Calculus Homework
Assignment 9
1. True, or false? As x → ∞
a. ln x = O(ln(ln x)).
b. ex + x = o(ex ).
c. 3 + 2 cos x = O(3).
d. ln x = o(x + ln(x)).
[like §7.6 #9,10]
Sol.
x)
a. True. Since lim ln(ln
= lim
ln x
x→∞
b. False. Since
c. True. Since
d. False. Since
x
lim e +x
x
x→∞ e
=
1
x→∞ ln x
x
lim e +1
x
x→∞ e
= 0.
= lim 1 +
5
3+2 cos x
≤ 3+2
= for
3
3
3
x
lim x+ln
=
lim
(x +
x→∞ ln x
x→∞
x→∞
1
ex
=1
x sufficiently large.
1) = ∞.
2. Which of the following functions grow faster than ln x as x →
∞? Which grow at the same rate as ln x? Which grow slower?
√
1
c. 3x
d. x− 2
a. esin x
b. log5 2 x
[like §7.6 #5]
Sol.
a. Since sin x ≤ 1, so esin x ≤ e. As x sufficiently large, we
sin x
have that eln x ≤ lnex , and thus
esin x
e
≤ lim
=0
x→∞ ln x
x→∞ ln x
Hence esin x grows slower than ln x as x → ∞.
b. Since
√
√
ln 2 x
ln 2 + 12 ln x
log5 2 x
lim
= lim ln 5 = lim
x→∞
x→∞ ln x
x→∞ (ln 5)(ln x)
ln x
ln 2
1 1
= lim
=
+
x→∞ (ln 5)(ln x)
2 ln 5
2 ln 5
√
Hence log5 2 x as the same rate as ln x as x → ∞.
c. Since
3x
(ln 3)3x
lim
= lim
= (ln 3) lim (x · 3x ) = ∞
1
x→∞ ln x
x→∞
x→∞
x
lim
Hence 3x grows faster than ln x as x → ∞.
d. Since
1
x− 2
1
lim
= lim √
=0
x→∞ ln x
x→∞
x · ln x
1
Hence x− x grows slower than ln x as x → ∞.
3. The half-life of polonium is 139 days, but your sample will not
be useful to you after 95% of the radioactive nuclei present on
the day the sample arrives has disintegrated. For about how
many days after the sample arrives will you be able to use the
polonium?
[like §7.5 #18]
Sol. Let y0 be the number of radioactive nuclei initially present
in the sample. Then the number y present at any later time t
will be y = y0 e−kt for some constant k. Since the half-life of
polonium is 139 days, then we have that
1
1
y0 = y0 e−139k ⇒ = e−139k
2
2
ln 2
Thus, k = 139 ≈ 0.00498667. Since the sample will not be
useful until 95% of the radioactive nuclei present in the sample
has decay, so we must have that t ≥ t1 where t1 satisfies
0.05y0 = y0 e−kt1 ⇒ ln(0.05) = −kt1
ln(0.05)
Hence t1 = ln(0.05)
= −0.00498667
≈ 600. Therefore, we are able to
−k
use the polonium after 600 days from the day that the sample
arrived.
√ 4. Given that α = csc−1 − 25 , find sin α, cos α, tan α, sec α
and cot α.
[like §7.7 #16]
√
√ Sol. Since α = csc−1 − 25 , so csc α = − 25 . Hence
√
2
1
1
sin α = − √ , cos α = ± √ , tan α = ∓2, sec α = ± 5, cot α∓
2
5
5
5. Evaluate
Z −2 the integrals,
3
a.
dx.
2
−3 x + 6x + 10
Z
−1
ecos 2x
√
b.
dx.
1 − 4x2
[like §7.7 #102,106]
Sol.
a.
Z −2
Z −2
3
dx
dx = 3
2
2
−3 x + 6x + 10
−3 (x + 3) + 1
Z 1
du
= 3
, by letting u = x + 3
2
0 u +1
1
3π
−1 = 3 tan u = 3(tan−1 1 − tan−1 0) =
4
0
2
2
−1
√
√
b. Let u = cos 2x, then du = −
dx = − 1−4x2 dx.
2
1−(2x)
Hence
Z
Z
−1
ecos 2x
−2
1
−1
√
ecos 2x · √
dx = −
dx
2
2
1 − 4x
1 − 4x2
Z
1
1
1
−1
eu du = − eu + C = − ecos 2x + C
= −
2
2
2
6. Find the limits,
a. lim x2 tan−1 x22 .
x→∞ √
2
b. lim+ 4x − 1 .
x→ 12 3 sec−1 2x
[like §7.7 #114,115]
Sol.
a.
2
lim x tan
x→∞
b.
lim+
x→ 12
√
−1
2
tan−1 (2x−2 )
=
lim
= lim
x→∞
x2 x→∞
x−2
4x2 − 1 = 1 lim
+
3 sec−1 2x
3 x→ 12
√ 4x
4x2 −1
√2
|2x| 4x2 −1
−4x−3
1+4x−4
−2x−3
2x4
=2
x→∞ x4 + 4
= lim
1
= 2 lim+ x|2x| =
1
3
3 x→ 2
7. Evaluate
the integrals,
Z
1
a.
dθ.
Z 3csc3θ − 2 2cot θ
3t + 2t + 30t + 27
b.
dt.
t2 + 9
0
[like §8.1 #51,60]
Sol.
a.
Z
b.
Z
0
3
Z
Z
sin θ
1
2 sin θ
dθ =
dθ
1 − 2 cos θ
2
1 − 2 cos θ
Z
1
du
=
by letting u = 1 − 2 cos θ
2
u
√
1
=
ln |u| + C = ln 1 − 2 cos θ + C
2
1
dθ =
csc θ − 2 cot θ
3t3 + 2t2 + 30t + 27
dt =
t2 + 9
Z 3
Z 3
(3t + 2)dt + 3
=
0
0
Z
3
3(t + 3) 3t + 2 + 2
dt
t +9
0
Z 3
t
dt
dt + 9
2
2
t +9
0 t +9
Z 3
du
dt
+
by letting u = t2 + 9
2
t
u
0
+1
3
3
Z
3 18
3t
+ 2t +
=
2
2 9
0
18
Z 1
39 3
dw
t
=
+ ln |u| + 3
by letting w =
2
2
2
3
0 w +1
9
1
39 3
3π
39 3
+ ln 2 + 3 tan−1 w =
+ ln 2 +
=
2
2
2
2
4
2
0
8. Find the area of the region bounded above by y = 4 cos x and
below by y = 3 sec x, − π6 ≤ x ≤ π6 .
[like §8.1 #87]
√
Sol. Solve 4 cos x = 3 sec x ⇒ cos2 x = 34 ⇒ cos x = ± 23 ⇒
x = ± π6 , since − π6 ≤ x ≤ π6 . This shows that the two curves
√
√
intersect at points ( π6 , 2 3) and (− π6 , 2 3).
Hence
Z
π
6
A =
Z
[4 cos x − 3 sec x]dx = 4
− π6
π
6
Z
cos xdx − 3
− π6
π
Z
6
= 4 sin x − 3
π
−6
Z √
π
6
− π6
π
6
sec xdx
− π6
tan x sec x + sec2 x
dx
tan x + sec x
3
du
by letting u = tan x + sec x
u
√1
3
√3
= 4 − 3 ln |u| = 4 − 3 ln 3
1
= 4−3
√
3