8.4 – exercÍcios – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + =...
TRANSCRIPT
8.4 – EXERCÍCIOS – pg. 344
Nos exercícios de 1 a 14, encontrar o comprimento de arco da curva dada.
1. 25 −= xy , 22 ≤≤− x
( )
( ) ..2642226
262651
1
2
2
2
2
2
2
2
2
cu
xdxdx
dxxfs
b
a
=+=
==+=
′+=
−−−
∫∫
∫
2. 132
−= xy , 21 ≤≤ x
31
3
2 −
=′ xy
dxxdxx
dx
x
xdx
x
s
31
2
1
21
32
2
13
2
322
13
2
.3
1.49
9
49
9
41
−
∫
∫∫
+=
+=+=
−
+=
−
+=
+
=
+=
−
∫
31342.927
11349
3
2.
18
1
23
49
.18
1
.6.496
1.
3
1
23
32
232
3
32
2
1
23
32
31
2
1
21
32
x
x
dxxx
3. ( ) 23
223
1xy += , 30 ≤≤ x
( ) xxy 2.22
3.
3
12
12+=′
( )
( )
( ) 1233
3
31
1
1
21
33
0
33
0
2
3
0
2
3
0
42
3
0
22
=+=+=+=
+=
++=
++=
∫
∫
∫
∫
xx
dxx
dxx
dxxx
dxxxs
4. 32
32
32
2=+ yx
=
=
tseny
tx
3
3
2
cos2
( )
..12
2.24cos.24
coscos.64
cos36cos364
2
0
22
0
2
0
2222
2
0
2424
cu
tsendtttsen
dttsentttsen
dtttsentsents
=
==
+=
+=
∫
∫
∫
ππ
π
π
5. 2
4
8
1
4
1
xxy += , 21 ≤≤ x
( ) 33 28
1
4
1 −−+=′ xxy
( ) dxxx
dxx
xx
dxx
xx
dxxx
xx
dxx
xs
∫
∫
∫
∫
∫
+=
++=
−+=
+−+=
−+=
2
1
26
3
2
1
6
126
2
1
6
36
2
1
3
36
2
1
2
3
3
144
1
16
1168
16
1.21
616
1
4
1.21
4
11
( )
( )
32
123
2
11
2.2
12
4
1
24.4
4
1
44
1
144
1
2
4
2
1
24
2
1
33
2
1
26
3
=
+−−=
+=
+=
+=
−
−
∫
∫
xx
dxxx
dxxx
6. y
yx4
1
3
1 3 += , 31 ≤≤ y
( )
2
4
2
2
22
4
14
4
1
.14
13.
3
1
y
y
yy
yyx
−=
−=
−+=′ −
( )4
24
4
48
4
484
4
482
2
4
16
14
16
1816
16
181616
16
18161
4
141
y
y
y
yy
y
yyy
y
yy
y
y
−=
++=
+−+=
+−+=
−+
( )
6
53
4
1
3
1
3.4
1
3
3
1.
4
1
3
4
1
4
14
16
14
3
3
1
13
3
1
22
3
1
2
4
3
1
4
24
=
+−−=
−+=
+=
+=
+=
−
−
∫∫
∫
yy
dyyydyy
y
dyy
ys
7. ( )xxeey
−+=2
1 de ( )1,0 a
+ −
2,1
1ee
( )xxeey
−−=′2
1
( )
( )
( )
dxee
dxee
dxee
dxeeee
dxees
xx
xx
xx
xxxx
xx
∫
∫
∫
∫
∫
−
−
−
−−
−
+−+=
+−+=
+−+=
+−+=
−+=
1
0
22
1
0
22
1
0
22
1
0
22
1
0
2
4
24
4
1
2
1
4
11
24
11
..24
11
4
11
( )
( )
( )
( )
( )
1
112
1
2
1
2
1
11
2
1
11
2
1
12
1
1
0
1
0
1
0
2
1
0
22
1
0
2
2
hsen
ee
ee
dxee
dxee
dxee
dxe
e
xx
xx
x
x
x
x
x
x
=
+−−=
+=
+=
+=
+=
++=
−
−
−
−
∫
∫
∫
∫
8. xy ln= , 83 ≤≤ x
xy
1=′
dxx
xdx
xs ∫∫
+=+=
8
3
28
3
2
111
( )
( ) dtttdx
tx
tx
tx
2.12
1
1
1
1
21
2
21
2
22
22
−=
−=
−=
=+
( ) ( )
( ) ( )∫ ∫
∫
∫
∫∫
+−+=
−+=
−=
−−
=+
=
11
1
11
1
1
.
