧Ò¸<\ ¸¥ƒ‰. g(x, y, z) = z 2
4
xy\ ì<t
g(x, y, z) = 0| ÃqXî Ì D–⌧ (x, y) 2 D| L
(0  x  4)
rg(x, y, z) = ( y, x, 2z) 6= (0, 0, 0)
t‰.
(iii) f (x, y)
f (4
x=4
t¿\ |¯ë¸ π⇠ï– Xt
4yt‡ 0  y  1x î⌅–⌧î
4y, y) = arctan (4y
4y 2 )
rf (x, y, z) = g(x, y, z) () (2x, 2y, 2z) =
(0  y  1)
| ÃqXî
0
t‡
f 0 (4
4y, y) =
4 8y
=0
1 + 16(y 1)2 y 2
(i)
1
1
t¿\ f (4 4y, y)î y = –⌧ ˘✓D ƒ‰.
2
2
⇡
¯ ✓@ f (2, 1/2) = arctan 1 = t‰.
4
»¿…<\ ⌅–⌧ l\ f X ✓‰D DPtÙt Ì R–⌧ f X
⇡
\◆✓@ t‡ \ü✓@ 0t‰.
4
–⌧ y =
3. ¸¥ƒ ·t¸ –⇣L¿X (\å)p¨| lX‹$.
2
f (x, y, z) = x2 + y 2 + z 2
( y, x, 2z)
2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
= 0tt x = y = z = 0xp tî ⌧}ptD ÃqX¿
Jî‰.
(ii) z = 0 tt
(a) x =
2x
=
y
=
2y
t‰. â, x2 = y 2 t‰.
x
ytt x = ±2, y ⌥ 2t‰.
(b) x = ytt tî ⌧}ptD Ãq‹¨ ⇠ ∆‰.
(iii) z 6= 0,
= 1tt xy = 0 t¿\ z = ±2 t‡ ⇣\ x, y = 0
t‰. 0|⌧,
f (±2, ⌥2, 0) = 8,
2
z
x
+
=1
3
2
sol) ¸¥ƒ ·t¸ –⇣L¿X \åp¨î h⇠
a)
x=y=0
b) z 2 = 4 + xy
sol) ¸¥ƒ ·t¸ –⇣L¿X \åp¨î h⇠
0  x  4t‡ y = 0x î⌅–⌧î
f (x, 0) = 0
y = z = 0,
p
f (± 2, 0, 0) = 2,
t‰.
(ii) f (x, y)
◆
t‰. 0|⌧,
<\Ä0 f X у⇣@ (0, 0)t‡ f (0, 0) = 0t‰. t⌧ RX
Ωƒ· – t \ , \å| p¨tÙt
(i) f (x, y)
2
x, 0, z
3
f (0, 0, ±2) = 4
t‰. ¯Ï¿\ ·t¸ –⇣⌅X \åp¨î 2t‰.
c) x + 3y 2z = 4
sol) ¸¥ƒ ·t¸ –⇣L¿X \åp¨î h⇠
– ·tX ) ›t ⌧}pt<\ ¸¥ƒ Ω∞X \ü✓X ë
f (x, y, z) = x2 + y 2 + z 2
z2
x2
X ⌧Ò¸<\ ¸¥ƒ‰. g(x, y, z) =
+
1\ ì<t
– ·tX ) ›t ⌧}pt<\ ¸¥ƒ Ω∞X \ü✓X ëX
3
2
g(x, y, z) = 0| ÃqXî Ì D–⌧ (x, y) 2 D| L
⌧Ò¸<\ ¸¥ƒ‰. g(x, y, z) = x + 3y 2z
4\ ì<t
✓
◆
g(x, y, z) = 0| ÃqXî Ì D–⌧ (x, y) 2 D| L
2
rg(x, y, z) = x, 0, z 6= (0, 0, 0)
rg(x, y, z) = (1, 3, 2) 6= (0, 0, 0)
3
132
SOLUTION
t¿\ |¯ë¸ π⇠ï– Xt
t¿\ ·t¸ –⇣⌅X \åp¨î
rf (x, y, z) = g(x, y, z) () (2x, 2y, 2z) =
| ÃqXî
0
2
,
y=
3
,
2
z=
y=
6
,
7
4
7
2
=) x = ,
7
(1, 3, 2)
– ·tX ) ›t ⌧}pt<\ ¸¥ƒ Ω∞X \ü✓X ë
X ⌧Ò¸<\ ¸¥ƒ‰. g(x, y, z) = xy + 2z
5\ ì<t
g(x, y, z) = 0| ÃqXî Ì D–⌧ (x, y) 2 D| L
=
z=
4
7
rg(x, y, z) = (y, x, 2) 6= (0, 0, 0)
t¿\ |¯ë¸ π⇠ï– Xt
t‰. 0|⌧,
f
✓
2 6
, ,
7 7
4
7
◆
t¿\ ·t¸ –⇣⌅X \åp¨î
2
=
p
rf (x, y, z) = g(x, y, z) () (2x, 2y, 2z) =
8
7
| ÃqXî
0
8/7t‰.
(±1, ±1, 2) ,
✓
5
0, 0,
2
◆
t‡
rg(x, y, z) = 2xyz 4 , x2 z 2 , 2x2 yz 6= (0, 0, 0)
f (±1, ±1, 2) = 6,
✓
◆
5
25
f 0, 0,
=
2
4
p
t¿\ ·t¸ –⇣⌅X \åp¨î 6t‰.
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = g(x, y, z)
2xyz 4 , x2 z 2 , 2x2 yz
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0 x, y, z, 6= 0t¥| Xp ›D ¨Xt
4. ‰L <L– ıX‹$.
4
2y = x2 z 2
2
1= x y
›–‰
5
t‡ x = y = 0t‰.
2
0|⌧, ¥¥¸ ⇣‰@
– ·tX ) ›t ⌧}pt<\ ¸¥ƒ Ω∞X \ü✓X ëX
⌧Ò¸<\ ¸¥ƒ‰. g(x, y, z) = x2 yz 2 4\ ì<t g(x, y, z) =
0| ÃqXî Ì D–⌧ (x, y) 2 D| L
1 = yz
2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
(ii) xy = 0tt z =
f (x, y, z) = x2 + y 2 + z 2
() (2x, 2y, 2z) =
(y, x, 2)
(i) xy 6= 0tt z 2 = 4t‡ z = 2x Ω∞ x = ±1, y = ±1t‰.
z = 2x Ω∞î •X¿ J‰.
d) x yz = 4
sol) ¸¥ƒ ·t¸ –⇣L¿X \åp¨î h⇠
x2 , y, z 2 D ÒXt
a) º
ABCX 8 ¥ D x, y, z| ` L, f (x, y, z) =
sin x sin y sin zX \◆✓D lX‹$.
sol) g(x, y, z) = x + y + z ⇡, 0 < x, y, z < ⇡\ ì<t, ⌧}
pt g(x, y, z) = 0D ÃqXî ⇣‰X —i D–⌧ (x, y) 2 D –
t
4 = 2y 2 = x2 = z 2
t¿\ ¥¥¸ ⇣‰@
⇣ p
p ⌘ ⇣ p
p ⌘
± 2, 1, ± 2 , ± 2, 1, ⌥ 2
\
5t‰.
f (x, y, z) = x2 + y 2 + z 2
=)
t‰.
p
e) xy + 2z = 5
2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä sol) ¸¥ƒ ·t¸ –⇣L¿X \åp¨î h⇠
x=
2
12 ∏¯ÑX \©
rg(x, y, z) = (1, 1, 1) 6= (0, 0, 0)
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = g(x, y, z)
8
>
<cos x sin y sin z =
() sin x cos y sin z =
>
:
sin x sin y cos z =
4⌧t‰. (y = ±k|‡ P‡ ⌧}pt– Ö) ⇣\
⇣ p
⇣ p
p ⌘
p ⌘
f ± 2, 1, ± 2 = f ± 2, 1, ⌥ 2 = 5
133
SOLUTION
| ÃqXî
0
12 ∏¯ÑX \©
2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä sol) g(x, y, z) = y 2 + z 2 2, h(x, y, z) = x + y 1\ Pê. pt
g(x, y, z) = 0, h(x, y, z) = 0D ÃqXî ⇣‰X —iD D\ Pt
(x, y, z) 2 D– t
sin x · sin(y z) = 0,
rg(x, y, z) = (0, 2y, 2z) 6= (0, 0, 0),
sin y · sin(x z) = 0,
sin z · sin(x
rh(x, y, z) = (1, 1, 0) 6= (0, 0, 0),
y) = 0,
=) x = y = z
⇣⇡ ⇡ ⇡⌘
t‰. 0|⌧ ⇣
, ,
| p¨Xt
3 3 3
⇣ ⇡ ⇡ ⇡ ⌘ 3p3
f
, ,
=
3 3 3
8
p
3 3
t‡ \◆✓
| –D L ⇠ à‰.
8
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = rg(x, y, z) + µrh(x, y, z)
8
>
<3 = · 0 + µ · 1,
=) 2 = · 2y + µ · 1,
>
:
1 = · 2z + µ · 0
| ÃqXî , µ 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<
\Ä0
µ = 3,
b) x⌘–X ⇠¿Ñt | ` L, º
X X X 8t \
⇠î º
D lX‹$.
sol) º
X x, y, z| »¸Ùî ¿D X, Y, Z| X‡ x⌘–X
⇠¿Ñt Rt| Xt, ¨xïY– Xt
X
Y
Z
=
=
= 2R
sin x
sin y
sin z
t‰. 0|⌧ º
=)
=)
2y =
2
2
4y +
1
=±
2
1,
2
2z = 1
2
4z = 8 2 = 2
t‰. 0|⌧ ¥¥¸ ⇣‰@
(2, 1, 1),
(0, 1, 1)
X X X 8tî
t‡
X + Y + Z = 2R(sin x + sin y + sin z)
<\ ¸¥ƒ‰. t L, x + y + z = ⇡,
h⇠ f @ g|
f (x, y, z) = sin x + sin y + sin z,
f (2, 1, 1) = 5,
0 < x, y, z < ⇡ t‰. t⌧
g(x, y, z) = x + y + z
⇡
\ XXt g(x, y, z) = 0D ÃqXî Ì D–⌧ (x, y, z) 2 D
tt
rg(x, y, z) = (1, 1, 1) 6= (0, 0, 0)
t¿\ |¯ë¸ π⇠ï– Xt
f (0, 1, 1) = 1
t¿\ \◆✓@ 5, \ü✓@ 1 t‰.
b) x y z = 1, x2 + y 2 = 1, f (x, y, z) = 2x + y + 3zX
\◆✓
sol) g(x, y, z)x y z + 1, h(x, y, z) = x2 + y 2 1\ Pê.
pt g(x, y, z) = 0, h(x, y, z) = 0D ÃqXî ⇣‰X —iD D\
Pt (x, y, z) 2 D– t
rf (x, y, z) = rg(x, y, z) () (cos x, cos y, cos z) = (1, 1, 1)
rg(x, y, z) = (1, 1, 1) 6= (0, 0, 0),
rh(x, y, z) = (2x, 2y, 0) 6= (0, 0, 0)
D ÃqXî 2 Rt t¨\‰. 0 < x, y, z < ⇡ t¿\ •\ t¿\ |¯ë¸ π⇠ï– Xt
⇡
Ω∞î x = y = z = t‰. ⇣\,
rf (x, y, z) = rg(x, y, z) + µrh(x, y, z)
3
8
>
⇣ ⇡ ⇡ ⇡ ⌘ 3p3
<2 = · 1 + µ · 2x,
f
, ,
=
=)
1 = · ( 1) + µ · 2y,
3 3 3
2
>
:
3 = · ( 1) + µ · 0
⇣⇡ ⇡ ⇡⌘
t¿\
, ,
–⌧ f \
⇠‡, 0|⌧ º
XX
3 3 3
| ÃqXî , µ 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<
X 8t p \
⇠î º
@ º
t‡ ¯ LX X X
\Ä0
8tî 3 3R t‰.
= 3, µ · 2x = 5, µ · 2y = 2
µ8⌧ 12.4 .
=) µ2 4x2 + µ2 4y 2 = 4µ2 = 29
1. ¸¥ƒ ⌧}pt– t f X \◆✓¸ \ü✓D lX‹$.
p
29
a) y 2 +z 2 = 2, x+y = 1, f (x, y, z) = 3x+2y +zX \◆✓¸
=) µ = ±
\ü✓
2
134
SOLUTION
t‰. 0|⌧ ¥¥¸ ⇣‰@
✓
◆
5
2
7
p , p ,p +1 ,
29
29
29
✓
5
2
p ,p ,
29
29
7
p +1
29
12 ∏¯ÑX \©
(ii) z 6= 0x Ω∞
(
2x = 2z + µ2 · x
2y = 2z + µ2 · y
◆
=) x = y
t‡, 0|⌧
✓
=) x = ±
◆
p
5
2
7
p , p , p + 1 = 29 + 3,
29
29
29
✓
◆
p
5
2
7
p ,p , p +1 =
f
29 + 3
29
29
29
f
t¿\ \◆✓@
p
t‰.
y=±
1/2,
y) = µ2(x
p
1/2,
y)
p
z=⌥ 2
0|⌧, ¥¥¸ ⇣‰@
⇣ p
⌘ ⇣ p
p
p
p ⌘
± 1/2, ⌥ 1/2, 0 , ± 1/2, ± 1/2, ⌥ 2
29 + 3 t‰.
t‡,
2. ¸¥ƒ P ·t ⌅– ®P àî ⇣‰ ⌘–
⇣D lX‹$.
p
=) 2(x
8⌧–⌧ îl⌧
⇣ p
⌘
p
f ± 1/2, ⌥ 1/2, 0 = 1,
⇣ p
p
p ⌘
f ± 1/2, ± 1/2, ⌥ 2 = 3
t¿\
–⇣–⌧ ⌘P ·tL¿X ⇣ •
L¥ ⇣‰@
⇣ p
p
p
p
p ⌘
L¥ ⇣¸ ⌧|
± 1/2, ⌥ 1/2, 0 , • < ⇣‰@ ± 1/2, ± 1/2, ⌥ 2
a) x + y + z = 0, x + y = 1, –⇣–⌧ •
<⇣
t‰.
sol) g(x, y, z) = x + y + z, h(x, y, z) = x2 + y 2 1\ Pt P
·t¸ –⇣–⌧X • L¥ ⇣ ✏ ⌧| < ⇣@ h⇠
b) 2x + y + 3z = 10, z = x2 + y 2 , –⇣–⌧ • L¥ ⇣
2
2
2
sol) t 8⌧X Ω∞ g(x, y) = 2x + y + 3(x2 + y 2 ) 10<\ P‡
f (x, y, z) = x + y + z
h⇠
f (x, y) = x2 + y 2
– ⌧}pt g(x, y, z) = 0, h(x, y, z) = 0t ¸¥ƒ Ω∞X \
✏ \åx ⇣¸ ⇡‰. pt g(x, y, z) = 0, h(x, y, z) = 0D ÃqX – ⌧}pt g(x, y) = 0t ¸¥LD LX \
⇠î ⇣D >î
î ⇣‰X —iD D\ Pt (x, y, z) 2 D– t
ɸ ⇡‰. pt g(x, y) = 0 D ÃqXî ⇣‰X —iD D\ Pt
(x, y) 2 D– t
rg(x, y, z) = (1, 1, 1) 6= (0, 0, 0),
rg(x, y) = (2 + 6x, 1 + 6y) 6= (0, 0)
rh(x, y, z) = (2x, 2y, 0) 6= (0, 0, 0),
t¿\ |¯ë¸ π⇠ï– Xt
2
2
rf (x, y) = rg(x, y)
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = rg(x, y, z) + µrh(x, y, z)
8
>
<2x = · 1 + µ · 2x,
=) 2y = · 1 + µ · 2y,
>
:
2z = · 1 + µ · 0
() (2x, 2y) = (2 + 6x, 1 + 6y)
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0 x, y 6= 0t‡ 6= 0 t¿\
x
6x + 2
=
=) x = 2y
y
6y + 1
=) 3y 2 + y 2 = 0
| ÃqXî , µ 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<
\Ä0
(i) z = 0x Ω∞
2, y =
t‰. ⇣\ z = x2 + y 2
0|⌧, ¥¥¸ ⇣‰@
= 0t¿\
2x = µ2x,
=) x =
2y = µ2y
1 ⇣î x =
4
2
,y =
3
3
<¿\ zå\î êŸ <\ ∞ ⌧‰.
( 2, 1, 5) ,
✓
4 2 20
, ,
3 3 9
◆
t‰. µ = 0tt x = y = 0xp tî ⌧}ptD ÃqX¿
t‡, h(x, y, z) = x2 + y 2 + z 2 \ Pt
Jî‰. µ 6= 0tt x = yt‡ 0|⌧
p
p
h ( 2, 1, 5) = 30,
x = ± 1/2, y = ⌥ 1/2
✓
◆
4 2 20
580
h
, ,
=
⇡ 7.1605
3
3
9
81
t‰.
135
SOLUTION
t¿\ –⇣–⌧ P ·tL¿X
•
L¥ ⇣@
t‰.
✓
4 2 20
, ,
3 3 9
◆
12 ∏¯ÑX \©
t¿\ –⇣–⌧ ! P ·tL¿X
p
25 5 119 60
,±
,
t‰.
53
106 54
•
L¥
⇣‰@
d) x + y + 2z = 2, z = x2 + y 2 , –⇣–⌧ • L¥ ⇣¸
c) x + 4z = 5, z 2 = x2 + 4y 2 , –⇣–⌧ • L¥ ⇣
2
2
2
•<⇣
sol) g(x, y, z) = x + 4z 5, h(x, y, z) = z
x
4y \ Pt
sol) t 8⌧X Ω∞ g(x, y) = x + y + 2(x2 + y 2 ) 2<\ P‡
P ·t¸ –⇣–⌧X • L¥ ⇣ ✏ ⌧| < ⇣@ h⇠
h⇠
f (x, y) = x2 + y 2
f (x, y, z) = x2 + y 2 + z 2
X ⌧}pt g(x, y) = 0| LX \
⇠î ⇣D >î ɸ ⇡‰.
– ⌧}pt g(x, y, z) = 0, h(x, y, z) = 0t ¸¥ƒ Ω∞X \
pt g(x, y) = 0 D ÃqXî ⇣‰X —iD D\ Pt (x, y) 2 D
✏ \åx ⇣¸ ⇡‰. pt g(x, y, z) = 0, h(x, y, z) = 0D ÃqX – t
î ⇣‰X —iD D\ Pt (x, y, z) 2 D– t
rg(x, y) = (1 + 4x, 1 + 4y) 6= (0, 0)
rg(x, y, z) = (1, 0, 4) 6= (0, 0, 0),
t¿\ |¯ë¸ π⇠ï– Xt
rh(x, y, z) = ( 2x, 8y, 2z) 6= (0, 0, 0),
rf (x, y) = rg(x, y)
() (2x, 2y) = (1 + 4x, 1 + 4y)
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = rg(x, y, z) + µrh(x, y, z)
8
>
<2x = · 1 + µ · ( 2x),
=) 2y = · 0 + µ · ( 8y),
>
:
2z = · 4 + µ · 2z
| ÃqXî 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<\Ä
0 x, y 6= 0t‡ 6= 0 t¿\
x
1 + 4x
=
=) x = y
y
1 + 4y
=) 3y 2 + y 2 = 0
=) x =
1, y =
1 ⇣î x =
1
1
,y =
2
2
| ÃqXî , µ 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<
t‰. ⇣\ z = x2 + y 2 <¿\ zå\î êŸ <\ ∞ ⌧‰.
\Ä0
0|⌧, ¥¥¸ ⇣‰@
(i) y = 0 x Ω∞ z 2 = x2 t¿\ x = 1, z = 1 ⇣î
✓
◆
1 1 1
5
5
(
1,
1,
2)
,
,
,
x=
,z = ,
2 2 2
3
3
1
t‡, h(x, y, z) = x2 + y 2 + z 2 \ Pt
(ii) µ =
x Ω∞
4
h ( 1, 1, 2) = 6,
✓
◆
x
z
1 1 1
3
2x = + , 2z = 4
h
, ,
=
2
2
2
2
2
4
3x
5z
✓
◆
=)
=
=
1 1 1
2
8
p
t¿\ –⇣–⌧ P ·tL¿X • L¥ ⇣@
, ,
,
25
60
5 119
2 2 2
=) 12x = 5z =) x =
,z =
,y = ±
• < ⇣@ ( 1, 1, 2) t‰.
53
53
106
t‰. 0|⌧ ¥¥¸ ⇣‰@
(1, 0, 1) ,
✓
5
5
, 0,
3
3
◆
,
!
p
25 5 119 60
,±
,
53
106 53
e) 2x + y 2z = 8, x2 + y 2 = 1, xy…t–⌧ • L¥ ⇣¸
⌧| < ⇣
sol) g(x, y, z) = 2x + y 2z 8, h(x, y, z) = x2 + y 2 1\ Pt
P ·t¸ xy…t –⌧X • L¥ ⇣ ✏ ⌧| < ⇣@ h⇠
f (x, y, z) = z 2
t‡
✓
◆
5
5
50
f (1, 0, 1) = 2, f
, 0,
=
,
3
3
9
!
p
25 5 119 60
375
f
,±
,
=
53
106 53
212
– ⌧}pt g(x, y, z) = 0, h(x, y, z) = 0t ¸¥ƒ Ω∞X \
✏ \åx ⇣¸ ⇡‰. pt g(x, y, z) = 0, h(x, y, z) = 0D ÃqX
î ⇣‰X —iD D\ Pt (x, y, z) 2 D– t
rg(x, y, z) = (2, 1, 2) 6= (0, 0, 0),
rh(x, y, z) = (2x, 2y, 0) 6= (0, 0, 0),
136
SOLUTION
13 ‰⌘ Ñ
sol) f (x, y) = xyî [0, 1] ⇥ [0, 1]–⌧ çt¿\,
Ñ •X‰.
X¨
Rij –⌧ ⇣ (xi , yj )|
Ït
t¿\ |¯ë¸ π⇠ï– Xt
rf (x, y, z) = rg(x, y, z) + µrh(x, y, z)
8
>
<0 = · 2 + µ · 2x,
=) 0 = · 1 + µ · 2y,
>
:
2z = · ( 2) + µ · 0
n X
n
X
Ì D–⌧
›\‰. ¯
n X
n
X
ij
x i yj x y =
i=1 j=1
i=1 j=1
n4
1 n(n + 1) n(n + 1)
·
·
n4
2
2
1
2
= 2 (n + 1)
4n
=
| ÃqXî , µ 2 Rt t¨\‰. ⌧}pt¸ ⌅ Ω) ›<
\Ä0
1. z = 0x Ω∞ 2x + y = 8, x2 + y 2 = 1D ΩXt ) ›
5x2 32x + 63 = 0D ªî‰. X¿Ã t ) ›@ ‰¸D t‰. n ! 1x ˘\D ËXt
¿¿ J<¿\ ®⌧t‰.
0
1
ZZ
n X
n
X
2. z =
6= 0tt
xydA = lim @
x i yj x y A
0 = 2 + µ2x,
0=
D
+ µ2y
p
2p
1p
± 5
5, y = ±
5, z =
5
5
2
i=1 j=1
1
1
= lim
(n + 1)2 =
2
n!1 4n
4
=) 2y = x
=) x = ±
n!1
8
t‰.
t‰.
2. Ì D = [0, 1] ⇥ [0, 1] ⇢ R2 –⌧ h⇠ f
⇠»‰.
(
1, y < x,
f (x, y) =
2, y x.
0|⌧, ¥¥¸ ⇣‰@
p
2p
1p ± 5
±
5, ±
5,
5
5
2
8
!
X
Ì D|
t‡,
f
2p 1p
5,
5,
5
5
f
2p
5,
5
p
5 8
2
p
1p
5,
5
t¿\ –⇣–⌧ P ·tL¿X
!
p
2p 1p
5 8
5,
5,
,
5
5
2
!
Rij = [xi 1 , xi ] ⇥ [yj 1 , yj ]
p
69
=
4 5,
4
!
p
5 8
69
=
+4 5
2
4
•
t⌘ Ñ
• < ⇣@
!
p
1p
5 8
5,
5
2
‰⌘ Ñ
µ8⌧ 13.1 .
1. Ì D = [0, 1] ⇥ [0, 1] ⇢ R2 |
Rij = [xi 1 , xi ] ⇥ [yj 1 , yj ]
xi =
i
,
n
yj =
j
n
yj =
j
n
(1  i, j  n)
f (x, y)dxdyX ✓D lX‹$.
(1  i, j  n)
sol) f (x, y)î {(0, 0)t + (1, 1)(1 t) | 0  t  1} –⌧ àçt
¿\ tî \\ 8tX · – àç⇣t Ïh(D X¯\‰.
