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Solutions Manual for Basic Engineering Mathematics
C. K. Chan, C. W. Chan and K. F. Hung
August 21, 2008
Contents
1 Complex Numbers 1
2 Linear Algebra 13
3 Infinite series, Power series and Fourier series 41
4 Partial Differentiation 53
5 Multiple Integrals 83
6 Vector Calculus 125
7 Differential Equations 157
8 Laplace and Fourier Transformations 177
9 Partial Differential Equations 191
i
Chapter 1
Complex Numbers
(1) z = 2− 3i, |z| =√
22 + 32 =√
13.
(2) By definition,1
z=
(
a
a2 + b2
)
+ i
( −b
a2 + b2
)
. Therefore,
z · 1z
= (a + ib)
(
a
a2 + b2+ i
−b
a2 + b2
)
= 1 + 0i = 1.
(3) (a)3 + i
2− i=
3 + i
2− i× 2 + i
2 + i=
(6− 1) + i(3 + 2)
22 + 12=
5 + 5i
5= 1 + i.
(b)2 +√
3i
1− 4i=
2 +√
3i
1− 4i× 1 + 4i
1 + 4i==
(
2− 4√
3)
+(
8 +√
3)
i
17
(c)1− i
1 +√
3i=
1− i
1 +√
3i× 1−
√3i
1−√
3i=
(
1−√
3)
−(
1 +√
3)
i
4
(4) Let z = a + ib and w = c + id. We have (z) = (a− ib) = a + ib = z. Since (z + w) =
(a + c) + i(b + d) = a+c−i(b+d) and z+w = (a−ib)+(c−id), we have (z + w) = z+w.
Moreover, observe that
z · w = (ac− bd) + i(ad + bc) = (ac− bd)− i(ad + bc)
and
z · w = (a− ib)(c− id) = (ac− bd) + i(−bc− ad).
We thus conclude that z · w = z · w. Finally, the last result follows from the identity
|z · w|2 = (z · w)(z · w) = z · w · z · w = zz · ww = |z|2 |w|2 .
(5) If z = a+ ib, then z + z = (a+ ib)+(a− ib) = 2a = 2 Re z and z− z = (a+ ib)− (a− ib) =
1
CHAPTER 1. COMPLEX NUMBERS
2ib = 2i Im z. Since |Re z|2 = a2 ≤ a2 + b2 = |z|2, we may take square root to obtain
|Re z| ≤ |z|.
(6) |z + w|2 = (z+w)(z + w) = (z+w)(z+w) = z z+w w+w z+z w = |z|2+|w|2+w z+w z =
|z|2 + |w|2 + 2 Re(w z). Since the earlier results together imply Re(w z) ≤ |Re(w z)| ≤|w z| = |w| |z| = |w| |z|, we obtain
|z + w|2 ≤ |z|2 + |w|2 + 2 |z| |w| = (|z|+ |w|)2.
The triangle inequality thus follows from taking square root of the above inequality.
(7) (a) Distance between(
12 , 1)
and the center =
√
(
12 − 0
)2+ (1− 1)2 = 1
2
Thus, 12 + i lies inside the circle.
(b) Distance between(
1, i2
)
and the center =
√
(1− 0)2 +(
12 − 1
)2=
√5
2 > 1
Thus, 1 + i2 lies outside the circle.
(c) Distance between(
12 ,
√2
2 i)
and the center =
√
(
12 − 0
)2+(√
22 − 1
)2= 0.57947
Thus, 12 +
√2
2 i lies inside the circle.
(d) Distance between(
−12 ,√
3i)
and the center =
√
(
−12 − 0
)2+(√
3− 1)2
= 0.88651
Thus, −12 +√
3i lies inside the circle.
(8) For z = cos θ + i sin θ, we have z · z = cos2 θ + sin2 θ = 1. Therefore,
1
z= z = cos θ − i sin θ. Therefore, z +
1
z= z + z = 2 cos θ. Furthermore, DeMoivre’s
Theorem implies z2 = cos 2θ + i sin 2θ and therefore z2 +1
z2= z2 + (z)2 = 2 cos 2θ.
(9) (a) (3+4i)2−2(x− iy) = x+ iy ⇒ −7+24i = 3x− iy. Therefore, x = −73 and y = −24.
(b) Note that
(
1 + i
1− i
)2
= −1 and1
x + iy=
x− iy
x2 + y2. Therefore,
(
1 + i
1− i
)2
+1
x + iy= 1 + i⇒ −1 +
(
x
x2 + y2
)
− i
(
y
x2 + y2
)
= 1 + i.
Therefore, we havex
x2 + y2= 2 and
y
x2 + y2= −1. Solving these equations and
noting that x and y can not be both equal to zero, we conclude that x = 25 and
y = −15 .
(c) (3 − 2i)(x + iy) = 2(x − 2iy) + 2i − 1 ⇒ x + 2y = −1 and −2x + 7y = 2. Solving
these equations, one obtains x = −1 and y = 0.
2
CHAPTER 1. COMPLEX NUMBERS
(10) Since√
3 + i = 2(
cos π6 + i sin π
6
)
and 1− i =√
2(
cos 7π4 + i sin 7π
4
)
, we conclude that
(√
3 + i)4
(1− i)3=
24(
cos 4π6 + i sin 4π
6
)
(√2)3 (
cos 21π4 + i sin 21π
4
)
= 4√
2
(
cos55π
12− i sin
55π
12
)
= 4√
2
(
cos7π
12− i sin
7π
12
)
.
(11) (a) Since−1− i
√3
2= cos 4π
3 + i sin 4π3 , one concludes from DeMoivre’s Theorem that
(
−1− i√
3
2
)3
=
(
cos4π
3+ i sin
4π
3
)3
= cos12π
3+ i sin
12π
3= 1.
(b)√
3− i = 2(
cos 11π6 + i sin 11π
6
)
1 + i =√
2(
cos π4 + i sin π
4
)
(√3− i
)4(1 + i)10
=
[
2
(
cos11π
6+ i sin
11π
6
)]4[√
2(
cosπ
4+ i sin
π
4
)]10
= 29
[
cos
(
4× 11π
6+ 10× π
4
)
+ i sin
(
4× 11π
6+ 10× π
4
)]
= 29
(
cos59π
6+ i sin
59π
6
)
= 29
(
cos11π
6+ i sin
11π
6
)
= 256√
3− 256i
(c) 1−√
3i = 2(
cos 5π3 + i sin 5π
3
)
1 +√
3i = 2(
cos π3 + i sin π
3
)
(
1−√
3i)4 (
1 +√
3i)8
=
[
2
(
cos5π
3+ i sin
5π
3
)]4[
2(
cosπ
3+ i sin
π
3
)]8
= 212
[
cos
(
4× 5π
3+ 8× π
3
)
+ i sin
(
4× 5π
3+ 8× π
3
)]
= 212
(
cos28π
3+ i sin
28π
3
)
= 212
(
cos4π
3+ i sin
4π
3
)
= −2048− 2048√
3i
(12) Since −32 = 32 (cos π + i sinπ) and 8i = 8(
cos π2 + i sin π
2
)
, the 5th roots of −32 are given
3
CHAPTER 1. COMPLEX NUMBERS
by
zk = 2
[
cos(2k + 1
5π) + i sin(
2k + 1
5π)
]
, k = 0, 1, 2, 3, 4;
while the cubic roots of 8i are given by
zk = 2
[
cos(4k + 1
6π) + i sin(
4k + 1
6π)
]
, k = 0, 1, 2.
(13) (a)
(
−8 + 8√
3i)1/4
=
[
16
(
−1
2+
√3
2i
)]1/4
=
16
[
cos
(
2kπ +2π
3
)
+ i sin
(
2kπ +2π
3
)]1/4
= 2
[
cos
(
kπ
2+
π
6
)
+ i sin
(
kπ
2+
π
6
)]
, where k = 0, 1, 2 and 3
z0 = 2(
cos π6 + i sin π
6
)
=√
3 + i
z1 = 2(
cos 2π3 + i sin 2π
3
)
= −1 +√
3i
z2 = 2(
cos 7π6 + i sin 7π
6
)
= −√
3− i
z3 = 2(
cos 5π3 + i sin 5π
3
)
= 1−√
3i
(b)
(
−32 + 32√
3i)1/6
=
[
64
(
−1
2+
√3
2i
)]1/6
=
64
[
cos
(
2kπ +2π
3
)
+ i sin
(
2kπ +2π
3
)]1/6
= 2
[
cos
(
kπ
3+
π
9
)
+ i sin
(
kπ
3+
π
9
)]
, where k = 0, 1, ..., 5
z0 = 2(
cos π9 + i sin π
9
)
= 1.8794 + 0.68404i
z1 = 2(
cos 4π9 + i sin 4π
9
)
= 0.34730 + 1.9696i
z2 = 2(
cos 7π9 + i sin 7π
9
)
= −1.5321 + 1.2856i
z3 = 2(
cos 10π9 + i sin 10π
9
)
= −1.8794− 0.68404i
z4 = 2(
cos 13π9 + i sin 13π
9
)
= −0.34730− 1.9696i
4
CHAPTER 1. COMPLEX NUMBERS
z5 = 2(
cos 16π9 + i sin 16π
9
)
= 1.5321− 1.2856i
(14) If z = x + iy, thenz
z=
z × z
z × z=
(z)2
|z|2.
If z → 0 along the real axis, one has z = x + i0 and thereforez
z→ 1. On the other hand,
if z → 0 along the imaginary axis, then z = 0 + iy andz
z→ −1. We thus conclude that
limz→0
z
zdoes not exist. When z0 6= 0, lim
z→z0
z
z=
z0
z0.
(15) (a) From DeMoivre’s theorem, we have
(cos x + i sinx)5 = cos 5x + i sin 5x.
On the other hand, by binomial theorem, we also have
(cos x + i sinx)5 =5∑
k=0
(
5
k
)
cosk x (i sinx)5−k
Therefore, comparing the real and imaginary parts yields
cos 5x = cos5 x− 10 cos3 x sin2 x + 5 cos x sin4 x,
sin 5x = sin5 x− 10 cos2 x sin3 x + 5 cos4 x sinx
(b)
cos x cos 5x = cos6 x− 10 cos4 x sin2 x + 5 cos2 x sin4 x
sinx sin 5x = sin6 x− 10 cos2 x sin4 x + 5 cos4 x sin2 x
Adding these equations yields
cos x cos 5x + sin x sin 5x = sin6 x + cos6 x− 5(cos2 x sin4 x + cos4 x sin2 x)
cos 4x = sin6 x + cos6 x− 5 cos2 x sin2 x
Therefore,
sin6 x + cos6 x = cos 4x + 5 sin2 x cos2 x = cos 4x +5
4sin2 2x
5
CHAPTER 1. COMPLEX NUMBERS
Since cos 4x = 1− 2 sin2 2x, we have
sin6 x + cos6 x = cos 4x +5
8(1− cos 4x) =
3
8cos 4x +
5
8=
3 cos 4x + 5
8
(16) Using the identityz
z=
z2
|z|2for any non-zero complex no. z, we have
1 + sin x + i cos x
1 + sin x− i cos x=
(1 + sin x + i cos x)2
|1 + sin x + i cos x|2
=(1 + sin x)2 − cos2 x + 2i cos x(1 + sinx)
(1 + sin x)2 + cos2 x
=
(
1 + 2 sinx + sin2 x− cos2 x)
+ 2i cos x(1 + sinx)
2 (1 + sin x)
= sinx + i cos x
= cos(π
2− x) + i sin(
π
2− x)
= ei(π2−x).
DeMoivre’s Theorem now gives
(
1 + sin x + i cos x
1 + sin x− i cos x
)n
=[
ei(π2−x)]n
= ein(π2−x)
= cos n(π
2− x) + i sinn(
π
2− x).
(17) If the given number is a + bi, we are looking for a number x + yi such that
(x + yi)2 = a + bi.
This is equivalent to the system of equations
x2 − y2 = a and 2xy = b.
From these equations we obtain
(x2 + y2)2 = (x2 − y2)2 + 4x2y2 = a2 + b2.
Hence we must have
x2 + y2 =√
a2 + b2,
6
CHAPTER 1. COMPLEX NUMBERS
where the square root is positive or zero. Together with the equation x2− y2 = a we find
x2 =a +√
a2 + b2
2and y2 =
−a +√
a2 + b2
2.
These equation yield, in general, two opposite value for x and two for y. But these values
cannot be combined arbitrarily, for the equation 2xy = b is not a consequence of these
equation. We must therefore be careful to select x and y so that their product has the
sign of b. This leads to the general equation
√a + bi = ±
(
√
r + a
2+ i(sgn b)
√
r − a
2
)
where r =√
a2 + b2.
(18) (a) Let w = z2. Then the given equation becomes w2 − 2w + 4 = 0. The quadratic
formula implies
z2 = w =2±√
4− 16
2= 1±
√3i.
Case z2 = 1 +√
3i: In polar form, 1 +√
3i = 2(
cos π3 + i sin π
3
)
. Therefore,
z =√
2
(
cos
( π3 + 2kπ
2
)
+ i sin
( π3 + 2kπ
2
))
, k = 0, 1.
Equivalently, the square root formula with r = 2 and sgn b = 1 implies
z = ±(
√
2 + 1
2+ i
√
2− 1
2
)
= ±(
√
3
2+ i
√
1
2
)
.
Case z2 = 1−√
3i: In polar form, 1−√
3i = 2(
cos −π3 + i sin −π
3
)
. Therefore,
z =√
2
(
cos
(−π3 + 2kπ
2
)
+ i sin
(−π3 + 2kπ
2
))
, k = 0, 1.
Equivalently, the square root formula with r = 2 and sgn b = −1 implies
z = ±(
√
2 + 1
2− i
√
2− 1
2
)
= ±(
√
3
2− i
√
1
2
)
.
(b) Let w = z3. Then the equation becomes w2 + 2w + 2 = 0. Solving the equation
yields
z3 = w =−2±
√4− 8
2= −1± i.
7
CHAPTER 1. COMPLEX NUMBERS
Case z3 = −1 + i: In polar form, −1 + i =√
2(
cos 3π4 + i sin 3π
4
)
. Therefore,
z = 21/6
(
cos
(
3π4 + 2kπ
3
)
+ i sin
(
3π4 + 2kπ
3
))
, k = 0, 1, 2.
Case z3 = −1− i: In polar form, −1− i =√
2(
cos −3π4 + i sin −3π
4
)
. Therefore,
z = 21/6
(
cos
(
−3π4 + 2kπ
3
)
+ i sin
(
−3π4 + 2kπ
3
))
, k = 0, 1, 2.
(c) Let w = z2. Then the given equation becomes w2 + 4w + 16 = 0. The quadratic
formula implies
z2 = w =−4±
√16− 64
2= −2± 2
√3i.
Case z2 = −2 + 2√
3i: In polar form, 1 +√
3i = 4(
cos 2π3 + i sin 2π
3
)
. Therefore,
z = 2
(
cos
(
2π3 + 2kπ
2
)
+ i sin
(
2π3 + 2kπ
2
))
, k = 0, 1.
Equivalently, the square root formula with r = 4 and sgn b = 1 implies
z = ±(
√
4− 2
2+ i
√
4 + 2
2
)
= ±(
1 + i√
3)
.
Case z2 = −2− 2√
3i: In polar form, −2− 2√
3i = 4(
cos −2π3 + i sin −2π
3
)
. There-
fore,
z = 2
(
cos
(
−2π3 + 2kπ
2
)
+ i sin
(
−2π3 + 2kπ
2
))
, k = 0, 1.
Equivalently, the square root formula with r = 4 and sgn b = −1 implies
z = ±(
√
4− 2
2− i
√
4 + 2
2
)
= ±(
1− i√
3)
.
(d) z4 = −1 = cos π + i sinπ. Therefore,
z = cos
(
π + 2kπ
4
)
+ i sin
(
π + 2kπ
4
)
, k = 0, 1, 2, 3.
(19) (a) ei π2 = cos π
2 + i sin π2 = i
8
CHAPTER 1. COMPLEX NUMBERS
(b) 4e−i π2 = 4
[
cos(
−π2
)
+ i sin(
−π2
)]
= −4i
(c) 8ei 7π3 = 8
(
cos 7π3 + i sin 7π
3
)
= 4 + 4√
3i
(d) 2e−i 3π4 = 2
[
cos(
−3π4
)
+ i sin(
−3π4
)]
= −√
2−√
2i
(e) 6ei 2π3 eiπ = 6ei 5π
3 = 6(
cos 5π3 + i sin 5π
3
)
= 3− 3√
3i
(f) ei π4 e−iπ = e−i 3π
4 =[
cos(
−3π4
)
+ i sin(
−3π4
)]
= −√
22 −
√2
2 i
(20) Let z = x + iy and define ez = ex (cos y + i sin y).
(a) If z = x + 0i, ez = ex (cos 0 + i sin 0) = ex
(b) |ez| =√
(ex cos y)2 + (ex sin y)2 = ex. In particular∣
∣eiy∣
∣ = e0 = 1.
(c) ez1+z2 = e(x1+x2)+i(y1+y2) = ex1+x2 (cos (y1 + y2) + i sin (y1 + y2)) = ez1ez2 .
(d) |ez| = ex 6= 0 for any z ∈ C. Therefore, ez 6= 0. Furthermore,
1
ez=
1
ex (cos y + i sin y)= e−x (cos y − i sin y) = e−x+i(−y) = e−z.
(e) Repeat application of (c) and (d), or by induction, yields enz = (ez)n for any integer
n.
(f)
d
dteiwt =
d
dtcos wt + i
d
dtsinwt = −w sinwt + iw cos wt = iw (cos wt + i sinwt)
= iweiwt.
(21) Since ei2πkt = cos (2πkt) + i sin (2πkt), the cosine function is an even function and the
sine function is an odd function, we have
n∑
k=−n
ei2πkt =n∑
k=−n
cos (2πkt) + in∑
k=−n
sin (2πkt) = 1 + 2n∑
k=1
cos (2πkt)
Recall that
1 + z + z2 + · · ·+ zn =1− zn+1
1− z.
9
CHAPTER 1. COMPLEX NUMBERS
Then
1 +n∑
k=1
(
eiθ)k
=1−
(
eiθ)n+1
1− (eiθ)=
1−(
ei(n+1)θ)
1− (eiθ)
=1− cos(n + 1)θ − i sin(n + 1)θ
1− cos θ − i sin θ
=1− cos(n + 1)θ − i sin(n + 1)θ
(1− cos θ)2 + sin2 θ
=[1− cos(n + 1)θ − i sin(n + 1)θ] [1− cos θ − i sin θ]
2(1− cos θ)
Comparing the real part of this equation yields
1 + cos θ + cos 2θ + · · ·+ cosnθ =(1− cos(n + 1)θ)(1− cos θ) + sin(n + 1)θ sin θ
4 sin2(θ/2).
Simplifying the right hand side of the equation, we have
(1− cos(n + 1)θ)(1− cos θ) + sin(n + 1)θ sin θ
4 sin2(θ/2)
=(1− cos(n + 1)θ)(2 sin2(θ/2)) + sin(n + 1)θ(2 sin(θ/2) cos(θ/2))
4 sin2(θ/2)
=2 sin2(θ/2)− 2 sin(θ/2)[cos(n + 1)θ sin(θ/2)− sin(n + 1)θ cos(θ/2)]
4 sin2(θ/2)
=1
2+
sin[(2n + 1)θ/2]
2 sin(θ/2)
Hencen∑
k=1
cos (kθ) = −1
2+
sin[(2n + 1)θ/2]
2 sin(θ/2).
Finally, by letting θ = 2πt, we have
n∑
k=−n
ei2πkt = 1 + 2n∑
k=1
cos (2πkt) = 1 + 2
[
−1
2+
sin[(2n + 1)πt]
2 sinπt
]
=sin[(2n + 1)πt]
sinπt.