1
1
2
2
2
21
221
2
2
tt
dtdt
dtt
t
dtt
t
dtt
t
tdx
x
xI
Cx
xx
Ct
tt
Cttt
dtt
dtt
t
+++
−+++=
++
−+=
++−−+=
+−
−+= ∫∫
11
11ln
2
11
1
1ln
2
1
1ln2
11ln
2
1
1
21
1
21
2
22
2
3ln
2
11
3
1ln
2
12
4
2ln
2
13
11
11ln
2
11
8
3
2
22
+=
−−+=
++
−+++=
x
xxs
9. ( )xseny ln1−= , 46
ππ≤≤ x
xsen
xy
cos−=′
∫
∫∫
∫
=
=+
=
+=
4
6
4
6
4
6
2
22
4
6
2
2
cos
cos
cos1
π
π
π
π
π
π
π
π
dxxec
sen
dxdx
xsen
xxsen
dxxsen
xs
x
..332
12ln
262
632ln
132
3.
2
22ln
3
32ln1
2
2ln
cotcosln 4
6
cu
xgec
−
−=
−
−=
−
−=
−−−=
−=π
π
10. 3xy = , de ( )0,00P ate ( )8,41P
21
2
3xy =′
( ) ..1101027
8
14.4
91
3
2.
9
4
23
4
91
9
4
4
91
23
4
0
23
4
0
cu
x
dxxs
−=
−
+=
+
=
+= ∫
11. 24 3 += xy de ( )2,00P ate ( )6,11P
21
21
62
3.4 xxy ==′
( )
( )
( ) ..1373754
1
137373
2.
36
1
23
361
36
1
361
1
0
23
1
0
cu
x
dxxs
−=
−=
+=
+= ∫
12. ( )16 3 2 −= xy de ( )0,10P ate ( )6,221P
31
31
4
3
2.6
−
−
=
=′
x
xy
dxxx
dxx
x
dx
x
dxxs
31
22
1
21
32
22
13
2
32
22
13
2
22
1
32
16
16
161
161
−−
−
−
−
−
∫
∫
∫
∫
+=
+=
+=
+=
( )
..1717254
17171818
172216
2
3
16
2
3
232
3
32
22
1
23
32
cu
x
−=
−=
−
+=
+
=
−
−
13. ( ) ( )3241 +=− xy de ( )2,30 −P ate ( )9,01P
( )
( )
( ) 21
23
23
42
3
41
41
+=′
++=
+=−
xy
xy
xy
( )
( )( )
...27
13131080
4
1310
3
2.
9
4
23
34
91
9
4
44
91
23
23
0
3
23
0
3
cu
x
dxxs
−=
−=
++=
++=
−
∫
14. 32 yx = , de ( )0,00P ate ( )4,81P
31
32
3
2 −
=′
=
xy
xy
dxxx
dx
x
x
dxxs
∫
∫
∫
−
−
+=
+=
+=
8
0
312
1
32
8
0 32
32
8
0
32
.3.49
9
49
9
41
31
32
3
2.9
49
−
=
+=
xdu
xu
( )
( ) ..8108027
1
4440403
2.
18
1
23
49
.18
1
8
0
23
32
cu
x
−=
−=
+
=
Nos exercícios de 15 a 21, estabelecer a integral que da o comprimento de arco da curva
dada.
15. 2xy = , 20 ≤≤ x
xy 2=′
∫ +=2
0
241 dxxs
16. x
y1
= de
4,
4
10P ate
4
1,41P
2
1
xy
−=′
∫
∫
∫
+=
+=
+=
4
41
2
4
4
41
4
4
4
41
4
1
1
11
dxx
x
dxx
x
dxx
s
17. 122 =− yx de ( )22,30 −P ate ( )22,30P
( ) ( ) yyxyx
yx
2.12
11
1
21
221
2
22
+=′⇒+=
+=
dyy
y
dyy
yy
dyy
ys
∫
∫
∫
−
−
−
+
+=
+
++=
++=
22
22
2
2
22
22
2
22
22
22
2
2
1
21
1
1
11
18. xey = , de ( )1,00P ate ( )2
1 ,2 eP
xey =′
dxesx
∫ +=
2
0
21
19. 122 −+= xxy , 10 ≤≤ x
22 +=′ xy
( )
dxxx
dxxx
dxxs
∫
∫
∫
++=
+++=
++=
1
0
2
1
0
2
1
0
2
584
4841
221
20. xy = , 42 ≤≤ x
21
2
1 −=′ xy
∫ +=
4
24
11 dx
xs
21. xseny 3= , π20 ≤≤ x
xy 3cos3=′
dxxs ∫ +=
π2
0
2 3cos91
Nos exercícios de 22 a 29, calcular o comprimento de arco da curva dada na forma
paramétrica.
22.