0|⌧ f î Ì D–⌧ Ñ •X‰.
t⌧
X ¨
Rij –⌧ ⇣ (xi , yj )| ›Xt n2 ⌧X ¨
n(n + 1)
Rij ⌘–⌧ y
xx Ì– t˘Xî ⇣‰@
⌧
2
t‡ t ⇣‰–⌧X h+✓@ 2t‰. ⇣\ ò8¿ Ìx y < x
(n 1)n
–⌧
⌧X ⇣‰–⌧X h+✓@ 1t‰. â
2
✓
◆
n X
n
n X
n
X
X
i j
1
f (xi , yj ) x y =
f
,
· 2
n
n
n
i=1 j=1
i=1 j=1
✓
◆
n
n
X X
X
X ✓i j◆ 1
i j
1
=
f
,
· 2+
f
,
· 2
n n
n
n n
n
i=1
i=1 i>j,
\ Ñ`XÏ f (x, y) Z=ZxyX ¨ÃiD ƒ∞X‡, n ! 1x ˘\
D ËXÏ t⌘ Ñ
i
,
n
D
L¥ ⇣,
2p
5,
5
xi =
\ Ñ`XÏ
Z Z f X ¨ÃiD ƒ∞X‡, n ! 1x ˘\D ËXÏ
t‰.
13
‰L¸ ⇡t
xydAX ✓D lX‹$.
D
137
ij,
1jn
1jn
n(n + 1) 1
(n 1)n 1
=2·
· 2+
· 2
2
n
2
n
3n + 1
=
2n
SOLUTION
t‰. n ! 1x ˘\D ËXt
ZZ
D
sol) xD» ¨– XXÏ
◆
Z 3Z 2
Z 2 ✓Z 3
y
y
(y + 1)x dydx =
(y + 1)x dx dy
0
1
n X
n
X
f (x, y)dA = lim @
f (xi , yj ) x y A
n!1
13 ‰⌘ Ñ
1
1
1
i=1 j=1
=
3n + 1
3
= lim
=
n!1
2n
2
=
t‰.
µ8⌧ 13.2 .
1. ‰L ⇠ı ÑD ƒ∞X‹$.
2. ëX ¡⇠ a–
Z 3Z 1
0
p
0
x + ydxdy =
0
=
Z 3 ✓Z 1
sol)
◆
0
0
2
(1 + y)3/2
3
◆
2 3/2
y
dy
3
3
4 5/2
y
15
0
t‰.
Z 1Z 1
0
(x3 y
Z 1Z 1
\ ò¿º ⇠ à‡
0  x  a,
0yx
0  x  a,
xya
Ì D2 î
\ ò¿º ⇠ à‰. 0|⌧
ZZ
ZZ
xy
x+y
dxdy +
dxdy
a
2
D
D
Z aZ1 x
Z xZ2 a
xy
x+y
=
dydx +
dydx
a
2
0 0
0 x
1
1
3
= a3 + a3 = a3
8
4
8
t‰.
3
(x y
3. xy…tX
lX‹$.
¨– XXÏ
12xy)dxdy =
2
Z 1 ✓Z 1
0
=
=
t‰.
Z 1 ⇣⇣
0
0
57
y
4
57
8
◆
(x y
12xy)dx dy
⌘
⌘
2
y
4
Z 1
3
6y
(4y
a)
ZZ
(x2
ÌD
‰L¸ ⇡t ¸¥LD L, t⌘ ÑX ✓D
xy + y 2 )dA, Dî xï, ¡
D
<\ Xϯx
x = 1¸ Ï<
y = x2
Ì
24y) dy

1
57 1 2
dy =
y
4 2
0
sol)
Ì D|
0  y  x2
D : 0  x  1,
\ ò¿º ⇠ à‰. 0|⌧
ZZ
Z 1Z x2
(x2 xy + y 2 )dA =
(x2
D
Z 3Z 2
1
x}
12xy)dxdy
=
c)
Ì D1 , D2 |
2
sol) xD»
0
1
Ì D1 @
x + ydx dy
0
Z 3✓

p
4
(1 + y)5/2
15
124 12 p
=
3
15
5
=
b)
t …tX
18
ln 3
D2 = {(x, y) 2 [0, a] ⇥ [0, a]|y
¨– XXÏ
p
1)dy
| Pê. xD» ¨| ¨©XÏ ‰L t⌘ ÑD ƒ∞X‹$.
ZZ
ZZ
xy
x+y
dxdy +
dxdy
2
D1 a
D2
0
Z 3Z 1
(3y+1
D1 = {(x, y) 2 [0, a] ⇥ [0, a]|y < x},
x + ydxdy
sol) xD»
1
1
t‰.
a)
Z 2
0
=
Z 1
0
(x4
0
(y + 1)xy dydx
23
=
140
1
138
xy + y 2 )dydx
1 5 1 6
x + x )dx
2
3
SOLUTION
t‰.
b)
13 ‰⌘ Ñ
t‰.
ZZ
(1+2x+3y)dxdy, Dî P Ï<
D
\ Xϯx
2
y=1 x ¸y=x
2
1
Ì
<\ Xϯx
1  x  1,
x
2
1y1
x
1 x2 1
Z 1
=
2x2
(2 + 4x
\ ò¿º ⇠ à‰. 0|⌧
ZZ p
x3 + 1dxdy =
D
4x3 )dx
ZZ
Ì
=
8
3
0
x2
p
⌘
x3 + 1 dx
1)
t‰.
x+y = 2\ Xϯx
D
f)
ZZ
xº
x
D
p
y 3 + 1dA, Dî ¡
x = 0, y = 2@ y = x\ Xϯ
Ì
Ì D|
D:
y2  x  2
2  y  1,
y
sol)
\ ò¿º ⇠ à‰. 0|⌧
ZZ
Z 1Z 2 y
2xdxdy =
2xdxdy
=
Z 1
(4
Ì D|
D : 0  x  2,
4y + y 2
ZZ
y 4 )dy
2
72
=
5
x
D
p
y 3 + 1dA =
=
2
xydxdy, Dî ¡
D 3
Xϯx Ì
2@ Ï<
y2 =
x + 4\
x
0
0
Z 2
3
Z 2
y+2x
Z 9 p
1
tdt
1 6
26
=
9
1 2p 3
y y + 1dy =
2
y2 + 4
\ ò¿º ⇠ à‰. 0|⌧
ZZ
Z 1 Z y2 +4
2
2
xydxdy =
xydxdy
3
D 3
2 y+2
◆
Z 1✓
1 5
4 2
=
y
3y 3
y + 4y dy
3
3
2
9
=
4
1 2p 3
y y + 1dy
2
t‰. y + 1 = t|‡ Xê. ¯Ït 3y 2 dy = dtt¿\
0
2  y  1,
0
0
y=x
Ì D|
D:
Z 2Z 2 p
x y 3 + 1dydx
Z 2Z y p
=
x y 3 + 1dxdy
t‰.
ZZ
xy2
\ ò¿º ⇠ à‰. 0|⌧
2 y2
D
sol)
0
Z 1⇣
2 p
= (2 2
9
y 2 = x@ ¡
2xdxdy, Dî Ï<
sol)
d)
Z 1Z x2 p
x3 + 1dydx
0
t‰.
c)
0  y  x2
D : 0  x  1,
1
=
y = x2
Ì
2
\ ò¿º ⇠ à‰. 0|⌧
ZZ
Z 1 Z 1 x2
(1 + 2x + 3y)dxdy =
(1 + 2x + 3y)dydx
D
y = 0, x = 1¸ ·
Ì D|
sol)
D:
x3 + 1dxdy, Dî ¡
D
Ì D|
sol)
e)
ZZ p
t‰.
ZZ
2
p
ex dxdy, Dî {(x, y) 2 R2 |0  y  1,
y  x  1}@
D
p
2
{(x, y) 2 R | 1  y  0, 1
1 + y  x  1}X i—i
sol) Ì D|
g)
D : 0  x  1,
139
x2
2x  y  x2
SOLUTION
\ ò¿º ⇠ à‰. 0|⌧,
ZZ
13 ‰⌘ Ñ
\ƒ ò¿º ⇠ à‰. 0|⌧
2
ex dxdy =
D
Z 1Z x2
x2 2x
0
=
Z 1Z 1
2
ex dydx
Z 1
p
0
x3
e dxdy =
y
0
x2
Z 1
2xe dx
=
1
1
= (e
3
0
=e
Z 1Z x2
3
ex dydx
0
3
x2 ex dx
0
t‰.
1)
t‰.
4. t⌘ ÑX
a)
Z ⇡Z ⇡
0
y
ÌD \‹X‡, ‰L ⇠ı ÑD ƒ∞X‹$.
sin x
dxdy
x
sol) t⌘ ÑX
a)
ÌD D|‡ Xt
D : 0  x  ⇡,
µ8⌧ 13.3 .
1. ‰L ⇠ı ÑX ✓D lX‹$.
Z 1Z 1Z 1
0
0yx
0
zex+y dxdydz
0
sol) xD»
¨– Xt
Z 1Z 1Z 1
\ƒ ò¿º ⇠ à‰. 0|⌧
Z ⇡Z ⇡
Z ⇡Z x
sin x
sin x
dxdy =
dydx
x
x
0 y
0 0
Z ⇡
=
sin xdx
0
0
ze
x+y
Ñ✓@
dxdydz =
0
0
=
0
=
Z 1Z 1
x2
b)
ÌD D|‡ Xt
Z 3 Z 4Z 2
1 1
1)2 dz
0
0x
p
y
Z 1Z 1
x3 sin(y 3 )dydx =
x2
0
Z 1 Z py
0
=
Z 1
0
=
¨– Xt
Z 3 Z 4Z 2
!
x3 sin(y 3 )dx dy
0
=
1 2
y sin y 3 dy
4
1
(1
12
=
cos(1))
=
(x + y 2 + z 3 )dxdydz
1
Z 3 ✓Z 4
1
Z 3
0
◆ ◆
(x + y 2 + z 3 )dx dy dz
2
48 + 6z 3 dz
= 312
3
ex dxdy
t‰.
y
sol) t⌘ ÑX
ÌD D|‡ Xt
D : 0  x  1,
0  y  x2
c)
Z ⇡Z 1Z p1+z2
0
140
0
0
3
◆
2 + 2(y + z )dy dz
1
1
p
1)2
Ñ✓@
1 1 0
Z 3 ✓Z 4 ✓Z 2
1
t‰.
Z 1Z 1
◆
1)dy dz
(x + y 2 + z 3 )dxdydz
sol) xD»
2z sin xdydzdx
◆
dx dy dz
0
\ƒ ò¿º ⇠ à‰. 0|⌧
0
z(e
Z 1
◆
t‰.
D : 0  y  1,
c)
zey (e
x+y
0
x3 sin(y 3 )dydx
sol) t⌘ ÑX
ze
0
Z 1✓Z 1
1
= (e
2
t‰.
0
0
0
=2
b)
Z 1✓Z 1✓Z 1
SOLUTION
sol) xD»
¨– Xt Ñ✓@
Z ⇡Z 1Z p1+z2
2z sin xdydzdx
0 0 0
!
p
Z
Z
Z
2
⇡
=
=
=
1
t‰.
1+z
2z sin xdy dz
0
0
0
0
13 ‰⌘ Ñ
0
◆
Z ⇡✓Z 1
p
sin x · 2z 1 + z 2 dz dx
!
b)
dx
ZZZ
x, y
Z ⇡Z 2 p
2
1  z  ex + 1}
sol) xD»
t sin xdtdx
Z0 ⇡ 1 p
2
=
(2 2 1) sin xdx
0 3
4 p
= (2 2 1)
3
D = {(x, y, z) 2 R3 |0  x  2, 0  y 
xdxdydz,
D
ZZZ
¨– Xt
xdxdydz =
D
0
Z 1Z yZ x
0
0
2xyzdzdxdy
0
sol) xD» ¨– Xt Ñ✓@
◆ ◆
Z 1Z yZ x
Z 1✓Z y✓Z x
2xyzdzdxdy =
2xyzdz dx dy
0
0
0
0
=
=
0
Z 1✓Z y
0
0
0
1 5
y dy
4
Z 1
c)
x ydx dy
0
ZZZ
2ydxdydz,
D
sol) xD»
t‰.
=
ydV,
D
D = {(x, y, z) 2 R |
2, x + y  z  8
sol) xD»
ZZZ
D
x
2
1  x  1, 0  y 
2
1
=
=
1
Z 1
52
3
x+y
y(8
2
x
x )
2x +
28
)dx
3
0
( 2x2
y
2
3
2
y=x ¸¡
0 
y = x+2
=
2ydzdydx
1 x2
0
Z 2 ✓Z x+2 ✓Z x+1
1
x2
Z 2 ✓Z x+2
1
0
Z 1 ✓Z 2
1
=
=
y }
¨– Xt
! !
Z 1 Z 2 Z 8 x2 y 2
ydV =
ydz dy dx
2
D = {(x, y, z) 2 R3 |(x, y) 2 R,
Z 2 Z x+2Z x+1
2. R X Ì D Dò@ ⇡t ¸¥LD L, ‰L º⌘ ÑX
✓D xD» ¨| t©XÏ lX‹$.
a)
x(ex
¨– Xt
3
3
Z 2Z x ⇣
z  x + 1}t‡ R@ xy…t–⌧ Ï<
\ Xϯx Ì
1
=
24
ZZZ
xdzdydx
y 1
t‰.
◆
3
0
⌘
y + 2)dy dx
0 0
◆
Z 2✓
x3
2 x2
2
=
x e + 2x
dx
2
0
Z 2
2
10
=
+
x2 ex dx
3
0
=
t‰.
d)
Z 2Z xZ ex2 +1
Z 2
x2
◆
◆
2ydz dy dx
0
◆
2y(x + 1)dy dx
(x + 1)((x + 2)2
x4 )dx
1
=
◆
y dy dx
513
20
t‰.
d)
ZZZ
y=1
141
(x + 2y + 3z)dxdydz,
D
x2 ¸ P …t z =
Dî P Ï<0et y = x2
1, z = x + 1\ Xϯx
ƒ
1,
Ì
SOLUTION
sol) xD» ¨– Xt
ZZZ
(x + 2y + 3z)dxdydz
g)
=
x2 1
1
=
x2 1
1
=
Z 1
◆
( 5x4
!
(x+3)dxdydz,
D
◆
!
D:
5 2
x + 5x + 2xy + 4y dy dx
2
z + 2 = 0<\ Xϯx
Ì Dî
p
1  x  1,
2
ZZZ
10x3 + 5x2 + 10x)dx
1
(x + 3)dxdydz =
D
=
=
Dî $ …t x = 0, y = 0, z = 0,
xydxdydz,
xydxdydz =
D
0
=
0
Z 1✓Z 1 x
0
Z 1
◆
p
x
◆
1 2 1
x + x)dx
2
6
D
sol) xD» ¨– XXÏ
ZZZ
(3y 2 + z)dxdydz
D
=
1
=
=
Z 1 ✓Z 2
1
Z 1
1
=
0
176
15
0
(x4
3y 2 + zdz
0
3y 2 (1
1
p
3 1
x2 dx
x2 dx
µ8⌧ 13.4 .
1. R2 –⌧ P Ï< x y 2 +1 = 0¸ x+y 2 2y = 11\ Xϯx
ÌX ◆t| lX‹$.
2  y  3,
ƒ
Ì
area(D) =
Z 1 Z 2 Z 1 x2
2x2 dx
ÌD D|‡ Xt
y2
y 2 + 2y + 11
1x
Dî xy…t¸ P …t y = 0, y = 2 \ ò¿º ⇠ à‰. ¯Ït DX ◆tî
x2 <\ Xϯx
✏ Ï<0et z = 1
2
t‰.
D:
(3y 2 + z)dxdydz,
p
◆
(x + 3)dz dydx
!
(x + 3)(y + 2)dy dx
2 2x2
4(x + 3)
sol) P Ï< <\ Xϯx
t‰.
ZZZ
0
0  z  y+2
y)dy dx
0
1 4 1 3
=
(
x + x
6
2
0
1
=
120
Z 1
p
=4 2
p
= 6 2⇡
xydz dy dx
0
xy(1
◆
2 2x2
1
Z 1
¨– Xt
Z 1✓Z 1 x✓Z 1 x y
2x2 ,
2
p
p Z 1
=4 2
(x + 3) 1
D
sol) xD»
ZZZ
p
Z 1 Z p2 2x2
1
x + y + z = 1\ Xϯx ¨t¥
t‰.
p
1
ZZZ
Ì
Z 1 Z p2 2x2 ✓Z y+2
1
t‰.
f)
2x2  y 
\ ò¿º ⇠ à‰. ¯Ï¿\
4
=
3
e)
Dî xy…t¸ ¿–0et 2x2 +y 2 =
sol) ¸¥ƒ
(x + 2y + 3z)dz dy dx
1
Z 1 Z 1 x2✓
ZZZ
2 ✏ …t y
D
Z 1 Z 1 x2✓Z x+1
13 ‰⌘ Ñ
=
y 2 +2y+11
1dxdy
2 y2 1
Z 3
2
( 2y + 2y + 12)dy
2
!
1
x2 ) + (1
2
10x2 + 9)dx
Z 3Z
!
=
dy dx
◆
2x2 + x4 )dy dx
125
3
t‰.
2. ¸¥ƒ º(–
ÌX Ä<| lX‹$.
a) {(x, y, z) 2 R3 |x
0, y
sol) ¸¥ƒ º(–
ÌD D|‡ Xt
D : 0  y  1,
142
0, x + 2y  2, 0  z  x2 + y 2 }
0x2
2y,
0  z  x2 + y 2
SOLUTION
<\ ò¿º ⇠ à‰. ¯Ït
Z 1Z 2 2yZ x2 +y2
vol(D) =
1dV
0 0
0
✓Z
◆
Z 1
2 2y
2
2
=
(x + y )dx dy
0
=
Z 1✓
t‰.
d) xy …t–⌧ P ¡ x = 2, y = 0¸ Ï< y = x2 <\ XÏ
¯x ƒ Ì R–⌧ X⌧ h⇠ f (x, y) = 4 yX ¯ò⌅X
Dò– à‡ xy…t ⌅– àî Ì
0
0
14 3
y + 10y 2
3
8y +
8
3
5
6
=
13 ‰⌘ Ñ
◆
sol) ¸¥ƒ º(–
dy
ÌD D|‡ Xt
0  y  x2 ,
D : 0  x  2,
0z4
y
\ ò¿º ⇠ à‰. ¯Ït
t‰.
vol(D) =
b) P …t z = 0¸ z = x + 4 ✏ ¿–0et x2 + 2y 2 = 4\
Xϯx ƒ Ì
0
=
sol) ¸¥ƒ º(– ÌD D|‡ Xt
r
r
1 2
D : 2  x  2,
2
x y 2
2
vol(D) =
2
=
Z 2
2
=
1 2
2x
Z x+4
p 1 2
2 2x 0
Z p 1 2
2
Z 2p
p
2x
1 2
x ,
2
=
2(x + 4)
2
p
= 8 2⇡
4
!
(4
0
y)dy dx
◆
1 4
x dx
2
4x2
0
0  z  x+4
112
=
15
1dV
e) ¨ 0et |x| + |y| = 1¸ P …t z = 0, 3x + z = 3<\
Xϯx Ì
¸¥ƒ ÌD D@ D0 X i—i<\ ò¿º ⇠ à‰. Ï0⌧
!
(x + 4)dy dx
p
Z 2 Z x2
Z 2✓
1dV
0
t‰.
1 2
2x
2
0
0
\ ò¿º ⇠ à‰. ¯Ït
Z 2 Z p2
Z 2Z x2Z 4 y
1  x  0,
D:
x2 dx
D0 : 0  x  1,
1  y  x + 1,
0z3
3x
1y1
0z3
3x
x
x
x,
t‰. ¯Ït
t‰.
c) R = {(x, y) 2 R2 |1  y  2, y  x  y 3 }–⌧ X⌧ h⇠
f (x, y) = ex/y X ¯ò⌅X Dò– à‡ xy…t ⌅– àî Ì
vol(D [ D0 )
Z 0 Z x+1 Z 3 3x
Z 1Z
=
dV +
sol 8⌧–⌧ ¸¥ƒ
=
ÌD D|‡ Xt
D : 1  y  2,
3
yxy ,
vol(D) =
1
=
y
Z 2 Z y3
1
=
0
Z 2
y
(yey
2
1
1
= e4
2
2e
(3
1
0ze
x/y
=
Z 0
x 1
6(1
1
\ ò¿º ⇠ à‰. ¯Ït
Z 2Z y3Z ex/y
1
x 1 0
Z 0 ✓Z x+1
=4+2=6
1dV
!
ex/y dx dy
ye)dy
0
x+1Z 3 3x
x 1
◆
Z 1 ✓Z
3x)dy dx +
x2 )dx +
Z 1
0
6(1
dV
0
x+1
(3
x 1
◆
3x)dy dx
x)2 dx
0
t‰.
f) 8 …t x + 2y + z = 3, 2x + y + z =
y = x2 <\ Xϯx Ì
1, y = 2@ Ï<0et
sol) ¸¥ƒ º(– ÌD D|‡ Xt
p
p
D:
2  x  2, x2  y  2,
2x y 1  z  3 x 2y
143
SOLUTION
\ ò¿º ⇠ à‰. ¯Ït
Z p2 Z 2Z 3 x 2y
vol(D) = p
1dV
=
p
Z p
4. ¡⇠ 0  a < b– t l⌅ [a, b]–⌧ h⇠ f î çt‡ f
t‰. ¡⇠ ↵ > 1– t ‰L Ò›t 1ΩhD Ùt‹$.
ZZ
Z b
x↵ dxdy =
x↵ f (x)dx
◆
y + 4)dy dx
x2
2
2
1 4
p ( x
2 2
112 p
=
2
15
=
t‰.
2 x2
2x y 1
Z p2 ✓Z 2
(x
x3
13 ‰⌘ Ñ
D
4x2 + 2x + 6)dx
a
Ï0⌧ D = {(x, y) 2 R2 | x 2 [a, b], 0  y  f (x)}t‰.
proof ) h⇠ f î çt¿\ xD»
t‰.
g) ¿–0et 2x + y = 2X ¥Ä– àî
z = x + 2, z = y 1\ Xϯx ƒ Ì
2
2
ZZ
Ì ⌘ P …t
↵
x dxdy =
D
=
\ ò¿º ⇠ à‰. ¯Ït
Z 1 Z p2 2x2 Z x+2
vol(D) =
1dV
p
=
=
1
2
2x2
1
p
2 2x2
y 1
Z 1 Z p2 2x2
Z 1 Z p2 2x2
p
1
(x
(x
1  z  x+2
!
a
0
Z b
[x↵ y]0
!
f (x)
dx
x↵ f (x)dx
t‰.
µ8⌧ 13.5 .
1. R2 X Ì D– t ‰L t⌘ ÑX ✓D ¿⇠¿X ı›D
t©XÏ lX‹$.
a)
y + 3)dy dx
x2 dx
Z b
x↵ dy dx
a
y + 3)dy dx
2 2x2
Z 1 p
p
=
2 2(x + 3) 1
1
p
= 3 2⇡
=
¨– XXÏ
!
Z b Z f (x)
a
sol) º(–<\ Xϯx ÌD D|‡ Xt
p
p
D : 1  x  1,
2 2x2  y  2 2x2 , y
0
ZZ
ex+y dxdy,
D
D = {(x, y) 2 R2 | |x| + |y|  1}
sol) u = x + y, v = x
t‡
x=
y\ Pt (u, v) 2 D⇤ = [ 1, 1] ⇥ [ 1, 1]
u+v
,
2
t‰. ⇣\
@(x, y)
= det
@(u, v)
t‰.
y=
✓1
2
1
2
u
1
2
1
2
v
2
◆
=
1
2
3. l⌅ [a, b]–⌧ h⇠ f î çt‡, ®‡ x 2 [a, b]– t t¿\ dxdy = 1 dudvÑD L ⇠ à‰. 0|⌧
2
f (x)
0t‰. ¡⇠ ↵ > 1– t ‰L Ò›t 1ΩhD Ù
Z
Z
Z 1Z 1
t‹$.
1 u
ZZ
Z b
x+y
e
dxdy =
e dudv
1
↵
↵+1
y dxdy =
f (x)
dx
D
1
1 2
D
a ↵+1
1
=e
Ï0⌧ D = {(x, y) 2 R2 | x 2 [a, b], 0  y  f (x)}t‰.
e
proof ) h⇠ f î çt¿\ xD» ¨– XXÏ
!
ZZ
Z b Z f (x)
y ↵ dxdy =
y ↵ dy dx
D
a
=
Z b
a
=
Z b
0
f (x)
1
y ↵+1
↵+1
0
1
f (x)↵+1 dx
↵
+
1
a
dx
t‰.
b)
ZZ
(2x + y)(x + 2y)dxdy,
D
D = {(x, y) 2 R2 |1  2x + y 
2, 0  x + 2y  3}
sol) u = 2x + y, v = x + 2y\ Pt (u, v) 2 D⇤ = [1, 2] ⇥ [0, 3]
t‡
2u v
u + 2v
x=
, y=
3
3
144
SOLUTION
t‰. ⇣\
@(x, y)
= det
@(u, v)
✓ 2
3
1
3
1
3
2
3
◆
=
1
3
1
dudvÑD L ⇠ à‰. 0|⌧
3
ZZ
Z 3Z 2
uv
9
(2x + y)(x + 2y)dxdy =
dudv =
3
4
D
0 1
t¿\ dxdy =
✓
(x 2y + 1)2
2
e)
dxdy, Dî $ ⇣
,
2
7
D (3x
✓
◆+
✓y + 2) ◆
1 4
5 1
,
,
,
D -”⇣<\ Xî ¨
7 7
7 7
ZZ
13 ‰⌘ Ñ
◆ ✓
◆
1
4 2
,
,
,
7
7 7
Ì
sol) u = y +3x, v = 2y x\ Pt (u, v) 2 D⇤ = [ 1, 2]⇥[ 1, 0]
t‡
2u v
u + 3v
x=
, y=
7
7
t‰. ⇣\
✓2
◆
t‰.