Integrating this result yields
∫ 1
0
sin[(2n + 1)πt]
sinπtdt =
∫ 1
0
[
1 + 2n∑
k=1
cos (2πkt)
]
dt = 1 + 2n∑
k=1
∫ 1
0cos (2πkt) dt = 1.
(22) Suppose w is a root of the polynomial P (z) = a0 +a1z +a2z2 + · · ·+anzn, i.e., P (w) = 0.
10
CHAPTER 1. COMPLEX NUMBERS
Then
P (w) = a0 + a1w + a2w2 + · · ·+ anwn = a0 + a1w + a2w2 + · · ·+ anwn
= a0 + a1w + a2w2 + · · ·+ anwn = P (w) = 0.
(23) Let z = x + yi, then
∣
∣
∣
∣
z − 3
z + 3
∣
∣
∣
∣
= 2
=⇒ |z − 3| = 2 |z + 3|=⇒ |(x− 3) + yi| = 2 |(x + 3) + yi|
=⇒√
(x− 3)2 + y2 = 2
√
(x + 3)2 + y2
=⇒ (x− 3)2 + y2 = 4[
(x + 3)2 + y2]
=⇒ x2 + 10x + y2 + 9 = 0
=⇒ (x + 5)2 + y2 = 16
(24) Let z = x + yi, then
|z + 3|+ |z − 3| = 10
=⇒ |(x + 3) + yi|+ |(x− 3) + yi| = 10
=⇒√
(x + 3)2 + y2 +
√
(x− 3)2 + y2 = 10
=⇒ (x + 3)2 + y2 + 2
√
(x + 3)2 + y2
√
(x− 3)2 + y2 + (x− 3)2 + y2 = 100
=⇒[
(x + 3)2 + y2] [
(x− 3)2 + y2]
=(
41− x2 − y2)2
=⇒ 64x2 + 100y2 = 1600
=⇒ x2
25+
y2
16= 1
(25) If z = a + bi and |z| = 1, then a2 + b2 = 1.
w =z − 1
z + 1=
(a + bi)− 1
(a + bi) + 1=
(a− 1) + bi
(a + 1) + bi=
(a− 1) + bi
(a + 1) + bi
(a + 1)− bi
(a + 1)− bi
=(a− 1) (a + 1) + b2 + 2bi
(a + 1)2 + b2=
(
a2 + b2)
− 1 + 2bi
(a + 1)2 + b2
=(1)− 1 + 2bi
(a + 1)2 + b2=
2bi
(a + 1)2 + b2
Therefore, w is a purely imaginary number.
11
Chapter 2
Linear Algebra
(1) (a) A =
0 −1 −2
1 0 −1
2 1 0
(b) A =
1 12
13
2 1 23
3 32 1
(c) A =
0 −1 −212 0 −1
223
13 0
(d) A =
e e2 e3
e2 e4 e6
e3 e6 e9
(2) The entry at the i-th row and the j-th column of BTAT is given by the product of i-th
row of BT =[
b1i b2i · · · bni
]
and the j-th column of AT =[
aj1 aj2 · · · ajn
]Twhich
is equal to∑
bki ajk. On the other hand, the entry at the i-th row and the j-th column
of (AB)T is the same as the the entry at the j-th row and i-th column of AB, and is
therefore given by∑
ajk bki.
If A−1 exists, the above result ⇒ AT(
A−1)T
=(
A−1A)T
= IT = I.
(3) (a)
BT =[
A(
ATA)−1
AT]T
=(
AT)T[
(
ATA)−1]T
AT = A[
(
ATA)T]−1
AT
= A(
ATA)−1
AT = B
13
CHAPTER 2. LINEAR ALGEBRA
(b)
B2 =[
A(
ATA)−1
AT]2
=[
A(
ATA)−1
AT] [
A(
ATA)−1
AT]
= A(
ATA)−1 (
ATA) (
ATA)−1
AT = A(
ATA)−1
AT = B
(4) (a) Let A =
[
1 0
1 1
]
and B =
[
1 1
0 1
]
.
(A + B)2 =
([
1 0
1 1
]
+
[
1 1
0 1
])2
=
[
5 4
4 5
]
A2 + 2AB + B2 =
[
1 0
1 1
]2
+ 2
[
1 0
1 1
][
1 1
0 1
]
+
[
1 1
0 1
]2
=
[
4 4
4 6
]
(b) Let A =
[
1 0
1 1
]
and B =
[
1 1
0 1
]
.
(A + B) (A−B) =
([
1 0
1 1
]
+
[
1 1
0 1
])([
1 0
1 1
]
−[
1 1
0 1
])
=
[
1 −2
2 −1
]
A2 −B2 =
[
1 0
1 1
]2
−[
1 1
0 1
]2
=
[
0 −2
2 0
]
(c) Let C =
[
0 1
0 0
]
6= 0.
C2 =
[
0 1
0 0
][
0 1
0 0
]
=
[
0 0
0 0
]
(5) (a)
2 6 −4 10
4 1 3 −2
1 −3 4 −7
r1 ←→ r3−−−−−−→
1 −3 4 −7
4 1 3 −2
2 6 −4 10
(b)
1 −3 4 −7
4 1 3 −2
2 6 −4 10
−4r1 + r2 −→ r2−−−−−−−−−−−−→
1 −3 4 −7
0 13 −13 26
2 6 −4 10
(c)
1 −3 4 −7
0 13 −13 26
2 6 −4 10
12r3 −→ r3−−−−−−−→
1 −3 4 −7
0 13 −13 26
1 3 −2 5
(6)
2x1 − x2 + 3x3 − 2x4 = 1 · · · · · · (1)
x2 − 2x3 + 3x4 = 2 · · · · · · (2)
4x3 + 3x4 = 3 · · · · · · (3)
4x4 = 4 · · · · · · (4)
14
CHAPTER 2. LINEAR ALGEBRA
From (4), x4 = 1.
From (3), 4x3 + 3 (1) = 3 =⇒ x3 = 0.
From (2), x2 − 2 (0) + 3 (1) = 2 =⇒ x2 = −1.
From (1), 2x1 − (−1) + 3 (0)− 2 (1) = 1 =⇒ x1 = 1.
(7)
1 2 1 3
3 −1 −3 −1
2 3 1 4
−3r1 + r2 −→ r2−−−−−−−−−−−−→
1 2 1 3
0 −7 −6 −10
2 3 1 4
−2r1 + r3 −→ r3−−−−−−−−−−−−→
1 2 1 3
0 −7 −6 −10
0 −1 −1 −2
r2 ←→ r3−−−−−−→
1 2 1 3
0 −1 −1 −2
0 −7 −6 −10
−7r2 + r3 −→ r3−−−−−−−−−−−−→
1 2 1 3
0 −1 −1 −2
0 0 1 4
−r2 −→ r2−−−−−−−→
1 2 1 3
0 1 1 2
0 0 1 4
−r3 + r2 −→ r2−−−−−−−−−−−→
1 2 1 3
0 1 0 −2
0 0 1 4
−r3 + r1 −→ r1−−−−−−−−−−−→
1 2 0 −1
0 1 0 −2
0 0 1 4
−2r2 + r1 −→ r1−−−−−−−−−−−−→
1 0 0 3
0 1 0 −2
0 0 1 4
=⇒ x =
x1
x2
x3
=
3
−2
4
(8)
A→
0 1 2
0 −2 −4
1 1 1
0 2 4
→
0 1 2
0 0 0
1 1 1
0 0 0
→
1 0 −1
0 1 2
0 0 0
0 0 0
= R.
Therefore, the augmented matrix [A |0 ] of Ax = 0 is reduced to [R |0 ]. Rx = 0 is of
the form
x1 −x3 = 0
x2 + 2x3 = 0, giving x1 = x3 and x2 = −2x3. Therefore, we have
x =[
t −2t t]T
where t is an arbitrary scalar as the solution.
(9) (a) With A =
1 −2 1
2 1 8
1 −12 −11
and b =
1
3
−1
, we may reduce the augmented matrix
15
CHAPTER 2. LINEAR ALGEBRA
of Ax = b as follows:
[A |b ] =
1 −2 1
2 1 8
1 −12 −11
∣
∣
∣
∣
∣
∣
∣
∣
1
3
−1
→
1 −2 1
0 5 6
0 −10 −12
∣
∣
∣
∣
∣
∣
∣
∣
1
1
−2
→
1 −2 1
0 1 65
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
115
0
.
We thus have
x1 −2x2 + x3 = 1
x2 + 65x3 = 1
5
, giving x2 = 15 − 6
5x3 and x1 = 1 + 2x2− x3.
Putting x3 = t, one obtains x2 = 15 − 6
5 t and x1 = 1+2(
15 − 6
5 t)
− t = 75 − 17
5 t. (b) With
A =
3 −2 2
2 1 −3
1 −10 18
and b =
4
5
−8
,
[A |b ] =
3 −2 2
2 1 −3
1 −10 18
∣
∣
∣
∣
∣
∣
∣
∣
4
5
−8
→
0 28 −52
0 21 −39
1 −10 18
∣
∣
∣
∣
∣
∣
∣
∣
28
21
−8
→
0 28 −52
0 1 −137
1 −10 18
∣
∣
∣
∣
∣
∣
∣
∣
28
1
−8
→
1 −10 18
0 1 −137
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
−8
1
0
→
1 0 −47
0 1 −137
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
2
1
0
.
We thus obtain x =[
2 + 47 t 1 + 13
7 t t]T
, where t is an arbitrary scalar.
(c) In this problem, A =
1 1 p
1 p 1
p 1 1
, b =
1
p
p2
.
[A |b ] =
1 1 p
1 p 1
p 1 1
∣
∣
∣
∣
∣
∣
∣
∣
1
p
p2
→
1 1 p
0 p− 1 1− p
0 1− p 1− p2
∣
∣
∣
∣
∣
∣
∣
∣
1
p− 1
p2 − p
→
1 1 p
0 p− 1 1− p
0 0 (1− p)(p + 2)
∣
∣
∣
∣
∣
∣
∣
∣
1
p− 1
p2 − 1
.
There are 3 cases:
16
CHAPTER 2. LINEAR ALGEBRA
(i) If p = 1, the above augmented matrix is reduced to
1 1 1
0 0 0
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
1
0
0
. Hence the sys-
tem of linear equations is reduced to one single equation x1 + x2 + x3 = 1. Thus
x =[
1− s− t s t]T
, where t and s are arbitrary scalars.
(ii) If p = −2, the augmented matrix becomes
1 1 −2
0 −3 3
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
1
−3
3
. Hence the correspond-
ing system of linear equations is inconsistent.
(iii) For p 6= 1, p 6= −2, the augmented matrix may be reduced to
1 1 p
0 1 −1
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1
1−1−p2+p
.
We conclude that the corresponding linear system has one and only one solution given by
x =[
(1+p)2
2+p1
2+p −1+p2+p
]T.
(10) (a)
1 1 1 1 1 1
−1 −1 0 0 1 −1
−2 −2 0 0 3 1
0 0 1 1 3 3
1 1 2 2 4 4
−r1 + r5 −→ r5
2r1 + r3 −→ r3
r1 + r2 −→ r2−−−−−−−−−−−−−→
1 1 1 1 1 1
0 0 0 0 2 0
0 0 2 2 5 3
0 0 1 1 3 3
0 0 1 1 3 3
r2 ←→ r5−−−−−−→
1 1 1 1 1 1
0 0 1 1 3 3
0 0 2 2 5 3
0 0 1 1 3 3
0 0 0 0 2 0
−r2 + r4 −→ r4
−2r2 + r3 −→ r3−−−−−−−−−−−−−−→
1 1 1 1 1 1
0 0 1 1 3 3
0 0 0 0 −1 −3
0 0 0 0 0 0
0 0 0 0 2 0
r4 ←→ r5
−r3 −→ r3−−−−−−−−−→
1 1 1 1 1 1
0 0 1 1 3 3
0 0 0 0 1 3
0 0 0 0 2 0
0 0 0 0 0 0
−2r3 + r4 −→ r4−−−−−−−−−−−−→
1 1 1 1 1 1
0 0 1 1 3 3
0 0 0 0 1 3
0 0 0 0 0 −6
0 0 0 0 0 0
(b) The system of linear equations is inconsistent. If it is consistent, then there are
three non-zero equations with five variables. Therefore, the system has infinitely
many solutions.
17
CHAPTER 2. LINEAR ALGEBRA
(11)
1 1 3 0
1 −2 0 0
−2 3 −1 0
−1 2 0 0
2 −3 1 0
−2r1 + r5 −→ r5
r1 + r4 −→ r4
2r1 + r3 −→ r3
−r1 + r2 −→ r2−−−−−−−−−−−−−−→
1 1 3 0
0 −3 −3 0
0 5 5 0
0 3 3 0
0 −5 −5 0
−15r5 −→ r5
13r4 −→ r4
15r3 −→ r3
−13r2 −→ r2
−−−−−−−−−−→
1 1 3 0
0 1 1 0
0 1 1 0
0 1 1 0
0 1 1 0
−r2 + r5 −→ r5
−r2 + r4 −→ r4
−r2 + r3 −→ r3−−−−−−−−−−−−−→
1 1 3 0
0 1 1 0
0 0 0 0
0 0 0 0
0 0 0 0
−r2 + r1 −→ r1−−−−−−−−−−−−−→
1 0 2 0
0 1 1 0
0 0 0 0
0 0 0 0
0 0 0 0
The reduced system of equations is :
x1 + 2x3 = 0 · · · · · · (1)
x2 + x3 = 0 · · · · · · (2)
Let x3 = t, then from (2), x2 = −x3 = −t.
From (1), x1 = −2x3 = −2t.
Therefore, x =
x1
x2
x3
=
−2t
−t
t
.
(12)
1 1 −1 1
2 3 a 3
1 a 3 2
−r1 + r3 −→ r3
−2r1 + r2 −→ r2−−−−−−−−−−−−−−→
1 1 −1 1
0 1 a + 2 1
0 a− 1 4 1
− (a− 1) r2 + r3 −→ r3−−−−−−−−−−−−−−−−−→
1 1 −1 1
0 1 a + 2 1
0 0 − (a + 3) (a− 2) − (a− 2)
(a) If
− (a + 3) (a− 2) = 0
− (a− 2) = 0, then the system is consistent with infinitely many solu-
tion. Therefore, a = 2.
(b) If − (a + 3) (a− 2) 6= 0, then the system is consistent with one and only one solution.
Therefore, a 6= 2 or a 6= −3.
(c) If
− (a + 3) (a− 2) = 0
− (a− 2) 6= 0, then the system is inconsistent. Therefore, a = −3.
(13)
1 −1 1 2
3 1 4a− 1 2
2 a a + 1 2
−2r1 + r3 −→ r3
−3r1 + r2 −→ r2−−−−−−−−−−−−−−→
1 −1 1 2
0 4 4a− 4 −4
0 a + 2 a− 1 −2
14r2 −→ r2−−−−−−−→
18
CHAPTER 2. LINEAR ALGEBRA
1 −1 1 2
0 1 a− 1 −1
0 a + 2 a− 1 −2
− (a + 2) r2 + r3 −→ r3−−−−−−−−−−−−−−−−−→
1 −1 1 2
0 1 a− 1 −1
0 0 − (a + 1) (a− 1) a
(a) If
− (a + 1) (a− 1) = 0
a = 0, then the system is consistent with infinitely many so-
lutions. However, the above system has no solution. Therefore, the system is not
possible to have infinitely mant solutions.
(b) If − (a + 1) (a− 1) 6= 0, then the system is consistent with one and only one solution.
Therefore, a 6= 1 or a 6= −1.
(c) If
− (a + 1) (a− 1) = 0
a 6= 0, then the system is inconsistent. Therefore, a = 1 or
a = −1.
(14) The augmented matrix
1 1 −1
−a −1 a
a2 1 −a
∣
∣
∣
∣
∣
∣
∣
∣
b1
b2
b3
may be reduced to
1 1 −1
0 a− 1 0
0 1− a2 a(a− 1)
∣
∣
∣
∣
∣
∣
∣
∣
b1
b2 + ab1
b3 − a2b1
———(*)
(a) If a 6= 0, 1, (*) is reduced to
1 1 −1
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
b1
b2+ab1a−1
b3+(a+1)b2+ab1a(a−1)
, which implies consis-
tency of the linear system. On the other hand, if the system is consistent for any
b1, b2 and b3, then it follows from (*) that a 6= 0, 1.
(b) If a = 0, (*) becomes
1 1 −1
0 1 0
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
b1
b3
b2 + b3
. It thus follows that the system is
consistent if and only if b2 + b3 = 0.
(c) If a = 1 and b = 0, (*) becomes
1 1 −1
0 0 0
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
0
0
0
. As such, we have one single
equation x1 + x2 − x3 = 0, and x =[
s− t t s]T
.
19
CHAPTER 2. LINEAR ALGEBRA
(15)
[A |b ] =
1 3 −2 5
5 8 −5 12
3 16 −11 28
∣
∣
∣
∣
∣
∣
∣
∣
b1
b2
b3
→
1 3 −2 5
0 −7 5 −13
0 7 −5 13
∣
∣
∣
∣
∣
∣
∣
∣
b1
b2 − 5b1
b3 − 3b1
→
1 3 −2 5
0 1 −57
137
0 0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
b1
5b1−b27
b3 + b2 − 8b1
.
Therefore, the system is consistent if and only if b3 + b2−8b1 = 0. When b = [1 4 4]T ,
the system is reduced to one with 2 linear equations in 4 unknowns:
x1 +3x2 −2x3 +5x4 = 1
x2 −57x3 +13
7 x4 = 17
.
Therefore, solutions are x =[
− t7 + 4s
7 + 47
5t7 − 13s
7 + 17 t s
]T.
(16)
1 1 1
α β γ
β γ α
γ α β
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1
1
1
→
1 1 1
α β γ
β γ α
α + β + γ α + β + γ α + β + γ
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1
1
3
→
1 1 1
α β γ
β γ α
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1
1
3− 3(α + β + γ)
→
1 1 1
0 β − α γ − α
0 γ − β α− β
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1− 3α
1− 3β
3− 3(α + β + γ)
.
Therefore, if α+β+γ 6= 1, then the system is inconsistent . Assume now that α+β+γ = 1.
20
CHAPTER 2. LINEAR ALGEBRA
The augmented matrix now takes the form
1 1 1
0 β − α γ − α
0 γ − β α− β
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1− 3α
1− 3β
0
.
Case (i) If α = β = γ = 13 , then the system is reduced to a single equation x1+x2+x3 = 3.
Therefore, it has infinitely many solutions.
Case (ii) If α = β 6= γ, then the above matrix may be reduced to
1 1 1
0 1 0
0 0 1
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
31−3βγ−β1−3αγ−α
0
.
In this case, the system has one and only one solution. Similar, we have the same
conclusion if β = γ 6= α or if α = γ 6= β.
Case (iii) If α, β and γ are all distinct, then
1 1 1
0 β − α γ − α
0 γ − β α− β
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
3
1− 3α
1− 3β
0
→
1 1 1
0 1 γ−αβ−α
0 0 δ
0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
31−3αβ−α
η
0
,
where δ =α2 + β2 + γ2 − αβ − βγ − γα
α− βand η = 1− 3β +
(1− 3α)(β − γ)
β − α.
Since δ =(α− β)2 + (β − γ)2 + (γ − α)2
2(α− β)6= 0, the system of linear equations has a unique
solution.
21
CHAPTER 2. LINEAR ALGEBRA
(17) (a)
1 a b
0 2 c
0 0 −1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 1 0
0 0 1
→
1 a b
0 1 c2
0 0 −1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 12 0
0 0 1
→
1 a b
0 1 c2
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 12 0
0 0 −1
→
1 a b
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 12
c2
0 0 −1
→
1 a 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 b
0 12
c2
0 0 −1
→
1 0 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 −a2 b− ac
2
0 12
c2
0 0 −1
Therefore,
1 a b
0 2 c
0 0 −1
−1
=
1 −a2 b− ac
2
0 12
c2
0 0 −1
.
