Solutions
Introduction
0.3 v̄[0,2] = 4/2 = 2, v̄[1,4] = 15/3 = 5 and
v̄[2,6] = 32/4 = 8.
0.5 The lower and upper bounds are 12.6 and
14.4 respectively.
1
1.23 y = −2x/k − 1
a) k = −2/3, b) impossible, c) impossible
1.25
a)
b)
y
y
1
1
3
0.7 distance: 4c
0.9 The lower and upper estimates are 40 m and
50 m respectively.
0.11 v̄[t0 ,t] = (t3 − t30 )/(3(t − t0 ))
= (t − t0 )(t2 + t0 t + t20 )/(3(t − t0 ))
= (t2 + t0 t + t20 )/3.
x
3
x
d)
y
y
1
0.15 a) 0.728 ft, b) v̄ = 3.64 ft/s
c) 0.615125 ft, v̄ = 12.3025 ft/s
x
-1
-1
c)
0.13 Since the velocity of A is greater than that
of B throughout the interval from t = 0 to t = 1,
and since the cars start from the same position,
we may infer that the distance between them is
increasing on the intervals [0, 0.25] and [0.75, 1].
Similarly, the distance is decreasing on [1, 1.25].
3
1
3
x
-1
-1
1.27 a) D(f ) = [1, 4], R(f ) = [0, 3],
b) f (2) = 0, c) x = 1
Chapter 2
Chapter 1
1.1 {1, 2, 3, 4} and {1, 2} are subsets.
1.3 −1 and 1/2
1.11 (f /g)(x) = 1/(x2 + 1)2 and R(f /g) =
(0, 1].
1.13
2.3 limx→1 f (x) = 2.5
2.11 a) 1, b) 2, c) 2, d) 1,
e) limx→0− f (x + 2) = limx→2− f (x) = 1,
f ) limx→−1− f (x2 ) = limx→1+ f (x) = 2
2.13 a)
y
12
10
8
y
6
2
4
2
1
x
x
0
1
2
3
4
1.15 [1, 2) ∪ [2, 3] = [1, 3], [1, 2] r {2} = [1, 2),
[1, 2] r {1, 2} = (1, 2), [0, 3] r [1, 3] = [0, 1),
R r (−∞, 0) = [0, ∞), [1, 2] ∩ (3/2, 3) = (3/2, 2],
[1, 2] ∩ [2, 3] = [2, 2], and (a, b) ∪ (b, a)
= ((a + b − |a − b|)/2, (a + b + |a − b|)/2)
1.17 a) D(f ) = R r {2}√
b) f (3) = 0, f (−3) = − 3/5
c) f (f (1)) = f (−1) = −2/3
1.19 The equation y 2 = x2 is equivalent to the
equation y = ±x. The latter equation does not
define y as a function of x on R because it assigns every value x 6= 0 two distinct values of
y.
1.21 f (x) = 4x + 1 and g(x) = −4x + 9.
0
0.5 1
2
3
4
b) R(f ) = [2, 11] r {3}, c) 3
2.15 If we define f (x) := 1 for x ≥ x0 and
f (x) := −1 for x < x0 , then limx→x0 f (x) does
not exist, but limx→x0 f (x)2 = 1. Consequently,
the answer in general is “no.”
2.17 a) Proceeding as in Exercise 2.16, we find
that limh→0 (g(1 + h) − g(1))/h = limh→0 (3 + h)
= 3.
b) limh→0 (g(t + h) − g(t))/h
= limh→0 (2t + h + 1) = 2t + 1
Chapter 3
3.7 Let 1 > ε > 0 and limn→∞ xn = 2. Then
there exists an N ∈ R such that |xn − 2| < ε/5
2
Solutions
for all n > N . This shows that |xn + 2| ≤
|xn − 2| + 4 ≤ ε/5 + 4 < 5 for all n > N , and
therefore, |x2n − 4| = |xn + 2||xn − 2| < 5ε/5 = ε
for all n > N . Hence limn→∞ x2n = 4 and, by
implication, limx→2 x2 = 4.
3.9 Assume that x0 is an accumulation point of
D and let ε > 0. Then there exists a sequence
(xn )∞
n=1 in D r {x0 } such that limn→∞ xn = x0
and, by implication, there exists an n such that
0 < |xn − x0 | < ε. Setting x := xn , it follows
that x ∈ D and 0 < |x − x0 | < ε. Conversely,
if for every ε > 0 there exists an x ∈ D such
that 0 < |x − x0 | < ε then, in particular, for
each n ∈ N there exists an xn ∈ D such that
0 < |xn − x0 | < 1/n. Since the latter inequality
clearly shows that limn→∞ xn = x0 , we may
infer that x0 is an accumulation point of D.
3.11 a) For ε > 1, we define δ := ε/7. If
0 < |x − 1| < δ, then |x| ≤ |x − 1| + 1 < 2,
and therefore, |x3 − 1| = |x − 1||x2 + x + 1| ≤
|x − 1|(|x|2 + |x| + 1) < 7δ = ε. Consequently,
limx→1 x3 = 1.
b) For ε > 0, we set δ := ε. If 0 < |x − 2| < δ,
then |(x2 − 4)/(x − 2) − 4| = |(x + 2) − 4| = |x −
2| < δ = ε. Thus, limx→2 (x2 − 4)/(x − 2) = 4.
c) For ε > 0, we set δ := ε. If√0 < |x − √
2| < δ
for some√x ∈ [−1, √
∞). Then | x + 1 − 3| =
|x − 2|/| x + 1 + √ 3| < |x −
√2| < δ = ε. Consequently, limx→2 x + 1 = 3.
and similarly,
√
√
n
x − n x0
1
d √
n
x
= lim
= √ n−1 .
n
x→x
dx
x − x0
0
n x0
x=x0
4.9 The line described by the equation y(x) =
f 0 (x0 )(x − x0 ) + f (x0 ) has slope f 0 (x0 ) and
passes through the point (x0 , f (x0 )) because
y(x0 ) = f 0 (x0 )(x0 − x0 ) + f (x0 ) = 0 + f (x0 ) =
f (x0 ). Thus, it must be the tangent line to the
graph of f at (x0 , f (x0 )). Furthermore, for the
function f (x) = x2 we find that
(x + ∆x)2 − x2
= 2x,
∆x→0
∆x
f 0 (x) = lim
and therefore, the tangent line to the graph of f
at the point (2, 4) is described by the equation
y(x) = 4(x − 2) + 4 = 4x − 4.
4.13
f ’(x)
x
4.15 a) v̄[0,4] = 0 and v̄[1,4] = 4/3,
b) y = 4(x − 1)/3 − 1 = 4x/3 − 7/3,
c) y(x) = −1.
d) It is positive for t ∈ (1, 2)∪(2, 4] and negative
for t ∈ [0, 1).
4.17 tangent line: y(x) = 2(x − 2) + 3 = 4x − 1
Chapter 4
10
4.1 v(t) = 2ct
8
4.5
6
f(x)
4
(x + ∆x)3 − x3
f (x) = lim
∆x→0
∆x
3x2 ∆x + 3x∆x2 + ∆x3
= lim
∆x→0
∆x
= lim (3x2 + 3x∆x + ∆x2 ) = 3x2 .
0
∆x→0
4.7
√
√
4
x − 4 x0
d √
4
x
= lim
x→x0
dx
x − x0
x=x0
1
= lim √ 3 √ 2 √
√ 4 2 √
x→x0 4 x + 4 x 4 x + 4 x √
x0 + 4 x0 3
0
1
= √ 3,
4
4 x0
2
x
0
-2
1
2
3
4.19 f (2.1) ≈ −0.85 and f (1.9) ≈ −1.15.
√
3
x
x2 + 15
6.9
3.9708
4.21
Hence f 0 (64) ≈ 0.029.
7.0
4.0000
7.1
4.0291
4.23 a) TRUE because
f (x) − f (2)
= 0.
x→2
x−2
f 0 (2) = lim
b) POSSIBLY TRUE/POSSIBLY FALSE because the stated limit condition is satisfied for
Solutions
f (x) = 0 as well as for g(x) = 1, but only f
assumes the value 0 at 2.
c) TRUE because
lim f (x) = f (2) + lim (f (x) − f (2))
x→2
x→2
f (x) − f (2)
lim (x − 2)
(x − 2) x→2
= f (2) + 0 · 0 = f (2).
= f (2) + lim
x→2
3
√
6.5 tan(π/6)√ = cot(π/3) = 1/ 3, tan(π/3) =
cot(π/6) = 3,
√ tan(π/4) = cot(π/4) = 1, and
cot(5π/6) = − 3.
6.7 The given equation cot(2α) = 1/2 implies
that (cot2 (α) − 1)/(2 cot(α)) = 1/2, or equivalently, cot2 (α) − cot(α) − 1 =√0. Solving for
cot(α) yields cot(α) = √
(1 + 5)/2. Consequently, tan(α) = 2/(1 + 5), and the√equation
of the line is therefore y = −2x/(1 + 5) + 1.
6.9 a) y = −(x − 1)/3 + 2, b) y = 2x + 5,
c) y = 3(x − 2)/2 − 1
4.25
f ’(x)
g’(x)
h’(x)
4.27 the instantaneous rate of change in the
person’s weight
6.11 If the light ray were traveling downward
as in Figure 6.2, it would be reflected in the
direction of the line that connects the point of
reflection with the origin. By symmetry, the reflected ray in Figure 6.8 must travel along the
same line but in the opposite direction—not toward the origin but away from it. Consequently,
the equation describing the path of the reflected
ray is y = 2((3/2)2 − 1/4)x/3 = 4x/3.
Chapter 5
Chapter 7
5.1 To construct y/x we mark two points D and
C on the line L in Figure 5.1 such that DP = y
and CP = 1. Furthermore, we mark a point B
on the x-axis at distance x from P and draw
a line segment from B to D. The point A is
now obtained as the intersection of the x-axis
with the line through C that runs parallel to
BD. Since the triangles 4ACP and 4BDP
are similar, it follows that x/y = BP /DP =
AP /CP = AP .
7.9 a) yes, f (x0 ) = 0,
b) limx→x0 f (x) = limx→x0 (x + x0 ) = 2x0 ,
c) f is continuous at x0 if and only if x0 =
0, because limx→x0 f (x) = f (x0 ) if and only if
x0 = 0.
d) Since f (x) = x + x0 for all x 6= x0 , it follows
that f is differentiable at x0 if and only if x0 =
0.
x
x
x
5.3 If x0 is the point where f assumes its minimum, then the difference quotient
4x0 e + 2e2
f (x0 + e) − f (x0 )
=
= 4x0 + 2e
e
e
must be zero for e = 0. Thus 4x0 = 0, and by
implication, the minimal value of f is f (x0 ) =
f (0) = −2.
Chapter 6
6.1 tan(α) = a/b = (a/c)/(b/c)
= sin(α)/ cos(α) and
cot(α) = b/a = cos(α)/ sin(α).
6.3 Since the hypotenuse of a √
right triangle with
two equal sides of length 1√is 2,
√ it follows that
sin(π/4) = cos(π/4) = 1/ 2 = 2/2.
7.11 a) TRUE because the assumption of continuity at x0 implies that x0 is contained in the
domain of f .
b) TRUE because f is continuous at x0 .
c) Again TRUE because f is continuous at x0 .
d) POSSIBLY TRUE/POSSIBLY FALSE. The
function f : R → R, f (x) := |x − x0 | is continuous but not differentiable at x0 .
7.13 g and h are continuous but f is not.
7.15 a) The statement is in general not true because it is for example not satisfied if f (x) := −1
for x ∈ [1, 2) and f (2) := 1.
b) This statement is true because a continuous function cannot “jump” from negative to
positive values and will pass through the x-axis
somewhere in the interval from 1 to 2.
c) The statement is in general not true because
it fails, for example, for the function f (x) :=
|x − 3/2|.
d) This statement is true because if f (1) and
4
Solutions
f (2) had different signs, then the assumed continuity of f would imply that there exists a point
z ∈ (1, 2) such that f (z) = 0.
v’(t)
v(t)
v’(0)=0.28
7.17 Yes because if f were differentiable, then
it would also be continuous (by Theorem 7.7).
7.19
t
t
y
T=354.17
T=354.17
Figure S.1: the graphs of v 0 (t) and v(t).
x
0
1
2
3
4
5
6
10.9 [0, 1] and [−1/2, 1/2]
Chapter 8
8.5 Since
q 0 (t) =
10.7 (f ◦ g)(4) = f (g(4)) = g(4)3 + g(4) + 1 =
√ 3 √
4 + 4 + 1 = 11
d
(m(t)v(t)) = m 0 (t)v(t) + m(t)v 0 (t),
dt
equation (8.11) implies that
m(t)v 0 (t) = F (t) − m 0 (t)v(t) + m 0 (t)(v(t) − u)
= F (t) − m 0 (t)u.
8.7 a) f (x)g(x)
b) (f (x + ∆x) − f (x))g(x)
c) (f (x + ∆x) − f (x))(g(x + ∆x) − g(x))
d) f (x)(g(x + ∆x) − g(x))
e) f (x + ∆x)g(x)
f ) (f (x + ∆x) − f (x))g(x + ∆x)
8.9 a) T = 85/0.24 s, and therefore, m(t) =
−85t/(85/0.24) + 350 + 85 = −0.24t + 435 kg.
b)
m(t)
435
350
t
T=354.17
c) m 0 (t) = −0.24 kg/s
d) v 0 (t) = 85 · 50/(435 · 85/0.24 − 85t)
= 50/(1812.5 − t)
e) See Figure S.1.
Chapter 10
10.3 (f 2 ) 0 = f 0 f + f f 0 = 2f f 0 and (f gh) 0 =
(f g) 0 h+(f g)h 0 = (f 0 g +f g 0 )h+f gh 0 = f 0 gh+
f g 0 h + f gh 0
10.11 If f and g are continuous, then
limx→x0 f (g(x)) = f (limx→x0 g(x)) = f (g(x0 ))
for all x0 ∈ E (that are not isolated points of
E). Thus, f ◦ g is continuous on E.
10.15 a) 4x3 − 6x2 ,
b) (3 − 3x2 )x2 + (3x − x3 + 2)2x,
c) 600(3x3 − 4)599 9x2 ,
d) 3000(3x − 4)999 ,
e) 100(x2 +(x2 +1)100 )99 (2x+100(x2 +1)99 2x),
f ) 20x(x2 + 1)9 (x − 2)10 + 10(x2 + 1)10 (x − 2)9 ,
g) ((1 − 4x)(3 − x) + (x − 2x2 + 7))/(3 − x)2 ,
h) (4x(x + 3) − (2x2 − 5))/(x + 3)2 ,
i) −12/x5 ,
j) (5x4 (x7 − 2) − 7x11 )/(x7 − 2)2 ,
−3
x−3
4 − x2 + 2x(x − 3)
k) −2
,
4 − x2
(4 − x2 )2
l) 3x2 − 10x − 3/x2
10.17 a) y = 5(x − 1) + 2,
b) y = −11(x − 2)/3 − 3, c) y = 0,
d) y = 5(x − 4)/8 + 4
10.19 a) (f + g) 0 (2) = 1, (f g) 0 (2) = 7,
(f /g) 0 (2) = 1,
b) y = 7(x − 2) − 1
10.21 Since f (x)g(x) = x for all x ∈ R,
it follows that f (0)g(0) = 0 and f 0 (0)g(0) +
f (0)g 0 (0) = 1. While the former of these equations implies that either f (0) = 0 or g(0) = 0,
the latter allows us to infer that either f (0) 6= 0
or g(0) 6= 0. Consequently, exactly one of the
values f (0) and g(0) must be equal to zero and
exactly one of them different from zero.
10.23 f20 = (f 0 ◦f1 )f 0 , f30 = (f 0 ◦f2 )(f 0 ◦f1 )f 0 ,
f40 = (f 0 ◦ f3 )(f 0 ◦ f2 )(f 0 ◦ f1 )f 0 , and in general
fn0 = (f 0 ◦ fn−1 ) · · · (f 0 ◦ f1 )f 0
10.25 Since d/dx y 3 = 3y 2 y 0 , we may infer that
the rate of change in y 3 is twelve times the rate
Solutions
5
of change in y for y = 2 (because then 3 · y 2 =
12).
Chapter 11
11.3 f assumes its global maximum at a and its
global minimum at x1 . Furthermore, the set of
all points at which f assumes a local maximum
is {a, x4 } ∪ [x2 , x3 ), and similarly, f assumes a
local minimum at all points in the set {x1 , b} ∪
(x2 , x3 ].
2
1
-4 -3 -2 -1
x
0
1
2
3
Figure S.2: graph for Exercise 11.29a.
11.5 In solving for x the equation f 0 (x) = 3x2 −
6x − 9 = 0 we find the critical points of f to be
at x = 3 and x = −1.
10
11.9 √
roots at x = 1, 3, 4, local maximum√
at x =
(8 − 7)/3, local minimum at x = (8 + 7)/3
4
10
f(x)
7
6
5
4
3
f(x)
8
6
2
x
f(x)
-2
-1
0
1
2
3
4
Figure S.3: graph for Exercise 11.29b.
5
x
0
1
2
3
4
5
-5
-10
-15
11.11 d3 /dx3 (x + 1)4 = 24(x + 1), f 00 (x) =
20x3 + 30x4 , and f (4) (x) = 120x + 360x2 .
11.15 roots
√ at x = 0, 1, 2, local maximum√at
x = 1 − 1/ 3, local minimum at x = 1 + 1/ 3,
point of inflection at x = 1
2
2x3 + x − 1
=
11.21 lim
x→−∞
3x3 − 4
3
x+1
11.23 lim+ 3
=∞
x→1 x − 1
11.25 a) concave up on (7/8, ∞), concave down
on (−∞, 7/8), point p
of inflectionpat x = 7/8
b) concave up on (− p
3/2, 0)∪( p
3/2, ∞), concave down on (−∞, − 3/2) ∪ (0,p 3/2), points
of inflection at x = 0 and x = ± 3/2√
c) concave
√ up on (−∞, −2/5 − 34/5) ∪
34/5,√
∞), concave down on (−2/5 −
(−2/5
+
√
34/5, −2/5√+ 34/5), points of inflection at
x = −2/5 ± 34/5
11.27 f is increasing on (500, ∞) and decreasing on (−∞, 500).
11.29 a) increasing on (−2, ∞), decreasing on
(−∞, 2), local minima at all points in [−2, 1];
see Figure S.2 for the graph.
b) increasing on (0, ∞), decreasing on (−∞, 0),
local minimum at x = 0; see Figure S.3 for the
graph.
11.31 a) maximum at x = −1,
b) points of inflection at x = 0 and x = 1,
c)
f(x)
x
-3
-2
-1
1
2
3
11.33 a) [2, 4] ∪ [6, 8], b) [0, 2] ∪ [4, 6] ∪ [8, 10],
c) [1, 3] ∪ [5, 7] ∪ [9, 10], d) [0, 1] ∪ [3, 4] ∪ [7, 9],
e) x = 0, 2, 4, 6, 8, 10, f ) x = 1, 3, 5, 7, 9,
g) global maximum at x = 0, global mimimum
at x = 1
√
11.35 a) √
increasing on ((14 − 3 3)/13, 1) ∪
3)/13), decreasing on (−∞, −1) ∪
(1, (14 + 3 √
(−1, (14 − 3 3)/13), concave up on (−1, 1) ∪
(2, ∞), concave down on (−∞, −1) ∪ (1, 2), asymptotes: x = 1, x = −1, and y = 0; see Figure S.4 for the graph.
b) decreasing on (−∞, −1) ∪ √
(1, ∞), increasing
√
on (−1, 1), concave up on√(− 3, 0)√∪ ( 3, ∞),
concave down on (−∞, − 3) ∪ (0, 3), asymptote: y = 0; see Figure S.5 for the graph.
c) increasing on (0, 2) ∪ (2, ∞) decreasing on
6
Solutions
the second derivative test implies that L assumes a minimum at is critical point.
f(x)
12.7 length of the spiral: L ≈ 28055895 mm;
storage capacity: 22444716000 bytes
5
12.9 infrared laser: L ≈ 5611179 mm;
blue light laser: L ≈ 28055895 mm
x
-2
2
12.11 The area in dependence on x is A(x) =
x(1√− x3 ), and the critical point of A is x0 =
the maximal area is
1/ 3 4. Consequently,
√
A(x0 ) = 3/(4 3 4).
4
-5
Figure S.4: graph for Exercise 11.35a.
12.13 dmin = 2 and P = (6/5, 8/5).
√
12.15 dmin = 50/7 and P = (−1, −1/7).
12.17 The distance from an arbitrary point
(x, x2 + 2) on the parabola to the line is
f(x)
0.5
d(x) =
x
-4
-2
2
4
-0.5
Figure S.5: graph for Exercise 11.35b.
(−∞, 2) ∪ (−2, 0); concave up on (−2, 2), concave down on (−∞, −2) ∪ (2, ∞), asymptotes:
x = 2, x = −2, and y = −1.
f(x)
x
-2
Since d obviously assumes its minimal value at
x0 = 1, it follows that the point on the parabola
closest to the line is (1, 3).
12.19 Denoting by h and r the height and radius of the cylindrical can respectively, it follows
that V = 355 = πr2 h. Consequently, the surface area is given by the equation
A = 2πr2 + 2πrh = 2πr2 +
710
.
r
Setting the derivative of dA/dr equal
to
zero and solving for r yields r =
p
3
355/(2π)
≈ 3.837 cm and, by implication,
p
h = 2 3 355/(2π) ≈ 7.674 cm.
5
-4
|x2 − 2x + 2|
(x − 1)2 + 1
√
√
=
.
5
5
2
4
-5
11.37 The largest slope is assumed at a critical
point of y 0 . Solving therefore for x the equation
y 00 = −6x + 6 = 0 yields x = 1, and since
y 000 = −6 < 0, we may conclude that y 0 assumes
its largest value at x = 1.
12.21 If we denote by x the distance along
opposite shore from the point nearest to
man to the point where he intends to leave
water, then the total time it takes to get to
car is
1
2p 2
x + 0.22 + (0.5 − x).
T (x) =
3
6
Setting the derivative T 0 (x)√equal to zero and
solving for x yields√x = 0.2/ √15, and the minimal time is T (0.2/ 15) = (2 15 + 5)/60 h.
Chapter 12
Chapter 13
12.5 Since
13.5 z1 ≈ −6.47, z2 ≈ 0.54, and z3 ≈ 1.43.
L 00 (x) = √
a2
b2
+p
> 0,
2
2
+x
b + (d − x)2
a2
the
the
the
the
13.7 x0 ≈ 1.49937 and dmin ≈ 2.52796.
13.9 There is only one root at z ≈ 1.37880.
Solutions
13.11
a)
f(x)
b)
1
1
3
0
-1
x
c)
f(x)
1
x
3
0
-1
0
f(x)
3
x
-1
13.13 For x0 = 1 Newton’s method breaks
down after one step, because x1 = 0 = f 0 (0) =
f 0 (x1 ) and x2 = x1 −f (x1 )/f 0 (x1 ) is undefined.
For x0 = 2 and x0 = 10, the corresponding sequences of values xn do not converge but alternate erratically between positive and negative
values.
13.15 (x0 , y0 ) ≈ (1.46557, 3.14790)
7
14.17 a) It is difficult to tell whether a left or
right sum gives a lower or upper estimate because f is increasing on [0, 2] but decreasing on
[2, 4].
b) We first determine an upper estimate for the
area under the graph over the interval [0, 2] with
a margin of error less than or equal to 0.05.
Given the error estimate
E ≤ |f (0) − f (2)|
we choose 8/0.05 = 160 as our value for n so
that ∆x = 2/160 = 1/80. The corresponding
upper estimate for the area under the graph is
represented by the following right sum:
160
X
k
1
k
4−
≈ 5.358.
80
80 80
Chapter 14
P7
k
k=−2 1/(k +3) = 49/20,
k=2 2 = 253,
P
P
5
4
k
k
i=2 2 = 6·2 ,
n=1 2n = 30,
n=−1 3 = 18.
14.3
P7
P3
14.5 n = 39.2/0.002 = 19600
14.7
f(x) upper estimate
left sum
f(x) lower estimate
right sum
x
x
a
b
8
2
= ≤ 0.05,
n
n
a
b
14.9 Solving for n the inequality
k=1
Now we can either repeat this process for the
area under the graph over the interval [2, 4] or,
in observing the graph to be symmetric with
respect to the line x = 2, we can multiply the
above result by 2 to infer that the total area
under the graph is less than 10.72 with a margin
of error less than 2 · 0.05 = 0.1.
14.19 Using the
√ formula in Exercise 14.18
with f (x) = 1 − x2 , n = 100, and xk =
−1 + p
k∆x = −1 + k/50, we may infer that
P100
∆x2 + (f (xk ) − f (xk−1 ))2 ≈ 3.14 repk=1
resents an estimate for the length of the upper half of the circle described by the equation x2 + y 2 = 1. Consequently, an estimate
for the entire circumference of the circle is C ≈
2 · 3.14 = 6.28.
