Solução_Calculo_Stewart_6e

F.
510 ¤ CHAPTER 11 PROBLEMS PLUS
TX.10
(c) The area of each of the small triangles added at a given stage is one-ninth of the area of the triangle added at the preceding
stage. Let a be the area of the original triangle. Then the area a n of each of the small triangles added at stage n is
a n = a ·
1
9 = a
n 9 . Since a small triangle is added to each side at every stage, it follows that the total area A n n added to the
figure at the nth stage is A n = s n−1 · a n =3· 4 n−1 a ·
9 = a · 4n−1
. Then the total area enclosed by the snowflake
n 32n−1 curve is A = a + A 1 + A 2 + A 3 + ···= a + a · 1
3 + a · 4
3 + a · 42
3 3 + a · 43
+ ···.Afterthefirst term, this is a
5 37 geometric series with common ratio 4 a/3
,soA = a +
9 1 − 4 9
= a + a 3 · 9
5 = 8a . But the area of the original equilateral
5
triangle with side 1 is a = 1 2 · 1 · sin π √
3
3 = 4 . So the area enclosed by the snowflake curve is 8 √
3
5 · 4 = 2 √ 3
5 .
7. (a) Let a =arctanx and b =arctany. Then, from Formula 14b in Appendix D,
tan(a − b) =
tan a − tan b tan(arctan x) − tan(arctan y)
=
1+tana tan b 1 + tan(arctan x) tan(arctan y) = x − y
1+xy
Now arctan x − arctan y = a − b =arctan(tan(a − b)) = arctan x − y
1+xy since −π 2