(b)
3 1 2
−1 3 −4
4 0 5
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 1 0
0 0 1
→
0 10 −10
−1 3 −4
0 12 −11
∣
∣
∣
∣
∣
∣
∣
∣
1 3 0
0 1 0
0 4 1
→
0 1 −1
−1 3 −4
0 12 −11
∣
∣
∣
∣
∣
∣
∣
∣
110
310 0
0 1 0
0 4 1
→
0 1 −1
−1 0 −1
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
110
310 0
− 310
110 0
−1210
410 1
→
0 1 0
−1 0 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
−1110
710 1
−1510
510 1
−1210
410 1
→
1 0 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1510 − 5
10 −1
−1110
710 1
−1210
410 1
.
22
CHAPTER 2. LINEAR ALGEBRA
(c)
−2 3 −1
−1 2 −1
−6 9 −4
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 1 0
0 0 1
→
0 −1 1
1 −2 1
0 −3 2
∣
∣
∣
∣
∣
∣
∣
∣
1 −2 0
0 −1 0
0 −6 1
→
0 1 −1
1 0 −1
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
−1 2 0
−2 3 0
3 0 −1
→
0 1 0
1 0 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
2 2 −1
1 3 −1
3 0 −1
→
1 0 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 3 −1
2 2 −1
3 0 −1
.
(18) Let ∆ = αδ − βγ.
Case (1) If α 6= 0, a simple calcultion shows that
[
α β
γ δ
]
→[
1 βα
γ δ
]
→[
1 βα
0 ∆α
]
.
If ∆ 6= 0, then the last matrix may be reduced to I by two more elementary row operations,
which implies that A is nonsingular. If ∆ = 0, then the reduced row echelon form of A
is of the form
[
1 βα
0 0
]
, meaning that A is singular.
Case (2) Suppose α = 0. When ∆ 6= 0, then both γ and β are non-zero so that
A =
[
0 β
γ δ
]
→[
γ δ
0 β
]
→[
1 δγ
0 β
]
→[
1 δγ
0 1
]
→ I.
If ∆ = 0, then either γ = 0 or β = 0. Hence either A =
[
0 β
0 δ
]
or A =
[
0 0
γ δ
]
.
Therefore, A is singular in both cases.
23
CHAPTER 2. LINEAR ALGEBRA
(19)
1 1 1
1 2 4
1 3 9
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
1 1 0
0 0 1
σ1→
1 1 1
0 1 3
1 3 9
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
−1 1 0
0 0 1
σ2→
1 1 1
0 1 3
0 2 8
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
−1 1 0
−1 0 1
σ3→
1 1 1
0 1 3
0 0 2
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
−1 1 0
1 −2 1
σ4→
1 1 1
0 1 3
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
−1 1 012 −1 1
2
σ5→
1 1 1
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
−52 4 −3
212 −1 1
2
σ6→
1 1 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
12 1 −1
2
−52 4 −3
212 −1 1
2
σ7→
1 0 0
0 1 0
0 0 1
∣
∣
∣
∣
∣
∣
∣
∣
3 −3 1
−52 4 −3
212 −1 1
2
.
For 1 ≤ j ≤ 7, let Ej be the elementary matrix obtained by applying σj to I. Let
Fj = E−1j . Since E7E6E5E4E3E2E1A = I, we conclude that A−1 = E7E6E5E4E3E2E1.
Therefore,
A = (E7E6E5E4E3E2E1)−1 = F1F2F3F4F5F6F7.
All Fj can be written down easily. For example,
F1 =
1 0 0
−1 1 0
0 0 1
−1
=
1 0 0
1 1 0
0 0 1
.
(20) (a) −2r1 + r2 −→ r2 : E1 =
1 0 0
−2 1 0
0 0 1
r2 + r3 −→ r3 : E2 =
1 0 0
0 1 0
0 1 1
−r3 −→ r3 : E3 =
1 0 0
0 1 0
0 0 −1
24
CHAPTER 2. LINEAR ALGEBRA
(b)
1 2 1
0 1 2
0 0 1
= E3E2E1A
=⇒
1 2 1
0 1 2
0 0 1
−1
= (E3E2E1A)−1 = A−1E−11 E−1
2 E−13
=⇒
1 2 1
0 1 2
0 0 1
−1
E3E2E1 = A−1E−11 E−1
2 E−13 E3E2E1 = A−1
A−1 =
1 −2 3
0 1 −2
0 0 1
1 0 0
0 1 0
0 0 −1
1 0 0
0 1 0
0 1 1
1 0 0
−2 1 0
0 0 1
=
11 −5 −3
−6 3 2
2 −1 −1
(21) det(A) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 2 3 4
0 −4 −8 −12
0 −8 −16 −24
13 14 15 16
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 2 3 4
0 −4 −8 −12
0 0 0 0
13 14 15 16
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 0.
(22) (a) i. x =
1 2 3
3 8 11
1 5 7
−1
−1
1
1
=
12
12 −1
−5 2 −172 −3
2 1
−1
1
1
=
−1
6
−4
.
ii. ∆ =
∣
∣
∣
∣
∣
∣
∣
∣
1 2 3
3 8 11
1 5 7
∣
∣
∣
∣
∣
∣
∣
∣
= 2, ∆1 =
∣
∣
∣
∣
∣
∣
∣
∣
−1 2 3
1 8 11
1 5 7
∣
∣
∣
∣
∣
∣
∣
∣
= −2, ∆2 =
∣
∣
∣
∣
∣
∣
∣
∣
1 −1 3
3 1 11
1 1 7
∣
∣
∣
∣
∣
∣
∣
∣
= 12,
∆3 =
∣
∣
∣
∣
∣
∣
∣
∣
1 2 −1
3 8 1
1 5 1
∣
∣
∣
∣
∣
∣
∣
∣
= −8.
Therefore, x1 =∆1
∆= −1, x2 =
∆2
∆= 6, x3 =
∆3
∆= −4.
25
CHAPTER 2. LINEAR ALGEBRA
(b) i. A−1 =
1 0 1 0
1 1 0 0
0 0 1 1
0 1 1 1
−1
=
0 1 1 −1
0 0 −1 1
1 −1 −1 1
−1 1 2 −1
x =
x1
x2
x3
x4
= A−1b =
0 1 1 −1
0 0 −1 1
1 −1 −1 1
−1 1 2 −1
1
1
1
0
=
2
−1
−1
2
ii. detA =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 1 0
1 1 0 0
0 0 1 1
0 1 1 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −1, detA1 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 1 0
1 1 0 0
1 0 1 1
0 1 1 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −2,
detA2 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 0
1 1 0 0
0 1 1 1
0 0 1 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 1, detA3 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 1 0
1 1 1 0
0 0 1 1
0 1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 1,
detA4 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 1 1
1 1 0 1
0 0 1 1
0 1 1 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −2
Therefore, x1 = ∆A1
∆A= −2
−1 = 2, x2 = ∆A2
∆A= −1, x3 = ∆A3
∆A= −1 and
x4 = ∆A4
∆A= 2.
(c) i. A−1 =
1 1 1 1
1 1 0 0
1 0 1 1
0 1 0 1
−1
=
−1 1 1 0
1 0 −1 0
2 −1 −1 −1
−1 0 1 1
x =
x1
x2
x3
x4
= A−1b =
−1 1 1 0
1 0 −1 0
2 −1 −1 −1
−1 0 1 1
1
1
1
1
=
1
0
−1
1
ii. ∆A =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 1
1 1 0 0
1 0 1 1
0 1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −1, ∆A1 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 1
1 1 0 0
1 0 1 1
1 1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −1
26
CHAPTER 2. LINEAR ALGEBRA
∆A2 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 1
1 1 0 0
1 1 1 1
0 1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 0, ∆A3 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 1
1 1 1 0
1 0 1 1
0 1 1 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 1
∆A4 =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 1 1 1
1 1 0 1
1 0 1 1
0 1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= −1
Therefore, x1 = ∆A1
∆A= −1
−1 = 1, x2 = ∆A2
∆A= 0, x3 = ∆A3
∆A= −1 and x4 =
∆A4
∆A= 1.
(23) (a) By Kirchoff’s Law, one obtains the circuit equations
i1 −i2 −i3 = 0
(R1 + R2 + R3) i1 + R4i2 = E1 + E2
R4i2 −R5i3 = E2
.
Cramer’s rule then yields
i1 =−E1(R4 + R5)− E2R5
∆, i2 =
−E1R5 − E2(R1 + R2 + R3 + R5)
∆
and
i3 =−E1R4 + E2(R1 + R2 + R3)
∆,
where ∆ = −R1R4 −R1R5 −R2R4 −R2R5 −R3R4 −R3R5 −R4R5.
(b) Using Kirchoff’s Law, currents in the circuit satisfy
i1 −i2 −i3 = 0
R2i2 −R3i3 = 0
(R1 + R4) i1 + R3i3 = −E0
.
Using Cramer’s rule, we obtain
i3 = −E0R2
R, i2 = −E0R3
Rand i1 = −E0(R2 + R3)
R,
where R = R1R2 + R1R3 + R2R3 + R2R4 + R3R4.
(24) Let x1, x2 and x3 be the numbers of batches of Milky, Extra Milky and Supreme respec-
27
CHAPTER 2. LINEAR ALGEBRA
tively. We therefore obtain
x1 + x2 + x3 = 6
x1 + 2x2 + 3x3 = 14
x1 + 4x2 + 9x3 = 36
.
Therefore x1 = 1, x2 = 2, x3 = 3.
(25) x1, x2, x3, x4 satisfy
−x1 + x4 = 100
x1 −x2 = 300
−x3 + x4 = 500
x2 −x3 = 100
.
Solving the system, we obtain x1 = x4 − 100, x2 = x4 − 400, x3 = x4 − 500, where x4 is
any integer ≥ 500.
(26) (a) The matrices are
[
0.8 0.37
0.2 0.63
]
and
[
0.75 0.15
0.25 0.85
]
. The combined network is repre-
sented by
[
0.75 0.15
0.25 0.85
][
0.8 0.37
0.2 0.63
]
=
[
0.63 0.372
0.37 0.628
]
.
(b)
[
v
w
]
=
[
0.75 0.15
0.25 0.85
][
0.8 0.37
0.2 0.63
][
x
y
]
=
[
0.63x + 0.372y
0.37x + 0.628y
]
.
(27) The system may be re-written as
R3 + R4 + R6 −R3 −R4
−R3 R2 + R3 + R5 −R5
−R4 −R5 R1 + R4 + R5
i1
i2
i3
=
E0
0
0
.
Using Cramer’s rule, one has i2 =∆2
∆and i3 =
∆3
∆, where ∆ is the determinant of the
coefficient matrix,
∆2 = E0 (R1R3 + R3R4 + R3R5 + R4R5) ,
∆3 = E0 (R2R4 + R3R4 + R3R5 + R4R5) .
Therefore, i2 = i3 ⇒ ∆2 = ∆3 ⇒ R1R3 = R2R4.
28
CHAPTER 2. LINEAR ALGEBRA
(28) Consider the homogeneous systems in 3 unknowns
a1 b1 c1
a2 b2 c2
a3 b3 c3
x1
x2
x3
=
0
0
0
and the non-homogeneous system in 2 unknowns
a1 b1
a2 b2
a3 b3
[
y1
y2
]
=
c1
c2
c3
If y =[
α β]T
is a solution of the nonhomogeneous system, then x =[
α β −1]T
is
a solution of the homogeneous system. Therefore, consistency of the nonhomogeneous
system implies existence of nontrivial solutions of the homogeneous system, which also
implies the coefficient matrix of the homogeneous system has zero determinant, i.e.,
det
a1 b1 c1
a2 b2 c2
a3 b3 c3
= 0.
(29) Elementary row operations give
1 3 −2 5
4 1 3 −2
−1 3 −4 7
∣
∣
∣
∣
∣
∣
∣
∣
0
0
0
→
1 0 1 −1
0 1 −1 2
0 0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
0
0
0
.
Solutions of the homogeneous system are[
−s + t s− 2t s t]T
, where s and t are arbi-
trary scalars. Taking s = 3, t = 1, we have
−2
1
4
−1
+
3
1
3
+ 3
−2
3
−4
+
5
−2
7
= 0.
Hence the given vectors are linearly dependent.
(30) Since ‖u± v‖2 = 〈u± v,u± v〉 = 〈u,u〉 + 〈v,v〉 ± 2 〈u,v〉, we deduce that ‖u + v‖2 +
‖u− v‖2 = 2 〈u, u〉+ 2 〈v,v〉 = 2 ‖u‖2 + 2 ‖v‖2.
29
CHAPTER 2. LINEAR ALGEBRA
(31)
w =
(〈v,u〉‖u‖2
)
u =(−4)× 1 + 5× (−2) + 6× 3
12 + (−2)2 + 32
1
−2
3
=
27
−47
67
.
Furthermore,
〈v −w,u〉 =
⟨
[
−4 5 6]T−[
27 −4
767
]T,[
1 −2 3]T⟩
=
⟨
[
−307
397
367
]T,[
1 −2 3]T⟩
= −30
7− 78
7+
108
7= 0.
(32) The jth row of QT is the same as the transpose of the jth column of Q. Therefore, the
entry at the jth row and the kth column of the product QTQ is just∑n
i=1 qij qik. This
implies (a)⇔ (b). The equivalence of (a) and (c) may be similarily proved by considering
the product QQT .
(33) det (A− λI) =
∣
∣
∣
∣
∣
∣
∣
∣
−2− λ 2 −3
2 1− λ −6
−1 0 −λ
∣
∣
∣
∣
∣
∣
∣
∣
= −λ3 − λ2 + 9λ + 9 = − (λ + 3) (λ− 3) (λ + 1)
Therefore, the eigenvalues of A are λ = 3, λ = −1 and λ = −3.
(34)
det (A− λI) =
∣
∣
∣
∣
∣
∣
∣
∣
1− λ 2 1
0 3− λ 1
0 5 −1− λ
∣
∣
∣
∣
∣
∣
∣
∣
= −λ3 + 3λ2 + 6λ− 8
= − (λ− 1) (λ + 2) (λ− 4)
Therefore, the eigenvalues of A are λ = 1, λ = 4 and λ = −2.
(35) When λ = 0, (A− λI)v = 0 =⇒
3 −1 −2 0
2 0 −2 0
2 −1 −1 0
→
1 0 −1 0
0 1 −1 0
0 0 0 0
and v =
1
1
1
r, where r ∈ R.
30
CHAPTER 2. LINEAR ALGEBRA
When λ = 1, (A− λI)v = 0 =⇒
2 −1 −2 0
2 −1 −2 0
2 −1 −2 0
→
1 −12 −1 0
0 0 0 0
0 0 0 0
and v =
12s + t
s
t
=
12
1
0
s +
1
0
1
t, where s, t ∈ R.
Thus, the eigenvectors are
1
1
1
,
12
1
0
and
1
0
1
.
Therefore, D =
0 0 0
0 1 0
0 0 1
and P =
1 12 1
1 1 0
1 0 1
.
AP =
3 −1 −2
2 0 −2
2 −1 −1
1 12 1
1 1 0
1 0 1
=
0 12 1
0 1 0
0 0 1
PD =
1 12 1
1 1 0
1 0 1
0 0 0
0 1 0
0 0 1
=
0 12 1
0 1 0
0 0 1
= AP
(36) When λ = 2, (A− λI)v = 0 =⇒
0 1 −1 0
1 0 −1 0
1 1 −2 0
→
1 0 −1 0
0 1 −1 0
0 0 0 0
and v =
1
1
1
r, where r ∈ R.
When λ = 1, (A− λI)v = 0 =⇒
1 1 −1 0
1 1 −1 0
1 1 −1 0
→
1 1 −1 0
0 0 0 0
0 0 0 0
and v =
−s + t
s
t
=
−1
1
0
s +
1
0
1
t, where s, t ∈ R.
Thus, the eigenvectors are
1
1
1
,
−1
1
0
and
1
0
1
.
31
CHAPTER 2. LINEAR ALGEBRA
Therefore, D =
2 0 0
0 1 0
0 0 1
and P =
1 −1 1
1 1 0
1 0 1
.
AP =
2 1 −1
1 2 −1
1 1 0
1 −1 1
1 1 0
1 0 1
=
2 −1 1
2 1 0
2 0 1
PD =
1 −1 1
1 1 0
1 0 1
2 0 0
0 1 0
0 0 1
=
2 −1 1
2 1 0
2 0 1
= A
(37) (a) The characteristic polynomial
f(λ) = det
−1− λ 2 −3
2 2− λ −6
−1 −2 1− λ
= −λ3 + 2λ2 + 20λ + 24 = (6− λ) (λ + 2)2 .
Eigenvalues are λ = 6 and λ = −2(double root).
(i) For λ = −2, (A − λI) =
1 2 −3
2 4 −6
−1 −2 3
. We use Gaussian elimination to solve
the homogeneous system
1 2 −3
2 4 −6
−1 −2 3
v = 0 and conclude that v1 = [3 0 1]T and
v2 = [−2 1 0]T are corresponding eigenvectors.
(ii) For λ = 6, (A− λI) =
−7 2 −3
2 −4 −6
−1 −2 −5
. Use Gaussian elimination to solve the homo-
geneous system
−7 2 −3
2 −4 −6
−1 −2 −5
v = 0 to obtain v3 = [−1 − 2 1]T as an eigenvector.
Finally, P =
3 −2 −1
0 1 −2
1 0 1
diagonalizes A, i.e., P−1AP =
−2 0 0
0 −2 0
0 0 6
.
32
CHAPTER 2. LINEAR ALGEBRA
(b)
f(λ) = det
0− λ −2 0
−2 1− λ −2
0 −2 2− λ
= −λ3 + 3λ2 + 6λ− 8 = (1− λ)(λ + 2)(λ− 4).
Eigenvalues are λ = 1, λ = −2, λ = 4.
(i) For λ = 1, (A − λI) =
−1 −2 0
−2 0 −2
0 −2 1
. Solving (A − λI)v = 0, we obtain v1 =
[−2 1 2]T as a corresponding eigenvector.
(ii) For λ = −2, (A − λI) =
2 −2 0
−2 3 −2
0 −2 4
. Gaussian elimination when applied to the
homogeneous system (A−λI)v = 0 gives v2 = [2 2 1]T as a corresponding eigenvector.
(iii) For λ = 4, (A − λI) =
−4 −2 0
−2 −3 −2
0 −2 −2
. We now solve the homogeneous system
(A − λI)v = 0 and obtain the third eigenvector v3 = [1 − 2 2]T . Therefore, P =
−2 2 1
1 2 −2
2 1 2
diagonalizes A, i.e., P−1AP =
1 0 0
0 −2 0
0 0 4
.
(c)
f(λ) = det
4− λ 2 2
2 1− λ −4
2 −4 1− λ
= −λ3 + 6λ2 + 15λ− 100 = − (λ + 4) (λ− 5)2 .
Eigenvalues of A are λ = 5 (double root), −4.
For λ = 5, (A − λI) =
−1 2 2
2 −4 −4
2 −4 −4
. We thus obtain two linearly independent
eigenvectors v1 = [2 0 1]T and v2 = [2 1 0]T by solving the homogeneous system
(A− λI)v = 0.
For λ = −4, (A−λI) =
8 2 2
2 5 −4
2 −4 5
. Therefore, v3 = [1 − 2 − 2]T is a correspond-
33
CHAPTER 2. LINEAR ALGEBRA
ing eigenvector.
Furthermore, putting P =
2 2 1
0 1 −2
1 0 −2
, we have P−1AP =
5 0 0
0 5 0
0 0 −4
.
(38) (i) (a) det (A− λI) =
∣
∣
∣
∣
∣
2− λ 3
−1 6− λ
∣
∣
∣
∣
∣
= λ2 − 8λ + 15 = (λ− 3) (λ− 5) = 0
Thus, the eigenvalues of A are λ = 3 and λ = 5.