1
@(x, y)
1
7
7
= det 1
=
3
@(u, v)
7
ZZ
7
7
1
c)
xydxdy, Dî $ ¡ y = x+3, y = x+1, y = x+2,
t¿\ dxdy = dudvÑD L ⇠ à‰. 0|⌧
D
7
y = x 2\ Xϯx ¨
Ì
ZZ
Z 0Z 2
(x 2y + 1)2
1 (1 v)2
1
dxdy
=
dudv =
2
2
(3x
+
y
+
2)
7
(u
+
2)
4
D
1
1
sol) u = x + y, v = x y\ Pt (u, v) 2 D⇤ = [ 2, 2] ⇥ [ 3, 1]
t‡
t‰.
u+v
u v
x=
, y=
ZZ
2
2
f)
xydxdy, D = {(x, y) 2 R2 |5  2x2 + y 2  8, 1 
t‰. ⇣\
D
✓1
◆
1
@(x, y)
1
x2 y 2  2, x 0, y 0}
2
2
= det 1
=
1
@(u, v)
2
2
2
sol) u = 2x2 + y 2 , v = x2 y 2 \ Pt (u, v) 2 D⇤ = [5, 8] ⇥ [1, 2]
t‡
1
r
r
t¿\ dxdy = dudvÑD L ⇠ à‰. 0|⌧
u+v
u 2v
2
x=
, y=
3
3
ZZ
Z 2Z 1 2
u
v2
t‰. ⇣\
xydxdy =
dvdu
8
!
D
2
3
1
1
p
p
@(x,
y)
6 A
= 3
= det 6p1A
1
p
@(u, v)
6 B
3 B
✓
◆
t‰.
1
u+v
u 2v
p
=
A=
,B =
3
3
12 AB
ZZ
x+y
d)
dxdy, Dî $ ¡ y = 2x 1, y = 2x 2,
1
y
t¿\ dxdy =
dudvÑD L ⇠ à‰. 0|⌧
D 2x
12 |xy|
y = 1 x, y = 4 x\ Xϯx ¨
Ì
ZZ
Z 8Z 2
1
1
(xy)dxdy
=
dvdu =
12
4
sol) u = x + y, v = 2x y\ Pt (u, v) 2 D⇤ = [1, 4] ⇥ [1, 2]
D
5 1
t‡
t‰.
u+v
2u v
x=
, y=
3
3
2. R3 X Ì D– t ‰L º⌘ ÑX ✓D ¿⇠¿X ı›D
t‰. ⇣\
✓1
◆
t©XÏ lX‹$.
1
@(x, y)
1
3
= det 32
=
1
@(u, v)
3
ZZZ
3
3
a)
xydxdydz, Dî Ï/ ⌧X …t y = 0, y = 1, z = 0,
1
D
t¿\ dxdy = dudvÑD L ⇠ à‰. 0|⌧
3
z = 1, z = 2x 4, z = 2x\ Xϯx …â!t¥
ZZ
Z 4Z 2
x+y
u
5
dxdy =
dvdu = ln 2
sol) u = y, v = z, w = 2x z\ Pt (u, v, w) 2 D⇤ =
2x
y
3v
2
D
1 1
[0, 1] ⇥ [0, 1] ⇥ [0, 4]t‡
t‰.
w+v
x=
, y = u, z = v
2
145
SOLUTION
t‰. ⇣\
0
0
@(x, y, z)
= det @1
@(u, v, w)
0
| | Qt|î É@ F X Ì¿Xt |Xå t¨\‰î
Ét‰. F X
1Ñh⇠‰@ ‰mh⇠t¿\ XÌ–⌧ ç
<\ ¯Ñ •X‡, XÌt º
¥Ä@ ⇡@ 4t¿\ Ω
@(x, y, z)
\∞t0 L8– Ì¿XX t¨ ÏÄî
6= 0 x¿|
@(u, v, w)
p¨Xt ©ÑX‰. t⌧
1
1
1
2
2
1
0A =
2
0
0
1
13 ‰⌘ Ñ
1
dudvdwÑD L ⇠ à‰. 0|⌧
2
ZZZ
Z 4Z 1Z 1
uw + uv
5
xydxdydz =
dudvdw =
4
4
D
0 0 0
t¿\ dxdydz =
ZZZ
x0 ) + v(x2
x0 ),
y = y0 + u(y1
y0 ) + v(y2
y0 )
\ ì<t ⇣ (x0 , y0 ), (x1 , y1 ), (x2 , y2 ) º
✓
◆
@(x, y, z)
x1 x0 x2 x0
det
= det
y1 y 0 y2 y0
@(u, v, w)
t‰.
b)
x = x0 + u(x1
= x 1 y2
Dî Ï/ ⌧X …t z = 0, z =
(x + y)dxdydz,
D
1, y z = 0, y
t¥
z = 1, x
sol) u = x z, v = y
[0, 1] ⇥ [0, 1] ⇥ [0, 1]t‡
t‰. ⇣\
z, w = z\ Pt (u, v, w) 2 D⇤ =
y = v + w,
0
1
@(x, y, z)
= det @0
@(u, v, w)
0
x 0 y2
x 2 y 1 + x 2 y 0 + x 0 y1
= area( ) 6= 0
z = 1\ Xϯx …â! t¿\ F î | | Qt‰.
z = 0, x
x = u + w,
x 1 y0
X -”⇣t¿\
0
1
0
z=w
b) ⌅X 8⌧–⌧ ¸¥ƒ h⇠ F – XX ÑïD t©XÏ ‰L
ÑD ƒ∞X‹$.
ZZ
ZZ
Iy =
x2 dxdy, Ix =
y 2 dxdy
D
D
sol) ⌅–⌧ X⌧ F – XX ÑïD
ZZ
Iy =
x2 dxdy
1
1
1A = 1
1
©Xt ƒ∞– Xt
D
=
t¿\ dudvdw = dxdydzÑD L ⇠ à‰. 0|⌧
ZZZ
Z 1Z 1Z 1
xydxdydz =
(u + v + 2w)dudvdw = 2
Z 1Z 1 u
0
(x0 + u(x1
0
✓
x0 ) + v(x2
◆
x0
dvdu
y0
◆
x0
y0
x
x0 x2
· det 1
y 1 y0 y2
✓
D
0 0 0
x
x0 x2
= det 1
y
y0 y 2
1
t‰.
⇣1
1
·
x0 2 + x0 (x1 + x2 2x0 )
2
3
3. …tX 8 ⇣ P0 (x0 , y0 ), P1 (x1 , y1 ), P2 (x2 , y2 )| -”⇣<\
1 ⇥
+
(x1 x0 )2 + (x1 x0 )(x2
Xî º
ÌD D|‡ X‡, ‰L <L– ıX‹$.
12
t‡ », ¿\
ZZ
a) 2(– Ì
= {(u, v) 2 R2 |u
0, v
0, u + v  1}–⌧
Ix =
y 2 dxdy
Ì D\X h⇠ F | ‰L¸ ⇡t XXê.
x0 ))2
x0 ) + (x2
x0 ) 2
⇤⌘
D
F (u, v) ⌘ (1
u
v)P0 + uP1 + vP2
=
t Ñ F : ! Dî F (0, 0) = P0 , F (1, 0) = P1 , F (0, 1) = P2 |
ÃqXî | | QÑD Ùt‹$.
proof ) ¸¥ƒ h⇠ F : ! D| ‰‹ ò¿¥t
✓ ◆ ✓
◆
u
x0 + u(x1 x0 ) + v(x2 x0 )
F
=
v
y0 + u(y1 y0 ) + v(y2 y0 )
t¿\ F î ò
X⌧ |(¿Xt‰. |(¿X
(u, v) 2
$ (x, y) 2 D
Z 1Z 1 u
0
0
✓
(y0 + u(y1
y0 ) + v(y2
◆
x0
dvdu
y0
◆
x0
y0
x
x0 x2
· det 1
y 1 y0 y 2
✓
x
x0 x2
= det 1
y1 y0 y 2
⇣1
1
y0 2 + y0 (y1 + y2 2y0 )
·
2
3
1 ⇥
+
(y1 y0 )2 + (y1 y0 )(y2
12
t‰.
146
y0 ))2
y0 ) + (y2
y0 ) 2
⇤⌘
SOLUTION
13 ‰⌘ Ñ
t¿\, lX$î
µ8⌧ 13.6 .
1. ˘å\ XXD t©XÏ ‰L
a)
Z 2Z p2x x2
p
0
2x x2
sol) ¸¥ƒ
2x x2
0  r  2 cos ✓
x2
8
|y|dydx =
(1
(x + y)2 )dxdy,
0, y
0}
(1
3+
Z ⇡/2 Z 2 cos ✓
⇡/2 0
r2 |sin ✓|drd✓
3y 2 <\ Xϯx
sol) ¸¥ƒ
Z ⇡/2
8
|sin ✓| cos3 ✓d✓
⇡/2 3
0  ✓  2⇡,
ZZZ
D = {(x, y) 2 R2 |1  x2 + y 2 
⇡
,
2
Ñ✓@
(x2 y 2 )dV =
D
ZZZ
4
0
(x + y) )dxdy =
D
=
0
1
0
3
2
Z ⇡/2
=
1dV,
D
x2
sol) ¸¥ƒ
Ñ✓@
Z ⇡/2Z 2
(r cos ✓ + r sin ✓)2 )rdrd✓
r5 (8
4r2 ) sin2 ✓ cos2 ✓drd✓
0
2
⇡
3
D = {(x, y, z) 2 R3 |x2 + (y
ÌD –0eå\\ \⌅Xt
0  r  2 sin ✓,
t¿\, lX$î
Ñ✓@
ZZZ
1dV =
15
(1 + sin(2✓))d✓
4
D
0z
Z ⇡Z 2 sin ✓ p
r 4
0
=
3
(10 + 3⇡)
8
1)2  1, 0  z 
y2 }
0  ✓  ⇡,
(1
Z 2⇡Z p2
t‰.
e)
p
1r2
Ì
ÌD –0eå\\ \⌅Xt
p
0  r  2, r2 (2 cos2 ✓+1)  z  8 r2 (1+2 sin2 ✓)
t¿\, lX$î
4
3
ÌD ˘å\\ \⌅Xt
2
4⇡
3
x2 y 2 dV, Dî ¿–Ï<t z = 3x2 + y 2 ¸ z =
D
=
0✓
ZZ
ZZZ
d)
t‰.
t¿\, lX$î
8 cos2 ✓d✓
t‰.
Ñ✓@
=
sol) ¸¥ƒ
p
ÌD ˘å\\ \⌅Xt
p
D
Z ⇡/3
2r3 cos2 ✓drd✓
0
0
=
4, x
Z ⇡/3Z 2
0
=
Z 2Z p2x x2
ZZ
D
|y|dydx
t¿\, lX$î
b)
(2x2 )dxdy =
=
⇡
⇡
✓ ,
2
2
0
ZZ
ÑX ✓D lX‹$.
Ñ✓@
p
4
r2
r2 drd✓
0
8
(3⇡
9
4)
t‰.
t‰.
c)
ZZ
2x2 dxdy,
D
sol) ¸¥ƒ
D = {(x, y) 2 R2 |x2 +y 2  4, 0  y 
ÌD ˘å\\ \⌅Xt
0✓
⇡
,
3
p
3x}
2. ë⇠ ✏ < 1– t D✏ = {(x, y) 2 R2 |✏2  x2 + y 2  1}|
X‡, p ëX ¡⇠| Xê. ✏ ! 0+ | L t⌘ Ñ
ZZ
1
dxdy
2 + y 2 )p/2
(x
D✏
⇠4Xî pX î⌅| lX‹$.
0r2
147
SOLUTION
sol) ¸¥ƒ
Ì D✏ D ˘å\\ \⌅Xt
0  ✓  2⇡,
t‰.
✏r1
t¿\ t⌘ Ñ@
ZZ
1
dxdy =
2 + y 2 )p/2
(x
D✏
Z 2⇡Z 1
0
13 ‰⌘ Ñ
b)
6
r1 p drd✓
ZZZ
x2
0
r
1 p
✏
ÌD –0eå\\ \⌅Xt
p
0  ✓  2⇡, 0  r  2, 2r2  z  6
t¿\, lX$î
2⇡ ⇥ 2 p ⇤1
drd✓ =
r
✏
2 p
2⇡
=
(1 ✏2 p )
2 p
ZZZ
t‰. 2 < pt¿\ 2 p < 0tp 0|⌧ ✏ ! 0+ | L
⌧∞\‰.
(case2) 1 p = 1x Ω∞.
Z 2⇡Z 1
0
✏
1
r1 p drd✓ = 2⇡ [ln(r)]✏ =
t‰. 0 < 2
Ñ✓@
2⇡
=
2⇡ ln(✏)
r3 sin2 ✓dzdrd✓
3r3 (2
r2 ) sin2 ✓drd✓
0
0
0
0
Z 2⇡Z p2
ZZZ
sol) ¸¥ƒ
t¿\, lX$î
ZZZ
<\ ⇠4\‰.
2
0  r  1,
2
6(x + y )dxdydz =
D
a)
…t x
(x + y )dxdydz,
Dî –0et x + y = 4@ P d)
z=3✏y
1\ Xϯx
D
sol) ¸¥ƒ
z=
2
ƒ
2
ƒ
Ì
0  r  2,
r cos ✓
0
=
0
Z 2⇡Z 2
0
(r )(r(sin ✓
D
Ì
p
0
148
0
6r3 dzdrd✓
2r
11
⇡
5
x2 + y 2 ¸ …t z = 1\ Xϯx
rz1
Ñ✓@
zdxdydz =
Z 2⇡Z 1Z 1
0
=
⇡
4
zrdzdrd✓
0
r
0
1
r(1
2
Z 2⇡Z 1
0
=
= 32⇡
Z 2⇡Z 1Z 3 r2
0  r  1,
D
cos ✓) + 4)drd✓
r2
ÌD –0eå\\ \⌅Xt
t¿\, lX$î
ZZZ
r cos ✓ 3
3
Dî z =
0  ✓  2⇡,
3  z  r sin ✓ + 1
t¿\, lX$î Ñ✓@
ZZZ
Z 2⇡Z 2Z r sin ✓+1
(x2 + y 2 )dxdydz =
(r3 )dzdrd✓
D
zdV,
sol) ¸¥ƒ
ÌD –0eå\\ \⌅Xt
0  ✓  2⇡,
ZZZ
2r  z  3
0
3. R3 X Ì D Dò@ ⇡t ¸¥LD L, ‰L º⌘ ÑX
t‰.
✓D ˘å\ XX ⇣î –0eå\ XXD t©XÏ lX‹$.
2
y2 ¸
Ñ✓@
=
2
x2
ÌD –0eå\\ \⌅Xt
0  ✓  2⇡,
2 p
0|⌧ t⌘ Ñt ⇠4Xî pX î⌅î 0 < p < 2t‰.
ZZZ
2r 2
6(x2 + y 2 )dxdydz, Dî Ï<t z = 3
D
p
–‘t z = 2 x2 + y 2 <\ Xϯx Ì
c)
2⇡ ⇥ 2 p ⇤1
r
✏
2 p
2⇡
=
(1 ✏2 p )
2 p
p t¿\
(y )dxdydz =
D
Z 2⇡Z p2Z 6 r2
t‰.
r1 p drd✓ =
✏
2
= 2⇡
t‰. 0|⌧ ✏ ! 0 | L Ñ✓@ ⌧∞\‰.
(case3) 1 < 1 px Ω∞.
0
r2
Ñ✓@
Ñ✓@
+
Z 2⇡Z 1
y 2 <\ Xϯx ֥
sol) ¸¥ƒ
✏
\ ò¿º ⇠ à‰.
(case1) 1 p < 1x Ω∞.
Z 2⇡Z 1
y 2 dxdydz, Dî P Ï<t z = 2x2 + 2y 2 ¸ z =
D
r2 )drd✓
SOLUTION
t‰.
e)
13 ‰⌘ Ñ
t‰.
ZZZ
(x + y + 2z)dxdydz,
D
Dî P …t y = 0, y = 2 ✏ b)
–0et x2 + z 2 = 1<\ Xϯx
⌅– àî ÄÑ
ƒ
z = r sin ✓,
Ì ⌘–⌧ xy …t y 2 + z 2  9, z  0}
sol) 8⌧–⌧ ¸¥ƒ
0  r  1,
0
=
Z ⇡Z 1
0
=
0
=
0y2
=
r(r cos ✓ + 2r sin ✓ + y)dydrd✓
c)
0
ZZZ
4, y
2r(r cos ✓ + 2r sin ✓ + 1)drd✓
8
+⇡
3
0
⇡/2 1
0
⇡/2
Z 2⇡Z ⇡
(⇢3 sin2 )d⇢d d✓
(20 sin2 )d d✓
D = {(x, y, z) 2 R3 |1  x2 + y 2 + z 2 
ydxdydz,
D
0, z
0}
sol) 8⌧–⌧ ¸¥ƒ
ÌD ltå\\ \⌅Xt
0  ✓  ⇡,
ZZZ p
x2 + y 2 + z 2 dxdydz, D = {(x, y, z) 2 R3 |x2 + y 2 +
z  4}
Z 2⇡Z ⇡ Z 3
0
4. R3 X Ì D– t ‰L º⌘ ÑX ✓D ltå\ XXD
t©XÏ lX‹$.
2
x2 + y 2 dxdydz
t‰.
t‰.
a)
1⇢3
= 10⇡ 2
D
=
 ⇡,
D
t‰. lX$î Ñ✓@ 0|⌧
ZZZ
(x + y + 2z)dxdydz
Z ⇡Z 1Z 2
⇡/2 
ÑD L ⇠ à‰. 0|⌧
ZZZ p
ÌD ⌅@ ⇡@ »\¥ å\\ XXXt
0  ✓  ⇡,
D = {(x, y, z) 2 R3 |1  x2 +
ÌD ltå\\ \⌅Xt
0  ✓  2⇡,
y=y
t⌥å XXXt x, zî ¿⇠ r, ✓X C 1 h⇠t‡,
✓
◆
@(x, z)
cos ✓
sin ✓
= det
=r
r sin ✓ r cos ✓
@(r, ✓)
t‰. 8⌧–⌧ ¸¥ƒ
x2 + y 2 dxdydz,
D
sol) –0eå\@ ¿⇠à º XXD å⌧\‰.
x = r cos ✓,
ZZZ p
=
ÑD L ⇠ à‰. 0|⌧
ZZZ p
 ⇡,
=
0
0
0
0
Z ⇡Z ⇡/2
0
0
(⇢3 sin2 sin ✓)d⇢d d✓
1
(
15
sin2 sin ✓)d d✓
4
t‰.
x2 + y 2 + z 2 dxdydz
Z 2⇡Z ⇡Z 2
Z 2⇡Z ⇡
0
15⇡
=
8
0⇢2
D
=
Z ⇡Z ⇡/2Z 2
0
ÌD ltå\\ \⌅Xt
0
1⇢2
D
=
0  ✓  2⇡,
 ⇡/2,
ÑD L ⇠ à‰. 0|⌧
ZZZ
ydxdydz
D
sol) 8⌧–⌧ ¸¥ƒ
0
⇢3 sin d⇢d d✓
0
4 sin d d✓
ZZZ
p
zdxdydz, Dî ⇠lt z = 6
D p
–‘t z = x2 + y 2 <\ Xϯx Ì
d)
sol) 8⌧–⌧ ¸¥ƒ
= 16⇡
0  ✓  2⇡,
149
x2
y 2 (z
ÌD ltå\\ \⌅Xt
0
 ⇡/4,
0⇢
p
6
0)¸
SOLUTION
ÑD L ⇠ à‰. 0|⌧
ZZZ
zdxdydz
g)
D
=
Z 2⇡Z ⇡/4Z p6
0
=
0
Z 2⇡Z ⇡/4
0
(⇢3 sin cos )d⇢d d✓
0
zdxdydz, D = {(x, y, z) 2 R3 |x2 + y 2 + (z
pD
3x2 + 3y 2 }
0
 ⇡/6,
=
=
0
0
sol) 8⌧–⌧ ¸¥ƒ
ÌD ltå\\ \⌅Xt
Z 2⇡Z ⇡/6
0}
0  ✓  ⇡,
0
 ⇡/4,
0  ✓  ⇡,
3⇡/4 
0⇢1
1)2 
 ⇡,
0⇢1
ÑD L ⇠ à‰. 0|⌧
ZZZ
p
cos( x2 + y 2 + z 2 )dxdydz
D
Z ⇡Z ⇡/4Z 1
=
0
+
0
0
Z ⇡Z ⇡
0
p
= (2
(⇢2 sin cos ⇢)d⇢d d✓
Z 1
(⇢2 sin cos ⇢)d⇢d d✓
3⇡/4 0
2)⇡(2 cos 1
sin 1)
t‰.
D
Z 2⇡Z ⇡/6Z 2 cos
0
z 2  0, y
0  ⇢  2 cos
ÑD L ⇠ à‰. 0|⌧
ZZZ
zdxdydz
0
x2 + y 2 + z 2 )dxdydz, D = {(x, y, z) 2 R3 |x2 +
y 2 + z 2  1, x2 + y 2
ÌD ltå\\ \⌅Xt
0  ✓  2⇡,
=
D
p
@
ZZZ
sol) 8⌧–⌧ ¸¥ƒ
cos(
(9 sin cos )d d✓
t‰.
1, z
ZZZ
0
9
= ⇡
2
e)
13 ‰⌘ Ñ
(⇢3 sin cos )d⇢d d✓
0
4(sin cos5 )d d✓
5. ë⇠ ✏ < 1– t D✏ = {(x, y, z) 2 R3 |✏2  x2 + y 2 + z 2  1}
| X‡, p ëX ¡⇠| Xê. ✏ ! 0+ | L º⌘ Ñ
ZZZ
1
dxdydz
2 + y 2 + z 2 )p/2
(x
D✏
⇠4Xî pX î⌅| lX‹$.
37
⇡
48
sol) ¸¥ƒ
ÌD✏ D ltå\\ \⌅Xt
t‰.
f)
ZZZ
0  ✓  2⇡,
(x2 + y 2 )dxdydz,
D
2
2
Dî
2
Ì {(x, y, z) 2 R3 |y
@ l x + y + (z + 1)  1X P—i
sol) 8⌧–⌧ ¸¥ƒ
ÌD ltå\\ \⌅Xt
⇡
3⇡
✓
,
4
4
⇡

2
 ⇡,
0⇢
2 cos
⇡
4
=
2
⇡
15
⇡
2
2 cos
0
(⇢4 sin3 )d⇢d d✓
✏⇢1
\ ò¿º ⇠ à‰.
(case1.) 2 p < 1x Ω∞.
Z 2⇡Z ⇡Z 1
0
0
D
=
 ⇡,
|x|} t¿\ º⌘ Ñ@
ZZZ
Z 2⇡Z ⇡Z 1
1
⇢2 p sin d⇢d d✓
dxdydz =
2
2
2 p/2
D✏ (x + y + z )
0
0 ✏
ÑD L ⇠ à‰. 0|⌧
ZZZ
(x2 + y 2 )dxdydz
Z 3⇡
Z ⇡Z
4
0
✏
4⇡ ⇥ 3 p ⇤1
⇢
✏
p
4⇡
=
(1 ✏3 p )
3 p
⇢2 p sin d⇢d d✓ =
3
t‰. 3 < pt¿\ 3 p < 0tp 0|⌧ ✏ ! 0+ | L
⌧∞\‰.
(case2.) 2 p = 1x Ω∞.
Z 2⇡Z ⇡Z 1
t‰.
0
150
0
✏
1
⇢2 p sin d⇢d d✓ = 4⇡ [ln(⇢)]✏ =
Ñ✓@
4⇡ ln(✏)
SOLUTION
13 ‰⌘ Ñ
7. ¡⇠ r > 0– t Dr = {(x, y, z) 2 R3 |x2 + y 2 + z 2  r2 }|
` L, ‰LX ÑD ƒ∞X‹$. ¯¨‡ lim Ir D lX‹$.
t‰. 0|⌧ ✏ ! 0+ | L Ñ✓@ ⌧∞\‰.
(case3.) 1 < 2 px Ω∞.
Z 2⇡Z ⇡Z 1
4⇡ ⇥ 3 p ⇤1
⇢2 p sin d⇢d d✓ =
⇢
✏
3 p
0
0 ✏
4⇡
=
(1 ✏3 p )
3 p
r!1
Ir =
ZZZ
sol) ¸¥ƒ
t‰. 0 < 3 p t¿\ Ñ✓@ 0<\ ⇠4\‰.
0|⌧ º⌘ Ñt ⇠4Xî pX î⌅î 0 < p < 3t‰.
Ir =
=
ZZZ
1⇢R
0
0⇢r
2
2
2
x2 + y 2 + z 2 e (x +y +z ) dxdydz
2
⇢3 sin e ⇢ d⇢d d✓
0
2
3
⇢ e ⇢ d⇢
2
e r (r2 + 1))
x2
y2
z2
Ì D = {(x, y, z) 2 R3 | 2 + 2 + 2  1}– t
a
b
c
‰LX º⌘ ÑD ƒ∞X‹$. Ï0⌧ a, b, cî ëX ¡⇠t‰.
