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Sample Calculus Problems
Part 1: Single Variable Functions
1
Part 2: Multi-Variable Functions
58
Part 3: Sequences and Series
90
Part 4: Vector Analysis
109
Last revision: September 10, 2014
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Part 1: Single Variable Functions
1.
Evaluate the limit
Solution:
Remark:
√
4 − 3x + 1
lim 2
x→5 x − 7x + 10
. (Do not use L'Hôpital's Rule.)
√
√
√
4 − 3x + 1
(4 − 3x + 1)(4 + 3x + 1)
√
lim 2
= lim
x→5 x − 7x + 10
x→5 (x2 − 7x + 10)(4 +
3x + 1)
16 − (3x + 1)
√
= lim
x→5 (x2 − 7x + 10)(4 +
3x + 1)
15 − 3x
√
= lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
3(5 − x)
√
= lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
−3
√
= lim
x→5 (x − 2)(4 +
3x + 1)
−3
√
=
(5 − 2)(4 + 3 ⋅ 5 + 1)
−3
=
3⋅8
1
=−
8
= is the most frequently used verb in mathematics. It was introduced in 1557 by
Robert Recorde to avoid the tedious repetition of the words `is equal to'. It is important to
use the equal sign correctly.
ˆ
To introduce , the phantom equal sign, to avoid the tedious repetition of the symbol
= is not a good idea. The solution above should not go like:
√
4 − 3x + 1
lim 2
x→5 x − 7x + 10
√
√
(4 − 3x + 1)(4 + 3x + 1)
√
lim
x→5 (x2 − 7x + 10)(4 +
3x + 1)
15 − 3x
√
lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
3(5 − x)
√
lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
1
ˆ
One must also not use other symbols, which have completely dierent meanings, in place
of =. The solution above should not go like:
√
√
√
4 − 3x + 1
(4 − 3x + 1)(4 + 3x + 1)
√
lim 2
⇒ lim
x→5 x − 7x + 10
x→5 (x2 − 7x + 10)(4 +
3x + 1)
15 − 3x
√
→ lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
3(5 − x)
√
lim
x→5 (x − 5)(x − 2)(4 +
3x + 1)
ˆ
The equal sign always stands between two things, although sometimes one of these things
are at the end of the previous line or at the beginning of the next line. The solution above
should not start like:
This begs the question: What is
ˆ
√
√
(4 − 3x + 1)(4 + 3x + 1)
√
= lim
x→5 (x2 − 7x + 10)(4 +
3x + 1)
√
√
(4 − 3x + 1)(4 + 3x + 1)
√
equal to lim
?
x→5 (x2 − 7x + 10)(4 +
3x + 1)
The equal sign can be used between two functions when we deal with identities , like
x2 − 1
=x+1
x−1
for all
x =/ 1
or when we deal with equations , like
Find all
x
such that
x2 = 4.
Therefore we can not just drop some of the limit signs in the solution above to make it
look like:
√
√
√
4 − 3x + 1 (4 − 3x + 1)(4 + 3x + 1)
√
lim 2
=
x→5 x − 7x + 10
(x2 − 7x + 10)(4 + 3x + 1)
⋮
−3
√
=
(x − 2)(4 + 3x + 1)
−3
√
=
(5 − 2)(4 + 3 ⋅ 5 + 1)
−3
=
3⋅8
1
=−
8
The equalities on the lines marked with
not equal to
−
1
8
7
are not correct.
because, for instance, if we let
−3
1
1
√
= =/ − .
8
(1 − 2)(4 + 3 ⋅ 1 + 1) 2
2
x = 1
then
7
7
−3
√
is
(x − 2)(4 + 3x + 1)
−3
√
=
(x − 2)(4 + 3x + 1)
2.
Let
m
Express
m
be the slope of the tangent line to the graph of
as a limit. (Do not compute
Solution:
(x0 , f (x0 ))
The slope
m
x2
x+2
y =
m.)
at the point
y = f (x)
of the tangent line to the graph of
(−3, −9).
at the point
is given by the limit
f (x) − f (x0 )
x→x0
x − x0
m = lim
or equivalently by the limit
f (x0 + h) − f (x0 )
.
h→0
h
m = lim
Therefore two possible answers are
x2
(−3 + h)2
− (−9)
− (−9)
x
+
2
−3
+
h
+
2
m = lim
= lim
.
x→−3
h→0
x − (−3)
h
3.
Suppose that
Solution:
lim f (x) =/ 0
and
lim g(x) = 0.
Assume that
lim
f (x)
g(x)
x→c
x→c
x→c
Show that
exists, and let
f (x)
x→c g(x)
lim
L = lim
x→c
f (x)
.
g(x)
does not exist.
Then by the product
rule for limits we obtain
lim f (x) = lim (
x→c
x→c
f (x)
f (x)
⋅ g(x)) = lim
⋅ lim g(x) = L ⋅ 0 = 0 .
x→c
g(x)
g(x) x→c
This contradicts the fact that
true:
4.
lim
x→c
f (x)
g(x)
lim f (x) =/ 0.
x→c
Therefore our assumption cannot be
does not exist.
lim f (x) = 0 and there exists a constant K such that ∣g(x)∣ ≤ K
x→c
interval containing c. Show that lim (f (x)g(x)) = 0.
Suppose that
some open
Solution:
for all
x =/ c
x→c
We have
∣f (x)g(x)∣ = ∣f (x)∣ ⋅ ∣g(x)∣ ≤ ∣f (x)∣ ⋅ K
in some open interval
around c. Therefore
−K∣f (x)∣ ≤ f (x)g(x) ≤ K∣f (x)∣ .
Now applying the Sandwich Theorem and using the fact that
the result.
3
lim ∣f (x)∣ = 0 we obtain
x→c
in
5.
Determine the following limits if
a.
x→0−
d.
x→0−
lim f (x) = A
x→0+
lim f (x2 − x)
b.
x→0−
lim (f (x3 ) − f (x))
e.
x→1−
and
lim f (x) = B .
x→0−
c.
lim (f (x2 ) − f (x))
lim f (x3 − x)
x→0+
lim f (x2 − x)
Solution: a.
If x < 0, then x2 > 0 and −x > 0. Therefore x2 − x > 0 for x < 0, and
2
x −x approaches 0 from the right as x approaches 0 from the left. lim− f (x2 − x) = A.
x→0
b.
Since
x2 > 0
the left. Hence
c.
For
as
x
d.
Since
left.
e.
x
6.
Let
(x − 1)2
x < 0, x2 approaches 0 from the right as x approaches
lim− (f (x2 ) − f (x)) = lim− f (x2 ) − lim− f (x) = A − B .
for
x→0
0 < x < 1,
x→0
we have
x3 < x
x→0
x3 − x < 0. So x3 − x approaches
3
Therefore lim f (x − x) = B .
+
and
approaches 0 from the right.
0 from the left
x→0
x3 < 0 for x < 0, x3 approaches 0 from the left as x approaches
3
3
Hence lim (f (x ) − f (x)) = lim f (x ) − lim f (x) = B − B = 0.
−
−
−
For
x→0
0<x<1
x→0
we have
x2 < x
approaches 1 from the left.
Q be the point
+ y 2 = 1 and the
0 from
0 from the
x→0
x2 − x < 0. x2 − x approaches
2
Hence lim f (x − x) = B .
−
and
0 from the left as
x→1
of intersection in the rst quadrant of the circle C1 with equation
circle C2 with equation x2 + y 2 = r 2 . Let R be the point where the line
passing through the points
as r → 0+ .
P (0, r)
and
Q
intersects the
4
x-axis.
Determine what happens to
R
Solution:
x2 + y 2 = r2 from (x − 1)2 + y 2√= 1 we obtain x = r2 /2,
in x2 + y 2 = r 2 gives us Q(r 2 /2,
r2 − r4 /4).
Subtracting
substituting this back
and
R(a, 0) be the coordinates of R and let S be the foot of the perpendicular from
Q to the x-axis. Since the triangles RSQ and ROP are similar we have
Let
a − r2 /2
a
√
=
r2 − r4 /4 r
and hence
a=
r3 /2
√
.
r − r2 − r4 /4
Then
lim+ a = lim+
r→0
r→0
r3 /2
√
r − r2 − r4 /4
√
r3 /2
⋅
(r
+
r2 − r4 /4))
r→0
r2 − (r2 − r4 /4)
√
= 2 lim+ (1 + 1 − r2 /4)
r→0
√
= 2 ⋅ (1 + 1 − 02 /4) = 4 .
= lim+ (
Therefore
7.*
R
approaches the point
(4, 0)
as
r → 0+ .
Use the formal denition of the limit to show that
Solution:
Given
ε>0
we want to nd
δ>0
lim
x→1/2
1
= 2.
x
such that
1
1
0 < ∣x − ∣ < δ Ô⇒ ∣ − 2∣ < ε .
2
x
(⋆)
We will do this in two dierent ways.
Solve the Inequality Method : First we solve
1
∣ − 2∣ < ε
x
for
x.
1
1
∣ − 2∣ < ε ⇐⇒ 2 − ε < < 2 + ε
x
x
The next step depends on whether
If
2 − ε > 0,
that is if
ε < 2,
2−ε
is positive, zero or negative.
then
2−ε<
1
1
1
< 2 + ε ⇐⇒
>x>
.
x
2−ε
2+ε
*Examples marked red are not part of the Fall 2014 syllabus.
5
If
2 − ε = 0,
that is if
ε = 2,
then
2−ε<
If
2 − ε < 0,
that is if
ε > 2,
2−ε<
1
1
< 2 + ε ⇐⇒ x >
.
x
2+ε
then
1
1
< 2 + ε ⇐⇒ x >
x
2+ε
6
or
−
1
>x.
ε−2
Next we choose
δ
in such a way that every
lies in the solution set of
1
∣ − 2∣ < ε,
x
x
and therefore the implication
all three cases choosing a delta such that
1
0 < ∣x − ∣ < δ
2
in (⋆) holds. In
satisfying the condition
0 < δ ≤ 1/2 − 1/(2 + ε) = ε/(4 + 2ε)
achieves
this.
The Estimation Method : Suppose
0 < ∣x − 1/2∣ < δ
for some
δ > 0.
Then
1
2∣x − 1/2∣ 2δ
∣ − 2∣ =
<
.
x
∣x∣
∣x∣
δ to satisfy δ ≤ 1/4. (Why 1/4?) Then
0 < ∣x − 1/2∣ < δ Ô⇒ 1/2 − δ < x < 1/2 + δ Ô⇒ 1/4 < x < 3/4 Ô⇒ 4 > 1/x > 4/3 Ô⇒
1/∣x∣ < 4 and therefore
1
2δ
∣ − 2∣ <
< 8δ .
x
∣x∣
At this point let us also decide to choose
ε > 0 if we choose δ to satisfy δ ≤ ε/8 (as well as δ ≤ 1/4) then
1
ε
we will have ∣ − 2∣ < 8δ ≤ 8 ⋅
= ε and (∗) will hold. In conclusion, any choice of δ
x
8
satisfying 0 < δ ≤ min{ε/8, 1/4} works.
Hence for a given
8.* Show that x→−3
lim (x4 + 7x − 17) = 43 using the formal denition of the limit.
Solution:
For any given
ε>0
we have to nd a
δ>0
so that for all
0 < ∣x − (−3)∣ < δ Ô⇒ ∣x4 + 7x − 17 − 43∣ < ε .
*Examples marked red are not part of the Fall 2014 syllabus.
7
x
we have
(x4 + 7x − 17) − 43 = x4 + 7x − 60 = (x + 3)(x3 − 3x2 + 9x − 20). Suppose
that 0 < ∣x − (−3)∣ < δ and δ ≤ 1. Then −4 ≤ −3 − δ < x < −3 + δ ≤ −2. In particular,
∣x∣ < 4. Therefore, using the Triangle Inequality, we obtain ∣x3 − 3x2 + 9x − 20∣ ≤
∣x∣3 + 3∣x∣2 + 9∣x∣ + 20 < 43 + 3 ⋅ 42 + 9 ⋅ 4 + 20 = 168. Now if we choose δ to satisfy
0 < δ ≤ min{ε/168, 1}, then we have
ε
∣x4 + 7x − 17 − 43∣ = ∣x4 + 7x − 60∣ = ∣x + 3∣ ⋅ ∣x3 − 3x2 + 9x − 20∣ < δ ⋅ 168 ≤
⋅ 168 = ε
168
whenever 0 < ∣x − (−3)∣ < δ . We are done.
We have
9.* Suppose that for all 0 < ε < 1,
∣x − 1∣ <
ε2
Ô⇒ ∣f (x) − 3∣ < ε
4
∣x − 1∣ <
ε
Ô⇒ ∣g(x) − 4∣ < ε .
35
and
Find a real number
δ>0
If we take
1
10
ε=
in
ε=
1
10
1
1
Ô⇒ ∣f (x) − 3∣ <
.
400
10
(⋆⋆) we get
in
∣x − 1∣ <
Therefore if
1
.
5
(⋆) we get
∣x − 1∣ <
If we take
(⋆⋆)
such that
∣x − 1∣ < δ Ô⇒ ∣f (x) + g(x) − 7∣ <
Solution:
(⋆)
1
,
400
∣x − 1∣ <
1
1
Ô⇒ ∣g(x) − 4∣ <
.
350
10
then we have
∣f (x) + g(x) − 7∣ = ∣f (x) − 3 + g(x) − 4∣ ≤ ∣f (x) − 3∣ + ∣g(x) − 4∣ <
by the Triangle Inequality. Hence we can take
Remark:
10.*
Let
In fact, any
⎧
⎪
1
⎪
⎪
⎪
⎪
f (x) = ⎨
⎪
⎪
⎪
⎪
⎪
⎩0
a.
Show that if
b.
Show that
√
9−4 2
δ≤
1225
if
x=
1
n
δ=
works.
where
n
is a positive integer,
otherwise.
c =/ 0
then
lim f (x)
x→0
1
.
400
lim f (x) = 0.
x→c
does not exist.
*Examples marked red are not part of the Fall 2014 syllabus.
8
1
1 1
+
=
10 10 5
Solution: a.
the closest to
n
Assume
c
c > 0.
Then there is a positive integer
among all real numbers dierent from
is a positive integer. (Why?) Let
0 < ∣x − c∣ < δ Ô⇒ x =/
1
n
δ = ∣c −
1
∣ > 0.
m
c
m
Then for any
for any positive integer
1/m
such that
and of the form
ε > 0,
1/n
is
where
we have
n
Ô⇒ f (x) = 0 Ô⇒ ∣f (x) − 0∣ = ∣0 − 0∣ = 0 < ε .
Therefore
Assume
lim f (x) = 0.
x→c
c < 0.
Take
δ = ∣c∣.
Then for any
ε > 0,
we have
1
for any positive integer n
n
Ô⇒ f (x) = 0 Ô⇒ ∣f (x) − 0∣ = ∣0 − 0∣ = 0 < ε .
0 < ∣x − c∣ < δ Ô⇒ x < 0 Ô⇒ x =/
Therefore
lim f (x) = 0.
b.
be a real number and assume that
Let
L
x→c
there exists a
δ>0
such that for all
lim f (x) = L.
Then for every
ε > 0,
∣L∣ = ∣0 − L∣ = ∣f (x) − L∣ < ∣L∣/2
is not
x→0
x,
0 < ∣x − 0∣ < δ Ô⇒ ∣f (x) − L∣ < ε .
If
L
is not 0, let
ε = ∣L∣/2 > 0.
Then there is a
δ>0
such that
0 < ∣x∣ < δ Ô⇒ ∣f (x) − L∣ < ∣L∣/2 .
Take
x = −δ/2.
Then
0 < ∣x∣ < δ
is true, but
true. We have a contradiction.
On the other hand, if
L = 0,
let
ε = 1/2.
Then there is a
δ>0
such that
0 < ∣x∣ < δ Ô⇒ ∣f (x)∣ < 1/2 .
If
n
but
is a positive integer satisfying
1 = ∣1∣ = ∣f (x)∣ < 1/2
Hence
11.
lim f (x)
x→0
n > 1/δ ,
take
x = 1/n.
Then
0 < ∣x∣ < δ
is true,
is not true. Again we have a contradiction.
cannot exist.
Show that the equation
x2 − 10 = x sin x
has a real solution.
Solution:
f (10) =
Consider the function f (x) = x2 − 10 − x sin x. Then f (0) = −10 < 0 and
2
10 − 10 − 10 sin(10) = 90 − 10 sin(10) ≥ 90 − 10 = 80 > 0. Note that f is
continuous on
the function
c
in
( 0, 10)
f
[ 0, 10].
Therefore we can apply the Intermediate Value Theorem to
[ 0, 10] for the value 0 to conclude that there is a point
f (c) = 0. This c is also a solution of the given equation.
on the interval
such that
9
12.
Consider the equation
1−
x2
= cos x .
4
a.
Show that this equation has at least one real solution.
b.
Show that this equation has at least two real solutions.
c.
Show that this equation has at least three real solutions.
Solution:
Let
f (x) = 1 −
correspond to the zeros of
As
f (0) = 0, x = 0
Now observe that
x2
− cos x.
4
f.
The solutions of the equation
1−
x2
= cos x
4
is one zero.
f (π/2) = 1 − π 2 /16 > 0
f (π) = 2 − π 2 /4 < 0
and
as
4 > π > 3.
As
f
is continuous on the entire real line, applying the Intermediate Value Theorem to
the function
f
[π/2, π]
on the interval
interval such that
f (c) = 0.
Finally, as the function
f
we conclude that there is a point
c
in this
This is our second zero.
is even, we have
f (−c) = f (c) = 0,
and
x = −c
is our third
zero.
13.
Show that at any moment there are two antipodal points on the equator of the Earth with
the same temperature.
Solution:
First we will make a mathematical model of the problem.
We will
θ. We
θ in radians,
= 7π/4, and
consider the equator as a circle, and use the longitude as our coordinate
choose the positive direction for
and let it take any real value.
θ = 15π/4,
among other values.
the point corresponding to
T (θ + 2π) = T (θ)
for all
θ.
θ.
want to show that there is a
θ
to correspond to the East, measure
So 45○ W corresponds to θ = −π/4, θ
Note that
We let
T (θ)
θ+π
T is a continuous
T (c + π) = T (c).
We will assume that
c
such that
corresponds to the antipode of
denote the temperature at
θ.
We have
function. We
f (θ) = T (θ + π) − T (θ). Note that since T is continuous, f is
continuous. Our quest to nd a c such that T (c + π) = T (c) is equivalent to nd a
c such that f (c) = 0. If f (0) = 0, then we let c = 0 and we are done. Suppose that
f (0) =/ 0. Observe that f (0) = T (π) − T (0) = T (π) − T (2π) = −f (π). In other words,
f (0) and f (π) have opposite signs. Now we apply the Intermediate Value Theorem
to f on the interval [0, π] for the value 0, and conclude that there is a point c in
[0, π] such that f (c) = 0. We are done.
Consider the function
Remark:
It is possible to show that at any moment there are two antipodal points on Earth
with the same temperature and the same pressure.
10
14.
Find all tangent lines to the graph of
y = x3
that pass through the point
(2, 4).
Solution:
As dy/dx = d(x3 )/dx = 3x2 , the equation of the tangent line through
3
3
2
a point (x0 , x0 ) on the graph is y − x0 = 3x0 (x − x0 ). This line passes through
(2, 4) exactly when 4 − x30 = 3x20 (2 − x0 ), or in other words, x30 − 3x20 + 2 = 0. We
x0 = 1 is a root of this polynomial. Therefore we have the factorization
√
(x0 − 1)(x20 − 2x0 − 2). The roots of the quadratic factor are x0 = 1 ± 3.
√
√
Therefore the tangent lines to y = x3 at the points (1, 1), (1 +
3,
10
+
6
3), and
√
√
(1 − 3, 10√− 6 3) pass through
(2,
4)
. The equations of these lines are y = 3x − 2,
√
√
√
y = (12 + 6 3)x − (20 + 12 3), and y = (12 − 6 3)x − (20 − 12 3), respectively.
observe that
x30 − 3x20 + 2 =
√
15.
Evaluate the limit
Solution:
√
lim
x→0
lim
x→0
1 + sin2 x2 − cos3 x2
.
x3 tan x
√
⎛⎛ 1 + sin2 x2 − 1 1 − cos3 x2 ⎞ x ⎞
1 + sin2 x2 − cos3 x2
+
⋅
= lim
x→0 ⎝⎝
x3 tan x
x4
x4
⎠ tan x ⎠
sin x2 2
1
x
1 − cos3 x2
= lim ((( 2 ) ⋅ √
)⋅
+
)
4
2
x→0
x
x
tan x
1 + sin x2 + 1
Now we observe that:
sin x2
=1
x→0 x2
1
1
1
lim √
=√
=
2 2
2 2
x→0
1 + sin x + 1
1 + sin 0 + 1 2
x
lim
=1
x→0 tan x
lim
11
and
1 − cos3 x2
1 − cos x2
=
lim
(
⋅ (1 + cos x2 + cos2 x2 ))
x→0
x→0
x4
x4
2 sin2 (x2 /2)
⋅ (1 + cos x2 + cos2 x2 ))
= lim (
x→0
x4
lim
2
1
sin(x2 /2)
= ⋅ (lim
) ⋅ lim(1 + cos x2 + cos2 x2 )
x→0
2 x→0 x2 /2
1
= ⋅ 12 ⋅ 3
2
3
= .
2
√
1 + sin2 x2 − cos3 x2
1 3
lim
= (1 ⋅ + ) ⋅ 1 = 2 .
3
x→0
x tan x
2 2
Therefore:
16.
Find the equation of the tangent line to the graph of
x = 1.
y = sin2 (πx3 /6)
at the point with
dy
= 2 sin(πx3 /6) ⋅ cos(πx3 /6) ⋅ 3πx2 /6 .
dx √
3π
dy
∣ = 2 sin(π/6) ⋅ cos(π/6) ⋅ π/2 =
.
dx x=1
4
Solution:
Since
y = sin2 (πx3 /6) ⇒
y∣x=1 = 1/4,
using the point-slope formula we nd the equation of the tangent
line as
1
y− =
4
√
3π
(x − 1)
4
or, after some reorganization,
√
y=
17.
Let
√
3π
1 − 3π
x+
.
4
4
⎧
⎪
1
⎪
⎪
2x + x2 sin( )
⎪
⎪
⎪
x
⎪
f (x) = ⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩0
a.
Find
b.
Show that
f ′ (x)
for all
f′
Therefore,
x.
is not continuous at
0.
12
if
x =/ 0,
if
x = 0.
Solution: a.
For
f ′ (x) =
for
x =/ 0.
For
x=0
x =/ 0 we compute the derivative using the rules of dierentiation:
d
(2x + x2 sin(1/x)) = 2 + 2x sin(1/x) + x2 cos(1/x) ⋅ (−1/x2 )
dx
we must use the denition of the derivative:
f (0 + h) − f (0)
2h + h2 sin(1/h)
= lim
h→0
h→0
h
h
1
= lim 2 + lim h sin( ) = 2 + 0 = 2
h→0
h→0
h
f ′ (0) = lim
lim h sin(1/h) = 0
whose proof uses the Sandwich (or
h→0
Squeeze) Theorem. Here is a recap of the proof: Since ∣ sin(1/h)∣ ≤ 1 for all h =
/0
Here we used the fact that
we have
∣h sin(1/h)∣ = ∣h∣ ⋅ ∣ sin(1/h)∣ ≤ ∣h∣
for all
1
−∣h∣ ≤ h sin( ) ≤ ∣h∣
h
As
lim ∣h∣ = 0 = lim(−∣h∣),
h→0
lim h sin(1/h) =
h→0
h→0
0.
To summarize:
b.
it
follows
h =/ 0.
for all
by
the
⎧
⎪
1
1
⎪
⎪
2 + 2x sin( ) − cos( )
⎪
⎪
⎪
x
x
⎪
f ′ (x) = ⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩2
lim f ′ (x).
h =/ 0 .
Sandwich
if
x =/ 0,
if
x = 0.
Theorem
that
(a)
and lim 2x sin(1/x) = 0 as in part
.
x→0
x→0
′
However lim cos(1/x) does not exist. It follows that lim f (x) does not exist and
x→0
x→0
hence f ′ is not continuous at 0.
Consider
x→0
We have
lim 2 = 2
Therefore
13
18.
Find
d2 y
∣
if y is a dierentiable function of x satisfying the equation x3 +2y 3 = 5xy .
dx2 (x,y)=(2,1)
Solution:
x3 + 2y 3 = 5xy
Ô⇒
d/dx
3x2 + 6y 2
dy
dy
= 5y + 5x
dx
dx
(⋆)
Ô⇒
x = 2, y = 1
12 + 6
Ô⇒
dy
dy
= 5 + 10
dx
dx
4
Ô⇒
dy
=7
dx
dy 7
=
dx 4
at
(x, y) = (2, 1)
Now we dierentiate the equation marked
14
(⋆)
with respect to
x
to nd the second
derivative.
3x2 + 6y 2
Ô⇒
dy
dy
= 5y + 5x
dx
dx
d/dx
2
dy
d2 y
dy
dy
d2 y
6x + 12y ( ) + 6y 2 2 = 5
+5
+ 5x 2
dx
dx
dx
dx
dx
Ô⇒
x = 2, y = 1,
dy 7
=
dx 4
2
d2 y
7
d2 y
7
12 + 12 ( ) + 6 2 = 10 ⋅ + 10 2
4
dx
4
dx
Ô⇒
d2 y 125
=
dx2
16
Remark:
An alternative approach is to solve
dierentiate this with respect to
19.
x
y′
at
from
(x, y) = (2, 1)
(⋆),
viz.
y′ =
y ′′ .
to nd
5y − 3x2
,
6y 2 − 5x
and then
A piston is connected by a rod of length 14 cm to a crankshaft at a point 5 cm away from
the axis of rotation of the crankshaft. Determine how fast the crankshaft is rotating when the
piston is 11 cm away from the axis of rotation and is moving toward it at a speed of 1200
cm/sec.
Solution:
Let
P (x, y)
and
Q(a, 0)
be the ends of the connecting rod as shown in
the picture. The axis of rotation of the crankshaft passes through the origin of the
xy -plane
and is perpendicular to it.