When λ = 3, (A− λI)v = 0 =⇒[
−1 3 0
−1 3 0
]
→[
1 −3 0
0 0 0
]
and v =
[
3
1
]
s, where s ∈ R.
When λ = 5, (A− λI)v = 0 =⇒[
−3 3 0
−1 1 0
]
→[
1 −1 0
0 0 0
]
and v =
[
1
1
]
t, where t ∈ R.
(b) Using the eigenvalues and eigenvectors, Therefore,
D =
[
3 0
0 5
]
and P =
[
3 1
1 1
]
AP =
[
2 3
−1 6
][
3 1
1 1
]
=
[
9 5
3 5
]
PD =
[
3 1
1 1
][
3 0
0 5
]
=
[
9 5
3 5
]
= AP
(c) Dn =
[
λn1 0
0 λn2
]
P−1=
[
3 1
1 1
]−1
=
[
12 −1
2
−12
32
]
A6 − 6A4 + 11A2
= PD6P−1 − 6PD4P−1 + 11PD2P−1 = P(
D6 − 6D4 + 11D2)
P−1
=
[
3 1
1 1
]([
36 0
0 56
]
− 6
[
34 0
0 54
]
+ 11
[
32 0
0 52
])[
12 −1
2
−12
32
]
=
[
−5562 17712
−5904 18054
]
34
CHAPTER 2. LINEAR ALGEBRA
(ii) (a)
det (A− λI) =
∣
∣
∣
∣
∣
∣
∣
∣
−1− λ 2 0
1 1− λ 0
0 0 1− λ
∣
∣
∣
∣
∣
∣
∣
∣
= −λ3 + λ2 + 3λ− 3
= − (λ− 1)(
λ−√
3)(
λ +√
3)
= 0
Thus, the eigenvalues of A are λ = 1, λ =√
3, and λ = −√
3.
When λ = 1, (A− λI)v = 0 =⇒ v =
0
0
1
r, where r ∈ R.
When λ =√
3, (A− λI)v = 0 =⇒ v =
√3− 1
1
0
s, where s ∈ R.
When λ = −√
3, (A− λI)v = 0 =⇒ v =
−√
3− 1
1
0
t, where t ∈ R.
(b) Using the eigenvalues and eigenvectors, Therefore,
D =
1 0 0
0√
3 0
0 0 −√
3
and P =
0√
3− 1 −√
3− 1
0 1 1
1 0 0
AP =
−1 2 0
1 1 0
0 0 1
0√
3− 1 −√
3− 1
0 1 1
1 0 0
=
0 3−√
3 3 +√
3
0√
3 −√
3
1 0 0
PD =
0√
3− 1 −√
3− 1
0 1 1
1 0 0
1 0 0
0√
3 0
0 0 −√
3
=
0 3−√
3 3 +√
3
0√
3 −√
3
1 0 0
35
CHAPTER 2. LINEAR ALGEBRA
(c) P−1=
0√
3− 1 −√
3− 1
0 1 1
1 0 0
−1
=
0 0 1√3
63+
√3
6 0
−√
36
3−√
36 0
A6 − 6A4 + 11A2
= P(
D6 − 6D4 + 11D2)
P−1
=
0√
3− 1 −√
3− 1
0 1 1
1 0 0
6 0 0
0 6 0
0 0 6
0 0 1√3
63+
√3
6 0
−√
36
3−√
36 0
=
6 0 0
0 6 0
0 0 6
(39)
det (A− λI) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
4− λ 2 0 4
0 2− λ −1 0
0 0 3− λ 3
0 4 0 7− λ
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= λ4 − 16λ3 + 89λ2 − 194λ + 120
= (λ− 1) (λ− 4) (λ− 5) (λ− 6) = 0.
Thus, the fourth eigenvalue of A is λ = 5.
When λ = 5, (A− λI)v = 0 =⇒
−1 2 0 4 0
0 −3 −1 0 0
0 0 −2 3 0
0 4 0 2 0
→
1 0 0 −3 0
0 1 0 12 0
0 0 1 −32 0
0 0 0 0 0
. There-
fore, v =
3
−12
32
1
s, where s ∈ R. and, the eigenvector associated with λ = 5 is
6
−1
3
2
.
36
CHAPTER 2. LINEAR ALGEBRA
(40)
det (A− λI) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1− λ 2 0 2
0 −2− λ −1 0
0 0 3− λ 3
0 0 0 −1− λ
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= λ4 − λ3 − 7λ2 + λ + 6
= (λ− 1) (λ− 3) (λ + 2) (λ + 1) = 0
Thus, the eigenvalues of A are λ = 1, λ = 3, λ = −1, and λ = −2.
When λ = 1, v =
1
0
0
0
. When λ = 3, v =
1
1
−5
0
. When λ = −1, v =
7
−3
3
−4
.
When λ = −2, v =
2
−3
0
0
.
(41) Assume A is a matrix with eigenvalue λ and eigenvector x, i.e., Ax = λx.
A2x = A (Ax) = A (λx) = λ (Ax) = λ (λx) = λ2x
(42) (a) For A =
[
0.7 0.4
0.3 0.6
]
, f(λ) = (λ− 0.3)(λ− 1). Therefore, eigenvalues are λ1 = 0.3
and λ2 = 1, with eigenvectors v1 =[
1 −1]T
and v2 =[
4 3]T
respectively.
Taking P =
[
1 4
−1 3
]
, we have A = P
[
0.3 0
0 1
]
P−1. Therefore,
Ak = P
[
0.3k 0
0 1
]
P−1 =
[
1 4
−1 3
][
0.3k 0
0 1
][
37 −4
717
17
]
=1
7
[
4 + 3× 0.3k 4− 4× 0.3k
3− 3× 0.3k 3 + 4× 0.3k
]
.
(b) If the Red Party is in power now, the probability that it is to be in power after k
elections is[
1 0]
Ak
[
1
0
]
=4 + 3× 0.3k
7.
If the Red Party is not in power now, the probability that it is to be in power after
k elections is[
0 1]
Ak
[
0
1
]
3− 3× 0.3k
7.
37
CHAPTER 2. LINEAR ALGEBRA
When k →∞, the respective probabilities are4
7and
3
7.
(43) (a)
[
1 2
1 0
][
an−1
an−2
]
=
[
an−1 + 2an−2
an−1
]
=
[
an
an−1
]
for n ≥ 3.
(b) A =
[
1 2
1 0
]
has eigenvalues λ1 = 2 and λ2 = −1, with eigenvectors v1 =
[
2
1
]
and v2 =
[
−1
1
]
respectively. Therefore, one obtains
A =
[
2 −1
1 1
][
2 0
0 −1
][
2 −1
1 1
]−1
and
Ak =
[
2 −1
1 1
][
2k 0
0 (−1)k
][
13
13
−13
23
]
=1
3
[
2k+1 + (−1)k 2k+1 − 2× (−1)k
2k − (−1)k 2k + 2× (−1)k
]
.
(c)
[
an
an−1
]
=
[
1 2
1 0
][
an−1
an−2
]
=
[
1 2
1 0
]2 [
an−2
an−3
]
= . . . =
[
1 2
1 0
]n−2 [
a2
a1
]
Therefore, an =2n − (−1)n
3for n ≥ 3.
(44) (a)
Eigenvalues Eigenvectors Orthonormal eigenvectors
λ = 3[
1 1 1]T [
1√3
1√3
1√3
]T
λ = 6[
1 1 −2]T [
1√6
1√6− 2√
6
]T
λ = 8[
1 −1 0]T [
1√2− 1√
20]T
Therefore, Q =
1√3
1√6
1√2
1√3
1√6
−1√2
1√3
−2√6
0
and D =
3 0 0
0 6 0
0 0 8
(b)
Eigenvalues Eigenvectors
λ = 7 (double root)[
1 −2 0]T
and[
0 2 1]T
λ = −2[
2 1 −2]T
38
CHAPTER 2. LINEAR ALGEBRA
[
1 −2 0]T
and[
0 2 1]T
are not orthogonal to each other, although each of them
is orthogonal to[
2 1 −2]T
. Using Gram Schmidt Process, one obtains 3 orthonor-
mal eigenvectors w1 =[
1√5− 2√
50]T
, w2 =[
43√
52
3√
55
3√
5
]T, w3 =
[
23
13 −2
3
]T.
Therefore, Q =
1√5
43√
523
− 2√5
23√
513
0 53√
5−2
3
and D =
7 0 0
0 7 0
0 0 −2
39
Chapter 3
Infinite series, Power series and
Fourier series
(1) (a) Convergent, Comparison with1
n3/2.
∑ 1√
n(n2 + 1)=∑ 1√
n3 + n≤∑ 1√
n3.
(b) Divergent, Comparison with1
n.
∑ π/4
n≤∑ tan−1 n
n
(c) Convergent, Comparison with1
n2.
∑ lnn
n3≤∑ n
n3=∑ 1
n2
(d) Convergent, Comparison with1
n3.
1
(n + 1)(n + 2)(n + 3)≤ 1
n3
(e) Convergent, ratio test
((n+1)!)2
(2(n+1))!
(n!)2
(2n)!
=(2n)!((n + 1)!)2
(2(n + 1))!(n!)2=
(n + 1)2
(2n + 1)(2n + 2)=
n + 1
4n + 2→ 1
4
41
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
(f) Convergent, ratio test
(n+1)!(n+1)n+1
n!nn
=(n + 1)!nn
n!(n + 1)n+1=
nn
(n + 1)n=
1(
n+1n
)n =1
(
1 + 1n
)n →1
e
(g) Convergent, ratio test
xn+1
(n+1)n+1
xn
nn
=xn+1nn
xn(n + 1)n+1=
xnn
(n + 1)n+1=
nn
(n + 1)n
x
n + 1→ 0
(h) Divergent, integral test
∫ ∞
a
dx
x lnx ln(lnx)= [ln(ln(lnx))]∞a
(i) Convergent, integral test
∫ ∞
a
en
9 + e2ndx =
[
1
3tan−1
(
ex
3
)]∞
a
(j) Divergent, integral test
∫ ∞
a
2ln(ln n)
n lnndx =
[
2ln(ln x)
ln 2
]∞
a
(2) (a) Conditionally.
The series is not absolutely convergent because x < sin−1 x for 0 < x ≤ 1 , i.e.
1
n≤ sin−1
(
1
n
)
.
Since sin−1 x is an increasing continuous function, we have
sin−1
(
1
n
)
> sin−1
(
1
n + 1
)
n→∞−−−→ 0.
By Leibniz test, the series converges conditionally.
(b) Absolutely
∑
∣
∣
∣
∣
(−1)nn3
en
∣
∣
∣
∣
=∑ n3
en,
which converges by ratio test.
42
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
(c) Absolutely
∑
∣
∣
∣
∣
(−1)n
2 + 2n + n2
∣
∣
∣
∣
=∑ 1
2 + 2n + n2≤∑ 1
n2,
which converges by Comparison test.
(d) Conditionally
∑
∣
∣
∣
∣
(−1)n
n lnn
∣
∣
∣
∣
=∑ 1
n lnn,
which diverges by Integral test. Since x lnx is an increasing function for x > 1,
1
n lnn>
1
(n + 1) ln(n + 1)
n→∞−−−→ 0.
By Leibniz test, the series converges conditionally.
(3) Consider the k-th partial sum:
sk =k∑
n=1
1
(4n− 1)(4n + 3)=
k∑
n=1
[
1
4(4n− 1)− 1
4(4n + 3)
]
=
(
1
12− 1
28
)
+
(
1
28− 1
44
)
+ · · ·+(
1
4(4n− 1)− 1
4(4n + 3)
)
=1
12− 1
4(4k + 3)
→ 1
12
(4) (a) [−1, 1), 0 < s ≤ 1; [−1, 1], s > 1
R = limn→∞
∣
∣
∣
∣
an
an+1
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
∣
1ns
1(n+1)s
∣
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
(
1 +1
n
)s∣∣
∣
∣
= 1
At x = 1, the series becomes∑ 1
ns which converges if s > 1 and diverges if s ≤ 1.
At x = −1, the series becomes the alternating series∑ (−1)n
ns , which converges
absolutely if s > 1 and conditionally if s ≤ 1.
(b) (−e, e)
R = limn→∞
∣
∣
∣
∣
an
an+1
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
∣
∣
n!nn
(n+1)!(n+1)n+1
∣
∣
∣
∣
∣
∣
= limn→∞
(
n + 1
n
)n
= e
43
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
At x = e, the series becomes∑ n!en
nn , which diverges because
an =n!en
nn=
n!
nn
(
1 + n + · · ·+ nn
n!+ · · ·
)
> 1
and therefore an does not converges to 0.
Similarly, at x = −e, the series becomes the alternating series∑ (−1)nn!en
nn , which
diverges because bn = (−1)nn!en
nn does not converge to 0.
(c) (−1, 1)
R = limn→∞
∣
∣
∣
∣
an
an+1
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
n
n + 1
∣
∣
∣
∣
= 1
At x = −1 or x = 1, the terms of the series are (−1)nn and n respectively, and do
not converge to 0.
(d) (−2, 0]
R = limn→∞
∣
∣
∣
∣
an
an+1
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
∣
(−1)n−1
n(−1)n
n+1
∣
∣
∣
∣
∣
= limn→∞
n + 1
n= 1
Translating to the left by 1 units, we are interested in the interval (−2, 0)
At x = −2, the series becomes∑ (−1)2n−1
n = −∑ 1n , which is the negative harmonic
series and diverges.
At x = 0, the series becomes∑ (−1)n−1
n , which is the alternating harmonic series and
converges.
(e) (−4, 0)
R = limn→∞
∣
∣
∣
∣
an
an+1
∣
∣
∣
∣
= limn→∞
∣
∣
∣
∣
∣
n2
2n
(n+1)2
2n+1
∣
∣
∣
∣
∣
= limn→∞
2n2
(n + 1)2= 2
Translating to the left by 2 units, we are interested in the interval (−4, 0).
At x = −4 or x = 1, the terms of the series are (−1)2n2 and n2 respectively, and do
not converge to 0.
(5) The Maclaurin Series of a function f(x) is given by
f(x) = f(0) + f ′(0)x +f ′′(0)
2!x2 +
f ′′′(0)
3!x3 +
f (4)(0)
4!x4 + · · ·
(a) 1− 2x + 2x2 − 43x3 + 2
3x4
Note:
e−2x =∞∑
n=0
(−2x)n
n!=
∞∑
n=0
(−2)n
n!xn
44
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
(b) 2x− 43x3
Note
sin(2x) =∞∑
n=0
(−1)n(2x)2n+1
(2n + 1)!
(c) x2 + x3 + x4
2! + x5
3!
Note
x2ex = x2∞∑
n=0
xn
n!=
∞∑
n=0
xn+2
n!
(d) ln 3 + 23x− 2
9x2 + 881x3 − 4
81x4
Note
ln(3 + 2x) = ln(1 + 2(1 + x)) =∞∑
n=1
(−1)n−1
n[2(1 + x)]n
(6) (a) Let z = ax. Then
ez =
∞∑
n=0
zn
n!=
∞∑
n=0
(ax)n
n!
(b) Let z = x2 . Then
cos z =∞∑
n=0
(−1)nz2n
(2n)!=
∞∑
n=0
(−1)nx2n
4n(2n)!
(c)
ex = 1 + x +x2
2!+
x3
3!+ · · ·+ xn
n!+ · · ·
e−x = 1− x +x2
2!− x3
3!+ · · ·+ xn
n!+ · · ·
cosh x =ex + e−x
2=
∞∑
n=0
x2n
(2n)!
(d)
ln |1 + x| = x− 1
2x2 +
1
3x3 − 1
4x4 + · · ·+ (−1)n−1
nxn + · · ·
ln |1− x| = −x +1
2x2 − 1
3x3 +
1
4x4 + · · ·+ (−1)n
nxn + · · ·
1
2ln
∣
∣
∣
∣
1 + x
1− x
∣
∣
∣
∣
=1
2(ln |1 + x| − ln |1− x|) =
∞∑
n=1
x2n−1
2n− 1
45
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
(7) (a)∑∞
n=1(−1)n+1
n (x− 1)n
lnx = ln(1 + (x− 1)) =∞∑
n=1
(−1)n+1
n(x− 1)n
(b)∑∞
n=0(−1)(x + 1)n
1
x= − 1
1− (x + 1)= −
∞∑
n=0
(x + 1)n =∞∑
n=0
(−1)(x + 1)n
(c)∑∞
n=0(−1)nπ2n
(2n)!
(
x− 12
)2n
sinπx = cos
[
π
(
x− 1
2
)]
=∞∑
n=0
(−1)n
(2n)!
[
π
(
x− 1
2
)]2n
(8) (a) Since |x| is an even function, all bn = 0.
a0 =1
π
∫ π
−π|x| dx =
1
π
∫ π
0xdx +
1
π
∫ 0
−π−xdx = π
an =1
π
∫ π
−π|x| cos nxdx = 2
1
π
∫ π
0x cos nxdx =
2
π
−1 + cos πn
n2=
2
π
(−1)n − 1
n2
Hence,
f (x) =π
2+
2
π
∞∑
n=1
(−1)n − 1
n2cos nx
Note: Choose f = x, g′ = cos nx and use integration by parts, we get
∫
x cos nxdx = xsinnx
n−∫
sinnx
ndx = x
sinnx
n+
cos nx
n2
(b)
f (x) = sin2 x− 2 cos3 x =1
2(1− cos 2x)− 2
(
cos2 x)
cos x
=1
2(1− cos 2x)− (1 + cos 2x) cos x
=1
2(1− cos 2x)− cos x− cos 2x cos x
=1
2(1− cos 2x)− cos x− 1
2(cos 3x + cos x)
=1
2− 3
2cos x− 1
2cos 2x− 1
2cos 3x
46
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
Note:
cos 2x = 1− 2 sin2 x = 2 cos2 x− 1
cos mx cos nx =1
2[cos (m + n)x + cos (m− n)x]
(c)
a0 =1
π
∫ π
0sin tdt =
2
π
For n 6= 1,
bn =1
π
∫ π
0sin t sinntdt
=1
π
∫ π
0−1
2[cos (1 + n) t− cos (1− n) t] dt
=
[
−1
2
sin (1 + n) t
1 + n+
1
2
sin (1− n) t
1− n
]π
0
= 0
For n = 1,
b1 =1
π
∫ π
0sin t sin tdt =
1
2π
∫ π
0(1− cos 2t) dt =
[
1
2π
(
t− 1
2sin 2t
)]π
0
=1
2
an =1
π
∫ π
0sin t cos ntdt =
1
π
∫ π
0
1
2[sin (1 + n) t + sin (1− n) t] dt
=
[
1
π
(
−1
2
cos (1 + n) t
1 + n− 1
2
cos (1− n) t
1− n
)]π
0
= − 1
π
cos πn + 1
n2 − 1= − 1
π
(−1)n + 1
n2 − 1
Hence, if n is even, i.e. n = 2m, a2m = − 1π
24m2−1
if n is odd an = 0
f (x) =1
π+
1
2sin t− 1
π
∞∑
m=1
2
4m2 − 1cos 2mt
Note:
sin mx sinnx = −1
2[cos (m + n)x− cos (m− n)x]
sinmx cos nx =1
2[sin (m + n)x + sin (m− n)x]
47
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
(9) (a)
a0 =2
π
∫ π
0x (π − x) dx =
π2
3
an =2
π
∫ π
0x (π − x) cos nxdx = −2
cos πn + 1
n2= −2
(−1)n + 1
n2
bn =2
π
∫ π
0x (π − x) sinnxdx = − 4
π
cos πn− 1
n3=
4
π
1− (−1)n
n3
The half range cosine series of f (x) is
f (x) =π2
6− 2
∞∑
n=1
(−1)n + 1
n2cos nx
=π2
3− 2
∞∑
m=1
(−1)2m + 1
4m2cos 2mx
=π2
6−
∞∑
m=1
1
m2cos 2mx
The half range sine series of f (x) is
f (x) =4
π
∞∑
n=1
1− (−1)n
n3sinnx =
8
π
∞∑
m=1
1
(2m− 1)3sin (2m− 1) x
(b)
a0 =4
π
∫ π/2
0cos xdx =
4
π
an =4
π
∫ π/2
0cos x cos 2nxdx = − 4
π
cos πn
4n2 − 1=
4
π
(−1)n−1
4n2 − 1
f (x) =2
π+
4
π
∞∑
m=1
(−1)n−1
4n2 − 1cos 2nx
(10) (a) Since |sinx| is an even function, all bn = 0.
