Solução_Calculo_Stewart_6e

F.
98 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
33. f(x) =x +2sinx has a horizontal tangent when f 0 (x) =0 ⇔ 1+2cosx =0 ⇔ cos x = − 1 2
⇔
x = 2π +2πn or 4π +2πn,wheren is an integer. Note that 4π and 2π are ± π units from π.Thisallowsustowritethe
3 3 3 3 3
solutions in the more compact equivalent form (2n +1)π ± π 3
, n an integer.
35. (a) x(t) =8sint ⇒ v(t) =x 0 (t) =8cost ⇒ a(t) =x 00 (t) =−8sint
(b) The mass at time t = 2π 3
has position x √3
2π
3 =8sin
2π
=8 3 2
and acceleration a √3
2π
3 = −8sin
2π
= −8 3 2
=4 √ 3,velocityv
2π
3 =8cos
2π
=8
3
− 1 2 = −4,
= −4 √ 3.Sincev
2π
3 < 0, the particle is moving to the left.
37. From the diagram we can see that sin θ = x/10 ⇔ x =10sinθ. Wewanttofind the rate
of change of x with respect to θ,thatis,dx/dθ. Taking the derivative of x =10sinθ,weget
dx/dθ = 10(cos θ). Sowhenθ = π 3 , dx
dθ =10cosπ 3 =10 1
2
=5ft/rad.
39. lim
x→0
sin 3x
x
3sin3x
=lim
x→0 3x
=3 lim
3x→0
sin 3x
3x
[multiply numerator and denominator by 3]
[as x → 0, 3x → 0]
sin θ
=3lim
θ→0 θ
[let θ =3x]
=3(1) [Equation 2]
=3
tan 6t sin 6t
41. lim
t→0 sin 2t =lim ·
t→0 t
sin 6t
=6lim
t→0 6t
1
cos 6t ·
t
6sin6t 1
=lim · lim
sin 2t t→0 6t t→0 cos 6t · lim
t→0
1
· lim
t→0 cos 6t · 1
2 lim
t→0
sin(cos θ)
sin lim cos θ
θ→0
43. lim =
θ→0 sec θ lim sec θ = sin 1 =sin1
1
θ→0
45. Divide numerator and denominator by θ. (sin θ also works.)
lim
θ→0
47. lim
x→π/4
49. (a)
(b)
sin θ
θ +tanθ =lim
θ→0
1 − tan x
sin x − cos x =
d
dx tan x = d
dx
d
dx sec x = d
dx
sin θ
θ
1+ sin θ
θ
lim
x→π/4
·
1
cos θ
sin x
cos x ⇒ sec2 x =
1
cos x
=
2t
sin 2t =6(1)· 1
1 · 1
2 (1) = 3
sin θ
lim
θ→0 θ
sin θ
1+lim lim
θ→0 θ θ→0
1 − sin x
· cos x
cos x
(sin x − cos x) · cos x =
lim
x→π/4
1
cos θ
cos x cos x − sin x (− sin x)
cos 2 x
=
2t
2sin2t
1
1+1· 1 = 1 2
cos x − sin x
(sin x − cos x)cosx =
lim
x→π/4
−1
cos x = −1
1/ √ 2 = −√ 2
= cos2 x +sin 2 x
. Sosec 2 x = 1
cos 2 x
cos 2 x .
(cos x)(0) − 1(− sin x)
⇒ sec x tan x = . Sosec x tan x = sin x
cos 2 x
cos 2 x .