E ≤ |f (1) − f (4)|3/n = 13.5/n ≤ 0.002
yields n ≥ 13.5/0.002 = 6750.
14.11 If s 0 (t) = v(t) = 5t2 , then s(t) = 5t3 /3 +
C and therefore s(0) = C. Consequently, the
total distance traveled is s(t) − s(0) = 5t3 /3.
14.13 The velocity of a segment of length
0.2 m at distance x from the center of rotation is approximately 2πx/2 = πx m/s, and the
corresponding kinetic energy is 0.1(πx)2 /2 J.
Consequently, the total kinetic
energy is apP6
proximately equal to 0.1 k=1 (0.2 · πk)2 /2 =
91π 2 /500 J.
14.15 An
P6upper estimate for the moment of inertia is k=1 0.1(0.2k)2 = 91/250.
Chapter 15
15.1
8
Solutions
Chapter 16
Pn−1
16.1 V ≈ k=0 π(R2 − x2k )(R + d)/n
16.3 With R = 0.02 m and d = 0.016 m we
obtain: m = 1000π0.024 · 0.0362 /3 ≈ 0.0326 kg.
P10
16.5 V ≈ k=1 8π(4−(−4+8k/10)2 /16)2 /10 ≈
339.12
P20
16.7 V ≈ k=1 πk/16 = 13.125π
16.9 a) A(x) = 4x,
P5
b) k=0 4k/9 = 20/3,
P6
c) k=1 4k/9 = 28/3,
P11
d) left sum: k=0 4k/36 = 22/3;
P12
right sum:
k=1 4k/36 = 26/3
16.11 Using aPright sum, we find the approxn
2
imation V ≈
k=1 16(4 − xk )/n. Setting for
example n = 100 and xk = −2 + k/25 yields the
numerical result V ≈ 42.66.
Chapter 17
P4
2
17.3 left sum:
k=1 (5xk − 5xk )/2 with xk =
1 + (k − 1)/2
P4
right sum: k=1 (5xk −5x2k )/2 with xk = 1+k/2
P4
2
midpoint sum:
k=1 (5xk − 5xk )/2 with xk =
1 + (2k − 1)/4.
17.11 Proof of c): if a ≤ c ≤ b, then the equation in c) follows directly from the addition of
areas between the graph of f and the x-axis over
the intervals [a, c] and [c, b]. Given this observation, the equation can also be established for
any other ordering of the boundaries of integration according to their magnitude. For instance,
if b ≤ a ≤ c, then
Z c
Z b
f (x) dx =
f (x) dx
−
c
b
Z a
Z c
=
f (x) dx +
f (x) dx
b
=−
a
Z
b
f (x) dx +
Z
a
or equivalently,
Z b
Z
f (x) dx =
a
c
f (x) dx +
a
c
f (x) dx,
a
Z
b
f (x) dx
c
as desired. All other possible orderings can be
handled in a similar fashion.
Proof
of d):R this follows
R a from c) because
Ra
a
f
(x)
dx
=
f
(x)
dx+
f (x) dx, and therea
a
a
Ra
fore, a f (x) dx = 0.
17.13 A rough estimate of the area under the
graph shows that the correct value must have
been 58.7.
17.15 a) 18 + 5 = 23, b) not enough information, c) 5 − 11 = −6, d) 5 + 11 = 16, e) not
enough information, f ) 18 + 5 − 11 = 12
R 10
R 10
R 10
17.17 −5 g(u) du = −5 f (u) du + −5 2 du =
4 + 30 = 34.
R0
R4
17.19 a) 0 f (x) dx = 1 + 7 = 8, 1 f (x) dx =
R2
−2, and 1 f (x) dx = 1 − 2 = −1.
b) If f (x) ≥ 0 for all x ∈ [1, 2], then −1 =
R2
f (x) dx ≥ 0 which is obviously impossible.
1
c) If f (x) < 3.5 for all x ∈ [0, 4], then 7 =
R4
R4
f (x) dx < 2 3.5 dx = 7 which again is false.
2
17.21
a) POSSIBLY TRUE/POSSIBLY FALSE: the
statement is true for f (x) = 1 and g(x) = 2, and
it is false for f (x) = arctan(x) and g(x) = 2.
b) POSSIBLY TRUE/POSSIBLY FALSE: the
statement is true for f (x) = 1 and g(x) = 2, and
it is false for f (x) = − arctan(x) and g(x) = 2.
c) TRUE: if f (x) ≤ g(x), then 0 ≤ g(x) −
R1
f (x), and therefore, 0 ≤ 0 (g(x) − f (x)) dx =
R1
R1
R1
g(x) dx − 0 f (x) dx. Hence 0 f (x) dx ≤
0
R1
g(x) dx.
0
17.23 The distance is zero.
Chapter 19
3
R3
19.7 14.12: A = 0 x3 dx = x4 /40 = 81/4,
5
R5
16.6: V = π 0 x dx = πx2 /20 = 25π/2,
1
R1
16.8: V = π 0 x6 dx = πx7 /70 = π/7,
2
R2
16.9: V = 0 4x dx = 2x2 0 = 8,
3
R3
16.10: V = π 0 x2 dx = πx3 /30 = 9π,
R2
16.11: V = −2 4(4 − x2 ) dx
2
= 4(4x − x3 /3)−2 = 128/3.
R2
2
19.9 −2 f 0 (x) dx = f (x)|−2
= 23 − ((−2)2 − 2 + 1) = 5
3
R3
19.11 a) 1 f1 (x) dx/2 = x2 /41 = 2,
3
R3
f (x) dx/2 = x3 /61 = 13/2,
1 2
3
R3
f (x) dx/2 = x4 /81 = 63/8,
1 R3
3
b) 1 fn (x) dx/2 = (3n+1 − 1)/(2n + 2)
√
19.13 a) Since the graph of 16 − x2 is symmetric withR respect
to the y-axis,
R 0 √ we may
x√
2
infer that 0 16 − t dt = −x 16 − t2 dt
for all x ∈ [−4, 4].
Thus −f (−x) =
Solutions
R0 √
R −x √
16 − t2 dt = −x 16 − t2 dt = f (x) as
− 0
desired.
b) f (0) = 0 p
√
c) f 0 (±2) = 16 − (±2)2 = 12
19.15 To determine the boundaries of integration, we solve for x the equation 2x/a2 −
x2 /a3 = 0. This yields x = 0 and x =
2a. Thus, the areaR enclosed by the parabola
2a
and the x-axis is 0 (2x/a2 − x2 /a3 ) dx =
2a
(x2 /a2 − x3 /(3a3 )) = 4/3.
0
19.17 a) x = 0, 4, 8, b) x = 2, 6, 10, c) x = 0,
d) x = 2, e) x = 1, 3, 5, 7, 9 (Note: all answers
are deduced from the observation that g 0 = f .)
19.19 f 0 (1) = 0 because f is constant.
19.21 We give a proof by contradiction: if f 0 (x)
0
all x ∈
were greater than or equal
R xto g0 (x)+1R for
x
R, then f (x) − f (0) = 0 f (x) ≥ 0 (g 0 (x) +
1) dx = g(x) − g(0) + x, or equivalently f (x) ≥
g(x) − g(0) + f (0) + x for all x ∈ R. Thus, in
contradiction to the assumption f ≤ g, it would
follow that f (x) > g(x) for all x > g(0) − f (0).
19.23 Since f 00 (x) ≤ 0, it follows that f 0 is
decreasing, and therefore,
Z y
Z x+y
f 0 (t) dt ≤
f 0 (t) dt
f (x + y) − f (x) =
x
0
= f (y) − f (0) = f (y).
Thus, f (x + y) ≤ f (x) + f (y) as desired.
Chapter 20
20.1 sequence of cubes: 1, 8, 27, 64, 125, 216, . . .
first differences: 7, 19, 37, 61, 91, . . .
second differences: 12, 18, 24, 30, . . .
third differences: 6, 6, 6, . . .
fourth differences: 0, 0, . . .
Furthermore, if a0 , a1 , . . . , an is a finite sequence
such that a0 = 0, then the sum of the first differences is (a1 −a0 )+(a2 −a1 )+· · ·+(an −an−1 ) =
(a1 +· · · +an )−(a0 +· · · +an−1 ) = an −a0 = an
as desired.
Chapter 21
21.5 The result follows by observing that
1
a
1
d 1
ln |ax + b| = ·
=
.
dx a
a ax + b
ax + b
21.7 a) D = (−∞, 2) ∪ (3, ∞), b) D = [−3, ∞),
c) D = R r {4}, d) D = (2, ∞)
9
21.9 a) 3 ln |x| + C, b) ln(1 + x2 ) + C,
c) −4 ln |1 − 5x|/5 + C, d) x + 4 ln |x − 3| + C,
e) x/2 − 3 ln |2x + 3|/4 + C, f ) ln |x3 − 3| + C
21.11 y = x − 1
21.13 Since d/dx x ln(x) = ln(x) + 1, it follows
that the slope of the normal line is −1/(ln(2) +
1). Consequently, the equation of the normal
line is y = −(x − 2)/(ln(2) + 1) + 2 ln(2).
21.15 Since ln(x) is not defined for x < 0, the
equation ln(x2 ) = 2 ln(x) cannot be valid for all
x ∈ Rr{0}. The correct statement is “ln(x2 ) =
2 ln |x| for all x ∈ R r {0}.”
Chapter 22
22.3 Since the number of elements in the domain is larger than the number of elements in
the set of values, it follows that f assigns the
same output value to at least two of the elements in the domain. Thus f is not invertible.
22.11 ea /ab = eln(e
a
/eb )
= eln(e
a
)−ln(eb )
= ea−b
22.13 a) max. domain: D(f ) = R, domain of
f −1 : D(f −1 ) = R,
b) D(f ) = R, D(f −1 ) = R,
c) D(f ) = R, not invertible,
d) D(f ) = R r {0}, D(f −1 ) = R r {1},
e) D(f ) = [5/3, ∞), D(f −1 ) = [0, ∞),
f ) D(f ) = R r {2}, D(f −1 ) = R r {1}
22.15 a) 2e2x , b) 3e3x−2 , c) 3e3x , d) 1, e) −1,
2
f ) 10xe5x , g) ex + xex , h) 2xe4x + 4x2 e4x ,
i) −4/(ex − e−x )2 , j) 2e2x /(e2x + 1)2
22.17 Since P 0 (t)/(2000 − P (t)) = 6/5, it follows that
Z
P 0 (t)
dt
− ln |2000 − P (t)| =
2000 − P (t)
Z
6
6t
=
dt =
+ C.
5
5
Using the initial value P (0) = 100, we find that
C = − ln(1900), and therefore
2000 − P (t) = ±1900e−6t/5 .
The sign on the right-hand side must be positive, because otherwise, for t = 0, the values on
the two sides do not match. Hence
P (t) = 2000 − 1900e−6t/5 .
10
Solutions
22.19
a)
b)
y
f(x)
1.0
y
8
6
12
4
8
0.5
x
0
1
x
-2
c)
-1
0
1
2
Figure S.6: graph for Exercise 22.23.
d)
y
y
10
20
8
6
10
4
2
-2
-1
x
0
1
2
x
-3
e)
-2
-1
0
1
f)
y
y
8
12
6
8
4
4
2
-1
0
1
x
-2
g)
-1
0
1
2
h)
y
-1
y
x
1
4
2
-2
3
-4
2
-6
1
-8
-2
0
-1
−bt
22.25 First we apply the exponential function
to both sides of the given inequality to ob√
tain the equivalent inequality (x + y)/2 ≥ xy.
(Note: this is permissible because the exponential function is increasing.) Since x and y are
positive, the latter inequality in turn can be
written in the form (x − y)2 ≥ 0, and the validity of the original equation has thus been established.
Chapter 23
x
-2
1
4
2
-1
x
0.0
23.3 proof of b): ax /ay = ex ln(a) /ey ln(a) =
e(x−y) ln(a) = ax−y ,
proof of d): a0 = e0 ln(a) = e0 = 1,
proof of f): (a/b)x = ex ln(a/b) = ex(ln(a)−ln(b)) =
ex ln(a) /ex ln(b) = ax /ay
23.7 d/dx 3x−1 = 3x−1 ln(3), d/dx (1/2)3x =
x
x
3(1/2)3x ln(1/2), and d/dx 22 = 22 2x ln2 (2).
x
0
1
2
−at
22.21 Since C (t) = K(−be +ae )/(a−b),
it is easy to see that C has a critical point at
t0 = ln(a/b)/(a − b). Furthermore,
K
b2 e−bt0 − a2 e−at0
a−b
a2
b2 e−bt0 K
1 − 2 e−(a−b)t0
=
a−b
b
2 −bt0
b e
a
K
=
1−
a−b
b
< 0 (because a > b > 0).
23.9 a) y = −2x , b) y = 2−x , c) y = 10 − 2x ,
d) y = 2−(x−4) , e) y = −2−x , f ) y = −2−x .
Graph of f :
f(x)
8
6
4
C 00 (t0 ) =
According to the second derivative test, we may
thus conclude that C assumes its maximal value
at t0 .
22.23 local minimum at x = e−1/2 , no local
maximum, point of inflection at x = e−3/2 (see
Figure S.6 for the graph of f ).
2
x
-2
-1
0
1
2
3
23.11 a) Denoting by x the ticket price in
dollars and by N (x) = N0 ax the number of
spectators in dependence on the ticket price,
the given information implies that N0 a25 =
N (25) = 200 and N0 ax+5 = N (x + 5) =
9N (x)/10 = 9N0 ax /10. Solving for a and N0
yields a = (9/10)1/5 and N0 = 200(9/10)−5 .
Thus N (x) = 200(9/10)(x−25)/5 .
b) The revenue R(x) is given by the equation
R(x) = xN (x) = 200x(9/10)(x−25)/5 .
Solutions
c) Since R 0 (x) = (1 + x ln(9/10)/5)N (x), it is
easy to see that R assumes its maximal value at
x = 5/ ln(10/9) ≈ $47.46.
23.13 a) smallest concentration: C(0) = 5000;
largest concentration:
C(4/ ln(2)) = 500(10 + 4 · 2−1/ ln(2) / ln(2)) =
5000 + 2000/(e ln(2));
b) t = 8/ ln(2)
23.15 It will take about 4.32 months, or approximately four months and 10 days.
23.17 Since the assumptions on f and g imply that limx→∞ 2f (x) = limx→∞ 3g(x) = ∞, we
may apply L’Hôpital’s rule to conclude that
2f (x)
2f (x) f 0 (x) ln(2)
ln(2)
.
= lim g(x) 0
=
g(x)
x→∞ 3
x→∞ 3
ln(3)
g (x) ln(3)
lim
23.19 If the current year with an income of
$40, 000 is counted as the first year, and if payments into the retirement fund are always made
at the end of each year then, at the end of n
years, the amount of money in the fund will be
4000
n−1
X
1.06k 1.03n−1−k =
k=0
4(1.06n − 1.03n )
.
0.00003
Setting n equal to 40 yields an amount of
$936, 490.67, and the value in current dollars
will be $936, 490.67/1.02540 = $348, 777.81.
Chapter 24
24.3 proof of b): loga (x/y) = ln(x/y)/ ln(a) =
ln(x)/ ln(a) − ln(y)/ ln(a) = loga (x) − loga (y),
proof of d): loga (1) = ln(1)/ ln(a) = 0
24.5 The diagrams below for a = 1/4, 1/2, 2, 4
illustrate that loga (x) is decreasing for 0 < a <
1 and increasing for 1 < a. In fact, for 0 <
a < 1 we have loga (x) = − log1/a (x), and the
corresponding graphs are therefore reflections of
each other across the x-axis.
2
log a(x)
11
24.7 d/dx xln(x)
= xln(x) 2 ln(x)/x,
√
√
2
2−x
2
2−x
ln(x2 −3)+
d/dx
√(x −3) 2 = (x −3) e (− e−1
2x( 2 − x)/(x − 3)), d/dx x = ex
24.9 a) 3/((3x − 1) ln(2)),
b) x1/ log2 (5)−1 / log2 (5),
2
c) (log5 (3x2 ))x (2x ln(log5 (3x2 )))
2
+x/(log5 (3x ) ln(5))), d) x1/ ln(4)−1 / ln(4),
e) −2, f ) 2x log10 (x2 ) + 2x/ ln(10),
g) x4x (4 ln(x) + 4), h) 2x(6x)4x−1
+x2 (6x)4x−1 (4 ln(6x) + (4x − 1)/x),
i) exe−1 + ex ,
j) (πxπ−1 log10 (πx) − xπ /(x ln(10)))/ log210 (πx)
24.11 Since fa (x) = log2 (x) + log2 (a), the
graphs of fa for different values of a are vertical
shifts of each other:
f a(x)
4
a=4
3
a=2
2
a=1
1
0
x
1
2
3
-1
-2
24.13 Using 100 subdivisions
along the interval
R2
[1, a], we obtain: 1 log2 (x) dx
P100
≈ k=1 log2 (1 + k/100)/100 ≈ 0.562,
R3
log (x) dx
1P 3
100
≈ k=1 2 log3 (1 + 2k/100)/100 ≈ 1.190,
R4
log (x) dx
1P 4
100
≈ k=1 3 log4 (1 + 3k/100)/100 ≈ 1.851,
R5
log (x) dx
1P 5
100
≈
R e k=1 4 log5 (1 + 4k/100)/100 ≈ 2.535,
ln(x) dx
1P
100
≈ k=1 (e−1) ln(1+(e−1)k/100)/100 ≈ 1.009.
24.15 Since f 0 (x) = 0 and f (e) = f (1/e) = e,
it follows that f (x) = e for all x ∈ (0, ∞) r {1}.
a=2
1
0
-1
a=4
x
1
2
3
a = 1/4
a = 1/2
-2
Chapter 25
25.1 no answer
25.3 The trapezoid is easily seen to be composed of a rectangle of height h and width b
and two right triangles of height h. Since the
lengths of the base sides of the two triangles
add up to a − b, it follows that the total area of
the trapezoid is hb + h(a − b)/2 = h(a + b)/2.
12
Solutions
Chapter 26
e)
y
1.5
26.1 Dividing both sides of the equation
cos2 (t) + sin2 (t) = 1 by cos2 (t) yields 1 +
tan2 (t) = 1/ cos2 (t). The second identity is derived in a very similar fashion via a division by
sin2 (t).
26.3 d/dt sin(3t2 ) = 6t cos(3t2 ),
d/dt ecos(t) = − sin(t)ecos(t) , and
d/dt cos3 (4t − 1) = −12 sin(4t − 1) cos2 (4t − 1).
26.5 a) sin(x) + x cos(x), b) 2 cos(x) sin(x),
c) 3e3x / cos2 (e3x ), d) 4e4x sin(2x)+
2e4x cos(2x), e) xcos(x) (cos(x)/x − sin(x) ln(x)),
f ) 2x cos(3x) − 3x2 sin(3x), g) 2x cos(x2 ),
h) −2−x ln(2) cos(5x) − 2−x 5 sin(5x),
i) 2 sin(e4x ) + 8xe4x cos(e4x ),
j) (4 cos(4x) − 3 sin(3x))/(sin(4x) + cos(3x))
26.7
a)
y
x
6
-4
-2
2
4
-0.5
b)
y
4
2
x
-6
-4
-2
2
4
6
2
4
6
-2
c)
y
1
x
-6
-4
-2
-1
d)
1
-4
-2
2
4
2
4
6
-1.5
f)
1
-6
-4
-2
y
x
6
-1
-2
Figure S.7: graphs for Exercise 26.7.
26.13 If x is measured in degrees, then
limx→0 sin(x)/x = π/180 and d/dx sin(x) =
π cos(x)/180.
Chapter 27
0.5
-6
x
-6
y
x
6
-6
-1
For e) and f) see Figure S.7.
26.9 dy/dx = tan(x) − cot(x)
26.11 In solving for cos(θ) the equation
Cb4
dT
Cb4 cos(θ)
= 4 2
− 4 2
,
dθ
R sin (θ)
r sin (θ)
we find that cos(θ) = (r/R)4 as desired.
27.3 sin(arcsin(−0.1)) = −0.1,
arcsin(sin(7π/6)) = −π/6,
arccos(cos(−π/6)) = π/6, and
arctan(tan(π)) = 0.
27.5 Using the quotient rule and the trigonometric theorem of Pythagoras, we obtain
√
√
3s2 ( 3 − 2 cos(θ) + 3 cos2 (θ))
d2 A
.
=
dθ2
2 sin3 (θ)
√
√
Since the quadratic equation 3 − 2x + 3x2 =
0 is easily seen to have no solutions in R, it
follows that d2 A/dθ2 > 0 and, by implication,
A assumes
√ a minimum at its critcal point θ =
arccos(1/ 3).
27.7 d/dx arctan(x) = 1/(1 + tan2 (arctan(x)))
= 1/(1 + x2 ) and d/dx arccot(x) =
1/(−1 − cot2 (arccot(x)) = −1/(1 + x2 ).
√
2
27.9 a) arcsin(x)
√ + x/ 1 − x3x,
2
b) 2 arcsin(x)/ 1 − x , c)√3e /(1 + e6x ),
2
d) 4e4x arcsin(2x) + 2e4x / 1 − 4x
√,
arccos(x)
e) x
(arccos(x)/x√− ln(x)/ 1 − x2 ),
f ) 2x arccos(3x) − 3x2 / 1 − 9x2 , √
− 2−x 10x/ 1 − 25x4 ,
g) −2−x ln(2) arccos(5x2 )√
4x
4x
h) 2 arcsin(e ) + 8xe / 1 − e8x , i) 0
27.11 a) 2 arctan(x) + C, b) 2 arcsin(x) + C,
c) 5 ln(1 + x2 )/2 + C, d)
√ 3 arctan(2x)/2 + C,
e) arcsin(2x) + C, f ) 2 1 − x2 + C
27.13 Only the expressions in b) and d) are well
defined.
Solutions
27.15 Setting f (x) := arctan(x) + arctan(1/x),
we find that
1
1
1
−
· 2 =0
f (x) =
2
2
1+x
1 + 1/x x
0
for all x 6= 0. Consequently, there are constants C, D ∈ R such that f (x) = C for all
x ∈ (0, ∞) and f (x) = D for all x ∈ (−∞, 0).
In evaluating f at, for example, −1 and 1, we
obtain C = arctan(1) + arctan(1) = π/2 and
D = arctan(−1) + arctan(−1) = −π/2 as desired.
27.17 Since cos(x) > sin(x) for all x ∈
(−3π/4, π/4), it follows that f 0 (x) = cos(x) −
sin(x) > 0 for all x ∈ [−π/4, π/4) and, by
implication, f is strictly increasing. Furthermore, following the outline given in the hint,
−1
(x) = arcsin(x2 − 1)/2 and
we find that f√
(f −1 ) 0 (x) = x/ 2x2 − x4 .
Chapter 28
28.1 δ must be between 0◦ and 42.5◦ .
28.3 As we trace the path of a light ray from the
tip of the pencil to the observer (see Figure S.8),
we notice that the refraction at the surface effects a decrease in the angle of observation and
thus causes the pencil to appear bent.
observer
13
29.9 Setting L := limx→∞ xa /bx , it follows that
ln(L) = lim (a ln(x) − x ln(b))
x→∞
a ln(x)
− ln(b) = −∞,
= lim x
x→∞
x
because b > 1 and ln(b) > 0. Consequently,
L = 0.
29.11 The limits are 3/2 and −1/2.
29.13 The limit is 1.
29.15 The limit is −1/2.
29.17 For k = 2, we find that
1 − cos(sin(x))
sin(sin(x)) cos(x)
= lim
x→0
xk
2x
sin(sin(x))
cos(sin(x)) cos(x)
1
= lim
= .
= lim
x→0
x→0
2x
2
2
lim
x→0
29.19 continuous compounding: $200e0.06·10 ≈
$364.42,
quarterly compounding: $200(1 + 0.06/4)4·10 ≈
$362.80
29.21 According to L’Hôpital’s rule, we may
infer that
ef (x)
f 0 (x)ef (x)
=
lim
= 2.346.
x→∞ eg(x)
x→∞ g 0 (x)eg(x)
lim
Consequently,
f 0 (x)
f 0 (x)ef (x) eg(x)
=
lim
x→∞ g 0 (x)
x→∞ g 0 (x)eg(x) ef (x)
2.346
= 1.
=
2.346
lim
angle of
observation
pencil
Chapter 30
R2
R8
2
30.3 0 xe(2x ) dx = 0 eu /4 du = (e8 − 1)/4,
R 2 2 x3
R8 u
3x e dx = 0 e du = e8 − 1
0
R π/4
R1
30.5 0 tan(x)/ cos2 (x) dx = 0 u du = 1/2
Figure S.8: a pencil immersed in water.
28.5 The minimal height is 0.5 tan(42.5◦ ) ≈
0.46 miles.