The point
P
where the rod is connected to
crankshaft moves on a circle with radius 5 cm and center at the origin. The point
Q where the rod is connected to the piston moves along the positive x-axis. θ
angle between the ray OP and the positive x-axis.
y
P (x, y)
14 c
m
θ
−5
5
Q(a, 0)
x
x2 + y 2 = 25
The question is:
a = 11 cm
and
da
dθ
= −1200 cm/sec Ô⇒
=?
dt
dt
15
is the
We have
x2 + y 2 = 52
(I)
and
(x − a)2 + y 2 = 142 .
(II)
At the moment in question a = 11 cm. Substituting this in (I) and (II) we obtain
x2 + y 2 = 52 and (x − 11)2 + y 2 = 142 . Subtracting the second equation from the
√ rst
2
2
2
gives 22x−11 = 5 −14 , and solving for x we get x = −25/11 cm. Then y = 20 6/11
cm.
Dierentiating (I) and (II) with respect to time
x
and
(x − a) ⋅ (
At the moment in question
√
y = 20 6/11
a = 11
t
we obtain
dx
dy
+y
=0
dt
dt
(III)
dx da
dy
− )+y
=0.
dt dt
dt
(IV)
cm,
da
= v = −1200
dt
cm/sec,
x = −25/11
cm and
cm. Substituting these in (III) and (IV) we get
−5
√ dy
dx
+4 6
=0
dt
dt
(V )
and
√ dy
dx
+ 20 6
= −146v .
(VI)
dt
dt
dx
dx 146
Subtracting 5 times (V) from (VI) we nd −121
= −146 v , and hence
=
v.
dt
dt 121
dy
365
√ v.
Substituting this back in (V) gives
=
dt 242 6
−146
Now we are ready to compute
dθ
.
dt
Since
tan θ = y/x,
dierentiation gives
dy
dx
x −y
dθ
sec2 θ
= dt 2 dt
dt
x
and using
sec2 θ = 1 + tan2 θ = 1 + (y/x)2
we obtain
dy
dx
dθ x dt − y dt
.
=
dt
x2 + y 2
√
25
20 6
dx 146
Plugging x = −
cm, y =
cm,
=
v,
11
11
dt 121
√
dθ
73
1460 6
√ v=
gives
=−
radian/sec.
dt
11
110 6
16
and
dy
365
√ v
=
dt 242 6
in this formula
Remark:
Remark:
2 ⋅ 5 x cos θ
√
1460 6
11
radian/sec is
√
1460 6 60
⋅
11
2π
rpm or approximately 3105 rpm.
This problem has a shorter solution if we use the law of cosines. Start with
= 142 and dierentiate with respect to t to obtain
x2 + 52 −
dθ 5 cos θ − 11
=
v.
dt
5 sin θ
Put
x = 11
cm in the rst equation to nd
cos θ = −5/11
and then
√
sin θ = 4 6/11.
Now
substituting these in the second equation gives the answer.
20.
Determine how fast the length of an edge of a cube is changing at the moment when the
length of the edge is 5 cm and the volume of the cube is decreasing at a rate of 100 cm3/sec.
Solution:
a denote the length of an edge of the cube, and V denote the volume
of the cube. Then we have V = a3 . Dierentiating with respect to time t gives
dV
da
dV
= 3a2
. Substituting
= −100 cm3/sec and a = 5 cm for the moment in
dt
dt
dt
da
4
question, we obtain
= − cm/sec. Therefore the length of the edge is decreasing
dt
3
4
at a rate of
cm/sec at that moment.
3
21.
Let
We measure the radius and the height of a cone with 1% and 2% errors, respectively. We
use these data to compute the volume of the cone. Estimate the percentage error in volume.
Solution:
r , h,
Let
and
V
be the radius, the height and the volume of the cone,
respectively.
V =
π 2
2π
π
dV
dr dh
r h Ô⇒ dV =
rhdr + r2 dh Ô⇒
=2
+
.
3
3
3
V
r
h
Since the error in
r
is 1% we have
∣
dr
∣ ≤ 1% .
r
Similarly
∣
dh
∣ ≤ 2% .
h
triangle inequality we obtain
∣
dV
dr dh
dr
dh
∣ = ∣2
+ ∣ ≤ 2 ∣ ∣ + ∣ ∣ ≤ 2 ⋅ 1% + 2% = 4% .
V
r
h
r
h
The error in volume is 4% .
17
Now using the
22.
A cone of radius 2 cm and height 5 cm is lowered point rst into a tall cylinder of radius
7 cm that is partially lled with water. Determine how fast the depth of the water is changing
at the moment when the cone is completely submerged if the cone is moving with a speed of 3
cm/s at that moment.
Solution:
Let
r
and
h
be the radius and the height of the part of the cone that is
under the water level, respectively. Let
and let
y
L
be the depth of the water in the cylinder
be the vertical distance from the tip of the cone to the bottom of the
cylinder. Let
V0
be the volume of the water.
Then
2
π
π 2
4π
V0 = π ⋅ 7 ⋅ L − r2 h = π ⋅ 72 ⋅ L − ( h) h = 49πL − h3
3
3 5
75
2
where we used the fact that
r/h = 2/5.
Now dierentiating this with respect to time
0=
t
gives
d
dL 4π 2 dh
V0 = 49π
− h
.
dt
dt 25 dt
In particular at the moment when the cone is completely submerged we have
cm and
49
h=5
dL
dh
=4
.
dt
dt
On the other hand, at the same moment
h = L − y Ô⇒
because
dy/dt = −3
dh dL
dh dL dy
=
−
Ô⇒
=
+3
dt dt dt
dt dt
cm/s as the cone is being lowered at a speed of 3 cm/s.
From these two equations we obtain
dL/dt = 4/15
cm/s. In other words, the depth
of the water is increasing at a rate of 4/15 cm/s at that moment.
18
23.
Find the absolute maximum value and the absolute minimum value of
on the interval
[−1, 6].
Remark: How to nd the absolute maximum and the absolute
continuous function f on a closed interval [ a, b] of nite length:
i. Compute
f (x) = x4/3 − x − x1/3
minimum values of a
f ′.
ii. Find the critical points of
iii. Add the endpoints
a
and
iv. Compute the value of
f
f
b
in
(a, b).
to this list.
at each point in the list.
v. The largest value is the the absolute maximum and the smallest value is the absolute
minimum of
Solution:
x=0
f
on
[a, b].
4
1
f ′ (x) = x1/3 −1− x−2/3 .
3
3
The derivative is not dened at
is a critical point. Next we solve f ′ (x)
x = 0, therefore
= 0.
4 1/3
1
x −1− x−2/3 = 0 we let z = x1/3 to obtain the equation 4z 3 −3z 2 −
3
3
1 = 0. Since z = 1 is a root, we have the factorization 4z 3 −3z 2 −1 = (z −1)(4z 2 +z +1).
As the quadratic factor has no real roots, z = 1 is the only solution. Therefore
x = z 3 = 13 = 1, which belongs to the interval [−1, 6], is the only other critical point.
In the equation
We have f (0) = 0, f (1) = −1, f (−1) = 3, and f (6) = 5
5 ⋅ 61/3 − 6 > 3 ⇐⇒ 5 ⋅ 61/3 > 9 ⇐⇒ 53 ⋅ 6 > 93 ⇐⇒ 725 > 721.
⋅ 61/3 − 6.
We conclude that the absolute maximum and minimum values of
on the interval [−1, 6] are 5 ⋅ 61/3 − 6 and −1, respectively.
24.
f (x) = x4/3 −x−x1/3
Find the absolute maximum and the absolute minimum values of
interval
f (x) =
[ 0, ∞).
Remark:
Observe that
x2
x+1
+x+9
on the
When looking for the absolute maximum and the absolute minimum values of a
continuous function on an interval that is not necessarily closed or of nite length, a modied
version of the algorithm above can be used.
ˆ
ˆ
In Step iv , if an endpoint does not belong to the interval, then we compute the appropriate
one-sided limit of the function at that point instead of the value of the function.
In Step v , if the largest value (which can be
∞)
occurs only as a limit, then we conclude
that there is no absolute maximum. Similarly for the smallest value.
Solution:
and
x = 2.
f (0) =
We compute
Only
x=2
1
1
, f (2) =
,
9
5
is in the
and
x2 + 2x − 8
. The roots of f ′ (x) = 0 are x = −4
(x2 + x + 9)2
interval [ 0, ∞). So our list is 0, 2, and ∞.
f ′ (x) = −
lim f (x) = 0.
x→∞
and there is no absolute minimum.
Since
19
1 1
> > 0,
5 9
the absolute maximum is
1
5
25.
Suppose that a function
f (x0 ) = 0
for some
Solution:
x0 > 0.
x0 > 0,
Since
f
f
f ′ (x) = f (x/2) for all x and f (0) = 1.
x1 such that 0 < x1 < x0 and f (x1 ) = 0.
satises
then there is
is dierentiable,
f
is continuous. Suppose that
Show that if
f (x0 ) = 0 for some
f on [ 0, x0 ] to conclude that there
f
(x
)
−
f
(0)
1
0
is c such that 0 < c < x0 and f ′ (c) =
= − < 0. Then f ′ (x) = f (x/2)
x0 − 0
x0
gives f (c/2) = f ′ (c) < 0. We also have f (0) = 1 > 0. We apply the Intermediate
Value Theorem to f on [ 0, c/2] to conclude that there is x1 such that 0 < x1 < c/2
and f (x1 ) = 0. As 0 < x1 < c/2 < c < x0 we are done.
Remark:
We apply the Mean Value Theorem to
With a little bit more work it can be shown that
f (x) > 1
26.
Show that if f is a twice-dierentiable function such that f (0)
f ′ (1) = 5, and f ′′ (x) ≥ 0 for all x, then f (x) ≥ 1/3 for all 0 ≤ x ≤ 1.
for all
x > 0.
= 1, f ′ (0) = −1, f (1) = 2,
Solution:
Let x1 < x2 be in [0, 1]. The Mean Value Theorem applied to the
′
function f on the interval [x1 , x2 ] says that there is a point c in (x1 , x2 ) such that
f ′ (x2 ) − f ′ (x1 )
= f ′′ (c). As f ′′ (c) ≥ 0 we have f ′ (x1 ) ≤ f ′ (x2 ). In particular, we
x2 − x1
have −1 = f ′ (0) ≤ f ′ (x) ≤ f ′ (1) = 5 for all x in (0, 1).
0 < x < 1.
Let
interval
[0, x], there exists a point c1
f ′ (c1 ) ≥ −1,
f on the
f (x) − f (0)
= f ′ (c1 ). Since
x−0
By the Mean Value Theorem applied to the function
in
(0, x) such that
it follows that
f (x) ≥ −x + 1
( Q)
0 < x < 1. Similarly, applying the Mean Value Theorem to the function
f on the interval [x, 1], we see that there exists a point c2 in (x, 1) such that
f (1) − f (x)
= f ′ (c2 ). Now using the fact that f ′ (c2 ) ≤ 5 we conclude that
1−x
for
f (x) ≥ 5x − 3
for
0 < x < 1.
5
[0, 1].
Adding
in
(QQ)
Remark:
times the inequality
(Q) to the inequality (QQ) we get f (x) ≥ 1/3 for all x
In fact, it can be shown that
f (x) > 1/3
20
for all
x
in
[0, 1].
Sketch the graph of
10 −1/3 10 2/3 10 −1/3
x
− x =
x (1 − x).
3
3
3
(−∞, 0) and (1, ∞).
Solution:
y′
<0
on
y = 5x2/3 − 2x5/3 .
y′ =
20
10 −4/3 20 −1/3
x
− x
= − x−4/3 (x + 1/2).
9
9
9
y ′′ < 0 on (−1/2, 0) and (0, ∞).
y ′′ = −
Therefore
Therefore
y′ > 0
y ′′ > 0
on
(0, 1),
and
(−∞, −1/2),
and
on
These give us the following table of signs and shapes.
x
y′
y ′′
∣
∣
∣
−1/2
∣
0∣
↑
−
+
"
27.
0
^
^
^
^
↑
−
−
inf.pt.
We compute the
minimum,
y -coordinates
1
+
−
"
0∣
∣
↑
loc.min.
−
−
loc.max.
of the important points to get
√
3
(0, 0)
for the local
(1, 3)√for the local maximum,
and (−1/2, 3 2) for the inection point.
√
3
2 > 1, we have 3 3 2 > 3. Also note that the function is continuous
Note that since
at
x = 0,
but
lim y ′ = lim+ (
x→0+
x→0
and
lim− y ′ = lim− (
x→0
Therefore
(0, 0)
or
x = 5/2.
10 −1/3
x (1 − x)) = −∞ .
3
is a cusp.
Finally we nd the
0
x→0
10 −1/3
x (1 − x)) = ∞
3
x-intercepts: y = 0 ⇒ 5x2/3 − 2x5/3 = 0 ⇒ 2x2/3 (5 − 2x) = 0 ⇒ x =
(The
graph is on the next page .)
21
Now we use these data to draw the graph:
Remark:
Here is the same graph drawn by Maple :
22
28.
Two corridors meet at a corner. One of the corridors is 2 m wide and the other one is 3
m wide. What is the length of the longest ladder that can be carried horizontally around this
corner?
Solution:
of
L
Length of the longest ladder will be equal to the absolute minimum value
in the picture.
y
2m
L
3m
x
Using the relation
3/x = y/2
we obtain
L = (x2 + 9)1/2 + (4 + y 2 )1/2 = (x2 + 9)1/2 + (4 + (6/x)2 )1/2 .
Therefore we have to minimize
2
L = (x2 + 9)1/2 ⋅ (1 + )
x
for
0<x<∞ .
We rst look at the critical points:
dL
2
2
= (x2 + 9)−1/2 ⋅ x ⋅ (1 + ) + (x2 + 9)1/2 ⋅ (− 2 ) = 0
dx
x
x
18
Ô⇒ x + 2 − (2 + 2 ) = 0 Ô⇒ x = 181/3 m Ô⇒ L = (22/3 + 32/3 )3/2
x
Since
m
lim L = ∞ and lim L = ∞, the value at the critical point is indeed the absolute
x→0+
minimum.
x→∞
Hence the length of the longest ladder that can be carried around this corner is
(22/3 + 32/3 )3/2 m.
Remark:
(22/3 + 32/3 )3/2
m is approximately
7.02
23
m.
29.
Find the maximum possible total surface area of a cylinder inscribed in a hemisphere of
radius 1.
Solution:
h=
Let S be the total surface area of the cylinder,
2
1/2
(1 − r ) . Hence we want to maximize
S = 2πr2 + 2πr(1 − r2 )1/2
for
S = 2πr2 + 2πrh.
We have
0 ≤ r ≤ 1.
1
h
r
1 dS
= 2r + (1 − r2 )1/2 − r2 (1 − r2 )−1/2 = 0 Ô⇒
2π dr
2r(1 − r2√
)1/2 = 2r2 − 1 Ô⇒ 4r2 (1 − r2 ) = (2r2 − 1)2 Ô⇒ 8r4 − 8r2 + 1 = 0 Ô⇒
8 ± 32
1
r2 =
. Note that 2r(1 − r 2 )1/2 = 2r 2 − 1 implies r 2 ≥
, and therefore
16
2
√ √
√
2+ 2
2+ 2
r2 =
and r =
.
4
2
√ √
√
2+ 2
We get S = (1 + 2)π at the critical point r =
, S = 0 at the endpoint r = 0,
2
√
√
2 > 1 we have 1 + 2 > 2, and the absolute
and S = 2π at the endpoint r = 1. As
First we nd the critical points:
maximum value occurs at the critical point.
The maximum possible surface area of the cylinder is
24
(1 +
√
2)π .
30.
A fold is formed on a 20 cm
×
30 cm rectangular sheet of paper running from the short
side to the long side by placing a corner over the long side. Find the minimum possible length
of the fold.
Solution:
ABCD be the sheet of paper and let P be the point on the edge
C is folded over. The fold runs from Q on the edge BC to R
on the edge CD . Let S be the projection of R on to the edge AB . Let L be the
length of the fold QR and let x = CQ.
AB
Let
where the corner
R
D
C
L
x
Q
x
A
S
B
P
P Q/P√
B = RP /RS by the similarity
of the triangles P BQ and RSP . Hence
√
2
2
RP = 20x/ x − (20 − x) = 20x/ 40x − 400 and L2 = RQ2 = RP 2 + P Q2 =
400x2 /(40x − 400) + x2 = x3 /(x − 10). The largest possible value of x is 20
cm. The smallest possible value of x occurs when RP = 30 cm; that is when
√
√
20x/ 40x − 400 = 30 or x = 45 − 15 5 cm.
Then
Therefore we want to minimize
Dierentiating
L
where
L2 = x3 /(x − 10)
2LdL/dx = 3x2 /(x − 10) − x3 /(x − 10)2
√x = 15 cm
L = 15 3 cm.
obtain
for
√
45 − 15 5 ≤ x ≤ 20.
dL/dx = 0 we
x = 15 cm we have
and setting
as the only critical point in the domain. For
cm we have
√
√
x =√20 cm we have L = 20 2 cm and at the endpoint x = 45 − 15 5
√
L = 15 18 − 6 5 cm.
√
15 3
cm is the smallest of these values, we conclude that this is the smallest
At the endpoint
Since
possible value for the length of the fold.
√
w ≤ 2 2ℓ/3, then the length
of the shortest possible fold is 3 3w/4. Therefore, for a A4 size paper with w = 210 mm and
ℓ = 297 mm, the shortest fold is approximately 273 mm; and for a letter size paper with w = 8.5
in and ℓ = 11 in, the shortest fold is approximately 11.04 in.
Remark:
If the question is posed for a
√
w×ℓ
sheet of paper with
25
31.
The Rubber Duck is a sculpture designed by Florentijn Hofman and constructed from
PVC. For the purposes of this question, we consider the Rubber Duck to consist of a spherical
head of radius
a
and a spherical body of radius
b.
The research shows that the cuteness
the Rubber Duck is given by
⎧
a
a
⎪
⎪
⎪
⎪
⎪ b (1 − b )(a + b)
K =⎨
⎪
⎪
⎪
⎪
⎪
⎩0
if
0 ≤ a < b,
if
0 < b ≤ a.
Find the dimensions of the cutest Rubber Duck with a total surface area of
Solution:
4πb2 =
The total surface area of the Rubber Duck is
400π m2 , giving b2 = 100 − a2 . Therefore,
⎧
a(100 − 2a2 )
⎪
⎪
⎪
⎪ 100 − a2
K =⎨
⎪
⎪
⎪0
⎪
⎩
if
4πa2 + 4πb2 .
400π m2 .
Hence
4πa2 +
√
0 ≤ a ≤ 5 2,
otherwise,
and we have to nd where the absolute maximum value of
K =
on the interval
√
0≤a≤5 2
100a − 2a3
100 − a2
occurs.
We rst nd the critical points. As
dK
(100 − 6a2 )(100 − a2 ) − (100a − 2a3 )(−2a)
,
=
da
(100 − a2 )2
dK /da = 0
gives
a4 − 250a2 + 5000 = 0. Now using the quadratic
√
√
250 ± 2502 − 4 ⋅ 5000
2
a =
= 125 ± 25 17 ,
2
26
formula we obtain
K
of
which gives us four solutions
√ √
√ √
√ √
√ √
a = 5 5 + 17, 5 5 − 17, −5 5 + 17, −5 5 − 17 .
√
5 √
2. On the
√
a = 5 5 − 17
Only the rst two of these are positive and the rst one is greater than
other hand,
√ √
√
0 < 5 5 − 17 < 5 2
as
3<
√
17 < 5.
We conclude that
is the only critical point.
We have
√
K = 10 5 −
for
√ √
a = 5 5 − 17.
As
K =0
√
√
17 − 3
17 ⋅ √
>0
17 − 1
at the endpoints
a=0
and
√
a=5 2
of the interval,
this is the absolute maximum value.
Therefore the cutest Rubber Duck has
32.
√ √
a = 5 5 − 17
m and
√√
b=5
17 − 1
m.
A dessert in the shape of a hemisphere with radius 1 dm is made by baking a cylindrical
cake of height
h,
and topping it with a spherical cap of ice cream and surrounding it with a
hemispherical ring of chocolate mousse as shown in the gure. If the cake costs 8/π /dm3 , the
ice cream costs 9/π /dm3 and the chocolate mousse costs 12/π /dm3 , determine the value of
h
for
(a) the least expensive and (b) the most expensive dessert that can be made.
You may use the fact that the volume of a hemispherical ring of height
h is 2πh3 /3.
Ice cream
(9/π
/dm3 )
Cake
(8/π
/dm3 )
Chocolate
mousse
1
Solution:
We have
Cost
=
12
⋅ (Volume
π
of the ring) +
8
⋅ (Volume
π
of the cylinder)
9
⋅ (Volume of the cap)
π
12 2π 3 8
9 2π
2π
=
⋅ h + ⋅ π(1 − h2 )h + ⋅ ( ⋅ 13 − h3 − π(1 − h2 )h)
π 3
π
π
3
3
+
= 6 − h + 3h3
27
(12/π
h
/dm3 )
and hence we want to nd the absolute maximum and the absolute minimum values
of
= 6 − h + 3h3
Cost
We rst nd the critical points. As
gives
h = 1/3
h = −1/3 dm.
the cost is 52/9
dm and
and at that point
for
d(Cost)/dh = −1 + 9h2 , setting this equal to zero
Only the rst of these are in the interval [0, 1],
.
Next we look at the endpoints of the interval.
dm gives Cost
=8
33.
h=1
h=0
dm gives Cost
=6
, and
h=1
.
Therefore the least expensive dessert has
has
0≤h≤1.
h = 1/3 dm, and the most expensive dessert
dm.
We want to build a greenhouse that has a half cylinder roof of radius
r
and height
r
mounted horizontally on top of four rectangular walls of height h as shown in the gure. We
have 200π m2 of plastic sheet to be used in the construction of this structure. Find the value
of
r
for the greenhouse with the largest possible volume we can build.
Solution:
We have
200π m2 = Total
Surface Area
1
1
= 2 ⋅ (r + 2r)h + 2 ⋅ πr2 + ⋅ 2πr ⋅ r
2
2
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¸
¹ ¹ ¹ ¹ ¹ ¹¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
walls
half-disks
and hence
h=
π 100
(
− r).
3 r
In particular,
0 < r ≤ 10.
Let
V
denote the volume of
the greenhouse. Then
1
V = r ⋅ 2r ⋅ h + ⋅ πr2 ⋅ r
´¹¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¶ 2
bottom
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
top
and substituting
h
in terms of
Maximize
r
in this, our problem becomes:
V =
π
(400r − r3 )
6
28
for
top
0 < r ≤ 10.
We rst nd the critical points in the interval
(0, 10].
d
π
20
V = (400 − 3r2 ) = 0 Ô⇒ r = ± √ m
dr
6
3
√
Since neither of these satises 0 < r ≤ 10 as
3 < 2, there are no critical points
our interval. Now we look at the endpoints. At r = 0 we have lim V = 0, and
r→0+
r = 10 we have V = 500π .
Therefore the maximum possible volume is
h=0
34.
Let
f
500π
m3 and occurs when
r = 10
in
at
m and
m.
be a continuous function.
x2
a.
Find
f (4)
if
b.
Find
f (4)
if
∫0
∫0
f (t) dt = x sin πx
f (x)
t2 dt = x sin πx
x.
for all
for all
x.
FTC1
2
x2
x
d
d
↓
Solution: a.
∫0 f (t) dt = x sin πx Ô⇒ dx ∫0 f (t) dt = dx (x sin πx) Ô⇒
f (x2 ) ⋅ 2x = sin πx + x ⋅ π cos πx. Now letting x = 2 we get f (4) = π/2.
f (x)
t3
f (x)3
b. ∫
t dt = x sin πx Ô⇒ ] = x sin πx Ô⇒
= x sin πx.
3 0
3
0
gives f (4) = 0.
f (x)
Remark:
2
One might ask if such functions exist. In part
(b), f (x) =
√
3
Hence
3x sin πx
x = 4
is the unique
function satisfying the given condition.
In part
(a),
anything for
√
√
sin π x π
√
f (x) =
+ cos π x for x > 0,
2
2 x
x < 0 so long as it is continuous.
29
and
f (0) = π
by continuity; but it can be
35.
Compute
d2 y
∣
dx2 (x,y)=(0,0)
if
y
x+y
∫0
Solution:
x
satisfying the equation:
x
we obtain:
is a dierentiable function of
e−t dt = xy
2
Dierentiating the equation with respect to
x+y
∫0
e−t dt = xy
2
Ô⇒
d/dx
x+y
d
d
2
e−t dt =
(xy)
∫
dx 0
dx
Ô⇒
FTC1
e−(x+y) ⋅
2
Ô⇒
d
d
(x + y) =
(xy)
dx
dx
e−(x+y) (1 +
2
dy
dy
)=y+x
dx
dx
(e)
Ô⇒
x = 0, y = 0
dy
= −1
dx
Now dierentiating (e) with respect to
x
at
(x, y) = (0, 0)
again we get:
e−(x+y) (1 +
2
dy
dy
)=y+x
dx
dx
Ô⇒
d/dx
d
dy
d
dy
2
(e−(x+y) (1 + )) =
(y + x )
dx
dx
dx
dx
Ô⇒
2
e
−(x+y)2
2
dy
dy dy
d2 y
2 d y
(−2(x + y))(1 + ) + e−(x+y)
=
+
+
x
dx
dx2 dx dx
dx2
Ô⇒
x = 0, y = 0, dy/dx = −1
d2 y
= −2
dx2
30
at
(x, y) = (0, 0)
36.
Suppose that
f
is a continuous function satisfying
f (x) = x ∫
for all
x,
and
c
0
f (t) dt + x3
f (c) = 1.
is a real number such that
Solution:
x
Express
f ′ (c)
in terms of
c
only.
