Solução_Calculo_Stewart_6e

F.
454 ¤ CHAPTER 10 PROBLEMS PLUS
TX.10
(b) dy
dt = (1 + t3 )(6t) − 3t 2 (3t 2 ) 6t − 3t4
=
(1 + t 3 ) 2 (1 + t 3 ) =0when 6t − 2 3t4 =3t(2 − t 3 )=0 ⇒ t =0or t = 3√ 2,sothereare
horizontal tangents at (0, 0) and 3 √ 2, 3√ 4 . Using the symmetry from part (a), we see that there are vertical tangents at
(0, 0) and 3 √ 4, 3√ 2 .
(c) Notice that as t →−1 + ,wehavex →−∞and y →∞.Ast →−1 − ,wehavex →∞and y →−∞.Also
y − (−x − 1) = y + x +1= 3t +3t2 +(1+t 3 ) (t +1)3 (t +1)2
= = → 0 as t →−1. Soy = −x − 1 is a
1+t 3 1+t 3 t 2 − t +1
slant asymptote.
(d) dx
dt = (1 + t3 )(3) − 3t(3t 2 )
= 3 − 6t3
dy
and from part (b) we have
(1 + t 3 ) 2 (1 + t 3 )
2
dt
d dy
Also d2 y
dx = dt dx
= 2(1 + t3 ) 4
2 dx/dt 3(1 − 2t 3 ) > 0 ⇔ t< √ 1 3 3
2 .
=
6t − 3t4
(1 + t 3 ) 2 .Sody dx = dy/dt
dx/dt = t(2 − t3 )
1 − 2t 3 .
So the curve is concave upward there and has a minimum point at (0, 0)
and a maximum point at 3 √ 2, 3√ 4 . Using this together with the
information from parts (a), (b), and (c), we sketch the curve.
3 3t 3t
(e) x 3 + y 3 2
3
=
+
= 27t3 +27t 6
= 27t3 (1 + t 3 )
= 27t3
1+t 3 1+t 3 (1 + t 3 ) 3 (1 + t 3 ) 3 (1 + t 3 ) and 2
3xy =3 3t 3t
2 27t 3
=
1+t 3 1+t 3 (1 + t 3 ) 2 ,sox3 + y 3 =3xy.
(f) We start with the equation from part (e) and substitute x = r cos θ, y = r sin θ. Then x 3 + y 3 =3xy
r 3 cos 3 θ + r 3 sin 3 θ =3r 2 cos θ sin θ. Forr 6= 0,thisgivesr =
by cos 3 θ,weobtainr =
1 sin θ
3
cos θ cos θ
1+ sin3 θ
cos 3 θ
(g) The loop corresponds to θ ∈ 0, π 2
, so its area is
A = π/2
0
r 2
2 dθ = 1 2
π/2
0
9
= lim
b→∞
2 −
1
(1 + 3 u3 ) −1 b
= 3 0 2
=
3secθ tan θ
1+tan 3 θ .
2 3secθ tan θ
dθ = 9 π/2 sec 2 θ tan 2 θ
1+tan 3 θ 2
0
(1 + tan 3 θ) dθ = 9 ∞
2 2 0
3cosθ sin θ
cos 3 θ +sin 3 . Dividing numerator and denominator
θ
⇒
u 2 du
(1 + u 3 ) 2 [let u =tanθ]
(h)Bysymmetry,theareabetweenthefoliumandtheliney = −x − 1 is equal to the enclosed area in the third quadrant,
plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is 1 , and since y = −x − 1 ⇒
2
1
r sin θ = −r cos θ − 1 ⇒ r = −
, the area in the fourth quadrant is
sin θ +cosθ
1 −π/4 2 2
1
3secθ tan θ
−
−
dθ CAS
= 1 2
sin θ +cosθ 1+tan 3 θ
2 . Therefore, the total area is 1 +2
1
2 2 =
3
. 2
−π/2