Chapter 29
29.1 f increases far more rapidly for large values of x than g.
29.7 The limits are 1 and 1/4.
30.7 H(x) = ln3 (x)/3 + C
30.9 increase in velocity: 20000 ln(2) mi/h
30.13 a) π 2 /32, b) − ln(3)/4,
c) sin4 (x)/4 + C, d) −1/(2 cos2 (t)) + C,
e) −1/ ln(ln(x)) + C, f ) − ln2 (x) + C,
g) ee −e, h) arcsin2 (x−1)/2+C, i) 2/3, j) 8/15,
k) ln | sin(x)|+C, l) − cot2 (x)/2−ln | sin(x)|+C
m) 2(x +√2)5/2 /5 − 4(x +√2)3/2 /3 + C,√
n) (4 − 2 2)/3, o) 2(1 + x − ln(1 + x)) + C
R5
R2
30.15 1 f (3x − 1) dx = 2 f (u)/3 du = 10/3
14
R π/2
30.17 Since 4 = 0 cos(x)f (2 sin(x)) dx =
R2
R2
f (u)/2 du, we may infer that 0 f (u) du = 8.
0
30.19 v 0 (0) > 0 if and only if
−g − 50m 0 (0)/m(0) > 0 if and only if
|m 0 (0)| > g/50
30.21 a) arctan(x) + C, b) ln(1 + x2 )/2 + C,
c) x − arctan(x) + C
30.23 a) F (x3 )/3 + C, b) F (ln(x)) + C,
c) F (sin(x3 ))/3 + C
Chapter 31
31.3 cosh2 (x) − sinh2 (x)
= (ex + e−x )2 /4 − (ex − e−x )2 /4
= (e2x + 2 + e−2x − e2x + 2 − e−2x )/4 = 1.
31.5 x = cosh(y) if and only if 2x = ey + e−y if
and only √
if 0 = (ey )2 − 2xey + 1 if and only if
y
e = x ± x2 − 1.
√
31.7 RSince
A = xy − B = x x2 − 1 − B,
x√ 2
B = 1 u − 1 du, and
√
t = arcosh(x) = ln(x + x2 − 1), we may infer
that
p
x2
d
(A − B − t) = x2 − 1 + √
dx
x2 − 1
√
p
1 + x/ x2 − 1
2
√
−2 x −1−
x + x2 − 1
p
x2 − 1
= − x2 − 1 + √
= 0.
x2 − 1
Having thus shown A − B − t to be constant
as a function of x, we may determine the value
of this constant by setting x equal to 1. This
yields A − B − t = 0 − 0 − 0 = 0. Consequently,
t = A − B as desired.
31.11 The result easily follows from the following observation:
√
√
(1 − 2)2
1
√
√
√
=
= (1 − 2)2 .
2
2
(1 + 2)
((1 + 2)(1 − 2))
31.17 In generalizing the calculation in Exercise 31.16, we find that h(x)+xh 0 (x) = g(h(x)),
or equivalently, h 0 (x)/(g(h(x))−h(x)) = 1/x as
desired.
31.19 In analogy to our calculation in Exercise 31.18, we differentiate both sides of the
equation f (x) = h(x) − ax − b to conclude that
h 0 (x) − a = f 0 (x) = g(h(x)), or equivalently,
h 0 (x)/(g(h(x)) + a) = 1 as desired.
Solutions
R0√
R1 √
31.21 a) 0 x 1 − x2 dx = − 1 u/2 du
= 1/3,
R1 √
R π/2
b) 0 x2 1 − x2 dx = 0 sin2 (t) cos2 (t) dt
R π/2
R π/2
= 0 sin2 (2t)/4 dt = 0 (1 − cos(4t))/8 dt
= π/16,
R0
R1 √
√
c) 0 x/ 1 − x2 dx = − 1 1/(2 u) du = 1,
R1 2 √
R π/2 2
d) 0 x / 1 − x2 dx = 0 sin (t) dt
R π/2
= 0 (1 − cos(2t))/2 dt = π/4
31.23 a) In √
applying (31.12) with c = 0, d =
2, f (x) = 1/ 4 + x2 , and g(t) = 2 sinh(t), we
obtain:
Z 2
Z arsinh(1)
√
1
√
dx =
dt = ln(1 + 2).
4 + x2
0
arsinh(0)
√
b) Here we use (31.13) with f (x) = 1/ 4 + x2 ,
g(t) = 2 sinh(t), g −1 (x) = arsinh(x/2), and
H(x) = x to infer that
Z
x
1
√
dx = arsinh
+ C.
2
4 + x2
31.25 a) sinh(x) cosh(y) + cosh(x) sinh(y)
= (ex+y +ex−y −e−x+y −e−(x+y) +ex+y −ex−y +
e−x+y − e−(x+y) )/4
= (ex+y − e−(x+y) )/2 = sinh(x + y).
b) The proof is completely analogous to that in
a).
c) Setting y equal to x in a) yields sinh(2x) =
2 sinh(x) cosh(x).
d) The first equality is an immediate consequence of the result in b) with x = y and the
second and third inequalities are easily deduced
from the hyperbolic theorem of Pythagoras.
R1 √
31.27 Given that 10 = 0 f ( 4 − x2 ) dx =
R π/6
2f (2 cos(t)) cos(t) dt, we may infer that
R0π/6
f (2 cos(t)) cos(t) dt = 5.
0
Chapter 32
32.1 Since a2 = b2 + c2 = 13, the equation of
the ellipse is x2 /13 + y 2 /4 = 1.
32.3 In adopting essentially the same notation
as in Figure 32.2, we denote by...
• (x0 , y0 ) the point at which the light ray is
reflected off the graph of the hyperbola.
• α the angle between the incoming light ray
and the tangent line to the hyperbola at
(x0 , y0 ), which, according to the general
law of reflection, is the same as the angle
Solutions
between this tangent line and the reflected
light ray.
• β the angle between the positive x-axis and
the tangent line at (x0 , y0 ). Note: this
choice of β represents a slight deviation
from our use of notation in the case of the
elliptic mirror, because in Figure 32.2 β denoted the angle between the tangent line
and the negative x-axis.
• γ the angle between the negative x-axis and
(the extension of) the reflected light ray.
• δ the angle between the positive x-axis and
the incoming light ray.
Since the upper branch of the hyperbola is the
graph of the function
bp 2
x − a2 ,
f (x) = y =
a
we may infer that
bx0
b2 x0
tan(β) = f 0 (x0 ) = p 2
= 2 .
a y0
a x0 − a2
Furthermore, elementary geometry shows that
α + δ = β and α + β + γ = π. Hence
15
Chapter 33
33.1 The value of c will be greater for an adult
because the larger body of the adult has a larger
surface area and thus experiences a greater force
of air resistance at a given speed.
R1
R1
33.3 −1 x/(x2 +4x+4) dx = −1 1/(x+2) dx−
R1
2/(x + 2)2 dx = ln(3) − 4/3
−1
√
R1
R1√
33.5
1/(x2 −2) dx = 0 2/(4(x− 2)) dx−
√
√
√
R1√ 0
2/(4(x + 2)) dx = ln( 2 − 1)/ 2.
0
g
vmax − v(t)
1
= 2 t
ln
33.7 −
2vmax vmax + v(t)
vmax
2g
vmax − v(t)
=−
t
if and only if ln
vmax + v(t)
vmax
vmax − v(t)
= e−2gt/vmax .
if and only if
vmax + v(t)
√
33.11 Using the substitution u := x with
2u du = dx in each
√ case, we obtain the following
results: a) 2√
ln | x − 1| − ln(x) + C, √
b) 2 arctan( x − 1) + C, c) 4 − 4 ln( 2 + 1).
33.13 a) Rearranging the terms in the given
identities concerning the preservation of the momentum and the (kinetic) energy yields
γ = π − 2β + δ.
Since the reflected light ray is described by the
equation
y = − tan(γ)(x − x0 ) + y0 ,
we need to prove that
y0
= tan(γ),
c − x0
or equivalently,
y0
= tan(π − 2β + δ)
c − x0
tan(δ)(1 − tan2 (β)) − 2 tan(β)
=
.
1 − tan2 (β) + 2 tan(β) tan(δ)
Using the fact that tan(δ) = y0 /(c + x0 ), the
term on the right-hand side of the equation
above is easily seen to be equal to
b4 x2
2b2 x0
y0
1 − 4 02 − 2
c + x0
a y0
a y0
.
b4 x20
2b2 x0
1− 4 2 + 2
a y0
a (c + x0 )
Finally, in performing almost exactly the same
steps as in the calculation on p.248 (with the
identity b2 = c2 − a2 in place of b2 = a2 − c2 ),
we arrive at the desired conclusion that tan(γ)
is indeed equal to y0 /(c − x0 ).
M (v − v0 ) = m(u0 − u),
M (v 2 − v02 ) = m(u20 − u2 ).
Since v 2 − v02 = (v − v0 )(v + v0 ) and u20 − u2 =
(u0 + u)(u0 − u), we may infer that
M (v − v0 ) = m(u0 − u),
v + v0 = u0 + u.
In solving these equations for v and u, we find
that
2mu0
(M − m)v0
+
,
M +m
M +m
2M v0
(m − M )u0
+
.
u=
m+M
m+M
v=
b) If u0 = 0, then the change in momentum is
M (v−v0 ) =
M (M − m)v0
2mM v0
−M v0 = −
.
M +m
m+M
c) Since the number of collisions N (t) is certainly proportional to the distance traveled,
and since the distance traveled is approximately
equal to v(t)∆t, it follows that there is a constant of proportionality λ such that N (t) =
λv(t)∆t.
16
Solutions
d) Since the total change in momentum over
the time span from t to t + ∆t is approximately
equal to the number of collisions multiplied by
the change in momentum effected by each of
them, we may use the results of b) (with v(t) in
place of v0 ) and c) to infer that
illustration for
Rn
m−1
f (x) dx ≥
Pn
k=m
f (k):
f(x)
2mM v(t)N (t)
p(t + ∆t) − p(t)
≈−
∆t
(m + M )∆t
2mM λ
v(t)2 .
=−
m+M
F (t) ≈
In defining c to be equal to 2mM λ/(m + M )
we arrive at the desired conclusion: F (t) ≈
−cv(t)2 .
33.15 a) v(t) = 2 tan(2t + arctan(2)) − 2 =
(6 tan(2t) + 2)/(1 − 2 tan(2t)),
2
b) v(t) = 4/(2 − et ) − 3
Chapter 34
34.1 3/8
34.3 210 1
9
9
9
9
130
34.5 9
+
+
+
=
0
1
2
3
2
512
20 X
32
1
≈ 0.89
34.9 32
k
2
k=12
550 X
1
1000
34.11 1000
≈ 0.9986
k
2
k=450
3 1 4
9
34.13 · 4
=
4 2 2
32
365!
≈ 0.507
34.15 1 −
36523 342!
Chapter 35
35.3
R n+1
Pn
illustration for k=m f (k) ≥ m f (x) dx:
f(x)
x
m
n+1
x
m-1
n
R
35.13 arcsin(x) dx√
R
= x arcsin(x) − R x/ √
1 − x2 dx
= x arcsin(x) + √1/(2 u) du
= x arcsin(x) + 1 − x2 + C
Qn
n
35.15 k=2 k k = e k=2 k ln(k)
2
≤ e(n+1) (ln(n+1)−1/2)/2−ln(4)+1
2
2
= (n + 1)(n+1) /2 e−(n+1) /4 e− ln(4) e1
2
√
(n+1) −(n+1)2 /4
= (1/4) n + 1
e
R1
35.17 a) 0 arctan(x) dx
R1
1
= x arctan(x)|0 − 0 x/(1 + x2 ) dx =
1
(x arctan(x)
− ln(1 + x2 )/2)0 =R π/4 − ln(2)/2,
R
b) x sinh(x) dx = x cosh(x) − cosh(x) dx
= xRcosh(x) − sinh(x) + C,
R
c) x3 ln(x) dx = x4 ln(x)/4 − x3 /4 dx =
x4 ln(x)/4 − x4 /16 + C,
R2
R4
2
d) 0 x3 ex dx = 0 ueu /2 du
R
4
4
= ueu /2|0 − 0 eu /2 du = (3e4 + 1)/2,
R 2
R
e) ln (x) dx = x ln2 (x) − 2 ln(x) dx
= xR ln2√(x) − 2x ln(x) + 2x + C,
f ) x x + 1 dx
R √
√
3
3
= 2x x + 1 /3 − 2 x + 1 /3 dx
√
√
3
5
= 2x x + 1 /3 − 4 x + 1 /15 + C
Pn
Qn
1/k
= k=4 ln(k)/k
35.19 a) ln
k=4 k
Rn
≤ 3 ln(x)/x dx = ln2 (n)/2 − ln2 (3)/2, and
Qn
√ ln(n) √ ln(3)
/ 3
. Furtherefore, k=4 k 1/k ≤ n
R n+1
Pn
thermore,
ln(x)/x dx =
k=4 ln(k)/k ≥ 4
ln2 (n + 1)/2 − ln2 (4)/2 and, by implication,
√
Qn
ln(n+1)
1/k
n + 1 P /2ln(4) .
k=4 kQ ≥
n
n
k ln(k) = k=1 ln2 (k)
b) ln
R
Pn k=1
n
= k=2 ln2 (k) ≥ 1 ln2 (x) dx
2
(n) − 2n ln(n) + 2n − 2, and therefore,
=
Qnn ln ln(k)
n ln(n) 2n
2 2n
k
≥
e /(e
k=1
n ).
Qn
n
2
c) Since ln
k=1 (k + 1)
R n+1
Pn
2
= k=1 ln(k + 1) ≤ 1
ln(x2 + 1) dx
R
n+1
n+1
−
2x2 /(1 + x2 ) dx
= x ln(x2 + 1)
P
1
1
Solutions
= (n + 1) ln(n2 + 2n + 2) − ln(2) − 2n
+2 arctan(n + 1) − 2 arctan(1) and
arctan(n
+ 1) < π/2, we may infer that
Qn
2
(k
+ 1)
k=1
≤ (n2 + 2n + 2)n+1 e2 arctan(n+1) /(2e2n+π/2 )
≤ (n2 + 2n + 2)n+1 eπ/2 /(2e2n )
< 5(n2 + 2n + 2)n+1 /(2e2n ).
R n+1
Pn
d) ln(n + 1) = 1
1/x dx ≤ k=1 1/k
R
Pn
n
= 1 + k=2 1/k ≤ 1 + 1 1/x dx = 1 + ln(n).
e) With ∆x := π/(2n) and xk := k∆x it follows
that R
Pn
π/2
1 = 0 sin(x) dx
k=1 sin(xk )∆x, and
Pn≤
therefore, 2n/π ≤ k=1 sin(πk/(2n)).
35.21 a) Since f (a) = a and f (b) = b if and
only if f −1 (a) = a and f −1 (b) = b, the reRb
sult of Exercise 35.20 implies that a f (x) dx +
R f −1 (b)
Rb
R b −1
f (x) dx = f −1 (a) f (t) dt + a f −1 (x) dx =
a
b
xf −1 (x)a = b2 − a2 as desired.
b) Here the given assumptions allow us to infer that f −1 (b) = a and f −1 (a) = b. Hence
b
R b −1
R f −1 (b)
f (x) dx = xf −1 (x)a − f −1 (a) f (t) dt =
a
Rb
Ra
0 − b f (x) dx = a f (x) dx.
35.23 The correct answer is c).
17
Rn
≥ n ln(n)/2 + ln(2) + 2 x ln(x) dx = n ln(n)/2
− ln(2) + n2 ln(n)/2 −n2 /4 + 1 or,
equivalently,
Qn
√ n2 +n
√
n2 −4
k
4
n
/ 2 e
.
k=1 k ≥
Pn
37.9 a) k=3 arctan(k + 1)
≤ Rarctan(n + 1)/2 + arctan(4)/2
n
+ 3 arctan(x + 1) dx
= (n + 3/2) arctan(n + 1) − 7 arctan(4)/2
+ ln((n2 + 2n + 2)/17)/2,
R n+1/2
Pn
arctan(x + 1) dx
k=3 arctan(k + 1) ≥ 5/2
= (n + 3/2) arctan(n + 3/2) − 7 arctan(7/2)/2
+ ln((4n2 + 12n + 13)/53)/2,
R n+1/2
Pn
b) k=5 k/(k 2 + 1) ≤ 9/2
x/(x2 + 1) dx
= ln((4n2 + 4n + 5)/85)/2
P
n
k/(k 2 + 1) ≥ 5/52 + n/(2(n2 + 1))
Rk=5
n
+ 5 x/(x2 + 1) dx = 5/52 + n/(2(n2 + 1))
+ ln((n2 + 1)/26)/2
Qn
ln(k)
37.11 a) Since ln
k=4 k
Pn
= R k=4 ln2 (k) ≤ ln2 (n)/2 + ln2 (4)/2
n
+ 4 ln2 (x) dx = (n + 1/2) ln2 (n)
−2n ln(n) + 2n − 7 ln(2) ln(4) + 8 ln(4) − 8,
we may infer that
n
Y
k ln(k) ≤
k=4
Chapter 36
1 34 6
1 33 6
1 32 6
36.3 2 · 4
+ 3· 3
+ 4· 2
4 4 2
4 4 3
4 4 4
≈ 0.46
Chapter 37
37.5 Trap(n1 , f, −2, 0) =
P20
−1/20 + k=0 ((−2 + k/10)3 /8 + 1)/10 ≈ 1.499,
Mid(n2 , f, 0, 3)
P30
= k=1 ((−1 + (2k − 1)/20)3 /8 + 1)/10 ≈ 5.530
37.7 Since the function f (x) = x ln(x) is easily
seenQ to be concave
may
Qnup, kwe
Pn infer that
n
k
k
k
ln
=
ln
=
k=1
k=2
k=2 k ln(k)
R n+1/2
x ln(x) dx
≤ 3/2
n+1/2
2
= (x ln(x)/2 − x2 /4)3/2
2
= (n + 1/2)2 ln(n + 1/2)/2 − (n + 1/2)
Qn /4 k
−9 ln(3/2)/8 + 9/16, and therefore, k=1 k
p
(n+1/2)2 9/16 9/8
2
9/8
≤ n + 1/2
e
2 / e(n+1/2) /43
p √ 9p
(n+1/2)2 √
9 √ (n+1/2)2
.
n + 1/2
/ 83 4e
= 8 2 e
Using now a trapezoid approximation
for the
Qn
k
lower estimate, we see that ln
k=1 k
n(n+1/2) ln(n) e2n−8
.
47 ln(2)−8 n2n
Furthermore,
R n+1/2 2
Pn
2
ln (x) dx
k=4 ln (k) ≥ 7/2
= (n + 1/2) ln2 (n + 1/2)
−2(n + 1/2) ln(n + 1/2) + 2(n + 1/2)
−7 ln2 (7/2)/2 + 7 ln(7/2) − 7
and, by implication,
n
Y
k ln(k) ≥
k=4
(n + 1/2)(n+1/2)(ln(n+1/2)−2)
.
e6−2n (7/2)7 ln(7/2)/2−7
Q
n
ek
b) In this case we have ln
k=1 e
R n+1/2 x
Pn
√
= k=1 ek ≤ 1/2
e dx = e(en − 1), and
√
Qn
k
n
therefore, k=1 ee ≤ e e(e −1) .
The
estimate is R obtained as follows:
Pn lower
n
k
n
e
≥
e
/2 + e/2 + 1 ex dx = 3en /2 − e/2
k=1
Qn
√ 3en −e
k
and, by implication, k=1 ee ≥ e
.
√
√
Qn
n
1/(k k)
1/(k
k)
37.13 k=1 e
= e k=1
≥
√
√ 5 √ (4−1/n)/√n
5/2−(2−1/(2n))/ n
e
= e / e
P
Chapter 38
38.3
R∞
0
∞
2−x dx = −2−x / ln(2)|0 = 1/ ln(2)
18
Solutions
38.5 Since
Z ∞
2
Chapter 39
1
dx =
x lnα (x)
Z
∞
ln(2)
1
du,
uα
39.5 The verification is completely analogous to
the one in Exercise 39.2.
(j)
we may infer that the integral is convergent for
α > 1 and divergent for α ≤ 1.
R1 √ 3
R1 √
3
38.17 0 x/ x2 + x dx ≤ 0 x/ x dx
R1
= 0 1/x3/2 dx < ∞ (convergent),
R1 2
R1
x /(x4 + x5 + x6 ) dx ≥ 0 x2 /(3x4 ) dx
0R
1
1/(3x2 ) dx = ∞ (divergent),
=
R1 0 √
R1
1/ x2 + x dx ≤ 0 1/x1/2 dx < ∞ (conver0
gent),
R 1/2 √
R 1/2
1/ x3 − x6 dx ≥ 0 1/x3/2 dx = ∞ (di0
vergent),
√
R2
R1 √
3
(x + 1)/ 3 x4 + x6 dx ≥ 0 1/ 2x4 dx
0R
1
≥ 0 1/(2x4/3 ) dx = ∞ (divergent)
38.19 a) 1, b) 1/2, c) ln(3)/2, d) π/2
38.21 a) Since f is strictly decreasing, it follows
Rb
Rb
that c f (x) dx ≥ c f (b) dx = f (b)(b − c), and
Rb
therefore, bf (b) ≤ cf (b) + c f (x) dx.
b) Since the limit of f (x) as x tends to ∞ is assumed to be zero, it
follows that Rlimb→∞ bf (b)
R∞
∞
≤ c limb→∞ f (b) + c f (x) dx = c f (x) dx.
c) According
R ∞ 0 ≤ limb→∞ bf (b)
R ∞to b), we have
≤ limc→∞ c f (x) dx = 0 f (x) dx−
R∞
R∞
Rc
limc→∞ 0 f (x) dx = 0 f (x) dx − 0 f (x) dx
= 0, and therefore limb→∞ bf (b) = 0 as desired.
d) f is invertible because it is assumed to be
strictly decreasing. Furthermore, since f (x) is
positive and limx→∞ f (x) = 0, we may infer
that limx→0+ f −1 (x) = limx→∞ f −1 (f (x)) =
limx→∞ x = ∞.
e) Using
R a the result of Exercise
R a35.20, it follows
that 0 f −1 (x) dx = limc→0+ c f −1 (x) =
a
R f −1 (a)
limc→0+ xf −1 (x) − limc→0+ −1
f (t) dt =
c
f
(c)
39.7 For 0 ≤ j ≤ n we have Pn,x0 (x0 )
Pn
= k=j f (k) (x0 )(x − x0 )k−j /(k − j)!
x=x0
= f (j) (x0 ) as desired.
Pn
39.11 Q2n (x) = k=0 (−1)k x2k /(2k)!
39.15 ln(1/2) ≈ P7,1 (1/2)
P7
k+1
(1/2)k /k ≈ −0.692 and, sim=
k=1 (−1)
ilarly ln(3/2) ≈ P7,1 (3/2) ≈ 0.405, ln(2) ≈
P7,1 (2) ≈ 0.760 and ln(3) ≈ P7,1 (3) ≈ 12.69.
Using a calculator, we notice that the errors
in the approximation are markedly smaller for
ln(1/2) and ln(3/2) than for ln(2) and ln(3).
39.17 Using the result of Exercise 39.8, it follows that P7,3 (x) = f (x).
Pn
39.19 a) Pn,2
= k=0 4 lnk (2)(x − 2)k /k!,
P(x)
n
b) Qn (x) = Pk=1 xk /(k − 1)!,
n
c) Pn,1 (x) =P k=0 (k + 1)(−1)k (x − 1)k ,
n
d) Qn (x) = k=0 xk
39.21 Here we use the formula cos2 (x) =
(cos(2x) + 1)/2 to infer that Q2n (x) = 1 +
P
n
k 2k−1 2k
x /(2k)!.
k=1 (−1) 2
Chapter 40
40.3 Setting f (x) := arctan(x), the mean value
theorem allows us to infer that
| arctan(x) − arctan(x0 )| = |f (x) − f (x0 )|
= |f 0 (t)(x − x0 )| = |x − x0 |/(1 + t2 ) ≤ |x − x0 |
as desired.
40.11 With f (x) := sin(x) we obtain:
limx→0 sin(x)/x
= limx→0 (Q1 (x) + f 00 (t)x2 /2)/x
= limx→0 (1 − sin(t)x/2) = 1.
af −1 (a) − limb→∞ f (b)f −1 (f (b))
Rb
+ limb→∞ f −1 (a) f (t) dt = − limb→∞ bf (b)
Rb
R∞
+ limb→∞ 0 f (t) dt = 0 f (x) dx.
R∞
R∞
38.23 a) e 1/(x ln2 (x)) dx = 1 1/u2 du = 1,
R∞
R∞
x
b) 0 e−e ex dx = 1 e−u du = 1/e, c) divergent, d) divergent
40.13 Since 211 /11! ≈ 0.51 · 10−4 , it follows
that the error in the approximation sin(2) ≈
Q10 (2) ≈ 0.90935 is less that 10−4 .