We have:
FTC1
x
x
x
d
d
↓
f (x) = ∫ f (t) dt + x ∫ f (t) dt + 3x3 = ∫ f (t) dt + xf (x) + 3x3
dx
dx 0
0
0
Now substituting
x = c and using the facts that f (c) = c ∫
c
0
f (t) dt + c3
and
f (c) = 1
we obtain:
f ′ (c) = ∫
Remark:
c
0
f (t) dt + cf (c) + 3c2 =
It can be shown that
f (x) = 2x(ex
2 /2
f (c) − c3
1
+ c + 3c2 = + c + 2c2
c
c
− 1) is the only function that satises the given
condition.
37.
x
Evaluate the limit
Solution:
3
∫ sin(xt ) dt
lim 0
x→0
x5
.
We rst make the change of variable
x
−1/3
∫0 sin(xt ) dt = x
3
t = x−1/3 u, dt = x−1/3 du,
x4/3
∫0
sin(u3 ) du
Now we have:
x4/3
x
3
x−1/3 ∫0 sin(u3 ) du
∫0 sin(xt ) dt
lim
= lim
x→0
x→0
x5
x5
x4/3
sin(u3 ) du
∫
= lim 0
x→0
x16/3
d x4/3
3
∫0 sin(u ) du
↓
dx
= lim
x→0
16/3 x13/3
L'H
sin(x4 ) ⋅ 4/3 x1/3
x→0
16/3 x13/3
sin(x4 ) ⋅ 4/3 x1/3
= lim
x→0
16/3 x13/3
1
sin(x4 )
= lim
4 x→0 x4
1
1
= ⋅1=
4
4
FTC1
↓
= lim
31
to obtain:
38.
Suppose that f is a twice-dierentiable function satisfying f (0) =
−8, f (2) = 5, f ′ (2) = −3, f ′′ (2) = 7 ; and also suppose that the function
g(x) =
has a critical point at
x = 2.
−2, f ′ (0) = 11, f ′′ (0) =
1 x
f (t) dt
x ∫0
Determine whether the critical point of
g
at
x=2
is a local
minimum, a local maximum or neither.
Solution:
We rst observe that
d
d 1 x
g(x) =
(
f (t) dt)
dx
dx x ∫0
FTC1
x
x
1
1d
f
(t)
dt
+
f (t) dt
∫
∫
x2 0
x dx 0
x
1
1
= − 2 ∫ f (t) dt + f (x)
x 0
x
↓
= −
and
as
g ′ (2) = 0
we
must
have
−1/4 ∫
2
2
0
f (t) dt + 1/2 f (2) = 0,
and
hence
∫0 f (t) dt = 2f (2) = 10.
Now dierentiating a second time we obtain that:
x
d
1
1
d2
g(x)
=
(
−
f
(t)
dt
+
f (x))
∫
dx2
dx
x2 0
x
FTC1
x
x
1 d
1
2
1 ′
f
(t)
dt
−
f
(t)
dt
−
f
(x)
+
f (x)
∫
∫
x3 0
x2 dx 0
x2
x
x
2
1
1
1
= 3 ∫ f (t) dt − 2 f (x) − 2 f (x) + f ′ (x)
x 0
x
x
x
↓
=
Substituting
x=2
in this we get
g ′′ (2) =
and conclude that
39.
Suppose that
f
g
1
1 2
1
3
f (t) dt − f (2) + f ′ (2) = − < 0
∫
4 0
2
2
2
has a local maximum at
x = 2.
is a continuous and positive function on
[0, 5],
and the area between the
y = f (x) and the x-axis for 0 ≤ x ≤ 5 is 8. Let A(c) denote the area between the graph
y = f (x) and the x-axis for 0 ≤ x ≤ c, and let B(c) denote the area between the graph of
dR
∣ = 7, nd
y = f (x) and the x-axis for c ≤ x ≤ 5. Let R(c) = A(c)/B(c). If R(3) = 1 and
dc c=3
f (3).
graph of
of
Solution:
We have
A(3) = B(3) = 4.
A(3) + B(3) = 8
and
32
R(3) = 1 Ô⇒ A(3) = B(3),
implying
As
A(c) = ∫
c
0
f (t) dt
dierentiating these with respect to
Calculus Part 1 we obtain A′ (c) =
A′ (3) = f (3) and B ′ (3) = −f (3).
and
B(c) = ∫
5
c
f (t) dt ,
c and using the Fundamental Theorem of
f (c) and B ′ (c) = −f (c). In particular,
On the other hand,
d
d A(c) A′ (c)B(c) − A(c)B ′ (c)
R(c) =
=
dc
dc B(c)
B(c)2
and letting
7=
Let
f
gives
d
A′ (3)B(3) − A(3)B ′ (3) f (3)B(3) + A(3)f (3) f (3)
R(c)∣ =
=
=
dc
B(3)2
B(3)2
2
c=3
and hence
40.
c=3
f (3) = 14.
be a continuous function and let
g(x) = ∫
Express
g ′′ (x)
f (x)
in terms of
Solution:
For
−1 < x < 1
g(x) = ∫
x
−1
= x∫
for
1
−1
−1 < x < 1.
we rewrite the denition of
f (t) (x − t) dt + ∫
x
−1
f (t) ∣x − t∣ dt .
f (t) dt − ∫
x
−1
1
x
g(x)
as follows:
f (t) (t − x) dt
f (t)t dt + ∫
1
x
f (t)t dt − x ∫
1
x
f (t) dt
Now we dierentiate using the FTC1 to obtain:
g ′ (x) = ∫
=∫
=∫
x
−1
x
−1
x
−1
x
x
d
d
f (t) dt −
f (t)t dt
∫
∫
dx −1
dx −1
1
1
1
d
d
+
f (t)t dt − ∫ f (t) dt − x
f (t) dt
∫
∫
dx x
dx x
x
f (t) dt + x
f (t) dt + xf (x) − f (x)x − f (x)x − ∫
f (t) dt − ∫
Using FTC1 again we compute
1
x
1
x
f (t) dt + xf (x)
f (t) dt
g ′′ (x)
to nd
x
1
d
d
f (t) dt −
f (t) dt
∫
∫
dx −1
dx x
= f (x) + f (x)
= 2f (x)
g ′′ (x) =
for
−1 < x < 1.
33
41.
Evaluate the limit
lim (n (
n→∞
1
1
1
+
+⋯+
)) .
2
2
(2n + 1)
(2n + 3)
(4n − 1)2
1
on the interval [2, 4]. If we divide this
x2
2
2k
interval into n subintervals of equal length
using the points xk = 2 +
, 0 ≤ k ≤ n,
n
n
2k − 1
and choose our sample points to be the midpoints ck = 2 +
, 1 ≤ k ≤ n, of these
n
Solution:
Consider the function
f (x) =
subintervals, the Riemann sum for these data becomes
n
n
k=1
k=1
∑ f (ck ) ∆xk = ∑ f (2 +
n
2k − 1 2
2n
)⋅ = ∑
n
n k=1 (2n + 2k − 1)2
and the denition of the denite integral gives
4 dx
2n
1 4
1 1 1
lim ∑
=∫
=− ] =− + = .
2
2
n→∞
x 2
4 2 4
2 x
k=1 (2n + 2k − 1)
n
Therefore
lim (n (
n→∞
42.
1
1
1
1
+
+⋯+
)) = .
2
2
2
(2n + 1)
(2n + 3)
(4n − 1)
8
Evaluate the following integrals.
a.
2
2
∫ x sin(x ) cos(x ) dx
b.
1 √
∫0 x 1 − x dx
Solution: a.
Let
u = sin(x2 ).
Then
du = 2x cos(x2 ) dx,
and
1 u2
1
1
2
2
x
sin(x
)
cos(x
)
dx
=
u
du
=
⋅
+ C = sin2 (x2 ) + C .
∫
∫
2
2 2
4
b.
Let
u = 1 − x.
1
∫0
Then
du = −dx,
In part
x = 0 ⇒ u = 1, x = 1 ⇒ u = 0.
Therefore:
0
1
√
x 1 − x dx = ∫ (1 − u)u1/2 (−du) = ∫ (u1/2 − u3/2 ) du
1
=[
Remark:
and
u3/2
3/2
0
−
1
u5/2
5/2
] =
0
2 2 4
− =
3 5 15
(a), if we let u = cos(x2 ), then we obtain the answer
1
2
2
2
2
∫ x sin(x ) cos(x ) dx = − 4 cos (x ) + C ;
34
1
sin(2x2 ), and then let u = sin(2x2 ),
2
1
2
2
2
∫ x sin(x ) cos(x ) dx = − 8 cos(2x ) + C .
and if we rst observe that
obtain the answer
sin(x2 ) cos(x2 ) =
then we
These are all correct answers. If we write
1
2
2
2
2
∫ x sin(x ) cos(x ) dx = 4 sin (x ) + C1
1
2
2
2
2
∫ x sin(x ) cos(x ) dx = − 4 cos (x ) + C2
1
2
2
2
∫ x sin(x ) cos(x ) dx = − 8 cos(2x ) + C3
then
43.
C2 = C1 +
1
4
Show that
C3 = C1 +
1
.
8
a
a
f (x)
∫0 f (x) + f (a − x) dx = 2
Solution:
Let
and
Let
u = a − x.
for any positive continuous function on
[0, a] .
a
f (x)
I =∫
dx .
0 f (x) + f (a − x)
Then
du = −dx, x = 0 ⇒ u = a, x = a ⇒ u = 0,
and
a
0
a
f (x)
f (a − u)
f (a − x)
I =∫
dx = ∫
(−du) = ∫
dx .
0 f (x) + f (a − x)
a f (a − u) + f (u)
0 f (x) + f (a − x)
Therefore
a
f (x)
f (a − x)
dx + ∫
dx
0 f (x) + f (a − x)
0 f (x) + f (a − x)
a f (x) + f (a − x)
dx
=∫
0 f (x) + f (a − x)
a
2I = ∫
=∫
and
I=
a
0
dx = a
a
.
2
(See
the remark on the next page .)
35
Remark:
Here is an explanation of what is going on with no integral signs: Consider the
x- and
y = f (x)/(f (x) + f (a − x)) is
rectangle with a vertex at the origin, and sides along the positive
lengths
a
and 1, respectively.
respect to the center
(a/2, 1/2)
The graph of
y -axes
with
symmetric with
of this rectangle, and therefore divides it into two regions of
equal area. Since the area of the rectangle is
a
I= .
2
the
a
36
and
I
is the area of the lower half, we have
44.
Let
R
y = x − x2 and
R about the x-axis.
be the region bounded by the parabola
volume of the solid generated by revolving
the
x-axis,
and let
V
a.
Express
V
as an integral using the disk method. (Do not compute! )
b.
Express
V
as an integral using the cylindrical shell method. (Do not compute! )
Solution:
The parabola intersects the
x-axis at x = 0 and x = 1.
1
be the
Therefore we have
1
V = π ∫ R(x)2 dx = ∫ (x − x2 )2 dx
0
0
for part
For part
(a).
(b), the cylindrical shell method gives a y-integral,
d
V = 2π ∫ (radius
c
of the shell)(height of the shell) dy
as the region is revolved about the
x-axis.
We have
c = 0
and
d = 1/4 ,
the
y-
coordinate of the highest point of the parabola. The radius of the shell is the vertical
distance from the red rectangle in the gure to the
x-axis,
which is
y.
The height
of the shell is the horizontal length of the rectangle; that is, the dierence between
the
x-coordinates
of the right and the left sides of the rectangle.
√
1
+
1 − 4y
equation y = x−x2 for x, we nd these values as x =
2
respectively. Hence
V = 2π ∫
1/4
0
y(
1+
√
and
√
1 − 4y 1 − 1 − 4y
−
) dy .
2
2
37
By solving the
x=
1−
√
1 − 4y
,
2
45.
R be the region bounded by the curve y 2 = x2 − x4 . Let V be the volume obtained by
rotating R about the x-axis. Let W be the volume obtained by rotating R about the y -axis.
Let
a.
Express
V
b.
Express
W
c.
Compute
using both the disk method and the cylindrical shells method.
V
Solution:
and
W.
By symmetry,
V
is 2 times the volume obtained by revolving the portion
x-axis. In the disk method, the red
vertical rectangles, when revolved about the x-axis, form the disks that are used in
√
the computation of V , and therefore the radii of the disks are given by
x2 − x4 .
of
R
using both the disk method and the cylindrical shells method.
lying in the rst quadrant about the
Hence:
1
V = 2 ⋅ π ∫ (radius
0
Again by symmetry,
of
R
W
2
of disk) dx
1 √
= 2 ⋅ π ∫ ( x2 − x4 )2 dx
0
is 2 times the volume generated by revolving the portion
lying in the rst quadrant about the
y -axis.
This time in the washer method,
the green horizontal rectangles, when revolved about the
that are used in the computation of
y -axis,
form the washers
W.
To nd the outer and the inner radii of the washers we have to solve y 2 =√
x2 −x4 for x.
2
2
2
2
2
Applying the quadratic formula to (x ) −x +y = 0 we obtain x = (1± 1 − 4y 2 )/2,
38
and this gives us
x=
√
(1 +
√
1 − 4y 2 )/2
and
√
x=
(1 −
√
1 − 4y 2 )/2
as the two nonnegative solutions. Hence:
1/2
((outer radius of washer)2 − (inner radius of washer)2 ) dy
¿
¿
Á 1 + √1 − 4y 2 2
Á 1 − √1 − 4y 2 2
1/2
Á
Á
À
À
= 2 ⋅ π∫
((
) −(
) ) dy
2
2
0
W = 2 ⋅ π∫
0
Now we consider the cylindrical shells method for both volumes.
about the
y -axis
When revolved
the red vertical rectangles generate the cylindrical shells that are
used in the computation of
√
by
x2 − x4 . Hence:
1
W = 2 ⋅ 2π ∫ (radius
0
W,
and therefore the heights of these shells are given
of shell)(height of shell) dx
When revolved about the
x-axis
1 √
= 2 ⋅ 2π ∫ x x2 − x4 dx
0
the green horizontal rectangles generate the
V . Therefore the heights
√
(1 − 1 − 4y 2 )/2. Hence:
cylindrical shells that are used in the computation of
√
of these shells are given by
V = 2 ⋅ 2π ∫
0
(1 +
√
1 − 4y 2 )/2 −
√
1/2
(radius of shell)(height of shell) dy
¿
¿
Á 1 + √1 − 4y 2 Á 1 − √1 − 4y 2
1/2
Á
Á
À
À
= 2 ⋅ 2π ∫
y(
−
) dy
2
2
0
Finally we compute the volumes. To compute
V
we use the integral we obtained
with the disk method,
x3 x5 1 4π
V = 2 ⋅ π ∫ (x − x ) dx = 2π [ − ] =
3
5 0 15
0
1
2
4
39
and to compute
W
we use the integral we obtained with the washer method
W = 2 ⋅ π∫
0
= 2π ∫
= 2π ∫
= π∫
1/2
(
1/2 √
0
π/2
0
π/2
1+
√
1 − 4y 2 1 −
−
2
√
1 − 4y 2
) dy
2
1 − 4y 2 dy
cos θ ⋅
1
cos θ dθ
2
cos2 θ dθ
0
1 + cos 2θ
dθ
2
0
θ sin 2θ π/2
= π[ +
]
2
4 0
π2
=
4
= π∫
π/2
where we made the change of variable
Remark:
46.
Compare this example with
y=
1
1
sin θ, dy = cos θ dθ.
2
2
Example 27 in Part 2.
Find the absolute maximum and the absolute minimum values of
interval
[−2, 2].
Solution:
f ′ (x) = 2xex + (x2 − 3)ex = (x2 + 2x − 3)ex .
f (x) = (x2 − 3)ex
We want to solve
on the
f ′ (x) = 0.
(x2 + 2x − 3)ex = 0 Ô⇒ x2 + 2x − 3 = 0 Ô⇒ x = 1, −3 .
Since
−3
does not belong to the interval
Hence we are going to compute
f
[−2, 2], the only
x = 1,−2, 2.
critical point is
x = 1.
at the points
f (1) = −2e, f (−2) = e−2 , and f (2) = e2 . Since e > 1 we have e2 > e−2 > −2e.
the absolute maximum is e2 and the absolute minimum is −2e.
40
Therefore
47.
Find the absolute maximum and the absolute minimum values of
Solution:
Let
y = x1/x .
Then
ln y =
obtain
ln x
.
x
x1/x .
Dierentiating this with respect to
x
we
1
1 dy d
d ln x x ⋅ x − ln x ⋅ 1 1 − ln x
1 − ln x
d 1/x
=
ln y =
=
x = x1/x
=
Ô⇒
.
2
2
y dx dx
dx x
x
x
dx
x2
Since 1 − ln x > 0 for 0 < x < e and 1 − ln x < 0 for x > e, the absolute maximum value
of x1/x occurs at x = e and is e1/e . x1/x has no absolute minimum value.
Remark:
Although the reasoning above does not require it, let us also look at what happens
at the endpoints of the domain. Since
L'H
ln x ↓
1/x
lim ln y = lim
= lim
=0,
x→∞
x→∞ x
x→∞ 1
we have
48.
lim x1/x = lim y = lim eln y = e0 = 1.
x→∞
x→∞
Evaluate the limit
Solution:
n
xk =
lim (
n→∞
∑ 2k/n
k=1
partition
x→∞
1
n
On the other hand,
lim x1/x = 0
x→0+
as
0∞ = 0.
1 n k/n
∑2 ).
n k=1
n
is a Riemann sum
∑ f (ck ) ∆xk
for
f (x) = 2x
on
[0, 1]
for the
k=1
k
, 0 ≤ k ≤ n,
n
and the sample points
ck =
k
, 1 ≤ k ≤ n.
n
Therefore,
1
1
2x
1 n
1
lim ( ∑ 2k/n ) = ∫ 2x dx =
] =
.
n→∞ n
ln 2 0 ln 2
0
k=1
Remark:
Alternatively, we can use the formula for the sum of a nite geometric series
n
∑ 2k/n =
k=1
2(n+1)/n − 21/n
21/n
=
,
21/n − 1
21/n − 1
and then
1
1 n k/n
21/n
t2t ↓
2t + t ⋅ ln 2 ⋅ 2t
=
lim
=
lim
=
lim
.
∑ 2 ) = n→∞
t
t
1/n
+
+
t→0
n k=1
ln 2 ⋅ 2
ln 2
n(2 − 1) t→0 2 − 1
L'H
lim (
n→∞
41
2
49.
Evaluate the limit
Solution:
form 1∞ .
Since
sin x 1/x
) .
x→0
x
sin x
1
lim
= 1 and lim 2 = ∞,
x→0 x
x→0 x
lim (
this is limit has the indeterminate
sin x
2
ln (
)
sin x 1/x
x
Let y = (
) . Then ln y =
. As x → 0, this will have the indeterminate
x
x2
0
form
and we can use L'Hôpital's Rule.
0
ln(sin x) − ln x
lim ln y = lim
x→0
x→0
x2
cos x 1
−
x cos x − sin x
↓
= lim sin x x = lim
x→0
x→0
2x
2x2 sin x
cos x − x sin x − cos x
− sin x
↓
= lim
= lim
2
x→0 4x sin x + 2x cos x
x→0 4 sin x + 2x cos x
sin x
−
−1
1
x
= lim
=
=−
x→0 sin x
4+2
6
4
+ 2 cos x
x
L'H
L'H
Here applications of L'Hôpital's Rule are indicated with
L'H
. Then
2
sin x 1/x
lim (
)
= lim y = lim eln y = e−1/6
x→0
x→0
x→0
x
using the continuity of the exponential function.
Remark:
The rst example of an application of L'Hôpital's Rule in Guillaume François
Antoine Marquis de L'Hôpital's book Analyse des Inniment Petits pour l'Intelligence des
Lignes Courbes of 1696:
There is a typo. Can you nd it?
42
50.
Evaluate
cos(2x) − e−2x
x→0
sin4 x
lim
2
.
Solution:
cos(2x) − e−2x
x4
cos(2x) − e−2x
lim
=
lim
(
⋅
)
x→0
x→0
x4
sin4 x
sin4 x
2
cos(2x) − e−2x
x 4
= lim
⋅
(lim
)
x→0
x→0 sin x
x4
2
cos(2x) − e−2x
= lim
x→0
x4
2
−2 sin(2x) + 4xe−2x
↓
= lim
x→0
4x3
2
2
−4 cos(2x) + 4e−2x − 16x2 e−2x
↓
= lim
x→0
12x2
2
2
8 sin(2x) − 48xe−2x + 64x3 e−2x
↓
= lim
x→0
24x
sin(2x)
8
2
2
= lim (
− 2e−2x + x2 e−2x )
x→0
3x
3
2
4
= −2+0=−
3
3
2
2
L'H
L'H
L'H
Remark:
It is easier to solve this problem using the Taylor series, which will be seen in Calculus
II:
cos(2x) − e−2x
x→0
sin4 x
2
lim
(2x)2 (2x)4 (2x)6
(−2x2 )2 (−2x2 )3
+
−
+ ⋯) − (1 + (−2x2 ) +
+
+ ⋯)
2!
4!
6!
2!
3!
= lim
4
x→0
x3
(x −
+ ⋯)
3!
56
4
− x4 + x6 + ⋯
45
= lim 3
2
x→0
x4 − x6 + ⋯
3
4 56 2
− + x +⋯
= lim 3 45
2
x→0
1 − x2 + ⋯
3
4
=−
3
(1 −
43
51.
Find the value of the constant
the limit for this value of
Solution:
sin(x + ax3 ) − x
x→0
x5
a for which the limit lim
a.
exists and compute
We have
sin(x + ax3 ) − x
x→0
x5
cos(x + ax3 )(1 + 3ax2 ) − 1
↓
= lim
x→0
5x4
− sin(x + ax3 )(1 + 3ax2 )2 + cos(x + ax3 )(6ax)
↓
= lim
x→0
20x3
3
− cos(x + ax )(1 + 3ax2 )3 − sin(x + ax3 ) ⋅ 3(1 + 3ax2 )(6ax) + cos(x + ax3 )(6a)
↓
.
= lim
x→0
60x2
lim
L'H
L'H
L'H
The numerator of the fraction inside this last limit goes to
the limit does not exist unless
a = 1/6.
If
a = 1/6
−1+6a as x → 0.
Therefore
then
sin(x + x3 /6) − x
x→0
x5
− cos(x + x3 /6)(1 + x2 /2)3 − sin(x + x3 /6) ⋅ 3(1 + x2 /2)x + cos(x + x3 /6)
= lim
x→0
60x2
lim
is the sum of
cos(x + x3 /6)(1 − (1 + x2 /2)3 )
x→0
60x2
cos(x + x3 /6)(3/2 + 3x2 /4 + x4 /8)
1
= − lim
=−
x→0
60
40
lim
and
− sin(x + x3 /6) ⋅ 3(1 + x2 /2)x
x→0
60x2
(1 + x2 /2)(1 + x2 /6)
sin(x + x3 /6)
1
⋅
lim
=− .
= − lim
3
x→0
x→0
x + x /6
20
20
lim
Hence
sin(x + x3 /6) − x
3
=− .
5
x→0
x
40
lim
Remark:
Once again there are shorter ways of doing this. If we use the Taylor series then
(x + ax3 )3 (x + ax3 )5
+
−⋯
3!
5!
1
1
1
= x + (a − ) x3 + (
− a) x5 + ⋯
6
120 2
sin(x + ax3 ) = (x + ax3 ) −
and it is immediate that the limit exists exactly when
44
a = 1/6
and then its value is
−3/40.
52.
Let b > a > 0 be constants. Find
(x − b)2 + y 2 = a2 about the y -axis.
Solution:
the area of the surface generated by revolving the circle
We have
Surface Area
= 2π ∫
d
c
¿
2
Á
dx
Á
À
x 1 + ( ) dy
dy
for a surface generated
√ by revolving a curve about the y -axis. We have x = b +
√
a2 − y 2 and x = b − a2 − y 2 for the right and left halves of the circle, respectively.
Then
¿
2
Á
dx
y
dx
a
Á
À
Ô⇒
.
= ∓√
1+( ) = √
dy
dy
a2 − y 2
a2 − y 2
Hence
√
a
dy
a2 − y 2 ) √
2
−a
a − y2
√
a
a
+ 2π ∫ (b − a2 − y 2 ) √
dy
−a
a2 − y 2
a
1
dy
= 4πab ∫ √
−a
a2 − y 2
a
Surface Area
= 2π ∫ (b +
a
y
= 4πab arcsin ( ) ]
a −a
π
π
= 4πab (arcsin 1 − arcsin(−1)) = 4πab ( − ( − )) = 4π 2 ab
2
2
Remark:
The surface generated by revolving a circle about a line (in the same plane) that
does not intersect it is called a torus.
53.
if
dy
= xy 2
dx
Solution:
We have
Find
y(1)
and
y(0) = 1.
dy
1 1
dy
dy
= xy 2 Ô⇒ 2 = x dx Ô⇒ ∫ 2 = ∫ x dx Ô⇒ − = x2 + C .
dx
y
y
y 2
Then
y(0) = 1 Ô⇒ −1 = C .
Hence
y=
2
2 − x2
45
. This in turn gives
y(1) = 2.
Remark:
54.
What would your answer be if the question asked
y(2)?
Nitrogen dioxide is a reddish-brown gas that contributes to air pollution, and also gives the
smog its color. Under sunlight it decomposes producing other pollutants, one of which is ozone.
As nitrogen dioxide decomposes, its density decreases at a rate proportional to the square of
the density. Suppose that the density of nitrogen dioxide
minutes and
6/25
grams per liter at time
t=3
Q
1/2 grams per liter at time t = 0
Q when t = 15 minutes. (Assume
is
minutes. Find
that no new nitrogen dioxide is added to the environment.)
Solution:
−k dt
and
dQ/dt = −kQ2 where k is a positive
integrating we obtain −1/Q = −kt + C where C
We have
constant. Then
dQ/Q2 =
is a constant.
t = 0 minutes we nd −1/(1/2) = −1/Q(0) = C , hence C = −2 liters per gram.
t = 3 min gives −1/(6/25) = −1/Q(3) = −k ⋅ 3 − 2 and hence k = 13/18
per gram per minute. Finally we obtain Q = 18/(13t + 36).