a0 =1
π
∫ π
−π|sinx| dx =
1
π
∫ π
0sinxdx +
1
π
∫ 0
−π− sinxdx =
4
π
an =1
π
∫ π
−π|sinx| cos nxdx = 2
1
π
∫ π
0sinx cos nxdx = − 2
π
cos πn + 1
−1 + n2
=2
π
1 + (−1)n
n2 − 1
48
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
Hence,
f (x) =2
π− 4
π
∞∑
m=1
1
4m2 − 1cos 2mx
(b)
a0 =1
π
∫ π
−πexdx =
1
π
(
eπ − e−π)
an =1
π
∫ π
−πex cos nxdx
=1
π
[(
n2
n2 + 1
)(
ex sinnx
n− ex cos nx
n2
)]π
−π
=1
π
[
n2
n2 + 1
(
eπ cos nπ
n2− e−π cos nπ
n2
)
]
=1
π
[
(−1)n
n2 + 1
(
eπ − e−π)
]
bn =1
π
∫ π
−πex sinnxdx
=
[
ex sinnx− n2
n2 − 1
(
ex sinnx
n− ex cos nx
n2
)]π
0
= −n
π
[
(−1)n
n2 + 1
(
eπ − e−π)
]
f (x) =1
2π
(
eπ − e−π)
+∞∑
n=1
1
π
[
(−1)n
n2 + 1
(
eπ − e−π)
cos nx
]
+∞∑
n=1
n
π
[
(−1)n
n2 + 1
(
eπ − e−π)
sinnx
]
=1
π
(
eπ − e−π)
[
1
2+
∞∑
n=1
(−1)n
n2 + 1(cos nx− n sinnx)
]
=2 sinhπ
π
[
1
2+
∞∑
n=1
(−1)n
n2 + 1(cos nx− n sinnx)
]
∫
ex cos nxdx = ex sinnx
n−∫
ex sinnx
ndx
= ex sinnx
n+ ex cos nx
n2+
∫
(
−ex cos nx
n2
)
dx
49
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
Therefore,∫
ex cos nxdx =
(
n2
n2 − 1
)(
ex sinnx
n− ex cos nx
n2
)
∫
ex sinnx
ndx = ex sinnx
n−∫
ex cos nxdx
= ex sinnx−(
n
n2 − 1
)(
ex sin nx
n− ex cos nx
n2
)
Note:
sinhx =ex − e−x
2
(11)
a0 =1
π
∫ π
0(x− π)2 dx +
1
π
∫ 2π
ππ2dx =
1
3π2 + π2 =
4
3π2
an =1
π
∫ π
0(x− π)2 cos nxdx +
1
π
∫ 2π
ππ2 cos nxdx
=2
π
− sinπn + πn
n3+
[
πsin nx
n
]2π
π
=2
n2
bn =1
π
∫ π
0(x− π)2 sinnxdx +
1
π
∫ 2π
ππ2 sinnxdx
=1
π
2 cos πn− 2 + n2π2
n3−[
πcos nx
n
]2π
π
=2
π
(−1)n − 1
n3+
π
n− π
n+
(−1)n π
n
=2
π
(−1)n − 1
n3+
(−1)n π
n
f (x) =2
3π2 +
∞∑
n=1
2
n2cos nx +
∞∑
n=1
(
2
π
(−1)n − 1
n3+
(−1)n π
n
)
sinnx
50
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
Put x = 0
π2 =2
3π2 + 2
∞∑
n=1
1
n2
π2
3= 2
∞∑
n=1
1
n2
∞∑
n=1
1
n2=
π2
6
Put x = π,
π2
2=
1
2
(
f(
π−)+ f(
π+))
=2π2
3+ 2
∞∑
n=1
cos nπ
n2
−π2
6= 2
∞∑
n=1
(−1)n
n2
∞∑
n=1
(−1)n−1
n2=
π2
12
Note:
∫
(x− π)2 cos nxdx = (x− π)2sinnx
n−∫
(2x− 2π)sinnx
ndx
= (x− π)2sinnx
n+ (2x− 2π)
cos nx
n2−∫
(
−2cos nx
n2
)
dx
= (x− π)2sinnx
n+ (2x− 2π)
cos nx
n2+ 2
sin nx
n3∫
(x− π)2 sinnxdx = − (x− π)2cos nx
n+
∫
(
(2x− 2π)cos nx
n
)
dx
= − (x− π)2cos nx
n+ (2x− 2π)
sin nx
n2−∫
2sinnx
n2dx
= − (x− π)2cos nx
n+ (2x− 2π)
sin nx
n2− 2
cos nx
n3
(12) (a)
a0 =1
π
∫ 0
−π−1dx = −1
an =1
π
∫ 0
−π− cos nxdx = −sinπn
πn= 0
bn =1
π
∫ 0
−π− sinnxdx =
1− cos πn
πn=
1− (−1)n
πn
51
CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES
f (x) = −1
2+
∞∑
n=1
1− (−1)n
πnsinnx
(b) Since x2 is an even function, all bn = 0.
a0 =
∫ 1
−1x2dx =
2
3
an =
∫ 1
−1x2 cos nπxdx =
[
x2 sinnπx
nπ+ 2x
cos nπx
n2π2+ 2
sinnπx
n3π3
]1
−1
=(−1)n 4
n2π2
Therefore,
f (x) =1
3+
∞∑
n=1
4
n2π2(−1)n cos nπx
Note: Choose f = x2 and g′ = cos nπx,
∫
x2 cos nπxdx = x2 sinnπx
nπ−∫
2xsinnπx
nπdx.
Choose f = 2x and g′ = sinnπx,
∫
2xsinnπx
nπdx = −2x
cos nπx
n2π2−∫
(
−2cos nπx
n2π2
)
dx
= −2xcos nπx
n2π2− 2
sinnπx
n3π3
Hence,∫
x2 cos nπxdx = x2 sin nπx
nπ+ 2x
cos nπx
n2π2+ 2
sin nπx
n3π3
52
Chapter 4
Partial Differentiation
(1) f (x, y) = 2x−3yx+y
Suppose that (x, y) approaches (0, 0) along the x-axis. Then
limx→0y=0
f (x, y) = limx→0
2x
x= 2.
Suppose that (x, y) approaches (0, 0) along the y-axis. Then
limy→0x=0
f (x, y) = limx→0
−3y
y= −3.
Since the limits are not equal, then the limit does not exist.
(2) w =√
x2 + y2
∂w∂x =
∂(√
x2+y2)
∂x =(
12
) (
x2 + y2)− 1
2 (2x) = x√x2+y2
∂w∂y = y√
x2+y2
Using polar coordinates of the point (x, y), i.e., x = r cos θ and y = r sin θ,
limr→0
r=√
x2+y2
x√
x2 + y2= lim
r→0
r cos θ√
(r cos θ)2 + (r sin θ)2= cos θ
and
limr→0
r=√
x2+y2
y√
x2 + y2= lim
r→0
r sin θ√
(r cos θ)2 + (r sin θ)2= sin θ.
53
CHAPTER 4. PARTIAL DIFFERENTIATION
Since both of the limits are dependent on θ, then there is no unique value for the derivatives
for w at (0, 0). Therefore, the derivatives do not exist.
(3) By the product formula, we have
∂w
∂t= e−
x2
4t
[
−1
2t−
32 +
x2
4t−
52
]
=∂2w
∂x2.
(4) (a) ∂w∂x = ex cos y + 2x + y, ∂w
∂y = −ex sin y + x − 2y, ∂2w∂x2 = ex cos y + 2, ∂2w
∂y2 =
−ex cos y − 2y, ∂2w∂y∂x = −ex sin y + 1 = ∂2w
∂x∂y .
(b) ∂w∂x = 2x
x2+y2−z2 , ∂w∂y = 2y
x2+y2−z2 , ∂w∂z = −2z
x2+y2−z2 , ∂2w∂x2 = 2(y2−x2−z2)
(x2+y2−z2)2, ∂2w
∂y2 =
2(x2−y2−z2)
(x2+y2−z2)2, ∂2w
∂z2 = −2(x2+y2+z2)
(x2+y2−z2)2, ∂2w
∂x∂y = −4xy
(x2+y2−z2)2, ∂2w
∂z∂y = 4yz
(x2+y2−z2)2, ∂2w
∂z∂x =4zx
(x2+y2−z2)2.
(c)
∂w
∂x= e2x+y cos (x− y) + 2e2x+y sin (x− y) = e2x+y [cos (x− y) + 2 sin (x− y)]
∂w
∂y= −e2x+y cos (x− y) + e2x+y sin(x− y) = e2x+y [sin(x− y)− cos (x− y)]
∂2w
∂x2
= 2e2x+y [cos (x− y) + 2 sin (x− y)] + e2x+y [− sin (x− y) + 2 cos (x− y)]
= e2x+y [4 cos (x− y) + 3 sin (x− y)]
∂2w
∂y2= e2x+y [sin (x− y)− cos (x− y)] + e2x+y [− cos (x− y)− sin (x− y)]
= −2e2x+y cos (x− y)
∂2w
∂x∂y= 2e2x+y [sin (x− y)− cos (x− y)] + e2x+y [cos (x− y) + sin (x− y)]
= e2x+y [3 sin (x− y)− cos (x− y)]
∂2w
∂y∂x
= e2x+y [cos (x− y) + 2 sin (x− y)] + e2x+y [sin (x− y)− 2 cos (x− y)]
= e2x+y [3 sin (x− y)− cos (x− y)]
54
CHAPTER 4. PARTIAL DIFFERENTIATION
(d) ∂w∂x = 1√
x2+2y2
(
12
) (
x2 + 2y2)−1/2
(2x) = xx2+2y2
∂w∂y = 1√
x2+2y2
(
12
) (
x2 + 2y2)−1/2
(4y) = 2yx2+2y2
∂2w∂x2 =
(x2+2y2)−x(2x)
(x2+2y2)2= 2y2−x2
(x2+2y2)2
∂2w∂y2 =
2(x2+2y2)−2y(4y)
(x2+2y2)2= 2x2−4y2
(x2+2y2)2
∂2w∂x∂y = −2y(2x)
(x2+2y2)2= −4xy
(x2+2y2)2
∂2w∂y∂x = −x(4y)
(x2+2y2)2= −4xy
(x2+2y2)2
(e)
∂w
∂x= 2e2x−y2
sin(x + y) + e2x−y2
cos(x + y) = e2x−y2
[2 sin(x + y) + cos(x + y)]
∂w
∂y= −2ye2x−y2
sin(x + y) + e2x−y2
cos(x + y) = e2x−y2
[cos(x + y)− 2y sin(x + y)]
∂2w
∂x2
= 2e2x−y2
[2 sin(x + y) + cos(x + y)] + e2x−y2
[2 cos(x + y)− sin(x + y)]
= e2x−y2
[4 cos (x + y) + 3 sin (x + y)]
∂2w
∂y2= −2ye2x−y2
[cos(x + y)− 2y sin(x + y)]
+e2x−y2
[− sin(x + y)− 2 sin(x + y)− 2y cos (x + y)]
= e2x−y2 [−4y cos (x + y) +(
4y2 − 3)
sin (x + y)]
∂2w
∂x∂y
= 2e2x−y2
[cos(x + y)− 2y sin(x + y)] + e2x−y2
[− sin(x + y)− 2y cos(x + y)]
= e2x−y2
[(2− 2y) cos (x + y)− (4y + 1) sin (x + y)]
∂2w
∂y∂x
= −2ye2x−y2
[2 sin(x + y) + cos(x + y)] + e2x−y2
[2 cos(x + y)− sin(x + y)]
= e2x−y2
[(2− 2y) cos (x + y)− (4y + 1) sin (x + y)] =∂2w
∂x∂y
(f) ∂w∂x = yx ln y + yexy ln y
55
CHAPTER 4. PARTIAL DIFFERENTIATION
∂w∂y = xyx−1 + xexy ln y + 1
yexy
∂2w∂x2 = (ln y)2 yx + y2exy ln y
∂2w∂y2 = x (x− 1) yx−2 + x2exy ln y + 2x
y exy − 1y2 exy
∂2w∂x∂y = yx−1 + xyx−1 ln y + exy ln y + xyexy ln y + exy
∂2w∂y∂x = xyx−1 ln y + yx−1 + exy (ln y + 1) + xyexy ln y = ∂2w
∂x∂y
(5) (a) ∂w∂x = sin(2y − 3z2), ∂w
∂y = 2x cos(2y − 3z2), ∂w∂z = −6xz cos(2y − 3z2).
(b) ∂w∂x = 2Ax + Dy + Fz, ∂w
∂y = 2By + Dx + Ez, ∂w∂z = 2Cz + Ey + Fx.
(c) ∂w∂x = −(2y)−sin x)
(y+cos x)2= 2y sin x
(y+cos x)2, ∂w
∂y = 2(y+cos x)−(2y)(1)
(y+cos x)2= 2 cos x
(y+cos x)2.
(d) ∂w∂x = yex2+y2
+ 2x2yex2+y2
, ∂w∂y = xex2+y2
+ 2xy2ex2+y2
.
(e) ∂w∂x =
(x2+2y2)−(x−y)(2x)
(x2+2y2)2= x2+2y2−2x2+2xy
(x2+2y2)2= 2y2−x2+2xy
(x2+2y2)2,
∂w∂y =
(x2+2y2)(−1)−(x−y)(4y)
(x2+2y2)2= −x2−2y2−4xy+4y2
(x2+2y2)2= 2y2−x2−4xy
(x2+2y2)2.
(6) 1R = 1
R1+ 1
R2+ 1
R3= R2R3+R1R3+R1R2
R1R2R3
R = R1R2R3
R2R3+R1R3+R1R2
∂R∂R1
= (R2R3+R1R3+R1R2)(R2R3)−(R1R2R3)(R3+R2)
(R2R3+R1R3+R1R2)2= (R2R3)2
(R2R3+R1R3+R1R2)2,
∂R∂R2
= (R1R3)2
(R2R3+R1R3+R1R2)2, ∂R
∂R3= (R1R2)2
(R2R3+R1R3+R1R2)2.
(7) (a) a2 = c2 + b2 − 2bc cos A =⇒ cos A = c2+b2−a2
2bc or A = cos−1(
c2+b2−a2
2bc
)
∂∂a (cos A) = ∂
∂a
(
c2+b2−a2
2bc
)
=⇒ − sinA∂A∂a = − a
bc =⇒ ∂A∂a = a
bc sin A ,
∂A∂b = c2−a2−b2
2b2c sin A, ∂A
∂c = b2−a2−c2
2bc2 sin A.
(b) sin Aa = sin B
b = sin Cc =⇒ a = b sin A
sin B
∂a∂A = b cos A
sin B , ∂a∂B = −b sin A cos B
(sin B)2.
(8) (a) ∂w∂x = 2xy + 2y2 sinxy + y cos x, ∂w
∂y = x2 − 2 cos xy + 2xy sin xy + sinx,
∂2w∂x2 = 2y + 2y3 cos xy − y sinx, ∂2w
∂y2 = 4x sin xy − 2x2y cos xy,
∂2w∂x∂y = ∂
∂x
(
∂w∂y
)
= 2x + 4y sin xy + 2xy2 cos xy + cos x,
∂2w∂y∂x = ∂
∂y
(
∂w∂x
)
= 2x + 4y sinxy + 2xy2 cos xy + cos x.
(b) ∂w∂x = x
x2+y2 , ∂w∂y = y
x2+y2 , ∂2w∂x2 = y2−x2
(x2+y2)2, ∂2w
∂y2 = x2−y2
(x2+y2)2,
∂2w∂x∂y = ∂
∂x
(
∂w∂y
)
= −2xy
(x2+y2)2, ∂2w
∂y∂x = ∂∂y
(
∂w∂x
)
= −2xy
(x2+y2)2.
(c) ∂w∂x = yxy−1, ∂w
∂y = xy lnx, ∂2w∂x2 = y (y − 1) xy−2, ∂2w
∂y2 = (lnx)2 xy
∂2w∂x∂y = ∂
∂x
(
∂w∂y
)
= xy−1(1 + y lnx), ∂2w∂y∂x = ∂
∂y
(
∂w∂x
)
= xy−1(1 + y lnx).
56
CHAPTER 4. PARTIAL DIFFERENTIATION
(d) w = x sin2 y + exy = 12x(1− cos 2y) + exy
∂w∂x = sin2 y + yexy = 1
2(1− cos 2y) + yexy, ∂w∂y = x sin 2y + xexy,
∂2w∂x2 = y2exy, ∂2w
∂y2 = 2x cos 2y + x2exy,
∂2w∂x∂y = ∂
∂x
(
∂w∂y
)
= sin 2y + exy + xyexy,
∂2w∂y∂x = ∂
∂y
(
∂w∂x
)
= sin 2y + exy + xyexy.
(e) ∂w∂x =
(y2+sin x+1)(2y−2x)−(2xy−x2)(cos x)
(y2+sin x+1)2, ∂w
∂y =(y2+sin x+1)(2x)−(2xy−x2)(2y)
(y2+sin x+1)2,
∂2w∂x2 =
2 cos2 x(2xy−x2)(y2+sin x+1)3
+sin x(2xy−x2)+4(x−y) cos x
(y2+sin x+1)2− 2
y2+sin x+1,
∂2w∂y2 =
8y2(2xy−x2)(y2+sin x+1)3
+ 2x2−12xy
(y2+sin x+1)2,
∂2w∂x∂y = ∂
∂x
(
∂w∂y
)
=4y cos x(2xy−x2)
(y2+sin x+1)3− (4y2−4xy)+2x cos x
(y2+sin x+1)2+ 2
y2+sin x+1,
∂2w∂y∂x = ∂
∂y
(
∂w∂x
)
=4y cos x(2xy−x2)
(y2+sin x+1)3− (4y2−4xy)+2x cos x
(y2+sin x+1)2+ 2
y2+sin x+1.
(9) f (x, y, z) = cos(
2xy + z2)
fx = −2y sin(2xy + z2)
fxy = −2[
sin(2xy + z2) + 2xy cos(2xy + z2)]
= −2 sin(2xy + z2)− 4xy cos(2xy + z2)
fxyz = −4z cos(2xy + z2) + 8xyz sin(2xy + z2)
fz = −2z sin(2xy + z2)
fzz = −4z2 cos(2xy + z2)− 2 sin(2xy + z2)
fzzx = 8yz2 sin(2xy + z2)− 4y cos(2xy + z2)
(10) ∂w∂x = −x
(x2+y2+z2)32
, ∂2w∂x2 = 2x2−y2−z2
(x2+y2+z2)52
, ∂w∂y = −y
(x2+y2+z2)32
, ∂2w∂y2 = 2y2−x2−z2
(x2+y2+z2)52
,
∂w∂z = −z
(x2+y2+z2)32
, ∂2w∂z2 = 2z2−x2−y2
(x2+y2+z2)52
.
