COLUMBIA UNIVERSITY
Math V1102
Calculus II
Spring 2014
Midterm II
04.10.2014
Instructor: S. Ali Altug
Name and UNI:
Question:
1
2
3
4
5
6
Total
Points:
6
5
5
12
12
0
40
Score:
Instructions:
• There are 6 questions on this exam.
• Please write your NAME and UNI on top of EVERY page.
• In order to get full credit you need to answer the first 5 questions correctly.
• The last question is a bonus question, and you do not have to answer it.
• Unless otherwise is explicitly stated SHOW YOUR WORK in every question.
• Please write neatly, and put your final answer in a box.
• No calculators, cell phones, books, notebooks, notes or cheat sheets are allowed.
• Some useful identities:
–
sin2 (θ) + cos2 (θ) = 1
–
tan2 (θ) + 1 = sec2 (θ)
–
sin(2θ) = 2 sin(θ) cos(θ)
–
cos(2θ) = cos2 (θ) − sin2 (θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ)
12.17.2013
Final
Name & UNI:
1. (6 points) For which values of p does the following integral converge? (Take p ∈ R)
∞
Z
1
dx
x(ln(x))p
Hint: There analyze 1 and ∞ separately.
Solution:
Following the hint we break the integral into two parts,
Z ∞
Z 2
Z ∞
dx
dx
dx
=
+
p
p
x(ln(x))
x(ln(x))p
1
1 x(ln(x))
2
= I1 + I2
Note that by substituting u = ln(x) we get
Z
Z
dx
du
=
x(ln(x))p
up
 (ln(x))p+1

 p+1
= ln(x)


ln(ln(x))
p 6= 0, 1
p=0
p=1
We will analyze the twhree cases p = 0, 1 and p 6= 0, 1 separately.
• p = 0. In this case,
Z
2
I2 =
1
dx
x
T
= lim ln(x)|2
T →∞
= lim (ln(T ) − ln(2))
T →∞
=∞
Hence the series diverges.
• p = 1. In this case,
Z
2
I1 =
1
dx
x(ln(x))p
2
= lim+ ln(ln(x))|T
T →1
= lim+ (ln(ln(2)) − ln(ln(T )))
T →1
=∞
Since I1 diverges, the integral diverges.
• p 6= 0, 1. In this case,
Z
2
dx
p
x(ln(x))
1
2
xp+1 = lim+
T →1 p + 1
I1 =
T
(ln(T ))p+1
(ln(2))p+1
= lim+ (
−
)
p+1
p+1
T →1
(
(ln(2))p+1
p>1
p+1
=
−∞
p<1
Page 2
12.17.2013
Final
Name & UNI:
Therefore I1 converges if and only if p > 1.
On the other hand,
Z
∞
dx
x(ln(x))p
2
T
xp+1 = lim
T →∞ p + 1 2
I2 =
(ln(T ))p+1
(ln(2))p+1
= lim (
−
)
T →∞
p+1
p+1
(
∞
p>1
=
(ln(2))p+1
− p+1
p<1
Therefore we see that I2 converges if and only if p < 1.
Combining the results for I1 and I2 we get that for p 6= 1 the integral never converges.
Therefore we conclude that the integral does not converge for any value of p.
Page 3
12.17.2013
Final
Name & UNI:
2. (5 points) Let n ∈ R denote an arbitrary constant, and −1 ≤ y ≤ 1. Find the orthogonal trajectories to
the the family of curves given by
y = sin(x + n)
Solution: For each n let us denote the slope of the tangent line to the curve y = sin(x + n) at the
point (x, y) by mn (x, y). Differentiating, we get
mn (x, y) = cos(x + n)
Then the slopes of the orthogonal trajectory, morth (x, y), must satisfy,
morth (x, y) = −
Since y = sin(x + n), cos(x + n) =
p
1
cos(x + n)
1 − y 2 . Therefore we get the differential equation
1
dy
= −p
dx
1 − y2
Note that this equation is separable. Separating the variables gives
Z p
Z
2
1 − y dy = − dx
Z
⇒ cos2 (u)du = −x + C
sin(2u) u
+ = −x + C
2
p4
y 1 − y 2 + arcsin(y)
⇒
= −x + C
2
⇒
Where in integrating
p
1 − y 2 we used the substitution y = sin(u).
Page 4
12.17.2013
Final
Name & UNI:
3. (5 points) Solve the following differential equation initial value problem, where 0 < x ≤
sin(x)y 0 + 2 cos(x)y − 1 = 0
,
y
π
2
=
π
2.