Letting
Now letting
liters
Using this we compute the density of nitrogen dioxide as
Q(15) = 6/77
grams per
liter after 15 minutes.
55.
f is a function that has a continuous second derivative and
f (0) = 4 , f (1) = 3 , f ′ (0) = 5 , f ′ (1) = 7, f ′′ (0) = 8 and f ′′ (1) = 11 . Show that:
Suppose that
that satises
1
′′
∫0 f (x)f (x) dx ≤ 1
Solution:
du =
We rst do an integration by parts with
′
f (x) dx, v = f ′ (x), to obtain:
1
1
u = f (x), dv = f ′′ (x) dx,
hence
1
′
2
∫0 f (x)f (x) dx = [f (x)f (x)] − ∫0 (f (x)) dx
′′
′
0
The rst term on the right is equal to f (1)f ′ (1) − f (0)f ′ (0) = 3 ⋅ 7
the second term is nonnegative as (f ′ (x))2 ≥ 0. The result follows.
Remark:
It can be shown that:
1
′′
∫0 f (x)f (x) dx < 0
56.
Evaluate the following integrals.
a.
3
∫ sin x sin 2x dx
b.
e ln x
∫1 √ dx
x
46
− 4 ⋅ 5 = 1,
and
Solution: a.
We use the identity
sin 2x = 2 sin cos x
to obtain
3
3
∫ sin x sin 2x dx = ∫ sin x 2 sin x cos x dx
= 2 ∫ sin4 x cos x dx
= 2 ∫ u4 du
2
= u5 + C
5
2
= sin5 x + C
5
after the substitution
u = sin x, du = cos x dx.
b.
u=
The substitution
√
√
x, du = dx/(2 x)
e
∫1
ln x
√ dx = 4 ∫
1
x
gives:
√
e
ln u du
√
e
= 4 [u ln u − u]
√
√ 1√
= 4 ( e ln e − e + 1)
√
=4−2 e
Remark:
There are other ways of doing these. Here are some:
Solution: a.
substitution
Use the double-angle formula
u = 1 − cos 2x, du = 2 sin 2x dx :
sin2 x = (1 − cos 2x)/2
and then the
3
1 − cos 2x
) sin 2x dx
∫ sin x sin 2x dx = ∫ (
2
1
= √ ∫ u3/2 du
4 2
1
= √ u5/2 + C
10 2
1
= √ (1 − cos 2x)5/2 + C
10 2
3
Or you can run this by simply saying
u = sin2 x, du = 2 sin x cos x dx = sin 2x dx
3
3/2
∫ sin x sin 2x dx = ∫ u du
2
= u5/2 + C
5
2
= sin5 x + C
5
47
and:
Or use the identity
sin3 x = (3 sin x − sin 3x)/4 and the trigonometric product to sum
formulas:
1
3
∫ sin x sin 2x dx = ∫ 4 (3 sin x − sin 3x) sin 2x dx
1
= ∫ (3 sin x sin 2x − sin 3x sin 2x) dx
4
1
= ∫ (3(cos x − cos 3x) − (cos x − cos 5x)) dx
8
1
= ∫ (cos 5x − 3 cos 3x + 2 cos x) dx
8
1
1
1
=
sin 5x − sin 3x + sin x + C
40
8
4
Or, if you are willing to go complex, use the identity
sin θ = (eiθ − e−iθ )/(2i) :
3
eix − e−ix e2ix − e−2ix
)
dx
sin
x
sin
2x
dx
=
(
∫
∫
2i
2i
1
=
(e5ix + e−5ix − 3e3ix − 3e−3ix + 2eix + 2e−ix ) dx
16 ∫
1 e5ix − e−5ix e3ix − e−3ix
eix − e−ix
= (
−
+2
)+C
16
5i
i
i
1
1
1
=
sin 5x − sin 3x + sin x + C
40
8
4
3
b.
Do integration by parts:
e ln x
e
√
∫1 √ dx = ∫1 ln x d(2 x)
x
e √ dx
√
e
= [2 x ln x]1 − ∫ 2 x
x
1
e dx
√
= 2 e − 2∫ √
1
x
√
√ e
= 2 e − 4 [ x]1
√
√
=2 e−4 e+4
√
=4−2 e
Or do the other integration by parts rst,
e
∫1
and then from
e 1
ln x
√ dx = ∫ √ d(x ln x − x)
1
x
x
e
1 e x ln x − x
x ln x − x
] + ∫
dx
=[ √
2 1
x3/2
x
1
√ e
1 e ln x
= 1 + ∫ √ dx − [ x]1
2 1
x
√
1 e ln x
= 2 − e + ∫ √ dx
2 1
x
√
1 e ln x
√ dx = 2 − e
∫
2 1
x
48
solve for
e ln x
√
∫1 √ dx = 4 − 2 e .
x
Or rst do the substitution
e
∫1
x = eu , dx = eu du,
and then do an integration by parts:
1 u
ln x
√ dx = ∫ √ eu du
0
x
eu
=∫
=∫
1
u eu/2 du
0
1
0
u d(2eu/2 )
1
= [2ueu/2 ]0 − 2 ∫
=
eu/2 du
0
1/2
u/2 1
2e − 4[e ]0
1/2
1/2
= 2e
− 4e
√
=4−2 e
57.
1
+4
Evaluate the following integrals:
√
x
a.
∫ e
b.
∫
c.
dx
∫ (x2 + 1)2
√
dx
1 − x2 dx
Solution: a.
We rst do a change of variable
∫ e
√
x
x = t2 , dx = 2tdt,
dx = 2 ∫ tet dt
and then do an integration by parts,
u = t, dv = et dt Ô⇒ du = dt, v = et :
= 2tet − 2 ∫ et dt
= 2tet − 2et + C
√
√ √
= 2 xe x − 2e x + C
b.
We use the trigonometric substitution
49
x = sin θ, −
π
π
≤θ≤ .
2
2
Then
dx = cos θ dθ
and
√
1 − x2 =
√
√
1 − sin2 θ = cos2 θ = ∣ cos θ∣ = cos θ
∫
√
as
cos θ ≥ 0
for
−
π
π
≤θ≤ .
2
2
1 − x2 dx = ∫ cos θ ⋅ cos θ dθ
= ∫ cos2 θ dθ
1
(1 + cos 2θ) dθ
2∫
1
1
= θ + sin 2θ + C
2
4
1
1
= θ + sin θ cos θ + C
2
2
1
1 √
= arcsin x + x 1 − x2 + C
2
2
=
1
x
√
1 − x2
c.
and
x = tan θ, −
We use the trigonometric substitution
x2 + 1 = tan2 θ + 1 = sec2 θ .
π
π
<θ< .
2
2
Then
dx
sec2 θ
=
∫ (x2 + 1)2 ∫ (sec2 θ)2 dθ
= ∫ cos2 θ dθ
=
=
=
=
=
1
(1 + cos 2θ) dθ
2∫
1
1
θ + sin 2θ + C
2
4
1
1
θ + sin θ cos θ + C
2
2
1
1
x
1
arctan x + ⋅ √
⋅√
+C
2
2
x2 + 1
x2 + 1
1
1
x
arctan x + ⋅ 2
+C
2
2 x +1
√
x2 + 1
1
50
x
dx = sec2 θ dθ
Remark:
Other methods can also be used. For instance, the integral in part
(b) can be done
using integration by parts.
58.
Evaluate the following integrals:
a.
1/2
√
∫−1/2
1/2
b.
∫−1/2
c.
∫0
∞
1−x
arcsin x dx
1+x
dx
√
x + 1 − x2
dx
1 + ex
Solution: a.
1/2
√
∫−1/2
We rst re-write the integral as follows:
1/2
1/2
1−x
1
−x
√
√
arcsin x dx + ∫
arcsin x dx
arcsin x dx = ∫
1+x
−1/2
−1/2
1 − x2
1 − x2
The rst integral on the right vanishes as the integrand is odd and the integration
interval is symmetric about the origin. We do integration by parts for the second
integral with
√
u = arcsin x
and
dv = √
1 − x2 :
1/2
∫−1/2
arcsin x √
−x
1 − x2
1/2
√
∫−1/2
1/2
c.
Let
√
dx,
and hence
1 − x2 ]
1/2
−1/2
−∫
du = √
1/2
−1/2
1
1 − x2
t = sin θ, dt = cos θ dθ.
π/6
dx
cos θ dθ
√
=∫
2
−π/6 sin θ + cos θ
x+ 1−x
π/6 cos θ(cos θ − sin θ)
=∫
dθ
−π/6
cos2 θ − sin2 θ
1 π/6 1 + cos 2θ − sin 2θ
= ∫
dθ
2 −π/6
cos 2θ
1 π/6
= ∫
(sec 2θ + 1 − tan 2θ) dθ
2 −π/6
π/6
1 1
1
= [ ln ∣ tan 2θ + sec 2θ∣ + θ − ln ∣ sec 2θ∣]
2 2
2
−π/6
√
π
1
= ln( 3 + 2) +
2
6
u = e−x + 1, du = −e−x dx.
Then
e−x dx
dx
du
−x
=
∫ 1 + ex ∫ e−x + 1 = − ∫ u = − ln ∣u∣ + C = − ln(1 + e ) + C
51
and
π
dx = √ − 1
2 3
1−x
π
arcsin x dx = √ − 1
1+x
2 3
We start by changing variables
∫−1/2
1 − x2
dx = [arcsin x
Therefore:
b.
−x
v=
and from this
∫0
∞
c dx
dx
=
lim
= − lim [ln(1 + e−x )]c0 = − lim (ln(1 + e−c ) − ln 2) = ln 2
∫
c→∞
c→∞
1 + ex c→∞ 0 1 + ex
follows.
Remark:
Here is another way of doing part
(c).
Let
u = ex , du = ex dx.
Then
dx
ex dx
=
∫ 1 + ex ∫ ex + (ex )2
du
=∫
u + u2
1
1
) du
=∫ ( −
u 1+u
= ln ∣u∣ − ln ∣1 + u∣ + C
= x − ln(1 + ex ) + C
and from this
∫0
∞
c dx
dx
= lim ∫
x
c→∞ 0 1 + ex
1+e
= lim [x − ln(1 + ex )]c0
c→∞
= lim (c − ln(1 + ec ) + ln 2)
c→∞
= lim (− ln(e−c ) − ln(1 + ec ) + ln 2)
c→∞
= lim (− ln(e−c + 1) + ln 2)
c→∞
= ln 2
follows.
59.
Evaluate the improper integral
∫0
where
a
∞
dx
(ax + 1)(x2 + 1)
is a positive constant.
Solution:
We have the partial fraction decomposition
1
a2
ax − 1
1
=
(
− 2
).
2
2
(ax + 1)(x + 1) a + 1 ax + 1 x + 1
Hence:
dx
1
a2
ax − 1
=
(
∫ (ax + 1)(x2 + 1) a2 + 1 ∫ ax + 1 dx − ∫ x2 + 1 dx)
=
a2
1
a
(a ln ∣ax + 1∣ − ln(x2 + 1) + arctan(x)) + C .
+1
2
52
Therefore:
∫0
∞
c
dx
dx
=
lim
∫
2
(ax + 1)(x + 1) c→∞ 0 (ax + 1)(x2 + 1)
c
1
a
= 2
lim [a ln ∣ax + 1∣ − ln(x2 + 1) + arctan(x)]
a + 1 c→∞
2
0
=
a2
1
a
lim (a ln(ac + 1) − ln(c2 + 1) + arctan c
+ 1 c→∞
2
− a ln 1 +
=
=
as
60.
ac + 1
lim √
=a
c→∞
c2 + 1
The curve
and
a
ac + 1
1
√
lim
ln
(
)
+
lim arctan c
a2 + 1 c→∞
a2 + 1 c→∞
c2 + 1
a2
π
.
2
lim arctan c =
is revolved about the
region between the curve and the
solid
π
1
(a ln a + )
+1
2
c→∞
y = 1/x, x ≥ 1,
a
ln 1 − arctan 0)
2
x-axis
for
x≥1
x-axis
to generate a surface
is revolved about the
x-axis
S
and the
to generate a
D.
a.
Show that
D
has nite volume.
b.
Show that
S
has innite area.
Solution: a.
Using the disk method we obtain
Volume
Since
b.
p = 2 > 1,
= π∫
∞
1
R(x)2 dx = π ∫
∞
1
dx
.
x2
this integral converges.
The surface area formula gives
Surface Area
We have
1
x
√
1+
= 2π ∫
∞
y
√
1
1 + (y ′ )2 dx = 2π ∫
∞
1
∞ dx
1 1
≥
≥
0
for x ≥ 1, and ∫
= ∞.
x4 x
x
1
1
x
√
1+
1
dx .
x4
The surface area is innite
by the Direct Comparison Test.
Remark:
We want to get the surface
S
painted for a reasonable, nite price. We oer the job
to Painter1 and Painter2.
Painter1 says: It cannot be done.
S
has innite area, it cannot be painted with
nite amount of paint.
Painter2 says: It can be done.
D
has nite volume. We can ll the inside of
with volume(D ) cubic units of paint and let the excess paint run out.
53
S
Whom should we believe?
61.
Let
n
be a nonnegative integer. Show that
Solution:
ˆ
ˆ
We use induction on
Let
n = 0.
Let
n>0
Then
∞
∫0
∞
∞
tn e−t dt = n! .
n.
e−t dt = 1 = 0! .
and assume that
∫0
∫0
∫0
∞
tn−1 e−t dt = (n − 1)! .
tn e−t dt = lim ∫
c→∞
c
tn e−t dt
0
= lim ([−tn e−t ]0 + n ∫
c
c
c→∞
=
c
lim [−tn e−t ]0
c→∞
= n∫
Integration by parts gives
∞
+ n∫
0
∞
tn−1 e−t dt)
tn−1 e−t dt
0
tn−1 e−t dt
0
= n ⋅ (n − 1)!
= n!
where
Remark:
lim cn e−c = 0
c→∞
x > 0.
applications of L'Hôpital's Rule.
The Gamma function is dened by
Γ(x) = ∫
for
n
can be seen after
∞
tx−1 e−t dt
0
It can be shown that the improper integral on the right converges if and only if
Note that 0 is also a bad point besides
∞
for
0 < x < 1.
A calculation similar to the one in the solution above shows that
This relation
Γ(x) = Γ(x + 1)/x
Γ(x + 1) = xΓ(x)
for
x > 0.
x > 0.
can be used repeatedly to dene the Gamma function for all
real numbers which are not nonpositive integers.
Γ(n + 1) = n! for all nonnegative integers n, we can use the Gamma function to dene the
factorials of all real numbers which are not negative integers by x! = Γ(x + 1). In particular,
√
∞
∞
1
π √
−1/2 −t
−u2
(− )! = Γ(1/2) = ∫ t e dt = 2 ∫ e du = 2 ⋅
= π.
2
2
0
0
Since
The volume of an
n=1
(the interval
The case
n=4
n-dimensional ball with radius r is π n/2 rn /(n/2)! . Check this formula for
[−r, r]), n = 2 (the disk with radius r), and n = 3 (the sphere with radius r).
will be seen in Calculus II.
54
62.
Show that
∫0
Solution:
∞
ln x
dx = 0 .
x2 + 1
First of all, by denition,
∫0
∞
1 ln x
∞ ln x
ln x
dx = ∫
dx + ∫
dx
2
2
x +1
x2 + 1
0 x +1
1
and the improper integral on the left converges if and only if both basic improper
integrals on the right converge.
Consider
∫1
∞
ln x
dx .
x2 + 1
∫1
∞
Note that
0≤
ln x
ln x
≤ 2
2
x +1
x
for
x ≥ 1.
On the other hand,
∞
ln x
−t
dx
=
∫0 te dt
x2
= lim ∫
c→∞
= lim ∫
c→∞
c
te−t dt
0
c
0
t d(−e−t )
c
= lim ([ − te−t ]0 + ∫
c→∞
=
=
c
e−t dt)
0
−t c
lim (−ce − [e ]0 )
c→∞
lim (−ce−c − e−c + 1)
c→∞
−c
=1
where we used the substitution
x = et , dx = et dt,
followed by the denition of the
improper integral, then an integration by parts and nally the limit
L'H
c ↓
1
=
lim
= 0.
c→∞ ec
c→∞ ec
lim ce−c = lim
c→∞
Therefore by the Direct Comparison Test the improper integral
converges.
Now the convergence of
1
∫0
1 ln x
∫0 x2 + 1 dx
∞
ln x
dx
x2 + 1
follows as
∞ ln u
1 ln(1/u)
−du
ln x
=
−
dx
=
⋅
∫1 1 + u2 du
∫∞ (1/u)2 + 1 u2
x2 + 1
where we used the change of variable
∫0
∫1
∞
x = 1/u, dx = −du/u2 ,
and this also gives:
1 ln x
∞ ln x
ln x
dx
=
dx
+
dx
∫
∫
x2 + 1
x2 + 1
0 x2 + 1
1
∞ ln x
∞ ln x
dx
+
dx = 0
= −∫
∫
x2 + 1
x2 + 1
1
1
55
Remark:
Instead of computing
comparison
0≤
1
ln x
≤ 3/2
2
x
x
for
x≥1
∫1
∞
ln x
dx,
x2
one can show its convergence using the
and the fact that
∫1
∞
dx
x3/2
is convergent as
p = 3/2 > 1.
√
ln x
1
≤
for x ≥ 1 can be seen as follows: Consider f (x) =
x − ln x on
x2
x3/2
√
√
′
[1, ∞). Then f (x) = 1/(2 x) − 1/x = ( x − 2)/(2x)
√ and x = 4 is the only critical point. Since
′
′
f (x) < 0 for x < 4 and f (x) > 0 for x > 4, f (4) = 4 − ln 4 = 2 − 2 ln 2 = 2(1 − ln 2) > 0 must be
the absolute minimum value of f on [1, ∞), and we are done.
Here the fact that
63.*
Determine whether the improper integral
Solution:
∫0
∞
ex
dx
− e−x
converges or diverges.
By denition,
∫0
∞
1
∞
1
dx
dx
=
+
∫
∫
x
−x
x
−x
x
e −e
e − e−x
0 e −e
1
and the given integral converges if and only if both of the integrals on the right
hand side converge.
1
Let us consider
dx
∫0 ex − e−x
rst. Since we have the linearization
ex − e−x ≈ (1 + x) − (1 − x) = 2x ,
centered at
x = 0,
we expect
1/(ex − e−x )
to behave like
1/(2x)
near the bad point
0, and therefore this integral to diverge.
In fact, we have
1
1
1/(ex − e−x )
x
↓
= lim+ x −x = lim+ x −x = .
x→0 e + e
x→0 e − e
1/x
2
L'H
L = lim+
x→0
1 dx
is divergent (because p = 1 ≥ 1), we conclude
0 < L < ∞ and ∫
x
0
1
∞
dx
dx
diverges
by
the
Limit
Comparison
Test.
Therefore
∫0 ex − e−x
∫0 ex − e−x
Since
that
diverges too.
Remark:
The other improper integral on the right hand side converges. We have
1
1/(ex − e−x )
= lim
= 1.
−x
x→∞
x→∞
e
1 − e−2x
L = lim
Since
0<L<∞
∫1
and
∞
we conclude that
c
e−x dx = lim ∫ e−x dx = lim [−e−x ]c1 = lim (−e−c + e−1 ) = e−1 < ∞ ,
c→∞ 1
c→∞
c→∞
∫1
∞
ex
dx
− e−x
converges by the Limit Comparison Test.
*Examples marked red are not part of the Fall 2014 syllabus.
56
64.*
Determine whether the improper integral
Solution:
∫0
∞
1 − e−1/x
√
dx
x
converges or diverges.
By denition,
∫0
∞
1 1 − e−1/x
∞ 1 − e−1/x
1 − e−1/x
√
√
√
dx = ∫
dx + ∫
dx
0
1
x
x
x
and the given integral converges if and only if both integrals on the right hand side
converge.
On one hand, since
1 − e−1/x
√
x
L = lim+
= lim+ (1 − e−1/x ) = 1
1
x→0
x→0
√
x
is a positive real number, and
1
∫0
1 − e−1/x
√
dx
x
1 dx
∫0 √
x
converges as
p = 1/2 < 1;
we conclude that
converges by the Limit Comparison Test.
On the other hand, since
1 − e−1/x
√
1 − e−1/x
e−t
1 − e−t ↓
x
= lim+
L = lim
= lim
= lim+
=1
x→∞
x→∞
1
t→0 1
t→0
1/x
t
x3/2
L'H
is a positive real number, and
∫1
∞
1 − e−1/x
√
dx
x
Hence
∫0
∞
∫1
∞
dx
x3/2
converges as
p = 3/2 > 1;
also converges by the Limit Comparison Test.
1 − e−1/x
√
dx
x
converges.
*Examples marked red are not part of the Fall 2014 syllabus.
57
we conclude that
Part 2: Multi-variable Functions
1.
Consider the point
P (3, −5, 1)
L ∶ x = 2t − 1 , y = −t + 2 , z = −2t ; −∞ < t < ∞ .
and the line
a.
Find the equation of the plane passing through
P
perpendicular to
b.
Find the equation of the plane passing through
P
and containing
Solution: a.
L.
L.
v = 2i − j − 2k . Therefore we can
take the normal vector of the plane to be n = 2i − j − 2k . Then the equation of the
plane is 2 ⋅ (x − 3) + (−1) ⋅ (y − (−5)) + (−2) ⋅ (z − 1) = 0 , or 2x − y − 2z = 9 .
b.
The line
L
is parallel to the vector
n to the plane containing the line L and the point P will be
# »
v, and it will also be perpendicular to P Q where Q is any point
on the line. We can take Q(−1, 2, 0), the point corresponding to t = 0 , and then
# »
P Q = −4i + 7j − k . Now we can take the normal vector v to be
A normal
perpendicular to
RR i
j kRRRR
# » RRRR
R
n = v × P Q = RRR 2 −1 −2 RRRR = 15i + 10j + 10k ,
RRR
RR
RR−4 7 −1 RRR
or in fact
n = 3i + 2j + 2k .
Then the equation of the plane is
3 ⋅ (x − 3) + 2 ⋅ (y − (−5)) + 2 ⋅ (z − 1) = 0 ,
or
2.
3x + 2y + 2z = 1 .
Consider the plane
P ∶ 3x − 4y + z = 10 ,
and the points
a.
Find the equation of the line passing through
b.
Find the equation of the plane passing through
Solution: a.
Since
n = 3i − 4j + k
P
P (2, 3, −1)
perpendicular to
P
and
Q
b.
P
P.
P.
is normal to the plane, it will be parallel to any
v = 3i − 4j + k ,
x = 3t + 2, y = −4t + 3, z = t − 1; −∞ < t < ∞ .
A plane perpendicular to
Q(1, 2, 2) .
perpendicular to
line perpendicular to the plane. Hence we can take
of the line is
and
will have a normal
n′
and the equation
perpendicular to the normal
n = 3i − 4j + k of P . Also a plane containing the points P and Q will have a normal
# »
n′ perpendicular to P Q = −i − j + 3k . Therefore we can take n′ to be
RR i
j
# » RRRR
R
n = n × P Q = RR 3 −4
RR
RRR−1 −1
′
58
kRRRR
R
1 RRRR = −11i − 10j − 7k ,
R
3 RRRR
or rather
n′ = 11i + 10j + 7k .
This gives the equation of the plane as
11 ⋅ (x − 2) + 10 ⋅ (y − 3) + 7 ⋅ (z − (−1)) = 0 ,
or
3.
11x + 10y + 7z = 45 .
Find a parametric equation of the line
L
that intersects both of the lines
L1 ∶ x = 2t − 1,
y = −t + 2,
z = 3t + 1
L2 ∶ x = s + 5,
y = 2s + 3,
z = −s
and
perpendicularly.
Solution: v1 = 2i − j + 3k and v2 = i + 2j − k are the velocity vectors of the lines L1
and
L2 ,
respectively. Hence,
is a velocity vector for
RRR i
j kRRRR
RR
R
v = RRRR2 −1 3 RRRR = −5i + 5j + 5k
RRR
R
RR1 2 −1 RRRR
L. So we may take v = i − j − k.
Then
RR i
j kRRRR
RR
RRR
R
n = v×v1 = RR1 −1 −1 RRRR = −4i − 5j + k
RRR
RRR
RR2 −1 3 RR
is normal to the plane P containing the lines L and L1 . As P1 (−1, 2, 1)
an equation of P is
is in
P,
−4 ⋅ (x − (−1)) + (−5) ⋅ (y − 2) + 1 ⋅ (z − 1) = 0 ,
or
4x + 5y − z = 5.
At the point
P0
of intersection of
P
and
L2 , s
satises:
4 ⋅ (s + 5) + 5 ⋅ (2s + 3) − (−s) = 5
Hence
s = −2.
Substituting this back in the equations of
Therefore an equation of
x = t + 3,
L
L2
gives
P0 (3, −1, 2).
is:
y = −t − 1 ,
z = −t + 2 ;
59
(−∞ < t < ∞) .
4.
c be a constant. Show the angle between the position and the velocity vectors along the
r = ect cos t i + ect sin t j , −∞ < t < ∞ , is constant.
Let
curve
Solution:
We have
r = ect cos t i + ect sin t j
and
v=
dr
= (c ect cos t − ect sin t)i + (c ect sin t + ect cos t)j .
dt
Then
∣r∣ = ((ect cos t)2 + (ect sin t)2 )1/2 = ect ,
√
∣v∣ = ((c ect cos t − ect sin t)2 + (c ect sin t + ect cos t)2 )1/2 = c2 + 1 ect ,
r ⋅ v = ect cos t ⋅ (c ect cos t − ect sin t) + ect sin t ⋅ (c ect sin t + ect cos t) = c e2t .
Therefore, if
θ
is the angle between
cos θ =
and we conclude that
θ
r
and
v,
we have
r⋅v
c e2t
c
√
=
=√
,
ct
2
ct
2
∣r∣ ∣v∣ e ⋅ c + 1 e
c +1
is constant.
60
5. a.
Show that
b.
Show that
xy 2
=0.
(x,y)→(0,0) x6 + y 2
xy
lim
does
6
(x,y)→(0,0) x + y 2
lim
Solution: a.