∂2w∂x2 + ∂2w
∂y2 + ∂2w∂z2 = 2x2−y2−z2
(x2+y2+z2)52
+ 2y2−x2−z2
(x2+y2+z2)52
+ 2z2−x2−y2
(x2+y2+z2)52
= 0
(x2+y2+z2)52
= 0
(11) (a) ∂w∂x = A (Ct + D), ∂2w
∂x2 = 0, ∂w∂t = C (Ax + B), ∂2w
∂t2= 0
∂2w∂t2− c2 ∂2w
∂x2 = (0)− c2 (0) = 0
57
CHAPTER 4. PARTIAL DIFFERENTIATION
(b)
∂w
∂x=(
Akekx −Bke−kx)(
Ceckt + De−ckt)
,
∂2w
∂x2=(
Ak2ekx + Bk2e−kx)(
Ceckt + De−ckt)
= k2(
Aekx + Be−kx)(
Ceckt + De−ckt)
,
∂w
∂t=(
Aekx + Be−kx)(
Cckeckt −Dcke−ckt)
,
∂2w
∂t2=(
Aekx + Be−kx)(
Cc2k2eckt + Dc2k2e−ckt)
= c2k2(
Aekx + Be−kx)(
Ceckt + De−ckt)
,
∂2w
∂t2− c2 ∂2w
∂x2= c2k2
(
Aekx + Be−kx)(
Ceckt + De−ckt)
−(
c2)
k2(
Aekx + Be−kx)(
Ceckt + De−ckt)
= 0
(c) ∂w∂x = g′ (x + ct), ∂2w
∂x2 = g′′ (x + ct), ∂w∂t = cg′ (x + ct), ∂2w
∂t2= c2g′′ (x + ct)
∂2w∂t2− c2 ∂2w
∂x2 =[
c2g′′ (x + ct)]
− c2 [g′′ (x + ct)] = 0
(12) ∂w∂x = 2x
x2+y2 , ∂w∂y = 2y
x2+y2 , ∂2w∂x2 =
2(y2−x2)(x2+y2)2
, ∂2w∂y2 =
2(x2−y2)(x2+y2)2
∂2w∂x2 + ∂2w
∂y2 =2(y2−x2)(x2+y2)2
+2(x2−y2)(x2+y2)2
= 0
(13) (a) fx (x, y) = (x+3y)(2)−(2x−y)(1)
(x+3y)2= 7y
(x+3y)2, fy (x, y) = −7x
(x+3y)2
L.H.S. = xfx (x, y) + yfy (x, y) = (x)[
7y
(x+3y)2
]
+ (y)[
−7x(x+3y)2
]
= 0 = R.H.S.
(b) fxx (x, y) = −14y
(x+3y)3, fyy (x, y) = 42x
(x+3y)3, fxy (x, y) = 7x−21y
(x+3y)3
L.H.S. = x2 −14y
(x+3y)3+ 2xy 7x−21y
(x+3y)3+ y2 42x
(x+3y)3= 0 = R.H.S.
or
∂∂x [xfx (x, y) + yfy (x, y)] = 0 =⇒ fx (x, y) + xfxx (x, y) + yfyx (x, y) = 0 ... (1)
∂∂y [xfx (x, y) + yfy (x, y)] = 0 =⇒ xfxy (x, y) + fy (x, y) + yfyy (x, y) = 0 ... (2)
x× (1) + y × (2) :
xfx (x, y) + x2fxx (x, y) + xyfyx (x, y) + xyfxy (x, y) + yfy (x, y) + y2fyy (x, y) = 0
[xfx (x, y) + yfy (x, y)] + x2fxx (x, y) + xyfyx (x, y) + xyfxy (x, y) + y2fyy (x, y) = 0
x2fxx (x, y) + xyfyx (x, y) + xyfxy (x, y) + y2fyy (x, y) = 0
58
CHAPTER 4. PARTIAL DIFFERENTIATION
(14) ∂F∂u = ∂f
∂x∂x∂u + ∂f
∂y∂y∂u = ∂f
∂x (2u) + ∂f∂y (2v) = 2u∂f
∂x + 2v ∂f∂y
∂F∂v = ∂f
∂x∂x∂v + ∂f
∂y∂y∂v = ∂f
∂x (−2v) + ∂f∂y (2u) = −2v ∂f
∂x + 2u∂f∂y
(
∂F
∂u
)2
+
(
∂F
∂v
)2
=
(
2u∂f
∂x+ 2v
∂f
∂y
)2
+
(
−2v∂f
∂x+ 2u
∂f
∂y
)2
= 4u2
(
∂f
∂x
)2
+ 4v2
(
∂f
∂y
)2
+ 4v2
(
∂f
∂x
)2
+ 4u2
(
∂f
∂y
)2
= 4(
u2 + v2)
[
(
∂f
∂x
)2
+
(
∂f
∂y
)2]
If u∂F∂u − v ∂F
∂v = 0, then u(
2u∂f∂x + 2v ∂f
∂y
)
− v(
−2v ∂f∂x + 2u∂f
∂y
)
= 0
=⇒ 2(
u2 + v2) ∂f
∂x = 0 =⇒ ∂f∂x = 0 or u = 0 or v = 0.
f (x, y) = g (y), which is dependent on y only. Therefore, f (x, y) is independent of x.
(15) Since ∂w∂x = y(y2−x2)
(x2+y2)2and ∂w
∂y = x(x2−y2)(x2+y2)2
, therefore x∂w∂x + y ∂w
∂y = 0.
Since
∂2w
∂x2=
2xy (x2 − 3y2)
(x2 + y2)3,
∂2w
∂y2=
2xy (y2 − 3x2)
(x2 + y2)3and
∂2w
∂x∂y=−x4 − y4 + 6x2y2
(x2 + y2)3,
therefore, x2 ∂2w∂x2 + 2xy ∂2w
∂x∂y + y2 ∂2w∂y2 = 0.
(16) Let w = g (u, v) with u = xy and v = z
y .
∂w∂x = ∂w
∂u∂u∂x + ∂w
∂v∂v∂x = gu (u, v)
(
1y
)
+ gv (u, v) (0) = 1ygu (u, v)
∂w∂y = ∂w
∂u∂u∂y + ∂w
∂v∂v∂y = gu (u, v)
(
− xy2
)
+ gv (u, v)(
− zy2
)
= − 1y2 [xgu (u, v) + zgv (u, v)]
∂w∂z = ∂w
∂u∂u∂z + ∂w
∂v∂v∂z = gu (u, v) (0) + gv (u, v)
(
1y
)
= 1ygv (u, v)
x∂w
∂x+ y
∂w
∂y+ z
∂w
∂z=
x
ygu (u, v)− 1
y[xgu (u, v) + zgv (u, v)] +
z
ygv (u, v) = 0
(17) ∂w∂x = ∂w
∂u∂u∂x + ∂w
∂v∂v∂x = ∂w
∂u (1) + ∂w∂v (1) = ∂w
∂u + ∂w∂v
∂w∂t = ∂w
∂u∂u∂t + ∂w
∂v∂v∂t = ∂w
∂u (c) + ∂w∂v (−c) = c∂w
∂u − c∂w∂v
∂2w
∂x2=
∂
∂x
∂w
∂u+
∂
∂x
∂w
∂v=
(
∂2w
∂u2
∂u
∂x+
∂w
∂v∂u
∂v
∂x
)
+
(
∂2w
∂u∂v
∂u
∂x+
∂2w
∂v2
∂v
∂x
)
=
(
∂2w
∂u2+
∂2w
∂v∂u
)
+
(
∂2w
∂v2+
∂2w
∂u∂v
)
=∂2w
∂u2+ 2
∂2w
∂u∂v+
∂2w
∂v2
59
CHAPTER 4. PARTIAL DIFFERENTIATION
∂2w
∂t2= c
∂
∂t
∂w
∂u− c
∂
∂t
∂w
∂v= c
(
∂2w
∂u2
∂u
∂t+
∂w
∂v∂u
∂v
∂t
)
− c
(
∂2w
∂u∂v
∂u
∂t+
∂2w
∂v2
∂v
∂t
)
= c
(
c∂2w
∂u2− c
∂2w
∂v∂u
)
− c
(
c∂2w
∂u∂v− c
∂2w
∂v2
)
= c2 ∂2w
∂u2− 2c2 ∂2w
∂u∂v+ c2 ∂2w
∂v2
If c2 ∂2w∂x2 = ∂2w
∂t2, then c2
(
∂2w∂u2 + 2 ∂2w
∂u∂v + ∂2w∂v2
)
= c2 ∂2w∂u2 − 2c2 ∂2w
∂u∂v + c2 ∂2w∂v2
=⇒ 4c2 ∂2w∂u∂v = 0 =⇒ ∂2w
∂u∂v = 0.
(18) ∂w∂x = dw
du∂u∂x = f ′ (u) (2x)
∂w∂y = dw
du∂u∂y = f ′ (u) (2y)
y ∂w∂x = 2xyf ′ (u)
x∂w∂y = 2xyf ′ (u) = y ∂w
∂x
(19) ∂w∂x = ∂w
∂u∂u∂x + ∂w
∂v∂v∂x = ∂w
∂u (1) + ∂w∂v (1) = ∂w
∂u + ∂w∂v
∂w∂y = ∂w
∂u∂u∂y + ∂w
∂v∂v∂y = ∂w
∂u (−1) + ∂w∂v (1) = −∂w
∂u + ∂w∂v
Thus,(
∂w∂x
)
(
∂w∂y
)
=(
∂w∂u + ∂w
∂v
) (
−∂w∂u + ∂w
∂v
)
= −(
∂w∂u
)2+(
∂w∂v
)2.
(20) ∂f∂x = ∂f
∂u × ∂u∂x + ∂f
∂v × ∂v∂x + ∂f
∂w × ∂w∂x = ∂f
∂u −∂f∂w
∂f∂y = ∂f
∂u × ∂u∂y + ∂f
∂v × ∂v∂y + ∂f
∂w × ∂w∂y = −∂f
∂u + ∂f∂v
∂f∂z = ∂f
∂u × ∂u∂z + ∂f
∂v × ∂v∂z + ∂f
∂w × ∂w∂z = −∂f
∂v + ∂f∂w
Thus, ∂f∂x + ∂f
∂y + ∂f∂z =
(
∂f∂u −
∂f∂w
)
+(
−∂f∂u + ∂f
∂v
)
+(
−∂f∂v + ∂f
∂w
)
= 0.
(21) ∂w∂x = g
(
xy
)
∂∂x
√xy +
√xy ∂
∂xg(
xy
)
= 12
√
yxg(
xy
)
+√
xy g′(
xy
)
∂w∂y = g
(
xy
)
∂∂y
√xy +
√xy ∂
∂yg(
xy
)
= 12
√
xy g(
xy
)
−√
x3
y3 g′(
xy
)
Thus,
x∂w
∂x+ y
∂w
∂y− w
= x
[
1
2
√
y
xg
(
x
y
)
+
√
x
yg′(
x
y
)]
+ y
[
1
2
√
x
yg
(
x
y
)
−√
x3
y3g′(
x
y
)
]
−[√
xyg
(
x
y
)]
= 0.
(22) With w = g(u) and u = x− ct, one has
∂w
∂x=
dw
du
∂u
∂x= g′(u),
60
CHAPTER 4. PARTIAL DIFFERENTIATION
∂w
∂t=
dw
du
∂u
∂t= −c g′(u),
∂2w
∂x2=
∂
∂x
(
∂w
∂x
)
=∂
∂x
(
g′(u))
=d
du
(
g′(u)) ∂u
∂x= g′′(u)
and∂2w
∂t2=
∂
∂t
(
∂w
∂t
)
=∂
∂t
(
−cg′(u))
= (−c)d
du
(
g′(u)) ∂u
∂t= (−c)2g′′(u).
We therefore conclude that∂2w
∂t2= c2 ∂2w
∂x2.
(23) With w = g(u) and u = x2 − t, one has
∂w
∂x=
dw
du
∂u
∂x= g′(u)2x
∂w
∂t=
dw
du
∂u
∂t= g′(u)(−1)
∂2w
∂x2
∂
∂x
(
∂w
∂x
)
= 2∂
∂x
(
xg′(u))
= 2g′(u) + 2x∂
∂x
(
g′(u))
= 2g′(u) + 4x2g′′(u)
and∂2w
∂t2=
∂
∂t
(
∂w
∂t
)
=∂
∂t
(
−g′(u))
= (−1)2g′′(u).
Hence the result.
(24) ∂w∂s = ∂w
∂x∂x∂s + ∂w
∂y∂y∂s = 4xy3s + 6x2y2s,
∂w∂t = ∂w
∂x∂x∂t + ∂w
∂y∂y∂t = 4xy3t− 6x2y2t.
(25) (a) f(x, y) =√
9x2 + y2
f ≃ ∂f∂x (x0,y0)
x + ∂f∂y (x0,y0)
y
∂f∂x =
(
12
) (
9x2 + y2)− 1
2 (9) (2x) = 9x√9x2+y2
∂f∂y =
(
12
) (
9x2 + y2)− 1
2 (2y) = y√9x2+y2
f(1.95, 8.1) ≃ f(2, 8) +f = f(2, 8) +∂f
∂x (x0,y0)x +
∂f
∂y (x0,y0)
y
= 10 + (1.8)(−0.05) + (0.8)(0.1) = 9.99
Note:√
9 (1.95)2 + (8.1)2 = 9.9916.
(b) Let z = f (x, y) = 13√
x2+y2. zx = −2x
3(x2+y2)4/3 and zy = −2y
3(x2+y2)4/3 .
dz = −2(x∆x+y∆y)
3(x2+y2)4/3 = −2[(2)(−0.1)+(2)(0.04)]
3[(2)2+(2)2]4/3 = 0.005
61
CHAPTER 4. PARTIAL DIFFERENTIATION
f (1.9, 2.04) ≃ f (2, 2) + dz = 12 + 0.005 = 0.505
Note: 13√
1.92+2.042= 0.50485.
(c) Let z = f (x, y) = 1
(x2+y2)3/2 , then ∂z∂x = −3x
(x2+y2)5/2 and ∂z∂y = −3y
(x2+y2)5/2 .
1
(4.102 + 2.952)3/2= f (4.10, 2.95) = f (4 + 0.1, 3− 0.05)
≃ f (4, 3) +∂z
∂x (4,3)∆x +
∂z
∂y (4,3)
∆y
=1
125+
(−12
3125
)
(0.1) +
( −9
3125
)
(−0.05) = 0.00776
Note: 1
(4.102+2.952)3/2 = 0.0077602.
(d) f (x, y, z) = 3√
x2 + y2 + z2
fx = 2x
3(x2+y2+z2)2/3 , fy = 2y
3(x2+y2+z2)2/3 , fz = 2z
3(x2+y2+z2)2/3
3√
4.982 + 1.052 + 0.962 = f (4.98, 1.05, 0.96) = f (5− 0.02, 1 + 0.05, 1− 0.04)
≃ f (5, 1, 1) +∂f
∂x (5,1,1)∆x +
∂f
∂y (5,1,1)
∆y +∂f
∂z (5,1,1)∆z
= 3 +2 [(5) (−0.02) + (1) (0.05) + (1) (−0.04)]
27= 2.9933
Note: 3√
4.982 + 1.052 + 0.962 = 2.9935.
(26) fx (x, y) = 2x + 2y − 4
fy (x, y) = 2x
f(0.99, 2.04) ≃ f(1, 2) +f = f(1, 2) + fx (x0, y0)x + fy (x0, y0)y
= (1) + (2) (−0.01) + (2) (0.04) = 1.06
Note: f(0.99, 2.04) = 1. 0593.
(27) (a)
∂w
∂ρ=
∂w
∂x
∂x
∂ρ+
∂w
∂y
∂y
∂ρ+
∂w
∂z
∂z
∂ρ
= (2x) (sinφ cos θ) + (2y) (sin φ sin θ) + (−2z) (cos φ)
= 2 (ρ sinφ cos θ) (sinφ cos θ) + 2 (ρ sinφ sin θ) (sinφ sin θ)− 2 (ρ cos φ) (cos φ)
= 2ρ sin2 φ cos2 θ + 2ρ sin2 φ sin2 θ − 2ρ cos2 φ = −2ρ(
cos2 φ− sin2 φ)
= −2ρ cos 2φ
62
CHAPTER 4. PARTIAL DIFFERENTIATION
(b)
∂w
∂θ=
∂w
∂x
∂x
∂θ+
∂w
∂y
∂y
∂θ+
∂w
∂z
∂z
∂θ
= (2x) (−ρ sinφ sin θ) + (2y) (ρ sinφ cos θ) + (−2z) (0)
= 2 (ρ sinφ cos θ) (−ρ sinφ sin θ) + 2 (ρ sinφ sin θ) (ρ sinφ cos θ)
= −2ρ2 sin2 φ sin θ cos θ + 2ρ2 sin2 φ sin θ cos θ = 0
(28) Let f (x, y) = 11+x−y , then fx = −1
(1+x−y)2and fx = 1
(1+x−y)2.
f (0, 0) = 1, fx (0, 0) = −1 and fy (0, 0) = 1
Using the Taylor’s formula,
f (x, y) ≃ f (x0, y0) + [(x− x0) fx (x0, y0) + (y − y0) fy (x0, y0)]
= 1 + [(x− 0) (−1) + (y − 0) (1)] = 1− x + y.
(29) ∆w ≃ ∂w∂x ∆x + ∂w
∂y ∆y + ∂w∂z ∆z. With w = x2ye3z, we have ∂w
∂x = −2, ∂w∂y = 1 and
∂w∂z = −3 at the point (1,−1, 0). When ∆x = 0.01, ∆y = −0.03 and ∆z = 0.02, the linear
approximation formula gives ∆w ≃ −0.11.
(30) Let f (x, y) =√
x2 + y2, then fx (x, y) = x√x2+y2
and fy (x, y) = y√x2+y2
.
f ≈ fx (x, y)x + fy (x, y)y = x(x)+y(y)√x2+y2
∣
∣
∣
ff
∣
∣
∣≈∣
∣
∣
∣
x(x)+y(y)√x2+y2
1√x2+y2
∣
∣
∣
∣
=∣
∣
∣
x(x)+y(y)x2+y2
∣
∣
∣= x2|x/x|+y2|y/y|
x2+y2 = (0.005)x2+(0.005)y2
x2+y2 =
0.005 = 0.5%
(31) T ≃ dT =∣
∣
∣
∂T∂l (l0,g0)
∣
∣
∣|l|+
∣
∣
∣
∂T∂g (l0,g0)
∣
∣
∣|g|
∂T∂l = 2π
√
1g
(
12√
l
)
= π√gl
, ∂T∂g = 2π
√l(
− 12g3/2
)
= −π√
lg3
T ≃ dT = π√(3.5)(9.79)
(0.05) +∣
∣
∣−π√
(3.5)
(9.79)3
∣
∣
∣(0.05) = 0.036428
(32) 1R = 1
R1+ 1
R2⇒ R = R1R2
R1+R2and
∆R ≃ ∂R
∂R1∆R1 +
∂R
∂R2∆R2 =
R22
(R1 + R2)2 ∆R1 +
R21
(R1 + R2)2 ∆R2.
When R1 = 3, ∆R1 = 0.2, R2 = 8 and ∆R2 = −0.5, we obtain
∆R ≃ 64
121× 0.2 +
9
121× (−0.5) = 0.0686.