=
=
2
3
ty
tx, 31 ≤≤ t
( ) ( )
( ) ..1313858527
1
131385853
2.
18
1
23
49
18
1
49
49
3
1
23
2
3
1
2
3
1
24
cu
t
dttt
dttts
−=
−=+
=
+=
+=
∫
∫
23. ( )
( )
−=
−=
ty
tsentx
cos12
2, [ ]π,0∈t
( )
tseny
tx
2
cos12
=′
−=′
( )
( )
( ) dtt
dtt
dttsentt
dttsents
∫
∫
∫
∫
−=
−=
++−=
+−=
π
π
π
π
0
0
0
22
0
22
cos12
cos222
coscos214
4cos14
( )
..8
108
0cos2
cos82
cos.2.4
2222
2222
cos122
0
0
0
2
0
cu
t
dtt
sen
dtt
sen
dtt
=
−−=
−−=−=
=
=
−=
∫
∫
∫
ππ
π
π
π
24.
=
−=
ty
tsenx
cos, [ ]π2,0∈t
ππ
π
2
cos
2
0
2
0
22
==
+= ∫
t
dttsents
25.
=
=
tty
tsentx
cos, [ ]π,0∈t
22
0
22
0
2
0
222222
1ln2
11
2
1ln2
11
2
1
cos2coscos2cos
ππππ
π
π
π
++++=
++++=
+=
+−+++=
∫
∫
tttt
dtt
dttsentttsentttsentsentttts
26.
−=
+=
1
23
ty
tx, [ ]2,0∈t
10210
91
2
0
2
0
==
+= ∫
t
dts
27.
=
=
2
3
2
1
3
1
ty
tx
, 20 ≤≤ t
=′
=′
ty
tx
2.2
1
3.3
1 2
dtttdttts ∫∫ +=+=
2
0
2
2
0
24 1
dttdu
tu
2
12
=
+=
( ) ( )
( ) ..1553
1
153
2.
2
1
23
1
2
11
23
2
0
23
22
0
21
2
cu
tdttts
−=
−=
+=+= ∫
28.
=
=
tseney
tex
t
t cos, 21 ≤≤ t
+=′
+−=′
tt
tt
etsentey
ettsenex
cos
cos
( ) ( )
( ) ..22
2
coscos
22
1
2
1
2
1
2222
cueee
dte
tsentetsentes
t
t
tt
−==
=
++−=
∫
∫
29.
−=
+=
tttseny
tsenttx
cos22
2cos2,
20
π≤≤ t
tsenty
ttx
2
cos2
=′
=′
∫
∫
===
+=
2
0
22
0
2
2
0
2222
..42
22
4cos4
π π
π
πcu
tdtt
dttsenttts
30. Achar o comprimento da hipociclóide
=
=
ty
tsenx
3cos4
4 3
, [ ]π2,0∈t
−=′
=′
tsenty
ttsenx
.cos3.4
cos.3.4
2
2
( )
..24
012
48
2.12.4
cos124
cos12cos124
2
0
2
2
0
2
0
24242
cu
tsen
dtttsen
dttsentttsens
=
−=
=
=
+=
∫
∫
π
π
π
31. Achar o comprimento da circunferência.
=
=
tsenay
tax cos, [ ]π2,0∈t
tay
senax
cos=′
−=′
..244
cos4
2
0
2
0
2222
cuaatdta
dttatsenas
π
π
π
===
+=
∫
∫
32. Calcular o comprimento da parte da circunferência que está no 1° quadrante
=
=
47
4cos7
tseny
tx
..4
7
4
7
4
7
4cos
4
7
44
7
2
0
2
0
2
0
2
2
2
2
cutdt
dttt
sens
π
ππ
π
===
+
=
∫
∫
Nos exercícios de 33 a 35, calcular a área da região limitada pelas seguintes curvas, dadas
na forma paramétrica.
33.
=
=
tseny
tx cos e
=
=
tseny
tx
2
1
cos
-1 1
-1.5
-1
-0.5
0.5
1
1.5
x
y
( ) ( )
..2
1
2
1
2
144
2
0
2
0
cu
dttsentsendttsentsenA
π
ππ
ππ
=
−=
−+−−= ∫∫
34.
=
=
tseny
tx
3
3
2
cos2 e
=
=
tseny
tx
2
cos2
-2 -1 1 2
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
x
y
( ) ( )
( )
−−=
−
−=
−−−−−=
∫
∫
∫ ∫
2
0
24
2
0
242
0
2
0
2
0
23
1122
.24
cos1222
1
2
1.44
cos3.2.2224
π
ππ
π π
πdttsentsen
dtttsentsent
dttsenttsendttsentsenA
( )
..2
5
2
38
2
34
2
164
4
3cos
4
184
482
548cos
6
484
486
5cos
6
1484
484
2
0
2
0
23
2
0
2
0
442
0
5
2
0
445
2
0
64
cu
t
dttsenttsen
dttsendttsenttsen
dttsendttsenttsen
dttsentsen
πππππ
π
π
π
π
π
π
π
π ππ
π
π
=−
=−=
−=
+−−=
−+−=
−
+−+=
−−=
∫
∫ ∫
∫∫
∫
35.