We have
0 ≤ y 2 ≤ x6 + y 2
0≤∣
for all
(x, y) =/ (0, 0) .
not exist.
for all
(x, y).
Hence
xy 2
y2
∣
=
∣x∣
⋅
≤ ∣x∣ ⋅ 1 = ∣x∣
x6 + y 2
x6 + y 2
Since
lim
(x,y)→(0,0)
∣x∣ = 0 ,
the Sandwich Theorem gives
xy 2
=0.
(x,y)→(0,0) x6 + y 2
lim
b.
The limit along the
x-axis
lim
(x,y)→(0,0)
x-axis
along the
whereas the limit along the
lim
xy
x⋅0
= lim 6
= lim 0 = 0 ,
2
x→0 x + 02
x→0
+y
x6
y=x
(x,y)→(0,0)
y=x
along the line
is
line is
xy
x⋅x
1
= lim 6
= lim 4
=1.
2
2
x→0
x→0
+y
x +x
x +1
x6
Since these two limits are dierent, the two-variable limit
lim
xy
+ y2
(x,y)→(0,0) x6
exist by the Two-Path Test.
Remark:
Let
a, b, c, d
be positive constants. It can be shown that the limit
∣x∣a ∣y∣b
(x,y)→(0,0) ∣x∣c + ∣y∣d
lim
exists if
6.
a b
+ >1,
c d
and does not exist otherwise.
Determine all values of the constant
α>0
for which the limit
x2 y 3
(x,y)→(0,0) ∣x∣3 + ∣y∣α
lim
exists.
Solution:
We observe that
lim
(x,y)→(0,0)
x = y3
along the curve
(y 3 )2 y 3
y
x2 y 3
1
=
lim
= lim ⋅ lim
.
3
α
3
3
α
y→0 ∣y ∣ + ∣y∣
y→0 ∣y∣ y→0 1 + ∣y∣α−9
∣x∣ + ∣y∣
61
does not
1
y→0 1 + ∣y∣α−9
lim
is
1
if
α>9
and
exist. Therefore the limit of
1/2
α = 9.
if
x2 y 3
∣x∣3 + ∣y∣α
consequently the two-variable limit
along the curve
x2 y 3
(x,y)→(0,0) ∣x∣3 + ∣y∣α
lim
Now we will show that the limit is 0 if
0≤∣
On the other hand
α < 9.
x = y3
y
y→0 ∣y∣
lim
does not
does not exist, and
does not exist either for
α ≥ 9.
We have
x2 y 3
(∣x∣/∣y∣α/3 )2
∣
=
⋅ ∣y∣3−α/3 ≤ ∣y∣3−α/3
∣x∣3 + ∣y∣α
(∣x∣/∣y∣α/3 )3 + 1
Here we used the fact that if t ≥ 1, then t2 ≤ t3 and hence
t2
≤
1
, and if 0 < t < 1, then t2 < 1 gives
≤ 1. Since
lim ∣y∣3−α/3 = 0 for
(x,y)→(0,0)
t3 + 1
t3 + 1
x2 y 3
α < 9,
lim
= 0 follows by the Sandwich Theorem.
(x,y)→(0,0) ∣x∣3 + ∣y∣α
for all
t2
7.
(x, y) =/ (0, 0).
Determine all values of the positive constant
lim
k
(x,y)→(0,0) (x2
for which the limit
∣x∣
+ y 2 )k
exists.
Solution:
Suppose
The limit exists for
k<
1
2
. We have
k<
1
2
, and does not exist for
0 ≤ x2 ≤ x2 + y 2
for all
(x, y),
1
2
k≥
.
and hence
k
∣x∣
x2
1−2k
0≤ 2
≤
∣x∣
(
) ≤ ∣x∣1−2k
(x + y 2 )k
x2 + y 2
for all
(x, y) =/ (0, 0) .
1 − 2k > 0,
Since
Therefore
lim
(x,y)→(0,0)
(x2
we have
∣x∣1−2k → 0
as
(x, y) → (0, 0).
∣x∣
=0
+ y 2 )k
by the Sandwich Theorem.
Suppose
k=
1
2
. Then
lim
(x,y)→(0,0)
x-axis
along the
and
lim
(x,y)→(0,0)
y -axis
along the
(x2
∣x∣
∣x∣
= lim 2
= lim 1 = 1 ,
2
1/2
x→0 (x + 02 )1/2
x→0
+y )
∣0∣
∣x∣
=
lim
= lim 0 = 0 .
(x2 + y 2 )1/2 y→0 (02 + y 2 )1/2 x→0
62
Since these limits are dierent, the two-variable limit
lim
(x,y)→(0,0) (x2
∣x∣
+ y 2 )1/2
does not
exist by the Two-Path Test.
k > 1/2 .
Suppose
Then
lim
(x,y)→(0,0)
x-axis
along the
(x2
∣x∣
∣x∣
= lim 2
= lim ∣x∣1−2k = ∞
2
k
2
k
x→0
x→0
+y )
(x + 0 )
does not exist. Therefore the two-variable limit
∣x∣
(x,y)→(0,0) (x2 + y 2 )k
lim
does not exist
either.
8.
Let
where
of
a
a
and
a.
as
⎧
xa y b
⎪
⎪
⎪
⎪
⎪
⎪ x4 + y 6
f (x, y) = ⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩0
and
b
b
(x, y) =/ (0, 0)
if
(x, y) = (0, 0)
are nonnegative integers. In each of
for which
f (x, y)
if
f
is continuous at
(0, 0).
b. f (x, y) goes to 1 as (x, y) approaches (0, 0)
(x, y) approaches (0, 0) along the line y = −x.
c.
limit
(a-e), determine whether there exist values
satises the given condition.
f (x, y) goes to 0 as (x, y) approaches (0, 0)
lim f (x, y) does not exist.
along the line
y = x,
and
goes to
−1
along any line through the origin, and the
(x,y)→(0,0)
d.
f (x, y) goes to 0 as (x, y) approaches (0, 0) along any line through
y -axis, and f (x, y) goes to 1 as (x, y) approaches (0, 0) along the y -axis.
e.
f (x, y)
fx (0, 0)
and
fy (0, 0)
Solution: a.
If
exist, and
a=4
and
f (x, y)
b = 1,
then
is not dierentiable at
f
is continuous at
the origin except the
(0, 0).
(0, 0).
This follows from
the Sandwich Theorem as
0 ≤ ∣f (x, y)∣ = ∣
for
(x, y) =/ 0
b.
Let
a=3
x4
x4 y
∣
≤
⋅ ∣y∣ ≤ 1 ⋅ ∣y∣ ≤ ∣y∣
x4 + y 6
x4 + y 6
implies that the limit of
and
b = 1.
lim
f (x, y)
at
(0, 0)
is
0 = f (0, 0).
Then
(x,y)→(0,0)
y=x
x4
1
= lim
=1
4
6
x→0 x + x
x→0 1 + x2
f (x, y) = lim f (x, x) = lim
x→0
along the line
and
lim
(x,y)→(0,0)
y=−x
−1
−x4
=
lim
= −1 .
x→0 1 + x2
x→0 x4 + x6
f (x, y) = lim f (x, −x) = lim
x→0
along the line
63
c.
Let
a=2
and
b = 3.
lim
(x,y)→(0,0)
y=mx
Then we have
m3 x5
m3 x
=
lim
=0
x→0 1 + m6 x2
x→0 x4 + m6 x6
f (x, y) = lim f (x, mx) = lim
x→0
along the line
as well as
lim
(x,y)→(0,0)
along the y -axis
f (x, y) = lim f (0, y) = lim 0 = 0 .
y→0
y→0
However,
lim
(x,y)→(0,0)
y=x2/3
1 1
x4
=
lim
= =/ 0
x→0 2
x→0 x4 + x4
2
f (x, y) = lim f (x, x2/3 ) = lim
x→0
along the curve
and hence the limit of
d.
Let
a=0
and
f (x, y)
b = 6.
lim
(x,y)→(0,0)
y=mx
at
(0, 0)
does not exist by the 2-Path Test.
Then we have
m6 x6
m6 x2
=
lim
=0
x→0 1 + m6 x2
x→0 x4 + m6 x6
f (x, y) = lim f (x, mx) = lim
x→0
along the line
and
lim
(x,y)→(0,0)
y -axis
y6
= lim 1 = 1 .
y→0 y 6
y→0
f (x, y) = lim f (0, y) = lim
y→0
along the
e.
If
a=1
and
b = 1,
continuous there.
then
f (x, y)
is not dierentiable at
(0, 0)
as it is not even
This can be seen by considering its limit along the line
which does not exist. On the other hand,
fx (0, 0)
and
fy (0, 0)
y =x
f is
are both 0 as
identically zero on both axes.
Remark:
The complete lists of ordered pairs
(a, b)
of nonnegative integers that satisfy the
given conditions are as follows:
a.
all
b.
(3, 1), (1, 3)
c.
(1, 4), (2, 3)
d.
(0, 6)
e.
(0, 7), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1)
(a, b)
with
3a + 2b > 12
64
9.
Assume that
yz 2 + z 3 = −1
z
and
Solution:
z
and
w
and
w
are dierentiable functions of
zw3 − xz 3 + y 2 w = 1 .
∂z
∂x
Find
at
x
and
x,
y
but
satisfying the equations
xw3 +
(x, y, z, w) = (1, −1, −1, 1) .
Dierentiating the equations with respect to
depend on
y
x and keeping in mind that
does not; we obtain
w3 + x ⋅ 3w2 wx + y ⋅ 2zzx + 3z 2 zx = 0 ,
and
zx w3 + z ⋅ 3w2 wx − z 3 − x ⋅ 3z 2 zx + y 2 wx = 0 .
Substituting
Solving for
x = 1, y = −1, z = −1, w = 1,
zx
Let
Suppose
z = f (x, y)
w
5zx + 3wx = −1
and
2zx + 2wx = 1 .
we nd
∂z
5
=−
∂x
4
10.
we get
at
(x, y, z, w) = (1, −1, −1, 1) .
be a dierentiable function such that
f (3, 3) = 1,
fx (3, 3) = −2,
fy (3, 3) = 11,
f (2, 5) = 1,
fx (2, 5) = 7,
fy (2, 5) = −3.
is a dierentiable function of
u
and
v
satisfying the equation
f (w, w) = f (uv, u2 + v 2 )
for all
(u, v).
Find
Solution:
∂w
∂u
at
(u, v, w) = (1, 2, 3).
Dierentiating the identity
f (w, w) = f (uv, u2 + v 2 )
with respect to
fx (w, w)
u
gives
∂w
∂w
∂(uv)
∂(u2 + v 2 )
+ fy (w, w)
= fx (uv, u2 + v 2 )
+ fy (uv, u2 + v 2 )
∂u
∂u
∂u
∂u
by the Chain Rule. Hence
(fx (w, w) + fy (w, w))
∂w
= fx (uv, u2 + v 2 ) v + fy (uv, u2 + v 2 ) 2u
∂u
which leads to
∂w
= 2fx (2, 5) + 2fy (2, 5)
∂u
after substituting (u, v, w) = (1, 2, 3).
Now using fx (3, 3) = −2, fy (3, 3) = 11,
fx (2, 5) = 7, and fy (2, 5) = −3, we conclude that
(fx (3, 3) + fy (3, 3))
∂w 8
=
∂u 9
at
(u, v, w) = (1, 2, 3) .
65
11.
Let
u = x + y + z , v = xy + yz + zx, w = xyz , and suppose that f (u, v, w) is
f (u, v, w) = x4 + y 4 + z 4 for all (x, y, z). Find fu (2, −1, −2).
a dierentiable
function satisfying
Solution:
We can take
Dierentiating
(x, y, z) = (1, −1, 2)
as this gives
(u, v, w) = (2, −1, −2).
f (u, v, w) = x4 +y 4 +z 4 with respect to x, y , z , respectively, we obtain:
fu ⋅ ux + fv ⋅ vx + fw ⋅ wx = 4x3
fu ⋅ uy + fv ⋅ vy + fw ⋅ wy = 4y 3
fu ⋅ uz + fv ⋅ vz + fw ⋅ wz = 4z 3
Now using
u = x + y + z , v = xy + yz + zx, w = xyz ,
these give:
fu ⋅ 1 + fv ⋅ (y + z) + fw ⋅ yz = 4x3
fu ⋅ 1 + fv ⋅ (x + z) + fw ⋅ xz = 4y 3
fu ⋅ 1 + fv ⋅ (x + y) + fw ⋅ xy = 4z 3
Substituting
(x, y, z) = (1, −1, 2)
we get:
fu + fv − 2fw = 4
fu + 3fv + 2fw = −4
fu − fw = 32
−2fu + 8fw = −16, and
6fu = 240. So fu (2, −1, −2) = 40.
Subtracting 3 times the rst equation from the second gives
adding 8 times the third equation to this gives
Remark:
for a given
f (u, v, w) = u4 − 4u2 + 2v 2 + 4uw is one. Also note that
corresponding (x, y, z) must be the roots of T 3 − uT 2 + vT − w = 0 and
There are such
(u, v, w),
the
f.
In fact,
hence is determined up to a permutation of its entries, making the answer independent of the
choice.
12.
Let
z = f (x, y)
be a twice-dierentiable function and
x = r cos θ, y = r sin θ .
Show that
∂ 2f ∂ 2f ∂ 2z 1 ∂z 1 ∂ 2z
+
=
+
+
.
∂x2 ∂y 2 ∂r2 r ∂r r2 ∂θ2
Solution:
If
z = F (x, y)
y,
then by the chain
∂z ∂F ∂x ∂F ∂y
=
⋅
+
⋅
= Fx ⋅ cos θ + Fy ⋅ sin θ ,
∂r ∂x ∂r ∂y ∂r
(i )
∂z ∂F ∂x ∂F ∂y
=
⋅
+
⋅
= Fx ⋅ (−r sin θ) + Fy ⋅ (r cos θ) .
∂θ ∂x ∂θ ∂y ∂θ
(ii )
rule we have
is a dierentiable function of
x
and
and similarly,
We use
(i ) with F
=f
to obtain
∂z
= fx ⋅ cos θ + fy ⋅ sin θ .
∂r
66
(
A)
Then
To compute
∂ 2z ∂
∂
= (fx ) ⋅ cos θ + (fy ) ⋅ sin θ .
2
∂r
∂r
∂r
∂
∂
(fx ) and
(fy ) we use (i ) with F = fx and F = fy ,
∂r
∂r
respectively:
∂ 2z
= (fxx cos θ + fxy sin θ) cos θ + (fyx cos θ + fyy sin θ) sin θ
∂r2
= fxx cos2 θ + 2fxy cos θ sin θ + fyy sin2 θ
Similarly, using
(ii ) with F
=f
(
B)
gives
∂z
= fx ⋅ (−r sin θ) + fy ⋅ (r cos θ) ,
∂θ
and dierentiating this with respect to
θ
again gives
∂ 2z ∂
∂
= (fx ) ⋅ (−r sin θ) + fx ⋅ (−r sin θ)
2
∂θ
∂θ
∂θ
∂
∂
+ (fy ) ⋅ (r cos θ) + fy ⋅ (r cos θ)
∂θ
∂θ
= (fxx (−r sin θ) + fxy r cos θ)(−r sin θ) + fx (−r cos θ)
+ (fyx (−r sin θ) + fyy r cos θ)(r cos θ) + fy (−r sin θ)
2
2
= fxx r sin θ − 2fxy r2 cos θ sin θ + fyy r2 cos2 θ
− r(fx cos θ + fy sin θ)
13.
where we used
(ii ) with F
Now if we add
(B), 1/r times (A), and 1/r2 times (C), we obtain fxx + fyy .
Suppose that
= fx
(x, y) =/ (0, 0).
5, fyy (3, 1) = −4.
Solution:
C)
F = fy .
f (x, y) is a twice-dierentiable function with continuous derivatives satisfying
f(
for all
and
(
Find
x2
x
y
, 2
) = f (x, y)
2
+ y x + y2
fxx (3/10, 1/10)
if
fx (3, 1) = −8, fy (3, 1) = 7, fxx (3, 1) = 2, fxy (3, 1) =
We have:
fx (x, y) =
y
∂
x
, 2
)
f( 2
2
∂x x + y x + y 2
= fx (
x2
x
y
∂
x
, 2
)⋅
)
( 2
2
2
+y x +y
∂x x + y 2
+ fy (
= fx (
x
y
∂
y
,
)
⋅
)
(
x2 + y 2 x2 + y 2 ∂x x2 + y 2
y
1 ⋅ (x2 + y 2 ) − x ⋅ 2x
x
,
)
⋅
x2 + y 2 x2 + y 2
(x2 + y 2 )2
+ fy (
x2
67
x
y
−2xy
, 2
)⋅ 2
2
2
+y x +y
(x + y 2 )2
Dierentiating this a second time we obtain:
fxx (x, y) =
x
y
y 2 − x2
∂
(fx ( 2
,
)
⋅
∂x
x + y 2 x2 + y 2 (x2 + y 2 )2
+ fy (
x2
y
−2xy
x
, 2
)⋅ 2
)
2
2
+y x +y
(x + y 2 )2
x
y
y 2 − x2
= fxx ( 2
,
)
⋅
(
)
x + y 2 x2 + y 2
(x2 + y 2 )2
+ fxy (
2
y
−2xy
y 2 − x2
x
,
)
⋅
⋅
x2 + y 2 x2 + y 2 (x2 + y 2 )2 (x2 + y 2 )2
+ fx (
x
y
∂
y 2 − x2
,
)
⋅
(
)
x2 + y 2 x2 + y 2 ∂x (x2 + y 2 )2
+ fyx (
x
y
y 2 − x2
−2xy
,
)
⋅
⋅ 2
2
2
2
2
2
2
2
x +y x +y
(x + y ) (x + y 2 )2
x
y
−2xy
+ fyy ( 2
, 2
)⋅( 2
)
2
2
x +y x +y
(x + y 2 )2
+ fy (
= fxx (
x
y
∂
−2xy
,
)
⋅
(
)
x2 + y 2 x2 + y 2 ∂x (x2 + y 2 )2
x
y
y 2 − x2
,
)
⋅
(
)
x2 + y 2 x2 + y 2
(x2 + y 2 )2
+ 2fxy (
2
2
x
y
y 2 − x2
−2xy
,
)
⋅
⋅ 2
2
2
2
2
2
2
2
x +y x +y
(x + y ) (x + y 2 )2
2
x
y
−2xy
+ fyy ( 2
, 2
)⋅( 2
)
2
2
x +y x +y
(x + y 2 )2
+ fx (
x
y
2x(x2 − 3y 2 )
,
)
⋅
x2 + y 2 x2 + y 2
(x2 + y 2 )3
+ fy (
Now letting
(x, y) = (3/10, 1/10)
x
y
2y(3x2 − y 2 )
,
)
⋅
x2 + y 2 x2 + y 2
(x2 + y 2 )3
gives:
fxx (3/10, 1/10) = fxx (3, 10) ⋅ (−8)2 + 2fxy (3, 10) ⋅ (−8) ⋅ (−6) + fyy (3, 10) ⋅ (−6)2
+ fx (3, 10) ⋅ 36 + fy (3, 10) ⋅ 52
= 2 ⋅ 64 + 2 ⋅ 5 ⋅ 48 + (−4) ⋅ 36 + (−8) ⋅ 36 + 7 ⋅ 52
= 540
68
14.
Let
P0 (3, 2)
f (x, y) = x3 y − xy 2 + cx2
c is a constant.
A = 2i + 5j.
where
in the direction of the vector
Find
c if f
increases fastest at the point
Solution:
∇f = (3x2 y −y 2 +2cx)i+(x3 −2xy)j Ô⇒ (∇f )P0 = (50+6c)i+15j . Since f
increases the fastest at P0 in the direction of (∇f )P0 , A = 2i + 5j must be a positive
multiple of (∇f )P0 . Hence (50 + 6c)/2 = 15/5 Ô⇒ c = −22/3 . Finally we check that
c = −22/3 gives (∇f )P0 = 6i + 15j = 3A which is indeed a positive multiple of A.
15.
Find a vector that is tangent to the intersection curve of the surfaces
z = xy
at the point
Solution:
are level
P0 (1, 2, 2).
f (x, y, z) = x2 +y 2 +z 2
surfaces of f and g .
Let
and
g(x, y, z) = xy −z .
x2 + y 2 + z 2 = 9
Then the given surfaces
∇f = 2xi + 2yj + 2zk Ô⇒ (∇f )P0 = 2i + 4j + 4k and ∇g = yi + xj − k Ô⇒
= 2i + j − k . (∇f )P0 is normal to the surface dened by x2 + y 2 + z 2 = 9 and
is normal to the surface dened by z = xy . Therefore,
We have
(∇g)P0
(∇g)P0
(∇f )P0 × (∇g)P0
RRR i
RR
= RRRR 2
RR
RRR 2
j k RRRR
R
4 4 RRRR = −8i + 10j − 6k ,
R
1 −1 RRRR
or any multiple of it, is tangent to both of these surfaces, and hence, to their curve
of intersection at
16.
Let
a
P0 .
be a constant. Find and classify all critical points of
Solution:
f (x, y) = x3 − 3axy + y 3 .
fx = 3x2 − 3ay = 0 and fy = −3ax + 3y 2 = 0. If a = 0,
=0⇒x=0
= 0 ⇒ y = 0, and (0, 0) is the only critical point.
If a =
/ 0, then the rst equation gives y = x2 /a, and substituting this in the second
equation we get x4 −a3 x = 0 whose solutions are x = 0 and x = a. Now using y = x2 /a,
we get (0, 0) and (a, a) as the critical points.
At a critical point
then 3x2
and 3y 2
We compute the discriminant:
∆=∣
fxx (a, a) = 6a, (a, a) is a local minimum for a > 0 and
local maximum for a < 0. On the other hand, ∆(0, 0) = −9a2 implies that (0, 0) is
saddle point for a =
/ 0.
As
∆(a, a) = 27a2
6x −3a
fxx fxy
∣=∣
∣
fyz fyy
−3a 6y
and
Now we look at the sole critical point
(0, 0)
a
a
a = 0. As ∆(0, 0) = 0,
restrict f (x, y) = x3 + y 3 to the
in the case
the second derivative test fails in this case. If we
we get f (x, 0) = x3 . Since this single variable function does not have a local
x-axis
x = 0, f (x, y) cannot have a local maximum
conclude that (0, 0) is a saddle point when a = 0.
maximum or minimum at
at
(0, 0)
either. We
69
or minimum
and
17.
Find the absolute maximum and minimum values of the function
on the unit disk D = {(x, y) ∶ x2 + y 2 ≤ 1}.
f (x, y) = 2x3 +2xy 2 −x−y 2
Solution:
We rst nd the critical points of f (x, y) in the interior of D . At a
critical point we have fx = 6x2 + 2y 2 − 1 = 0 and fy = 4xy − 2y = 0. The second
equation implies that
obtain
√
y = ±1/ 6
y=0
or
x = 1/2.
Substituting these into the rst equation we
in the rst case, and no solution in the second case. Therefore
√
(x, y) = (1/ 6, 0)
interior of D .
the critical points are
points lie in the
and
√
(−1/ 6, 0).
Note that both of these
D, that is, the unit circle x2 + y 2 = 1. We can
≤ 1√. These solutions correspond to the upper
and lower semicircles.
Then f (x, ± 1 − x2 ) = x2 + x − 1 for −1 ≤ x ≤ 1, and
√
d
1
f (x, ± 1 − x2 ) = 2x + 1 = 0 ⇒ x = − . This gives the critical points (x, y) =
dx
2
√
√
(−1/2, 3/2) and (−1/2, − 3/2) of the restriction of f to the boundary of D. We
must also include the endpoints x = −1 and x = 1, in other words, the points
(x, y) = (1, 0) and (−1, 0) in our list.
Now we look at the boundary of
√
solve y as y = ± 1 − x2 , −1 ≤ x
Hence the absolute maximum and the absolute minimum of
f
on
D
occur at some
of the points
√
√
√
√
(1/ 6, 0), (−1/ 6, 0), (−1/2, 3/2), (−1/2, − 3/2), (1, 0), (−1, 0) .
The values of
f
at these points are
1
−
3
√
2 1
,
3 3
√
2
5
5
, − , − , 1 , −1 ,
3
4
4
respectively. Therefore the absolute maximum value is 1 and the absolute minimum
value is
Remark:
−5/4.
Here are two more ways of dealing with the critical points of the restriction of
the boundary of
D.
70
f
to
In the rst one we parametrize the boundary, which is the unit circle, by
−∞ < t < ∞.
x = cos t, y = sin t,
Then
d
d
1
f (cos t, sin t) = (cos2 t + cos t − 1) = −2 cos t sin t − sin t = 0 ⇒ cos t = −
dt
dt
2
√
√
and these give us (x, y) = (−1/2,
3/2), (−1/2, − 3/2), (1, 0), (−1, 0).
or
sin t = 0 ,
g(x,y)
In the second we use the Lagrange Multipliers Method for the boundary
O2
³¹¹ ¹ ¹ ·¹¹ ¹ ¹ µ
x2 + y 2 = 1.
⎧
⎧
f = λgx ⎫
6x2 + 2y 2 − 1 = λ 2x
⎪
⎪
⎪
⎪
⎪
⎪
⎪ x
⎪
⎪
∇f = λ∇g
4xy − 2y = λ 2y
} Ô⇒ ⎨ fy = λgy ⎬ Ô⇒ ⎨
g = 1
⎪
⎪
⎪ g = 1 ⎪
⎪
⎪
⎪
⎪
⎪
x2 + y 2 = 1
⎩
⎭
⎩
O
O1
O2
O3
y = 0 or x = (1 +λ)/2. If y = 0, then 3 gives x = ±1 (and λ = ±5/2). On√the other hand,
if x = (1 + λ)/2, then from 1 and 3 , 4x2 − 2λx + 1 = 0 ⇒ x = −1/2 and y = ± 3/2. Therefore
√
√
the points (x, y) = (−1/2,
3/2), (−1/2, − 3/2), (1, 0), (−1, 0) are added to the list.
O
gives
18.