38.25 Since the inequality ln(f (x))/x ≥ 2.416
can be written in the form
(x) ≤ e−2.416x ,
R ∞ 1/f
−2.416x
dx is obviand since the integral 1 e
ously convergent, Theorem
38.7
allows
us to inR∞
fer that the integral 1 1/f (x) dx is convergent
as well.
f (t) = g 0 (t) =
40.17 In applying theR mean value theorem to
x
the function g(x) := a f (u) du, we may conclude that there exists a t in (a, b) such that
1
g(b) − g(a)
=
b−a
b−a
Z
b
f (x) dx.
a
40.19 p(x) = x2 = f (x). In general, for f (x) =
Solutions
αx2 + βx + γ we find that
f (b) − f 0 (a)(b − a) − f (a)
(x − a)2
(b − a)2
+ f 0 (a)(x − a) + f (a)
p(x) =
α(b + a) + β − (2αa + β)
(x − a)2
b−a
+ (2αa + β)(x − a) + αa2 + βa + γ
=
19
for all sufficiently large values of n, and the comparison test, as stated in Theorem 41.7,
P∞ therefore allows us to infer that the series k=0 ak is
convergent.
c) Setting an := 1, we easily observe that the
given condition does not guarantee the series to
be convergent.
d) This condition guarantees the convergence of
the series by the ratio test.
= α(x − a)2 + 2αa(x − a) + βx + αa2 + γ
= αx2 + βx + γ = f (x).
In other words, the second order secant polynomial of a second order polynomial is the polynomial itself.
Chapter 41
41.17 If 0 < |q| < 1, then ln |q| < 0 and, by
implication, limn→∞ |q|n = limn→∞ en ln |q| =
limx→−∞ ex = 0. Hence limn→∞ q n = 0 whenever −1 < q < 1 (the case q = 0 is trivial).
Furthermore, for q = 1 we have limn→∞ q n =
limn→∞ 1 = 1, and for q > 1, it follows that
limn→∞ q n = limn→∞ en ln(q) = limx→∞ ex =
∞. Finally, if q ≤ −1, then the limit of q n does
not exist because q n ≥ 1 for even values of n
and q n ≤ −1 for odd values of n.
P∞
41.19 k=2 1/5k = 1/(1−1/5)−1−1/5 = 1/20
41.25 Since limn→∞ |an+1 /an |
= limn→∞ (n + 1)!nn /(n!(n + 1)n+1 )
= limn→∞ nn /(n + 1)n = limn→∞ 1/(1 + 1/n)n
= 1/e < 1,Pthe ratio test allows us to infer that
k
the series ∞
k=1 k!/k is convergent.
P∞
P∞
41.27 k=m q k = q m k=0 q k = q m /(1 − q)
41.29 Since the ratio of the side lengths of two
successive equilateral triangles that are cut out
is obviously 1/2, it follows that the total area of
the remaining figure is
√
√ ∞ k
3
3X 1
3
=
A=
−
.
4
4
2
6
√
k=1
41.31 a) 6/5, b) 1/90, c) 1, d) 1/(e3 − 1),
e) 1/24
P∞
41.33 a) Since the series k=1 1/k is divergent,
and since limn→∞ 1/n = 0, it follows that the
given condition does not guarantee the series to
be convergent.
b) The condition in b) implies that |an | < 1/n2
Chapter 42
42.3 The convergence at x = x0 is obvious. If x 6= x0 , then the assumption λ =
∞ in conjunction
with (42.1) implies that
p
limn→∞ n |an (x − x0 )n | = ∞. Consequently,
Theorem
P∞ 41.20b allows us to infer that the series k=0 ak (x − x0 )k is divergent.
42.5 radius of convergence:
√
r = 1/λ = 1/ limn→∞ n 3n = 1/3;
interval of convergence: (2/3, 4/3)
42.7 Since
limn→∞ |an+1 (x − x0 )n+1 /an (x − x0 )n |
= λ|x−x0 | for all x 6= x0 , Theorem
P∞ 41.23 allows
us to conclude that the series k=0 ak (x − x0 )k
is convergent whenever λ|x − x0 | < 1 and divergent whenever λ|x−x0 | > 1. In other words, the
series is convergent for x ∈ (x0 − 1/λ, x0 + 1/λ)
and divergent for x ∈ R r [x0 − 1/λ, x0 + 1/λ].
By implication, the radius of convergence is
r = 1/λ.
42.9 radius of convergence:
r = limn→∞ (n + 1)n+1 n!/(nn (n + 1)!)
= limn→∞ (1 + 1/n)n = e;
interval of convergence: (−2 − e, −2 + e)
42.11 According to L’Hôpital’s rule, we have
limx→∞ ln(x1/x ) = limx→∞ ln(x)/x
= limx→∞
√1/x = 0 and, by implication,
limn→∞ n n = e0 = 1.
This proves that the two series have thepsame radius of convergence because
limn→∞ n |nan | =
p
√
n
limn→∞ n n · limn→∞ |an |.
42.21 a) and c) are power series
but b) and d)
√
are not. In b) it is the term x that is inconsistent with the general representation√of a power
series, and in d) it is the exponent k because
√
k is not always an integer.
3
42.23
− 3x2 + 5 for x ∈ R,
P∞a) x2k+1
/(2k + 1)! for x ∈ R,
b) P k=0 x
∞
c) Pk=0 (−1)k x4k+5 /(2k + 1)! for x ∈ R,
∞
d) k=0 (−1)k 22k x2k+1 /(2k + 1)! for x ∈ R,
20
Solutions
P∞
e) P k=0 (−1)k 22k x2k /(2k)! for x ∈ R,
∞
f ) k=0 (−8)k x3k for x ∈ (−1/2,
√ 1/2),
√
P∞ −k−2 2k
g) Pk=0 2
x for x ∈ (− 2, 2),
∞
h)P k=3 k(k − 1)(k − 2)xk−3 /6 for x ∈ (−1, 1),
∞
i) Pk=0 22k x2k+1 /(2k + 1)! for x ∈ R,
∞
j) Pk=0 (−1)k xk+2 for x ∈ (−1, 1),
∞
k) P k=0 (−1)k x6k+2 /(2k + 1)! for x ∈ R,
∞
l) k=0 (−2)k xk for x ∈ (−1/2, 1/2)
42.25 a) The interval of convergence
is
√
n
n
=
(−1, 1) in
each
case
because
lim
n→∞
√
√
n
limn→∞ n2 = limn→∞ n 1 = 1 (see also Exercise 42.11).
P∞
b) For x = 1 the corresponding series k=1 1/k
is divergent by the integral test. For x = −1 we
have
∞
X
(−1)k
1 1 1
− + ± ...
k
2 3 4
k=1
∞ ∞
X
X
1
1
1
−
=−
,
=
2k 2k − 1
2k(2k − 1)
b) Using the results in a), we may infer that
∞
X
1
d
2
=
=
k(k − 1)xk−2
(1 − x)3
dx (1 − x)2
k=2
=
∞
X
∞
X
k 2 xk−2 −
k=2
kxk−2 ,
k=2
and therefore,
∞
X
k 2 xk = x +
k=1
∞
X
k 2 xk
k=2
∞
∞
k=2
k=1
X
X
2x2
2x2
k
=x+
+
kx
=
+
kxk
3
3
(1 − x)
(1 − x)
2x2
x
x2 + x
=
+
=
.
(1 − x)3
(1 − x)2
(1 − x)3
= −1 +
k=1
k=1
this series is Rconvergent because
∞
1/(2x(2x
− 1)) dx ≤ 1 1/(2x − 1)2 dx =
1
1/2 < ∞.
c) Here the convergence at x = ±1 follows
from the criterion stated in Exercise 41.8 in
conjunction with the Rintegral test because
∞
|(±1)n /n2 | = 1/n2 and 1 1/x2 dx < ∞.
d) The series is divergent at x = ±1 by the
sequence test because (±1)n does not approach
zero as n tends to ∞.
R −x3
dx
42.27
P∞ a) e k 3k+1
/(k!(3k
+ C,
= R k=0 (−1) x
P + 1))
k
x
/(k!k)
+ C,
b) R ex /x dx = ln(x) + ∞
k=1
c) Parctan(x2 )/x dx
∞
= k=0 (−1)k x4k+2 /((2k + 1)(4k + 2)) + C
Substituting 1/3 for x yields
(1/9 + 1/3)/(8/27) = 3/2.
b) Using the formula in Theorem 42.6, we obtain
1
limn→∞ |an+1 /an |
α(α − 1) · · · (α − n + 1)(n + 1)! = lim n→∞
α(α − 1) · · · (α − n)n!
n+1
= 1.
= lim n→∞ α − n r=
c) Given the definition of g in a), it follows that
(1 + x)g 0 (x) =
∞
X
α(α − 1) · · · (α − k + 1)
k=1
+
(k − 1)!
k=1
=α+
it follows that
k=1
x
kx =
(1 − x)2
+
k!
∞
X
α(α − 1) · · · (α − k + 1)k
k=1
=α 1+
for all x ∈ (−1, 1). Substituting 1/2 for x yields
P
∞
k
2
k=1 k/2 = (1/2)/(1 − 1/2) = 2.
k!
= αg(x),
k!
xk
xk
∞
X
α(α − 1) · · · (α − k + 1)
k=1
xk−1
xk
∞
X
α(α − 1) · · · (α − k + 1)(α − k)
k=1
k
(k − 1)!
∞
X
α(α − 1) · · · (α − k + 1)
k=1
∞
X
k 2 /3k =
f (k) (x) = α(α − 1) · · · (α − k + 1)(1 + x)α .
42.29 a) Since
∞
k=1
42.31 a) Here we only need to observe that the
kth derivative of f is
and
R∞
X
1
d 1
=
=
kxk−1 ,
(1 − x)2
dx 1 − x
P∞
x
k
!
Solutions
or equivalently,
α
g 0 (x)
=
for all x ∈ (−1, 1).
g(x)
1+x
d) Integrating both sides of the resulting equation in c) with the substitution u = g(x) on the
left yields
ln |g(x)| = α ln |1 + x| + C = α ln(1 + x) + C.
Setting x equal to zero, we find that 0 = ln(1) =
ln |g(0)| = ln(1) + C = C, and therefore, g(x) =
±(1 + x)α . The sign on the right-hand side of
this equation must be positive because g(0) =
1 > 0. Thus, we have shown that
g(x) = (1 + x)α for all x ∈ (−1, 1)
as desired.
e) If α = n, then the product
α(α − 1) · · · (α − k + 1)
is zero for all k > n. Given the definition of
n
k as stated in Chapter 34, the result in d)
therefore allows us to infer that
n
x
n
n
(x + y) = y 1 +
y
!
n
X n(n − 1) · · · (n − k + 1) xk
n
1+
=y
k!
yk
k=1
n n k
X
X
n x
n k n−k
n
x y
=y
=
.
k yk
k
k=0
k=0
21
and the general solution is y = x/(ln(x2 ) + C).
b) The equation p
is separable, and the general
solution is y = ± − ln(C − 3x2 )
43.17 y = C|x|−b/a
43.19 a) 7 ln |y| − 3y = x + C,
2
b) y = 2 arctan(Cesin(x) ), f ) y = 2 + Cex /2+x
43.21 b) y/x + 2 ln |y/x − 1| = ln |x| + C,
c) y = x tan(ln |x| + C)
Chapter 44
44.1 The derivative of Q(t) as given in (44.5)
is −Q0 e−t/RC /RC = −Q(t)/RC. Thus Q 0 (t) +
Q(t)/RC = 0.
44.3 Using the result of Exercise 44.2 and setting C := y0 eA(x0 ) , it follows that y(x) =
y0 eA(x0 )−A(x) = Ce−A(x) is a solution of the
differential equation in (44.10). Furthermore,
y(x0 ) = y0 eA(x0 )−A(x0 ) = y0 as desired.
44.5
y 0 (x) + a(x)y(x) = f (x)eA(x) e−A(x)
Z
− a(x)
f (x)eA(x) dx + D e−A(x)
Z
+ a(x)
f (x)eA(x) dx + D e−A(x)
= f (x)
as desired.
44.7 y(x) =
Chapter 43
43.9p
solution of y 2 y 0 − 2xy 2 y 0 − 1 = 0:
3
y = C − 3 ln |1 − 2x|/2;
solution of y 0 = xy − xy 0 :
y = Cex /(1 + x)
2 0
3
43.11
p solution of y y = x , y(1) = 0:
y = 3 3(x4 − 1)/4;
solution of sin(y)y 0 = cos(x), y(π/6) = π:
y = arccos(− sin(x) − 1/2)
43.13 Since z 0 /(z 2 −2z) = z 0 /(2z−4)−z 0 /(2z),
it follows that ln |z − 2|/2 − ln |z|/2 = ln |x| + D
or, equivalently, (z − 2)/z = ±e2D x2 . Thus, in
defining C to be equal to ±e2D , we find that
z(1 − Cx2 ) = 2, and therefore, z = 2/(1 − Cx2 )
as desired.
43.15 a) The equation is homogeneous because
(xy−2y 2 )/x2 = y/x−2(y/x)2 = z−2z 2 = F (z),
R
x
1
tet
2
/2
2
dt + 2e1/2 e−x /2
2
= 1 + e(1−x )/2
Verification: since y 0 /(1−y) = x, it follows that
− ln |1 − y| = x2 /2 + C, and the initial condition
y(1) = 2 implies that C = −1/2. Hence y − 1 =
2
2
e(1−x )/2 , or equivalently, y = 1 + e(1−x )/2 as
desired.
44.9 Using integration by parts, it follows that
Z
t
0
t
sin(ωτ )eτ /RC dt = RC sin(ωτ )eτ /RC − ωRC
0
Z
t
cos(ωτ )eτ /RC dτ
0
t
= RC sin(ωt)et/RC − ωR2 C 2 cos(ωτ )eτ /RC 2
2
−ω R C
2
0
Z
t
τ /RC
sin(ωτ )e
0
dτ,
22
Solutions
and therefore,
Z
t
sin(ωτ )eτ /RC dt
0
=
RC(et/RC (sin(ωt) − ωRC cos(ωt)) + ωRC)
.
1 + ω 2 R2 C 2
Multiplying both sides by E0 /R yields the desired result in (44.16).
44.11 a) y(x) = x2 (3x + D)e−x ,
b) y(x) = ln(x) + D/ ln(x),
c) y(x) = − cos(x) + (sin(x) + D)/x
44.13 Introducing the substitution u = 1/y, it
follows that y 0 = −u 0 /u2 , and the corresponding differential equation in u is u 0 −3u/x = −x2 .
Solving for u yields u(x) = (− ln(x)+D)x3 , and
therefore, y(x) = 1/((− ln(x) + D)x3 ).
√
44.15√a) Q(t) = (cos(10t) + 35 sin(10t))/72 −
e−10t/ 35 /72.
√
b) Since Q(t) = sin(10t + α)/12 − e−10t/ 35 /72
for α = arcsin(1/6), it follows that the phase
shift is arcsin(1/6)/10.
√
√
c) I(t) = 5 cos(10t+α)/6+5e−10t/ 35 /(36 35)
and Is (t) = 5 cos(10t + α)/6.
d)
Is(t)
1.0
0.5
t
0.0
0.5
1.0
1.5
-0.5
-1.0
44.17
A sin(t + α) = 2 sin(t)
√ − 3 cos(t) for A =
√
13 and α = − arcsin(3/ 13), and
√
A cos(t + β) = 2 sin(t)√− 3 cos(t) for A = 13
and β = π + arcsin(2/ 13).
44.19 a) y = x + C, b) 4 ln |1 + 3y|/9 − y/3 =
x2 /2 + C, c) y = Cex , d) y = −1/(x + C),
e) y = x/(1 + Cx), f ) y = x arcsin(Cx),
g) y = e2x + Dex , h) y = x2 /2 + C
p
2πk for some in|z| = n |w| and nφ = θ +p
teger k. This shows that n |w|ei(θ+2πk)/n is
an nth root of z for all integers k. Furthermore, the roots corresponding to values k between 0 and n − 1 are distinct because the corresponding angles (θ + 2πk)/n are distinct in
the complex plane, and if j is an integer outside the range from 0 to n − 1, then a division
of j by n with a remainder k ∈ {0, . . . , n − 1}
allows us to infer the existence of an integer
p such that j = pn + k and, by implication,
e(θ+2πj)/n = e(θ+2πk)/n+2πp = e(θ+2πk)/n . The
latter observation shows that there are no nth
roots other than those corresponding to values
k ∈ {0, . . . , n − 1}.
iπ/4
)(z − ei3π/4
45.21 z 4 + 1 = (z − e√
√ )(z −
i5π/4
i7π/4
2
)(z −e
) = (z − 2z +1)(z 2 + 2z +1)
e
45.23 real part: 5/17; imaginary part: 14/17
45.25 x5 −32 = (x−2)(x2 −4 cos(2π/5)+4)(x2 −
4 cos(4π/5) + 4)
45.27 x4 + x2 + 1 = (x2 − ei2π/3 )(x2 − ei4π/3 ) =
(x2 − x + 1)(x2 + x + 1)
45.29 ei7π/20 , ei(7π/20+2π/5) , ei(7π/20+4π/5) ,
ei(7π/20+6π/5) , ei(7π/20+8π/5)
45.31 r = 5, θ = π + arctan(3/4)
45.33 The
√ real and imaginary parts are 512/243
and 512 3/243 respectively, because
√ 10
√
(1 − i/ 3)10 = 2ei5π/6 / 3
= 210 ei50π/6 /35√= 1024ei2π/6 /243
= 1024(1/2 + i 3/2)/243.
45.35 Since f1 f2 = u1 u2 −v1 v2 +i(u1 v2 +v1 u2 ),
we may apply the product rule (for ordinary
real-valued functions) to infer that
(f1 f2 ) 0 = (u1 u2 − v1 v2 ) 0 + i(u1 v2 + v1 u2 ) 0
= u10 u2 + u1 u20 − v10 v2 − v1 v20
+i(u10 v2 + u1 v20 + v10 u2 + v1 u20 )
= (u10 + iv10 )(u2 + iv2 ) + (u1 + iv2 )(u20 + iv20 )
= f10 f2 + f1 f20 as desired.
Chapter 46
Chapter 45
45.1 z1 +z2 = 4−2i, z1 z2 = 7−4i, and z1 −z2 =
−4i.
p
√
2
2
45.3
√ z z̄ = a + b + i(ba − ab)
2
2
= a + b = |z|
45.11 z 4 + 2z 2 + 1 = (z + i)2 (z − i)2
45.19 If z = |z|eiφ is an nth root of w = |w|eiθ ,
then |w|eiθ = z n = |z|n einφ . Consequently,
46.5 According to Theorem 46.4, the assumption that y1 and y2 are solutions of (46.7) implies that the scalar multiples C1 y1 and C2 y2 are
solutions of (46.7) as well. Since Theorem 46.4
also asserts the sum of any two solutions to be
a solution, we may infer that C1 y1 + C2 y2 , too,
is a solution of (46.7).
46.7 x = (1 · 5 + 2 · 3)/(2 · 5 + 3 · 1) = 11/13
and x = (2 · 2 − 1 · 1)/(2 · 5 + 3 · 1) = 3/13.
Solutions
Verification: 2x − 3y = 22/13 − 9/13 = 1 and
x + 5y = 11/13 + 15/13 = 2.
46.15 Since
Z x0
−f (t)y2 (t)
C1 (x0 ) =
dt = 0,
Wy1 ,y2 (t)
x
Z x0 0
f (t)y1 (t)
C2 (x0 ) =
dt = 0,
W
y1 ,y2 (t)
x0
it follows that y0 (x0 ) = 0. To verify the second
equality, we apply the first fundamental theorem of calculus to infer that
f (x)y2 (x)
f (x)y1 (x)
and C20 (x) =
.
C10 (x) = −
Wy1 ,y2 (x)
Wy1 ,y2 (x)
This yields
y00 (x0 ) = C10 (x0 )y1 (x0 ) + C1 (x0 )y10 (x0 )
+ C20 (x0 )y2 (x0 ) + C2 (x0 )y20 (x0 )
=−
f (x0 )y2 (x0 )y1 (x0 ) f (x0 )y1 (x0 )y2 (x0 )
+
Wy1 ,y2 (x0 )
Wy1 ,y2 (x0 )
= 0.
−5x
46.21 a) y(x) = C1 e√
+ C2 xe−5x , √
3x
11x) + C2 e3x√sin( 11x),
b) y(x) = C1 e cos(
√
−(4+ 18)x
c) y(x) = C1 e
+ C2 e−(4− 18)x
R
2
2
2
2
46.23 a) y2 (x) = ex e2x /e2x dx = xex ,
2
and the general solution is y(x) = C1 ex +
2
C2 xex .
R
b) y2 (x) = e2xp e−x−ln |2x+1|/2 /e4x dx
R
= e2x e−5x / |2x + 1| dx, and the general solution is y(x) = C1 e2x + C2 y2 (x).
46.25 The assumptions y1 (x0 ) = y2 (x0 ) and
y10 (x0 ) = y20 (x0 ) imply that y1 and y2 are solutions of the same initial value problem. Since
the solution of an initial value problem is unique
(by Theorem 46.2), we may infer that y1 (x) =
y2 (x) for all x ∈ I.
46.27 a) The function y0 (x) = 0 (for all x ∈ R)
satisfies the initial conditions y(0) = y 0 (0) =
0 and also solves the homogeneous equation
(46.7). Since solutions of initial value problems
are unique, it follows that y(x) = y0 (x) = 0 for
all x ∈ R and, by implication, y(1) = y 0 (2) = 0.
b) Consider the nonhomogeneous differential
equation y 00 = 2. A solution of this equation that satisfies the initial conditions y(0) =
y 0 (0) = 0 is y(x) = x2 . Since in this case
y(1) 6= 0 6= y 0 (2), we may infer that the conclusion in a) is not permissible under the altered
assumptions in b).
23
√
46.29 Q(t)
√ 10t) and, by implication,
√ = 3 cos(
I(t) = −3 10 sin( 10t).
46.31 Since the exponential function does not
have any roots, the result in Exercise 46.30 allows us to conclude that Wy1 ,y2 (x) 6= 0 for all
x ∈ I if only Wy1 ,y2 differs from zero at a single
point x0 ∈ I, which is precisely the statement
of Theorem 46.9a.
Chapter 47
47.1 To verify that yn = 5/4 − (−3)n /4 is a
solution of the initial value problem, we observe
that
yn+2 + 2yn+1 − 3yn =
5 1
5 1
15 3
− (−3)n+2 + − (−3)n+1 −
+ (−3)n
4 4
2 2
4
4
9
3
3
= − (−3)n + (−3)n + (−3)n = 0
4
2
4
and
5 1
− = 1,
4 4
5 3
y1 = + = 2.
4 4
y0 =
47.3 a) yn = 5n /6 + 11(−1)n /6,
n
n
b) yn = −2/2
√ n + 2n/2 , √ n
c) yn = 5 cos(αn)
+ 3 5 sin(αn), where
√
α = arccos(−2/ 5).
47.5 The answer is c).
47.7 a) yn+2 − 5yn+1 + 6yn = 0,
b) yn+2 − 4 cos(3)yn+1 + 4yn = 0.
47.9 yn = 3(−2)n − 5n(−2)n /2
Chapter 48
48.13 a) L[f ](s) = 2/((s + 1)2 + 4) for s > −1,
b) L[f ](s) = 24e−2s /s5 for s > 0,
c) L[f ](s) = a/(s2 − a2 ) for s > |a|.
48.17 In replacing t with t + T in the defining
equation f (t + T ) = f (t), we obtain f (t + 2T ) =
f (t + T ) = f (t). Replacing again t with t + T
yields f (t + 3T ) = f (t + T ) = f (t), and continuing in this manner we finally arrive at the
conclusion that f (t + kT ) = f (t) for all nonnegative integers k.
48.29 If f ∈ Ea and g ∈ Eb for some a, b ∈ R,
then both f and g are piecewise continuous on
every interval [0, c] and, by implication, so is
24
Solutions
f g. Furthermore, there are constants M, N ∈ R
such that |f (t)e−s1 t | ≤ M for all s1 > a and
|g(t)e−s2 t | ≤ N for all s2 > b. Since for all
s > a + b we can find s1 > a and s2 > b such
that s = s1 + s2 , it follows that |f (t)g(t)e−st | =
|f (t)es1 t ||g(t)es2 t | ≤ M N for all s > a + b and
all t ∈ [0, ∞). Hence f g ∈ Ea+b .
