Respostas livro calculo a 6°
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8.4 – EXERCÍCIOS – pg. 344Nos exercícios de 1 a 14, encontrar o comprimento de arco da curva dada. 1. y = 5 x − 2 , − 2 ≤ x ≤ 2
s=
∫ a b
1 + f ′( x ) dx
2 2
= =
∫
−2
2
1 + 5 dx =
∫
−2
2
2
26 dx = 26 x
−2
26 (2 + 2) = 4 26 u. c.
2. y = x
2
3
−1 , 1 ≤ x ≤ 2
y′ =
2 −13 x 3
2
s=
∫
1 2
1+
4 9x
2 3
dx = ∫
1 1 2
2
9x
2
3
+4
2 3
dx
9x
2 = ∫ 9x 3 + 4 1
dx .
1
1 − 13 . x dx 3
2 −1 2 1 1 = . ∫ 9 x 3 + 4 . 6 . x 3 dx 3 6 1
2
9x 23 + 4 32 1 = . 3 18 2
2
1 3 3 3 2 2 2 1 2 23 9 x + 4 − 13 2 = 1 9 . 2 3 + 4 − 13 3 = . 27 18 3
3. y =
3 1 (2 + x 2 ) 2 , 0 ≤ x ≤ 3 3
y′ =
1 1 3 . (2 + x 2 ) 2 . 2 x 3 2
s= = = =
∫
0
3
1 + x 2 2 + x 2 dx 1 + x 2 + x 4 dx
2
(
)
∫
0 3 0
3
∫ (x ∫(
0 2 3 2
+ 1 dx
3
)
x3 33 x + 1 dx = +x = + 3 = 12 3 3 0
)
4. x
3
+y
2
3
=2
2
3
x = 2 cos 3 t y = 2 sen 3t π s = 4 ∫ 36 cos 4 t sen 2 t + 36 sen 4 t cos 2 t dt
0
2
π
= 4 ∫ 6. sen 2 t cos 2 t cos 2 t + sen 2 t dt
0
2
(
)
sen 2 t = 24 ∫ sen t . cos t dt = 24 . 2 0
2
π
π
2
0
= 12 u. c.
5. y =
1 4 1 x + 2 , 1≤ x ≤ 2 4 8x
y′ =
1 3 1 x + (− 2 ) x −3 4 8
s= = = = =
2
∫
1
2
1 1 + x 3 − 3 dx 4x 1 + x 6 − 2x3. 1 + x 6 − 2x3. 1 1 + dx 3 16 x6 4x 1 dx 16 x 6
2
∫
1
2
∫
1
2
∫
1 2 1
2
8 x 6 + 16 x12 + 1 dx 16 x 6
6 3
1 ∫ 4 x (4 x
6
+1
)
2
dx
= =
1 ∫ 4 x (4 x
3 1 2
+ 1 dx
)
2
1 3 −3 ∫ 4 x + x dx 41
(
)
1 x 4 x −2 = 4 . + 4 4 2 1 = =
2
1 4 1 1 2 − −1 + 2 4 2.2 2 123 32
6. x =
1 3 1 y + , 1≤ y ≤ 3 3 4y
x′ =
1 1 . 3 y 2 + (− 1) . y − 2 3 4 1 = y2 − 2 4y = 4y4 −1 4y2
4 y 4 −1 16 y 8 − 8 y 4 + 1 16 y 4 + 16 y 8 − 8 y 4 + 1 =