63
CHAPTER 4. PARTIAL DIFFERENTIATION
(33) ∆A ∼= ∂A∂a ∆a + ∂A
∂b ∆b + ∂A∂θ ∆θ = b sin θ
2 ∆a + a sin θ2 ∆b + ab cos θ
2 ∆θ. Therefore, one has
|∆A| ≤ 200
2sin
π
3|∆a|+ 150
2sin
π
3|∆b|+ 150× 200
2cos
π
3|∆θ|
≤ 43.75√
3 + 7500× 1.5× π
180≃ 272.1
(34) Let S = f (x, y) = xx−y , then ∂S
∂x = −y
(x−y)2and ∂S
∂y = x(x−y)2
.
|∆S| ≃ |dS| =∣
∣fx(9,5)
∣
∣ |∆x|+∣
∣fy(9,5)
∣
∣ |∆y|
=
∣
∣
∣
∣
− 5
(9− 5)2
∣
∣
∣
∣
|0.05|+∣
∣
∣
∣
9
(9− 5)2
∣
∣
∣
∣
|0.1| = 0.071875 ≤ 0.072
(35) Let Q = f (P, R, V, L) = πPR4
5V L , then ∂Q∂P = πR4
5V L = QP , ∂Q
∂R = 4πPR3
5V L = 4QR , ∂Q
∂V =
− 1V
πPR4
5V L = −QV , and ∂Q
∂L = − 1L
πPR4
5V L = −QL
∣
∣
∣
∣
∆Q
Q
∣
∣
∣
∣
≃∣
∣
∣
∣
dQ
Q
∣
∣
∣
∣
=1
Q
[(
Q
P
)
∆P +
(
4Q
R
)
∆R +
(
−Q
V
)
∆V +
(
−Q
L
)
∆L
]
=
∣
∣
∣
∣
∆P
P
∣
∣
∣
∣
+ 4
∣
∣
∣
∣
∆R
R
∣
∣
∣
∣
+
∣
∣
∣
∣
−∆V
V
∣
∣
∣
∣
+
∣
∣
∣
∣
−∆L
L
∣
∣
∣
∣
= (0.5%) + 4 (0.25%) + (0.15%) + (0.3%) = 1.95%
(36) (a) fx (x, y) = −2x cos(
y2 − x2)
fx (1, 1) = −2
fy (x, y) = 2y cos(
y2 − x2)
fy (1, 1) = 2
(b)
f (x, y) ≈ f (1, 1) + [(x− 1) fx (1, 1) + (y − 1) fy (1, 1)]
= (0) + (x− 1) (−2) + (y − 1) (2) = −2x + 2y
f(
0, 12
)
≈ −2 (0) + 2(
12
)
= 1
(c) fxx (x, y) = −2 cos(
y2 − x2)
− 4x2 sin(
y2 − x2)
fxx (1, 1) = −2
fyy (x, y) = 2 cos(
y2 − x2)
− 4y2 sin(
y2 − x2)
fyy (1, 1) = 2
fxy (x, y) = 4xy sin(
y2 − x2)
fyx (x, y) = 4xy sin(
y2 − x2)
= fxy (x, y)
64
CHAPTER 4. PARTIAL DIFFERENTIATION
fxy (1, 1) = 0
f (x, y) ≈ f (1, 1) + [(x− 1) fx (1, 1) + (y − 1) fy (1, 1)]
+1
2!
[
(x− 1)2 fxx (1, 1) + 2 (x− 1) (y − 1) fxy (1, 1) + (y − 1)2 fyy (1, 1)]
= (0) + (x− 1) (−2) + (y − 1) (2) +1
2
[
(x− 1)2 (−2) + 0 + (y − 1)2 (2)]
= −2x + 2y − (x− 1)2 + (y − 1)2
f(
0, 12
)
≈ −2 (0) + 2(
12
)
− (0− 1)2 +(
12 − 1
)2= 1
4
Note: f(
0, 12
)
= sin(
14
)
= 0.2474.
(37) We calculate the partial derivatives of f(x, y) at (0, 1) as follows:
fx(0, 1) = 14 , fy(0, 1) = −1
4 , fxx(0, 1) = 14 , fyy(0, 1) = 1
4 , fxy(0, 1) = −14 .
Since f(0, 1) = 12 , Taylor’s Formula then yields
f(x, y) ∼= 1
2+
1
4x− 1
4(y − 1) +
1
2
[
1
4x2 − 1
2x(y − 1) +
1
4(y − 1)2
]
.
(38) (a) Solving the equations fx = 4x3 − 4y = 0 and fy = 4y3 − 4x = 0, one obtains
(0, 0), (1, 1) and (−1,−1) as the critical points of the function. Using the second
order derivative test with fxx = 12x2, fxy = −4 and fyy = 12y2, we conclude that
H = fxxfyy − f2xy = 144x2y2 − 16. At (0, 0), H = −16 < 0. We therefore conclude
that (0, 0) is a saddle point. At (1, 1) and (−1,−1), H = 128 > 0 and A = 12 > 0.
As such, both are relative minimum points.
(b) Solving the equations fx = 12x + 6y − 6x2 = 0 and fy = 6x + 6y = 0, one obtains
(0, 0) and (1,−1) as the critical points. Since fxx = 12− 12x, fxy = 6 and fyy = 6,
we conclude that H = fxxfyy − f2xy = 72(1 − x) − 36. At (0, 0), H = 36 > 0 and
fxx = 12 > 0. Thus (0, 0) is a relative minimum. At (1,−1), H = −36 < 0 and
therefore (1,−1) is a saddle point.
(c) f (x, y) = 9x3 − 4xy + 13y3
fx = 27x2 − 4y fy = −4x + y2 fxx = 54x fyy = 2y fxy = fyx = −4
fx = 0 =⇒ y = 274 x2
fy = 0 =⇒ x = 14y2
=⇒ (x, y) = (0, 0) or(
49 , 4
3
)
At (0, 0), H = (0) (0)− (−4)2 = −16 < 0 A = 0.
At(
49 , 4
3
)
, H = (24)(
83
)
− (−4)2 = 48 > 0 A = 24 > 0.
Thus, (0, 0) is a saddle point and(
49 , 4
3
)
is a local minimum.
65
CHAPTER 4. PARTIAL DIFFERENTIATION
(d) Solve
∂w∂x = 3x2 − 96y = 0
∂w∂y = 192y2 − 96x = 0
to obtain two critical points (0, 0) and (8, 2) for
the given function. As ∂2w∂x2 = 6x, ∂2w
∂y2 = 384y and ∂2w∂x∂y = −96, one obtains
A B C H Conclusion
(0, 0) 0 −96 0 −9216 saddle point
(8, 2) 48 −96 768 27648 relative minimum
(e) Solve
∂w∂x = 2x + 4 = 0
∂w∂y = −6y + 6y2 − 12 = 0
to obtain critical points (−2, 2) and (−2,−1)
for the given function. As ∂2w∂x2 = 2, ∂2w
∂y2 = 12y − 6 and ∂2w∂x∂y = 0, one obtains
A B C H Conclusion
(−2, 2) 2 0 18 36 relative minimum
(−2,−1) 2 0 −18 −36 saddle point
(f) fx = 3x2 − 12, fy = 3y2 − 3, fxx = 6x, fyy = 6y and fxy = fyx = 0.[
fx = 0 =⇒ x = ±2
fy = 0 =⇒ y = ±1
]
=⇒ (x, y) = (2, 1) or (2,−1) or (−2, 1) or (−2,−1)
At (2, 1), H = (12) (6)− (0)2 = 96 > 0 A = 12 > 0.
At (2,−1), H = (12) (−6)− (0)2 = −96 < 0 A = 12 > 0.
At (−2, 1), H = (−12) (6)− (0)2 = −96 < 0 A = −12 < 0.
At (−2,−1), H = (−12) (−6)− (0)2 = 96 > 0 A = −12 < 0.
Thus, (2,−1) and (−2, 1) are saddle points, (2, 1) is a local minimum, and (−2,−1)
is a local maximum.
(g) Solve
fx = 6xy − 6x = 0
fy = 3x2 + 3y2 − 6y = 0
to obtain four critical points (0, 0), (0, 2) and
(±1, 1) for the given function. As fxx = 6y − 6, fyy = 6y − 6 and fxy = fyx = 6x,
thenf H A Nature
(0, 0) 2 36 −6 relative maximum
(0, 2) −2 36 6 relative minimum
(±1, 1) 0 −36 saddle point
66
CHAPTER 4. PARTIAL DIFFERENTIATION
(h) fx = y − 64x2 , fy = x− 27
y2 , fxx = 128x3 , fyy = 54
y3 and fxy = fyx = 1
fx = 0 =⇒ y = 64x2
fy = 0 =⇒ x = 27y2
=⇒ (x, y) =(
163 , 9
4
)
At(
163 , 9
4
)
, H =(
2732
) (
12827
)
− (1)2 = 3 > 0 A = 2732 > 0.
Thus,(
163 , 9
4
)
is a local minimum.
(i) Since
fx = cos x− cos(x + y) = 0
fy = cos y − cos(x + y) = 0,
we conclude that cosx−cos y = 0, or sin(x−y
2
)
sin(x+y
2
)
= 0. Therefore, x−y = 2nπ
or x+y = 2mπ for some integers m, n. As x, y are required to satisfy 0 < x+y < 2π,
the second possibility is ruled out, and we have x = y + 2nπ for some integer n.
Substituting this into the equation cosx− cos(x+ y) = 0, one has cos y− cos 2y = 0,
or 2 cos2 y − cos y − 1 = 0. Solving this quadratic equation, one obtains cos y = −12
or cos y = 1.
If cos y = 1, then y = 2kπ and x = (2k+2n)π where n, k are integers. This however
implies that x + y = (4k + 2n)π, which is impossible because 0 < x + y < 2π.
If cos y = −12 , then y = 2π
3 + 2kπ or y = 4π3 + 2lπ for some integers k and l. Let us
consider the following two cases:
i. When y = 2π3 + 2kπ, one has x = 2π
3 + (2k + 2n)π and therefore x + y =4π3 +(4k +2n)π. The condition 0 < x+ y < 2π forces 4k +2n = 0, or n = −2k.
As such, the critical points are given by (2π3 − 2kπ, 2π
3 + 2kπ) for any integer k.
ii. When y = 4π3 + 2kπ, one has x = 4π
3 + (2k + 2n)π and therefore x + y =8π3 + (4k + 2n)π. Once again, the condition 0 < x + y < 2π ⇒ −2k − 4
3 < n <
−2k − 13 , or n = −2k − 1. We therefore conclude that the critical points are
(4π3 − 2kπ − 2π, 4π
3 + 2kπ) for any integer k.
A simple calculation shows that fxx = − sinx+sin(x+ y), fyy = − sin y +sin(x+ y)
and fxy = sin(x + y). At (2π3 − 2kπ, 2π
3 + 2kπ), fxx = fyy = −√
3 and fxy =
−√
32 . Therefore, those critical points are relative maximum. On the other hand, at
(4π3 −2kπ−2π, 4π
3 +2kπ), one has fxx = fyy =√
3 and fxy =√
32 . We thus conclude
that those critical points are relative minimum.
(39) The cost C is given by C = 9xy + 16z(x + y), with xyz = 36. With z = 36xy , we conclude
that the cost is given by C = 9xy + 576y + 576
x . By solving equations ∂C∂x = 0 and ∂C
∂y = 0,
one obtains x = 4 and y = 4 as the critical point of C (as a function of two variables x,
y), and second order derivative test confirms that (4, 4) is a relative minimum. As such,
the optimal dimensions of the box are 4m×4m×9m.
67
CHAPTER 4. PARTIAL DIFFERENTIATION
(40) Solving the equations fx = y + 2− 2x = 0 and fy = x− 1
y = 0, we obtain (12 , 2) is the only
critical point of the function f(x, y) = xy + 2x − lnx2y. Since fxx = 2x2 , fyy = 1
y2 and
fxy = 1, we have fxx(12 , 2) fyy(
12 , 2)−
[
fxy(12 , 2)
]2= 8× 1
4 − 1 = 3 > 0. The point is thus
a relative minimum.
(41) (a) R (2) =[
8, 2, 2√
6]
R′ (t) =[
3t2, 1,√
6t]
R′ (2) =[
12, 1, 2√
6]
Equation of the line tangent: x−812 = y−2
1 = z−2√
62√
6.
(b)
Arc length =
∫ 3
0
∣
∣R′ (t)∣
∣ dt =
∫ 3
0
√
(3t2)2 + (1)2 +(√
6t)2
dt
=
∫ 3
0
√
9t4 + 6t2 + 1 dt
=
∫ 3
0
(
3t2 + 1)
dt =[
t3 + t]3
0= 30
(42) (a) 2 (x− 1) + 3 (y − 2)− 1 (z − 1) = 0 =⇒ 2x + 3y − z = 7
(b) 3 (x + 3)− 2 (y + 1) + 2 (z − 1) = 0 =⇒ 3x− 2y + 2z = −5
(c) 3 (x− 2)− (y + 2)− 2 (z − 3) = 0 =⇒ 3x− y − 2z = 2
(43) (a) Since 22 + 2× 12 = 6, P0 = (2, 1, 6) is a point on the surface. Putting
f(x, y, z) = x2 + 2y2 − z,
one has ∇f = 2xi + 4yj − k. Thus ∇f(2, 1, 6) = 4i + 4j − k, and this is a vector
normal to the surface at P0. Finally, equation of the tangent plane is given by
4(x− 2) + 4(y − 1)− (z − 6) = 0,
or 4x + 4y − z = 6.
(b) Since z(0, 0, 1) = (0) cos 0 − (0) e0 + 1 = 1, P0 = (0, 0, 1) is a point on the surface.
Putting
f(x, y, z) = x cos y − yex + 1− z,
one has ∇f = (cos y − yex) i + (−x sin y − ex) j − k. Thus ∇f(0, 0, 1) = i− j − k,
and this is a vector normal to the surface at P0.
68
CHAPTER 4. PARTIAL DIFFERENTIATION
Finally, equation of the tangent plane is given by
(x− 0)− (y − 0)− (z − 1) = 0,
or x− y − z = 1.
(44) (a) x = 1 + 3t, y = 2− 2t.
(b) x = 1 + t, y = 2 + 3t, z = 3− 2t.
(c) x = −3 + t, y = 2− 2t, z = 2 + t.
(45) (a) Let f (x, y, z) = xy + yz + zx− 11, then ∇f (x, y, z) = 〈y + z, x + z, x + y〉 and
∇f (1, 2, 3) = 〈5, 4, 3〉 .
Equation of the tangent plane 5 (x− 1)+4 (y − 2)+3 (z − 3) = 0 i.e., 5x+4y+3z = 22
(b) Let f (x, y, z) = x3 + y3 + z3 − 18, then ∇f (x, y, z) =⟨
3x2, 3y2, 3z2⟩
and
∇f (−2, 3,−1) = 〈12, 27, 3〉 .
Equation of the tangent plane 12 (x + 2)+27 (y − 3)+3 (z + 1) = 0 i.e., 4x+9y+z =
18
(c) Let f (x, y, z) = zey sinx− 1, then ∇f (x, y, z) = 〈zey cos x, zey sin x, ey sinx〉 and
∇f(π
2, 0, 1
)
= 〈0, 1, 1〉 .
Equation of the tangent plane 0(
x− π2
)
+ (y − 0) + (z − 1) = 0 i.e., y + z = 1
(d) Let f (x, y, z) = x2 + y2 − xyz − 7, then ∇f (x, y, z) = 〈2x− yz, 2y − xz,−xy〉 and
∇f (−2, 3,−1) = 〈−1, 4, 6〉 .
Equation of the tangent plane (−1) (x + 2)+4 (y − 3)+6 (z + 1) = 0 i.e., −x+4y +
6z = 8
(46) Equation of the cone is x2 + y2 − z2 = 0. At P = (x0, y0, z0), the normal vector to the
cone is given by n = ∇(
x2 + y2 − z2)
at P= 2x0 i + 2y0 j− 2z0 k.
(a) Equation of the tangent plane at P is n • (r− r0) = 0, or x0(x− x0) + y0(y − y0) +
z0(z− z0) = 0. Since x20 + y2
0 − z20 = 0, the equation is satisfied when x = y = z = 0.
Therefore, the tangent plane passes through the origin.
69
CHAPTER 4. PARTIAL DIFFERENTIATION
(b) Equation of the normal line is r− r0 = tn, where t is any real parameter. This
vector equation also takes the form x = x0 + x0t, y = y0 + y0t, z = z0 + z0t. When
x = y = 0, we have t = −1. This implies z = 2z0. Hence the normal line intersects
the z−axis at (0, 0, 2z0).
(47) The normal vector to the ellipsoid at (1, 1, 3) is given by
n = ∇(
x2 + 2y2 + 3z2)
at (1,1,3)= 2i + 4 j+18k.
As such, equation of the tangent plane is 2(x−1)+4(y−1)+18(z−3) = 0, or x+2y+9z =
30.
(48) Tangent vector =(
drdt
)
at t= 14
=(
i + 12√
tj)
at t= 14
= i + j. When t = 14 , x = 1
4 and y = 12 .
Vector equation of the normal line is given by T • (r− r0) = 0, where T = i + j is the
tangent vector, r =xi + yj and r0=14 i + 1
2 j.
(49) (a) By (45) b., ∇f (−2, 3,−1) = 〈12, 27, 3〉, then the normal line to the surface at
(−2, 3,−1) is x+212 = y−3
27 = z+13 or x = −2 + 12t, y = 3 + 27t and z = −1 + 3t.
(b) By (45) c., ∇f(
π2 , 0, 1
)
= 〈0, 1, 1〉, then the normal line to the surface at(
π2 , 0, 1
)
isx−π/2
0 = y1 = z−1
1 or x = π2 , y = t and z = 1 + t.
(c) By (45) d., ∇f (−2, 3,−1) = 〈−1, 4, 6〉, then the normal line to the surface at
(−2, 3,−1) is x+2−1 = y−3
4 = z+16 or x = −2− t, y = 3 + 4t and z = −1 + 6t.
(50) (a) r =⟨
t2, 3t, 0⟩
, ∇r (x, y, z) = 〈2t, 3, 0〉 and ∇r (1, 3, 0) = 〈2, 3, 0〉.x = 1 + 2s, y = 3 + 3s and z = 0.
(b) r =⟨
2t2, 3t, t3⟩
, ∇r (x, y, z) =⟨
4t, 3, 3t2⟩
and ∇r (2, 3, 3) = 〈4, 3, 3〉.x = 4 + 4s, y = 3 + 3s and z = 3 + 3s.
(c) r = 〈cos t, sin t, t〉, ∇r (x, y, z) = 〈− sin t, cos t, 1〉 and ∇r(
12 ,
√3
2 , π3
)
=⟨
−√
32 , 1
2 , 1⟩
.
x = 12 −
√3
2 s, y =√
32 + 1
2s and z = π3 + s.
(51) (a) r =⟨
t,−1, e2t⟩
, ∇r (x, y, z) =⟨
1, 0, 2e2t⟩
and ‖v‖ = ‖∇r (x, y, z)‖ =√
1 + 4e4t.
(b) r =⟨√
92 t2, 3t, t3
⟩
, ∇r (x, y, z) =⟨√
18t, 3, 3t2⟩
and ‖v‖ =√
18t2 + 9 + 9t4 =
3(
t2 + 1)
.
(52) Arc length of a curve is given by the integral
∫ β
α
√
(
dx
dt
)2
+
(
dy
dt
)2
+
(
dz
dt
)2
dt.
70
CHAPTER 4. PARTIAL DIFFERENTIATION
(a) arc length =∫ π0
√
(− sin t)2 + (1 + cos t)2dt =∫ π0
√2 + 2 cos tdt = 4.
(b) arc length =∫ 40
√
(t)2 +(√
2t + 1)2
dt =∫ 40
√t2 + 2t + 1dt =
∫ 40 (t + 1)dt = 12.
(c) arc length =∫ 2π0
√
(−a sin t)2 + (a cos t)2 + k2dt = 2π√
a2 + k2.
(53) (a) ∇f (x, y) =⟨
2x, 6y2⟩
, ∇f (2,−1) = 〈4, 6〉 and b‖b‖ = 1√
5〈1,−2〉 =
⟨
1√5,− 2√
5
⟩
.
∇f (2,−1) • b‖b‖ = 〈4, 6〉 •
⟨
1√5,− 2√
5
⟩
= −8√5
(b) ∇f (x, y) = 〈ey cos x, ey sinx〉, ∇f(
π3 , 0)
=⟨
12 ,
√3
2
⟩
and b‖b‖ =
⟨
45 ,−3
5
⟩
.