48.31
2e−2s − 2e−3s + 1
e−s − 4e−2s
+
L[f ](s) =
−3s
s(1 − e )
s2 (1 − e−3s )
48.33 a) −d3 /ds3 2/(s2 + 4),
b) d2 /ds2 (s + 3)/((s + 3)2 + 4),
c) (d2 /ds2 (s + 3)/((s + 3)2 + 4))/s,
d) Γ(3/2)/(s − 2)3/2 ,
e) e−2(s−1) (s − 1)/((s − 1)2 + 1),
f ) e−2s /s
48.35 a) te−t , b) e5t /6 − e−t /6
c) The inverse transform does not exist because
6 0.
lims→∞ F (s) = ∞ =
Rt
d) et sin(3t)/3, e) 0 sin(τ ) dτ
f ) The inverse transform does not exist, because
lims→∞ F (s) = 1 6= 0.
g) The inverse transform does not exist, because
F (s) does not converge to zero as s tends to ∞.
h) The inverse transform does not exist, be6 0.
cause lims→∞
( F (s) = ∞ =
0
for 0 ≤ t < 2,
i) f (t) = R t−2 τ
e
cos(τ
)
dτ
for
t ≥ 2.
0
48.37
Z t
L eat
e−aτ f (τ ) dτ (s)
0
Z t
−aτ
=L
e
f (τ ) dτ (s − a)
0
L[e−at f (t)](s − a)
=
s−a
48.39 For the first equation
form is
(
0
f (t) =
et−3 cos(2(t − 3))
and for the second it is
(
0
f (t) =
et cos(2(t − 3))
L[f ](s)
.
s−a
the inverse trans-
=
for 0 ≤ t < 3,
for t ≥ 3
for 0 ≤ t < 3,
for t ≥ 3.
48.41 If L[f ](s) were equal to sin(s)/s, it would
follow that
L[f 0 ](s) = sL[f ](s) − f (0) = sin(s) − f (0),
but this is impossible because sin(s) − f (0) does
not converge to zero as s tends to ∞.
Chapter 49
1
5
5
−
+
and
x−1 x−2 x−3
−x/2 + 5/2
4x − 1
1/2
.
+ 2
g(x) = 2
+
(x − 4x + 5)2
x − 4x + 5 x − 1
7/4
3i/2
3/4
49.9 f (z) =
−
−
2
z − i (z − i)
z+i
49.13 a) L[y](s) = s/((s2 +4)(s2 −1))+1/(s−1)
and y(t) = − cos(2t)/5 + e−t /10 + 11et /10.
b) L[y](s) = 1/((s − 1)(s − 2)2 ) and
y(t) = te2t − e2t + et .
c) L[y](s) = 1/(s2 ((s − 2)2 + 1))
+(s − 5)/((s − 2)2 + 1) and
y(t) = 21e2t cos(t)/25−72e2t sin(t)+t/5+4/25.
49.17 a) L[y](s) = 1/((s − 1)(s + 4))
+2/((s + 2)(s + 4)) − 3/(s + 4) and
y(t) = −21e−4t /5 + et /5 + e−2t .
b) L[y](s) = 1/((s + 1)(s2 + 1)) + 1/(s + 1) and
y(t) = 3e−t /2 − cos(t)/2 + sin(t)/2.
c) L[y](s) = 1/(2(s − 2)2 ) + 1/(2(s2 − 4)) and
y(t) = te2t /2 − e−2t /8 + e2t /8.
1/200
7/2000
3/16
+
−
49.19 f (x) =
x + 1 (x − 1)2
x−1
37x/100 + 23/20 23x/125 + 11/20
−
−
(x2 + 4x + 5)2
x2 + 4x + 5
49.21 a) L[y](s) = (s + 1)/((s + 1)2 + 1)2 +
3(s + 1)/((s + 1)2 + 1) and y(t) = te−t sin(t)/2 +
3e−t cos(t).
b) L[y](s) = 1/((s + 1)2 + 1)2 + 2/((s + 1)2 + 1)
and y(t) = 5e−t sin(t)/2 − te−t cos(t)/2.
c) L[y](s) = 2e−2s /(s(s + 3)(s − 1)) −
2e−4s /(s(s + 3)(s − 1)), and setting g(t) :=
et /2 + e−3t /6 − 2/3, it follows that
(
0
for 0 ≤ t < 2,
y(t) =
g(t − 2) for t ≥ 2
(
0
for 0 ≤ t < 4,
−
g(t − 4) for t ≥ 4.
49.7 f (x) =
d) L[y](s) = −3e−3s /(s2 (s+1))+3e−2s /(s2 (s+
1)2 ), and setting g(t) := 3te−t + 6e−t + 3t − 6
and h(t) := 3e−t + 3t − 3, we may infer that
(
0
for 0 ≤ t < 2,
y(t) =
g(t − 2) for t ≥ 2
(
0
for 0 ≤ t < 3,
−
h(t − 3) for t ≥ 3.
49.23 L[I](s) = −1/(50(s + 1)((s + 4)2 + 4))
and
Solutions
I(t) = e−4t cos(2t)/650 + 3e−4t sin(2t)/1300 −
e−t /650.
49.25 a) Using the results and the notation of
Example 49.15, it is not difficult to show that
2
1 + 2/(αs)
2
500 1 +
−
+
αs α2 s2
1 − e−αs
L[I](s) =
2
2
(s + b) + c
Note: in this representation of Is (t) all terms are
either constant or periodic with period α and, in
particular, the apparently divergent term 200t
has been replaced with the periodic term 200(t−
α[t/α]).
0.05
Is(t)
0.04
0.03
and
9 −bt
3
3
e sin(ct) +
I(t) = − e−bt cos(ct) +
25
100
25
1
+ 200t − (Φ(t) − Φ(t − α([t/α] + 1)))
4
Z t
1
−
(Φ(τ ) − Φ(τ − α([τ /α] + 1))) dτ.
2α 0
Rt
b) In evaluating the integral 1/(2α) 0 Φ(τ ) dτ
we obtain several terms that converge to zero (as
t tends to ∞) minus the constant 0.0168115 (approximately). Consequently, the steady state
solution is
3
+ 200t
25
1
+ Φ(t − α([t/α] + 1)) − 0.0168115
4
Z t
1
+
Φ(τ − α([τ /α] + 1)) dτ.
2α 0
Is (t) =
Since Φ(t − α([t/α] + 1) is periodic with period
α, it follows that
Z t
Φ(τ − α([τ /α] + 1)) dτ
0
25
=
Z
+
Z
α[t/α]
Φ(τ − α([τ /α] + 1)) dτ
0
t
Φ(τ − α([τ /α] + 1)) dτ
α[t/α]
= [t/α]
Z
0
Φ(τ ) dτ +
−α
= −0.0004[t/α] +
Z
Z
t−α([t/α]+1)
Φ(τ ) dτ
−α
t−α([t/α]+1)
Φ(τ ) dτ.
−α
Since 0.0004/(2α) = 200α, we may conclude
that
3
+ 200(t − α[t/α])
25
1
+ Φ(t − α([t/α] + 1)) − 0.0168115
4
Z t−α([t/α]+1)
1
Φ(τ ) dτ.
+
2α −α
Is (t) =
0.02
0.01
0.00
t
0.001
0.002
-0.01
-0.02
-0.03
Chapter 50
50.5
program exercise;
var k: integer;
y,dx: real;
begin
y:=0;
dx:=0.1;
writeln(y);
for k:=0 to 11 do
begin
y:=y+sqrt(1-sqr(y))*dx;
writeln(y)
end
end.
50.7
program exercise;
var k: integer;
y,dx: real;
begin
y:=0;
dx:=0.1;
writeln(y);
for k:=0 to 11 do
begin
y:=y+sqrt(1-sqr(y))*dx
-y*sqr(dx)/2;
writeln(y)
end
end.
50.9
program exercise;
var k: integer;
y,ys,w,dx: real;
0.003
26
Solutions
begin
y:=0;
w:=1;
dx:=pi/30;
writeln(y,’ ’,w);
for k:=0 to 14 do
begin
ys:=y;
y:=y+w*dx;
w:=w-ys*dx;
writeln(y,’ ’,w)
end
end.
50.11
program fallingbodies;
const gM=3.98E14;
R=4E7/(2*pi);
d=1.2E7;
var k: integer;
t,y,ys,w,dt: real;
begin
t:=0;
y:=d+R;
w:=0;
dt:=100;
while (y>=R) do
begin
writeln(t,’ ’,y);
t:=t+dt;
ys:=y;
y:=y+w*dt
-gM*sqr(dt)/(2*sqr(y));
w:=w-gM*dt/sqr(ys)
+gM*w*sqr(dt)/(sqr(ys)*ys);
end
end.
50.13 a) y(x) = 97e−7x /392+7ex /8−x/7−6/49
and y(2) ≈ 6.057.
b) Using the equations
yk+1 := yk + wk ∆x,
wk+1 := wk + (xk − 6wk + 7yk )∆x,
we find the approximation y(2) ≈ 5.478, and
the error is |6.057 − 5.478| = 0.579.
c) Using the equations
1
yk+1 :=yk + wk ∆x + (xk − 6wk + 7yk )∆x2 ,
2
wk+1 :=wk + (xk − 6wk + 7yk )∆x
1
+ (1 − 6xk + 43wk − 42yk )∆x2 ,
2
we find the approximation y(2) ≈ 5.735, and
the error is |6.057 − 5.735| = 0.322.
2
50.15 a) yk+1 := yk + eyk /20 ∆x
2
+yk eyk /10 ∆x2 /20 yields y(2) ≈ 3.6567.
b) yk+1 := yk + (sin(yk2 ) + xk )∆x
+(2yk (sin(yk2 ) + xk ) cos(yk2 ) + 1)∆x2 /2 yields
y(2) ≈ 2.1912.
c) Using the equations
yk+1 := yk + wk ∆x
1
+ (xk − wk sin(yk ) − yk )∆x2 ,
2
wk+1 := wk + (xk − wk sin(yk ) − yk )∆x
1
+ (1 − wk − wk2 cos(yk ))∆x2
2
1
− sin(yk )(xk − wk sin(yk ) − yk )∆x2 ,
2
we find that y(2) ≈ 0.8531.
Chapter 51
51.5 c2 = 0, c5 = −c2 /(4 · 5) = 0, c8 = −c5 /(7 ·
8) = 0, and in general c3k+2 = 0 for all k ≥ 0.
P∞
51.7 a) Setting y(x) = k=0 ck xk and assuming the general initial conditions y(0) = z1 ,
y 0 (0) = z2 , we find that c0 = z1 , c1 = z2 ,
c2 = z1 + z2 /2, c3 = z2 /3, and in general
ck+2 =
(k + 1)ck+1 + 2ck − 2ck−1 + 4ck−2
(k + 2)(k + 1)
for all k ≥ 2. Using this general equation
to determine c4 , c5 , c6 , c7 , and c8 , we notice
that there are coefficients ak and bk such that
c2k = z1 /k! + ak z2 and c2k+1 = bk z2 . Given
this observation, it is convenient to generate one
particular solution y1 by setting z1 := 1 and
z2 := 0. This yields c2k = 1/k!, c2k+1 = 0, and
y1 (x) =
∞
X
x2k
k!
k=0
2
= ex .
(The reader may easily verify that this function
does indeed satisfy the given differential equation.) Using the formula on p.363, it follows
that a second solution is
Z x
2
2
et−2t dx,
y2 (x) = ex
0
and the general solution is y(x) = C1 y1 (x) +
C2 y2 (x).
Solutions
b) Following exactly the same approach as in
a), it is not difficult to see that c0 = z1 , c1 = z2 ,
c2 = 5z1 /2, c3 = z2 , and in general
ck+2 =
(k + 5)ck + 2ck−2
(k + 1)(k + 2)
27
=
for all k ≥ 2. Consequently, the general solution
is
y(x) = z1 + z2 (x − 1)
∞
X
(−1)k (z2 − z1 )
(x − 1)k
+
k−1
for all k ≥ 2. As we use this general equation
to determine ck for a few more even and odd
indices, we notice that c2k+1 = z2 /k! and c2k =
ak z1 for some coefficients ak ∈ R. Setting z1 :=
0 and z2 := 1, we find the solution
y1 (x) =
∞
X
x2k+1
k=0
k!
=x
∞
X
x2k
k=0
k!
(−1)k (z2 − z1 ) (−1)k (z2 − z1 )
−
k−1
k
−
k=2
∞
X
k=2
(−1)k (z2 − z1 )
(x − 1)k
k
= z1 + z2 (x − 1)
+ (z2 − z1 )(x − 1) ln(x)
2
= xex .
+ (z2 − z1 )(ln(x) − (x − 1))
= z1 x + (z2 − z1 )x ln(x).
The corresponding second solution is
2
y2 (x) = xex
Chapter 52
Z
x
0
−3t2 /2
e
t2
dt,
and the general solution is y(x) = C1 y1 (x) +
C2 y2 (x).
c) Since the general solution is to be represented
as a power series about x0 = 1, we rewrite the
given equation as follows:
0 = x2 y 00 (x) − xy 0 (x) + y(x)
2
= (x − 1)2 y 00 (x) + 2(x − 1)y 00 (x) + y 00 (x)
− (x − 1)y 0 (x) − y 0 (x) + y(x).
Assuming the general initial conditions
y(0) =
P∞
z1 , y 0 (0) = z2 and substituting k=0 ck xk for
y(x), we find that c0 = z1 , c1 = z2 , and in
general
ck (k − 1)2 + ck+1 (2k 2 + k − 1)
(k + 2)(k + 1)
for all k ≥ 0. As we use this recursive equation
to determine ck for the first few values k ≥ 2,
we quickly notice that
ck =
p00 = 6/7 p01 = 1/7
p10 = 2/7 p11 = 5/7
52.7 a) 26/50, b) (25/51) · (26/50) + (26/51) ·
(25/50)+(25/51)·(24/50), c) (25/51)·(26/50)+
(26/51) · (25/50), d) 12/50
52.9
pAA = p
pBA = p
pAB = 1 − p
pBB = 1 − p
Chapter 53
00
= ((x − 1) + 1) y (x)
− ((x − 1) + 1)y 0 (x) + y(x)
ck+2 = −
52.1
(−1)k (z2 − z1 )
k(k − 1)
53.1 a12 = 2, a31 = −0.1, and a24 = 8.
−2 20 6
53.9 BA = −5 8 −6
−4 22 3
53.11 If we assume A to be an m×n matrix and
B to be a k × l matrix, then the size conditions
can be stated as follows: a) n = k, b) l = m,
c) n = k and l = m, d) m = n = k = l.
53.15 If we assume that A is an m × n matrix,
B a k × l matrix, and C an i × j matrix, then
the size conditions can be stated as follows: a)
n = k and l = i, b) n = k = i and l = j, c)
m = k and n = l = i.
n
0 0
a
53.21 Dn = 0 bn 0
0
0 cn
3 −9 6
53.23 AB =
4 2 15
53.25 A+B and AB−BA are not well defined.
28
Solutions
53.27 Answers may vary, but one possible answer is
2 0 0
A = 0 1 0 .
0 1 1
53.29 From the assumption |g(x) − g(y)| <
λ|x − y| we may infer that |g k+1 (x) − g k (x)| <
λk |g(x) − x| for all nonnegative integers k
and allPx ∈ R. Since 0 < λ < 1, the
∞
k
series
k=0 λ |g(x) − x| is convergent, and
according
to
the
comparison test, the series
P∞
k+1
k
(g
(x)
−
g
(x)) must be convergent as
k=0
well. In other words, the limit of g n (x) =
Pn−1
x + k=0 (g k+1 (x) − g k (x)) as n tends to ∞ exists and is finite. Denoting this limit by z (i.e.,
z = limn→∞ g n (x)) and observing that the condition |g(x) − g(y)| < λ|x − y| implies that g is
continuous (this is a trivial consequence of Theorem 7.8), we may conclude that z is a fixed
point of g because
g(z) = g lim g n (x) = lim g n+1 (x) = z.
n→∞
2
1/3
If b = 1, then x = 1/3.
1
2/3
2
5/3
If b = 2, then x = −1/3.
3
1/3
54.19 In denoting the entries of Qij (λ) by qkl
and those of Qij (λ)A by bkl , it follows that
(
m
X
if k 6= j,
akl
bkl =
pkq aql =
+
a
if k = j
λa
il
jl
q=1
as desired.
1 0
1
54.23
and
−2 1
−1
tinct reduction matrices.
Chapter 54
54.3 a) x1 = 8/5, x2 = 2/5, b) x1 = 1, x2 = 1,
x3 = 2
54.7 The matrix is not in standard reduced
form because the rule, which says that in each
leading column all but one entry must be zero, is
violated. To obtain a standard reduced matrix
we change the entries a12 and a33 to zero:
1 0 1 0
0 1 3 0 .
0 0 0 4
54.11 a) no solutions, b) x1 = 1 − 3t1 − 10t2 ,
x2 = 2 + t1 + 3t2 , x3 = t1 , x4 = 1, x5 = −t2 ,
x6 = t2 , c) x1 = 4, x2 = 7, x3 = 2
54.15 good: 27 out of 58, average: 19 out of 58,
poor: 12 out of 58.
1
1/3
54.17 If b = 1, then x = −2/3.
0
2/3
are two dis-
54.25 In denoting the entries of C, Idm , and
Idm C by cij , dij and bij respectively, it follows
that
m
X
bij =
dik ckj = dii cij = cij
k=1
n→∞
To prove that z is unique, we assume that
g(y) = y for some y ∈ R. Since x in the preceding argument was an arbitrary real number, it
follows that z = limn→∞ g n (y) = limn→∞ y = y
as desired.
53.31 Answers may vary.
0
1/2
as desired. The remaining equations can be verified in a similar fashion.
54.29 A reduction matrix is
−1/2 1/2
1/2
B = 3/2 −1/3 −1/2 ,
1
0
−1
and the product BA is easily seen to be equal
to Id. Thus Ais indeed invertible.
1/6 2/3 −1/6
54.31 A−1 = 1/6 −1/3 5/6
1/6 −1/3 −1/6
−1 −1 1
1
0
54.33 B = 1
2
1 −1
54.35 In reducing the augmented matrix below,
we find the same soultions as in Exercise 54.34.
1 0 1 1 2 1
−1 1 −1 2 0 3 .
1 1 0 −1 1 0
54.37
B =Q21 (−2)Q31 (−1)Q32 (−1)P2 (1/3)
· P3 (−1)Q13 (−1)Q12 (1)
4/3 −2/3 −1
1
= −2/3 1/3
1
0
−1
Solutions
1
1
1 0 −1
−1 0
1 0 0
−3
−3
−2
1 3
54.39 A−1 =
.
−1 −2 −4 1 3
−1 −1 0 0 1
54.41 a) Since A is the reduction matrix of
A−1 , we need to find a product of elementary
matrices that reduces
0 −1 1
1 −1
A−1 = 1
−1 0
1
to the identity matrix. This yields
A =Q21 (−1)Q31 (1)Q32 (1)P2 (−1)Q23 (1)
· Q13 (1)R12
b) Following the same approach as in a), we find
that
A =Q31 (7)Q21 (−2)Q32 (3)P2 (−1)P1 (17)
· Q13 (−2)Q12 (−7)R13 .
54.43 If Ax1 = Ax2 = b, then A(x1 − x2 ) =
Ax1 − Ax2 = b − b = 0 as desired.
1/2 0 0 0 0
0
1 0 0 0
0
0 1 0 0
54.45 a) B =
0
0 0 1 0
0
0 0 0 1
1 0 0 0 0
−1 1 0 0 0
b) C =
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 0 0 0 0
0 1 0 0 0
c) D =
0 0 2 0 0
0 0 0 1 0
0 −2 0 0 1
54.47 a) Yes, because the invertibility of A
guarantees by definition the existence of exactly
one solution.
b) Yes, because the assumption b = 0 implies
that x = 0 is a solution.
c) Yes, for the same reason as in b).
d) Yes, because the given assumption concerning the reduced matrix of A implies that A is
invertible, and the argument in a) therefore applies here as well.
e) Yes, for the same reason as in b).
54.49 The condition in c) guarantees the existence of infinitely many solutions, because if
29
A is not invertible, then the equation Ax = b
either has no solution or infinitely many solutions. Since the assumption b = 0 implies that
there is at least one solution, namely x = 0,
we may therefore infer that the number of solutions is infinite. The same argument applies
also to e) because the stated assumption concerning the reduced matrix of A implies that A
is not invertible.
Chapter 55
55.11 To apply the standard method we assign
clockwise currents to each of the four loops in
the network as follows: loop ABC: I1 ; loop
BCD: I2 ; loop CDE: I3 ; and loop ACE: I4 .
The corresponding system of linear equations is
represented by the matrix equation
40 −10
0
−10
0
I1
−10 40 −15
I2 0
0
= ,
0
−15 65 −20 I3 0
I4
−10
0
−20 30
5
and the solutions are I1 = 107/1406 A, I2 =
1/19 A, I3 = 63/703 A and I4 = 177/703 A. Using the probabilistic approach, we find the following values for the long term expectations of
the permissible states: qAB = 703/8376, qAC =
703/4188, qBA = 163/2792, qBC = 163/1396,
qBD = 163/2094, qCA = 38/349, qCB = 38/349,
qCD = 76/1047, qCE = 19/349, qDB = 21/349,
qDC = 21/349, qDE = 21/698, qEC := 0, and
qED := 0. To verify that these results are consistent with those obtained via the standard
method we can now compute the ratio of the
currents for any pair of connections in the circuit. Taking, for example, the connections from
A to C and B to D, we find that
I4 − I1
247
qAC − qCA
=
=
qBD − qDB
74
I2
as desired.
Chapter 56
1
1
56.3
4
3
3
2
1
0
3
7
4
6
56.5 a) (A + B)t = [aij + bij ]t = [aji + bji ] =
[aji ] + [bji ] = At + Bt ,
30
b) (λA)t = [λaij ]t = [λaji ] = λ[aji ] = λAt ,
c) (At )t = ([aij ]t )t = [aji ]t = [aij ] = A.
56.13 4 − 6 = −2
56.15 The results are easily verified by observing that I(1, 2, 3) = 0, I(1, 3, 2) = 1, I(2, 1, 3) =
1, I(2, 3, 1) = 2, I(3, 1, 2) = 2, and I(3, 2, 1) =
3.
56.19 −2
56.29 x1 = −22/5, x2 = −29/5, and x3 = −3.
√
√
56.35 λ1 = 2 + 3 and λ2 = 2 − 3.
56.37 det(2A) = 2n · 3
Qn
56.39 det(A) = (−1)[n/2] k=1 an−k+1,k
(where [x] denotes, as usual, the greatest iunteger less than or equal to x).
56.41 a) If A is invertible, and if v is a characteristic vector corresponding to a characteristic
value λ of A, then Av = λv, or equivalently,
v = λA−1 v. Since v 6= 0 (by the definition of
a characteristic vector), it follows that λ cannot
be zero.
b) Given the result of a), we may infer that the
equation Av = λv is equivalent to the equation (1/λ)v = A−1 v. In other words, 1/λ is a
characteristic value of A−1 .
56.43 In applying Theorem 56.22 to the first
column, we may infer the existence of constants a0 , . . . , an−1 (represented by cofactors)
such that P (x) = a0 + a1 x + · · · + an−1 xn−1 . In
other words, P (x) is a polynomial of degree less
than or equal to n − 1. Furthermore, P (xi ) = 0
for all i ∈ {2, . . . , n} because P (xi ) is the determinant of a matrix in which the first column equals the ith column. Since the values
x1 , . . . , xn are assumed to be distinct, it follows
that x2 , . . . , xn are n − 1 distinct roots of P ,
and since the degree of P is less than or equal
to n − 1, we may conclude that P (x1 ) 6= 0 (because a polynomial of degree less than or equal
to n − 1 cannot have more than n − 1 distinct
roots).
a b
56.45 If A =
is invertible, then
c d
1
d −b
−1
.
A =
ad − bc −c a
56.47 a) If Av = λv and Aw = λw, then
A(v + w) = Av + Aw = λv + λw = λ(v + w),
and the assumption v + w 6= 0 therefore implies
that v + w is a characteristic vector of A corresponding to the characteristic value λ.
Solutions
b) If v is a characteristic vector and µ 6= 0, then
µv 6= 0 and A(µv) = µAv = µλv = λ(µv).
Thus, µv is a characteristic vector of A corresponding to the characteristic value λ.
56.49 a) Yes, because A is invertible whenever
det(A) 6= 0.
2
2
b) Yes, because
√ if det(A) = det(A ) = 2, then
det(A) = ± 2 6= 0.
c) Yes, because if det(A) det(B) = det(AB) =
2, then det(A) must be different from zero.
d) No, because the zero matrix 0n×n is not invertible despite the fact that det(0n×n + Idn ) =
1 6= 0.
Chapter 57
57.5 0 = λ0 − λ0 = λ(0 + 0) − λ0 = λ0 + (λ0 −
λ0) = λ0 + 0 = λ0.
57.9 d) If Ax = 0 and Ay = 0, then
A(x + y) = Ax + Ay = 0 + 0 = 0 and
A(λx) = λAx = λ0 = 0 as desired.
f ) If y1 and y2 are in U , then there are vectors
x1 , x2 ∈ Fn such that Ax1 = y1 and Ax2 = y2 .