8. 3(–
Ix =
Iz =
ZZZ
ZZZ
2
2
(y + z )dxdydz,
Iy =
D
ZZZ
(z 2 + x2 )dxdydz,
D
(x2 + y 2 )dxdydz
D
sol) x = au, y = bv, z = cw\ Pt (u, v, w) 2 D⇤ = {(u, v, w) 2
R3 |u2 + v 2 + w2  1}t‡ ⇣\
0
1
Ñ✓@
a 0 0
@(x, y, z)
= det @0 b 0A = abc
@(u, v, w)
0 0 c
t‰. 3 < pt¿\ 3 p < 0tp 0|⌧ R ! 1| L
4⇡
\ ⇠4\‰.
p 3
(case2.) 2 p = 1x Ω∞.
Z 2⇡Z ⇡Z R
R
⇢2 p sin d⇢d d✓ = 4⇡ [ln(r)]1 = 4⇡ ln(R)
0
 ⇡,
t‰. ¯Ï¿\ r ! 1| L Ir î 2⇡\ ⇠4\‰.
1
\ ò¿º ⇠ à‰.
(case1.) 2 p < 1x Ω∞.
Z 2⇡Z ⇡Z R
4⇡ ⇥ 3 p ⇤R
⇢2 p sin drd✓ =
⇢
1
3 p
0
0 1
4⇡
=
(R3 p 1)
3 p
0
0
1
dxdydz
2 + y 2 + z 2 )p/2
(x
DR
Z 2⇡Z ⇡Z R
=
⇢2 p sin d⇢d d✓
0
Z r0
= 2⇡(1
t¿\ º⌘ Ñ@
ZZZ
0
Dr
p
Z 2⇡Z ⇡Z r
0
ÌDR D ltå\\ \⌅Xt
 ⇡,
2
t¿\
= 4⇡
0
2
Ì Dr D ltå\\ \⌅Xt
⇠4Xî pX î⌅| lX‹$.
0  ✓  2⇡,
2
x2 + y 2 + z 2 e (x +y +z ) dxdydz
0  ✓  2⇡,
6. ë⇠ R > 1– t DR = {(x, y, z) 2 R3 |1  x2 + y 2 + z 2 
R2 }| X‡, p ëX ¡⇠| Xê. R ! 1| L º⌘ Ñ
ZZZ
1
dxdydz
2 + y 2 + z 2 )p/2
(x
DR
sol) ¸¥ƒ
Dr
p
t¿\ dxdydz = (abc)dudvdwÑD L ⇠ à‰. \∏ D⇤ | lt
å\\ ò¿¥t
1
t‰. 0|⌧ R ! 1| L Ñ✓@ ⌧∞\‰.
(case3.) 1 < 2 px Ω∞.
Z 2⇡Z ⇡Z R
4⇡ ⇥ 3 p ⇤R
⇢2 p sin drd✓ =
⇢
1
3 p
0
0 1
4⇡
=
(R3 p 1)
3 p
0  ✓  2⇡,
0
t‰. ¯Ït
ZZZ
ZZZ
x2 dxdydz =
D
=
t‰. 0 < 3 p t¿\ Ñ✓@ R ! 1| L ⌧∞\‰. 0|⌧
º⌘ Ñt ⇠4Xî pX î⌅î p > 3t‰.
151
0⇢1
(a2 u2 ) · (abc)dudvdw
D⇤
Z 2⇡Z ⇡Z 1
0
=
 ⇡,
0
4⇡a3 bc
15
0
a3 bc⇢4 sin3 cos2 ✓d⇢d d✓
SOLUTION
t‡
ZZZ
y 2 dxdydz =
D
=
ZZZ
tp
ZZZ
z 2 dxdydz =
D
=
0
3
4
b ac⇢ sin
3
2
sin ✓d⇢d d✓
ZZZ
sol) 8⌧X
‰⌧¿⇠ ·t@ x2 + y 2 + z 2 = 16 ⌘ x  0 t‡
p
|z|  2 2 x ÄÑD ò¿∏‰.
e) X(u, ✓) = (u2 cos ✓, u2 sin ✓, u)
0
4⇡b3 ac
15
y = z 2 | zïD 0 <\ å⌅
sol) 8⌧X ‰⌧¿⇠ ·t@ ·
XÏ ª@ ·tt‰.
D⇤
(c2 w2 ) · (abc)dudvdw
0
0
Z 2⇡Z ⇡Z 1
0
=
(b2 v 2 ) · (abc)dudvdw
Z 2⇡Z ⇡Z 1
0
=
D⇤
14 ·t
c3 ab⇢4 sin cos2 d⇢d d✓
4⇡c3 ab
15
f) X(u, ✓) = (u, (1
[0, 2⇡]
u) cos ✓, (1
u) sin ✓),
sol) 8⌧X ‰⌧¿⇠ ·t@ · y = 1
å⌅XÏ ª@ ·tt‰ (Ë, 0  x  1.)
(u, ✓) 2 [0, 1] ⇥
x| xïD 0 <\
g) X( , ✓) = (3 cos cos ✓, 3 cos sin ✓, 4 sin )
t‰. 0|⌧
◆
✓ 2
◆
✓ 2
b + c2
a + c2
Ix = 4abc
⇡, Iy = 4abc
⇡,
15
15
✓ 2
◆
a + b2
Iz = 4abc
⇡
15
sol) 8⌧X ‰⌧¿⇠ ·t@ ¿–
å⌅XÏ ª@ ·tt‰.
x2 z 2
+
= 1D zïD 0 <\
9 16
t‰.
h) X( , ✓) = ((3 + cos ) cos ✓, (3 + cos ) sin ✓, sin )
14
sol) 8⌧X ‰⌧¿⇠ ·t@ – (x 3)2 + z 2 = 1| zïD 0
<\ å⌅XÏ ª@ ·t(torus)t‰.
·t
µ8⌧ 14.1 .
1. ‰LX ›<\ \⌅⌧ ‰⌧¿⇠ ·tD $ÖX‹$.
2. ‰L ·tD ‰⌧¿⇠| t©XÏ \⌅X‹$.
a) X(u, v) = (u
a) ¡
v, u + v, 1
3u + 8v)
sol) 8⌧X ‰⌧¿⇠ ·t@ h⇠ f (x, y) =
¯ò⌅| ò¿∏‰.
x = 2y = 3z@ 2x = y = z| ÏhXî …t
11
5
¡ X )•°0X x D µt
x + y + 1X sol) 8⌧– ¸¥ƒ ·t@
2
2
Lh⇠ ‹x
2x 10y + 9z = 0
ÑD L ⇠ à‰. t| ‰⌧¿⇠·t<\ \⌅Xt
✓
◆
2
10
X(u, v) = u, v,
u+ v
y2
z2
9
9
sol) 8⌧X ‰⌧¿⇠ ·t@ 3(–¡–⌧ ¿–
+
= 1D
4
9
ÃqXî ⇣‰X êËt‰. (t ·tD ¿–0et|‡ \‰.)
t‰.
b) X(u, v) = (u, 2 cos v, 3 sin v)
c) X(u, v) = (3u cos v, u, 3u sin v)
sol) 8⌧X ‰⌧¿⇠ ·t@ ·
å⌅XÏ ª@ ·tt‰.
b) ⇣ (1, 3, 1)D ¿ò‡, (1, 1, 1)– ⇠¡x …t
y =
1
z| yïD 0 <\ sol) 8⌧– ¸¥ƒ ·t@ 1(x
3
d) X( , ✓) = (4 sin  cos ✓, 4 sin sin ✓, 4 cos )
⇡ 3⇡
⇡ 3⇡
( , ✓) 2
,
⇥
,
4 4
2 2
x
y
1)
1(y
z=
1
3)
1(z + 1) = 0 â,
ÑD L ⇠ à‰. t| ‰⌧¿⇠·t<\ \⌅Xt
X(u, v) = (u, v, u
152
v + 1)
SOLUTION
14 ·t
t‰.
t‰.
c) ·t z = y 2
i) ·t x = 9
àî ÄÑ
p
y 2 + z 2 ⌘–⌧ P …t x = 2@ x = 4 ¨t–
sol) 8⌧– ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
sol) 8⌧–⌧ ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
✓
◆
1
1
X(u, ✓) = u, u sin ✓, u cos ✓
(2  u  4, 0  ✓  2⇡)
9
9
X(u, v) = (u, v, v 2 )
t‰.
d) ·t x2 + (y
t‰.
4)2 = 16
j) x2 + 2y 2 + 3z 2 = 6 (z
sol) 8⌧– ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
X(u, v) = (4 sin u, 4 + 4 cos u, v),
(0  u  2⇡)
t‰. (sin h⇠@ cos h⇠î ¸0h⇠t¿\ ⌅X| ‘hƒ
⇡@ ·tD \⌅\‰.)
e) ·t x2
y2 + z2 = 1
0)x ·t
sol) 8⌧–⌧ ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
p
p
p
X( , ✓) = ( 6 sin cos ✓, 3 sin sin ✓, 2 cos ),
⇣
⌘
⇡
0   , 0  ✓  2⇡
2
t‰.
sol) 8⌧–⌧ ¸¥ƒ ·t@ |˝ ·t(hyperboloid of one
sheet)tp ‰⌧¿⇠·t@
3. ‰L ‰⌧¿⇠ ·t ⌅X ⇣–⌧ ⌘…tX ) ›D lX‹$.
X(u, ✓) = ((cosh u)(cos ✓), sinh u, (cosh u)(sin ✓)) , (0  ✓  2⇡)
t‰.
f) xy…t ⌅X ¯ò⌅ y =
å⌅\ ·t
p
x
(0  x  1)D x ïD ⌘Ï<\
y4
X(u, ✓) = (9u2
u4 ) cos ✓, u, (9u2
( 3  u  3,
u4 ) sin ✓ ,
0  ✓  2⇡)
sol) 8⌧–⌧ ¸¥ƒ ‰⌧¿⇠ ·t X–
Xu (u, v) = (1, 1, 8u),
t
Xv (u, v) = ( 1, 1, 0)
t‰. X(1, 1) = (0, 2, 4)t¿\ ¸¥ƒ ⇣–⌧ ⌘…tX ) ›@
8x 8(y 2) + 2(z 4) = 0 â,
( 3  y  3)D y ïD
sol) 8⌧–⌧ ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
(0, 2, 4)
(Xu ⇥ Xv )(u, v) = ( 8u, 8u, 2)
(0  u  1, 0  ✓  2⇡)
t‰.
g) xy…t ⌅X ¯ò⌅ x = 9y 2
⌘Ï<\ å⌅\ ·t
v, u + v, 4u2 ),
t¿\
sol) 8⌧–⌧ ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
X(u, ✓) = (u2 , u cos ✓, u sin ✓)
a) X(u, v) = (u
4x + 4y
z=4
v),
(1, 1, 3)
t‰.
b) X(u, v) = (u2 , v 2 , 4u
sol) 8⌧–⌧ ¸¥ƒ ‰⌧¿⇠ ·t X–
Xu (u, v) = (2u, 0, 4),
t‰.
t
Xv (u, v) = (0, 2v, 1)
t¿\
h) y =
(Xu ⇥ Xv )(u, v) = ( 8v, 2u, 4uv)
2@ y = 2¨t– àî –0et x2 + z 2 = 9
sol) 8⌧–⌧ ¸¥ƒ ·tD ‰⌧¿⇠·t<\ \⌅Xt
X(u, ✓) = (3 cos ✓, u, 3 sin ✓)
t‰. X(1, 1) = (1, 1, 3)t¿\ ¸¥ƒ ⇣–⌧ ⌘…tX ) ›@
8(x 1) + 2(y 1) + 4(z 3) = 0 â,
0  ✓  2⇡, 2  u  2
4x + y + 2z = 3
153
SOLUTION
14 ·t
µ8⌧ 14.2 .
¸¥ƒ ·tX ◆t| lX‹$.
t‰.
c) X(r, ✓) = (r cos ✓, r sin ✓, r),
r = 2,
sol) 8⌧–⌧ ¸¥ƒ ‰⌧¿⇠ ·t X–
Xr (r, ✓) = (cos ✓, sin ✓, 1),
⇡
4
✓=
1. X(u, v) = (u + v, u v, v),
sol) ·t XX t å dSî
t
dS =
X✓ (r, ✓) = ( r sin ✓, r cos ✓, 0)
0  u  1,
p
0v1
6dudv
t¿\, lX‡ê Xî ◆tî
ZZ
Z 1Z 1 p
Xr (r, ✓) ⇥ X✓ (r, ✓) = ( r cos ✓, r sin ✓, r)
dS =
6dudv
X
p p
p0 0
t‰. X(2, ⇡/4)
=p
( 2, p2, 2)t¿\
= 6
p
p ¸¥ƒ ⇣–⌧ ⌘…tX
) ›@
2(x
2)
2(y
2) + 2(z 2) = 0 â,
t‰.
p
p
2x
2y + 2z = 0
t¿\
2. X(r, ✓) = (r cos ✓, r sin ✓, ✓), 0  ✓  2⇡,
sol) ·t XX t å dSî
p
dS = 1 + r2 drd✓
t‰.
d) X(u, v) = (u2 , 4u sin v, u cos v),
u = 1,
t¿\, lX‡ê Xî ◆tî
ZZ
Z 2⇡Z 1 p
Xv (u, v) = (0, 4u cos v, u sin v)
dS =
1 + r2 drd✓
sol) 8⌧–⌧ ¸¥ƒ ‰⌧¿⇠ ·t X–
Xu (u, v) = (2u, 4 sin v, cos v),
v=⇡
t
X
t¿\
(Xu ⇥ Xv )(u, v) = ( 4u, 2u2 sin v, 8u2 cos v)
t‰. X(1, ⇡) = (1, 0, 1)t¿\ ¸¥ƒ ⇣–⌧ ⌘…tX ) ›
@ 4(x 1) 8(z + 1) = 0 â,
t‰.
x + 2z =
e) X(u, v) = (4uv, 2u sin v, v cos u),
0
0
Z 2⇡ ⇣ p
⌘ 1
1
=
r 1 + r2 + sinh 1 (r)
d✓
2
0
0
⇣p
⌘
=⇡
2 + sinh 1 (1)
1
3. …t x + 2y + 3z = 12 ⌘–⌧ Ì {(x, y, z) 2 R3 |x 0, y
0, z 0}– ⌧\⌧ ÄÑ
sol) ¸¥ƒ ·tD
✓
◆
12 x 2y
12 x
X(x, y) = x, y,
, 0  x  12, 0  y 
3
2
t‰.
u = 0,
sol) 8⌧–⌧ ¸¥ƒ ‰⌧¿⇠ ·t X–
v=⇡
t
Xu (u, v) = (4v, 2 sin v, v sin u), Xv (u, v) = (4u, 2u cos v, cos u)
t¿\
\ ò¿¥t ·t XX t å dSî
p
14
dS =
dxdy
3
t¿\, lX‡ê Xî ◆tî
(Xu ⇥ Xv )(u, v)
ZZ
= (2uv cos v sin u + 2 cos u sin v,
4v(cos u + u sin u), 8u(v cos v
sin v))
t‰. X(0, ⇡) = (0, 0, ⇡)t¿\ ¸¥ƒ ⇣–⌧ ⌘…tX ) ›@
y=0
t‰.
0r1
t‰.
154
dS =
X
Z 12Z 122 x p
14
dydx
3
p0 0Z 12
14
12 x
=
dx
3
2
p 0
= 12 14
SOLUTION
14 ·t
4. …t 2x + 3y + 5z = 10⌘–⌧ –0e x2 + y 2  16X ¥Ä t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
⇣î Ωƒ– àî ÄÑ
p
dS
=
4x2 + 4y 2 + 1dxdy
sol) ¸¥ƒ ·tD
X
x2 +y 2 11
✓
◆
Z 2⇡Z p11 p
10 2x 3y
2
2
X(x, y) = x, y,
, x + y  16
=
r 4r2 + 1drd✓
5
0
0
Z 2⇡Z 45 p
1
\ ò¿¥t ·t XX t å dSî
=
tdtd✓
8
0
1
p
⌘
p
1⇣
38
=
135 5 1 ⇡
dS =
dxdy
6
5
t‰.
t¿\, lX‡ê Xî ◆tî
p
ZZ
ZZ
7. ·t z = xy ⌘–⌧ –0e x2 + y 2  9X ¥Ä ⇣î Ωƒ–
38
dS =
dxdy
àî ÄÑ
X
x2 +y 2 16 5
Z 2⇡Z 4 p
sol) ¸¥ƒ ·tD
38
=
rdrd✓
X(x, y) = (x, y, xy) , x2 + y 2  9
5
0
0
16 p
\ ò¿¥t ·t XX t å dSî
=
38⇡
p
5
dS = x2 + y 2 + 1dxdy
t‰.
5. Ï<t z = 16 x
sol) ¸¥ƒ ·tD
2
t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
dS =
2
y ⌘–⌧ xy …t ⌅Ω– àî ÄÑ
X
=
X(x, y) = x, y, 16
x
2
y
2
,
2
X
=
x2 +y 2 16
Z 2⇡Z 4 p
r
0
x + y  16
=
p
4x2 + 4y 2 + 1dxdy
4r2 + 1drd✓
1)⇡
p
x2 + y 2 D ÃqXî ÄÑ
dS = 16 sin d d✓
t¿\, lX‡ê Xî ◆tî
ZZ
Z 2⇡Z ⇡/4
dS =
16 sin d d✓
X
0
0
p
= 16⇡(2
2)
y 2 ⌘–⌧ …t z =
y2 ,
p
2
(10 10
3
X( , ✓) = (4 sin cos ✓, 4 sin sin ✓, 4 cos ) ,
⇡
0   , 0  ✓  2⇡
4
\ ò¿¥t ·t XX t å dSî
t‰.
x2
r2 + 1drd✓
0
8. lt x2 + y 2 + z 2 = 16 ⌘–⌧ z
sol) ¸¥ƒ ·tD
0
X(x, y) = x, y, 9
x2 + y 2 + 1dxdy
t‰.
Z 2⇡Z 65 p
1
=
tdtd✓
8
0
1
⇣
⌘
p
1
=
65 65 1 ⇡
6
6. Ï<t z = 9 x2
sol) ¸¥ƒ ·tD
r
0
2
\ ò¿¥t ·t XX t å dSî
p
dS = 4x2 + 4y 2 + 1dxdy
t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
dS =
x2 +y 2 9
Z 2⇡Z 3 p
p
2 ⌅Ω– àî ÄÑ
x2 + y 2  11
\ ò¿¥t ·t XX t å dSî
p
dS = 4x2 + 4y 2 + 1dxdy
t‰.
9. lt x2 + y 2 + z 2 = 12 ⌘–⌧ z x2 + y 2 D ÃqXî ÄÑ
sol) ¸¥ƒ ·tD
⇣p
⌘
p
p
X( , ✓) =
12 sin cos ✓, 12 sin sin ✓, 12 cos
,
⇡
0   , 0  ✓  2⇡
6
155
SOLUTION
12. –0et x2 + z 2 = 4X ¥Ä– àî ·t y = x2 + z 2
sol) ¸¥ƒ pt‰D ˘à å⌅XÏ › Xt, â y@ z| 
∏¥ t
D –0et x2 + y 2 = 4@ ·t z = x2 + y 2 <\
XXÏ tt` ⇠ à‰. t⌧ ¸¥ƒ ·tD
\ ò¿¥t ·t XX t å dSî
dS = 12 sin d d✓
t¿\, lX‡ê Xî ◆tî
ZZ
dS =
X
X(x, y) = x, y, x2 + y 2 , x2 + y 2  4
Z 2⇡Z ⇡/6
0
0
= 12⇡(2
12 sin d d✓
p
3)
\ ò¿¥t ·t XX t å dSî
p
dS = 4x2 + 4y 2 + 1dxdy
t‰.
10. lt x2 + y 2 + z 2 = 4z ⌘–⌧ z x2 + y 2 D ÃqXî ÄÑ
sol) ¸¥ƒ ltD ltå\\ ò¿¥t ⇢ = 4 cos t‰. t⌧
¸¥ƒ ·tD
X( , ✓) = 4 cos sin cos ✓, 4 cos sin sin ✓, 4 cos2
⇡
0   , 0  ✓  2⇡
6
dS =
0
0
y2 ,
1¸ x2 + y 2  9| Ãq
1  x2 + y 2  9
\ ò¿¥t ·t XX t å dSî
p
dS = 4x2 + 4y 2 + 1dxdy
=
1x2 +y 2 9
Z 2⇡Z 3 p
r
0
t‰.
1
x2 t‡ x
y2 x
2y) ,
y
Ì ⌅Ω– ìx …t
x2 ,
x
y2
\ ò¿¥t ·t XX t å dSî
p
dS = 6dxdy
11. ·t z = x2 y 2 ⌘–⌧ x2 + y 2
Xî ÄÑ
sol) ¸¥ƒ ·tD
X
0
X(x, y) = (x, y, 6 + x
t‰.
t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
dS =
Z 2⇡Z 2 p
=
r 4r2 + 1drd✓
0
8|sin 2 |d d✓
= 4⇡
X(x, y) = x, y, x2
x2 +y 2 4
p
4x2 + 4y 2 + 1dxdy
Z 2⇡Z 17 p
1
=
tdtd✓
8
0
1
⌘
1 ⇣ p
= ⇡ 17 17 1
6
13. xy…t–⌧ y
x + 2y + z = 6
sol) ¸¥ƒ ·tD
t¿\, lX‡ê Xî ◆tî
X
X
t‰.
dS = 8|sin 2 |d d✓
Z 2⇡Z ⇡/6
t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
dS =
,
\ ò¿¥t ·t XX t å dSî
ZZ
14 ·t
p
4x2 + 4y 2 + 1dxdy
4r2 + 1drd✓
Z 2⇡Z 37 p
1
=
tdtd✓
8
0
5
p ⌘
1 ⇣ p
= ⇡ 37 37 5 5
6
t¿\, lX‡ê Xî ◆tî
ZZ
Z 1Z px p
dS =
6dydx
X
0 x2
p
6
=
3
t‰.
p
p
14. z = 8x + y 3/2 3 y (0  x  1, 9  y  27)X ¯ò⌅
sol) ¸¥ƒ ·tD
⇣
p
p ⌘
X(x, y) = x, y, 8x + y 3/2 3 y , 0  x  1, 9  y  27
\ ò¿¥t ·t XX t å dSî
✓
◆
3 p
1
dS =
y+ p
dxdy
2
y
t¿\, lX‡ê Xî ◆tî
◆
ZZ
Z 1Z 27 ✓
3 p
1
dS =
y+ p
dydx
2
y
X
0 9
p
= 90 3 36
156
SOLUTION
µ8⌧ 14.3 .
t‰.
✓
¸¥ƒ ·t S@ ‰h⇠ f –
◆2
1
1
+ y 2  t‡ z
2
4
x2 + y 2 + z 2 = 1
sol) ¸¥ƒ ·tD
⇣
p
X(x, y) = x, y, 1
✓
◆2
1
1
x
+ y2 
2
4
15.
14 ·t
x
0x ÄÑ– ⌧\⌧ lt
x2
t t Ñ
lX‹$.
f (x, y, z)dS|
S
p
1. f (x, y, z) = 2 x2 + y 2 ; Sî ‰⌧¿⇠·t X(r, ✓) =
(r cos ✓, r sin ✓, ✓), (0  ✓  2⇡, 0  r  1)
⌘
y2 ,
sol)
Xr (r, ✓) = (cos ✓, sin ✓, 0)
X✓ (r, ✓) = ( r sin ✓, r cos ✓, 1)
\ ò¿¥t ·t XX t å dSî
r
1
dS =
dxdy
1 x2 y 2
t‡
dS = ||(Xr ⇥ X✓ )(r, ✓)|| drd✓ =
t¿\, lX‡ê Xî ◆tî
ZZ
ZZ
1
p
dS =
1 2
1
2
X
(x 2 ) +y  4 1 x2
Z ⇡/2 Z cos ✓
r
p
=
drd✓
1 r2
⇡/2 0
Z ⇡/2 Z 1
1
p dtd✓
=
2
2
t
⇡/2 sin ✓
Z ⇡/2
(1 |sin ✓|)d✓
=
y2
p
r2 + 1drd✓
t‰. f (X(r, ✓)) = 2r t¿\
ZZ
f (x, y, z)dS
ZS Z
=
f (X(r, ✓)) ||(Xr ⇥ X✓ )(r, ✓)|| drd✓
dxdy
[0,1]⇥[0,2⇡]
=
Z 2⇡Z 1
0
2r
0
t‰.
⇡/2
=⇡
ZZ
p
p
4
r2 + 1drd✓ = ⇡(2 2
3
2. f (x, y, z) = x3 y 2 z; Sî …t z = 3x
Ì [0, 2] ⇥ [0, 1]⌅Ω– ⌧\⌧ ÄÑ
2
t‰.
2y
1)
1 ⌘–⌧ xy…tX
sol) ·t S| ‰⌧TXt
16. xy…tX 8⇣ (0, 0), (0, 1), (1, 1)D -”⇣<\ Xî º
Ì ⌅Ω– ⌧\⌧ ·t z = 3 + 2x2 + 3y
sol) ¸¥ƒ ·tD
X(x, y) = x, y, 3 + 2x2 + 3y ,
0  x  1,
0
=
0
t‰.
x)
1 ⇣ p
=
5 10
24
p
16x2 + 10dx
p
26 + 30 sinh
(0  x  2, 0  y  1)
Xx (x, y) = (1, 0, 3),
Xy (x, y) = (0, 1, 2)
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
x
(1
1)
t‡
t¿\, lX‡ê Xî ◆tî
ZZ
Z 1Z 1 p
dS =
16x2 + 10dydx
Z 1
2y
t‰. ¯Ït
xy1
\ ò¿¥t ·t XX t å dSî
p
dS = 16x2 + 10dxdy
X
X(x, y) = (x, y, 3x
1
p
(2 10/5)
⌘
p
14dxdy
t‰. f (X(x, y)) = x3 y 2 (3x 2y 1) t¿\
ZZ
f (x, y, z)dS
ZS Z
=
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
[0,2]⇥[0,1]
=
Z 1Z 2 p
0
t‰.
