HOMEWORK #7 – SOLUTIONS Page 195 6) g ( t ) = 4 sec t + tan t
g ' ( t ) = 4 sec t tan t + sec 2 t
16) y = x 2 sin x tan x
y ' = 2x sin x tan x + x 2 cos x tan x + x 2 sin x sec 2 x =
2x sin x tan x + x 2 sin x + x 2 sin x sec 2 x =
(
x sin x 2 tan x + x + x sec 2 x
30) )
f ( x ) = sec x ⇒ f ' ( x ) = sec x tan x ⇒
f '' ( x ) = sec x sec 2 x + tan x sec x tan x =
(
)
sec 3 x + tan 2 x sec x = sec x sec 2 x + tan 2 x ⎛π⎞
f '' ⎜ ⎟ = 2 2 + 12 = 3 2
⎝ 4⎠
(
)
36) s = 2 cos t + 3sin t, t ≥ 0 a) v = s = −2 sin t + 3cos t
a = v = −2 cos t − 3sin t
b) (see graph) c) s = 2 cos t + 3sin t = 0 ⇒ t = 2.55359 sec. (TI − 89 ) d) v = −2 sin t + 3cos t = 0 ⇒ t = 0.982794 sec. (TI − 89 )
e) s (.982794 ) = 3.60555cm.
s = 0 or a = 0 ⇒ 2 cos t + 3sin t = 0 ⇒
t = nπ − .588003, n ∈I +
40) sin 4x
x ⎞
6x ⎞
⎛ sin 4x
⎛ 4 sin 4x
= lim ⎜
⋅
= lim ⎜
⋅
⎟
⎟=
x→0 sin 6x
x→0 ⎝
x
sin 6x ⎠ x→0 ⎝ 4x
6 sin 6x ⎠
sin 4x 1 6x
1
2
4 lim
⋅
= 4 ⋅1⋅ ⋅1 =
x→0
4x 6 sin 6x
6
3
lim
cos θ − 1
cos θ − 1
0
θ
= lim
= = 0 42) lim
θ →0
θ →0
sin θ
sin θ
1
θ
Page 203 8) (
F ( x ) = 4x − x 2
(
)
100
F ' ( x ) = 100 4x − x
) ⋅ ( 4 − 2x ) or 200 ( 2 − x ) ( 4x − x )
2 99
12) −2
1
f ( t ) = 3 1 + tan t ⇒ f ( t ) = (1 + tan t ) 3 ⋅ sec 2 t =
3
2
sec t
3 3 (1 + tan t )
20) 2
2 99
( ) x +2
1
y ' = ( x + 1) ⋅ ( x + 2 ) ⋅ 2x + x + 2 ⋅ 2x =
3
2x ( x + 1)
+ 2x ( x + 2 ) =
3( x + 2 )
2x ( x + 1) + 6x ( x + 2 ) 8x + 14x
2x ( 4x + 7 )
=
=
3( x + 2 )
3( x + 2 )
3( x + 2 )
y = x2 + 1
3
2
2
−2
2
2
2
2
2
3
3
1
2
3
3
2
2
2
2
3
2
3
2
2
2
3
2
3
2
2
2
24) y = 101− x ⇒ y ' = 101− x ⋅ ln10 ⋅ ( −2x ) = −2x ln10 ⋅101− x 30) 4
⎛ y2 ⎞
⎛ y 2 ⎞ ⎡ ( y + 1) 2y − y 2 ⋅1 ⎤
G ( y) = ⎜
⇒ G' = 5⎜
⋅⎢
⎥=
⎝ y + 1 ⎟⎠
⎝ y + 1 ⎟⎠ ⎢⎣
( y + 1)2
⎥⎦
40) 76) ⎛ y2 ⎞
5⎜
⎝ y + 1 ⎟⎠
4
⎡ 2y 2 + 2y − y 2 ⎤
⎛ y2 ⎞
⎢
⎥ = 5⎜
2
⎝ y + 1 ⎟⎠
⎢⎣ ( y + 1)
⎥⎦
⎡ y10 + 2y 9 ⎤ 5y 9 ( y + 2 )
5⎢
=
6 ⎥
y
+
1
(
)
( y + 1)6
⎢⎣
⎥⎦
y = sin ( sin ( sin x ))
y ' = cos ( sin ( sin x )) ⋅ cos ( sin x ) ⋅ cos x
f ( x ) = xe− x
f ' ( x ) = −xe− x + e− x = e− x (1 − x )
f '' ( x ) =  = e− x ( x − 2 )
f ''' ( x ) = e− x ( 3 − x )
f ( 4 ) ( x ) = e− x ( x − 4 )
∴ f (1000 ) ( x ) = e− x ( x − 1000 )
4
⎡ y 2 + 2y ⎤
=
⎢
2 ⎥
⎢⎣ ( y + 1) ⎥⎦
78) s = A cos (ω t + δ ) a) v = s = −A sin (ω t + δ ) ⋅ ω = −Aω sin (ω t + δ ) b) −Aω sin (ω t + δ ) = 0 ⇒ sin (ω t + δ ) = 0 ⇒
nπ − δ
,
ω
assuming A ≠ 0 and ω ≠ 0
ω t + δ = nπ ⇒ t =