Consequently, y1 + y2 is in U as well because
A(x1 + x2 ) = Ax1 + Ax2 = y1 + y2 . Furthermore, λy1 is in U for all λ ∈ F, because
A(λx1 ) = λAx1 = λy1 .
j) If f and g are in U , then (f + g) 0 + a(f + g) =
(f 0 + af ) + (g 0 + ag) = 0 + 0 = 0. Furthermore,
(λf ) 0 + a(λf ) = λ(f 0 + af ) = 0. Consequently,
f + g and λf are in U as well, and U therefore
is a subspace.
57.27 If v = λ1 v1 + · · · + λn vn and w = µ1 v1 +
· · · + µn vn , then v + w = (λ1 + µ1 )v1 + · · · +
(λn + µn )vn and
λ1 + µ 1
λ1
µ1
.
.
.
..
Sβ (v + w) =
= .. + ..
λn + µ n
λn
µn
= Sβ (v) + Sβ (w).
Furthermore, for λ ∈ F we have λv = λλ1 v1 +
· · · + λλn vn , and therefore
λλ1
λ1
..
..
Sβ (λv) = . = λ . = λSβ (v).
λλn
λn
57.29 a) The vectors are linearly independent
because
1 1 2
2 0 1 = 7 6= 0.
3 2 1
Solutions
b) The vectors are linearly dependent, because
R2 is two-dimensional.
c) Since the first three components of the given
vectors are identical with those of the vectors in
a), we may conclude that the vectors are linearly
independent.
57.31 If p(x) = b0 + b1 x + b2 x2 and q(x) =
c0 + c1 x + c2 x2 are polynomials in P2 , then
31
in the complex vector space Cn , and therefore
a1 + ib1 = · · · = an + ibn = 0. Hence a1 = · · · =
an = b1 = · · · = bn = 0 as desired.
Chapter 58
58.1
T (p + q)(x)
m1 x100 (t) = − k1 x1 (t) + k2 (x2 (t) − x1 (t))
m2 x200 (t) = − k2 (x2 (t) − x1 (t))
= T ((b0 + c0 ) + (b1 + c1 )x + (b2 + c2 )x2 )
= ((b0 + c0 ) + 2(b2 + c2 ))
+ (3(b1 + c1 ) − (b2 + c2 ))x + 3(b0 + c0 )x2
m3 x300 (t)
= (b0 + 2b2 ) + (3b1 − b2 )x + 3b0 x2
+ (c0 + 2c2 ) + (3c1 − c2 )x + 3c0 x2
= T (p)(x) + T (q)(x),
58.7 Setting
0
0
A(t) :=
−1
0
0
and for a given scalar λ ∈ R, we find that
T (λp)(x) = T (λb0 + λb1 x + λb2 x2 )
= (λb0 + 2λb2 ) + (3λb1 − λb2 )x + 3λb0 x2
= λ((b0 + 2b2 ) + (3b1 − b2 )x + 3b0 x2 )
= λT (p)(x).
This proves that T is a linear transformation
from P2 to P2 . Furthermore, setting γ := β :=
{1, x, x2 }, the matrix representing T is readily
seen to be
1 0 2
Aβγ = 0 3 −1 .
3 0 0
57.33 Since the real numbers are a subset of
the complex numbers, it follows that the scalar
multiplication of a vector in Cn with a real number is well defined. Given this observation, the
validity of the vector space axioms is easily verified. Furthermore, since every complex number
can be written in the form a · 1 + b · i, it follows that Cn (as a real vector space) is spanned
by the vectors e1 , . . . , en , ie1 , . . . , ien . Consequently, if we can show these vectors to be linearly independent, we will be able to conclude
that {e1 , . . . , en , ie1 , . . . , ien } is a basis and that
the dimension of Cn (as a real vector space)
is 2n. So let us assume that a1 , . . . , an and
b1 , . . . , bn are real numbers such that
a1 e1 + · · · + an en + b1 ie1 + · · · + bn ien = 0.
Then
(a1 + ib1 )e1 + · · · + (an + ibn )en = 0
+ k3 (x3 (t) − x2 (t))
= − k3 (x3 (t) − x2 (t))
1
0
0
0
2
0
1
0
0
0
0
0
0
0
0
0
0
−t2
,
1
0
it follows that
0
u1 (t)
u1 (t)
0
u20 (t)
u2 (t) 0
0
u3 (t) = A(t) u3 (t) + et ,
0
u4 (t)
u4 (t) 0
u50 (t)
u5 (t)
3
where u1 = x1 , u2 = x10 , u3 = x100 , u4 = x2 and
u5 = x20 .
58.9 The equivalent system is c).
58.11 In denoting by A(x) the matrix
0
0
..
.
1
0
..
.
0
1
..
.
...
...
..
.
0
0
..
.
0
−a0 (x)
0
−a1 (x)
0
−a2 (x)
...
...
1
−an−1 (x)
,
a first order system equivalent to the given differential equation is
u1 (x)
u10 (x)
u2 (x)
u20 (x)
.. = A(x) .. +
.
.
un0 (x)
un (x)
0
0
..
.
f (x)
where u1 = y, u2 = y 0 , ..., un = y (n−1) .
,
32
Solutions
Chapter 59
59.3 If x(t) = (x1 (t), x2 (t)) is a solution of
the system (59.6), then x100 (t) = x20 (t) =
−b(t)x1 (t) − a(t)x2 (t) = −b(t)x1 (t) − a(t)x10 (t).
Thus, y(t) := x1 (t) is a solution of the differential equation (59.4). Conversely, if y is a solution of (59.4), then x(t) := (y(t), y 0 (t)) is a
solution of (59.6).
59.5 If x is a solution of (59.7), then
Cx10 (t)
Cx10 (t)
(Cx) 0 (t) = ... = ... = Cx 0 (t)
Cxn0 (t)
b) No, because if X(0) = −Id, then X(0)2 =
Id, but X(0) 6= Id.
c) Yes, because if X(0)−1 = Id, then X(0) =
(X(0)−1 )−1 = Id−1 = Id.
59.17 a) The definition of the determinant implies that
d
det(X(t))
dt
X
d
=
δ(σ)x1j1 · · · xnjn
dt
σ=(j1 ,...,jn )∈Sn
X
0
=
δ(σ)x1j
x · · · xnjn + . . .
1 2j2
σ=(j1 ,...,jn )∈Sn
Cxn0 (t)
= CA(t)x(t) = A(t)(Cx(t)).
X
+
0
δ(σ)x1j1 · · · xn−1,jn−1 xnj
=
n
σ=(j1 ,...,jn )∈Sn
Thus, Cx is a solution of (59.7) as well.
59.7 Each column in the matrix
f (t) g(t)
f 0 (t) g 0 (t)
is a solution of (59.6) because f and g are solutions of (59.4). Furthermore, since f (t0 ) = 1,
f 0 (t0 ) = 0, g(t0 ) = 0, and g 0 (t0 ) = 1 (see the
proof of Theorem 46.9), we may infer that
f (t0 ) g(t0 )
= Id.
f 0 (t0 ) g 0 (t0 )
Consequently, the matrix above is indeed the
standard fundamental matrix Xt0 (t) of the system (59.6).
59.11 X(t) is a fundamental matrix because
each of its column vectors is a solution of the
system and det(X(0)) = −1 6= 0. Furthermore,
the solution of the initial value problem is
2
x(t) = X(t)X(0)−1
3
2 cos(2t) − 3 sin(2t)
=
.
3 cos(2t) + 2 sin(2t)
59.13 x2 (t)
59.15 As in Exercise 59.14, the first part of
each statement simply guarantees each column
in X(t) to be a solution of the system. Consequently, we only need to examine in each
case whether the second condition implies that
X(0) = Id.
a) Yes, because the assumption X(0)ei = ei implies that X(0) = X(0)Id = X(0)(e1 , . . . , en ) =
(e1 , . . . , en ) = Id.
0
x11
x21
..
.
xn1
...
...
..
.
...
0 x11
x1n
..
x2n .. + · · · + .
xn−1,1
. x0
xnn
n1
...
..
.
...
...
.
xn−1,n 0
xnn
x1n
..
.
0
0
=
Since X (t) = A(t)X(t), it follows that xki
P
n
a
x
and,
by
implication,
j=1 kj ji
0
0 x11 . . . x1n
x21 . . . x2n ..
.. ..
.
.
. xn1 . . . xnn Pn
Pn
j=1 a1j xj1 . . .
j=1 a1j xjn x21
...
x2n
=
.
.
..
..
..
.
...
xnn
xn1
xj1 . . . xjn n
x21 . . . x2n X
a1j .
=
.. = a11 det(X(t)).
..
..
.
. j=1
xn1 . . . xnn Similarly, we find that
x11 . . . x1n 0
0 x21 . . . x2n
..
.. =a22 det(X(t))
..
.
.
. xn1 . . . xnn ..
.
x11
...
x1n ..
.. .
..
.
. =ann det(X(t)),
xn−1,1 . . . xn−1,n 0
x0
...
xnn
n1
Solutions
Since the inverse of the matrix (v1 , v2 , v3 ) is
easily seen to be
and therefore,
d
det(X(t)) =
dt
n
X
aii (t) det(X(t)),
λk1
Bk = ...
0
...
..
.
...
0
..
.
λkn
for all nonnegative integers k, it follows that
P∞ k
0
k=0 λ1 /k! . . .
..
..
..
eB =
.
.
P∞ . k
0
...
k=0 λn /k!
λ
1
...
0
e
..
.. .
.
..
= .
.
b) eA
1
0
=
0
0
it follows that
0 3/2 −9/2
6
eAt = et 0 −2
0 −1
3
0 −4 8
1 5/2 −7/2
0 .
+ e2t 0 3 −6 + e3t 0 0
0 1 −2
0 0
0
−21et /2 + 16e2t − 9e3t /2
14et − 12e2t
b) x(t) =
t
2t
7e − 4e
60.23 a) eA
0 −1/2 3/2
1
−2 ,
[γij ] = 0
1 5/2 −7/2
Chapter 60
0
i=1
as desired.
Pn
b) Setting a(t) := i=1 aii (t), the result in a)
implies that det(X(t)) is a solution of the differential equation y 0 = a(t)y. Consequently, we
can find a constant C ∈ R such that det(X(t)) =
CeA(t) . Since the exponential function is never
zero, we may infer that det(X(t)) is different
from zero for all t unless C = 0, in which
case it is equal to zero for all t. In particular, if det(X(t0 )) 6= 0 for a single point t0 , then
det(X(t)) 6= 0 for all t.
60.5 Since
33
...
1
= 0
0
1 4
1 2
0 1
0 0
eλn
1 4
1 2
0 1
25/6
5
2
1
60.25 eA(t−2) =
1 t − 2 t2 − 3t + 2 2t3 /3 − 5t2 /2 + 14/3
0
1
2t − 4
2t2 − 7t + 6
0
0
1
2t − 4
0
0
0
1
60.27 a) The characteristic values of A are
λ1 = 1, λ2 = 2, λ3 = 3, and corresponding
characteristic vectors are
−3
−4
1
v1 = 4 , v2 = 3 , v3 = 0 .
2
1
0
60.29 Let v be a characteristic vector of A corresponding to the characteristic value λi . Then
every column vector of Pi (A) is a scalar multiple of v (see Theorems 60.20 and 60.19), and
the familiar properties of the determinant therefore imply that det(Pi (A)) = 0 (see Theorem 56.17b, c, e).
60.31 To give an example of how the entries
cij of eAt can be computed, we will examine
the entry c25 (assuming, of course, that n ≥
5). In taking a few powers of A, we readily
observe that the entry in the second row and
fifth column of Ak assumes the value 0 for k < 3
and the value k3 λk−3 for k ≥ 3. Thus,
c25 =
∞ k−3 k
X
k λ
t
k=3
3
k!
=
∞
t3 X (λt)k
t3 eλt
=
.
3!
k!
3!
k=0
In general, we find that
(
0
cij = j−i λt
t e /(j − i)!
if i > j,
if i ≤ j.
60.33 Since e2A v = Idv + 2Av + 22 A2 v/2 +
23 A3 v/6 + 24 A4 v/24 = v − 2v + 2v − 4v/3 +
2v/3 = v/3, we may infer that the components
of e2A v are x1 = 1/3, x2 = 1/3 and x3 = 2/3.
60.35
x(t) = 2(t−1)
+ 4e3(t−1)
−e
1 2(t−1)
−e
+ e3(t−1) + 2(t − 1)e3(t−1)
3
2(t−1)
e
− e3(t−1) + 2(t − 1)e3(t−1)
34
Chapter 61
61.3 x1 (t) = 1/4−8et +35e2t /4−2tet −5te2t /2,
x2 (t) = −1/2+3tet −5e2t /2+tet , x3 (t) = 1/4−
10et + 35e2t /4 − 3tet − 5te2t /2.
61.5 x2 (t) ≈ −0.06eλ1 t + 1.35eλ2 t + 0.49eλ3 t −
et − e2t + 0.22, x3 (t) ≈ 0.02eλ1 t − 2.13eλ2 t +
0.22eλ3 t + et + e2t − 0.11.
61.7 y(x) = −9 cos(2x)/40 − 11 sin(2x)/20 +
8ex /5 − x2 /4 − x/2 − 3/8
61.9 x1 (t) = −e2t /3 + 46e3t /27 − t/9 − 10/27,
x2 (t) = −e2t /3+35e3t /81−13t/27+23te3t /27−
8/81, x3 (t) = e2t /3 − 34e3t /81 + 2t/27 +
23te3t /27 + 7/81.
Chapter 62
62.3 L[x1 ](s) = −(3s + 13)/(2(s2 + 5s + 4)) +
5/(2(s + 1)),
L[x2 ](s) = (2s − 1)/(s2 + 5s + 4),
L[x3 ](s) = (s + 14)/(s2 + 5s + 4) and
x1 (t) = 5e−t /2 − 3e−2t cos(t)/2 − 7e−2t sin(t)/2,
x2 (t) = 2e−2t cos(t) − 5e−2t sin(t)/2,
x3 (t) = e−2t cos(t) − 12e−2t sin(t).
62.7 a) L[x1 ](s) = 2/(s − 2) − 1/(s + 1),
L[x2 ](s) = −1/(s + 1), L[x3 ](s) = 0 and
x1 (t) = −e−t + 2e2t , x2 (t) = −e−t , x3 (t) = 0.
b) L[x1 ](s) = 1/(s − 3) + 1/(s + 1) + 2/(s + 1)2 ,
L[x2 ](s) = 1/(s − 3) + 3/(s + 1)2 , L[x3 ](s) =
1/(s − 3) − 1/(s + 1) + 1/(s + 1)2 and
x1 (t) = e−t + e3t + 2te−t , x2 (t) = e3t + 3te−t ,
x3 (t) = −e−t + e3t + te−t .
62.9 L[x1 ](s) = (2s + 1)/(6(s2 + 1)2 )
−(s − 1)/(9(s2 + 1)) − (8s + 1)/(9(s2 + 4)),
L[x2 ](s) = (2s + 1)/(6(s2 + 1)2 )
+(s − 1)/(9(s2 + 1)) + (8s + 1)/(9(s2 + 4)) and
x1 (t) = − cos(t)/9 − t cos(t)/12 − 8 cos(2t)/9 +
7 sin(t)/36 + t sin(t)/6 − sin(2t)/18,
x2 (t) = cos(t)/9 − t cos(t)/6 + 8 cos(2t)/9 +
sin(t)/18 + t sin(t)/3 + sin(2t)/18.
Chapter 63
63.9 With the substitutions u1 = I1 , u2 = I10 ,
u3 = I2 , u4 = I20 , u5 = I3 , and u6 = I30 the
network is described by the equations
Solutions
u1
(C1 + C2 )u3
R2 u4
R2 u6
−
−
+
,
L2 C1
L2 C1 C2
L2
L2
u50 = u6 ,
u40 =
u60 =
R2 u4
u5
(R1 + R2 )u6
−
−
.
L3
C3 L3
L3
Setting λ1 := −433856.14, λ2 := −89369.25,
λ3 := −5389.48, λ4 := −1784.48, a :=
−18133.65, b := 23771.07, and ω := 1000 and
using the formula in (61.24), we find the following solutions:
I1 (t) ≈ 7.47 · 10−12 eλ1 t + 0.00014eλ2 t −
0.0015eλ3 t − 0.00044eλ4 t + 0.0018 cos(ωt) −
0.000056eat cos(bt) + 0.00051 sin(ωt) − 1.48 ·
10−6 cos2 (bt) sin(ωt)+0.000098eat sin(bt)−1.48·
10−6 sin(ωt) sin2 (bt) A,
I2 (t) ≈ 1.09 · 10−9 eλ1 t + 6.98 · 10−6 eλ2 t −
0.00071eλ3 t − 0.00036eλ4 t + 0.00086 cos(ωt) +
0.00021eat cos(bt) + 0.00034 sin(ωt) − 3.72 ·
10−6 cos2 (bt) sin(ωt)−0.000018eat sin(bt)−3.72·
10−6 sin(ωt) sin2 (bt) A, and
I3 (t) ≈ −6.03 · 10−9 eλ1 t + 4.66 · 10−6 eλ2 t −
0.00068eλ3 t + 0.00045eλ4 t + 0.00012 cos(ωt) +
0.00011eat cos(bt) − 0.00012 sin(ωt) − 2.03 ·
10−6 cos2 (bt) sin(ωt)−0.000011eat sin(bt)−2.03·
10−6 sin(ωt) sin2 (bt) A.
Chapter 64
64.7 The functions in the first row of U(x)
are y1 (x) = cos(x), y2 (x) = x cos(x), y3 (x) =
x2 cos(x), y4 (x) = sin(x), y5 (x) = x sin(x),
y6 (x) = x2 sin(x), and the solution is y(x) =
2 cos(x)+x cos(x)/2−x2 cos(x)/2−3 sin(x)/2+
2x sin(x).
64.15 y(x) = 9e−x /25 + 11ex /25 + cos(2x)/5 +
x cos(2x)/20 − 13 sin(2x)/200
64.17 y(x) = C1 e2x + C2 xe2x + C3 e2x cos(x) +
C4 e2x sin(x)
64.19 From the transform L[y](s) = 10/(s − 2)
−33/(s − 2)2 + (99 − 19s)/(21((s − 2)2 + 1)) +
1/(2(s − 1)) we find the same solution as in Exercise 64.18.
Chapter 65
u10 = u2 ,
65.1 See Figure S.9
u1
R1 u2
u3
E0
−
+
+
,
L1 C1
L1
L1 C1
L1
u30 = u4 ,
65.15 π/4
√
65.21 37/ 89
u20 = −
65.23 4x + y − 3z = −3
Solutions
z
= v1 w1 + v2 w2 = v · w.
c) According
to b),√ we may infer that
√
kAvk = Av · Av = v · v = kvk.
d) Since v/kvk is a vector of length 1, there
must exist an angle φ ∈ [0, 2π) such that
v/kvk = cos(φ)e1 + sin(φ)e2 .
Using the
definition of A, we thus obtain:
cos(θ) cos(φ) − sin(θ) sin(φ)
Av = kvk
sin(θ) cos(φ) + cos(θ) sin(φ)
cos(θ + φ)
= kvk
.
sin(θ + φ)
3
2
1
x
y
0
-1
-2 -1
0
1
1
2
0
-1 -2
Figure S.9: graph for Exercise 65.1.
65.27 The vector v × w points downward in
the direction that a right screw would move in
if turned in the direction from v to w.
z
x
w
0
v
35
y
v w
-1
Consequently, Av is a counterclockwise rotation
of v by θ. Note: the fact that Av is a rotation
of v by θ also follows from the result in c) in
conjunction with the observation that the angle
between Av and v is
cos(θ)kvk2
Av · v
= arccos
arccos
kAvkkvk
kvk2
= θ.
0
1
65.29
√
2
2
1
-2
0
3
e1 e2
65.31 v × w = −1 3
2
1
e3 −6
0 = −2
−2
−7
65.33 cube: b), rectangular box: a)
65.35 L intersects only the lines given in c) and
d).
√
65.37 38
65.39 If θ is the angle between v and w, then
Pw (v) = kvk cos(θ)w/kwk = (v · w)w/kwk2 .
65.41 (a + b)/2 + (c − (a + b)/2)/3
= (a + c)/2 + (b − (a + c)/2)/3
= (b + c)/2 + (a − (b + c)/2)/3
= (a + b + c)/3
65.43 a sphere of radius 2 centered at (−1, 3, 2)
√
65.45 arccos(−1/ 84)
√
65.47 7/ 18
65.49 a)
cos(θ) sin(θ)
cos(θ) − sin(θ)
t
AA=
sin(θ) cos(θ)
− sin(θ) cos(θ)
1 0
=
,
0 1 w1
v
=
b) A 1 · A
v
w
2
2 cos(θ)v1 − sin(θ)v2
cos(θ)w1 − sin(θ)w2
·
sin(θ)v1
cos(θ)v2
sin(θ)w1
cos(θ)w2
However, this alternative argument does not allow us to draw any conclusions concerning the
orientation of the rotation.
Chapter 66
66.3 Using the given values for α, β, and m, it
follows that
3mg
− cos(β)
−19.21
≈
,
Fa =
sin(β)
58.86
2 sin(β)
3mg
cos(β)
19.21
Fb =
≈
,
sin(β)
58.86
2 sin(β)
mg
− cos(α)
−18.80
Fp =
≈
,
sin(α)
19.62
2 sin(α)
mg
cos(α)
18.80
≈
.
Fr =
sin(α)
19.62
2 sin(α)
Furthermore, according to Newton’s third fundamental law of mechanics, the magnitude of
the forces at A and B is k − Fa k = k − Fb k =
61.92 kg · m/s2 .
p
66.5 d = 100 − g 2 /4 ≈ 8.71 m
66.7 The force increases, and the relevant explanation is given at the end of Example 66.1.
Chapter 67
67.29 Since t is the first component of r(t), as
defined in Example 67.11, we may infer that
36
Solutions
y
t1 = t2 whenever r(t1 ) = r(t2 ). Thus r is simple.
5
4
67.31 Using a computer to solve for λ the equation
2λ
15 = L =
sinh
δg
δgd
2λ
15λ
≈
sinh
9.81
7 · 9.81
15λ
3
a)
2
1
-4
-3
-2
-1
x
1
2
3
-1
b)
-2
c)
yields λ ≈ 1.993. Hence
4
-3
-4
f (x) ≈ 1.524 cosh(0.656x) − 7.653,
s
2
kg · m
d
≈ 10.01
.
kFk = λ 1 + f 0
2
s2
Figure S.10: graphs for Exercise 67.39.
y
x
2π
67.35 If the length of the rope is given to be L,
then λ is the solution of the equation
2λ
sinh
1=
δgL
δgd
2λ
.
y
x
In setting x := δgL/2λ and y := δgd/2λ, this
equation can be rewritten in the form x =
sinh(y). As L approaches d, or equivalently,
as x approaches y, the equation x = sinh(y)
is transformed into the equation x = sinh(x).
Since the only solution of the latter equation is
x = 0, it follows that x = δgL/2λ must approach zero as L approaches d, or equivalently,
λ must tend p
∞. By implication, the magnitude kFk = λ 1 + f 0 (d/2)2 = λ cosh(δgd/2λ)
must increase to ∞ as well. This observation
is intuitively not surprising, because one would
expect the force needed to stretch a rope of positive density into a perfectly straight line to be
infinite.
3 cosh(t)
3/ cos(t)
67.37 r(t) =
or r(t) =
4 sinh(t)
4 tan(t)
(answers may vary).
67.39 See Figure S.10.
2
1
67.41 tangent line: s(t) = 4 + t 4
0
4
R 2π q
(1 + cos(θ))2 + sin2 (θ) dθ
67.43 a) L = 0
R 2π
= 2 0 | cos(θ/2)| dθ = 8 (see Figure S.11).
R π/2 q
b) L = −π/2 cos2 (θ) + sin2 (θ) dθ
R π/2
= −π/2 dθ = π (see Figure S.11).
1
Figure S.11: graphs for Exercise 67.43a, b.
Rπ √
Rπ√
c) L = 0 θ4 + 4θ2 dθ = 0 θ θ2 + 4 dθ
√
R π2 +4 √
3
u/2 du = ( π 2 + 4 − 8)/3
= 4
y
x
2
π
67.45 If we denote by θ the angle between the
positive x-axis and the line segment from the
center of the circle at (1/2, 0) to the center of
the wheel, then a parameterization of the path
of P is
1/2 + cos(θ) + cos(2θ)/2
r(θ) :=
.
sin(θ) + sin(2θ)/2
Furthermore, using the double angle formulae
for sine and cosine, we find that
cos(θ)(1 + cos(θ))
r(θ) =
.
sin(θ)(1 + cos(θ))
Since this representation of r is identical with
the one given in Example 67.32, we may infer
Solutions
69.11
C
=
0.