157
0
14x3 y 2 (3x
2y
1)dxdy =
46 p
14
15
SOLUTION
14 ·t
3. f (x, y, z) = x2 + y 2 + z; Sî …t x 2y z = 5⌘–⌧ 5. f (x, y, z) = x2 + 2yx + zx; Sî (1, 0, 0), (0, 2, 0), (0, 0, 3)D
xy…tX 8⇣ (0, 0), (0, 1), (1, 1)D -”⇣<\ Xî º
X 8 -”⇣<\ Xî º
¥Ä@ Ωƒ
⌅Ω– ⌧\⌧ ÄÑ
sol) Sî …t 6x+3y +2z 6 = 0⌅X ·tt¿\ t| t©XÏ
‰⌧TXt
sol) ·t S| ‰⌧TXt
X(x, y) = (x, y, x
2y
5)
(0  x  y, 0  y  1)
✓
X(x, y) =
t‰. ¯Ït
Xx (x, y) = (1, 0, 1),
x, y, 3
3
y
2
3x
◆
y
+ 1, 0  y  2)
2
(0  x 
t‰. ¯Ït
Xy (x, y) = (0, 1, 2)
Xx (x, y) = (1, 0, 3),
t‡
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
t‰. f (X(x, y)) = x2 + y 2 + x
ZZ
f (x, y, z)dS
2y
p
Xy (x, y) = (0, 1, 3/2)
6dxdy
t‡
5 t¿\
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
S
=
Z 1Z y
0
=
0
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
Z 1Z y p
0
6(x2 + y 2 + x
2y
5)dxdy =
0
t‰. f (X(x, y)) =
8p
6
3
ZZ
t‰.
=
=
X(x, y) = (x, y, 4
x
f (x, y, z)dS
Z 2Z
Z 2Z
0
sol) ·t S| ‰⌧TXt
y)
(0  x  4, 0  y  4
t‰. ¯Ït
x)
2x2 t¿\
S
0
4. f (x, y, z) = yz; Sî …t x + y + z = 4 ⌘–⌧ ⌧1Ñ– ÄÑ
1
xy + 3x
2
7
dxdy
2
y
2 +1
0
y
2 +1
0
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
7
2
✓
1
xy + 3x
2
◆
21
2x2 dxdy =
8
t‰.
6. f (x, y, z) = x2 y 2 ; Sî …t z = 2x y + 3 ⌘–⌧ –0et
x2 + y 2 = 16X ¥Äò Ωƒ– àî ÄÑ
Xx (x, y) = (1, 0, 1),
Xy (x, y) = (0, 1, 1)
sol) ·t S| ‰⌧TXt
t‡
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
t‰. f (X(x, y)) = y(4 x
ZZ
f (x, y, z)dS
p
3dxdy
X(x, y)
= (x, y, 2x
y) t¿\
y + 3) (
t‰. ¯Ït
p
16
y2  x 
p
16
y2 ,
S
Z 4Z 4 x
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dydx
p
Z 4Z 4 x p
32 3
=
3y(4 x y)dydx =
3
0 0
=
0
Xx (x, y) = (1, 0, 2),
0
Xy (x, y) = (0, 1, 1)
t‡
t‰.
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
158
p
6dxdy
4  y  4)
SOLUTION
t‰. f (X(x, y)) = x2 y 2 t¿\
ZZ
f (x, y, z)dS
S
ZZ
=
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
14 ·t
t‰. ¯Ït
X ( , ✓) = (3 cos cos ✓, 3 cos sin ✓, 3 sin ),
X✓ ( , ✓) = ( 3 sin sin ✓, 3 sin cos ✓, 0)
t‡
x2 +y 2 16
=
Z 4 Z p16 y2 p
p
4
=
=
0
0
0
dS = ||(X ⇥ X✓ )( , ✓)|| d d✓ = 9 sin d d✓
y 2 )dxdy
t‰. f (X( , ✓)) = 9(sin2
cos2 ) t¿\
ZZ
f (x, y, z)dS
S
ZZ
=
f (X( , ✓)) ||(X ⇥ X✓ )( , ✓)|| d d✓
16 y 2
Z 2⇡Z 4 p
0
6(x2
Z 2⇡Z 4 p
6((r cos ✓)2
(r sin ✓)2 )rdrd✓
6r3 (cos2 ✓
sin2 ✓)drd✓ = 0
[0, ⇡
2 ]⇥[0,2⇡]
t‰.
=
7. f (x, y, z) = 2z 2 ; Sî lt x2 + y 2 + z 2 = 4 ⌘–⌧
{(x, y, z) 2 R3 |x 0, y 0, z 0}– ⌧\⌧ ÄÑ
sol) ·t S| ‰⌧TXt
X( , ✓)
(0 
= (2 sin cos ✓, 2 sin sin ✓, 2 cos )

⇡
⇡
, 0✓ )
2
2
=
Z 2⇡Z ⇡2
0
0
0
0
Z 2⇡Z ⇡2
9(sin2
cos2 )9 sin d d✓
81 sin (sin2
cos2 )d d✓ = 54⇡
t‰.
p
9. f (x, y, z) = x + y + z; Sî –‘t z =
x2 + y 2 ⌘–⌧
2
2
–0et x + y = 1X ¥Äò Ωƒ– àî ÄÑ
sol) ·t S| ‰⌧TXt
t‰. ¯Ït
X(✓, z) = (z cos ✓, z sin ✓, z)
X ( , ✓) = (2 cos cos ✓, 2 cos sin ✓, 2 sin ),
(0  ✓  2⇡, 0  z  1)
t‰. ¯Ït
X✓ ( , ✓) = ( 2 sin sin ✓, 2 sin cos ✓, 0)
X✓ (✓, z) = ( z sin ✓, z cos ✓, 0),
t‡
Xz (✓, z) = (cos ✓, sin ✓, 1)
dS = ||(X ⇥ X✓ )( , ✓)|| d d✓ = 4 sin d d✓
t‡
t‰. f (X( , ✓)) = 8 cos2 t¿\
ZZ
f (x, y, z)dS
ZS Z
=
f (X( , ✓)) ||(X ⇥ X✓ )( , ✓)|| d d✓
dS = ||(X✓ ⇥ Xz )(✓, z)|| d✓dz =
=
0
32 cos2 sin d d✓ =
0
16
⇡
3
[0,2⇡]⇥[0,1]
=
t‰.
Z 1Z 2⇡ p
0
8. f (x, y, z) = x2 + y 2
⌅Ω ⇠lt (z 0)
z 2 ; Sî lt x2 + y 2 + z 2 = 9 ⌘–⌧
2z 2 (cos ✓ + sin ✓ + 1)d✓dz =
0
p
2 2
⇡
3
t‰.
10. f (x, y, z) = y 2 z 2 ; Sî ·t z 2 = x2 + y 2 ⌘–⌧ z = 1¸
z = 4 ¨t– àî ÄÑ
sol) ·t S| ‰⌧TXt
X( , ✓)
= (3 sin cos ✓, 3 sin sin ✓, 3 cos )
2zd✓dz
t‰. f (X(✓, z)) = z(cos ✓ + sin ✓ + 1) t¿\
ZZ
f (x, y, z)dS
ZS Z
=
f (X(✓, z)) ||(X✓ ⇥ Xz )(✓, z)|| d✓dz
⇡
[0, ⇡
2 ]⇥[0, 2 ]
Z ⇡2 Z ⇡2
p
(0 
⇡
 , 0  ✓  2⇡)
2
sol) ·t S| ‰⌧TXt
159
X(✓, z) = (z cos ✓, z sin ✓, z)
(0  ✓  2⇡, 1  z  4)
SOLUTION
14 ·t
sol) ·t S1 , S2 , S3 , S4 , S5 , S6 |
t‰. ¯Ït
S1 = {(x, y, 0) 2 R3 | 0  x  1, 0  y  1},
X✓ (✓, z) = ( z sin ✓, z cos ✓, 0),
S2 = {(x, y, 1) 2 R3 | 0  x  1, 0  y  1},
Xz (✓, z) = (cos ✓, sin ✓, 1)
S3 = {(x, 0, z) 2 R3 | 0  x  1, 0  z  1},
t‡
dS = ||(X✓ ⇥ Xz )(✓, z)|| d✓dz =
p
S4 = {(x, 1, z) 2 R3 | 0  x  1, 0  z  1},
S5 = {(0, y, z) 2 R3 | 0  y  1, 0  z  1},
2zd✓dz
S6 = {(1, y, z) 2 R3 | 0  y  1, 0  z  1},
2
4
t‰. f (X(✓, z)) = z sin ✓ t¿\
ZZ
<\ XXt S = S1 [ S2 [ S3 [ S4 [ S5 [ S6 î p
·tt‡
ZZ
ZZ
ZZ
ZZ
f dS =
f dS +
f dS +
f dS
S
S1
S2
S3
ZZ
ZZ
ZZ
+
f dS +
f dS +
f dS
f (x, y, z)dS
ZZ
=
f (X(✓, z)) ||(X✓ ⇥ Xz )(✓, z)|| d✓dz
S
[0,2⇡]⇥[1,4]
=
Z 4Z 2⇡ p
1
2z 5 sin2 ✓d✓dz =
0
p
1365 2
⇡
2
S4
t‰.
S5
S6
t‰. t⌧ t ÑX ✓‰D 0\ lXê.
1. ·t S1 D ‰⌧T Xt
11. f (x, y, z) = x; Sî ·t x = y 2 + z 2 ⌘–⌧ –0et
y 2 + z 2 = 4X ¥Äò Ωƒ– àî ÄÑ
X(x, y) = (x, y, 0)
(0  x  1, 0  y  1)
t‰ ¯Ït
sol) ·t S| ‰⌧TXt
X(✓, r) = (r2 , r cos ✓, r sin ✓)
Xx (x, y) = (1, 0, 0)
Xy (x, y) = (0, 1, 0)
(0  ✓  2⇡, 0  r  2)
t‡
t‰. ¯Ït
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy = dxdy
X✓ (✓, r) = (0, r sin ✓, r cos ✓),
t‰. f (X(x, y)) = x2 + 2y 2 t¿\
ZZ
f (x, y, z)dS
S1
ZZ
=
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
Xr (✓, r) = (2r, cos ✓, sin ✓)
t‡
dS = ||(X✓ ⇥ Xr )(✓, r)|| d✓dr = r
t‰. f (X(✓, r)) = r2 t¿\
p
4r2 + 1d✓dr
[0,1]⇥[0,1]
=
ZZ
0
f (x, y, z)dS
ZS Z
=
f (X(✓, r)) ||(X✓ ⇥ Xr )(✓, r)|| d✓dr
Z 2Z 2⇡
0
0
r
3
p
4r2 + 1d✓dr =
x2 + 2y 2 dxdy = 1
0
t‰.
2. ·t S2 D ‰⌧T Xt
[0,2⇡]⇥[0,2]
=
Z 1Z 1
X(x, y) = (x, y, 1)
p
⇡
(1 + 391 17)
60
(0  x  1, 0  y  1)
t‰ ¯Ït
Xx (x, y) = (1, 0, 0)
t‰.
Xy (x, y) = (0, 1, 0)
12. f (x, y, z) = x2 + 2y 2 ; Sî T = {(x, y, z) 2 R3 |0  x 
1, 0  y  1, 0  z  1}X Ωƒ
160
t‡
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy = dxdy
‹
SOLUTION
14 ·t
5. ·t S5 D ‰⌧T Xt
t‰. f (X(x, y)) = x2 + 2y 2 t¿\
ZZ
f (x, y, z)dS
S2
ZZ
=
f (X(x, y)) ||(Xx ⇥ Xy )(x, y)|| dxdy
t‰ ¯Ït
[0,1]⇥[0,1]
=
Z 1Z 1
0
2
(0  y  1, 0  z  1)
X(y, z) = (0, y, z)
Xy (y, z) = (0, 1, 0)
2
x + 2y dxdy = 1
Xz (y, z) = (0, 0, 1)
0
t‡
t‰.
3. ·t S3 D ‰⌧T Xt
X(x, z) = (x, 0, z)
dS = ||(Xy ⇥ Xz )(y, z)|| dydz = dydz
(0  x  1, 0  z  1)
t‰. f (X(y, z)) = 2y 2 t¿\
ZZ
f (x, y, z)dS
ZS5Z
=
f (X(y, z)) ||(Xy ⇥ Xz )(y, z)|| dydz
t‰ ¯Ït
Xx (x, z) = (1, 0, 0)
Xz (x, z) = (0, 0, 1)
[0,1]⇥[0,1]
t‡
=
dS = ||(Xx ⇥ Xz )(x, z)|| dxdz = dxdz
0
2
t‰. f (X(x, z)) = x t¿\
ZZ
f (x, y, z)dS
S3
ZZ
=
f (X(x, z)) ||(Xx ⇥ Xz )(x, z)|| dxdz
=
0
2y 2 dydz =
0
6. ·t S6 D ‰⌧T Xt
X(y, z) = (1, y, z)
t‰ ¯Ït
1
x dxdz =
3
0
Xy (y, z) = (0, 1, 0)
Xz (y, z) = (0, 0, 1)
4. ·t S4 D ‰⌧T Xt
X(x, z) = (x, 1, z)
t‡
(0  x  1, 0  z  1)
dS = ||(Xy ⇥ Xz )(y, z)|| dydz = dydz
t‰ ¯Ït
t‰. f (X(y, z)) = 1 + 2y 2 t¿\
ZZ
f (x, y, z)dS
S6
ZZ
=
f (X(y, z)) ||(Xy ⇥ Xz )(y, z)|| dydz
Xx (x, z) = (1, 0, 0)
Xz (x, z) = (0, 0, 1)
t‡
[0,1]⇥[0,1]
dS = ||(Xx ⇥ Xz )(x, z)|| dxdz = dxdz
t‰. f (X(x, z)) = x2 + 2 t¿\
ZZ
f (x, y, z)dS
ZS4Z
=
f (X(x, z)) ||(Xx ⇥ Xz )(x, z)|| dxdz
[0,1]⇥[0,1]
Z 1Z 1
2
x + 2dxdz =
0
0
(0  y  1, 0  z  1)
2
t‰.
=
2
3
t‰.
[0,1]⇥[0,1]
Z 1Z 1
Z 1Z 1
7
3
=
Z 1Z 1
0
1 + 2y 2 dydz =
0
5
3
t‰.
¯Ï¿\
ZZ
t‰.
t‰.
161
f dS = 1 + 1 +
S
1 7 2 5
+ + + =7
3 3 3 3
SOLUTION
13. f (x, y, z) = xz; T
–0et y 2 + z 2 = 16¸ P …t
x = 0, x + y = 5\ Xϯx Ì| L, Sî T X Ωƒ
15 °0•
t‰ ¯Ït
Xy (y, z) = ( 1, 1, 0)
Xz (y, z) = (0, 0, 1)
sol) ·t S1 , S2 , S3 |
t‡
S1 = {(0, y, z) 2 R3 | y 2 + z 2  16},
S2 = {(x, 4 sin ✓, 4 cos ✓) 2 R3 | 0  ✓  2⇡, 0  x  5
y, y, z) 2 R3 | y 2 + z 2  16},
S3 = {(5
<\
dS = ||(Xy ⇥ Xz )(y, z)|| dydz =
4 sin ✓},
S1
S2
2dydz
t‰. f (X(y, z)) = z(5 y) t¿\
ZZ
f (x, y, z)dS
ZS4Z
=
f (X(y, z)) ||(Xy ⇥ Xz )(y, z)|| dydz
XXt S = S1 [ S2 [ S3 î p
‹·tt‡
ZZ
ZZ
ZZ
ZZ
f dS =
f dS +
f dS +
f dS
S
p
y 2 +z 2 16
S3
=
t‰. t⌧ t ÑX ✓‰D 0\ lXê.
Z 4 Z p16 z2
4
1. ·t S1 D ‰⌧T Xt
p
z(5
y)dydz = 0
16 z 2
t‰.
X(r, ✓) = (0, r cos ✓, r sin ✓)
(0  r  4, 0  ✓  2⇡)
¯Ï¿\
ZZ
t‰. ¯Ït f (X(r, ✓)) = 0 t¿\
ZZ
f (x, y, z)dS = 0
f dS = 0 + 0 + 0 = 0
S
t‰.
S1
t‰.
15
2. ·t S2 D ‰⌧T Xt
µ8⌧ 15.1 .
1. V (x, y) = xy | L ¯ò∏∏ °0• rV | xy …t– ¯
º<\ ò¿¥‹$.
sol) ¸¥ƒ °0•X ¯º@ ‰L¸ ⇡‰.
X(r, ✓) = (r, 4 sin ✓, 4 cos ✓)
(0  ✓  2⇡, 0  x  5
°0•
4 sin ✓)
t‰ ¯Ït
Xr (r, ✓) = (1, 0, 0)
X✓ (r, ✓) = (0, 4 cos ✓, 4 sin ✓)
t‡
dS = ||(Xr ⇥ X✓ )(r, ✓)|| dxdy = 4dxdy
t‰. f (X(r, ✓)) = 4r cos ✓ t¿\
ZZ
f (x, y, z)dS
S2
=
=
Z 2⇡Z 5 4 sin ✓
0
0
0
0
Z 2⇡Z 5 4 sin ✓
f (X(x, y)) ||(Xr ⇥ X✓ )(r, ✓)|| drd✓
16r cos ✓drd✓ = 0
t‰.
2. ‰L– ¸¥ƒ °0• F X ⌧∞ r · F ¸ å⌅ r ⇥ F |
lX‹$.(ıÃ Ö‹\‰.)
3. ·t S3 D ‰⌧T Xt
X(y, z)
= (5
y, y, z) (
p
16
z2  y 
p
a) F (x, y, z) = (y, z, x)
16
z 2 , 4  z  4)
162
SOLUTION
15 °0•
t‰.
sol)
r · F = 0,
b) f (x, y, z) = xy 2 + yz 2 + zx2 , (x, y, z) 2 R3
r ⇥ F = ( 1, 1, 1).
proof ) h⇠ f X ¯ò∏∏î
b) F (x, y, z) = (xy, yz, zx)
rf = 2xz + y 2 , 2xy + z 2 , 2yz + x2
sol)
t‡ rf X å⌅@
r · F = x + y + z,
r ⇥ (rf ) = (2z
r ⇥ F = ( y, z, x).
2z, 2x
2x, 2y
2y) = 0
t‰.
c) ✓
F (x, y, z)
◆
1 2
2
2 1
2
2
2 1
2
2
2
=
(x + y + z ), (x + y + z ), (x + y + z )
2
3
4
c) f (x, y, z) = xy 2 z 3 , (x, y, z) 2 R3
proof ) h⇠ f X ¯ò∏∏î
sol)
2
1
r · F = x + y + z,
2
✓ 3
1
2
r⇥F =
y
z, z
2
3
d) F (x, y, z) =
✓
rf = y 2 z 3 , 2xyz 3 , 3xy 2 z 2
1 2
x, x
2 3
t‡ rf X å⌅@
◆
r ⇥ (rf ) = (6xyz 2
y .
2yz
2xz
2xy
, 2
, 2
2
2
2
2
2
x + y + z x + y + z x + y2 + z2
◆
6xyz 2 , 3y 2 z 2
3y 2 z 2 , 2yz 3
2yz 3 ) = 0
t‰.
d) f (x, y, z) = ln (x2 + y 2 + z 2 ), (x, y, z) 6= (0, 0, 0)
sol)
proof ) h⇠ f X ¯ò∏∏î
4xyz
✓
◆
r·F =
,
2x
2y
2z
(x2 + y 2 + z 2 )2
rf
=
,
,
✓
◆
x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2
4x3
4y(x2 z 2 )
4z 3
r⇥F =
,
,
.
(x2 + y 2 + z 2 )2 (x2 + y 2 + z 2 )2 (x2 + y 2 + z 2 )2
t‡ rf X å⌅@
3. ‰L ‰h⇠ f –
a) f (x, y, z) =
p
r ⇥ (rf )
✓
4yz
4yz
=
,
2
2
2
2
2
(x + y + z )
(x + y 2 + z 2 )2
4xz
4xz
,
2
2
2
2
2
(x + y + z )
(x + y 2 + z 2 )2
◆
4xy
4xy
=0
(x2 + y 2 + z 2 )2
(x2 + y 2 + z 2 )2
t r ⇥ (rf ) = 0 ÑD Ùt‹$.
x2 + y 2 + z 2 , (x, y, z) 6= (0, 0, 0)
proof ) h⇠ f X ¯ò∏∏î
p
rf =
x
x2 + y 2 + z 2
t‡ rf X å⌅@
=
r ⇥ (rf )
✓
,p
y
z
x2 + y 2 + z 2
,p
x2 + y 2 + z 2
yz
yz
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
xz
(x2 + y 2 + z 2 )3/2
xy
(x2 + y 2 + z 2 )3/2
,
xz
,
(x2 + y 2 + z 2 )3/2
◆
xy
=0
(x2 + y 2 + z 2 )3/2
!
t‰.
4. r · F = 0 t‡ r · G = 0 t| Xê. ‰L °0• ⌘ ⌧∞t
0x °0•D >‡ ¯⌥¿ Jî Ω∞î ⇠@| ⌧‹X‹$.
(a) F + G
sol) °0• F | F = (P, Q, R), °0• G | G = (S, T, U )|‡
Pt(P, Q, R, S, T, U î h⇠) r · F = 0 t‡ r · G = 0t¿\
r · F = Px + Qy + Rz = 0,
r · G = Sx + T y + U z = 0
163
SOLUTION
15 °0•
@f
1 2
t‰.
t‡
6=
x sin y t¿\ F = rf | ÃqXî f t¨\
@y
2
°0• F + G = (P + S, Q + T, R + U ) t¿\ tX ⌧∞D
‰î
– ®⌧⌧‰. ¯Ï¿\ ¸¥ƒ °0• F î ¯ò∏∏
ltÙt
°0•t D»‰.
r · (F + G) = (P + S)x + (Q + T )y + (R + U )z
= (Px + Sx ) + (Qy + Ty ) + (Rz + Uz )
6. ‰L Ò›D ùÖX‹$.
= (Px + Qy + Rz ) + (Sx + Ty + Uz )
=0
a) r(cf + g) = crf + rg, (cî ¡⇠)
proof )
t‰.
r(cf + g) = ((cf + g)x , (cf + g)y , (cf + g)z )
(b) F ⇥ G
sol) °0• F ⇥ G X ⌧∞@ 0t D»‰.
(⇠@) °0• F , G |
F = (0, x, 0), G = (0, 0, 1) t|‡
Pê ¯Ït
= (cfx + gx , cfy + gy , cfz + gz )
= (cfx , cfy , cfz ) + (gx , gy , gz )
= c (fx , fy , fz ) + (gx , gy , gz )
= crf + rg.
F ⇥ G = (0, x, 0) ⇥ (0, 0, 1) = (x, 0, 0)
b) r(f g) = f rg + grf
proof )
t‡ tX ⌧∞@
r · (F ⇥ G) = 1 6= 0
r(f g) = ((f g)x , (f g)y , (f g)z )
t‰.
= (fx g + f gx , fy g + f gy , fz g + f gz )
(c) (F · G)F
sol) °0• (F · G)F X ⌧∞@ 0t D»‰.
(⇠@) °0• F , G |
F = (x, 0, z), G = (1, 0, 0) t|‡
Pê ¯Ït
= (f gx , f gy , f gz ) + (fx g, fy g, fz g)
= f (gx , gy , gz ) + g (fx , fy , fz )
= f rg + grf.
✓ ◆
f
grf f rg
=
(Ë, ÑXX x– t g(x) 6= 0t‰.)
g
g2
t‡ tX ⌧∞@
proof )
r · ((F · G)F ) = 2x x = x 6= 0
✓ ◆
✓ ◆ ✓ ◆ ✓ ◆ !
f
f
f
f
r
=
,
,
t‰.
g
g x
g y
g z
✓
◆
fx g f gx fy g f gy fz g f gz
=
,
,
5. F (x, y) = (x cos y, y sin x)\ ¸¥ƒ °0• F î ¯ò∏∏
g2
g2
g2
°0•t DÿD Ùt‹$.
✓
◆ ✓
◆
fx g fy g fz g
f gx
f gy
f gz
=
,
,
+
, 2 , 2
g2 g2 g2
g2
g
g
proof ) °0• F
¯ò∏∏ °0•t$t F = rf | ÃqX
g (fx , fy , fz ) f (gx , gy , gz )
î h⇠ f t¨X|
=
✓ \‰. F ◆= rf | ÃqXî f t¨\‰‡
g2
@f @f
Xt F (x, y) =
,
t¿\
grf f rg
@x @y
=
.
g2
@f
@f
= x cos y,
= y sin x
@x
@y
d) r ⇥ (F + G) = r ⇥ F + r ⇥ G
proof ) F = (P, Q, R), G = (S, T, U )|‡ Xê.