=
=
2ty
tx e
+=
+=
ty
tx
31
1
1 2
-2
-1
1
2
3
4
x
y
23
2
−=
=
xy
xy
2=x e 1=x
( )( ) 24,2
11,1
=→
=→
t
t
( )( ) 14,2
01,1
=→
=→
t
t
3
7
3
.1.
2
1
3
2
1
2
1
=
=
= ∫
t
dttA
( )
2
5
23
..31
1
0
2
1
0
1
=
+=
+= ∫
tt
dttA
..6
1
3
7
2
5cuA =−=
36. Calcular a área da arte da circunferência
tseny
tx
2
cos2
=
=
que está acima da reta 1=y
( )
( )62
3cos3cos21,3
22,0
π
π
=∴=∴=→
=→
ttt
t
( )
6
334
6
6332
2
3
3
22
3
2
1
62
2
1
232
1
62
22
12
2
2cos14
22
6
2
6
2
6
2
1
+=
++−=
++−=
−−−=
+−−−=
−−=
−−=
−=
∫
∫
πππ
ππ
ππ
ππππ
π
π
π
π
π
π
sensen
tsentdtt
dttsentsenA
..6
334
6
36334
36
334
cu
A
−=
−+=
−+
=
π
π
π
37. Calcular a área da região delimitada pela elipse
-3 -2 -1 1 2 3
-1
1
x
y
=
=
tseny
tx cos3
( )
..4
3
3
3
0
2
2
0
2
1
cu
dttsen
dttsentsenA
π
π
π
=
=
−=
∫
∫
auA .34
3.4 π
π==
38. Calcular a área da região limitada à direita pela elipse
=
=
tseny
tx
2
cos3 e a esquerda pela
reta 2
33=x
6cos
2
3cos3
2
33 π=∴∴= ttt
( )
..34
3
22
3.
2
3
2
34
1
6.
2
16
24
1
2
16
632
6
0
0
6
2
0
6
1
au
sen
tsent
dttsendttsentsenA
−=−=
−=
−=
=−= ∫∫
ππ
ππ
π
ππ
39. Calcular a área da região entre as curvas
=
=
tseny
tx
2
cos4 e
=
=
tseny
tx cos
-4 -3 -2 -1 1 2 3 4
-2
-1
1
2
x
y
( ) ππ
8244
0
2
1 =−= ∫ dttsentsenA
π
ππ
=
+−=−= ∫
0
2
0
2
22
1cos
2
14 tttsendttsentsenA
..78 auA πππ =−=
40. Calcular a área entre o arco da hipociclóide
=
=
tseny
tx
3
3
3
cos3, [ ]
2,0 π∈t
e a reta 3=+ yx
( )
( )
( )
ππ
π
π
π
π
π
32
27
2.
6
1.27
16
1cos
16
1cos
24
1cos
6
127
27
127
cos27
.cos3.33
2
0
35
0
2
64
2
0
2
4
2
0
2
4
3
0
2
3
1
==
+−−=
−=
−=
=
−=
∫
∫
∫
∫
tttsenttsenttsen
dttsentsen
dttsentsen
dtttsen
dttsenttsenA
..32
27144
32
27
2
9auA
ππ
−=−=
41. Calcular a área delimitada pela hipociclóide
=
=
ty
tsenx
3
3
cos4
4
-4 -3 -2 -1 1 2 3 4
-4
-3
-2
-1
1
2
3
4
x
y
( )
( )
( )
..2
3
coscos48
cos1cos48
cos48
cos..3.4.cos4
2
0
64
2
0
24
2
0
24
22
0
3
au
dttt
dttt
dttsent
dtttsentA
π
π
π
π
π
=
−=
−=
=
=
∫
∫
∫
∫
auA .62
3.4 π
π==
42. Calcular a área da região S , hachurada na figura 8.12
( )( )tky
tsentkx
cos1−=
−=
( ) ( )
( )
( )
..3
3.
2.2
12
2
1cos
2
12
coscos21
cos1
cos1cos1
2
2
2
2
0
2
2
0
22
2
0
22
2
0
auk
k
k
ttsenttsentk
dtttk
dttk
dttktkA
π
π
ππ
π
π
π
π
=
=
+=
++−=
+−=
−=
−−=
∫
∫
∫