MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH
SOLUTIONS
18.01 Calculus, Fall 2014
Professor: Jared Speck
Problem 1. Consider the following curve in parametric form:
x = t + ln t,
y = t − ln t.
Find the arc length of the portion of the curve corresponding to 1 ≤ t ≤ 2.
Hint: To evaluate the arc length integral, first make the inverse trig substitution t = tan u. Then
multiply the top and bottom by sin u and make the substitution v = cos u. Finally, use partial
fractions to evaluate the v integral.
Problem 2. Consider the following curve in the plane:
1
x2 y = .
2
Let a ≥ 1 be a number. The portion of the curve with 1 ≤ x ≤ a is revolved around the x−axis
to generate a solid of revolution. Show that the surface area of the solid does not become infinite
when a → ∞.
Problem 3. Compute the Taylor series (at the base point 0) for the function
1
.
(1 − x)3
Problem 4. Compute the Taylor series (at the base point 0) for the function
f (x) = arcsin x.
In particular, compute the first four non-zero terms in the series and illustrate the general pattern
for the coefficients.
Problem 5. Find all numbers p ≥ 0 such that the series
∞
X
1
ln 1 + p
k
k=1
converges.
Problem 6. Find all numbers p ≥ 0 such that the series
∞
X
k=2
1
k(ln k)p + 1
converges.
1
2
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 7. Show that if x is not a number of the form x =
sec2 x =
∞
X
π
2
+ mπ for some integer m, then
(sin x)2k .
k=0
2
Problem 8. Let f (x) = ex . Compute
f (2014) (0),
where f (2014) (x) is the 2014th derivative of f (x).
Hint: Taylor (The good kind of Taylor, not the Swift kind · · · )
Problem 9. Consider the following curve in the plane, described in polar form:
r = ln θ.
Find the area of the region in the second quadrant (x ≤ 0, y ≥ 0) that is trapped between the
x−axis, the y−axis, and the curve.
Problem 10. Compute
ex + cos2 x
.
x→∞
ex
lim
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
3
Solutions
Problem 1. Consider the following curve in parametric form:
x = t + ln t,
y = t − ln t.
Find the arc length of the portion of the curve corresponding to 1 ≤ t ≤ 2.
Hint: To evaluate the arc length integral, first make the inverse trig substitution t = tan u. Then
multiply the top and bottom by sin u and make the substitution v = cos u. Finally, use partial
fractions to evaluate the v integral.
Solution: We first compute that
dx
1
=1+ ,
dt
t
dy
1
=1− ,
dt s t
2 2
dx
dy
ds =
+
dt
dt
dt
r
√
1
= 2 1+ 2
t
√
√ t2 + 1
= 2
.
t
The arc length is therefore
Z
Arc Length =
√ Z
ds = 2
1
2
√
t2 + 1
dt.
t
To evaluate the integral, we first make the inverse trig substitution t = tan u, dt = sec2 u, use
tan2 u + 1 = sec2 u, multiply the top and bottom by sin u, and use sin2 = 1 − cos2 u, which leads to
Z √2
Z √ 2
t +1
sec u
dt =
sec2 u du
t
tan u
Z
1
=
du
sin u cos2 u
Z
sin u
du
=
2
sin u cos2 u
Z
sin u
=
du.
(1 − cos2 u) cos2 u
We then make the substitution v = cos u, dv = − sin u du, which leads to
Z
Z
sin u
1
du = −
dv
2
2
(1 − cos u) cos u
(1 − v 2 )v 2
Z
1
=−
dv.
2
v (1 − v)(1 + v)
4
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
To evaluate the v integral, we first make a partial fraction decomposition:
1
A B
D
C
= + 2+
+
.
2
v (1 − v)(1 + v)
v
v
1−v 1+v
The cover up method yields B = 1, C = 12 , D = 21 . Then plugging the value v = 2 into both sides
of the decomposition yields that A = 0. Hence,
1
1
1
1
1
1
= 2+
+
,
v 2 (1 − v)(1 + v)
v
2 1−v
2 1+v
and
Z
1
1 1 1 − v dv = + ln +c
−
v 2 (1 − v)(1 + v)
v 2
1 + v
1
1 1 − cos u =
+ ln
+c
cos u 2 1 + cos u √
1 − (1 + t2 )−1/2 1
+c
= 1 + t2 + ln 2
1 + (1 + t2 )−1/2 √
1 + t2 − 1 √
1
= 1 + t2 + ln √
+ c.
2
2 1 + t + 1
In our calculations above, we used a right triangle to deduce that
cos u = (1 + t2 )−1/2 .
√
Finally, we reinsert the factor of 2 and the bounds of integration to conclude that
√
√ Z 2 t2 + 1
Arc Length = 2
dt
t
1
√
)t=2
(
2
√ √
1 1 + t − 1 2
1 + t + ln √
= 2
2 1 + t2 + 1 t=1
)
√
√
(
√ √
√
1 5 − 1 1 2 − 1 5 − 2 + ln √
= 2
− ln √
2 5 + 1 2 2 + 1
√
√
)
(
√ √
√
1 2 + 1 1 5 + 1 = 2
5 − 2 + ln √
− ln √
.