@f
t‰. ¯¨‡
| x– t Ä
ÑXt
¯Ït F + G = (P + S, Q + T, R + U ) t‡
@x
Z
Z
r ⇥ (F + G)
@f
1
dx = x cos ydx = x2 cos y + C
@x
2
=((R + U )y (Q + T )z , (P + S)z (R + U )x ,
(F · G)F = ((x, 0, z) · (1, 0, 0))F = xF = (x2 , 0, zx)
t‰. t⌧ t| y– t⌧ ∏¯ÑXê. ¯Ït
✓
◆
1 2
@
x cos y + C
1 2
2
=
x sin y
@y
2
c) r
(Q + T )x
=(Ry + Uy
=(Ry
(P + S)y )
Qz
Qz , Pz
Tz , Pz + Sz
Rx , Qx
=r ⇥ F + r ⇥ G
164
Rx
Py ) + (Uy
U x , Qx + T x
Tz , Sz
Py
U x , Tx
Sy )
Sy )
SOLUTION
t‰.
L, F
sol) ·
e) r ⇥ (f F ) = f (r ⇥ F ) + rf ⇥ F
proof ) F = (P, Q, R) |‡ Xê.
¯Ït f F = f (P, Q, R) = (f P, f Q, f R)t‡
r ⇥ (f F ) =((f R)y
(f Q)z , (f P )z
(f R)x , (f Q)x
fz Q
f Qz ,
fz P + f Pz
fx R
f Rx , fx Q + f Qx
+ (fy R
=f (Ry
+ (fy R
f Q z , f Pz
f Rx , f Qx
fz Q, fz P
Qz , Pz
fz Q, fz P
t‰. 0|⌧, F
Z
(f P )y )
fy P
C
1
f Py )
=
Z 3
1
fy P )
=
Z 3
(t, t2 , 0) · (1, 2t, 0)dt
(t + 2t3 )dt = 44
1
Py )
fx R, fx Q
1t3
t Öê– \ |@
Z 3
F · ds =
F (C(t)) · C 0 (t)dt
f Py )
fx R, fx Q
Rx , Qx
\ |(work)X l0| lX‹$.
C| Öê ¿¡x êË| Xt
C(t) = (t, t2 , 0),
=(fy R + f Ry
=(f Ry
15 °0•
t‰.
fy P )
=f (r ⇥ F ) + rf ⇥ F
2. ‰L · –
lX‹$.
t‰.
t °0• F (x, y, z) = (x, y, z)X
7. F (x, y, z) = (xz, 1, xy 3 z)t‡ f (x, y, z) = x2 y| L ‰LD a) C(t) = (t, 2t, 3t), 0  t  1
lX‹$.
sol)
(ıÃ Ö‹\‰.)
Z
Z 1
F · ds =
F (C(t)) · C 0 (t)dt
C
0
(a) rf
Z 1
sol) rf = (2xy, x2 , 0).
=
(t, 2t, 3t) · (1, 2, 3)dt
0
(b) r ⇥ F
sol) r ⇥ F =(3xy 2 z, x
=
y 3 z, 0).
14tdt = 7.
0
b) C(t) = (sin t, cos t, 0), 0  t  2⇡
sol)
Z
Z 2⇡
F · ds =
F (C(t)) · C 0 (t)dt
(c) F · (rf )
sol) F · (rf ) = 2x2 yz + x2 .
(d) F ⇥ (rf )
sol) F ⇥ (rf ) = ( x3 y 3 z, 2x2 y 4 z, x3 z
Z 1
C
2xy).
0
=
Z 2⇡
0
8. ¯Ñ •\ °0• F – XÏ r ⇥ F î F @ ⇠¡x ? ¯⌥
‰t ùÖX‡, D»|t ⇠@| ⌧‹X‹$.
sol) ⇠¡t D»‰.
(⇠@) °0• F | F = (1, 0, y 2 ) t|‡ Pê. ¯Ït
r ⇥ F = (2y, 0, 0)
(sin t, cos t, 0) · (cos t,
sin t, 0)dt
= 0.
c) C(t) = (t2 , 2t, 3t3 ),
1t1
sol)
Z
Z 1
F · ds =
F (C(t)) · C 0 (t)dt
C
1
=
t‡
Z 1
1
(r ⇥ F ) · F = (2y, 0, 0) · (1, 0, y 2 ) = 2y 6= 0
=
Z 1
(t2 , 2t, 3t3 ) · (2t, 2, 9t2 )dt
(4t + 2t3 + 27t5 )dt
1
t¿\ r ⇥ F @ °0• F î ⇠¡t D»‰.
= 0.
µ8⌧ 15.2 .
1. òX •(force field) F (x, y, z) = (x, y, z)– Xt ¥§ Öê
3. ‰L
x = 1 Ä0 x = 3 L¿ Ï< y = x2 , z = 0 D 0| ¿¡ D
165
ÑD ƒ∞X‹$,
ÑD
SOLUTION
a)
Z
sol)
xdy
ydx,
C(t) = (sin t, cos t),
C
Z
xdy
ydx =
C
Z 2⇡
0
=
Z 2⇡
sol)
0  t  2⇡
sin t · ( sin tdt)
15 °0•
Z
cos t · (cos tdt)
ydx + (3y 2
x)dy + zdz =
C
Z 1
t5 dt + (3t10
0
=
Z 1
(15t14
4t5 )dt =
0
( 1)dt =
2⇡.
t) · (5t4 dt)
1
.
3
0
Z
b)
sol)
Z
xdx + ydy,
C(t) = (sin ⇡t, cos ⇡t),
C
Z 2
xdx + ydy =
C
c)
6. ‰DÏ¥ ‰⌧¿⇠·
0
0t2
0
a)
Z °0• F · ⌅X ÑXX ⇣ C(t)–⌧ C (t)– ⇠¡tt
F · ds = 0ÑD Ùt‹$.
C
proof ) °0• F
tt
sin ⇡t · (⇡ cos ⇡tdt) + cos ⇡t · ( ⇡ sin tdt)
x2 dx
xy 3 dy + dz, Cî ⇣ ( 1, 0, 1)Ä0 (1, 0, 1)L¿X
t¿\
C
Ï< z = x2 , y = 0
sol) ¸¥ƒ · Cî
Z
1t1
\ ò¿º ⇠ à‰. 0|⌧
x2 dx
C
F · ds =
b) ç °0• F
tt
C(t) = (t, 0, t2 ),
Z
Z
xy 3 dy + dz =
C
=
Z 1
1
Z 1
t2 dt
F (C(t)) · C 0 (t) = 0
F (C(t)) · C 0 (t) = 0t‰.
C
·
⌅X ÑXX ⇣ C(t)–⌧ C 0 (t)@ …â
Z
F · ds =
C
Z
C
kF kds
ÑD Ùt‹$. (F
C(t)–⌧ C 0 (t)@ …âÑ@ F (C(t)) =
0
(t)C (t), (t) > 0D X¯\‰.)
proof ) ds = kC 0 (t)kdt t¿\
0 + 2tdt
Z
(t2 + 2t)dt
1
=
⌅X ÑXX ⇣ C(t)–⌧ C 0 (t)@ ⇠¡
·
= 0.
Z
t ‰L <L– ıX‹$.
C–
2
3
C
F · ds =
=
Z
F (C(t)) · C 0 (t)dt
Z b
a
t‰.
=
Z b
(t)C 0 (t) · C 0 (t)dt
(t)kC 0 (t)kkC 0 (t)kdt
a
4. F (x, y, z)Z = (z, 2x, 3y)t‡ C(t) = (t, t2 , 2t)
L,
F · ds| ƒ∞X‹$.
Ñ
C
sol)
Z
C
F · ds =
=
=
Z 1
0
F (C(t)) · C 0 (t)dt
0
(2t, 2t, 3t2 ) · (1, 2t, 2)dt
Z 1
Z 1
(10t2 + 2t)dt =
0
5.
Ñ
Z
ydx + (3y 2
C
C(t) = (t, t5 , 0)
=
(0  t  1) |
=
Z b
Z
a
kF (C(t))kkC 0 (t)kdt
C
kF kds
1Ω\‰.
7. · CX 8t
M | L,
13
.
3
x)dy + zdz | ƒ∞X‹$. Ï0⌧
ÑD Ùt‹$.
proof ) T‹ àt
lt‡, ¥§ ¡⇠ M > 0t t¨XÏ kF k 
Z
C
F · ds  M l
ÄÒ›– Xt
|F (C(t)) · C 0 (t)|  kF (C(t))kkC 0 (t)k
(0  t  1) t‰.
166
SOLUTION
15 °0•
Z
1ΩX¿\
Z
C
Z b
F · ds =
a
Z b
F (C(t)) · C 0 (t)dt
Z b
a
|F (C(t)) · C 0 (t)| dt
a
kF (C(t))kkC 0 (t)kdt
M
kC 0 (t)kdt = M l


Z b
a
=
‰H\ · t‡ T
Z
Z
C
T · ds =
=
=
Z
Z
C
C4
F · ds =
=
T · ds|
2tdt = 0,
Z 1
F (C3 (t)) · C30 (t)dt
1
Z 1
( 1)dt =
Z 1
F (C4 (t)) · C40 (t)dt
2,
C
C 0 (t)
· C 0 (t)dt
0 (t)k
kC
a
C1
F · ds +
=2+0
Z b
1
Z 1
( 2t)dt = 0,
1
t¿\
Z
Z
F · ds =
Z b
Z
C2
F · ds +
2+0=0
Z
C3
F · ds +
Z
C4
F · ds
t‰.
kC 0 (t)k2
dt
0
a kC (t)k
a
F (C2 (t)) · C20 (t)dt
1
Z 1
1
C(t)X 8t| l\ Pt
Z b
C3
F · ds =
=
CX Ë⌅⌘ °0| L
ƒ∞X‹$.
C 0 (t)
sol) T =
t¿\ ·
kC 0 (t)k
Z 1
1
t¿\ ùÖt DÃ⌧‰.
8. C
C2
F · ds =
10.
kC 0 (t)kdt = l
Ñ
Z
2xyzdx + x2 zdy + x2 ydz| ƒ∞X‹$. Ï0⌧ C
C
î ‹ë⇣ (1, 1, 1)¸ ]⇣ (1, 2, 3)D áî Ë⌧· t‰.
sol) · C| ‰L¸ ⇡t XXt
t‰.
0  t  1,
C(t) = (1, t + 1, 1 + 2t),
9. °0• F = (z + 2xy, x , 3xz )–
XÏ $ ⇣
lX‡ê Xî
Ñ@
(±1, ±1, 0), (±1, ⌥1, 0)D -”⇣<\ ¿î ¨
D 0tî
Z
Z 1
Ñ@ 0ÑD Ùt‹$.
2
2
2xyzdx
+
x
zdy
+
x
ydz
=
(4t + 3)dt = 5
proof ) ¸¥ƒ ¨
· D C(t)| Xê. · C1 , C2 , C3 , C4
C
0
| ‰L¸ ⇡t XXt,
t‰.
C1 (t) = (1, t, 0),
1  t  1,
µ8⌧ 15.3 .
C2 (t) = ( t, 1, 0),
1  t  1,
2
2
1. ¯Ñ •\ º¿⇠ ‰h⇠ f rf (x, y, z) = (2xyzex , zex ,
C3 (t) = ( 1, t, 0),
1  t  1,
2
yex )| Ãq\‰. f (0, 0, 0) = 1| L f (1, 1, 2)| lX‹$.
C4 (t) = (t, 1, 0),
1  t  1,
sol)
2
f (x, y, z) = yzex + C, Cî ¡⇠
lX‡ê Xî
Ñ@
3
Z
C
F · ds =
Z
C1
F · ds +
2
Z
C2
F · ds +
2
Z
C3
F · ds +
Z
2
C4
F · ds
C1
F · ds =
=
Z 1
1
Z 1
2
2
f (x, y, z) = yzex + 1
<\ ò¿º ⇠ à‰. ƒ∞– Xt
Z
2
\ ì<t f î (2xyzex , zex , yex )X ÏP\ h⇠t‰. ⇣,
f (0, 0, 0) = 1 t¥| X¿\ C = 1t‡,
t‰. 0|⌧ f (1, 1, 2) = 2e + 1 t‰.
F (C1 (t)) · C10 (t)dt
1dt = 2,
2. F (x, y, z) = (xy, y, z 2 )| L F = rf | ÃqXî ‰h⇠ f
t¨Xî ?
1
167
SOLUTION
15 °0•
sol) r ⇥ F = (0, 0, x) 6= (0, 0, 0)t¿\ ⌅ ptD ÃqXî a) F (x, y) = (xy, y)t‡ CîZ‹ë⇣ (0, 0)¸ ]⇣ (2, 8)D ¿ò
h⇠ f î t¨X¿ Jî‰.
î · y = 2x2 t‰.
Ñ
F · ds| ƒ∞X‹$.
3. F (x, y, z) = (2xyz + sin x, x2 z, x2 y)| L ‰L <L– ıX‹
$.
sol) ·
C(t) = (t, 2t2 ),
\
Z
(Cî ¡⇠)
cos x + C,
b)
C
2
0
Z 2
(2t3 , 2t2 ) · (1, 4t)dt
10t3 dt = 40
t‰.
=
t, cos ⇡t, t), 0  t  2t‰.
sol) F î Ùt•t‡
(sin2 ⇡t +
b) ⌅X
sol) ⌅
\
C(2) = (2, 1, 2)
F · ds = f (C(2))
f (C(0))
= f (2, 1, 2)
f (0, 1, 0)
0t2
XXt
Z
t¿\
=9
Ñ@ (0, 0)¸ (2, 8)| áî Ω\– XtXî ?
Ñ@ Ω\– XtXî
Ñt‰. · C1 D
C1 (t) = (t, 4t),
C(0) = (0, 1, 0),
C
Z 2
0
F · ds| lX‹$. Ï0⌧ C(t)
Z
C
F · ds =
=
\ Pt F = rf t‰.
Z
0t2
XXt
a) F = rf | ÃqXî ‰h⇠ f | lX‹$.
sol)
f (x, y, z) = x2 yz
C
C|
C
F · ds =
=
Z 2
0
Z 2
(4t2 , 4t) · (1, 4)dt
(4t2 + 16t)dt =
0
128
3
t¿\ a)–⌧ ƒ∞\ ✓¸ |XX¿ Jî‰.
cos 2
6. ‰L ⌘ ¥§ °0•t ‰h⇠X ¯ò∏∏ °0•x¿ ⇣Ë
X‡, t| ÃqXî Ω∞ ¯ ‰h⇠| lX‹$.
t‰.
Z
4. F (x, y, z) = (e sin y, e cos y, 2z)| L
Ñ
F · dsX ✓ a) F (x, y) = (x, y)
sol)
C
p
p
D lX‹$. Ï0⌧ C(t) = ( t, t3 , exp( t)), 0  t  1t‰.
x2
y2
sol)
+
+ C,
f
(x,
y)
=
2
2
f (x, y, z) = ex sin y + z 2
\ ì<t rf = F t‰.
\ Pt rf = F t¿\ F î Ùt•t‰. ⇣\,
x
x
C(0) = (0, 0, 1),
b) F (x, y) = (y, x)
sol) F = (P, Q)\ ì<t
C(1) = (1, 1, e)
@Q
=
@x
t¿\
Z
C
F · ds = f (C(1))
f (C(0))
= f (1, 1, e)
= e sin 1 + e
f (0, 0, 1)
2
1
1 6= 1 =
@P
@y
t¿\ F î ¯ò∏∏ °0•t D»‰.
c) F (x, y) = (ex cos y, ex sin y)
sol) F = (P, Q)\ ì<t
@Q
= ex sin y 6=
@x
t‰.
5. ‰L <L– ıX‹$.
(C î ¡⇠)
ex sin y =
t¿\ F î ¯ò∏∏ °0•t D»‰.
168
@P
@y
SOLUTION
7. ‰L Z°0•@ Ùt•ÑD Ùt‹$. ¯¨‡ ¸¥ƒ ·
15 °0•
<\ Pt rf = F t¿\ F î R3 {(0, 0, 0)}–⌧ Ùt•t‰.
r0 = (x0 , y0 , z0 )\ ì‡ t⌧ r0 –⌧ 1L¿ Öê ¿¡ ‰‡
XÏ
F · ds| lX‹$.
` L F
Öê– \ |@ Ω\– 4 X‰. 0|⌧, Öê ·
C
C(t) = (x0 + t, y0 + t, z0 + t)| 0|⌧ ¿¡ ‰
X‡
a) F (x, y) = (xy 2 + 3x2 y, x3 + x2 y), Cî (1, 1)¸ (0, 2)| áî ƒ∞tƒ ¯ ✓@ Ÿ|` Ét‰. 0|⌧ lX‡ê Xî |@
Ñ
Z
sol)
F · ds = lim f (r0 + M ) f (r0 )
M !1
r0 !1
x2 y 2
3
f (x, y) =
+x y
2
2
=
kr
0k
\ Pt rf = F t¿\ F î Ùt•t‰. ⇣\, CX ‹ë⇣¸
]⇣@
(1, 1), (0, 2) t¿\
t‰.
Z
3
F · ds = f (0, 2) f (1, 1) =
µ8⌧ 15.4 .
ZZ
2
C
1. ‰LX
°0• F X ·t S– \ t Ñ
F · ndS|
S
t‰.
lX‹$.
(ƒ‰x –t ∆<t •@ x•<\ ⌅¸\‰.)
b) F (x, y) = (xy 2 , x2 y), Cî x = t3 1, y = t6 t, 0  t  1x
‰⌧¿⇠ ·
a) F (x, y, z) = (2x, y, 2z), Sî …t x + 2y + z = 2
å\…
sol)
2 2
t–
Xt
ò∞
º
ÄÑ
x y
f (x, y) =
sol) ¸¥ƒ ·tD
2
x
\ Pt rf = F t¿\ F î Ùt•t‰. ⇣\, CX ‹ë⇣¸
X(x, y) = (x, y, 2 x 2y) , 0  x  2, 0  y  1
2
]⇣@
( 1, 0), (0, 0) t¿\
Z
C
\ Pt
C
F · ds = f (0, 0)
f ( 1, 0) = 0
Xx ⇥ Xy = (1, 2, 1)
t‰.
t‡ •t x•t¥| X¿\
c) F (x, y) = (2x sin y, x2 cos y 3y 2 ), C(t) = (2t , t2 ),
t1
sol)
f (x, y) = x2 sin y y 3
1 
ndS = (1, 2, 1)dxdy
t‰. 0|⌧
ZZ
Z 2Z 1
F · ndS =
\ Pt rf = F t¿\ F î Ùt•t‰. ⇣\, CX ‹ë⇣¸
]⇣@
(1/2, 1), (2, 1) t¿\
Z
t‰.
F · ds = f (2, 1) f (1/2, 1)
S
0
=
(2x, y, 4
2x
0
Z 2Z 1
0
x
2
0
x
2
(4
2y)dydx =
4y) · (1, 2, 1)dydx
10
3
C
=
t‰.
15
sin 1
4
b) F (x, y, z) = ( y, x, z), Sî lt x2 + y 2 + z 2 = 4X ⌧ 1
Ñ– ÄÑ
sol) ¸¥ƒ ·tD
X( , ✓) = (2 sin cos ✓, 2 sin sin ✓, 2 cos ),
2r
⇡
⇡
8. òX •(force field) F (x, y, z) =
((x, y, z) 6= (0, 0, 0))–
0  , 0✓
krk3
2
2
XXÏ ¥§ Öê ⇣ r0 2 R3 {(0, 0, 0)}Ä0 1L¿ tŸX
‰. F t Öê– \ |(work)D πt ÑD t©XÏ lX‹$. \ Pt
Ï0⌧ r(x, y, z) = (x, y, z)t‰.
X ⇥ X✓ = (4 sin2 cos ✓, 4 sin2 sin ✓, 4 cos sin )
sol) h⇠ f |
t‡ •t x•t¥| X¿\
2
f (x, y, z) = p
, (x, y, z) 6= (0, 0, 0)
ndS = (4 sin2 cos ✓, 4 sin2 sin ✓, 4 cos sin )d d✓
x2 + y 2 + z 2
169
SOLUTION
t‰. 0|⌧
ZZ
=
\ Pt
S
F · ndS
0
0
Z ⇡2 Z ⇡2
Xx ⇥ Xy = (2x, 2y, 1)
t‡ •t x•t¥| X¿\
( 2 sin sin ✓, 2 sin cos ✓, 2 cos )
2
ndS = (2x, 2y, 1)dxdy
2
· (4 sin cos ✓, 4 sin sin ✓, 4 cos sin )d d✓
Z ⇡2 Z ⇡2
=
8 cos2 sin d d✓
0
=
15 °0•
t‰. 0|⌧
ZZ
ZZ
F · ndS =
0
S
4
⇡
3
=
t‰.
=
c) F (x, y, z) = (x, y, z), Sî lt x2 + y 2 + z 2 = 9
sol) ¸¥ƒ ·tD
=
 ⇡,
x2
0
0
0
0
Z 2⇡Z p2
y 2 )2 )(2x, 2y, 1)dxdy
r(4r2 + (2
r2 )2 )drd✓
(r5 + 4r)drd✓
32
⇡
3
t‰.
X( , ✓) = (3 sin cos ✓, 3 sin sin ✓, 3 cos ),
0
(2x, 2y, (2
x2 +y 2 2
Z 2⇡Z p2
0  ✓  2⇡
b) F (x, y, z) = (y 3 , x3 , ez ), Sî –0et x2 + y 2 = 1,
z3
sol) ¸¥ƒ ·tD
\ Pt
X ⇥ X✓ = (9 sin2 cos ✓, 9 sin2 sin ✓, 9 cos sin )
X(✓, z) = (cos ✓, sin ✓, z) ,
t‡ •t x•t¥| X¿\
0  ✓  2⇡,
1
1z3
\ Pt
ndS = (9 sin2 cos ✓, 9 sin2 sin ✓, 9 cos sin )d d✓
t‰. 0|⌧
ZZ
S
=
X✓ ⇥ Xz = (cos ✓, sin ✓, 0)
t‡ •t x•t¥| X¿\
ndS = (cos ✓, sin ✓, 0)dzd✓
F · ndS
Z 2⇡Z ⇡
0
t‰. 0|⌧
ZZ
Z 2⇡Z 3
F · ndS =
(sin3 ✓, cos3 ✓, ez ) · (cos ✓, sin ✓, 0)dzd✓
(3 sin cos ✓, 3 sin sin ✓, 3 cos )
0
· (9 sin2 cos ✓, 9 sin2 sin ✓, 9 cos sin )d d✓
Z 2⇡Z ⇡
=
27 sin d d✓
0
S
0
=
0
0
= 108⇡
=4
Z 2⇡
(sin3 ✓ cos ✓ + cos3 ✓ sin ✓)dzd✓
1
sin ✓ cos ✓d✓
0
t‰.
=0
2. ‰LX
1
Z 2⇡Z 3
°0• F X ·t S–
\t Ñ
lX‹$.
(ƒ‰x –t ∆<t •@ x•<\ ⌅¸\‰.)
ZZ
S
F · ndS|
c) F (x, y, z) = (ex sin y, yz, x2 ), Sî ¡¨
⌅– àî Ï< 0et z = 4 y 2
sol) ¸¥ƒ ·tD
a) F (x, y, z) = (2x, 2y, z 2 ), Sî xy…t ⌅ΩX Ï<t z =
2 x2 y 2 , z 0
sol) ¸¥ƒ ·tD
X(x, y) = x, y, 2
x2
y2 ,
t‰.
X(x, y) = x, y, 4
y2 ,
0  x  1,
\ Pt
x2 + y 2  2
Xx ⇥ Xy = (0, 2y, 1)
170
Ì [0, 1] ⇥ [0, 2]
0y2
SOLUTION
t‡ •t x•t¥| X¿\
15 °0•
<\ ‰⌧T ` ⇠ à‡, ⇣\
ndS = (0, 2y, 1)dxdy
t‰. 0|⌧
ZZ
Z 1Z 2
F · ndS =
(ex sin y, y(4
S
0
=
t‡ •t x•t¥| X¿\
y 2 ), x2 ) · (0, 2y, 1)dydx
0
Z 1Z 2
0
(x
2
Xx ⇥ Xy = (4x, 2y, 1)
4
ndS = (4x, 2y, 1)dxdy
2
2y + 8y )dydx
0
t‰. 0|⌧,
46
=
5
ZZ
t‰.
S1
F · ndS =
=
3. H•t z = xy –0et x + y = 4– Xt⌧ ò∞ ƒ
·tD S| ` L, SX ⌅Ω ï °0 )•<\ S| µ\ °0•
F (x, y, z) = (y, x, z)X ë(flux)D lX‹$.
sol) ¸¥ƒ ·tD
2
X(x, y) = (x, y, xy),
2
2
=
ZZ
ZZ
x +y 4
0
X ⇤ (x, y) = (x, y, x2 + 2y 2 )
<\ ‰⌧T ` ⇠ à‡, ⇣\
Xx⇤ ⇥ Xy⇤ = ( 2x, 4y, 1)
t‰. 0|⌧ lX‡ê Xî °0•X …@
ZZ
ZZ
F · ndS =
(y, x, xy) · ( y, x, 1)dxdy
S
x2 +y 2 4
ZZ
=
( y 2 x2 + xy)dxdy
t‡ •t x•t¥| X¿\
ndS = (2x, 4y, 1)dxdy
x2 +y 2 4
=
r3 (cos ✓ sin ✓
t‰. 0|⌧,
1)drd✓
ZZ
0
8⇡
t‰.