8
C
=
C
-2
=
6
C=
2
C=
1
-2
=
C
=
level curves of f
69.25
x
-1
x
C
=
-1
C
2
C
because r1 − r2 is obviously parallel to F12 .
√
68.17 t = (6 + 3 4 + 2g)/g
2
= 0 + (r1 − p) × F12 + 0 + (r2 − p) × F21
= (r1 − r2 ) × F12 = 0
=
=
= r10 × m1 r10 + (r1 − p) × m1 r100
+ r20 × m2 r20 + (r2 − p) × m2 r200
y
C
1
M
0
0.
6
=
C
y
C
68.11 Since
d
kr − r0 k2 = 2(r − r0 ) · r 0 = 0,
dt
we may conclude that kr − r0 k is constant and,
by implication, the curve described by r is contained in a sphere centered at r0 .
68.13 Since d/dt n·r = n·r 0 = 0, it follows that
n · r is constant. Therefore, n · r = n · r(0) or,
equivalently, n·(r−r(0)) = 0. This shows that r
traces out a curve that is contained in the plane
described by the equation n · (r − r(0)) = 0.
68.15 Using Newton’s second and third laws,
we see that
m1 r100 + m2 r200
F12 + F21
=
C 00 =
m1 + m2
m1 + m2
F12 − F12
=
= 0,
m1 + m2
and similarly,
x
69.13
=
and, by implication,
r !
r 2g π g 0
kr (10)k ≈
≈ 0.086 m/s.
sin 10
20 6 3 0.
4
=
0.
2
y
C
p
(π/40) cos( 2g/3 t)
, it
68.9 With r(t) ≈
−3/2
follows that
p
p
−(π/20) g/6 sin(10 2g/3)
r 0 (10) ≈
0
69.1 F = −γmM (r0 − r)/kr0 − rk3
=
Chapter 68
Chapter 69
C
that the curve described by r is indeed the polar
graph shown in Figure 67.9.
√
67.47 center: (1/6, −1/6, 1/3); radius: 210/6
67.49 a) not parallel and no intersection,
b) parallel, but no intersection,
c) not parallel, but intersection at the point
(6/5, −13/5, 7/5)
67.51 c) and d) are simple because the ycoordinate 3t3 is invertible on R, and b) is
simple as well because the given x- and ycoordinates are both invertible on (0, 3). By
contrast, a) is not simple, because r(−1) = r(1).
37
level curves of g
y
x
1
3
69.31 With h(x, y) = x, f (x) = x, and g(y) =
e
y 3 , we find that Fe (x) = ln |x| and G(y)
=
2
−1/(2y ). Consequently, the equation of the
flow lines is ln |x| + 1/(2y 2 ) = C.
69.33 a)
z
C=1
C=0
C = -1
38
Solutions
b)
69.39 a)
=
0
y
C
2
2
C
=
1
1
1/2
C=
C=
x
-2
c)
c)
-1
0
1
2
y
3
C=2
2
C=1
1
C=0
π/2
x
C = -1
-1
d) C = −1
C=0
69.41
a)
b)
c)
d)
e)
f)
C=1
69.35
graph
contour diagram
a
b
b
d
c
a
d
c
69.37 a) If P is an antiderivative of g/f , then
P (x) − y = C is the xy-equation of the flow
lines.
b) Here the equation is x − Q(y) = C, where Q
is an antiderivative of f /g.
69.43 Proceeding exactly as in Exercise 69.42,
we obtain
4t + 1
r(t) = −e−3t
4t − 3
(answers may vary).
69.45 b) f (x, y) := x − ey , c) f (x, y) := y − ex
(answers may vary).
69.47 b) and d).
Solutions
Chapter 71
71.5 ∂f /∂x = 2xyz 2 cos(x2 y),
∂f /∂y = x2 z 2 cos(x2 y), and
∂f /∂z = 2z sin(x2 y).
71.7 equation: −2x + 2y + z = 2, perpendicular
vector: n = −2e1 + 2e2 + e3
71.23 If f and g are continuous at r0 , then f (r)
and g(r) approach f (r0 ) and g(r0 ), respectively,
as r approaches r0 . Consequently, the sum of
f (r) and g(r) must approach the sum of f (r0 )
and g(r0 ). In other words, f + g is continuous
at r0 . Analogous arguments also serve to establish the continuity of λf , f g and f /g. Furthermore, if r is a continuous parameterization,
then r(t) approaches r(t0 ) as t approaches t0 .
Consequently, the assumption of continuity on
f allows us to infer that f ◦ r is continuous, because as t approaches t0 the values f (r(t)) must
approach f (r(t0 )). This completes the proof of
Theorem 71.21.
Turning our attention now to Theorem 71.22, we first observe that f clearly is continuous at a point r0 ∈ D if and only if
lim |f (r0 + u) − f (r0 )| = 0.
u→0
Since
|f (r0 + u) − f (r0 )| =
|f (r0 + u) − f (r0 ) − ∇f (r0 ) · u + ∇f (r0 ) · u|,
the assumption of differentiability on f implies
that
lim |f (r0 + u) − f (r0 )| = lim |∇f (r0 ) · u| = 0,
u→0
u→0
as desired.
√
71.27 ∂u f (1, −3) = −3/ 17
71.31 V (r) = −γmM/kr − r0 k
71.33 f (x, y) = x2 y 2 − x2 /2 + C
71.35 Only G is a gradient field.
71.37 f (r) = − ln(krk) + C
71.43 a) yexy sin(x) + exy cos(x) + 2xy,
b) xexy sin(x) + x2 ,
c) yz cos(xyz) cos(x2 y 2 z 2 )
−2xy 2 z 2 sin(xyz) sin(x2 y 2 z 2 ),
d) xy cos(xyz) cos(x2 y 2 z 2 )
−2x2 y 2 z sin(xyz) sin(x2 y 2 z 2 )
71.45 Here the condition to be satisfied is
−∂f /∂x
1
−∂f /∂y k 1 ,
1
2
39
or equivalently, −2∇f = e1 + e2 . Since
2
y −1
∇f =
,
2yx + 2
it follows that −2(y 2 − 1) = 1 and −2(2yx +
2) = 1, and the points whose coordinates simultaneously satisfy both of√ these equations
√
are√ easily seen
to be (−5 2/4, 1/ 2) and
√
(5 2/4, −1/ 2).
71.47 ∂u f = 0. Intuitive explanation: since the
curve described by the parameterization r(θ) =
(R cos(θ), R sin(θ)) is a level curve of f and since
u = r 0 (θ)/kr 0 (θ)k is tangent to this level curve,
it is natural to expect the directional derivative
of f in the direction of u to be zero.
71.49 Setting F(x, y) := (x − y)e1 + (x + y)e2
and denoting by θ the angle between F(x, y) and
r := (x, y), it follows that
cos(θ) =
1
F(x, y) · r
= √ ,
kF(x, y)kkrk
2
and therefore, θ = π/4. Applying now the techniques developed in Chapter 60 to solve the initial value problem
0 x
1 −1
x
x(0)
1
=
,
=
,
1 1
y0
y
y(0)
−1
we find the parameterization
t
x(t)
e (cos(t) + sin(t))
r(t) =
= t
.
y(t)
e (sin(t) − cos(t))
√
71.51 ∂u f = 18/ 5
71.53 In each case the flow lines are to be drawn
in such a way that they meet the given level
curves at a right angle at all points.
71.55 a) ∂u f (r) = ∇f (r) · u = k∇f (r)k ≥ 0.
b) ∇f is a vector in R2 , but the graph of f is
a subset of R3 and, by implication, any vector
tangent to the graph must be in R3 as well.
c) Same mistake as in b).
d) The derivative of a multivariable function is
a vector (the gradient of f ), not a number.
e) The stated conditions imply that
|f (r + u) − f (r) − ∇f (r) · u|
kuk
|∇f (r) · u|
= lim k∇f (r)k| cos(θ)|
= lim
u→0
u→0
kuk
√
= lim 5| cos(θ)|,
0 = lim
u→0
u→0
40
where θ is the angle between ∇f (r) and u. However, the limit of | cos(θ)| does not exist because
the value of θ is entirely arbitrary (u may approach 0 from any direction).
f ) ∂u f (r(t)) = ∇f (r(t))·r 0 (t) = r 0 (t)·r 0 (t) ≥ 0.
g) If r parameterizes a level curve, the function f ◦ r is constant and ∂u f (r(t)) = ∇f (r(t)) ·
r 0 (t) = d/dt f (r(t)) = 0 6= −2.
h) On an equipotential surface the potential energy remains constant and so does the sum of
the potential and kintetic energies (by the law
of the conservation of energy). Consequently, in
contradiction to the stated assumption K(1) =
2 6= 4 = K(3), the kinetic energy must remain
constant as well.
71.57
71.59 Using the well known values R ≈ 1.5 ·
1011 m, T ≈ 365.25 · 24 · 3600 s, and γ ≈
6.672 · 10−11 m3 s−2 kg−1 , it follows that M ≈
4π 2 R3 /(γT 2 ) ≈ 2.005 · 1030 kg.
Chapter 72
72.1 If either q > 0 and Q > 0 or q < 0 and Q <
0, then qQ > 0, and F(r) points in the direction
from q to Q, thus modeling an attractive force.
Otherwise qQ will be negative, and F(r) will
represent a repellent force pointing away from
Q.
72.13 With r(t) = (3t, 1, 2 − t) for t ∈ [0, 1] the
value of the path integral turns out to be 24.
72.15 a) zero, b) negative, c) negative, d) positive, e) negative, f ) zero
72.17
= −exy + C,
R a) V (x, y)
−3
b) C F · dc = e − e2 ,
c) −e2 + C + 10 = V (1, 2) + 10 = V (−1, 3) +
2
−3
2
mv
p /2 = −e + C + 2v , and therefore, v =
−3
2
(e − e + 10)/2.
R2
72.19 a) 1 F(r(t)) · r 0 (t) dt = 0
Solutions
b) Setting α := arctan(3/2), we define
√
13 cos(t)
√
s(t) :=
13 sin(t)
for t ∈ [α, α + 2π] and
2(t − α − 2π + 1)
s(t) :=
3(t − α − 2π + 1)
for t ∈ [α+2π, α+2π +1]. Then s(α) = (2, 3) =
r(1), s(α + 2π + 1) = (4, 6) = r(2), and
Z
α+2π+1
F(s(t)) · s 0 (t) dt
α
=
Z
α+2π
13 dt +
α
Z
α+2π+1
0 dt = 13 6= 0.
α+2π
Rb
72.21 a) FALSE because a F(r(t)) · r 0 (t) dt =
Rb 0
kr (t)k2 dt ≥ 0.
a
b) FALSE for the same reason as in a).
c) POSSIBLY TRUE/POSSIBLY
FALSE: set√
ting r(t) := (cos(t), sin(t))/ 2π for t ∈ [0, 2π]
and s(t) := (cos(2t), sin(2t)) for t ∈ [0, π], it
is easy to see that r and s describe closed flow
lines of the vector fields F(x, y) := (−y, x) and
G(x, y) := 2(−y, x) respectively. However, the
corresponding path integral is equal to 1 only
for F and r, but not for G and s.
d) FALSE because there are no closed flow lines
in gradient fields.
e) POSSIBLY TRUE/POSSIBLY FALSE: the
examples in c) apply also in this case.
f ) FALSE because the path integral of a gradient field over a closed curve is always equal to
zero.
g) FALSE: F and r (or G and s) in c) provide
a counterexample.
h) TRUE because the path integral of a gradient field over a closed curve is indeed always
equal to zero.
i) FALSE: the vector field F(x, y) = (y, 0) defined on R2 serves as a counterexample.
72.23 the parameterization in b)
Chapter 73
73.3 See Figure S.12.
73.13 Since the solutions of the quadratic
equa√
tion ax2 + bx + c = 0 are (−b ± b2 − 4ac)/2a,
it follows that real solutions exist if and only if
b2 − 4ac ≥ 0 or, equivalently, ac − b2 /4 ≤ 0.
Solutions
global maximum
local maximum
global minimum
41
From the equality of the second components, we
may infer that either y = 0 and, by implication,
x = ±a, or λ = b2 . In the latter case, the
equality of the first components in the vector
equation above implies that x = a2 x0 /(a2 − b2 ).
Given the constraint g(x, y) = x2 /a2 + y 2 /b2 =
1, it is easy to see that the corresponding value
of y satisfies the equation
a2 x20
2
2
.
y =b 1− 2
(a − b2 )2
Figure S.12: graph for Exercise 73.3.
Furthermore, for both equations in (73.12) it
is the case that ac = ∂ 2 f /∂x2 (r0 )∂ 2 f /∂y 2 (r0 )
and b2 /4 = (∂ 2 f /∂y∂x(r0 ))2 . Consequently,
the term shown in (73.13) is indeed the common discriminant of the equations in (73.12).
√
73.15 The √
critical points are (0, 0), (0, 1/ 2)
and (0, −1/ 2).√Furthermore, since
√ Df (0, 0) =
8 > 0, Df (0, 1/ 2) = Df (0, −1/ 2) = −14 <
0 and ∂ 2 f /∂x2 (x, y) = −2(y 4 − y 2 + 2) < 0,
we may√conclude that f has saddle points at
(0, ±1/ 2) and a local maximum at (0, 0).
73.21 Setting f (x, y) := (x − x0 )2 + (y − y0 )2
and g(x, y) := y − mx − b yields
2(x − x0 )
−λm
= ∇f = λ∇g =
.
λ
2(y − y0 )
In eliminating λ and using the constraint
g(x, y) = 0, we find that
(x − x0 )/m = −(y − y0 ) = y0 − mx − b,
or equivalently, x = (x0 − mb + my0 )/(1 + m2 ).
The corresponding value for y is (mx0 + m2 y0 +
b)/(1 + m2 ). Substituting x and y back into the
defining equation for f yields
f (x, y)
(my0 − mb − m2 x0 )2
(mx0 + b − y0 )2
+
2
2
(1 + m )
(1 + m2 )2
(mx0 + b − y0 )2
=
.
1 + m2
=
Finally, in taking the square root we find the
same formula for the minimal distance as in
Chapter 12.
73.25 In setting f (x, y) := (x − x0 )2 + y 2 and
g(x, y) = x2 /a2 + y 2 /b2 , we obtain
2(x − x0 )
2x/a2
=λ
.
2y
2y/b2
Substituting the values for x and y 2 back into
the defining equation for f yields
x20
2
.
f (x, y) = b 1 − 2
a − b2
To determine the minimal distance, we need to
compare this value of f with f (a, 0) = (a − x0 )2
and f (−a, x0 ) = (a + x0 )2 . In fact, we only
need to decide whether the value of f above is
less than or greater than (a − x0 )2 because the
assumption x0 > 0 implies that (a − x0 )2 <
(a + x0 )2 . Since
x20
2
2
(a − x0 ) − b 1 − 2
a − b2
2
b2
a2
a
−
x
−
,
= 2
0
a − b2
a
and since the term on the right-hand side of
this equation is less than or greater than zero
depending on whether a < b or a > b, we may
infer that the minimal distance from the ellipse
to the point (x0 , 0) is
(
if a < b,
a − x0
dmin =
2
2
2
2
b (1 − x0 /(a − b )) if a > b.
73.27 Setting
f (a, b, c) := (1 − (a − b + c))2
+ (−1 − (a + b + c))2
+ (0 − (4a + 2b + c))2
+ (6 − (25a + 5b + c))2 ,
it is easy to see that the equation ∇f = 0 is
equivalent to the matrix equation
643 131 31
a
150
131 31 7 b = 28 .
31
7
4
c
6
42
Solutions
Multiplying both sides with the inverse of the
matrix on the left yields a = 5/12, b = −49/60,
and c = −3/10 (see Figure S.13 for the graph).
Note: the symmetric arrangement of the coefficients in the matrix above is not coincidental—
think about it.
y
10
8
6
4
2
x
-2
2
4
∂2
f ((2n + 1)π/2, (2m + 1)π/2) =
2
∂x
(
−1 if n and m are both even or both odd,
1
otherwise,
6
-2
Figure S.13: graph for Exercise 73.27.
73.29 Since a parameterization of the plane is
s
,
t
r(s, t) :=
D/C − As/C − Bt/C
it follows that the minimal distance is the minimum of f (s, t) := kr(s, t) − pk, where p =
x0 e1 + y0 e2 + z0 e3 . In setting the gradient of f
equal to 0 we find that
AD + (B 2 + C 2 )x0 − ABy0 − ACz0
,
A2 + B 2 + C 2
BD + (A2 + C 2 )y0 − BAx0 − BCz0
t=
,
A2 + B 2 + C 2
s=
and the corresponding value of f is
dmin = f (s, t) =
1)π/2, (2m + 1)π/2), where n and m are understood to be arbitrary integers. Since Df (x, y) =
sin2 (x) sin2 (y) − cos2 (x) cos2 (y), we may infer
that f has a saddle point at each point of the
form (nπ, mπ) because Df (nπ, mπ) = −1 < 0.
Furthermore, f assumes a local extremum at
each point of the form ((2n+1)π/2, (2m+1)π/2)
because Df ((2n + 1)π/2, (2m + 1)π/2) = 1 > 0.
Since
|Ax0 + By0 + Cz0 − D|
√
.
A2 + B 2 + C 2
73.31 Since the area of the triangle is easily seen
to be (bx+ay +ab)/2, we define f (x, y) := (bx+
ay + ab)/2, and the constraint is represented by
the equation g(x, y) := x2 /a2 + y 2 /b2 = 1. In
solving for x, y, and λ the equations ∇f (x, y) =
λ∇g(x,
√ y) = 1, we find that x =
√ y) and g(x,
Consequently,
the maxia/ 2 and y = b/
√ 2. √
√
mal area is f (a/ 2, b/ 2) = ( 2 + 1)ab/2.
73.33 a) The critical points are (0, 0), (2, 0)
and (3, 0). Since Df (0, 0) < 0, Df (2, 0) > 0,
Df (3, 0) < 0, and ∂ 2 /∂x2 f (2, 0) > 0, we may
infer that f has a local minimum at (2, 0) and
saddle points at (0, 0) and (3, 0).
b) The set of critical points consists of the
points of the form (nπ, mπ) and ((2n +
it follows that f assumes a local maximum at
((2n + 1)π/2, (2m + 1)π/2) whenever n and m
are both even or both odd and otherwise a local
minimum.
73.35 There are no local extrema, but f
has saddle points at (0, 0, 0) and all points
of the form (2, y, 1), (−2, y, 1), (2, y, −1) and
(−2, y, −1) (where y is understood to be an arbitrary real number).
73.37 If S = (x0 , y0 , z0 ), E = (x1 , y1 , z1 ), and
R = (x, y, 0) then the distance that the light ray
travels from S to E via R is
q
D(x, y) = (x − x0 )2 + (y − y0 )2 + z02
q
+ (x − x1 )2 + (y − y1 )2 + z12 .
Setting
q
(x − x0 )2 + (y − y0 )2 + z02 ,
q
f1 (x, y) := (x − x1 )2 + (y − y1 )2 + z12 ,
f0 (x, y) :=
it is not difficult to show that
(x − x0 )/f0 (x, y)
∇D(x, y) =
(y − y0 )/f0 (x, y)
(x − x1 )/f1 (x, y)
.
+
(y − y1 )/f1 (x, y)
Consequently, (x, y) is a critical point of D only
if
y − y0
x − x0
f0 (x, y)
=
=−
,
x − x1
f1 (x, y)
y − y1
that is, only if
(x − x0 )(y − y1 ) − (y − y0 )(x − x1 ) = 0.
Solutions
43
Given this result, we may conclude that the
third component of the vector
x − x0
x − x1
n = y − y0 × y − y1
−z0
−z1
e) Since the first and second components of the
particle’s velocity vector are not affected by the
passage through the potential barrier, we may
infer that x2 = v1+ T and y2 = v2+ T . Given this
observation, it follows that
is zero. In other words, n is parallel to the xyplane and, by implication, the plane containing
S, E, and R is perpendicular to the xy-plane.
73.39 a) According to (73.31) and (73.32), we
may infer that
m
m
kv+ k2 + V+ = kv+ − we3 k2 + V− ,
2
2
or equivalently,
m(x1 − x2 )
mx1
+
= mv1+ − mv1+ = 0,
t+
T − t+
2(V+ − V− )
= w2 + 2v3+ w.
m
Solving for w yields
r
2(V+ − V− )
+
,
w = −v3 ± (v3+ )2 +
m
and similarly,
my1
m(y1 − y2 )
+
= mv2+ − mv2+ = 0.
t+
T − t+
In other words, the first two components of
∇f (x1 , y1 , t+ ) are equal to zero. Furthermore,
since
krx1 ,y1 − r0 k = kr1 − r0 k = kv+ kt+ ,
kr2 − rx1 ,y1 k = kr2 − r1 k = kv− kt− ,
we may infer that
mkr2 − rx1 ,y1 k2
mkrx1 ,y1 − r0 k2
−
2
2(T − t+ )
2t2+
m
m
= kv− k2 − kv+ k2
2
2
= V+ − V− (by (73.31)).
but since w must be negative, it follows that
r
2(V+ − V− )
+
w = −v3 − (v3+ )2 +
.
m
b) Since T = t+ + t− , we obtain
Z T
Z t+ m
kv+ k2 − V+ dt
L(t) dt =
2
0
0
Z T
m
+
kv− k2 − V− dt
2
t+
m
=
kv+ k2 − V+ t+
2
m
kv+ + we3 k2 − V− t− .
+
2
c) Performing a calculation analogous to the one
in b), we obtain
Z T
mkrx,y − r0 k2
Lx,y,τ (t) dt =
− τ V+
2τ
0
mkr2 − rx,y k2
− (T − τ )V− .
+
2(T − τ )
d) In denoting the first and second components
of r2 by x2 and y2 respectively, we find ∇f to
be equal to
mx m(x − x2 )
+
τ
T −τ
my
m(y
−
y
)
2
.
+
τ
T
−
τ
2
2
mkr2 − rx,y k
mkrx,y − r0 k
−
+ V− − V+
2(T − τ )2
2τ 2
Consequently,
the third component of
∇f (x1 , y1 , t+ ) is zero as well, and (x1 , y1 , t+ )
has thus been shown to be a critical point of f .
Chapter 74
74.7 For n = 1, we have
m1
X
lim
m1 →∞
∆x1,i1 →0 i1 =1
f (ri1 )∆x1,i1 = lim
m
X
m→∞
∆xi →0 i=1
R1R2R3
f (xi )∆xi .
xy 2 z 3 dz dy dx = 27
√
√
R
3
74.25
2(4 4 − x2 − 2 4 − x2 /3) dx
R
= R (32 cos2 (t) − 64 cos4 (t)/3) dt
= (16 cos(2t)/3 + 8t − 8 cos(4t)/3) dt
= 8 sin(2t)/3 + 8t − 2 sin(4t)/3 + C
= 16 sin(t) cos(t)/3 + 8t
−8 sin(t) cos(t)(2 cos2 (t) − 1)/3
√ +C
= 8 arcsin(x/2) + (x3 + 2x) 4 − x2 /3 + C
R 2 R √4−y2
74.27 −2 √ 2 (x2 + y 2 ) dx dy
− 4−y
R1R1√
74.29 0 y 1 − x2 dx dy
R 1 R π/2
= 0 arcsin(y) (cos(2t) + 1)/2 dt dy
74.19
0
0
0
44
Solutions
R1
π/2
(sin(t) cos(t) + t)/2|arcsin(y) dy
p
R01
= 0 (−y 1 − y 2 − arcsin(y) + π/2)/2 dy
R 1 R √1−x2 R √1−x2 −y2
74.31 −1 −√1−x2 √ 2 2 dz dy dx
− 1−x −y
R 1 R √1−x2 p
√
= −1 − 1−x2 2 1 − x2 − y 2 dy dx
R 1 R π/2
2(1 − x2 ) cos2 (t) dt dx
−1 −π/2
R1
= −1 π(1 − x2 ) dx = 4π/3
=
= 1/3
74.37 By symmetry, the first and second components of the force on q must be zero. Using the given values for q, d, and δ, and setting
∆x := ∆y := 0.1 and xk := yk := −2 + 0.1k, an
approximation for the third component is obtained as follows:
40 40
1
1 XX
F3 ≈
q
3 ∆x∆y
πε0 i=1 j=1
x2i + yj2 + 4
≈
1.04687
.
