Math 126
Name:
Fall 2016
Score:
/41
Show all your work
Dr. Lily Yen
No Calculator permitted in this part. Read the questions carefully. Show all your work
and clearly indicate your final answer. Use proper notation.
Problem 1: For each sequence or series below, determine if it converges; find the limit or
sum if possible, or prove that it diverges to ∞, or −∞, or does not exist.
√
2 − 5n n
a. an =
. Find the limit of the sequence.
3+n
√
√
√
3/2 (2/n3/2 −5)
3/2 −5
3/2
n
an = 2−5n
= 2−5n
= n n(3/n+1)
= n 2/n
→ n(−5/1) → −∞ as n → ∞.
3+n
3+n
3/n+1
Test 3
Score:
nπ ∞
ln n
b.
+ cot
. Find the limit of the sequence.
n
n + 1 n=1
If you consider the graphs
of y = ln(x) and y = x, it is clear that
/2
However, limn→∞ cot
−∞.
nπ
n+1
ln(n)
n
→ 0 as n → ∞.
= limx→π− cot(x) = −∞, so the given sequence tends to
Score:
c.
∞
X
2(5n )
/3
.
32n
n=0
P∞ 2(5n ) P∞
5 n
n=0 32n =
n=0 2( 9 ) is a geometric series
5
2
2
< 1, so it converges to 1−5/9
= 4/9
= 9/2.
9
with first term 2 and common ratio
Score:
/3
d.
∞
X
2
.
2−1
n
n=2
P
P∞
P∞
∞
2
2
1
1
n=2 n2 −1 =
n=2 (n−1)(n+1) =
n=2 ( n−1 − n+1 ) =
( 11 − 13 ) + ( 12 − 41 ) + ( 13 − 15 ) + ( 14 − 16 ) + · · · = 11 + 21 =
3
2
Score:
e.
∞
X
(−10)n
n=0
n!
. Determine both convergence and absolute convergence of the series.
an+1 (−10)n+1 n! −10 n
=
=
=
If an = (−10)
,
then
n!
an
(−10)n (n+1)!
n+1
given sequence converges absolutely by Ratio Test.
10
n+1
→ 0 < 1 as n → ∞, so the
Score:
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/3
Math 126
/4
Math 126
Fall 2016
Show all your work
Dr. Lily Yen
Calculators permitted in this part.
Problem 2: Consider the infinite series
Test 3
Name:
∞
X
(−1)n−1 n
n2 + 1
n=1
a. Evaluate the N th partial sum for the first four N ’s. Provide exact answers only.
1
X
(−1)n−1 n
n=1
2
X
n=1
3
X
n=1
4
X
n2
+1
=
1
2
(−1)n−1 n
1 2
1
= − =
2
n +1
2 5
10
(−1)n−1 n
1 2
3
2
= − +
=
2
n +1
2 5 10
5
Score:
/2
(−1)n−1 n
1 2
3
4
14
= − +
−
=
b. With your calculator, plot the
20 N ’s. You must specify
n2 N
+th
1 partial
2 sum
5 for
10 the17first 85
n=1
your TI entries.
0.5
0.4
0.3
0.2
0.1
n
4
8
12
16
20
Score:
/2
c. Approximate this infinite series with the 100th partial sum and find the error bound.
Provide a 6-decimal place accuracy.
S100 ≈ 0.264636.
Since the series is alternating, the size of the error is bounded by the size of the
100 101 101
= 10202
≈ 0.0099
absolute value of the first omitted term: (−1)
1012 +1 Score:
/2
Problem 3: For the function f (x) = ex + x2 , set up the integral expression for its arc length
for x ∈ [1, 4]. Approximate the integral accurate to 4 decimal places.
Z 4p
Z 4p
0
2
1 + (f (x)) dx =
1 + (e2 + 2x)2 dx ≈ 66.9916
1
1
Score:
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Math 126
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Problem 4: Use separation of variables to find the general solution of the following differential equation.
p
1 − y2
y0 =
1 + x2
√
R
R 1
1−y 2
dy
If dx = 1+x2 , then √ 1 2 dy = 1+x
2 dx, so arcsin(y) = arctan(x) + C, so
1−y
y = sin(arctan(x) + C)
Using that sin(u + v) = sin(u) cos(v) + cos(u) sin(v), that sin(arctan(x)) =
1
cos(arctan(x)) = √1+x
2 , you can rewrite the solution as
y=
√ x
,
1+x2
and that
x cos(C) + sin(C)
√
1 + x2
Score:
/3
Problem 5: Find the orthogonal trajectories of y = k ln(x), for arbitrary constants k and
positive x. Sketch at least three members of each family (the function and its orthogonal
trajectory). Label each curve with its parameter value.
y
ln(x)
and y 0 = xk . Thus the orthogonal trajectories satisfy
R
R
dy
= − k1 x = − ln(x)
x, so y dy = − x ln(x) dx, so, using integration by parts,
dx
y
R
R
1 2
y = − 12 x2 ln(x) + 12 x2 x1 dx = − 12 x2 ln(x) + 21 x dx = − 12 x2 ln(x) + 14 x2 + C. Hence
2
If y = k ln(x), then k =
q
y = ± −x2 ln(x) + 12 x2 + 2C
y
k=3
4
3
k=2
2
k=1
1
x
1
2
3
C=1
C=5
C = 10
k = −4
Score:
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Math 126
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Problem 6: A stomach flu spread through Capilano University. Let y(t) be the fraction of
the population at Cap that has gotten the flu at time t days from first observation. Assume
that the rate at which the flu spreads is proportional to the product of the fraction y of the
population that has the flu and the fraction 1 − y that has not yet contracted the flu.
a. Write down the differential equation satisfied by y in terms of a proportionality constant
k.
dy
= ky(1 − y)
dt
b. Find k, assuming that 10 % of the population at Cap got the flu at t = 0, and 40 %
got the flu at t = 2 days.
R 1
R
If dy
=
ky(1
−
y),
then
dy
=
k dt, so by partial fractions,
y(1−y)
R 1dt 1
R
+ 1−y dy = k dt, so ln(y) − ln(1 − y) = kt + C.
y
If t = 0, then y = 0.10, so C = ln(0.10) − ln(0.90) = −2.197. If t = 2, then y = 0.40,
so ln(0.40) − ln(0.60) = 2k − 2.197, so k = 0.8959.
c. Using the assumption of the previous part, determine when three quarters of the population at Cap will have the flu. Provide a 2-decimal place accuracy.
If y = 0.75, then ln(0.75) − ln(0.25) = 0.90t − 2.20, so t = 3.68 days.
Score:
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Math 126
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