S2
F · ndS =
=
4. Ï<t z = x2 + 2y 2 ¸ z = 12 2x2 y 2 <\ Xϯx Ö¥
X Ωƒ| S| ` L, x• ï °0 )•<\ S| µ\ °0•
F (x, y, z) = (x, y, z)X ëD lX‹$.
sol) ¸¥ƒ –·t Sî ·t
S1 : z = 12
2x2
y2
S2 : z = x2 + 2y 2 ,
x2 + y 2  4
\ tË¥ƒ p
‹·tt‰. 0|⌧ lX‡ê Xî
ZZ
ZZ
ZZ
F · ndS =
F · ndS +
F · ndS
S
S1
\ l` ⇠ à‰. <
=
2x
ZZ
x2 +y 2 4
(x, y, x2 + 2y 2 ) · (2x, 4y, 1)dxdy
(x2 + 2y 2 )dxdy
x2 +y 2 4
Z 2⇡Z 2
3
r (1 + sin2 ✓)drd✓
0
0
t¿\, lX‡ê Xî
…@
ZZ
ZZ
…@
t‰.
2
ZZ
= 12⇡
S2
S1 –⌧î ·tD
X(x, y) = (x, y, 12
r(12 + r2 (cos2 ✓ + 1))drd✓
0
= 60⇡
ndS = ( y, x, 1)dxdy
0
(12 + 2x2 + y 2 )dxdy
t‰. t⌧ S2 –⌧î ·tD
0D Ãqt| X¿\
=
y 2 ) · (4x, 2y, 1)dxdy
x2 +y 2 4
Z 2⇡Z 2
Xx ⇥ Xy = ( y, x, 1)
Z 2⇡Z 2
2x2
2
\ Pt
t‡ •t n · k
(x, y, 12
x2 +y 2 4
2
y )
171
S
F · ndS =
S1
F · ndS +
ZZ
= 60⇡ + 12⇡ = 72⇡
S2
F · ndS
SOLUTION
16
°0•X
µ8⌧ 16.1 .
1. ‰L– ¸¥ƒ Z
–·
Ñ ¨
16 °0•X
Ñ ¨
| ªî‰.
2
2
3
d) P (x, y) = xex +y +2ex , Q(x, y) = ln(x2 +y 2 )+sinh(y 2 )+1;
Cî ˘å\\ \‹⌧ Ì {(r cos ✓, r sin ✓)|1  r  2, 0  ✓ 
t©XÏ
Ñ
P dx + Qdy| lX‹$.
⇡/3}X Ωƒ
C
sol) 8⌧–⌧ ¸¥ƒ ÌD D\ ì‡ ¯∞ ¨| ©Xt
2
2
2
◆
Z
ZZ ✓
a) P (x, y) = x y, Q(x, y) = x
y ; Cî 8 ⇣
@P
@Q
P dx + Qdy =
+
dxdy
(0, 0), (2, 0), (0, 2)| -”⇣<\ Xî º
XX
@y
@x
C
sol) Ì D| ‰L¸ ⇡t XXt
Z ZD
2
2
2x
=
( 2yxex +y + 2
)dxdy
D = {(x, y)|0  x  2, 0  y  2 x},
x
+ y2
D
Z ⇡3 Z 2
2
8⌧–⌧ ¸¥ƒ · C– t C = @D t‰. t⌧ ¯∞ ¨|
=
( r3 sin(2✓)er + 2 cos ✓)drd✓
©Xt
0
1
◆
Z
ZZ ✓
p
9e4
@P
@Q
= 3
P dx + Qdy =
+
dxdy
8
@y
@x
C
D
Z 2Z 2 x
| ªî‰.
=
( x2 + 2x)dydx
ëX )•D
C
0
4
=
3
| ªî‰.
» L, Green
¨|
0
e) P (x, y) = 7y esin x , Q(x, y) = 15x sin(y 3 +8y); Cî ⌘Ït
(2, 5)t‡ ⇠¿Ñt 4x –
sol) Ì D| ‰L¸ ⇡t XXt
D = (x, y)|(x 2)2 + (y + 5)2  16 ,
2
b) P (x, y) =⇣ex +⌘ y ⇣sin x, ⌘ Q(x, y) = x cos y + ln(y 2 + 3); Cî
⇡
⇡ ⇡
8⌧–⌧ ¸¥ƒ · C– t C = @D t‰. t⌧ ¯∞ ¨|
8 ⇣ (0, 0), 0,
,
,
D -”⇣<\ Xî º
XX
6
6 6
©Xt
sol) Ì D| ‰L¸ ⇡t XXt
◆
Z
ZZ ✓
n
@P
@Q
⇡
⇡o
+
dxdy
P
dx
+
Qdy
=
D = (x, y)|0  x  , x  y 
,
@y
@x
6
6
C
D
ZZ
8⌧–⌧ ¸¥ƒ · C– t C = @D t‰. t⌧ ¯∞ ¨|
=
( 7 + 15)dxdy
©Xt
D
Z
Z
◆
Z
ZZ ✓
@P
@Q
=8
dD = 128⇡
P dx + Qdy =
+
dxdy
D
@y
@x
C
D
Z ⇡6 Z ⇡6
| ªî‰.
=
( sin x + cos y)dydx
p0 x
f) P (x, y) = x3 y 3 , Q(x, y) = x3 + y 3 ; C = C1 [ C2 , C1 @ –
3 1
⇡
=
⇣D ⌘Ï<\ X‡ ⇠¿Ñt 3x –t‡ C2 î –⇣D ⌘Ï<\
2
2 12
X‡ ⇠¿Ñt 1x –
| ªî‰.
sol) Ì D| ‰L¸ ⇡t XXt
D = (x, y)|1  x2 + y 2  9 ,
c) P (x, y) = x2 + y 2 + 2, Q(x, y) = x2 y 2 + 1; Cî ˘å\\
\‹⌧ Ì {(r cos ✓, r sin ✓)|1  r  3, 0  ✓  ⇡/4}X Ωƒ
8⌧–⌧ ¸¥ƒ · C– t C = @D t‰. t⌧ ¯∞ ¨|
sol) 8⌧–⌧ ¸¥ƒ ÌD D\ ì‡ ¯∞ ¨| ©Xt
©Xt
◆
Z
ZZ ✓
@P
@Q
◆
Z
ZZ ✓
P dx + Qdy =
+
dxdy
@P
@Q
@y
@x
+
dxdy
P dx + Qdy =
C
Z ZD
@y
@x
C
Z ZD
=
( 2y + 2x)dxdy
=
(3y 2 + 3x2 )dxdy
D
Z ⇡4 Z 3
D
Z 2⇡Z 3
=
2r2 (cos ✓ sin ✓)drd✓
=
3r3 drd✓
0
1
0
1
52 p
=
( 2 1)
= 120⇡
3
172
SOLUTION
| ªî‰.
sol) ¸¥ƒ
2. ‰L– ¸¥ƒ
C 1 , C2 , C3 |
t‰.
p
4
C1 (t) = (t, t4 ), 0  t  2,
p
p
4
4
C2 (t) = ( 2 t, 2), 0  t  2,
C3 (t) = (0, 2
t),
d) ‰⌧¿⇠· x(t) = t2 , y(t) = t3 9t (0  t  3)@ xï<\
Xϯx Ì
sol) ¸¥ƒ ÌX Ωƒ| @D | Xê. · C1 , C2 |
0t2
Ì DX ◆tî ¯∞ ¨– Xt
area(D)
Z
=
xdy
Z@D
Z
=
xdy +
=
C1
4
Z p
C1 (t) = (t2 , t3
C2 (t) = (9
Z
xdy +
C2
\ ì<t
4t4 dt + 0 + 0 =
0
=
b) ‰⌧¿⇠· x(t) = 2 sin t, y(t) = cos 2t (0  t  ⇡/2)@ t‰.
¡ y = 1, x = 0<\ Xϯx Ì
sol) ¸¥ƒ ÌX Ωƒ| @D | Xê. · C1 , C2 , C3 |
3. ·
C1 (t) = (2 cos t,
C2 (t) = (0, 1
0t
@D
Z 3
t2 (3t2
xdy +
C1
9)dt + 0 =
C1
=
Z ⇡2
xdy +
C2
xdy
C2
Z 3
(3t4
9t2 )dt =
0
324
5
‰L¸ ⇡t ¸¥LD L
Ñ
Z
y
x
dx + 2
dy
2 + y2
x
x
+
y2
C
a) Cî Ë⌅– x2 + y 2 = 1
sol) µ8⌧ 15.3X 8àX a)– Xt
2
Z
Z
| lX‹$.
0t2
Ì DX ◆tî ¯∞ ¨– Xt
area(D)
Z
=
xdy
Z@D
Z
=
xdy +
C
⇡
,
2
0  t  2,
t),
C3 (t) = (t, 1),
\ ì<t
cos 2t)
0t9
t, 0),
0
t‰.
0  t  3,
9t)
Ì DX ◆tî ¯∞ ¨– Xt
area(D)
Z
Z
=
xdy =
xdy
C3
4
8p
2
5
2
Ñ ¨
ÌD D\ Pt ¯∞ ¨– Xt
Z
area(D) =
xdy
Z@D
⇡
=
sin 2t(cos tdt)
Z0 ⇡
4
=
2 cos2 t sin tdt =
3
0
ÌX ◆t| lX‹$.
a) y = x4 @ y = 2, yï<\ Xϯx Ì
(8⌧ ÙD) x 0x pt î
sol) ¸¥ƒ ÌX Ωƒ| @D | Xê. ·
\ ì<t
16 °0•X
Ñ@ 2⇡\ ¸¥ƒ‰.
2
b) Cî ¿– x4 + y9 = 1
sol) ¿–D C, Ë⌅–D C1 <\ ì<t ¯∞ ¨– Xt(9@
Ω\– 4 \
Ñ– Xt)
Z
Z
y
x
y
x
dx
+
dy
=
dx + 2
dy
2 + y2
2 + y2
2 + y2
x
x
x
x
+
y2
C
C1
xdy
C3
2 cos t(2 sin 2t)dt + 0 + 0
0
=
Z ⇡2
0
8 cos2 t sin tdt =
8
3
1ΩX‡ ¯ ✓@ 2⇡ t‰.
4. C
L,
t‰.
c) x(t) = sin 2t, y(t) = sin t (0  t  ⇡) <\ Xϯx
Ì
Ï<
Ñ
y = x2 ¸ ¡ y = 2\ Xϯx
Z
2x2 ydx + (x2 + y 2 )dy
C
173
ÌX Ωƒ|
SOLUTION
| ‰L– ¸¥ƒ P
¿ )ï<\ lX‹$.
C1
=
Z p2
Z
p (2t
2
16 p
=
2
5
t‰.
5
4
3
+ 2t + 2t )dt +
16 p
2=
3
32 p
2
15
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
C
Z
=
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
C1
Z
+
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
C2
Z
+
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
C3
Z
+
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
2x2 ydx + (x2 + y 2 )dy
C2
Z p2
p
=
2
4t dt
2
=
C4
dt +
Z a
Z 0
(6at + 5a2 )dt +
0
= a + 8a3
(10at
a2
1)dt + 0
a
(6a3 + a) = 2a3
t‰.
b) Green ¨| t©XÏ lX‹$.
sol) ¸¥ƒ ¨
ÌD D\ ì<t ¯∞ ¨– Xt
Z
ì<t ¸¥ƒ · C– t C = @D t‰. t⌧ ¯∞ ¨|
©Xt
◆
Z
ZZ ✓
@P
@Q
P dx + Qdy =
+
dxdy
@y
@x
C
Z ZD
=
( 2x2 + 2x)dxdy
D
Z p2 Z 2
p
Z a
0
b) Green ¨| t©XÏ lX‹$.
sol) Ì D|
n
o
p
p
D = (x, y) |
2  x  2, x2  y  2
=
Ñ ¨
\ ì<t
a)
ÑX X| t©XÏ lX‹$.
sol) · C1 , C2 |
p
p
C1 (t) = (t, t2 ),
2t 2
p
p
C2 (t) = ( t, 2),
2t 2
\ ì<t
Z
2x2 ydx + (x2 + y 2 )dy
C
Z
Z
=
2x2 ydx + (x2 + y 2 )dy +
16 °0•X
( 2x2 + 2x)dydx
P dx + Qdy =
C
=
=
ZZ ✓
Z ZD
◆
dxdy
4ydxdy
Z aZD a
0
@P
@Q
+
@y
@x
4ydxdy
0
= 2a3
| ªî‰.
2 x2
32 p
2
15
µ8⌧ 16.2 .
1. ⌧∞
Z Z ¨| t©XÏ Ì DX Ωƒ @D| µ\ °0• F X
flux
F · ndS| lX‹$.
| ªî‰.
@D
5. C ⇣(0, 0)¸ (a, a)D fi-¿⇣<\ ¿î ¨
|‡ ` L,
Ñ
Z
(1 + 10xy + y 2 )dx + (6xy + 5x2 )dy
C
| ‰L– ¸¥ƒ P
¿ )ï<\ lX‹$.
a)
ÑX X| t©XÏ lX‹$.
sol) · C1 , C2 |
C1 (t) = (t, 0),
C2 (t) = (a, t),
C3 (t) = ( t, a),
C4 (t) = (0, t),
0ta
0ta
at0
at0
X Ωƒ
a) F (x, y, z) = (x, y, z); Dî lt x2 + y 2 + z 2 = 4X ¥Ä
sol) r · F (x, y, z) = 3 t¿\, ⌧∞
ZZ
@D
F · ndS =
ZZZ
¨– Xt
3dV = 32⇡
D
t‰.
b) F = (yz 6 , xz 3 , z 3 ); Dî Ï<t z = x2 + y 2 ¸ …t z = 2\
Xϯx Ì
174
SOLUTION
16 °0•X
Ñ ¨
sol) r · F (x, y, z) = 3z 2 t¿\, ⌧∞ ¨@ –0e å\ ¿XD f) F (x, y, z) = (x2 + y + 2, x2 y + ezx , 2z); Dî Ï<t
¨©Xt
z = x2 + y 2 ¸ …t z = 2y\ Xϯx Ì
ZZ
ZZZ
F · ndS =
3z 2 dV
sol) r · F (x, y, z) = 2x 3 t¿\, ⌧∞ ¨@ –0e å\
@D
D
Z 2⇡Z p2Z 2
¿XD ¨©Xt
=
3rz 2 dzdrd✓ = 12⇡
ZZ
ZZZ
0
0
r2
F · ndS =
(2x 3)dV
@D
D
t‰.
Z Z
Z
⇡
2 sin ✓
2r sin ✓
=
2
c) F (x, y, z) = (2x + eyz , 4xy 2 + exz + x log(1 + z), 10z
Dî x2 + y 2  z  yx Ì
r(2r cos ✓
0
8xyz);
=
3)dzdrd✓
r2
0
3
⇡
2
sol) r · F (x, y, z) = 12 t¿\, ⌧∞ ¨@ –0e å\ ¿XD t‰.
¨©Xt
2
ZZ
ZZZ
g) F (x, y, z) = (xy 2 + ez , yz 2 + cos(xz), zx2 + sin(xy)); Dî
F · ndS =
12dV
lt å\ƒ–⌧ 0  ⇢  1 + cos , 0   ⇡, 0  ✓  ⇡x
@D
D
Z ⇡Z sin ✓Z r sin ✓
Ì
=
12rdzdrd✓
0
0
r2
sol) r · F (x, y, z) = ⇢2 t¿\, ⌧∞
ZZ
ZZZ
F · ndS =
⇢2 dV
3
= ⇡
8
t‰.
@D
d) F (x, y, z) = (x + 2y z, 2x + y + 5z, x + 2y 3z); Dî …t
x + 2y + 3z = 6¸ 8 å\…t<\ Xϯx Ì
sol) r · F (x, y, z) = 1 t¿\, ⌧∞
¨©Xt
ZZ
ZZZ
F · ndS =
( 1)dV
@D
=
D
Z 6Z 3
0
0
Z 2
1
2x
2y
3
D
=
Z ⇡Z ⇡Z 1+cos
0
0
⇢4 sin d⇢d d✓ =
0
2. ¸¥ƒ °0•¸
Ì D–
t
ZZ
F dV | ®P ¡⌘ ƒ∞XÏ ⌧∞ ¨
( 1)dzdydx =
a) F (x, y, z) = (x, y, z); Dî
6
32
⇡
15
t‰.
¨@ ¡På\ ¿XD
1
3x
¨| ¨©Xt
@D
F · ndS @
ZZZ
D
r·
1ΩhD Ùt‹$.
Ì 1  x2 + y 2 + z 2  9
0
t‰.
sol) < r · F (x, y, z) = 3t¿\,
ZZZ
ZZZ
2
yz 2
2
zx2
2
xy 2
e) F (x, y, z) = (x + e , y + e , z + e ); Dî –0et
r · F dV =
3dV
D
D
x2 + y 2 = 1¸ …t z = 0¸ z = 8 x\ Xϯx Ì
Z Z Z
2⇡
⇡
3
=
sol) r · F (x, y, z) = 2(x + y + z) t¿\, ⌧∞
å\ ¿XD ¨©Xt
ZZ
ZZZ
F · ndS =
2(x + y + z)dV
@D
¨@ –0e
Z 2⇡Z 1Z 8 r cos ✓
0
=
255
⇡
4
0
0
3⇢2 sin d⇢d d✓ = 104⇡
1
t‰.
·t DX Ωƒ @Dî ‰L¸ ⇡t ò¿º ⇠ à‰.
D
=
t‰.
0
@D = @D1 [ @D2 ,
@D1 = {(x, y, z) 2 R3 |x2 + y 2 + z 2 = 1},
2r(r(cos ✓ + sin ✓) + z)dzdrd✓
0
@D2 = {(x, y, z) 2 R3 |x2 + y 2 + z 2 = 9}.
@D1 D
X( , ✓) = (sin cos ✓, sin sin ✓, cos ), ( , ✓) 2 [0, ⇡] ⇥ [0, 2⇡]
175
SOLUTION
\ ‰⌧T Xt
X ⇥ X✓ = (sin
2
cos ✓, sin
2
t‰.
·t DX Ωƒ @Dî ‰L¸ ⇡t ò¿º ⇠ à‰.
sin ✓, sin cos )
t‡ ·tX ï °0 n1 t x•t ⇠¥| X¿\
n1 dS = ( sin
t‰. 0|⌧,
ZZ
@D1
=
ZZ
2
cos ✓,
sin
2
sin ✓,
@D =
6
[
@Dn ,
n=1
sin cos )d d✓
@D1 : x = 1, 0  y  2, 0  z  3,
@D2 : y = 2, 0  x  1, 0  z  3,
@D3 : x = 0, 0  y  2, 0  z  3,
F · n1 dS
@D4 : y = 0, 0  x  1, 0  z  3,
@D5 : z = 3, 0  x  1, 0  y  2,
(sin cos ✓, sin sin ✓, cos )
@D1
· ( sin2 cos ✓, sin2 sin ✓,
Z 2⇡ Z ⇡
=
sin d d✓
0
=
16 °0•X
@D6 : z = 0, 0  x  1, 0  y  2
sin cos )d d✓
@D1 Ä0 ‹ët⌧ @D6 L¿
0
n1 (x, y, z) = (1, 0, 0)
4⇡
n5 (x, y, z) = (0, 0, 1)
⇤
X ( , ✓)
t‰. <
= (3 sin cos ✓, 3 sin sin ✓, 3 cos ), ( , ✓) 2 [0, ⇡] ⇥ [0, 2⇡]
n6 (x, y, z) = (0, 0, 1)
(0  u  2, 0  v  3)
\ ‰⌧T ` ⇠ à‰. ¯Ït
ZZ
ZZ
F · n1 dS =
X ⇤ ⇥ X✓⇤ = (9 sin2 cos ✓, 9 sin2 sin ✓, 9 sin cos )
t‡ ·tX ï °0 n2 t x•t ⇠¥| X¿\
@D1
n2 dS = (9 sin2 cos ✓, 9 sin2 sin ✓, 9 sin cos )d d✓
t‰. 0|⌧,
ZZZ
ZZ
F · n2 dS =
n4 (x, y, z) = (0, 1, 0)
@D1 –⌧î F · n1 = x2 tp @D1 D
X(u, v) = (1, u, v)
\ ‰⌧T Xt
°0î
n2 (x, y, z) = (0, 1, 0)
n3 (x, y, z) = ( 1, 0, 0)
t‰. t⌧ @D2 D
@D2
X x•Ë⌅ï
=
x2 dS
@D1
Z 2Z 3
0
1dvdu = 6
0
@D2 –⌧î F · n2 = y 2 tp @D2 |
(3 sin cos ✓, 3 sin sin ✓, 3 cos )
@D2
2
X(u, v) = (u, 2, v) (0  u  1, 0  v  3)
· (9 sin cos ✓, 9 sin2 sin ✓, 9 sin cos )d d✓
Z 2⇡ Z ⇡
\ ‰⌧T ` ⇠ à‰. ¯Ït
=
27 sin d d✓
ZZ
ZZ
0
0
= 108⇡
F · n2 dS =
y 2 dS
@D2
t‰. 0|⌧ lX‡ê Xî t Ñ
ZZ
ZZ
ZZ
F · ndS =
F · n1 dS +
@D
@D1
D ª‡, ⌧∞ ¨
=
@D2
0
F · n2 dS = 104⇡
t‰.
@D3 –⌧î F · n3 =
1ΩhD L ⇠ à‰.
b) F (x, y, z) = (x2 , y 2 , z 2 ); Dî 8 å\…t¸ …t x = 1, y =
2, z = 3\ Xϯx Ì
sol) < r · F (x, y, z) = 2(x + y + z)t¿\,
ZZ
ZZZ
F · ndS =
2(x + y + z)dV
@D
D
=
Z 1Z 2Z 3
0
0
@D2
2(x + y + z)dzdydx = 36
Z 3Z 1
x2 tp @D3 |
X(u, v) = (0, u, v)
(0  u  2, 0  v  3)
\ ‰⌧T ` ⇠ à‰. ¯Ït
ZZ
ZZ
F · n3 dS =
@D3
t‰.
@D4 –⌧î F · n4 =
176
x2 dS = 0
@D3
y 2 tp @D4 |
X(u, v) = (u, 0, v)
0
4dudv = 12
0
(0  u  1, 0  v  3)
Ñ ¨
SOLUTION
\ ‰⌧T ` ⇠ à‰. ¯Ït
ZZ
ZZ
F · n4 dS =
@D4
16 °0•X
\ ‰⌧T Xt
X✓ ⇥ Xz = (3 cos ✓, 3 sin ✓, 0)
y 2 dS = 0
@D4
t‡ ·tX ï °0 n1 t x•t ⇠¥| X¿\
t‰.
@D5 –⌧î F · n5 = z 2 tp @D5 |
n1 dS = (3 cos ✓, 3 sin ✓, 0)d✓dz
(0  u  1, 0  v  2)
X(u, v) = (u, v, 3)
\ ‰⌧T ` ⇠ à‰. ¯Ït
ZZ
ZZ
F · n5 dS =
@D5
=
z 2 dS
@D5
Z 2Z 1
0
t‰.
@D6 –⌧î F · n6 =
9dudv = 18
=
0
\ ‰⌧T ` ⇠ à‰. ¯Ït
ZZ
ZZ
F · n6 dS =
Z 2⇡Z 2
0
81(cos4 ✓ + sin4 ✓) + 6(cos ✓ + sin ✓)dzd✓ = 243⇡
0
t‰. t⌧ @D2 |
(0  u  1, 0  v  2)
@D6
t‰. 0|⌧,
ZZ
F · n1 dS
Z@D
Z1
=
(27 cos3 ✓ + 2, 27 sin3 ✓ + 2, z 3 ) · (3 cos ✓, 3 sin ✓, 0)d✓dz
@D1
z 2 tp @D6 |
X(u, v) = (u, v, 0)
X ⇤ (r, ✓) = (r cos ✓, r sin ✓, 2)(r, ✓) 2 [0, 3] ⇥ [0, 2⇡]
\ ‰⌧TXt
Xr⇤ ⇥ X✓⇤ = (0, 0, r)
2
z dS = 0
@D6
t‡ ·tX ï °0 n2 t x•t ⇠¥| X¿\
t‰. 0|⌧
ZZ
Ñ ¨
n2 dS = (0, 0, r)drd✓
F · ndS =
@D
t¿\, ⌧∞ ¨
6 ZZ
X
n=1
@Dn
F · ni dS = 36
t‰. 0|⌧,
ZZ
1ΩhD L ⇠ à‰.
@D2
F · n2 dS =
Z 2⇡ Z 3
0
8rdzd✓
0
= 72⇡
c) F (x, y, z) = (x3 + 2, y 3 + 2, z 3 ); Dî –µt x2 + y 2 = 9@ t‰. ⇣\ @D |
3
…t z = 0, z = 2\ Xϯx Ì
X ⇤⇤ (r, ✓) = (r cos ✓, r sin ✓, 0), (r, ✓) 2 [0, 3] ⇥ [0, 2⇡]
sol) < r · F (x, y, z) = 3(x2 + y 2 + z 2 )t¿\,
ZZ
ZZZ
F · ndS =
3(x2 + y 2 + z 2 )dV
@D
=
D
Z 2⇡Z 3Z 2
0
0
3r(r2 + z 2 )dzdrd✓ = 315⇡
\ ‰⌧TXt
Xr⇤⇤ ⇥ X✓⇤⇤ = (0, 0, r)
t‡ ·tX ï °0 n3 t x•t ⇠¥| X¿\
0
t‰.