O
Three hemispheres with radiuses
1, x and y , where 1 ≥ x ≥ y ≥ 0, are stacked on top of each
h.
other as shown in the gure. Find the largest possible value of the total height
Solution:
We want to maximize
h(x, y) =
√
1 − x2 +
√
x2 − y 2 + y
on the closed and bounded region
D = {(x, y) ∶ 0 ≤ y ≤ x ≤ 1}.
We rst nd the critical points of
h
hx = − √
x
1 − x2
+√
x
x2 − y 2
From the second equation we obtain
used the fact that x > 0 and
1−2y 2 = 2y 2 −y 2 Ô⇒ 3y 2 = 1
critical point of
h
is
in the interior of
=0
and
D.
hy = − √
We want
y
x2 − y 2
+ 1 = 0.
y 2 = x2 −y 2 Ô⇒ x2 = 2y 2 Ô⇒ x =
√
2y where we
y > 0. Now√substituting this in the rst
√ equation gives
. So the only
Ô⇒
√ y = 1/√ 3 as y > 0 and hence x = 2/3
√
√
(x, y) = ( 2/3, 1/ 3) and it lies in D as 0 ≤ 1/ 3 ≤ 2/3 ≤ 1.
Now we consider the restriction of
h
to the boundary of
71
D.
y =0
1 − x2 + x
Side 1 : On the bottom edge of the triangle, we √
have
therefore we are considering the function
h(x, 0) =
0 ≤ x ≤ 1,
0 ≤ x ≤ 1.
and
for
and
x
d
1
h(x, 0) = − √
+ 1 = 0 Ô⇒ x2 = 1 − x2 Ô⇒ 2x2 = 1 Ô⇒ x = √
dx
2
1 − x2
x > 0. Taking √
the endpoints x = 0
points (x, y) = (1/ 2, 0), (0, 0), (1, 0).
as
and
x=1
into account, Side 1 gives us the
Side 2 : On the right edge of the triangle, √
we have x = 1 and
h(1, y) = 1 − y 2 + y for 0 ≤
0 ≤ y ≤ 1, and therefore
y
≤ 1. As in the case of
√
(x, y) = (1, 1/ 2), (1, 0), (1, 1).
we are considering the function
Side 1 this leads to the points
Side 3 : On the top edge of the triangle, we
√ have y = x and 0 ≤ x ≤ 1, and therefore
we are considering the function h(x, x) =
1 − x2 + x for 0 ≤ x ≤ 1. Once again as in
the case of Side 1 we obtain the points
Now we nd the values of
h
√
√
(x, y) = (1/ 2, 1/ 2), (0, 0), (1, 1).
at these seven points:
f (0, 0) = f (1, 0) = f (1, 1) = 1
√
√
√
√
√
f (1/ 2, 0) = f (1, 1/ 2) = f (1/ 2, 1/ 2) = 2
√
√
√
f ( 2/3, 1/ 3) = 3
Therefore the maximum possible total height of the three hemispheres is
Remark:
√
3.
Here is a single variable argument which solves the problem for any number
√
k for some k ≥ 1.
Then for k + 1 hemispheres, where the second one from the bottom has radius r , the maximum
√
√
√
kr + 1 − r2 for 0 ≤ r ≤ 1. The
possible height will be H(r) =
√ value of H(r) is 1 and k at
k + 1 at its only critical point
the √
endpoints r = 0 and r = 1, respectively; and its value is
r = k/(k + 1).√ Therefore we conclude inductively that the maximum possible height for n
n for n ≥ 1.
hemispheres is
hemispheres: Suppose the maximum possible height for
72
k
hemispheres is
19.
Find the absolute maximum and the absolute minimum of
sphere x2 + y 2 + z 2 = 1.
Solution:
z 2 − 1.
We will use the Lagrange Multipliers Method. Let
ˆ
ˆ
ˆ
O2
If
and
z = 0,
O3
we obtain
then
the points
O3
gives
(±1, 0, 0).
fx
fy
fz
g
=
=
=
=
z = 4λ2 z ,
y = 0,
λgx
λgy
λgz
0
⎧
⎫
3x2
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
z
⎬ Ô⇒ ⎨
y
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
2 + y2 + z2
⎪
⎪
x
⎩
⎭
and hence
and then
O4
z=0
gives
or
λ = 1/2
x = ±1.
O O
O
O
O1 O3 3x = −x√ y√= −z
on the unit
g(x, y, z) = x2 + y 2 +
Then
⎧
⎪
⎪
⎪
⎪
⎪
∇f = λ∇g
} Ô⇒ ⎨
g = 0
⎪
⎪
⎪
⎪
⎪
⎩
From
f (x, y, z) = x3 + yz
O1
O2
O3
O4
=
=
=
=
λ 2x
λ 2y
λ 2z
1
or
λ = −1/2.
In this case we have
λ = 1/2, then from 1 and 3 , 3x√2 = x and y = z . Therefore we either
have x = 0 in which case y = z = ±1/ 2 by 4 , or we have x = 1/3 and then
√
√
y = z = ±2/3 again by 4 . In this case the critical points are (0, ±1/ 2, ±1/ 2)
and (1/3, ±2/3, ±2/3).
If
If
λ = −1/2,
then from
and
,
2
and
. A reasoning similar to
(0, ±1/ 2, ∓1/ 2) and (−1/3, ±2/3, ∓2/3).
√
√
Hence the critical points are (±1, 0, 0), (0, ±1/ 2, ±1/ 2), (1/3, ±2/3, ±2/3),
√
√
(0, ±1/ 2, ∓1/ 2), (−1/3, ±2/3, ∓2/3) , and the values of f at these points are
±1, 1/2, −1/2, 13/27, −13/27 , respectively. Therefore the absolute maximum is 1 and
the absolute minimum is −1.
the previous case gives the points
73
20.
Evaluate the following integrals:
1
1
a.
3
∫0 ∫√y sin(πx ) dx dy
b.
2 −x
∬R y e dA
2
√
2y−y 2
2
c.
∫0 ∫0
d.
∫0
∞
where
∫0
R = {(x, y) ∶ 0 ≤ y ≤ x}
xy
dx dy
+ y2
x2
∞
(x2
dy dx
+ y 2 )2 + 1
Solution: a.
We will rst express the iterated integral as a double integral and
√
x-integral goes from x = y to x = 1 as
gure in the xy -plane. Then the y -integral
then reverse the order of integration. The
shown by the red line segments in the
goes from
y=0
to
y = 1.
Therefore the intervals of the x-integral trace out the region R bounded by the
parabola y = x2 , the line x = 1, and the x-axis. Note that the x-integrals are always
from left to right in the interval
1
0 ≤ y ≤ 1.
Hence we have
1
3
3
∫0 ∫√y sin(πx ) dx dy = ∬R sin(πx ) dA .
y -integral rst.
integration for the y -
Now we express this double integral as an iterated integral with the
The green line segment in the gure shows the interval of
integral which goes from y = 0 to y = x2 . Then:
74
1
1
3
3
∫0 ∫√y sin(πx ) dx dy = ∬R sin(πx ) dA
=∫
=∫
=∫
x2
1
∫0
0
1
0
1
0
sin(πx3 ) dy dx
y=x2
sin(πx3 ) y]y=0 dx
sin(πx3 )x2 dx
1
1
cos(πx3 )]0
3π
2
=
3π
=−
b.
We integrate with respect to
y
rst to obtain
∞
x
2 −x
2 −x
∬R y e dA = ∫0 ∫0 y e dy dx
1 ∞
2
= ∫ x3 e−x dx
3 0
1 ∞
= ∫ te−t dt
6 0
1
=
6
2
2
where we used the integration by parts
∫0
∞
te−t dt = ∫
∞
0
t d(−e−t )
= lim ([−te−t ]c0 + ∫
c→∞
c
e−t dt)
0
c
= − lim c − lim [e−t ]c0
c→∞ e
c→∞
1
↓
= − lim c − lim (e−c − 1)
c→∞ e
c→∞
=1
L'H
in the last step.
c.
This time we will use the polar coordinates. To do so we rst determine the
region of integration
R.
√
x√= 0 to x = 2y − y 2 as shown
2y − y 2 Ô⇒ x2 = 2y − y 2 Ô⇒
by the red line segment in the gure. Since x =
√
x2 + (y − 1)2 = 12 , x = 2y − y 2 gives the right semicircle of the circle x2 + (y − 1)2 = 1.
On the other hand, the y -integral goes from y = 0 to y = 2. Therefore R is right half
of the disk x2 +(y −1)2 ≤ 1. Note that the polar equation of the circle x2 +(y −1)2 = 1
is r = 2 sin θ , and to obtain the right semicircle we vary θ from θ = 0 to θ = π/2.
In the iterated integral the
x-integral
goes from
75
Hence:
2
√
2y−y 2
∫0 ∫0
xy
xy
dx dy = ∬ 2
dA
2
+y
R x + y2
π/2
2 sin θ r cos θ ⋅ r sin θ
=∫
r dr dθ
∫
r2
0
0
x2
=∫
=∫
=∫
π/2
2 sin θ
∫0
0
π/2
0
π/2
sin θ cos θ r dr dθ
sin θ cos θ [
r2 r=2 sin θ
]
dr dθ
2 r=0
2 sin3 θ cos θ dθ
0
π/2
sin4 θ
=
]
2 0
1
=
2
d.
Again we use the polar coordinates. This time the integration region is
{(x, y) ∶ x ≥ 0 and y ≥ 0},
∫0
∞
∫0
that is, the rst quadrant.
∞
dy dx
1
=
dA
∬
(x2 + y 2 )2 + 1
R (x2 + y 2 )2 + 1
π/2
∞ r dr dθ
=∫
∫
r4 + 1
0
0
76
R=
π/2
1
r=c
lim [arctan(r2 )]r=0 dθ
c→∞
2
0
π/2 π
=∫
dθ
4
0
π2
=
8
=∫
21.
Evaluate the double integral
Solution:
the line
Let
y=x
R′
1
∬R (x2 + y 2 )2 dA
where
R is the region shown in the gure.
be the portion of the region lying in the rst quadrant between
and the
x-axis.
By symmetry we have:
1
1
∬R (x2 + y 2 )2 dA = 8 ∬R′ (x2 + y 2 )2 dA
π/4
2 sec θ
1
= 8∫
r dr dθ
∫
(r2 )2
0
sec θ
π/4
1 r=2 sec θ
= 8∫
[− 2 ]
dθ
2r r=sec θ
0
= 3∫
π/4
cos2 θ dθ
0
1 + cos 2θ
dθ
2
0
3
sin 2θ π/4
= [θ +
]
2
2 0
3π 3
=
+
8 4
= 3∫
π/4
77
22.
r2 = 2 cos(2θ)
in the plane is the base of a solid right
√
2 − r2 . Find the cylinder's volume.
sphere z =
The region enclosed by the lemniscate
cylinder whose top is bounded by the
Solution:
Let
R be the the region enclosed by the lemniscate in the rst quadrant.
By symmetry the volume is
4∬
R
√
2 − r2 dA
= 4∫
√
2 cos 2θ
π/4
0
∫0
√
√
2 − r2 r dr dθ
r= 2 cos 2θ
1
4 π/4 3/2
[− (2 − r2 )3/2 ]
(2 − (2 − 2 cos 2θ)3/2 ) dθ
= 4∫
dθ = ∫
3
3 0
0
r=0
√
√
π/4
8 2
2 2π 32 π/4 3
(1 − (2 sin2 θ)3/2 ) dθ =
=
−
sin θ dθ
∫
3
3 √ 3 ∫0
0
√
2 2π 32 π/4
2 2π 32 1
2
=
−
(1 − cos2 θ) sin θ dθ =
−
√ (1 − u ) du
∫
∫
3
3 0
3
3 1/ 2
√
√
1
3
2 2π 32
u
2 2π 32 2
5
=
−
[u − ] √ =
−
( − √ )
3
3
3 1/ 2
3
3 3 6 2
√
√
2 2π 64 40 2
=
−
+
.
3
9
9
π/4
78
23.
Let
D be the region in space bounded by
y = x2 on the sides, and the xy -plane
the plane
y+z = 1
on the top, the parabolic
V of the
region D in terms of iterated integrals with orders of integration (a) dz dy dx and (b) dx dy dz .
cylinder
at the bottom. Express the volume
Solution:
79
1
∭D (x2 + y 2 + z 2 )2 dV where D is the region bounded by the
√
cylinder x2 + y 2 = 1 on the sides and by the hemisphere z =
4 − x2 − y 2 at the bottom. Express
24.
Consider the triple integral
this integral in terms of iterated integrals in
a.
Cartesian coordinates,
b.
cylindrical coordinates,
c.
spherical coordinates, and
d.
evaluate the integral in the coordinate system of your choice.
Solution:
x2 + y 2
We observe that the projection of
≤ 1.
D
to the
xy -plane
is the unit disk
A vertical line passing through a point of the unit disk enters the region
at a point on the hemisphere
and remains in the region from there on. The equations
√
√
of the hemisphere z =
4 − x2 − y 2 and the cylinder x2 + y 2 = 1 are z = 4 − r2 and
r = 1,
respectively, in cylindrical coordinates. The answers to parts
(a),
√
1−x2
∞
1
1
1
dV
=
√
∭D (x2 + y 2 + z 2 )2
∫−1 ∫−√1−x2 ∫ 4−x2 −y2 (x2 + y 2 + z 2 )2 dz dy dx ,
and
(b),
2π
1
∞
1
1
∭D (x2 + y 2 + z 2 )2 dV = ∫0 ∫0 ∫√4−r2 (r2 + z 2 )2 r dz dr dθ ,
immediately follow from these observations.
To do part
ˆ
ˆ
ˆ
(c) we further observe that
the equations of the hemisphere and the cylinder are
ρ=2
and
ρ sin ϕ = 1 ,
respectively,
a ray starting at the origin enters the region at a point on the hemisphere and
leaves the region at a point on the cylinder, and
such a ray intersects the region exactly when
through a point on the intersection circle when
0 ≤ ϕ ≤ π/6.
ϕ = π/6.)
(The ray passes
Therefore,
2π
π/6
csc ϕ 1
1
ρ2 sin ϕ dρ dϕ dθ .
∭D (x2 + y 2 + z 2 )2 dV = ∫0 ∫0 ∫2
4
ρ
80
We use this last iterated integral for the computation
(d):
2π
π/6
csc ϕ 1
1
sin ϕ dρ dϕ dθ
∭D (x2 + y 2 + z 2 )2 dV = ∫0 ∫0 ∫2
ρ2
2π
π/6
1 ρ=csc ϕ
=∫
[−
]
sin ϕ dϕ dθ
∫0
ρ ρ=2
0
2π
π/6
1
=∫
(− sin ϕ + ) sin ϕ dϕ dθ
∫
2
0
0
2π
π/6
1 − cos 2ϕ sin ϕ
=∫
+
) dϕ dθ
∫0 (−
2
2
0
2π
ϕ sin 2ϕ cos ϕ ϕ=π/6
=∫
[− +
−
]
dθ
2
4
2 ϕ=0
0
√
√
2π
3
3 1
π
−
+ ) dθ
=∫
(− +
12
8
4
2
0
√
π
3
= π (1 − −
)
6
4
81
25.
be the region in space bounded on the top by the sphere x2 + y 2 + z 2 = 2 and on the
bottom by the paraboloid z = x2 + y 2 . Express the volume V of D in terms of iterated integrals
Let
in the
D
(a) Cartesian, (b) cylindrical, and (c) spherical coordinates.
Solution:
Note that the projection of the curve of intersection of the sphere and
x2 + y 2 = 1 in the xy -plane. This curve also bounds
the paraboloid is the unit circle
the projection of the solid
D
V =∫
to the
Hence we have
√
2−x2 −y 2
√
1−x2
1
−1
xy -plane.
∫−√1−x2 ∫x2 +y2
and
V =∫
0
2π
1
dz dy dx
√
2−r2
∫0 ∫r 2
r dz dr dθ .
D through the sphere if the
angle it makes with the positive z -axis is less than π/4, whereas it does so through
the paraboloid if this angle is between π/4 and π/2. Hence:
Now observe that a ray starting at the origin leaves
V =∫
0
2π
√
2
π/4
∫0
∫0
ρ2 sin ϕ dρ dϕ dθ
+∫
2π
0
π/2
∫π/4 ∫0
82
cos ϕ/ sin2 ϕ
ρ2 sin ϕ dρ dϕ dθ
26.
Consider the iterated integral
π/2
∫0
2−r2
1
∫0 ∫r
dz dr dθ
in cylindrical coordinates.
a.
Change the order of integration into
b.
Express the integral in spherical coordinates with order of integration
Solution: a.
z = 2−
x2
−
y2.
z = r
is the cone
z 2 = x2 + y 2 ,
and
z = 2 − r2
dϕ dρ dθ .
is the paraboloid
These surfaces intersect along a circle that is also the curve of
intersection of the cylinder
0 ≤ θ ≤ π/2,
dr dz dθ.
r=1
and the horizontal plane
the integration region
D
z = 1.
Since
0≤r≤1
and
is the region in the rst octant bounded by
the paraboloid on the top and the cone at the bottom.
When we integrate with respect to
r
rst, we will be moving along a curve on
83
which
z
and
θ
are constant. These two conditions describe a horizontal plane and a
z -axis, respectively. Therefore the r-integration is along a ray
perpendicular to the z -axis like the blue ones in the gure. Such a ray starts on the
z -axis in D, and leaves D through the cone if 0 ≤ z ≤ 1, and through the paraboloid
if 1 ≤ z ≤ 2. Therefore,
plane containing the
π/2
∫0
2−r2
1
∫0 ∫r
dz dr dθ = ∫
π/2
1
z
∫0 ∫0 dr dz dθ
0
+∫
where we used the fact that
b.
We have
π/2
∫0
as
z = 2 − r2
dV = r dz dr dθ
(a).
and
2−r2
1
∫0 ∫r
ρ
and
θ
r ≥ 0 Ô⇒ r =
D
in cylindrical coordinates where
D
2
√
2−z
∫1 ∫0
0
dz dr dθ = ∭
Since we want to integrate with respect to
on which
π/2
√
dr dz dθ
2−z.
1
dV
r
is the region described in part
ϕ rst, we will be moving along the curves
are constant. These two conditions describe a sphere with center
z -axis, respectively. Therefore the
ϕ-integration takes place along a vertical semicircle subtended by a diameter along
the z -axis and with center at the origin like the green ones in the gure. Such a
semicircle starts on the positive z -axis in D , and leave the region of integration
√
√
intersecting the cone for 0 ≤ ρ ≤
2 and the paraboloid for 2 ≤ ρ ≤ 2. The upper
half of the cone z = r has the equation ϕ = π/4 in spherical coordinates. On the
√
2
2
other hand, z = 2 − r Ô⇒ ρ cos ϕ = 2 − (ρ sin ϕ) Ô⇒ cos ϕ = (1 +
4ρ2 − 7)/(2ρ) for
a point on the paraboloid as ϕ ≤ π/4 . Therefore,
at the origin and a half-plane whose spine is the
1
1
∭D r dV = ∭D ρ sin ϕ dV
=∫
√
2
π/2
0
∫0
π/4
∫0
+∫
where we substituted
ρ dϕ dρ dθ
π/2
0
2
∫√2 ∫0
dV = ρ2 sin ϕ dϕ dρ dθ
coordinates.
84
√
arccos((1+ 4ρ2 −7)/(2ρ))
ρ dϕ dρ dθ
for the volume element in spherical
27.
The region
about the
R
z -axis
to obtain a solid
iterated integrals in
Solution:
z 2 = y 2 − y 4 in the right half of the yz -plane is rotated
the xyz -space. Express the volume V of D in terms of
bounded by the curve
D
in
(a) the Cartesian, (b) the cylindrical and (c) the spherical coordinates.
Since the surface bounding the the solid
D
is obtained by revolving a
z -axis, its equation in the cylindrical coordinates will not depend
on θ . Since r = y in the yz -plane and the curve has the equation z 2 = y 2 − y 4 , we
conclude that z 2 = r 2 −r 4 is the equation of the surface in the cylindrical coordinates.
curve about the
From this
V =∫
0
follows as the projection of
D
2π
1
√
r2 −r4
∫0 ∫−√r2 −r4 r dz dr dθ
to the
xy -plane
is the unit disk.
The equation z 2 = r 2 − r 4 in the cylindrical coordinates transforms to z 2 = (x2 +
y 2 ) − (x2 + y 2 )2 in the Cartesian coordinates, and once again using the fact that the
projection of
D
to the
xy -plane
is the unit disk, one now obtains:
√
1−x2
1
√
(x2 +y 2 )−(x2 +y 2 )2
V =∫ ∫ √
∫−√(x2 +y2 )−(x2 +y2 )2 dz dy dx
−1 − 1−x2
The spherical coordinates require a little bit more work. First note that the equation
z 2 = r2 −√
r4 in the cylindrical coordinates now gives (ρ cos ϕ)2 = (ρ sin ϕ)2 − (ρ sin ϕ)4 ,
or ρ =
sin2 ϕ − cos2 ϕ/ sin2 ϕ, in the spherical coordinates. Next note that for
2
sin ϕ − cos2 ϕ = − cos 2ϕ to be nonnegative, ϕ must be between π/4 and 3π/4 in the
interval
[0, π].
Therefore:
V =∫
0
Remark:
2π
√
sin2 ϕ−cos2 ϕ/ sin2 ϕ
3π/4
∫π/4
∫0
Compare this example with
ρ2 sin ϕ dρ dϕ dθ
Example 45 in Part 1.
85
28.
0
Evaluate the iterated integral
2y+3
∫−1 ∫−y
x+y
ex−2y dx dy .
2
(x − 2y)
Solution:
29.
Let
(u, v)
and
Solution:
(x, y)
be two coordinate systems. Show that
By denition,
∂(x, y) xu xv
∣ = xu yv − xv yu
=∣
∂(u, v) yu yv
and
∂(u, v) ux uy
∣ = ux vy − uy vx .
=∣
∂(x, y) vx vy
86
∂(x, y) ∂(u, v)
⋅
= 1.
∂(u, v) ∂(x, y)
Therefore,
∂(x, y) ∂(u, v)
⋅
= (xu yv − xv yu )(ux vy − uy vx )
∂(u, v) ∂(x, y)
= xu ux yv vy + xv vx yu uy − xu uy yv vx − xv ux yu vy
= xu ux yv vy + xv vx yu uy − xu uy yv vx − xv ux yu vy
+ xv vx yv vy + xu ux yu uy − xv vx yv vy − xu ux yu uy
= (xu ux + xv vx ) (yu uy + yv vy )
− (xu uy + xv vy ) (yu ux + yv vx )
= xx yy − xy yx
=1⋅1−0⋅0
=1
30.
Evaluate the double integral
Solution:
∬R e
x2 /y
dA
where
R = {(x, y) ∶ x2 ≤ y ≤
√
x }.
u = y 2 /x, v = x2 /y . In this coordinate
G = {(u, v) ∶ 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1} .
We change the coordinates to
system the region of integration becomes
We have
1
∂(u, v) ux uy
∂(x, y)
1
−y 2 /x2 2y/x
∣=∣
=− ,
=∣
∣ = −3 Ô⇒
=
2
2
2x/y −x /y
∂(x, y) vx vy
∂(u, v) ∂(u, v)
3
∂(x, y)
and the change of variables formula gives
x2 /y
∬R e
dx dy = ∬ ev ∣
G
1
1
1
e−1
∂(x, y)
∣ du dv = ∫ ∫ ev ∣ − ∣ dv du =
.
∂(u, v)
3
3
0
0
87
Remark:
any
We can multiply dierentials. The rules are
uv -coordinate
du du = 0 = dv dv
and
dv du = −du dv
in
system. For instance, when changing from Cartesian to polar coordinates,
we have
dx dy = d(r cos θ) d(r sin θ)
= (cos θ dr − r sin θ dθ)(sin θ dr + r cos θ dθ)
= −r sin2 θ dθ dr + r cos2 θ dr dθ
= r(sin2 θ + cos2 θ) dr dθ
= r dr dθ
and in
Example 30 we have
du dv = d(y 2 /x) d(x2 /y)
= (2y/x dy − y 2 /x2 dx)(2x/y dx − x2 /y 2 dy)
= 4 dy dx + dx dy
= −4 dx dy + dx dy
= −3 dx dy ,
hence
1
dx dy = − du dv .
3
This can be used to keep track of how the area element changes under
a coordinate change, but note that the sign of the factor in front must be corrected by hand so
that it is positive on the region of integration.
31.
C
Consider the transformation
T ∶ x=
such that the inequality
u
v
,y=
u+v+1
u+v+1
. Show that there is a constant
∂(x, y)
∬G ∣ ∂(u, v) ∣ du dv ≤ C
holds for all regions
G
contained in the rst quadrant of the
uv -plane.
u
v
u+v
≥ 0, y =
≥ 0 and x + y =
≤1
u+v+1
u+v+1
u+v+1
x
y
for u ≥ 0 and v ≥ 0. We also have u =
and v =
. Therefore T maps
1−x−y
1−x−y
the rst quadrant of the uv -plane into the triangle R = {(x, y) ∶ x+y ≤ 1, x ≥ 0, y ≥ 0}
Solution:
Observe that
x=
in a one-to-one manner.
88
Hence:
∂(x, y)
∬G ∣ ∂(u, v) ∣ du dv = ∬T (G) dx dy = (Area
Remark:
of
T (G)) ≤ (Area
of
R) =
1
2
One can also do a straightforward computation.
1
∂(x, y) xu xv
(v + 1)/(u + v + 1)2
−u/(u + v + 1)2
=∣
∣=∣
∣=
2
2
−v/(u + v + 1)
(u + 1)/(u + v + 1)
∂(u, v) yu yv
(u + v + 1)3
and hence
∞
∞
du dv
∂(x, y)
1 ∞ dv
1
∣
∣
du
dv
≤
=
∬G ∂(u, v)
∫0 ∫0 (u + v + 1)3 2 ∫0 (v + 1)2 = 2 .
32.
Compute the Jacobian
∂(x, y, z)
∂(ρ, ϕ, θ)
where
(ρ, ϕ, θ) are the spherical coordinates and (x, y, z)
are the Cartesian coordinates.