2 2 − 1 2 5 − 1
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
5
Problem 2. Consider the following curve in the plane:
1
x2 y = .
2
Let a ≥ 1 be a number. The portion of the curve with 1 ≤ x ≤ a is revolved around the x−axis
to generate a solid of revolution. Show that the surface area of the solid does not become infinite
when a → ∞.
Solution: We first compute that
1
y = x−2 ,
2
dy
= −x−3 ,
dx s
ds =
1+
dy
dx
2
dx =
√
1 + x−6 dx.
The surface area of the region of interest is
Z
Z
thin conical strip circumference × conical strip slant height = 2πy ds.
Hence, the surface area is
a
√
1
2π x−2 1 + x−6 dx.
2πy ds =
2
1
1
√
√
√
Now when x ≥ 1, we have 1 + x−6 ≤ 1 + 1−6 = 2. Hence, the above integral is
Z a
√
x−2 dx
≤π× 2×
√ −1 1a
= 2π −x 1
√
1
.
= 2π 1 −
a
√
Now as a → ∞, the above term converges to the finite number 2π.
Z
a
Z
6
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 3. Compute the Taylor series (at the base point 0) for the function
1
.
(1 − x)3
Solution: We first note that
1
1 d2 1
.
=
(1 − x)3
2 dx2 1 − x
The standard formula for the geometric series yields that
∞
X
1
=
xk = 1 + x + x 2 + x3 + x4 + · · · .
1 − x k=0
Twice differentiating the above series term-by-term and multiplying by 1/2, we deduce that
∞
∞
1 d2 X d2 k 1 X
1
=
x =
k(k − 1)xk−2
(1 − x)3
2 dx2 k=0 dx2
2 k=2
= 1 + 3x + 6x2 + · · · .
The above series is the desired Taylor series.
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
7
Problem 4. Compute the Taylor series (at the base point 0) for the function
f (x) = arcsin x.
In particular, compute the first four non-zero terms in the series and illustrate the general pattern
for the coefficients.
1
d
arcsin x = √1−x
Solution: Our strategy is to first compute the Taylor series for dx
2 and to then
integrate that series term-by-term to recover the Taylor Series for arcsin x. More precisely, we first
use the fundamental theorem of calculus to deduce that
Z x
dy
p
arcsin x =
.
1 − y2
0
We will compute the Taylor series for the integrand √ 1
1−y 2
in two steps. In the first step, we compute
the Taylor series for the function
1
√
= (1 − u)−1/2 .
1−u
Differentiating with respect to u, we deduce that
1
d
(1 − u)−1/2 = (1 − u)−3/2 ,
du
2 2
1
d
3
(1 − u)−1/2 =
(1 − u)−5/2 ,
2
du
2
2
1
3
5
d3
−1/2
(1
−
u)
=
(1 − u)−7/2 ,
du3
2
2
2
···
Setting u = 0 in the above formulas, we deduce that
d
1
(1 − u)−1/2 |u=0 = ,
du
2 1
3
d2
−1/2
(1
−
u)
|
=
,
u=0
du2
2
2
d3
3
5
1
−1/2
(1 − u)
|u=0 =
,
3
du
2
2
2
··· .
From the Taylor formula
f (u) = a0 + a1 u + a2 u2 + a3 u3 + · · · ,
f (n) (0)
an =
,
n!
and the above computations, it follows that
1
1
1
1×3
1
1×3×5
1
√
=1+ × u+
× u2 +
× u3 + · · · .
2 1!
2 × 2 2!
2 × 2 × 2 3!
1−u
8
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
We now substitute u = y 2 to deduce the Taylor series for the integrand √ 1
1−y 2
:
1
1
1
1×3
1
1×3×5
1
p
= 1 + × y2 +
× y4 +
× y6 + · · · .
2 1!
2 × 2 2!
2 × 2 × 2 3!
1 − y2
Integrating this series term-by-term, we conclude that
Z x
dy
p
arcsin x =
1 − y2
0
Z x
1
1 2 1×3
1 4 1×3×5
1 6
1+ × y +
× y +
× y + · · · dy
=
2 1!
2 × 2 2!
2 × 2 × 2 3!
0
1
1 1 3 1×3
1 1 5 1×3×5
11 7
= y+ ×
y +
×
y +
×
y + · · · |y=x
y=0
2 1! 3
2 × 2 2! 5
2 × 2 × 2 3! 7
1
1 1 3 1×3
1 1 5 1×3×5
11 7
=x+ ×
x +
×
x +
×
x + ··· .
2 1! 3
2 × 2 2! 5
2 × 2 × 2 3! 7
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
9
Problem 5. Find all numbers p ≥ 0 such that the series
∞
X
1
ln 1 + p
k
k=1
converges.
Solution: We first claim that for large k, ln 1 + k1p ∼ k1p . To verify this claim, we use L’Hôpital’s
∞
form to deduce that
rule in the ∞
1
× (−p)k −p−1
ln 1 + k1p
1+ k1p
lim
= lim
1
k→∞
k→∞
(−p)k −p−1
kp
1
= lim
k→∞ 1 + 1p
k
= 1.