·t DX Ωƒ @Dî ‰L¸ ⇡t ò¿º ⇠ à‰.
n3 dS = (0, 0, r)drd✓
t‰. 0|⌧,
ZZ
@D = @D1 [ @D2 [ @D3 ,
2
2
@D1 : x + y = 9, 0  z  2,
@D2 : x2 + y 2  9, z = 2,
@D3 : x2 + y 2  9, z = 0.
<
@D1 D
X(✓, z) = (3 cos ✓, 3 sin ✓, z), (✓, z) 2 [0, 2⇡] ⇥ [0, 2]
@D3
F · n3 dS =
ZZ
0drd✓ = 0
@D3
t‰. 0|⌧ lX‡ê Xî t Ñ
ZZ
F · ndS =
Z Z@D
ZZ
ZZ
F · n1 ds +
F · n2 dS +
@D1
177
@D2
@D3
F · n3 dS = 315⇡
SOLUTION
D ª‡, ⌧∞ ¨
1ΩhD L ⇠ à‰.
16 °0•X
| L, @D| µ\ F X flux
ZZ
@D
Ñ ¨
F · ndS| lX‹$. Ï0–⌧
n@ ·t @DX x•Ë⌅ï °0t‰.
⇡
⇡
3. ˘å\\ ⌧ ⌧ · r = cos ✓, z = sin 2✓

✓

\
2
2
Xϯx ÌD zïD ⌘Ï<\ å⌅XÏ ª@ å⌅¥X Ä<| sol)
r · F (x, y, z) = 5(x2 + y 2 + z 2 )
⌧∞ ¨| t©XÏ lX‹$.
t¿\ ⌧∞ ¨– Xt
ZZ
ZZZ
F · ndS =
5(x2 + y 2 + z 2 )dV
sol) å⌅¥X ¯º@ ‰L¸ ⇡‰.
@D
D
=
Z 2⇡Z 2Z 4
5r(r2 + z 2 )dzdrd✓
0
+
0 2
Z 2⇡Z 4Z 4
0
=
2
5r(r2 + z 2 )dzdrd✓
r
1360
3104⇡
⇡+
= 1488⇡
3
3
t‰.
å⌅¥
Xt
(¿Xî
ÌD D, tX Ωƒ| S| P‡ S| ‰⌧T 5. 3(– ı⌅X ÑXX ⇣ r = (x, y, z)¸ ‡ ⌧ ⇣ r0 =
(x0 , y0 , z0 )– t °0• F ‰L¸ ⇡t ¸¥8 à‰‡ Xê.
X(✓, )
h ⇡ ⇡i
= (cos ✓ cos , sin ✓ sin , sin 2✓) , (✓, ) 2
,
⇥ [0, 2⇡]
2 2
F (x, y, z) =
= ( 2 cos 2✓ cos ✓ cos , 2 cos 2✓ cos ✓ sin ,
r0
(x x0 , y
= p
r 0 k3
( (x x0 )2 + (y
y0 , z
z0 )
y0 )2 + (z
z0 ) 2 ) 3
‰L <L– ıX‹$.
t‰. ¯Ït
X✓ ⇥ X
r
kr
a) r 6= r0 | L r · F (r) = 0ÑD Ùt‹$.
sin ✓ cos ✓)
t‡ ·tX ï °0 n t x•t ⇠¥| X¿\
sol)
ndS = (2 cos 2✓ cos ✓ cos , 2 cos 2✓ cos ✓ sin , sin ✓ cos ✓)d✓d
f (x, y, z) = (x
x0 )2 + (y
y0 )2 + (z
z0 ) 2
1
(x, y, z) t|‡ XXt r·F = 1 t|‡ Pê. ¯Ït
3
t¿\ ⌧∞ ¨– Xt å⌅¥X Ä<î
1
F = p 3 (x x0 , y y0 , z z0 )
ZZZ
ZZZ
ZZ
( f)
1dV =
r · F dV =
F · ndS
D
S
t‰. \∏
Z ZD
1
=
(cos ✓ cos , cos ✓ sin , sin 2✓)·
✓
◆ 1 · (pf )3 (x x ) · 3(pf )2 · p1 · f
S 3
0
x
@ (x x0 )
2 f
p 3
p 6
=
(2 cos 2✓ cos ✓ cos , 2 cos 2✓ cos ✓ sin , sin ✓ cos ✓)d✓d
@x
( f)
( f)
Z 2⇡Z ⇡2
2
1
3(x
x
)2
=
cos4 ✓d✓d
p 3
p 50
=
⇡ 3
( f)
( f)
0
2
t‰. \∏, °0• F | F =
=
⇡2
2
t‰.
p
4. D –‘t z = x2 + y 2 ¸ P …t z = 2, z = 4\ Xϯx
Ìt‡, °0• F
t‡ », ¿\
✓
◆
@ (y y0 )
1
p 3
= p 3
@y
( f)
( f)
F (x, y, z) = (x(x2 + y 2 + z 2 ), y(x2 + y 2 + z 2 ), z(x2 + y 2 + z 2 ))
178
@
@z
✓
(z z0 )
p
( f )3
◆
1
= p 3
( f)
3(y y0 )2
p
( f )5
3(z z0 )2
p
( f )5
SOLUTION
t‰. 0|⌧
16 °0•X
§†l§
@ (x x0 )
@ (y y0 )
@ (z z0 )
p
p
p
+
+
@x ( f )3
@y ( f )3
@z ( f )3
3
3
p
= p 3
(x x0 )2 + (y y0 )2 + (z
( f)
( f )5
3
3
p
= p 3
=0
( f)
( f )3
¨– Xt
ZZ
r · F (r) =
z0 ) 2
b) ⌘Ït r0 t‡ ⇠¿Ñt r > 0x l| D| ` L,
ZZ
Z S
=
(r ⇥ F ) · ndS
@S
F · ds
0
(0, 6 cos t, 9 sin t) · ( 3 sin t, 3 cos t, 0)dt
Z 2⇡
=
t‰.
Z 2⇡
=
18 cos2 tdt = 18⇡
0
@D
F ·ndS
t‰.
| lX‹$. Ï0–⌧ n@ lt @DX x• Ë⌅ï °0t‰.
sol) ¸¥ƒ lt D|
x = x0 + r sin cos ✓,
y = y0 + r sin sin ✓,
0
 ⇡,
0  ✓  2⇡
Ñ ¨
b) F (x, y, z) = (xz + z 2 + 3, y 2 + x + 1, xye x
z = z0 + r cos , z = 4 x2 y 2 (z 0)t‡ n · k > 0
2
y2
); Sî Ï<t
sol) ·t SX Ωƒ @Sî x2 + y 2 = 4, z = 0t¿\ ‰L¸ ⇡t
‰⌧T` ⇠ à‰.
\ ‰⌧T Xt DX x• Ë⌅ï °0î
n = (sin cos ✓, sin sin ✓, cos )
x(t) = (2 cos t, 2 sin t, 0),
t‰. 0|⌧
(r sin sin , r sin sin ✓, r cos )
r3
· (sin cos ✓, sin sin ✓, cos )
1
= 2
r
§†l§
F (x, y, z) · n =
¨– Xt
ZZ
=
t‰. ⇣\ ⇠¿Ñt rx lX â◆t 4⇡r2 <\ ¸¥¿î ÉD
¡0Xt lX‡ê Xî t Ñ@
ZZ
ZZ
1
F · ndS =
dS = 4⇡
2
@D
@D r
Z S
@S
=
(0  t  2⇡)
(r ⇥ F ) · ndS
F · ds
Z 2⇡
(3, 4 sin2 t + 2 cos t + 1, 2 sin 2te 4 )
Z 2⇡
2 cos t(4 sin2 t + 2 cos t + 1)
0
t‰.
=
· ( 2 sin t, 2 cos t, 0)dt
6 sin t dt = 4⇡
0
c) t Ω∞– ⌧∞ ¨
©⇠î¿ UxX‡ ¯ t | $ÖX t‰.
‹$.
sol) t 8⌧X Ω∞ ⌧∞ ¨| ©` ⇠ ∆‰. ⌧∞ ¨î °
0• F C1 x Ì–⌧ ©⇠îp, ¸¥ƒ 8⌧–⌧ F lX c) F (x, y, z) = (x2 y + z 2 , exy cos y + 5, z 2
⌘Ïx r0 –⌧ X⇠¿ J<¿\ ¨X ptD D⌅à ©qX¿ y 2 = z 2 + x2 (2  y  3)t‡ n · j < 0
ª\‰.
µ8⌧ 16.3 .
1. §†l§
¨| t©XÏ
ZZ
x2 + y 2 ); Sî –‘t
sol) ·t SX Ωƒ @Sî ‰L¸ ⇡t l1⇠¥ à‰.
S
(r ⇥ F ) · ndS| lX‹$.
@S = @S 1 [ @S 2
@S 1 = {(x, y, z) 2 R3 |x2 + z 2 = 4, y = 2}
a) F (x, y, z) = (z, 2x, 3y); Sî x + y + z = 9X ⌅Ω ⇠lt
(z 0)t‡ n · k 0
2
2
2
sol) ·t SX Ωƒ @Sî x2 + y 2 = 9, z = 0t¿\ ‰L¸ ⇡t
‰⌧T` ⇠ à‰.
x(t) = (3 cos t, 3 sin t, 0),
@S 2 = {(x, y, z) 2 R3 |x2 + z 2 = 9, y = 3}
¯¨‡ @S 1 , @S 2 |
‰⌧T Xt
@S 1 (t) = (2 cos t, 2, 2 sin t),
@S 2 (t) = (3 cos t, 3, 3 sin t),
(0  t  2⇡)
179
0  t  2⇡
0  t  2⇡
SOLUTION
t¿\ §†l§
=
=
ZZ
¨– Xt
(r ⇥ F ) · ndS
Z
Z
F · ds =
F · ds +
Z
@S
Z 2⇡
t¿\ §†l§ ¨– Xt
ZZ
(r ⇥ F ) · ndS
S
Z
Z
Z
=
F · ds =
F · ds +
S
@S1
2
2
@S2
4 cos t
(8 cos t + 4 sin t, e
16 °0•X
F · ds
=
@S
Z 2⇡
(8, 4 cos2 t
@S2
F · ds
4 sin2 t + 1, e8 sin t cos t )
0
· ( 2 sin t, 2 cos t, 0)dt
Z 2⇡
+
(11, 4 cos2 t 4 sin2 t + 4, e16 sin t cos t )
cos 2 + 5,
0
4 sin2 t 4 cos2 t + 4) · ( 2 sin t, 0, 2 cos t)dt
Z 2⇡
+
(27 cos2 t + 9 sin2 t, e9 cos t cos 3 + 5,
0
· ( 2 sin t, 2 cos t, 0)dt
Z 2⇡
=
( 16 sin t 8 cos3 t + 8 cos t sin2 t
0
9 sin2 t 9 cos2 t + 9) · ( 3 sin t, 0, 3 cos t)dt
Z 2⇡
=
( 16 sin t cos2 t 8 sin3 t 8 cos t sin2 t
0
+
0
+ 8 cos3 t 8 cos t)dt
Z 2⇡
+
( 81 sin t cos2 t
@S1
Ñ ¨
Z 2⇡
( 22 sin t + 8 cos3 t
2 cos t)dt
8 cos t sin2 t + 8 cos tdt = 0
0
t‰.
3
2
27 sin t + 27 cos t sin t
0
3
27 cos t
e) F (x, y, z) = (yexz , xeyz , z 2 exy ); S î Ï<t z =
x2 + y 2 4(z  0)t‡ n · k < 0
+ 27 cos t)dt = 0
sol) ·t SX Ωƒ @Sî x2 + y 2 = 4, z = 0t¿\ ‰L¸ ⇡t
‰⌧T` ⇠ à‰.
t‰.
x(t) = (2 cos t, 2 sin t, 0),
(0  t  2⇡)
§†l§ ¨– Xt
ZZ
d) F (x, y, z) = (x + y + z + 3, x
y +z ,e
); Sî –0
2
2
(r ⇥ F ) · ndS
et x + y = 4( 1  z  2)t‡ n@ –0etX e )•
S
Z
=
F · ds
2
2
2
2
2
2
2xyz
@S
=
sol) ·t SX Ωƒ @Sî ‰L¸ ⇡t l1⇠¥ à‰.
Z 2⇡
0
=
Z 2⇡
( 2 sin t, 2 cos t, 0) · ( 2 sin t, 2 cos t, 0)dt
4dt = 8⇡
0
@S = @S 1 [ @S 2
t‰.
@S 1 = {(x, y, z) 2 R3 |x2 + y 2 = 4, z =
1}
@S 2 = {(x, y, z) 2 R3 |x2 + y 2 = 4, z = 2}
¯¨‡ @S 1 , @S 2 |
¨| t©XÏ
Z
C
F · ds| lX‹$.
a) F (x, y, z)
=
(xy, x2 + y 2 , xz); Cî
(0, 0, 4), (0, 1, 4), (1, 0, 4)| -”⇣<\ Xî º
X Ωƒ\, ⌅–⌧ ¸ L ⇠‹ƒ )•
‰⌧T Xt
@S 1 (t) = (2 cos t, 2 sin t, 1),
@S 2 (t) = (2 cos t, 2 sin t, 2),
2. §†l§
0  t  2⇡
sol) ·
Cî ·t
S = {(x, y, z) 2 R3 | 0  x  1
0  t  2⇡
180
8
y, 0  y  1, z = 4}
⇣
Ì
SOLUTION
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
16 °0•X
Ñ ¨
Xx (x, y) = (1, 0, 0)
t¿\ §†l§ ¨– Xt
Z
ZZ
ZZ
1
p ( 4x + 2y)dS
F · ds =
(r ⇥ F ) · ndS =
5
C
S
S
Z Z p 2
Xy (x, y) = (0, 1, 0)
t‰.
(0  x  1
X(x, y) = (x, y, 4)
y, 0  y  1)
t‰. ¯Ït
3
=
n=
3
(Xx ⇥ Xy )(x, y)
= (0, 0, 1)
||(Xx ⇥ Xy )(x, y)||
9 y
p
( 4x + 2y)dxdy = 0
9 y2
c) F (x, y, z) = (y + 3z, x + 3z, x + 3y); Cî lt x2 + y 2 + z 2 = 4
@ …t x + y + z = 0X ıµÄÑ<\, ⌅–⌧ ¸ L ⇠‹ƒ )•
t‡
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy = dxdy
t‰. ⇣\ °0• F X curlt
sol) ·
S = {(x, y, z) 2 R3 |x + y + z = 0, x2 + y 2 + z 2  4}
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
r ⇥ F (x, y, z) = (0, z, x)
X(x, y) = (x, y, x
t¿\ §†l§ ¨– Xt
Z
ZZ
ZZ
F · ds =
(r ⇥ F ) · ndS =
xdS
C
S
=
Z 1Z 1 y
0
Cî ·t
0
(x2 + y 2 + xy  2)
t‰. ¯Ït
Xx (x, y) = (1, 0, 1)
Xy (x, y) = (0, 1, 1)
S
xdxdy =
y)
1
6
n=
(Xx ⇥ Xy )(x, y)
1
= p (1, 1, 1)
||(Xx ⇥ Xy )(x, y)||
3
D ªî‰. ⇣\ °0• F X curl@
t‰.
r ⇥ F (x, y, z) = (0, 2, 0)
b) F (x, y, z) = (x2 y 2 , y 2 z 2 , z 2 x2 ); Cî …t z = 2y + 5 \ ltƒ‰. ·t S ⇠¿Ñt 2x –⇣ÑD ¡0Xê. t⌧
@ –0et x2 + y 2 = 9X ıµÄÑ<\, ⌅–⌧ ¸ L ⇠‹ƒ §†l§ ¨– Xt
Z
ZZ
ZZ
)•
2
p dS
F · ds =
(r ⇥ F ) · ndS =
3
C
S
S
p
2
8
3
sol) · Cî ·t
= p · 4⇡ =
⇡
3
3
3
2
2
S = {(x, y, z) 2 R |x + y  9, z = 2y + 5}
t‰.
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
X(x, y) = (x, y, 2y + 5)
(x2 + y 2  9)
t‰. ¯Ït
Xx (x, y) = (1, 0, 0)
d) F (x, y, z) = (x 2y + z, 2x 3y z, x 2y + z); Cî –0
et (x 1)2 + y 2 = 1¸ Ï<t z = 4 x2 y 2 X ıµÄÑ<\,
⌅–⌧ ¸ L ⇠‹ƒ )•
sol) ·
Xy (x, y) = (0, 1, 2)
n=
(Xx ⇥ Xy )(x, y)
1
= p (0, 2, 1)
||(Xx ⇥ Xy )(x, y)||
5
t¿\
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
p
Cî ·t
S = {(x, y, z) 2 R3 |z = 4
x2
y 2 , (x
1)2 + y 2  1}
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
X(x, y) = (x, y, 4
x2
y2 )
((x
1)2 + y 2  1)
t‰. ¯Ït
5dxdy
Xx (x, y) = (1, 0, 2x)
Xy (x, y) = (0, 1, 2y)
t‰. ⇣\ °0• F X curlt
n=
r ⇥ F (x, y, z) = (2z, 2x, 2y)
181
(Xx ⇥ Xy )(x, y)
1
=p
(2x, 2y, 1)
2
||(Xx ⇥ Xy )(x, y)||
4x + 4y 2 + 1
SOLUTION
t‡
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy =
t‰. ⇣\ °0• F X curlt
p
16 °0•X
t‰.
· @S1 , @S2 , @S3 , @S4 |
4x2 + 4y 2 + 1dxdy
@S1 = {(t, 0, 5) 2 R3 | 0  t  2},
@S2 = {(2, t, 5) 2 R3 | 0  t  3},
r ⇥ F (x, y, z) = ( 1, 2, 4)
@S3 = {(2
t¿\ §†l§ ¨ ✏ xD» ¨– Xt
Z
ZZ
ZZ
2x + 4y + 4
p
F · ds =
(r ⇥ F ) · ndS =
dS
4x2 + 4y 2 + 1
C
S
S
Z 2Z p1 (x 1)2
=
( 2x + 4y + 4)dydx
p
=
0
@S4 = {(0, 3
<\
Z
1 (x 1)2
0
Z ⇡Z 2 cos ✓
( 2r cos ✓ + 4r sin ✓ + 4)rdrd✓ = 2⇡
0
t‰.
3. ‰L °0•¸ ·t–
¡⌘ ƒ∞t⌧ §†l§
t
¨
S
(r ⇥ F ) · ndS@
Z
1ΩhD UxX‹$.
a) F (x, y, z) = (x + y + z, x2 + y 2 + z 2 , xyz); Sî ¨
0  x  2, 0  y  3, z = 5t‡, n · k > 0
sol) ·
Z 2
0
@S
t, 3, 5) 2 R3 | 0  t  2},
t, 5) 2 R3 | 0  t  3},
XXt @S=@S1 [ @S2 [ @S3 [ @S4 t‰. ¯Ït
Z
Z
F · ds =
F · ds +
F · ds
@S
@S1
@S2
Z
Z
+
F · ds +
F · ds
=
ZZ
Ñ ¨
F · ds|
+
@S3
@S4
(t + 5)dt +
Z 2
Z 3
0
(t
10)dt +
0
(t2 + 29)dt
Z 3
( 25
(t
3)2 )dt
0
=12 + 96 18 84 = 6
ZZ
ZZ
Ì t‰. ¯Ï¿\
(r ⇥ F ) · ndS =
S
@S
F · ds t 1Ω\‰.
@Sî ·t
b) F (x, y, z) = (xy + 1, yz + 2, xz + 5); Sî –0et x2 + y 2 = 9
(0  z  3)t‡ n@ –0etX e )•
S = {(x, y, 5) 2 R3 |0  x  2, 0  y  3}
sol) ·
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
X(x, y) = (x, y, 5)
(0  x  2, 0  y  3)
t‰. ¯Ït
S = {(x, y, z) 2 R3 |x2 + y 2 = 9, 0  z  3}
X Ωƒt‰. t⌧ ·t S| ‰⌧TXt
X(✓, z) = (3 cos ✓, 3 sin ✓, z)
Xx (x, y) = (1, 0, 0)
Xy (x, y) = (0, 1, 0)
(0  ✓  2⇡, 0  z  3)
t‰. ¯Ït
(Xx ⇥ Xy )(x, y)
n=
= (0, 0, 1)
||(Xx ⇥ Xy )(x, y)||
X✓ (✓, z) = ( 3 sin ✓, 3 cos ✓, 0)
Xz (✓, z) = (0, 0, 1)
t‡
n=
dS = ||(Xx ⇥ Xy )(x, y)|| dxdy = dxdy
t‰. ⇣\ °0• F X curlt
(X✓ ⇥ Xz )(✓, z)
= (cos ✓, sin ✓, 0)
||(X✓ ⇥ Xz )(✓, z)||
t‡
r ⇥ F (x, y, z) = (xz
2z, 1
ZZ
ZZ
yz, 2x
1)
t¿\
S
@Sî ·t
(r ⇥ F ) · ndS =
=
t‰. ⇣\ °0• F X curlt
(2x
S
Z 3Z 2
0
dS = ||(X✓ ⇥ Xz )(✓, z)|| d✓dz = 3d✓dz
(2x
1)dS
1)dxdy
r ⇥ F (x, y, z) = ( y, z, x)
t¿\
0
r ⇥ F (X(✓, z)) = ( 3 sin ✓, z, 3 cos ✓)
=6
182
SOLUTION
S
(r ⇥ F ) · ndS =
Z 3Z 2⇡
0
3( 3 sin ✓ cos ✓
z sin ✓)d✓dz
sol) (8⌧$X) °0• F X curlt (0, 0, 0)t D»‰. 0|⌧
ÑXX ·tD °D| Xîp tî |⇠⇠YX ⇠ D ⇠¥⌅‰.
0
=0
t‰.
· @S1 , @S2 |
6. –·t S p
)ï<\ Ò›
sin t, 0) 2 R3 | 0  t  2⇡},
@S1 = {(cos t,
@S2 = {(cos t, sin t, 3) 2 R3 | 0  t  2⇡},
@S1
=
Z 2⇡
0
+
Z 2⇡
sin t cos t)
ZZ
S
¿
(r ⇥ F ) · ndS = 0
a) S ⇣¯î ƒ ÌD D|‡ ` L, S@ D– ⌧∞ ¨|
©XÏ ⌅X Ò›D ùÖX‹$.
proof ) r · (r ⇥ F ) = 0 t¿\ ⌧∞ ¨– Xt
ZZ
ZZZ
(r ⇥ F ) · ndS =
0dV = 0
@S2
( sin t(1
‹·tt|‡ Xê. tL ‰L P
D ùÖX‹$.
<\ XXt @S=@S1 [ @S2 t‰. ¯Ït
Z
Z
Z
F · ds =
F · ds +
F · ds
@S
Ñ ¨
D ƒ∞X‹$. Ï0–⌧ Cî C(t) = (cos t, sin t, sin 2t)(0  t 
2⇡)t‰.
t‡
ZZ
16 °0•X
2 cos t)dt
S
( sin t(sin t cos t + 1) + cos t(3 sin t + 2))dt
D
t‰.
0
=0+0=0
b) @S@ S– §†l§ ¨| t©XÏ ⌅X Ò›D ùÖX‹$.
proof ) S –·ttt SX Ωƒ t¨X¿ Jî‰. â, @S = ;
t‰. ¯Ï¿\
(r ⇥ F ) · ndS =
F · ds t‰.
t‰. 0|⌧ §†l§ ¨– Xt
S
@S
ZZ
Z
(r
⇥
F
)
·
ndS
=
F · ds = 0
2
2
4. ·t Sî Ï<t z = 4x + y (z  4)t‡, Ë⌅ï °
S
@S
0• n@ n · k
0D ÃqXƒ] ›⇠»‰‡
Xê. °0•
ZZ
t‰.
x2 +y 2
F (x, y, z) = (4xy, 2xz, xe
)– XÏ
(r ⇥ F ) · ndS|
ZZ
ZZ
lX‹$.
sol) ·t SX Ωƒ @Sî 4x2 + y 2 = 4, z = 4t¿\ t| ‰⌧T
Xt
x(t) = (cos t, 2 sin t, 4),
0  t  2⇡
t¿\ §†l§ ¨– Xt
ZZ
Z
(r ⇥ F ) · ndS =
F · ds
=
Z 2⇡
0
=
Z 2⇡
@S
2
(8 sin t cos t, 8 cos t, cos te1+3 sin t ) · ( sin t, 2 cos t, 0)dt
8 sin2 t cos t + 16 cos2 tdt = 16⇡
0
t‰.
5.
Z
Ñ
2
(x2 +y 2 +e x )dx+(y 2 +z 2 +sin y)dy+(z 2 +y 2 +ln(z 2 +e2 ))dz
C
183
참고문헌
1. 권희대 외 8명, 미분적분학, 청문각, 2012
2. 김성기, 김도한, 계승혁, 해석개론, 제 2 개정판, 서울대학교출판문화원, 2011
3. 이익권, 이현대, 최광석, 미적분학, 경문사, 2020
4. J.Stewart, Calculus Metric Version 8E, CENAGE LEARNING, 2016
5. Marsden & Tromba, 벡터해석학 제 6판, 신한출판미디어, 2017
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