Solution:
We have
x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ.
Therefore
RRx x x RR
∂(x, y, z) RRRR ρ ϕ θ RRRR
= RR yρ yϕ yθ RRR
RR
∂(ρ, ϕ, θ) RRRR
RR zρ zϕ zθ RRR
RRRsin ϕ cos θ ρ cos ϕ cos θ −ρ sin ϕ sin θRRR
RR
RR
= RRRR sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ RRRR
RRR
RRR
−ρ sin ϕ
0
RR cos ϕ
RR
sin ϕ sin θ ρ cos ϕ sin θ
= (−ρ sin ϕ sin θ) ∣
∣
cos ϕ
−ρ sin ϕ
− (ρ sin ϕ cos θ) ∣
sin ϕ cos θ ρ cos ϕ cos θ
∣
cos ϕ
−ρ sin ϕ
+0⋅∣
sin ϕ cos θ ρ cos ϕ cos θ
∣
sin ϕ sin θ ρ cos ϕ sin θ
= (−ρ sin ϕ sin θ)(−ρ sin θ)(sin2 ϕ + cos2 ϕ)
− (ρ sin ϕ cos θ)(−ρ cos θ)(sin2 ϕ + cos2 ϕ)
+0
2
2
2
= ρ sin ϕ (sin θ + cos θ)
= ρ2 sin ϕ
where we used the cofactor expansion with respect to the third column.
89
Part 3: Sequences and Series
1. Let a1 = 1, a2 = a3 = 2, a4 = a5 = a6 = 3, a7 = a8 = a9 = a10 = 4,
an ∶ 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, . . . . What is a2015 ?
and so on.
That is,
Solution:
an = k if 1 + 2 + ⋯ + (k − 1) < n ≤ 1 + 2 + ⋯ + k . In other words, an = k if
k(k − 1)/2 < n ≤ k(k + 1)/2. Since for k = 63, k(k − 1)/2 = 1953 and k(k + 1)/2 = 2016,
we have a2015 = 63.
Remark:
2.
A more explicit formula
√
8n + 1 − 1
an = ⌈
⌉
2
can be obtained.
Curve 1 is an equilateral triangle with unit sides. For
n ≥ 2, Curve n is obtained from Curve n−1
by replacing the middle third of every edge with the other two sides of the outward pointing
Ln be the
Ln , An , lim Ln
equilateral triangle sitting on it. Let
length of Curve n and
region enclosed by Curve n . Find
and
Solution:
Let
en
and
dn
n→∞
An
be the area of the
lim An .
n→∞
denote the number of edges and the length of each edge
en = 4en−1 and dn = dn−1 /3 for n ≥ 2, and e1 = 3
and d1 = 1, we nd that en = 3 ⋅ 4n−1 and dn = 1/3n−1 for n ≥ 1. It follows that
Ln = en dn = 3 ⋅ (4/3)n−1 for n ≥ 1, and lim Ln = lim 3 ⋅ (4/3)n−1 = ∞.
of Curve n , respectively.
Since
n→∞
On the other hand, the
region is obtained by adjoining
en−1
equilateral triangles
(n − 1) region. Therefore,
√
√
3 (dn−1 /3)2
3 1 2
An = An−1 + en−1 ⋅
= An−1 + 3 ⋅ 4n−2 ⋅
(
)
4
4 3n−1
of side length
dn−1 /3
nth
n→∞
and then
to the
st
1
4 n−2
4 n−3
An = √ (( ) + ( ) + ⋯ + 1) + A1
9
4 3 9
√
3
A1 =
, this gives
4
√
√
3 3
4 n−1
3
An =
(1 − ( ) ) +
20
9
4
for
n ≥ 2.
Since
for
n ≥ 1.
We obtain
√
√
√
3 3
3 2 3
lim An =
+
=
.
n→∞
20
4
5
90
Remark:
Curve 1 :
Curve 2 :
Curve 3 :
Curve 4 :
t = 0 sec, takes out 1 ball at t = 1/2 sec,
takes out 1 ball at t = 7/8 sec, puts 3 balls at t = 15/16 sec, takes out
A magician puts 1 ball into an empty box at
t = 3/4 sec,
t = 31/32 sec, and
box at t = 1 sec.
puts 2 balls at
1 ball at
in the
so on. Then she challenges you to guess how many balls there are
You ask for the advice of your friends.
Friend1 says: The net result of
2k + 1st
and
2k + 2nd
steps for
k≥1
is to put at
least one more ball into the box. There are innitely many balls in the box at
t=1
sec.
ball1, ball2, ball3, and so on
ball1 at t = 0
sec, takes out ball1 at t = 1/2 sec, puts ball2 and ball3 at t = 3/4 sec, takes out
ball2 at t = 7/8 sec, puts ball4, ball5, and ball6 at t = 15/16 sec, takes out ball3
at t = 31/32 sec, and so on. For any given k , ballk is not in the box at time t = 1
Friend2 says: Imagine that the balls are labeled as
with invisible ink which only the magician can see. Then she puts
sec, because it was taken out at time
box at time
t=1
t = 1 − 1/22k−1
sec.
What do you think?
91
sec. There are no balls in the
3.
Let the sequence
{an }
be dened by
a1 = 1
and
an =
sequence converges and nd its limit.
Solution:
for
n ≥ 1.
Show that the
0 < 1 + an < 1 + an+1 and an+1 =
1/(1 + an ) > 1/(1 + an+1 ) = an+2 . Similarly, if an > an+1 > 0, then 1 + an > 1 + an+1 > 0
and an+1 = 1/(1 + an ) < 1/(1 + an+1 ) = an+2 . Since a1 = 1 > 1/2 = a2 , it follows that
Observe that if
0 < an < an+1 ,
1
1 + an−1
then
1 = a1 > a3 > a5 > ⋯ > a6 > a4 > a2 =
1
.
2
∞
Therefore, the sequence {a2n }n=1 is increasing and bounded from above by 1, and
∞
the sequence {a2n−1 }n=1 is decreasing and bounded from below by 1/2. By the
Monotonic Sequence Theorem, then both of these sequences are convergent.
lim a2n−1 = L and lim a2n = M . Taking the limit of a2n+1 = 1/(1 + a2n ) as n → ∞
n→∞
n→∞
we obtain L = 1/(1 + M ), and taking the limit of a2n = 1/(1 + a2n−1 ) as n → ∞ we
Let
From M + M L = 1 and LM +
∞
Therefore the sequence {an }n=1 converges to L = M .
obtain
M = 1/(1 + L).
L=1
it follows that
M = L.
√
√
L2 + L − 1 = 0 gives L = ( 5 − 1)/2
√ or L = −( 5 + 1)/2. Since 0 < an
n ≥ 1, we must have 0 ≤ L . Hence L = ( 5 − 1)/2 is the limit and
√
5−1
lim an =
.
n→∞
2
Finally,
Remark:
This result is sometimes expressed symbolically in the form:
√
5−1
=
2
1
1
1+
1
1+
1
1+
1
1+
1+
Remark:
1
1+⋯
xn = Fn /Fn+1 for n ≥ √1 where Fn ∶ 1, 1, 2, 3, 5, 8, . . .
showed that lim Fn /Fn+1 = ( 5 − 1)/2 .
Note that
sequence. So we
Remark:
for all
is the Fibonacci
n→∞
It can be showed with a little bit more work that any sequence satisfying the given
√
√
(√5 − 1)/2 if a1 =/ −(1 + 5)/2 and a1 =/ −Fn+1 /Fn for any n ≥ 1 .
5)/2 , then the sequence is constant; and if a1 = −Fn+1 /Fn for
On the other hand, if a1 = −(1 +
some n ≥ 1, then an = −1 and an+1 is undened.
recursion relation converges to
92
4.
Let the sequence
{xn }
be dened by
x0 = 2
and
xn =
xn−1
1
+
2
xn−1
a.
Find the limit of this sequence assuming it exists.
b.
Show that the limit exists.
Solution: a.
xn =
We assume that the limit
xn−1
1
+
2
xn−1
L = lim xn
n ≥ 1 Ô⇒ xn xn−1 =
for
for
n ≥ 1.
exists. Then
x2n−1
+1
2
for
n≥1
x2n−1
L2
+ 1 Ô⇒ L2 =
+ 1 Ô⇒ L2 = 2
2
2
√
√
L = 2 or L = − 2.
Ô⇒ lim(xn xn−1 ) = lim
as
lim xn−1 = lim xn = L .
Therefore
x0 = 2 > 0 and if xn > 0 then xn+1 = xn /2 + 1/xn > 0 + 0 = 0. Hence by induction
√ xn > 0
for all n ≥ 0. It follows that L = lim xn ≥ 0. We conclude that the limit is
2.
b.
In part
(a) we proved that xn > 0 for all n ≥ 0.
Therefore
{xn } is bounded from
below.
Now we will show by induction on
ˆ
ˆ
Let
n = 0.
Let
n>0
n
that
√
2 < xn+1 < xn
x0 = 2, x1 = 3/2, and x21 = 9/4 > 2,
√
assume that
2 < xn+1 < xn . Then
Since
and
xn+1 − xn+2 =
and hence
xn+2 < xn+1 .
and since
xn+2 > 0,
we have
n ≥ 0.
√
2 < x1 < x0 .
x2 − 2
xn+1
1
−
= n+1
>0
2
xn+1
2xn+1
On the other hand,
2
x2n+2
for all
2
x2 − 2
xn+1
1
−2=(
+
) − 2 = ( n+1
) >0
2
xn+1
2xn+1
we have
xn+2 >
√
2.
Therefore the sequence is decreasing.
Since the sequence is bounded from below and decreasing, we conclude that it
converges by the Monotonic Sequence Theorem.
93
5.
For each of the series in
the
nth
partial sum
a.
sn
(a-d), determine whether there exists a positive integer n such that
of the series satises the condition
∞
1
∑ n
n=1 2
b.
∞
c.
∑5
n
n=1
Solution:
2014 ≤ sn ≤ 2015.
n
∞
2999
)
∑(
n=1 3000
d.
∞
1
n=1 n
∑
Note that all terms of these series are positive, and therefore, their partial
sums form increaisng sequences. This will be used repeatedly in the following.
a.
We have
∞
1
1
1/2
= 1 < 2014
<
=
∑
k
n
1 − 1/2
n=1 2
k=1 2
n
sn = ∑
for all
n ≥ 1,
where we used the geometric series sum formula, and hence there are
no partial sums lying between 2014 and 2015.
b.
This time we have
n
4
k=1
k=1
sn = ∑ 5k ≤ ∑ 5k = 5 + 25 + 125 + 625 = 780 < 2014
for
n ≤ 4,
and
n
5
k=1
k=1
sn = ∑ 5k ≥ ∑ 5k = 5 + 25 + 125 + 625 + 3125 = 3905 > 2015
for
n ≥ 5.
c.
Every term of this series is between 0 and 1, hence its partial sums increase in
Therefore no
sn
lies between 2014 and 2015.
steps smaller than 1 starting with
∞
s1 = 2999/3000 < 2014.
Moreover,
n
2999
2999/3000
) =
= 2999 > 2015
∑(
1 − 2999/3000
n=1 3000
by the geometric series sum formula.
satisfying the condition
d.
Hence there is at least one partial sum
2014 ≤ sn ≤ 2015.
The same reasoning as in part
sn
(c) works, this time with the observation that
∞
1
= ∞ > 2015 ,
n=1 n
∑
to give the existence of at least one partial sum
sn ≤ 2015.
94
sn
satisfying the condition
2014 ≤
6.
Determine whether each of the following series converges or diverges.
∞
a.
d.
g.
∑ (21/n − 21/(n+1) )
n=1
∞
nn
n=1 n!
∑
∞
1
n=3 n ln n ln(ln n)
Hence
b.
We have
lim sn = 1 .
n→∞
n ln n
n
n=2 3
e.
1
√
n sin n
n=1 n +
∞
sin n
∑ 2
n=1 n
c.
∑
∞
f.
∑
h.
∑
Solution: a.
∞
b.
n
n
k=1
k=1
∞
π
∑ (−1)n+1 cos ( )
n
n=1
∞ n
n
5 −2
∑ n
n
n=1 7 − 6
n
sn = ∑ (21/k − 21/(k+1) ) = ∑ 21/k − ∑ 21/(k+1) = 2 − 21/(n+1) .
Therefore the series converges and
∞
k=1
∑ (21/n − 21/(n+1) ) = 1 .
n=1
We have
an = n ln n/3n
and
∣an+1 ∣
(n + 1) ln(n + 1)/3n+1 1
n+1
ln(n + 1)
= lim
=
⋅
lim
⋅
lim
n→∞ ∣an ∣
n→∞
n ln n/3n
3 n→∞ n n→∞ ln n
ρ = lim
1
ln(x + 1) ↓ 1
1/(x + 1) 1
1
= ⋅ lim
⋅ 1 ⋅ lim
= ⋅1= .
x→∞
3
ln x
3 x→∞ 1/x
3
3
L'H
=
Since
c.
ρ = 1/3 < 1 ,
∞
the series
n ln n
n
n=2 3
∑
an = (−1)n+1 cos(π/n),
lim ∣an ∣ = lim cos(π/n) = cos 0 = 1 .
n→∞
n→∞
by the nth Term Test.
d.
Since
We have
an = nn /n!
converges by the Ratio Test.
we
have
Therefore
∣an ∣ = cos(π/n)
lim an =/ 0 , and
n→∞
for
n ≥ 2,
and
the series diverges
and
n
∣an+1 ∣
(n + 1)n+1 /(n + 1)!
1
ρ = lim
= lim
= lim (1 + ) = e .
n
n→∞ ∣an ∣
n→∞
n→∞
n /n!
n
Since
e.
∞
the series
nn
n=1 n!
∑
diverges by the Ratio Test.
We observe that
Since
√
1/(n + n sin n)
L = lim
= lim
n→∞
n→∞
1/n
0<L<∞
1
√
n sin n
n=1 n +
∑
∞
ρ = e > 1,
1
1
=
=1.
sin n 1 + 0
1+ √
n
∞
and the harmonic series
1
n=1 n
∑
diverges, we conclude that the series
diverges by the Limit Comparison Test.
95
∞
f.
∑ (5/7)n
The geometric series
have
L = lim
series
n→∞
∞ n
5 − 2n
∑
n=1
g.
Since
on
n=1
(5n − 2n )/(7n − 6n )
7n − 6n
ln x
(1, ∞).
5n /7n
converges as
r = 5/7 Ô⇒ ∣r∣ = 5/7 < 1.
1 − (2/5)n 1 − 0
=
= 1.
n→∞ 1 − (6/7)n
1−0
= lim
Since
We also
0 < L < ∞,
the
converges by the Limit Comparison Test.
(0, ∞) , ln(ln x) is an increasing function
positive on [3, ∞) . Therefore x ln x ln(ln x) is
is an increasing function on
Both
ln x
ln(ln x) are
[3, ∞) as it is the product of three positive and increasing
implies that 1/(x ln x ln(ln x)) is a positive and decreasing
and
positive and increasing on
functions. This in turn
function on
[3, ∞) .
Since it is also continuous, we can apply the Integral Test. The
improper integral
∫3
∞
∞
dx
du
=∫
x ln x ln(ln x)
ln(ln 3) u
diverges, where we used the change of variable
∞
the series
h.
1
∑
n=3 n ln n ln(ln n)
Hence
diverges.
∞
Consider the series
u = ln(lnx), du = dx/(x ln x) .
1 + sin n
.
n2
n=1
∑
Since
0 ≤ 1 + sin n ≤ 2
for
n ≥ 1,
we have
∞
1 + sin n
2
1
≤
for n ≥ 1 . The p-series ∑
converges as p = 2 > 1 . Therefore
2
2
2
n
n
n=1 n
∞
∞
∞
2
1
1 + sin n
the series ∑
=
2
converges. Then the series ∑
converges by the
∑
2
2
n2
n=1 n
n=1 n
n=1
∞
∞
1 + sin n
1
Direct Comparison Test. Since both ∑
and ∑
converge, the series
2
2
n
n=1
n=1 n
∞
sin n
∑ 2 , which is their dierence, also converges.
n=1 n
0≤
Remark:
Other tests can be used too. Here are some examples:
∞
1
2
n=1 n
In part
(a), the Limit Comparison Test with the series
In part
(b), the Limit Comparison Test with the geometric series
In part
(d), the Root Test also works where we use the fact that n→∞
lim
96
∑
also works.
∞
1
n
n=1 2
∑
also works.
1/n
(n!)
n
=
1
e
.
Remark:
The following was a bonus problem on Moodle in Spring 2010 Math 102 course.
Problem K: In class we showed that 0.12 = 0.121212. . .
= 12/99 as an application
of the geometric series.
x = 0.121212 . . . , then 100x =
therefore x = 12/99 . Of course
We learn the following trick in elementary school: If
12.121212 . . . ;
their dierence gives
99x = 12
and
when we see this in elementary school, no one talks about convergence.
That is what we are going to do in this problem. We will forget about convergence
as we know it, and take a trip to Tersonia.
In Tersonia after they teach the students about integers and rational numbers,
they come to the decimal representations of numbers. As you know our decimal
expansions have the form
±dn dn−1 . . . d2 d1 d0 .d−1 d−2 . . .
where each
di
is in
{0, 1, . . . , 9}.
We can have innitely many nonzero digits after
the decimal point, but we must have only nitely many nonzero digits before the
decimal point. In Tersonia they do just the opposite. Their decimal expansions
have the form
. . . t3 t2 t1 t0 .t−1 t−2 . . . t−n
where each
ti
is in
{0, 1, . . . , 9}.
Note that there is no minus sign. They can have
innitely many nonzero digits before the decimal point, but they can only have
nitely many digits after the decimal point.
Take a few minutes to convince yourself that Tersonians can add and multiply
their decimal expansions just like we do.
Why no minus sign? Well, because Tersonians don't need it. Negative numbers
are already there.
Then
y = 12.0 = . . . 121212.0 .
y = −12/99 . So in fact 12.0
For instance, consider the number
100y = . . . 121200.0
and
−99y = 12 .
Therefore
is a negative number.
Here are some problems from Tersonian Elementary School Mathematics Book :
Part
(e)
a.
1
=?
2
e.
Find two nonzero numbers
b.
1
=?
3
A
and
c.
1
=?
7
B
such that
was later turned into a programming challenge.
d.
A B = 0.
A Java applet that computes
last n digits of A and B when their last digits are
http://www.fen.bilkent.edu.tr/otekman/calc2/ters.html .
the
97
−1 = ?
given
can
be
found
at
7.
∞
Determine whether the sum of the series
Solution:
Let
bn =
4n
n!(n + 1)!
for
(−4)n
n=0 n!(n + 1)!
∑
n ≥ 0.
is positive or negative.
Then:
ˆ b >0 n≥0
ˆ lim b = lim n!(n4+ 1)! = lim 4n! ⋅ lim (n +1 1)! = 0
for all
n
.
n
n
n→∞
n→∞
n
n→∞
n→∞
as both limits are 0, the rst
one being one of the Useful Limits .
ˆb
n ≥ bn+1 ⇐⇒
4n
4n+1
≥
⇐⇒ (n + 2)(n + 1) ≥ 4 ⇐⇒ n ≥ 1.
n!(n + 1)! (n + 1)!(n + 2)!
Therefore the series satises the conditions of the Alternating Series Test.
In
particular, it converges. Moreover,
∞
(−4)n
4
42
43
44
4 4 4
=1−
+
−
+
−⋯=1−2+ − +
−⋯
1!2! 2!3! 3!4! 4!5!
3 9 45
n=0 n!(n + 1)!
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ¯
S=∑
s3 =−1/9
and, by the Alternating Series Estimate,
1/45,
Remark:
and
S
In other words,
∣S − (−1/9)∣ <
is negative.
∞
(−4)n
n=0 n!(n + 1)!
∑
∣S − s3 ∣ < b4 .
b4
is
1
J1 (4),
2
∞
(−1)n (x/2)2n+1
is the
n!(n + 1)!
n=0
approximately −0.03302166401.
where
the rst kind of order 1, and its value is
J1 (x) = ∑
98
Bessel function of
8.
∞
Find the radius of convergence of the power series
∑ (−1)n
n=0
Solution:
xn
.
(2n + 1)(n2 + 1)
We will give two dierent solutions.
Solution 1 : Here we will use the Ratio Test.
We have
an = (−1)n
xn
(2n + 1)(n2 + 1)
ρ = lim
n→∞
and
∣an+1 ∣
∣an ∣
∣(−1)n+1
= lim
n→∞
xn+1
∣
(2n+1 + 1)((n + 1)2 + 1)
xn
∣
(2n + 1)(n2 + 1)
∣(−1)n
∣x∣n+1
(2n+1 + 1)((n + 1)2 + 1)
= lim
n→∞
∣x∣n
(2n + 1)(n2 + 1)
1 + 2−n
n2 + 1
= lim (
⋅
)∣x∣
n→∞ 2 + 2−n (n + 1)2 + 1
∣x∣
=
.
2
If
∣x∣ < 2 ,
then
ρ = ∣x∣/2 < 1
Test. On the other hand, if
and the power series converges absolutely by the Ratio
∣x∣ > 2 ,
then
ρ = ∣x∣/2 > 1
and the power series diverges
by the Ratio Test.
It follows by the denition of the radius of convergence that
R = 2.
Solution 3 : Here we will use the radius of convergence formulas.
We have
cn =
(−1)n
,
(2n + 1)(n2 + 1)
and the radius of convergence formula gives
∣cn+1 ∣
1/((2n+1 + 1)((n + 1)2 + 1))
1
= lim
= lim
n→∞
R n→∞ ∣cn ∣
1/((2n + 1)(n2 + 1))
n2 + 1
1 + 2−n
1
⋅
= lim (
)
=
.
n→∞ 2 + 2−n (n + 1)2 + 1
2
Therefore
R = 2.
99
9.
∞
∑ (−1)n
Consider the power series
n=0
a.
x2n+1
(2n + 1)(n2 + 1)
.
Find the radius of convergence of the power series.
b.
Determine whether the power series converges or diverges at the right endpoint of its
interval of convergence.
c.
Determine whether the power series converges or diverges at the left endpoint of its
interval of convergence.
Solution: a.
We have
an = (−1)n
x2n+1
(2n + 1)(n2 + 1)
and
ρ = lim ∣an ∣1/n
n→∞
1/n
∣x∣2n+1
= lim ( n
)
n→∞ (2 + 1)(n2 + 1)
∣x∣2 ⋅ ∣x∣1/n
= lim
n→∞ 2 ⋅ (1 + 2−n )1/n ⋅ n1/n ⋅ (1 + n−2 )1/n
2
=
If
∣x∣ <
√
2,
Root Test;
Root Test.
b.
At
x=
∣x∣
.
2
2
ρ = ∣x∣
√ /2 < 1 and the power series
and if ∣x∣ >
2√
, then ρ = ∣x∣2 /2 > 1 and the
Therefore R =
2.
then
√
2
converges absolutely by the
power series diverges by the
we have
√
∞
2n+1
x
(
2)2n+1
n
=
(−1)
∑ (−1)n n
∑
(2 + 1)(n2 + 1) n=0
(2n + 1)(n2 + 1)
n=0
√ ∞
2n
= 2 ∑ (−1)n n
.
(2 + 1)(n2 + 1)
n=0
∞
∞
Consider
the
corresponding
absolute
value
series
2n
.
Since
n
2
n=0 (2 + 1)(n + 1)
∞
1
p-series ∑ 2 with p = 2 > 1
n=1 n
∑
2n
1
for all n ≥ 1 and the
<
(2n + 1)(n2 + 1) n2
∞
2n
converges, ∑
converges by the Direct Comparison Test; and then
n
2
n=0 (2 + 1)(n
√ + 1)
the power series at x =
2 converges absolutely by the Absolute Convergence Test.
0<
c.
At
√
x=− 2
we have
√
∞
x2n+1
(− 2)2n+1
n
= ∑ (−1) n
∑ (−1) n
(2 + 1)(n2 + 1) n=0
(2 + 1)(n2 + 1)
n=0
√ ∞
2n
= 2 ∑ (−1)n+1 n
.
(2 + 1)(n2 + 1)
n=0
∞
n
100
This is just
−1 times the series we considered in part (b), and therefore it converges
absolutely.
10.
Determine the radius of convergence and the interval of convergence of the power series
∞
xn
.
n+1
n=0 3n + (−1)
∑
Also determine the type of convergence at each point of the interval of convergence.
Solution:
As
cn =
1
3n + (−1)n+1
, the formulas for the radius of convergence gives
∣cn+1 ∣
∣1/(3(n + 1) + (−1)n+2 )∣
3 + (−1)n+1 /n
1
= lim
= lim
=
lim
= 1.
R
∣cn ∣
∣1/(3n + (−1)n+1 )∣
3 + 3/n + (−1)n+2 /n
Therefore
∣x∣ > 1.
At
x=1
R = 1.
Hence we have absolute convergence for
we have
∣x∣ < 1
and divergence for
∞
∞
xn
1
=
.
∑
n+1
n+1
n=0 3n + (−1)
n=0 3n + (−1)
∑
Since
1/(3n + (−1)n+1 )
1
1
= lim
=
n→∞
n→∞ 3 + (−1)n+1 /n
1/n
3
L = lim
∞
is a positive real number and the harmonic series
diverges by the Limit Comparison Test at
At
x = −1
we have
x = 1.
1
n=1 n
∑
diverges, the power series
∞
∞
xn
(−1)n
=
.
∑
n+1
n+1
n=0 3n + (−1)
n=0 3n + (−1)
∑
Let
i.
ii.
iii.
un =
1
3n + (−1)n+1
.
1
> 0 for all n ≥ 1 .
3n + (−1)n+1
0 < 3n + (−1)n+1 ≤ 3n + 1 < 3n + 2 ≤ 3(n + 1) + (−1)n+2
un > un+1 for n ≥ 1 .
1
= 0.
lim un = lim
n→∞
n→∞ 3n + (−1)n+1
un =
for all
n ≥ 1.
Hence
∞
It follows that the series
(−1)n
n+1
n=0 3n + (−1)
∑
converges by the Alternating Series Test.
