COLUMBIA UNIVERSITY
Math V1102
Calculus II
Spring 2014
Practice Exam II
04.04.2014
Instructor: S. Ali Altug
Name and UNI:
Question:
1
2
3
4
Total
Points:
6
6
12
8
32
Score:
Instructions:
• There are 4 questions on this exam.
• Please write your NAME and UNI on top of EVERY page.
• 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.
12.17.2013
Final
Name & UNI:
1
1. (6 points) Find the orthogonal trajectory to the family of curves y = x n , where n = 1, 2, 3, · · · .
1
Solution: Let mn (x, y) be the slope of the tangent line of the curve y = x n at the point (x, y).
Then by differentiating we see,
1
xn
mn (x, y) =
nx
Since we are looking for orthogonal trajectories to this family, at each point (x, y) the slope,
morth (x, y), of the tangent to any curve in the orthogonal trajectory should satisfy,
mn (x, y)morth (x, y) = −1
for every n. Therefore we need to have,
morth (x, y) = −
nx
1
xn
We would like to eliminate the dependence of this last equation on n and get an equation in terms
1
of x and y alone. For this we use the relation y = x n . Note that,
ln(x)
n
ln(x)
⇒n=
ln(y)
1
y = x n ⇒ ln(y) =
Therefore we have,
−
nx
x
1
n
=−
x ln(x)
y ln(y)
Hence our orthogonal trajectory satisfies the differential equation,
x ln(x)
dy
=−
dx
y ln(y)
We realize that this equation is separable. Then separating the variables we get,
Z
Z
y ln(y)dy = − x ln(x)dx
By integration by parts we get,
Z
x ln(x)dx =
x2
2
ln(x) −
1
2
+C
Therefore the orthogonal trajectory is given by the family of equations
y2
1
x2
1
ln(y) −
=−
ln(x) −
+C
2
2
2
2
2. (6 points) Solve the following initial value problem, where 0 ≤ x ≤
y 0 cos(x) + y + sin(x) = 0
,
Solution: Rewriting the equation by dividing it by cos(x) gives,
y 0 + y sec(x) + tan(x) = 0
Page 2
π
2.
y(0) = 0.
12.17.2013
Final
Name & UNI:
Notice that this is a linear equation of first order. Therefore we will use an integrating factor. An
integrating factor is given by,
R
e sec(x)dx
R
Since sec(x)dx = ln | sec(x) + tan(x)|, and both sec(x) and tan(x) are positive for 0 ≤ x ≤ π2 , we
see that the integrating factor is,
e
R
sec(x)dx
= sec(x) + tan(x)
Multiplying the whole equation with sec(x) + tan(x) then gives,
(sec(x) + tan(x))y 0 + sec(x)(sec(x) + tan(x))y + tan(x)(sec(x) + tan(x)) = 0
Which is,
((sec(x) + tan(x))y)0 = − tan(x)(sec(x) + tan(x))
Hence we get (I strongly recommend you go through the computation of the integrals below),
Z
Z
(sec(x) + tan(x))y = − tan(x) sec(x)dx − tan2 (x)dx
= − sec(x) + x − tan(x) + C
Which implies,
y=
x+C
−1
sec(x) + tan(x)
Finally since y(0) = 0, we get that C = 1 and the solution is given by,
y=
x+1
−1
sec(x) + tan(x)
3. Determine if the following series converge or diverge. (Please clearly state which method you are using.)
(a) (4 points)
n
∞ X
1
1−
n
n=1
(b) (4 points)
∞
X
nn
(n + 1)n+1
n=1
(c) (4 points)
∞
X
1
n ln(n)
n=3
Solution:
(a) Diverges since,
lim
n→∞
1
1−
n
n
=
1
6= 0
e
(b) Diverges. To see this first rewrite the series as
n
∞
∞ X
X
n
1
1
1
n
=
1
n+1
n + 1 n=1 1 + n n + 1
n=1
Page 3
12.17.2013
Final
Now limit compare with the series
do this).
P∞
1
n=1 n
Name & UNI:
(note that both series have positive terms so we can
1
1
n
n
=
e
1 + n1 n + 1
lim
n→∞
P1
Since
n diverges (by the p-test), by limit comparison test we conclude that our series also
diverges.
Note: For exercise try using the ratio or the root tests to see why we did not use them.
(c) Note that for n > 3 n ln(n) is positive, decreasing and continuous so we can apply the integral
test.
Z ∞
dx
∞
= lim ln(ln(x))|3 = ∞
N
→∞
x
ln(x)
3
Therefore by the integral test we see that our series also diverges.
4. Determine if the following series absolutely converge, conditionally converge or diverge.
(a) (4 points)
∞
X
(−1)n n!
en
n=1
(b) (4 points)
∞
X
1
(−1) e sin
n
n=1
n
1
n
Solution:
(a) Diverges. By the ratio test we have,
(−1)n+1 (n + 1)! (−1)n n! = lim n + 1 = ∞
lim
/
n→∞ en+1
en n→∞ e
Therefore we see that the series diverges.
(b) Converges conditionally. Let us first show that the series converges. For this note that for x ≥ 1,
and
1 1
d 1
ex = − 2 ex < 0
dx
x
d
1
1
1
sin
= − 2 cos
<0
dx
x
x
x
1
Therefore both functions are decreasing and hence their product is also decreasing. Moreover e n
and sin n1 are both positive for n ≥ 1 hence the series is alternating. Finally,
1
1
lim e n sin
=0
n→∞
n
Therefore by the alternating series test we conclude that the series converges.
To show that it converges
conditionally
we show that the sum of the absolute value of the
P∞ only
1
terms of the series, n=1 e n sin n1 , diverge. For that we limit compare it with n1 (note that the
terms are positive so we can do this).
1
1
1
1
lim e n sin
/
lim n sin
=1
n→∞
n→∞
n
n =
n
P1
P 1
Since
e n sin n1 diverges. Hence the original series
n diverges (by the p-test) the series,
converges only conditionally.
Page 4