πε0
Alternatively, we can choose xk and yk to be the
midpoints of the subintervals in the partition of
[−2, 2] along the x- and y-axes. So in setting
xk := yk := −2 + 0.1(2k − 1)/2 we find the
(presumably improved) estimate
R π/2 R 1
y sin(xy) dx dy
74.43 a) 0
0
R π/2
= 0 (1 − cos(x)) dx = π/2 − 1,
R 1 R π/2
b) 0 0 y sin(xy) dy dx
R1
= 0 (−π cos(πx/2)/(2x) + sin(πx/2)/x2 ) dx
= limc→0 − sin(πx/2)/x|1c = π/2 − 1
74.45 a) 1/2, b) 1/4, c) 1/8, d) 1/2n
74.47
of f , it follows that
R 1 R 1 Given the definition
P∞
f (x, y) dx dy = n=1 (2n /2n −2n /2n ) = 0
0 0R R
P∞
1 1
and 0 0 f (x, y) dy dx = 1 + n=1 (−2n /2n +
2n+1 /2n+1 ) = 1. However, this surprising observation does not contradict the result stated in
74.10 because f is not bounded on [0, 1] × [0, 1].
74.49 First row beginning from the left:
R 2 R 2−x/2
f (x, y) dy dx,
R02 Rx/2
R1 R1
2
f (x, y) dy dx − −1 0 f (x, y) dy dx,
−2 0
R 1 R |y|
f (x, y) dx dy.
−1 −|y|
Second row beginning from the left:
R 1 R 1−|x|
f (x, y) dy dx,
−1 −1+|x|
R 3π/4 R sin(x)
f (x, y) dy dx,
− cos(x)
R3Rx
R−π/4
1Rx
f (x, y) dy dx + 1 √x2 −1 f (x, y) dy dx.
0 0
74.51 a)
z
1
40 40
1
1 XX
F3 ≈
q
3 ∆x∆y
πε0 i=1 j=1
x2i + yj2 + 4
≈
1.04735
.
πε0
74.39 a)
y
1
1
x
z
b)
z
4
y
3
2
2
2
x
1
c)
0
z
y
2
x
R
b) R S f (r) dA = 0·2+1·2+2·2+3·2+4·8 = 44,
c) S f (r) dA = 0 · 25/2 + 1 · 25/2 = 25/2
R
74.41
R a) S f (r) dA = 12,
b) RS f (r) dA = 81 for S = [0, 3] × [0, 3] × [0, 3]
andR S f (r) dA = 288 for S = [0, 4]×[0, 4]×[0, 4].
c) S f (r) dA = 3n3 (n − 1)/2
y
-1
2
x
1
d) See Figure S.14.
Solutions
z
45
c)
y
1
2
1/2
3
π/4
x
y
1
x
d)
1
y
1
Figure S.14: graph for Exercise 74.51d.
74.53
left to right:
R 2 R 1 RFrom
1−y
f (x, y, z) dz dy dx,
R04 R01 R0 1−|y|
f (x, y, z) dz dy dx,
0 −1 |y|−1
√
R 3 R 2 R 1+ 1−y2
√ 2 f (x, y, z) dz dy dx.
0 0
1−
x
0
e)
1−y
1
z
Chapter 75
75.3
R3R0
75.5
R 2π R ∞
0
−π/2
0
0
r2 (cos(θ) − sin(θ)) dr dθ = 18
r/(1 + r2 )2 dr dθ = π
y
75.7 From left to right:
R π/6 R 2/ cos(θ)
f (r cos(θ), r sin(θ))r dr dθ,
R0π/2 R02 cos(θ)
f (r cos(θ), r sin(θ))r dr dθ.
0
0
√
75.13 r = 5, θ = arctan(2), z = 3
R 2π R 2 R 3
75.15 0 0 2 (r2 + z)r dr dz dθ = 75π
75.17
R 2 R 2π R z/2
0
0
0
x
f ) The region is a torus with centerline radius
2 and inner radius 1.
f (r cos(θ), r sin(θ), z)r dr dθ dz
75.27 a)
y
2
1
x
0
b)
1
2
3
y
3
R1Rπ
75.29 a) 0 0 f (g(s, t)) ds dt,
R1R1
b) 0 0 6f (g(s, t)) ds dt,
R1R2
c) 0 0 f (g(s, t)) ds dt,
R π/4 R 1
d) 0
sf (g(s, t))/ cos2 (t) ds dt,
R 1 R 1 R0 2π
e) 0 0 0 f (g(θ, r, z))rz 2 dθ dr dz,
R 1 R 2π R 2π
f ) 0 0 0 f (g(θ, φ, r))r(2r+cos(φ)) dθ dφ dr
R π/2 R 1
75.31 a) −π/2 0 r2 cos(θ) dr dθ,
R π R −1/ cos(θ) 3 2
b) 3π/4 0
r sin (θ) dr dθ,
R π/2 R 1/ cos(θ)
c) −π/2 0
r/(1 + r2 ) dr dθ,
R π/2 R cos(θ) 3
d) −π/2 0
r dr dθ
2
1
x
0
1
2
R α/2 R 2π R R 2
ρ sin(φ) dρ dθ dφ
75.33 V = 0
0
0
= 2πR3 (1 − cos(α/2))/3. Furthermore, for α =
2π we have cos(α/2) = −1, and therefore, the
volume of a ball of radius R is 4πR3 /3.
46
Solutions
75.35 a)
z
1
y
x
1
1
R 2π R 1 R 1−r2
r3 cos(θ) sin(θ) dz dr dθ
0
0 0
R π/2
R g(φ) 4
R 2π
c) 0 0
ρ cos(θ) sin(θ) sin3 (φ) dρ dφ dθ
0
with
q
− cos(φ) + 1 + 3 sin2 (φ)
.
g(φ) =
2 sin2 (φ)
b)
R
R
πhR2 /3, W x dx dy dz = W y dx dy dz =
R 2π R R R h
R
and W z dx dy dz = 0 0 hr/R zr dz dr dθ
πh2 R2 /2. Consequently, the center of mass
at (0, 0, 3h/4).
p
R z(t) √
76.3 Since −2 −z(t) = 0 1/ −z dz
p
Rt 0
Rtp
z (t)/ −z(t) dt = − 0 4g/3 sin(α) dt
0p
− 4g/3 sin(α)t, we may infer that z(t)
−g sin2 (α)t2 /3 as desired.
75.39
Setting h(θ, r, z) := f (r cos(θ), r sin(θ), z), we
findR theR following
integrals:
2π 2 R 4
a) 0 0 r2 h(θ, r, z)r dz dr dθ,
R 2π R π/2 R cos(z)
h(θ, r, z)r dz dr dθ,
b) 0 −π/2 0
R 2π R 2 R √2−r2
c) 0 0 0
h(θ, r, z)r dz dr dθ,
R 2π R 1 R 3
d) 0 0 3r h(θ, r, z)r dz dr dθ.
75.41 From left to right:
Jg (x, ρ, σ) = 3(1 − x/2)2 (1 − ρ),
Jg (y, ρ, θ) = ρf (y)2√
,
Jg (θ, ρ, σ) = 5σ 2 ρ/ 2.
75.43
cylindrical coordinates:
R 2π R 1 In
R 2−2r
f (g(θ, r, z))r dz dr dθ,
0
0 0
where g(θ, r, z) represents the transformation to
cylindrical coordinates.
In spherical coordinates:
R 2π R π/2 R 1/(cos(φ)+2 sin(φ))
h(θ, φ, ρ) dρ dφ dθ,
0
0
0
where h(θ, φ, ρ) = f (g(θ, φ, ρ))ρ2 sin(φ). Here
g(θ, φ, ρ) represents the transformation to
spherical coordinates.
Chapter 76
76.1 Using cylindrical coordinates, we find
R 2π R R R h
R
that W dx dy dz = 0 0 hr/R r dz dr dθ =
=
=
=
76.5 In performing an integration in spherical coordinates, we easily determine the moment of inertia I to have the value mR2 /5.
The corresponding kinetic energy is K(t) =
3mkr 0 (t)k2 /5. Using these results, a calculation analogous to that for a rolling wheel (as
discussed in Chapter 76) yields
75.37
4s cos(π/6) − 2t sin(π/6)
a) g(s, t) =
=
4s sin(π/6) + 2 cos(π/6)t
√
2 3s√− t
for (s, t) ∈ [0, 1] × [0, 1] (answers
2s + 3t
may vary),
b) Jg = 8,
√
√
R1R1
c) 0 0 8(2 3s − t)(2s + 3t) ds dt
0,
=
is
5g
sin(α) cos(α)t2 ,
12
5g
z(t) = − sin2 (α)t2 .
12
y(t) =
Chapter 77
77.3
R 2π R √2
0
Z
0
re−r
2
/2
/(2π) dr dθ = 1 − e−1
2
810
e−(x−800) /800
√
dx ≈ 0.533
20 2π
780
77.13 Using the result of Exercise 77.12 with
= 0 and σ1 =R 1, we obtain:
√
Rµ1µ+σ
2
1
p(x) dx = −1 e−x /2 / 2π dx ≈ 0.68,
µ−σ
√
R2
R µ+2σ
2
p(x) dx = −2 e−x /2 / 2π dx ≈ 0.95,
µ−2σ
√
R µ+3σ
R3
2
p(x) dx = −3 e−x /2 / 2π dx ≈ 0.997.
µ−3σ
77.11
Chapter 78
78.3
1
1−1
0−1
p(s, t) = 0 + s 2 − 0 + t 2 − 0
2
3−2
4−2
1−t
= 2s + 2t (answers may vary).
2 + s + 2t
x
78.9 p(x, y) = y for any point
x2 + y 2
2
(x, y) ∈ R whose coordinates satisfy the inequality x2 + y 2 ≤ 1.
Solutions
a cos(θ) sin(φ)
78.13 p(θ, φ) = b sin(θ) sin(φ) for (θ, φ) ∈
c cos(φ)
[0, 2π] × [0, π] (answers may vary).
78.15 Example 78.2: the plane is oriented in
the direction of v × w = ∂p/∂s × ∂p/∂t.
Example 78.5: the positive normal vector
rh cos(θ)/R
∂
∂
p×
p = rh sin(θ)/R
∂θ
∂r
−r
induces a downward orientation toward the outside of the cone.
Example 78.7: since the z-component of the
positive normal vector
−∂f /∂x
∂
∂
p×
p = −∂f /∂y
∂x
∂y
1
is positive, it follows that the surface represented by the graph of f is oriented upward.
Example 78.10: in this case the positive normal
vector
R cos(θ)
∂
∂
p×
p = R sin(θ)
∂θ
∂z
0
induces an orientation toward the outside of the
cylinder.
78.23 Using the parameterization
1 + cos(θ) sin(φ)
p(φ, θ) = 2 + sin(θ) sin(φ)
1 + cos(φ)
with
cos(θ) sin(φ)
∂
∂
p×
p = sin(φ) sin(θ) sin(φ) ,
∂φ
∂θ
cos(φ)
it follows that
∂
∂
p×
p
F(p(φ, θ)) ·
∂φ
∂θ
1 + cos(θ) sin(φ)
cos(θ) sin(φ)
· sin(θ) sin(φ)
0
= sin(φ)
0
cos(φ)
= cos(θ) sin2 (φ) + cos2 (θ) sin3 (φ),
and therefore,
Z
F · dA
S(p)
Z 2π
=
0
Z
π
F(p(φ, θ)) ·
0
∂
4π
∂
p×
p dφ dθ =
.
∂φ
∂θ
3
47
78.29 Example 78.5:
Z 2π Z R r 2 2
r h
A=
+ r2 dr dθ
R2
0
0
p
= πR h2 + R2 .
Example 78.10:
Z 2π Z
A=
0
h
R dz dθ = 2πRh.
0
78.31 Substituting the components of p into
the given equation for the plane yields
2r cos(θ) + r sin(θ) + r = 2,
or equivalently,
r=
2
.
2 cos(θ) + sin(θ) + 1
Consequently, a parameterization for the curve
of intersection is
cos(θ)
2
sin(θ)
r(θ) =
2 cos(θ) + sin(θ) + 1
1
for θ ∈ [0, 2π]. Since the angle between the
normal vector n = 2e1 +e2 +e3 and the positive
z-axis (i.e., the axis of symmetry
of the cone) is
√
easily seen to be arccos(1/ 6) ≈ 65.9◦ > 45◦ ,
the curve described by r must be a hyperbola.
78.33
surface
area
√ 3
paraboloid
π( 17 − 1)/6
spherical cap
2πRh
√
√
solid of revolution 2π( 2 + ln(1 + 2))
78.35 First row on the left: positive;
first row on the right: zero;
second row on the left: zero;
second row on the right: positive.
Chapter 79
79.3 div F(r) = 5krk2
79.7 The parameterization of...
r cos(θ)
...the upper annulus is p1 (r, θ) = r sin(θ)
2
for (r, θ) ∈ [1, 2] × [0, 2π].
r cos(θ)
...the lower annulus is p2 (θ, r) = r sin(θ)
0
48
Solutions
for (θ, r) ∈ [0, 2π] × [1, 2].
2 cos(θ)
...the outer cylinder is p3 (θ, z) = 2 sin(θ)
z
for (θ, z) ∈ [0, 2π] × [0, 2].
cos(θ)
...the inner cylinder is p4 (z, θ) = sin(θ)
z
for (z, θ) ∈ [0, 2] × [0, 2π].
The corresponding
normal vectors are
positive
0
∂
∂
p1 ×
p1 = 0,
∂r
∂θ
r
0
∂
∂
p2 ×
p 2 = 0 ,
∂θ
∂r
−r
2 cos(θ)
∂
∂
p3 ×
p3 = 2 sin(θ) ,
∂θ
∂z
0
− cos(θ)
∂
∂
p4 ×
p4 = − sin(θ) ,
∂z
∂θ
0
and therefore,
Z
Z 2 Z 2π r cos(θ)
0
r sin(θ) · 0 dθ dr
F · dA =
∂D
1
0
0
r
Z 2π Z 2 r cos(θ)
0
r sin(θ) · 0 dθ dr
+
0
1
0
−r
Z 2 Z 2π 2 cos(θ)
2 cos(θ)
2 sin(θ) · 2 sin(θ) dθ dz
+
0
0
z
0
Z 2π Z 2 cos(θ)
− cos(θ)
sin(θ) · − sin(θ) dz dθ
+
0
0
0
0
= 0 + 0 + 16π − 4π = 12π.
Since div F(r) is easily seen to be equal to 2
for all r, we find the same value as above by
integrating the divergence over D:
Z 2π Z 2 Z 2
Z
div F(r) dV =
2r dz dr dθ = 12π.
D
0
1
0
79.9 a) The surface of V consists of the
following six faces:
S1 = {0} × [0, 1] × [0, 1], S2 = {1} × [0, 1] × [0, 1],
S3 = [0, 1] × {0} × [0, 1], S4 = [0, 1] × {0} × [0, 1],
S5 = [0, 1] × [0, 1] × {0}, S6 = [0, 1] × [0, 1] × {1}.
On S1 , S2 , S3 , and S4 the positive normal
vector is parallel to either the x- or the y-axis.
Consequently, the dot product of F(r) with
the positive normal vector is zero on each of
these four faces because F(r) points in the
direction of the z-axis and is thus perpendicular
to the positive normal. Furthermore, on S5
we have z = 0 and, by implication, F(r) = 0
for all r ∈ S5 . In order to determine the
remaining flux integral over S6 we introduce
the parameterization
x
p(x, y) := y for (x, y) ∈ [0, 1] × [0, 1].
1
In computing the positive normal vector, we
find that ∂p/∂x × ∂p/∂y = e3 , and therefore,
Z
Z 1Z 1
F · dA =
dx dy = 1.
0
∂V
0
Since div F(r) = 1, it also follows that
Z
Z 1Z 1Z 1
div F(r) dV =
dx dy dz = 1
0
V
0
0
as predicted by the divergence
theorem.
2 cos(θ) sin(φ)
b) In setting p(φ, θ) := 2 sin(θ) sin(φ) for
2 cos(φ)
(φ, θ) ∈ [0, π] × [0, 2π], we find the positive normal vector to be ∂p/∂φ × ∂p/∂θ = 2 sin(φ)p,
and therefore,
Z 2π Z π
Z
F · dA =
2kpk2 p · p sin(φ) dφ dθ
∂V (p)
=
Z
0
2π
0
Z
0
π
32 sin(φ) dφ dθ = 128π.
0
Using the result of Exercise 79.3, this value is
also found for the integral of the divergence:
Z
Z
div F(r) dV =
5krk2 dV
V
=
Z
2π
0
Z
π
0
V
Z
2
5ρ4 sin(φ) dρ dφ dθ = 128π.
0
c) In this case the divergence theorem does not
apply, because F is not defined on the entire
region V . Note: kF(x, y, z)k tends to ∞ as x
approaches 2.
79.11 a) div F is a scalar, while G is a vector.
b) The divergence operator cannot be applied
to multivariable functions, only to vector fields.
79.13 Setting F(r) := v, we may infer that
div F(r) = 0, and therefore, Exercise 79.12 implies that div(f v) = v · ∇f .
Solutions
79.15 Setting f (x, y, z) := (x2 + y 2 + z 2 )/2 and
g(x, y, z) := (x3 + y 3 + z 3 )/3, it follows that
F = ∇f and G = ∇g. Consequently, the result
of Exercise 79.14 implies that
Z
(F × G) · dA
∂D(p)
Z
=
div(∇f × ∇g) dV = 0.
D
79.17 Using the results of Exercises 79.10 and
79.16, it follows that
div(f ∇f ) = ∇f · ∇f + f div ∇f
2
∂2
∂2
∂
= k∇f k2 + f
f
+
f
+
f
∂x2
∂y 2
∂z 2
= k∇f k2 .
Chapter 80
1−x
80.3 curl F = y − 1
0
80.5 Given the parameterization
r cos(θ)
p(r, θ) := r sin(θ)
2−r
for (r, θ) ∈ [1, 2] × [0, 2π], we find that
r cos(θ)
∂
∂
p×
p = r sin(θ) ,
∂r
∂θ
r
and therefore,
Z
curl F · dA
S(p)
=
Z
=
Z
2π
Z
0
2
1
2π Z
0
1
r cos(θ)
1 · r sin(θ) dr dθ
1
r
2
(r cos(θ) + r sin(θ) + r) dr dθ = 3π.
1
The same value we also obtain from the sum
of the path integrals of F along the oriented
boundary curves C1 and C2 :
Z
Z
F · dc +
F · dc
C1
C2
49
0
−2 sin(t)
2 cos(t) · 2 cos(t) dt
=
0
2 sin(t)
0
Z 2π
1
− sin(t)
cos(t) · − cos(t) dt
+
0
− sin(t)
0
Z 2π
Z 2π
4 cos2 (t) dt +
(− sin(t) − cos2 (t)) dt
=
Z
2π
0
0
= 4π − π = 3π.
80.9 a) F is not a gradient field because
curl F = 2ze1 − 2xe2 − 2ye3 6= 0.
b) F is a gradient field because curl F = 0.
c) F is not a gradient field because curl F =
2ze1 − 2xe2 − 2ye3 6= 0.
80.11 Using the parameterization
2 cos(θ)
p(θ, z) := 2 sin(θ)
z
for (θ, z) ∈ [0, π] × [0, 2], the positive normal
vector is
2 cos(θ)
∂
∂
p×
p = 2 sin(θ) .
∂θ
∂z
0
Since curl F = 2ye1 + 2ze2 + 2xe3 , we find that
Z
curl F · dA
S(p)
Z π
=
Z
0
=8
2
(8 sin(θ) cos(θ) + 4z sin(θ)) dz dθ
0
Z
π
sin(θ) dθ = 16.
0
The boundary curve of the semicylinder described by p consists of four segments C1 , C2 ,
C3 , and C4 that may be described by the following parameterizations:
2 cos(t)
r1 (t) := 2 sin(t) for t ∈ [0, π],
0
−2
r2 (t) := 0 for t ∈ [0, 2],
t
−2 cos(t)
r3 (t) := 2 sin(t) for t ∈ [0, π],
2
2
r4 (t) := 0 for t ∈ [0, 2].
2−t
50
Solutions
The corresponding path integral is
Z
Z
Z
Z
F · dc +
F · dc +
F · dc +
F · dc
C1
C2
C3
C4
Z π
0
−2 sin(t)
4 cos2 (t) · 2 cos(t) dt
=
0
0
4 sin2 (t)
2
Z 2 t
0
4 · 0 dt
+
0
0
1
Z π
4
2 sin(t)
4 cos2 (t) · 2 cos(t) dt
+
0
0
4 sin2 (t)
Z 2 (2 − t)2
0
4 · 0 dt
+
0
0
−1
Z π
8 cos3 (t) dt + 0
=
0
Z π
+
(8 sin(t) + 8 cos3 (t)) dt + 0 = 16,
A.2 sin(2α) = sin(α + α) = sin(α) cos(α) +
cos(α) sin(α) = 2 sin(α) cos(α), cos(2α) =
cos(α + α) = cos(α) cos(α) − sin(α) sin(α) =
cos2 (α) − sin2 (α), and the alternative representations of the double angle formula for cosine
are easily deduced from the trigonometric theorem of Pythagoras. Furthermore,
tan(2α) =
2 tan(α)
tan(α) + tan(α)
=
.
1 − tan(α) tan(α)
1 − tan2 (α)
A.3 Considering the diagram in Figure S.15, it
follows that
c2 = h2 + (b cos(γ) − a)2
= b2 sin2 (α) + b2 cos2 (γ) − 2ab cos(γ) + a2
= a2 + b2 − 2ab cos(γ),
as desired. With regard to the law of sine, the
0
α
as desired.
80.13 Using Stokes’ theorem and the result of
Exercise 80.12b, we may infer that
Z
(f ∇g + g∇f ) · dc
C
Z
=
curl(f ∇g + g∇f ) · dA
S
Z
= (∇f × ∇g + ∇g × ∇f ) · dA
ZS
=
0 · dA = 0.
b
c
h
β
γ
a
Figure S.15: obtuse triangle.
same diagram also allows us to infer that
h/b
h/c
sin(π − β)
sin(β)
sin(γ)
=
=
=
=
,
c
c
b
b
b
S
80.15 Using the first of Maxwell’s equations, we
may infer that
curl E = −
∂
∂
B = − (B0 + tB1 ) = −B1
∂t
∂t
is independent of t.
and similarly, it can be shown that sin(α)/a =
sin(β)/b.
A.4 r2 = 3r2 sin(θ) cos(θ)
√
A.6 (0, 2): r = 2, θ = π/2,
√ (1, 4): r = 17, θ =
−3):
arctan(4), (−2, 2): r = 8, θ = 3π/4, (4,√
r = 5, θ = 2π−arctan(3/4), (−2, −4): r = 20,
θ = π + arctan(2)
Appendix A
A.1
cot(α) + cot(β)
1
=
cot(α + β)
cot(α) cot(β) − 1
tan(α) tan(β) cot(α) + cot(β)
=
·
tan(α) tan(β) cot(α) cot(β) − 1
tan(α) + tan(β)
=
1 − tan(α) tan(β)
tan(α + β) =
Appendix B
r s
B.1
and s = q/n, then
√ a a =
√ a) If r = p/mmn
√
mn
mn
rmn
smn
pn
qm
pn+qm
a
=
a a
=
a
=
√a pn+qm
mn
a
= ar+s . The second identity concerning the quotient of ar and as immediately
follows from the first by observing that 1/as =
a−s .
b) 82/3 = 4 and 16−3/4 = 1/8.
Solutions
Appendix C
51
x2
y2
+√
=1
13 + 7
13 − 2
C.2 If we set u := x − x0 and v := y − y0 , then
the given equation is equivalent to the equation
u2 /a2 + v 2 /b2 = 1 which in turn describes an
ellipse in standard position in a uv-coordinate
system centered at (x0 , y0 ).
C.1 √
=
=
=
=
C.3 y 2 = 4x/5
C.4 Given the equation in (C.2), we may infer
that
p
p
(x − c)2 + y 2 = (x + c)2 + y 2 ± 2a,
and therefore,
p
(x − c)2 = (x + c)2 ± 4a (x + c)2 + y 2 + 4a2 .
Rearranging the terms and squaring both sides
yields
16a2 ((x + c)2 + y 2 ) = (4a2 + 4xc)2 ,
or equivalently,
(a2 − c2 )x2 + a2 y 2 = a2 (a2 − c2 ).
Thus,
x2
y2
−
= 1.
a2
c2 − a2
C.5
2x2
2y 2
√
−√
=1
11 − 117
117 − 9
Appendix D
√
D.1 The roots are at x = (3 ± 89)/8, and the
global minimum is at (3/8, −19/2).
√ !
√ !
3 − 89
3 + 89
x−
D.2 f (x) = 4 x −
8
8
Appendix E
E.2
P1 For2 n = 1 the equation is valid because
k=1 k = 1 = 1·2·3/6. Assuming the equation
to be valid for a given n, we find that
n+1
X
k=1
k 2 = (n + 1)2 +
n(n + 1)(2n + 1)
6
(n + 1)(6n + 6 + n(2n + 1))
6
(n + 1)(2n2 + 7n + 6)
6
(n + 1)(n + 2)(2n + 3)
6
(n + 1)((n + 1) + 1)(2(n + 1) + 1)
,
6
= (n + 1)2 +
n
X
k=1
k2
as desired.
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