Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

106 0 CHAPTER 11 PROBLEMS PLUS
an = a · _.!:_ = .!::... Since a small triangle is added to each side at every stage, it follows that the total area An added to the
9n 9n
figure at the ntl1 stage is An = Sn-1 ·an = 3 · 4n- 1 · 9~,
4n-1
= a· 32 n_ 1
. Then the total area enclosed by the snowflake
. 2 3
curve is A = a + A1 + A2 + As + · · · = a + a · ~ + a · 3
; + a · : 5
+ a · : 7
+ · · · . After the first term, this is a
. . . h . 4 A a/3 a 9 8a B I f h . . I 'I I
geometric senes w1t common ratlo !) , so. = a'+ 1
_ .1 = a + 3 · 5
= S. ut t 1e area o .t e ongma eqUI atera
9 .
. I . h 'd 1 . 1 1 . 7r J3 S h I d b h wfl k . 8 J3 2
tnang e w1t s1 e IS a = '2 · · sm J3
3 = 4 . o t e area enc ose y t e sno a e curve IS 5 · =
4
T .
7. (a) Let a= arctanx and b =arctan y. Then, from Formula 14b in Appendix D,
(b) tana-tanb tan(arctan x)-tan(arctany) x-y
tan a - = 1 +tan a tan b = 1 + tan(arctanx) tan(arctan y) = 1 + xy
. x-y
Now arctan x - arctan y = a - b = arctan( tan( a - b)) = arctan - 1
-- since --t < a - b < ¥.
+xy
(b) From part (a) we have
120 1 28,561
120 t 1 - t Wi - 239 t 28,441 t 1 7r
arc t an 119
- arc an 2311
-.arc .an
120 1
= arc an 28
_ 561
= arc an = 4
1 + Wi . 239 28,441
(c) Replacing y by-yin the formula of part (a), we get arctanx +arctan y = arctan 1
x + y . So
-xy
1 + 1
4 arctan 1 2( 1 1) 2 5 5 2 t 5 t 5 t 5
5 = arctan 5 + arctan 5 = arctan
1 1
- arc an 12 - arc an 12 + arc an 12
1-5. 5
..!!... +..!!...
12 12
= arctan
5 5
= arctan g~
1 -12'12
Thus, from part (b), we have 4 arctan t - arctan 2 i 9
= arctan ~ i~ - arctan 2
; 9
= "i.
x 3 · x 5 x 7 x 9 x 11
(d) From Example 7 in Section 11.9 we have arctanx = x- 3 + 5 - 7+ 9 -~ +· ··,so
11 1 1 1 1 1
arctan 5 = 5 - 3 · 53 + 5 · 55 - 7 · 57 + 9 · 59 - 11 · 511 + .. .
This is an alternating series and the size of the terms decreases to 0, so by the Alternating Series Estimation Theorem,
the sum lies between ss and s6, that is, 0.197395560