Therefore, by limit comparison,
∞
X
1
ln 1 + p
k
k=1
converges if and only if
∞
X
1
kp
k=1
converges. By the integral comparison test, this latter series converges if and only if p > 1.
10
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 6. Find all numbers p ≥ 0 such that the series
∞
X
1
k(ln k)p + 1
k=2
converges.
Solution: We first note that
1
1
∼
,
p
k(ln k) + 1
k(ln k)p
as k → ∞, so by limit comparison, it suffices to investigate the convergence of
∞
X
1
.
p
k(ln
k)
k=2
By the integral comparison test, it suffices to investigate the convergence of
Z ∞
Z M
dx
dx
= lim
.
p
M →∞ x=2 x(ln x)p
x=2 x(ln x)
, which leads to
To this end, we make the substitution u = ln x, du = dx
x
1 1−p
Z
Z
dx
du
u , if p 6= 1,
1−p
=
=
p
p
ln u,
if p = 1.
x(ln x)
u
1
1−p
(ln x) , if p 6= 1,
1−p
=
ln(ln x),
if p = 1.
Hence, we compute that
Z
M
lim
M →∞
x=2
dx
=
x(ln x)p
1
(ln 2)1−p ,
p−1
if p > 1,
limit does not exist, if p ≤ 1.
Thus,
∞
X
k=2
converges if p > 1 and diverges if 0 ≤ p ≤ 1.
1
k(ln k)p + 1
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 7. Show that if x is not a number of the form x =
sec2 x =
∞
X
π
2
11
+ mπ for some integer m, then
(sin x)2k .
k=0
Solution: As long as x is not of the form x = π2 +mπ for some integer m, we have that | sin x| < 1.
Thus, under this assumption, it follows that the geometric series
∞
∞
X
X
2k
(sin x) =
[sin2 x]k
k=0
k=0
converges to
1
1
=
= sec2 x.
2
cos2 x
1 − sin x
12
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
2
Problem 8. Let f (x) = ex . Compute
f (2014) (0),
where f (2014) (x) is the 2014th derivative of f (x).
Hint: Taylor (The good kind of Taylor, not the Swift kind · · · )
Solution: Using Taylor’s formula, we simply have to expand
f (x) = a0 + a1 x + a2 x2 + · · ·
and use the relation
an =
f (n) (0)
n!
(with n = 2014).
To this end, we first recall the expansion for eu :
u2 u3 u4
+
+
+ ···
eu = 1 + u +
2!
3!
4!
We then substitute u = x2 to deduce that
x4 x6 x8
2
e x = 1 + x2 +
+
+
+ ···
2!
3!
4!
1
From this series expansion, it follows that a2014 = 1007!
and hence
2014!
1007!
= 2014 × 2013 × 2012 × · · · × 1009 × 1008.
f (2014) (0) = 2014! × a2014 =
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
13
Problem 9. Consider the following curve in the plane, described in polar form:
r = ln θ.
Find the area of the region in the second quadrant (x ≤ 0, y ≥ 0) that is trapped between the
x−axis, the y−axis, and the curve.
Solution: We first note that the second quadrant corresponds to
standard formula for the area under a curve in polar coordinates:
Z
Z
1 θ2 2
1 π
Area =
r dθ =
(ln θ)2 dθ.
2 θ1
2 π/2
π
2
≤ θ ≤ π. We then recall the
To evaluate the integral, we first use the integration by parts relations u = ln θ, du =
dv = ln θ dθ, v = θ ln θ − θ to deduce that
Z
Z
Z
2
(ln θ) dθ = u dv = uv − v du
Z
2
= θ(ln θ) − θ ln θ − (ln θ − 1) dθ
dθ
,
θ
= θ(ln θ)2 − 2θ ln θ + 2θ.
Inserting the factor of 21 and the bounds of integration, we conclude that the area of the region of
interest is
θ=π
Z
Z
1 π
1
2
2
(ln θ) dθ = u dv = θ(ln θ) − θ ln θ + θ
2 π/2
2
θ=π/2
h π i2 π π π
π
π
ln
+ ln
+ .
= (ln π)2 − π ln π −
2
4
2
2
2
2
14
MATH 18.01 - FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS
Problem 10. Compute
ex + cos2 x
.
x→∞
ex
lim
” type limit. If you try to use L’Hôpital’s rule repeatedly, then you
Solution: This is a “ ∞
∞
deduce that
ex + cos2 x
ex − 2 sin x cos x
lim
=
lim
x→∞
x→∞
ex
ex
x
e − sin(2x)
= lim
x→∞
ex
x
e − 2 cos(2x)
= lim
x→∞
ex
x
e + 4 sin(2x)
= lim
x→∞
ex
= ··· .
This will go on indefinitely and hence you will not solve the problem in this fashion.
To solve the problem, we avoid L’Hôpital’s rule altogether and instead multiply the top and
bottom of the original fraction by e−x . After performing some algebra, we see that we have to
compute
lim 1 + e−x cos2 x.
x→∞
This limit is 1 because e−x goes to 0 as x → ∞, while cos2 x is never larger than one.