π
4
Solution: We start by rewriting the equation by dividing both sides by sin(x).
y 0 + 2 cot(x)y −
1
=0
sin(x)
Note that this equation is linear and first order, therefore we can use an integrating factor to solve
the equation. An integrating factor is given by
e2
R
cot(x)dx
= e2 ln | sin(x)|
Since 0 < x ≤ π4 , sin(x) is positive hence | sin(x)| = sin(x). Therefore the integrating factor we
calculated is e2 ln(sin(x)) = sin2 (x). Now multiplying the equation by sin2 (x) then gives,
sin2 (x)y 0 + 2 sin(x) cos(x)y − sin(x) = 0
⇒(sin2 (x)y)0 = sin(x)
⇒ sin2 (x)y = − cos(x) + C
Finally since y
π
2
=
π
4
we see that C =
π
4,
and the solution is
y = − cot(x) csc(x) +
Page 5
π
csc2 (x)
4
12.17.2013
Final
Name & UNI:
4. Determine if the following series converge absolutely, conditionally or diverge.
(a) (4 points)
∞
X
n cos(nπ)
2n
n=1
(b) (4 points)
∞
X
(n!)2
(2n)!
n=1
(c) (4 points)
∞
X
(−1)n
ln(n)
n=2
Solution:
(a) We use ratio test.
n+1
(n + 1) cos((n + 1)π) n cos(nπ) n(−1)n = lim (n + 1)(−1)
lim /
/
n→∞ n→∞
2n+1
n
2n+1
n (n + 1) n
= lim
/ n
n→∞ 2n+1
2
n+1
lim
n→∞ 2n
1
=
2
Since
1
2
< 1 we conclude that the series converges absolutely.
(b) Once again we use the ratio test.
((n + 1)!)2 (n!)2
(n + 1)!(n + 1)!(2n)!
/
= lim
n→∞ (2(n + 1))! (2n)!
n→∞
n!n!(2n + 2)!
(n + 1)2
= lim
n→∞ (2n + 2)(2n + 1)
1
=
4
lim
Since
1
4
< 1 we conclude that the series converges absolutely.
1
1
(c) This series is alternating. Note that since ln(x) is increasing ln(x)
is decreasing, and limn→∞ ln(n)
=
0. Therefore the alternating series test tells us that the series converges.
P∞
1
To answer if it converges absolutely or conditionally we check if the series n=2 ln(n)
converges.
1
Note that the terms of this series are positive and decreasing, and ln(x) is continuous. Therefore
R ∞ dx
the integral test says that the series converges if and only if 2 ln(x)
converges. Now note that
for x > 2
1
1
ln(x) < x ln(x) ⇒
>
ln(x)
x ln(x)
Hence,
Z
2
∞
dx
>
ln(x)
Z
2
∞
1
x ln(x)
However by problem 1 we know
P∞that 1the right hand side diverges, hence the left hand side
diverges and hence the series, n=2 ln(n)
diverges.
n
P∞
Therefore n=2 (−1)
ln(n) converges conditionally.
Page 6
12.17.2013
Final
Name & UNI:
5. Determine if the following series converge or diverge.
(a) (4 points)
∞
X
1
n
ln(n)
ln(ln(n))
n=3
(b) (4 points)
∞
X
1
tan
n
n=1
(c) (4 points)
n
∞ X
1
1− 2
n
n=1
Solution:
(a) The function x ln(x) ln(ln(x)) is increasing and continuous (non-zero) and positive for x > 3
1
therefore x ln(x) ln(ln(x))
is decreasing, continuous and positive. Hence we can use the integral
test.
Z ∞
1
∞
= ln(ln(ln(x)))|3 = ∞
x ln(x) ln(ln(x))
3
Hence our series diverges.
1
(b) For n > 1 tan
n > 0, therefore we can use limit comparison test. We compare the series with
P∞
1
n=1 sin n (which we know from class that diverges).
tan n1
1
= lim
=1
lim
n→∞ sin 1
n→∞ cos 1
n
n
P∞
P∞
Since n=1 sin n1 diverges, so does n=1 tan n1 .
(c) Since 1 −
1
n2
≥1−
1
n
for n ≥ 1, we get
1−
1
n2
n
≥
1−
1
n
n
n
n
n
P∞
Both 1 − n12 and 1 − n1 are also positive. Moreover we know that the series n=1 1 − n1
diverges (we did this in class, and it is also in the practice midterm), by comparison test we see
that the series diverges.
Page 7
12.17.2013
Final
Name & UNI:
6. (5 points (bonus)) Determine if the following series converges or diverges.
∞
X
n=1
Hint: ln(x) <
√
1
ln(x)ln(ln(x))
x.
Solution:
√
p
2
Using the hint we see that ln(ln(x)) < ln(x). Therefore ln(x)ln(ln(x)) = e(ln(ln(x))) < e( ln(x)) =
1
eln(x) = x. Therefore ln(x)ln(ln(x))
> x1 . Hence by the comparison theorem the series diverges.
Page 8