Its absolute value series is the same as the series
∞
1
n+1
n=1 3n + (−1)
1+ ∑
and we showed
that this series diverges. Hence the power series converges conditionally at
101
x = −1 .
R = 1,
To summarize, the radius of convergence is
is
[−1, 1),
converges conditionally at
11.
the interval of convergence
the power series converges absolutely at every point of
x = −1 .
(−1, 1),
and it
∞
Consider the power series
a.
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
x .
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
f (x) = 1 + ∑
Show that the radius of convergence of the power series is
b.
R = 1.
Determine the behavior of this power series at the endpoints of its interval of
convergence.
c.
Show that
d.
Solve this dierential equation to show that
e.
Show that
2(1 − x)f ′ (x) = f (x)
for
∣x∣ < 1.
f (x) = √
1
1−x
for
∣x∣ < 1.
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) x2n+1
⋅
2n + 1
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
arcsin x = x + ∑
for
∣x∣ < 1.
Solution: a.
We have
cn =
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
for
n ≥ 1.
We use the radius of
convergence formula
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)(2n + 1)
1
∣cn+1 ∣
2n + 1
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)(2n + 2)
= lim
= lim
= lim
=1
n→∞
n→∞ 2n + 2
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)
R n→∞ ∣cn ∣
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
to obtain
b.
At
R = 1.
x=1
∞
we have the series
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
1+ ∑
. Since
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) 3 5
2n − 1 1
1
= ⋅ ⋅⋯⋅
⋅
>
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
2 4
2n − 2 2n 2n
∞
for
1
n=1 n
n > 1 and the harmonic series ∑
diverges, we conclude that the series at
x=1
diverges by the Direct Comparison Test.
At
x = −1
un =
∞
we obtain the alternating series
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)
.
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
1 + ∑ (−1)n
n=1
We have
(i ) un > 0
102
for all
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1)
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
n ≥ 0,
and we also have
with
(ii )
un > un+1
for all
n≥0
as
un /un+1 = (2n + 2)/(2n + 1) > 1
for
n ≥ 0.
On the other
hand,
12 ⋅ 32 ⋅ 52 ⋅ ⋯ ⋅ (2n − 1)2 1 ⋅ 3 3 ⋅ 5
(2n − 3)(2n − 1) 2n − 1 1
1
= 2 ⋅ 2 ⋅⋯⋅
⋅
⋅
<
2
2
2
2
2
2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
2
4
(2n − 2)
2n
2n 2n
√
for n > 1. Therefore 0 < un < 1/ 2n for n > 1, and the Sandwich Theorem gives
(iii ) lim un = 0 . We conclude that the series at x = −1 converges by the Alternating
u2n =
n→∞
Series Test. The convergence is conditional as we have already seen that the absolute
value series diverges.
c.
For
∣x∣ < 1
we have
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
x .
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
f (x) = 1 + ∑
Dierentiating this we get
f ′ (x) =
for
∣x∣ < 1 .
1 ∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n−1
+∑
x
2 n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n − 2) ⋅ 2
Therefore
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
x
n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n − 2)
2xf ′ (x) = x + ∑
and
2f ′ (x) = 1 +
for
∣x∣ < 1 .
∞
3
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n + 1) n
x+ ∑
x
2
n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
Taking the dierence of these two, we obtain
∞
1
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) 2n + 1
x+ ∑
⋅(
− 1) xn
2
2n
n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n − 2)
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) 1 n
1
=1+ x+ ∑
⋅
x
2
n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n − 2) 2n
∞
1
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
x
=1+ x+ ∑
2
n=2 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
=1+ ∑
x
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
= f (x)
2(1 − x)f ′ (x) = 1 +
for
∣x∣ < 1.
103
d.
For
∣x∣ < 1,
f ′ (x)
1
=
f (x) 2(1 − x)
1
Ô⇒ ln ∣f (x)∣ = − ln ∣1 − x∣ + C for some constant C
2
A
Ô⇒ f (x) = √
for some constant A .
1−x
2(1 − x)f ′ (x) = f (x) Ô⇒
Now substituting
e.
x=0
gives
We have
√
for
∣x∣ < 1 .
Substituting
√
for
∣x2 ∣ < 1 .
1 = f (0) = A.
Therefore,
f (x) = √
1
1−x
for
∣x∣ < 1.
∞
1
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
=1+ ∑
x
1−x
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
x2
for
1
1 − x2
x
gives
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) 2n
x
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
=1+ ∑
Integrating this we obtain
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) x2n+1
⋅
+C
2n + 1
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
arcsin x = x + ∑
for
∣x∣ < 1 .
Substituting
x=0
gives
0 = arcsin 0 = C ,
and hence
C = 0.
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) x2n+1
⋅
2n + 1
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
arcsin x = x + ∑
for
∣x∣ < 1 .
Remark:
It can be shown that
√
for
−1 ≤ x < 1 ,
and
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) n
1
=1+ ∑
x
1−x
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
∞
1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ (2n − 1) x2n+1
⋅
2n + 1
n=1 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ (2n)
arcsin x = x + ∑
for
∣x∣ ≤ 1 .
104
Thus
12.
∞
Consider the power series
n(2x − 1)3n+1
5n
n=1
∑
.
a.
Find the radius of convergence of the power series.
b.
Find the sum of the power series explicitly.
Solution: a.
n(2x − 1)3n+1
5n
an =
gives
∣an+1 ∣
∣an ∣
∣(n + 1)(2x − 1)3(n+1)+1 /5n+1 ∣
= lim
n→∞
∣n(2x − 1)3n+1 /5n ∣
n + 1 ∣2x − 1∣3
= lim
⋅
n→∞ n
5
∣2x − 1∣3
=
.
5
ρ = lim
n→∞
ρ < 1 and the series converges by the Ratio Test; and
ρ > 1√ and the series diverges by the Ratio Test. Since
3
1
5
∣2x − 1∣3 /5 < 1 ⇐⇒ ∣x − ∣ <
, it follows that the radius of convergence of the
2
2
√
3
5/2 .
power series is R =
If
if
∣2x − 1∣3 /5 < 1 ,
∣2x − 1∣3 /5 > 1 ,
b.
then
then
We know that
∞
∑ xn =
n=0
for
∣x∣ < 1 .
Dierentiating this we obtain
∞
∑ nxn−1 =
n=1
for
∣x∣ < 1 .
1
1−x
Now we replace
x
with
1
(1 − x)2
(2x − 1)3 /5
to get
∞
n(2x − 1)3(n−1)
=
5n−1
n=1
∑
for
∣(2x − 1)3 /5∣ < 1 .
∞
for
∣x − 1/2∣ <
√
3
2
(2x − 1)3
(1 −
)
5
Finally we multiply by
n(2x − 1)3n+1
=
5n
n=1
∑
1
(2x − 1)4 /5
(2x − 1)4 /5
2
(1 −
(2x − 1)3
)
5
5/2 .
105
=
to obtain
(2x − 1)4
5
⋅
4 (4x3 − 6x2 + 3x − 3)2
13.
Consider the function dened by:
∞
xn
n=0
n! 2n(n−1)/2
f (x) = ∑
a.
Find the domain of
b.
Evaluate the limit
c.
Show that
d.
Show that
f.
f (x) − ex
x→0 1 − cos x
3
f (2) < e + .
2
lim
.
f (−2) < 0 .
Solution: a.
Since
cn =
1
n! 2n(n−1)/2
, the radius of convergence formula gives
1
∣cn+1 ∣
1/((n + 1)! 2(n+1)n/2 )
1
= lim
= lim
= lim
=0.
n(n−1)/2
n→∞
n→∞ (n + 1)2n
R n→∞ ∣cn ∣
1/(n! 2
)
Therefore
b.
R = ∞.
This means that the domain of
Using
∞
xn
n=0
n! 2n(n−1)/2
f (x) = ∑
=1+x+
f
is
(−∞, ∞) .
x2
x3
+ 3
+⋯
2 ⋅ 2! 2 ⋅ 3!
we obtain
f (x) − ex
= lim
x→0 1 − cos x
x→0
lim
c.
(1 + x +
x2
x3
x2 x3
+ 3
+ ⋯) − (1 + x +
+
+ ⋯)
2 ⋅ 2! 2 ⋅ 3!
2! 3!
x2 x4
+
− ⋯)
2! 4!
1
1 7
7 3
− x2 −
− −
x −⋯
x−⋯
4
48
4
48
= lim
= lim
1 4
1 2
x→0 1
x→0 1
x2 −
x +⋯
−
x +⋯
2
24
2 24
1
−
1
= 4 =− .
1
2
2
1 − (1 −
We have
∞
2n
n=0
n! 2n(n−1)/2
f (2) = ∑
On the other hand,
∞
1
n=0
n! 2n(n−3)/2
=∑
∞
=1+2+
2 1 ∞
1
.
+ +∑
n(n−3)/2
2! 3! n=4 n! 2
1 1 ∞ 1
1
=1+1+ + + ∑
.
2! 3! n=4 n!
n=0 n!
e=∑
106
Since
n(n − 3) > 0
∞
1
n=4
n! 2n(n−3)/2
∑
d.
for
n ≥ 4,
∞
1
n=4 n!
<∑
. Therefore,
we have
1
n! 2n(n−3)/2
3
f (2) − e < .
2
<
1
n!
for
n ≥ 4.
Hence
This time we have
∞
(−2)n
n(n−1)/2
n=0 n! 2
∞
(−1)n
=∑
n(n−3)/2
n=0 n! 2
2 ∞
(−1)n
=1−2+ + ∑
2! n=3 n! 2n(n−3)/2
∞
(−1)n
=∑
n(n−3)/2
n=3 n! 2
1
1
1
1
=− + 2
− 5
+ 9
−⋯.
3! 2 ⋅ 4! 2 ⋅ 5! 2 ⋅ 6!
f (−2) = ∑
As
(n + 1)2n−1 > 1 for n ≥ 3 , we have −
too. Therefore,
14.
2
Estimate
f (−2) < −
−x
∫0 e dx
Solution:
2
1
n! 2n(n−3)/2
1
1
5
+ 2
= − < 0.
3! 2 ⋅ 4!
32
with error less than
We have
ex = 1 + x +
for all
x.
2
x,
∫0
i.
ii.
iii.
un =
< 0 for n ≥ 3
0.01.
x2
xn
+⋯+
+⋯
2!
n!
x4
x2n
+ ⋯ + (−1)n
+⋯
2!
n!
and integration gives
2
Let
1
(n + 1)! 2(n+1)(n−2)/2
Therefore,
e−x = 1 − x2 +
for all
+
e−x dx = 2 −
22n+1
.
n! (2n + 1)
2
23 25
22n+1
+
+ ⋯ + (−1)n
+⋯.
3 2! 5
n! (2n + 1)
Then:
22n+1
> 0 for all n ≥ 0 .
n! (2n + 1)
2n2 − 3n − 1 > 0 for n ≥ 2 Ô⇒ (n + 1)(2n + 1) > 22 (2n + 1)
un > un+1 for n ≥ 2 .
4n
2
lim un = lim ( ⋅
) = 0.
n→∞
n→∞ n! 2n + 1
un =
107
for
n ≥ 2.
Hence
The series satises the conditions of Alternating Series Test for
11!) = 32768/3586275 ≈ 0.009 < 0.01,
n ≥ 2.
Since
it follows by the Alternating Series Estimate
that the sum
2
∫0
e−x dx ≈ 2 −
2
2
approximates
15.
219
221
12223758182
23 25
+
+⋯−
+
=
≈ 0.89
3 2! 5
9! 19 10! 21 13749310575
−x
∫0 e dx
2
with error less than
0.01.
∞
Find the exact value of
Solution:
1
n
n=0 4 (2n + 1)
∑
.
On one hand we have
∞
∞
1
(1/4)n ∞ (1/2)2n
(1/2)2n+1
=
=
=
2
∑
∑
∑
n
n=0 2n + 1
n=0 2n + 1
n=0 2n + 1
n=0 4 (2n + 1)
3
5
(1/2)
(1/2)
= 2 (1/2 +
+
+ ⋯) .
3
5
∞
∑
On the other hand we have
∞
ln(1 + x) = ∑ (−1)n+1
n=1
for
−1 < x ≤ 1 .
In particular,
x = 1/2
xn
x2 x3 x4 x5
=x−
+
−
+
−⋯
n
2
3
4
5
gives
3
1
(1/2)2 (1/2)3 (1/2)4 (1/2)5
ln ( ) = ln (1 + ) = 1/2 −
+
−
+
−⋯,
2
2
2
3
4
5
and
x = −1/2
gives
1
1
(1/2)2 (1/2)3 (1/2)4 (1/2)5
ln ( ) = ln (1 − ) = −1/2 −
−
−
−
−⋯.
2
2
2
3
4
5
Therefore
∞
1
n=0
4n (2n + 1)
∑
223 /(23⋅
3
1
= ln ( ) − ln ( ) = ln 3 .
2
2
108
Part 4: Vector Analysis
1.
Find the value of the line integral
‰
(3x2 y 2 + y) dx + 2x3 y dy
C
where
C
is the cardioid
Solution:
r = 1 + cos θ
We use the Green's Theorem
‰
C
where
R
parameterized counterclockwise.
M dx + N dy = ∬ (
R
∂N ∂M
−
) dA
∂x
∂y
is the region enclosed by the simple closed curve
C.
Therefore
M
M
N
N
‰ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·
¹¹¹¹¹¹¹¹¹¹¹¹ µ
³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ
¬
¬
∂
∂
(3x2 y 2 + y) dx + 2x3 y dy = ∬ ( ( 2x3 y ) − (3x2 y 2 + y)) dA
∂y
R ∂x
C
= ∬ (6x2 y − (6x2 y + 1)) dA
R
= − ∬ dA
= −∫
= −∫
R
2π
1+cos θ
∫0
0
2π
0
[
r dr dθ
r2 r=1+cos θ
]
dθ
2 r=0
1 2π
= − ∫ (1 + 2 cos θ + cos2 θ) dθ
2 0
1 2π
1 + cos 2θ
=− ∫
(1 + 2 cos θ +
) dθ
2 0
2
1 3
= − ⋅ ⋅ 2π
2 2
3π
=− .
2
Remark We used the Circulation-Curl Form
of the Green's Theorem, but the computation is
exactly the same with the Flux-Divergence Form :
‰
C
M dy − N dx = ∬ (
R
109
∂M ∂N
+
) dA
∂x
∂y
In fact, when expressed in terms of components and coordinates, both forms of the Green's
Theorem can be summarized and most easily remembered as
‰
C
where
ω = A dx + B dy
Remark following Example 30 in Part 2.
Indeed, if
∂M
∂M
∂N
∂N
dx +
dy) dx + (
dx +
dy) dy
∂x
∂y
∂x
∂y
ω = M dy − N dx,
then
dω = dM dy − dN dx = (
=
∂M
∂M
∂N
∂N
dx +
dy) dy − (
dx +
dy) dx
∂x
∂y
∂x
∂y
∂N
∂M
∂N
∂M ∂N
∂M
dx dy −
dy dx =
dx dy +
dx dy = (
+
) dx dy
∂x
∂y
∂x
∂y
∂x
∂y
dx dx = 0 = dy dy
dx dy
now we must always get a
2.
The multiplication of
∂M
∂N
∂M
∂N
∂N ∂M
dy dx +
dx dy = −
dx dy +
dx dy = (
−
) dx dy ,
∂y
∂x
∂y
∂x
∂x
∂y
=
where we used
dω = dA dx + dB dy .
then
dω = dM dx + dN dy = (
and if
R
is a dierential form and
dierentials here is as described in the
ω = M dx + N dy ,
ω = ∬ dω
and
dy dx = −dx dy .
Unlike in the case of change of variables,
under the double integral and we must not get rid of the sign.
‰
F ⋅ dr
Evaluate
where
C
F=
and
C
−y
x
i+ 2
j
2
+ 9y
4x + 9y 2
4x2
is the unit circle parametrized in the counterclockwise direction.
Solution:
Observe that
curl
at all points
Consider
∂
x
∂
−y
( 2
)−
( 2
)
2
∂x 4x + 9y
∂y 4x + 9y 2
1
8x2
1
18y 2
= 2
−
+
−
=0
4x + 9y 2 (4x2 + 9y 2 )2 4x2 + 9y 2 4x2 + 9y 2
F=
(x, y) =/ (0, 0).
C0 ∶ r = (cos t)/2 i + (sin t)/3 j, 0 ≤ t ≤ 2π , the counterclockwise
4x2 + 9y 2 = 1. Let R be the region lying inside the
parametrization of the ellipse
unit circle and outside this ellipse.
Since curl
F=0
at all points of
Green's Theorem,
‰
‰
‰
C0
Therefore by the generalized
−y dx + x dy
2
2
C0 4x + 9y
t=2π −(sin t)/3 d((cos t)/2) + (cos t)/2 d((sin t)/3)
=∫
t=0
cos2 t + sin2 t
1 2π
1
π
= ∫
dt = ⋅ 2π = .
6 0
6
3
F ⋅ dr =
F ⋅ dr =
C
R, ∬ curl F dA = 0 .
R
110
3.
Find the surface area of the parameterized torus
r = (a + b cos u) cos v i + (a + b cos u) sin v j + b sin u k ,
0 ≤ u ≤ 2π , 0 ≤ v ≤ 2π ,
Solution:
where
0<b<a
are constants.
We have:
ru = −b sin u cos v i − b sin u sin v j + b cos u k
rv = −(a + b cos u) sin v i + (a + b cos u) cos v j
ru × rv = −b(a + b cos u) cos v cos u i − b(a + b cos u) sin v cos u j − b(a + b cos u) sin u k
∣ru × rv ∣ = b(a + b cos u)(cos2 v cos2 u + sin2 v cos2 u + sin2 u)1/2 = b(a + b cos u)
Therefore
Surface Area
= ∬ ∣ru × rv ∣ dA = ∬ b(a + b cos u) dA
=∫
where
R
2π
0
R = {(u, v) ∶ 0 ≤ u ≤ 2π
R
2π
∫0
and
b(a + b cos u) du dv = ab ⋅ 2π ⋅ 2π = 4π 2 ab
0 ≤ v ≤ 2π}.
111
4.
x2
Find the area of the portion
+ y 2 + z 2 = 4.
Solution:
S
of the cylinder
x2 + y 2 = 2y
that lies inside the sphere
F (x, y, z) = x2 + y 2 − 2y . Then the cylinder is given by F (x, y, z) = 0.
We will use the projection to the yz -plane. This projection is 2-1 from the portion of
the cylinder inside the sphere to the region R given by the inequalities 0 ≤ 2y ≤ 4−z 2
in the yz -plane, and by symmetry
Let
Surface Area of
where
p = i.
∣∇F ∣
S = 2∬
dA
R ∣∇F ⋅ p∣
We have
∇F = 2xi + (2y − 2)j
∣∇F ∣ = (4x2 + (2y − 2)2 )1/2 = (4x2 + 4y 2 − 8y + 4)1/2 = 2
∇F ⋅ p = ∇F ⋅ i = 2x
on
S
and therefore
Surface Area
∣∇F ∣
2
1
dA = 2 ∬
dA = 2 ∬
dA
R ∣∇F ⋅ p∣
R ∣2x∣
R ∣x∣
√
1
2y − y 2 on S
= 2∬ √
dA
as ∣x∣ =
R
2y − y 2
√
√
2
4−2y
2
1
4 − 2y
√
= 2∫ ∫ √
dz dy = 2 ⋅ 2 ∫ √
dy
2
0
− 4−2y
0
2y − y
2y − y 2
2 1
√
√
√
= 4 ⋅ 2 ∫ √ dy = 4 2 ⋅ 2 2 = 16 .
y
0
= 2∬
112
5.
Consider the parametrized surface
the area of the portion of the surface
√
S ∶ r = u2 i + 2 uvj + v 2 k, −∞ < u < ∞ , 0 ≤ v < ∞ .
S that lies inside the unit ball x2 + y 2 + z 2 ≤ 1.
Find
√
x = u2 , √
y = 2uv and z = v 2 on the surface. Therefore,
x2 + y 2 + z 2 ≤ 1 means (u2 )2 + ( 2uv)2 + (v 2 )2 ≤ 1. In other words, (u2 + v 2 )2 ≤ 1, or
u2 + v 2 ≤ 1. Hence the part of the surface lying inside the sphere is the image of the
region R, the upper half of the unit disk, in the uv -plane.
Solution:
We have
To compute the surface area we rst compute:
√
ru = 2ui + 2vj
√
rv = 2uj + 2vk
√
√
ru ×ru = 2 2v 2 i − 4uvj + 2 2u2 k
√
√
√
∣ru ×ru ∣ = ((2 2v 2 )2 + (4uv)2 + (2 2u2 )2 )1/2 = 2 2(u2 + v 2 )
Therefore,
Surface Area
= ∬ ∣ru ×ru ∣ du dv
R
√
= 2 2 ∬ (u2 + v 2 ) du dv
√
= 2 2∫
R
π
0
π
1
2
∫0 r ⋅ r dr dθ
1
= √ ∫ dθ
2 0
π
=√
2
where we used the polar coordinates in the
Remark:
uv -plane.
There is a shorter way of solving this problem which does not use Calculus. The
given parametrization maps the upper half-uv -plane onto the half-cone given by the equation
y 2 = 2xz ,
and the conditions
x≥0
and
z≥0
u-axis).
positive x- and
in a one-to-one manner (except on the
This half-cone has its vertex at the origin, its axis lies along the bisector of the
z -axes, and it has an opening angle of 45○ . The portion of this half-cone cut o by the unit
sphere is the lateral surface of a right cone with slant height
with area
√
πrℓ = π/ 2.
113
ℓ=1
and radius
√
r = 1/ 2,
hence
6.
Verify Stokes's Theorem for the vector eld F = y i + z k and the surface S , where S is the
z = x2 +y 2 satisfying z ≤ 3, with the unit normal vector eld n pointing
portion of the paraboloid
away from the
z -axis.
Solution:
C
is bounded by the curve
C
=∫
=∫
t=2π
t=0
2π
0
C∶r=
√
3 cos t i −
√
3 sin t j + 3 k, 0 ≤ t ≤ 2π .
Note that this parametrization is consistent with the direction of n. The circulation
of F around C is
‰
‰
F ⋅ dr = (y i + z k) ⋅ (dx i + dy j + dz k)
S
√
√
√
(− 3 sin t i + 3 k) ⋅ (d( 3 cos t) i + d(− 3 sin t) j + d(3) k)
√
√
(− 3 sin t)(− 3 sin t) dt = 3 ∫
2π
0
sin2 t dt = 3π
f (x, y, z) = z − x2 − y 2 , and
∇f = −2x i − 2y j + k points in the opposite direction to n, so we will choose the
minus sign from ± in the ux integral. The projection of S into the xy -plane is the
disk R = {(x, y) ∶ x2 + y 2 ≤ 3}. Finally,
RRR i
j
k RRRR
RR
RR
RR
RRR
RRR
RR
RR ∂
∂
∂
RRR = −k .
∇ × F = RRR
RR ∂x ∂y ∂z RRR
RRR
RRR
RR
RR
RRR
R
0
z RRRR
RR y
The ux of ∇ × F across S is
±∇f
∬S (∇ × F) ⋅ n dσ = ∬R F ⋅ ∣k ⋅ ∇f ∣ dA
On the other hand, the paraboloid is a level surface of
= ∬ (−k) ⋅ (2x i + 2y j − k) dA
R
√
= ∬ dA = Area of R = π( 3)2 = 3π .
R
‰
Hence
∬S (∇ × F) ⋅ n dσ =
F ⋅ dr.
C
7.
Verify Divergence Theorem for the vector eld
{(x, y, z) ∶ x2 + y 2 + z 2 ≤ 4} .
Solution:
F = xz i + yz j + z 3 k
and the region
∇ ⋅ F = ∂(xz)/∂x + ∂(yz)/∂y + ∂(z 3 )/∂z = z + z + 3z 2 = 2z + 3z 2 ,
2
∭D ∇ ⋅ F dV = ∭D (2z + 3z ) dV
=∫
2π
0
π
2
2
2
2
∫0 ∫0 (2ρ cos ϕ + 3ρ cos ϕ) ρ sin ϕ dρ dϕ dθ
π
96
cos2 ϕ) sin ϕ dϕ dθ
(8 cos ϕ +
∫
5
0
0
64 2π
128π
=
dθ =
.
∫
5 0
5
=∫
2π
114
and
D =
S = {(x, y, z) ∶ x2 + y 2 + z 2 = 4}, we
divide it as the upper hemisphere S1 = {(x, y, z) ∶ x2 + y 2 + z 2 = 4 and x ≥ 0} and the
lower hemisphere S2 = {(x, y, z) ∶ x2 + y 2 + z 2 = 4 and x ≤ 0}, and project both onto
the disk R = {(x, y) ∶ x2 + y 2 ≤ 4} in the xy -plane.
To compute the outward ux through the sphere
The sphere is a level surface of
in the same direction as
n,
f (x, y, z) = x2 + y 2 + z 2 . ∇f = 2x i + 2y j + 2z k
the outward pointing unit normal vector eld on
we will choose the plus sign from
±
in the ux integral.
For the upper hemisphere we have
±∇f
∬S F ⋅ n dσ = ∬R F ⋅ ∣k ⋅ ∇f ∣ dA
1
= ∬ (xz i + yz j + z 3 k) ⋅
R
2x i + 2y j + 2z k
dA
2z
= ∬ (x2 + y 2 + z 3 ) dA
R
= ∬ (x2 + y 2 + (4 − x2 − y 2 )3/2 ) dA
=∫
=∫
R
2π
2
2
2 3/2
∫0 (r + (4 − r ) ) r dr dθ
0
2π
0
(4 +
32
64π
) dθ = 8π +
.
5
5
A similar computation for the lower hemisphere gives
64π
∬S F ⋅ n dσ = −8π + 5 .
2
Therefore
∬S F ⋅ n dσ = ∬S F ⋅ n dσ + ∬S F ⋅ n dσ
2
1
64π
64π
128π
= (8π +
) + (−8π +
)=
,
5
5
5
and
points
∬S F ⋅ n dσ = ∭D ∇ ⋅ F dV
.
115
S,
so