∇f(
π3 , 0)
• b‖b‖ =
⟨
12 ,
√3
2
⟩
•⟨
45 ,−3
5
⟩
= 4−3√
310
(c) ∇f (x, y, z) = 〈yzexy, xzexy, exy〉, ∇f (2, 1,−1) =⟨
−e2,−2e2, e2⟩
and
b‖b‖ =
⟨
1√3, 1√
3, 1√
3
⟩
.
∇f (2, 1,−1) • b‖b‖ =
⟨
−e2,−2e2, e2⟩
•⟨
1√3, 1√
3, 1√
3
⟩
= − 2√3e2
(54) (a) ∇T (x, y) = 〈ex cos y − ey sinx,−ex sin y + ey cos x〉 and ∇T (0, 0) = i + j, then
‖∇T (x, y)‖ =√
(1)2 + (1)2 =√
2.
(b) −∇T (0, 0) = −i− j
(55) ∂T∂x = 2x and ∂T
∂y = −2y.
dxdt = ∂T
∂x = 2xdydt = ∂T
∂y = −2y=⇒
x (t) = C1e2t
y (t) = C2e−2t
With initial condition (−2, 1), x (t) = −2e2t and y (t) = e−2t.
(56) ∇f(1,−1, 0) = 2i + 2j− k. Therefore
Duf(1,−1, 0) = ∇f(1,−1, 0) • u
‖u‖ = (2i + 2j− k) • 2i− 3j + 6k√
22 + (−3)2 + 62= −8
7.
The direction along which f increases most rapidly at the point (1,−1, 0) is 2i + 2j− k,
and the direction along which f decreases most rapidly is −2i − 2j + k. The rate of
increase/decrease is given by
∇f(1,−1, 0) • ± (2i + 2j− k)
‖± (2i + 2j− k)‖ = ±3.
(57) To minimize f(x, y, z) = (x− 1)2 + y2 + (z + 1)2 subject to g = 2x− y + 5z − 3 = 0, we
71
CHAPTER 4. PARTIAL DIFFERENTIATION
use Lagrange Multiplier Method:
fx = λgx
fy = λgy
fz = λgz
g = 0
⇒
2(x− 1) = 2λ
2y = −λ
2(z + 1) = 5λ
2x− y + 5z = 3
From the first 3 equations, we conclude that x = λ + 1, y = −λ2 and
z = −1 + 5λ2 . Substitution into the last equation yields λ = 2
5 , which implies that the
minimum point is (75 ,−1
5 , 0). Hence the minimum distance is equal to
√
(7
5− 1)2 +
(
−1
5
)2
+ (0 + 1)2 =
√
6
5.
(58) Label the pentagon as ABCDE where ABE is an isoceles triangle with AB = AE = x,
BE = 2y and BC = ED = z. We wish to minimize P = 2x + 2y + 2z, subject to the
constraint g = y√
x2 − y2 + 2yz − 1000 = 0. Lagrange Multiplier Method yields the
equations
2 = λxy√x2−y2
2 = λ
(
√
x2 − y2 − y2√x2−y2
+ 2z
)
2 = λ2y
y√
x2 − y2 + 2yz = 1000
.
The third equation implies that λ = 1y . Substitution into the first and the second
equation yields y =√
3x2 , z =
(√
3+1)x
2 and√
x2 − y2 = x2 . These, together with the
fourth equation, give
x =
√
4000
6 + 3√
3, y =
√
1000
2 +√
3, z =
(
1 +√
3)
√
1000
6 + 3√
3.
(59) We minimize the function f(x, y) = (x − a)2 + y2, subject to the constraint g(x, y) =
y2 − 4bx = 0, where a > 0, b > 0.
Lagrange Multiplier method leads to
2(x− a) = λ (−4b)
2y = λ2y
y2 − 4bx = 0
.
There are two cases:
y = 0: The third equation yields x = 0 as well. Therefore, (0, 0) is the point on the
72
CHAPTER 4. PARTIAL DIFFERENTIATION
parabola which minimizes f(x, y). As such, the shortest distance from (a, 0) to the
parabola is given by√
(0− a)2 + 02 = a.
y 6= 0: The second equation implies λ = 1. Substituting λ = 1 into the first equation,
we obtain x = a − 2b. Therefore y = ±√
4b(a− 2b). To summarize, there are
two points on the parabola which minimize f(x, y), i.e., (a − 2b,√
4b(a− 2b)) and
(a − 2b, −√
4b(a− 2b)). The shortest distance from (a, 0) to the parabola is given
by 4b(a− b).
Note: You may prove that case (b) occurs if and only if a > 2b by considering the
intersection of the circle (x− a)2 + y2 = a2 and the parabola y2 = 4bx.
(60) We apply Lagrange Multiplier Method to the function T = 2x2 + 5y2 + 11z2 + 20xy −4xz + 16yz, subject to g = x2 + y2 + z2 − 1 = 0, to obtain the equations
4x + 20y − 4z = λ2x
20x + 10y + 16z = λ2y
−4x + 16y + 22z = λ2z
x2 + y2 + z2 = 1
The first three equations are equivalent to an eigenvalue problem
4 20 −4
20 10 16
−4 16 22
x
y
z
= λ
x
y
z
.
Solving the characteristic equation
∣
∣
∣
∣
∣
∣
∣
∣
4− λ 20 −4
20 10− λ 16
−4 16 22− λ
∣
∣
∣
∣
∣
∣
∣
∣
= 0, one obtains eigenvalues
λ = 9, −9 and 18. Corresponding eigenvectors are obtained as follows: when λ = −9,
x
y
z
=
2t
−2t
t
; when λ = 9,
x
y
z
=
2s
s
−2s
; when λ = 18,
x
y
z
=
r
2r
2r
. We
then substitute these solutions into the constraint and obtain t = ±13 , s = ±1
3 , r = ±13 .
Therefore, the critical points of the function T on the sphere are (23 , −2
3 , 13), (−2
3 , 23 , −1
3),
(23 , 1
3 , −23), (−2
3 , −13 , 2
3), (13 , 2
3 , 23) and (−1
3 , −23 , −2
3). Substituting these points into T ,
we conclude that (13 , 2
3 , 23) and (−1
3 , −23 , −2
3) are the hottest points on the sphere, with
maximum temperature = 18.
(61) We minimize H = R1i21 + R2i
22 + R3i
23 + R4i
24, subject to the constraint g = i1 + i2 + i3 +
i4 − I = 0. Lagrange Multiplier Method gives 2Rkik = λ for k = 1, 2, 3, 4, from which it
73
CHAPTER 4. PARTIAL DIFFERENTIATION
follows that ik = λ2Rk
for each k. Substitution into the constraint gives
λ
2=
I1
R1+ 1
R2+ 1
R3+ 1
R4
,
from which it follows that ik = RRk× I for k = 1, 2, 3, 4, where 1
R = 1R1
+ 1R2
+ 1R3
+ 1R4
.
(62) We maximize A = 4xy, subject to the constraint g = 9x2 + 4y2 − 36 = 0. Lagrange
Multiplier Method yields
4y = λ18x
4x = λ8y
9x2 + 4y2 = 36
. We may eliminate λ from the first
two equations to obtain 4y2 = 9x2. Substitution into the third equation ⇒ x =√
2
and y = 3√2. As a result, the largest rectangle that can be inscribed in the ellipse
has vertices at (√
2, 3√2), (−
√2, 3√
2), (−
√2,− 3√
2) and (
√2,− 3√
2). Its area is given by
A = 4×√
2× 3√2
= 12.
(63) We wish to minimize S = 2πrh + 2πr2, subject to g = πr2h − 1000 = 0. Lagrange
Multiplier Method gives
2πh + 4πr = λ2πrh
2πr = λπr2
πr2h = 1000
. The first two equations together yield
h = 2r. Substituting this into the last equation, we conclude that r = 3
√
500π ≃ 5.42 and
h = 2× 3
√
500π ≃ 10.84.
(64) Tx = 2x− 1, Ty = 4y, Txx = 2, Tyy = 4 and Txy = Tyx = 0.
Tx = 0 =⇒ x = 12 Ty = 0 =⇒ y = 0
H = fxx
(
12 , 0)
fyy
(
12 , 0)
−[
fxy
(
12 , 0)]2
= 8 > 0 A = fxx = 2 > 0
Thus, f (x, y) has a local minimum (coldest point) at(
12 , 0)
and f(
12 , 0)
= −14 .
It is evident that the maximum occurs at the boundary. Then we have to use the
Lagrange Multipliers method to find the maximum point.
Let f (x, y) = x2 − x + 2y2 and g (x, y) = x2 + y2 − 1.
2x− 1 = 2λx
4y = 2λy
x2 + y2 = 1
=⇒ (x, y, λ) =(
±1, 0, 12
)
or(
−12 ,±
√3
2 , 2)
For (±1, 0), f (1, 0) = 0 and f (−1, 0) = 2.
For(
−12 ,±
√3
2
)
, f(
−12 ,±
√3
2
)
= 2.25.
Thus, the maximum (hottest point) is at(
−12 ,±
√3
2
)
and f(
−12 ,±
√3
2
)
= 2.25.
74
CHAPTER 4. PARTIAL DIFFERENTIATION
(65) Let f (A, B, C) =∑4
i=1 (Axi + Byi + C − zi)2, then fA = 2
∑4i=1 (Axi + Byi + C − zi) xi,
fB = 2∑4
i=1 (Axi + Byi + C − zi) yi and fC = 2∑4
i=1 (Axi + Byi + C − zi).
fA = 0 =⇒ 2∑4
i=1 (Axi + Byi + C − zi)xi = 0 =⇒ 2A + B + 2C = 0
fB = 0 =⇒ 2∑4
i=1 (Axi + Byi + C − zi) yi = 0 =⇒ A + 2B + 2C + 2 = 0
fC = 0 =⇒ 2∑4
i=1 (Axi + Byi + C − zi) = 0 =⇒ 2A + 2B + 4C + 1 = 0
Solving the system of linear equations, A = 12 , B = −3
2 and C = 14 .
Thus, the ”fitted” plane is z = 12x− 3
2y − 14 .
(66) fx = −2x + 2, fy = −2y + 2, fxx = −2, fyy = −2 and fxy = fyx = 0.
fx = 0
fx = 0=⇒
x = 1
y = 1
H = fxx (1, 1) fyy (1, 1)− [fxy (1, 1)]2 = 4 > 0 A = fxx = −2 < 0
Thus, f (x, y) has a maxima at (1, 1) and f (1, 1) = 4.
(67) Let f (x, y, z) = (x− 1)2 + (y − 1)2 + (z + 2)2 and g (x, y, z) = x − 2y + 5z − 4, then
fx = 2 (x− 1), fy = 2 (y − 1), fz = 2 (z + 2), gx = 1, gy = −2 and gz = 5.
fx = λgx
fy = λgy
fz = λgz
x− 2y + 5z = 4
=⇒
2 (x− 1) = λ
2 (y − 1) = −2λ
2 (z + 2) = 5λ
x− 2y + 5z = 4
=⇒
x = 12λ + 1
y = −λ + 1
z = 52λ− 2
x− 2y + 5z = 4
=⇒(
12λ + 1
)
− 2 (−λ + 1) + 5(
52λ− 2
)
= 15λ− 11 = 4 =⇒ λ = 1
Thus, x = 32 , y = 0, z = 1
2 and f (x, y, z) = 152 . The minimum distance is
√302 .
(68) Let f (x, y, z) = x2 + y2 + z2 and g (x, y, z) = x2 + y2 − z2 − 1, then fx = 2x, fy = 2y,
fz = 2z, gx = 2x, gy = 2y and gz = −2z.
fx = λgx
fy = λgy
fz = λgz
x2 + y2 − z2 = 1
=⇒
2x = 2λx
2y = 2λy
2z = −2λz
x2 + y2 − z2 = 1
=⇒
x (λ− 1) = 0
y (λ− 1) = 0
z (λ + 1) = 0
x2 + y2 − z2 = 1
While λ 6= 0, z = 0 and x2 + y2 = 1. If x = 0, then y = ±1. If y = 0, then x = ±1.
Thus, f (0,±1, 0) = 1 and f (±1, 0, 0) = 1. The minimum distance is 1.
(69) Let f (x, y, z) = xyz and g (x, y, z) = x + y + z2 − 16, then fx = yz, fy = xz, fz = xy,
gx = 1, gy = 1 and gz = 2z.
75
CHAPTER 4. PARTIAL DIFFERENTIATION
fx = λgx
fy = λgy
fz = λgz
x + y + z2 = 16
=⇒
yz = λ ... (1)
xz = λ ... (2)
xy = 2λz ... (3)
x + y + z2 = 16 ... (4)
From (1) and (2), yz = xz =⇒ z (y − x) = 0 =⇒ y = x as z 6= 0.
Again, from (1) and (2), x = y = λz ... (5).
From (3) and (5),(
λz
) (
λz
)
= 2λz =⇒ λ = 2z3 ... (6) as λ 6= 0.
For (4) ,(
λz
)
+(
λz
)
+ z2 = 16 =⇒ 2λz + z2 = 16 =⇒ 4z2 + z2 = 16 =⇒ z = 4√
5.
So, λ = 1285√
5and x = y = 128
5√
5/ 4√
5= 32
5 .
Thus, f (x, y, z) =(
325
) (
325
)
(
4√5
)
= 4096125
√5.
(70) Let f (x, y, z) = x − 2y + 5z and g (x, y, z) = x2 + y2 + z2 − 36, then fx = 1, fy = −2,
fz = 5, gx = 2x, gy = 2y and gz = 2z.
1 = 2λx
−2 = 2yλ
5 = 2λz
x2 + y2 + z2 = 36
=⇒ (x, y, z, λ) =
(√
65 ,−2
√
65 ,√
30,√
524
)
(
−√
65 , 2√
65 ,−√
30,√
524
)
f(√
65 ,−2
√
65 ,√
30)
=(√
65
)
− 2(
−2√
65
)
+ 5(√
30)
= 6√
30 = 32.863
f(
−√
65 , 2√
65 ,−√
30)
= −6√
30 = −32.863
Thus, the maximum value of f is 32.863 and the minimum value of f is −32.863.
(71) Let x, y and z be respectively the length, the width and the height of the box. We wish
to minimize f(x, y, z) = 2xy + 8[xy + 2xz + 2yz], subject to the constraint g(x, y, z) =
xyz − 36 = 0. Using the Lagrange Multiplier method, we solve the equations
10y + 16z = λyz
10x + 16z = λxz
16x + 16y = λxy
xyz = 36
to obtain x = y = 3
√
2885 ≃ 3.862 and z = 10
16x = 58
3
√
2885 ≃ 2.414.
Physical consideration indicates that this must be a minimum. Thus the cost for building
such a box is approximately $447.48.
76
CHAPTER 4. PARTIAL DIFFERENTIATION
(72) We wish to minimize f(x, y) = x2 +(y +2)2, subject to the constraint g(x, y) = x2−2x+
y2 = 0. Using the Lagrange Multiplier method, we simplify the equations
2x = λ(2x− 2)
2(y + 2) = λ2y
x2 − 2x + y2 = 0
to conclude that y = 2x− 2 and 5x2 − 10x + 4 = 0.
Thus x = 1 − 1√5, y = − 2√
5or x = 1 + 1√
5, y = 2√
5. For mininum, we have x = 1 − 1√
5,
y = − 2√5.
The shortest distance is given by
√
(
1− 1√5
)2+(
− 2√5
+ 2)2
=√
5− 1 ≃ 1.236.
(73) Using the Lagrange Multiplier method, we solve the equations
1 = 2λx
3 = 2λy
5 = 2λz
x2 + y2 + z2 = 1
to obtain two critical points (x, y, z, λ) = ±(
1√35
, 3√35
, 5√35
, 2√35
)
. Thus the maximum
value is
f
(
1√35
,3√35
,5√35
)
=√
35
and the minimum value is
f
(
− 1√35
,− 3√35
,− 5√35
)
= −√
35.
(74) We minimize f(x, y, z) = (x + 1)2 + (y − 2)2 + (z − 3)2, subject to
g(x, y, z) = x + 2y − 3z − 4 = 0
h(x, y, z) = 2x− y + 2z − 5 = 0.
77
CHAPTER 4. PARTIAL DIFFERENTIATION
By the Lagrange Multiplier method, we solve the equations
2(x + 1) = λ + 2µ
2(y − 2) = 2λ− µ
2(z − 3) = −3λ + 2µ
x + 2y − 3z = 4
2x− y + 2z = 5
to obtain x = 3715 , y = 49
15 , z = 2515 . Physical consideration indicates that this must be a
minimum. Therefore, the shortest distance is given by
√
(
37
15+ 1
)2
+
(
49
15− 2
)2
+
(
25
15− 3
)2
≃ 3.9243.
(75) Let f (x, y, z) = x2 + y2 + z2 and g (x, y, z) = xyz − 1, then fx = 2x, fy = 2y, fz = 2z,
gx = yz, gy = xz and gz = xy.
2x = yz
2y = xz
2z = xy
xyz = 1
=⇒ (x, y, z) =
(1, 1, 1)
(−1,−1, 1)
(−1, 1,−1)
(1,−1,−1)
f (1, 1, 1) = f (−1,−1, 1) = f (−1, 1,−1) = f (1,−1,−1) = 3
Therefore, the minimum distance is√
3 and the maximum distance is ∞.
(76) We differentiate the equation x2 − 2yw + w3 + 1 = 0 with respect to x to obtain
2x− 2y∂w
∂x+ 3w2 ∂w
∂x= 0, (4.0.1)
which implies ∂w∂x = −2x
3w2−2y. If we now differentiate (4.0.1) with respect to x, we obtain
2− 2y∂2w
∂x2+ 6w
(
∂w
∂x
)2
+ 3w2 ∂2w
∂x2= 0.
From this equation, one concludes that
∂2w
∂x2= −2 + 6w
(
∂w∂x
)2
3w2 − 2y.
(77) Let F (x, y, z) = x2 + y2 − z2 − 2x (y + z) = 0. Then Fx = 2x − 2y − 2z, Fy = 2y − 2x
and Fz = −2z − 2x.
78
CHAPTER 4. PARTIAL DIFFERENTIATION
Thus, ∂z∂x = −Fx
Fz= x−y−z
x+z and ∂z∂y = −Fy
Fz= y−x
x+z .
∂2z
∂y2=
(x + z) (1)− (y − x)∂z∂y
(x + z)2=
(x + z)− (y − x)y−xx+z
(x + z)2
=(x + z)2 − (y − x)2
(x + z)3=
(y + z) (2x− y + z)
(x + z)3.
∂2z
∂y∂x=
∂
∂y
(
∂z
∂x
)
=(x + z)
(
−1− ∂z∂y
)
− (x− y − z) ∂z∂y
(x + z)2
=(x + z)
(
−1− y−xx+z
)
− (x− y − z) y−xx+z
(x + z)2
=x2 − 3xy − 2xz + y2 − z2
(x + z)3.
∂2z
∂x∂y=
∂
∂x
(
∂z
∂y
)
=(x + z) (−1)− (y − x)
(
1 + ∂z∂x
)
(x + z)2
=(x + z) (−1)− (y − x)
(
1 + x−y−zx+z
)
(x + z)2
=(x + z)2 (−1)− (y − x) (2x− y)
(x + z)3
=x2 − 3xy − 2xz + y2 − z2
(x + z)3.
(78) Let F (x, y, w) = x + y + xyw −w3, then Fx = 1 + yw, Fy = 1 + xw and Fw = xy − 3w2.
∂w∂x = − Fx
Fw= − 1+yw
xy−3w2 and ∂w∂y = − Fy
Fw= − 1+xw
xy−3w2
∂2w
∂x2=
∂
∂x
(
1 + yw
3w2 − xy
)
=
(
3w2 − xy)
∂∂x (1 + yw)− (1 + yw) ∂
∂x
(
3w2 − xy)
(3w2 − xy)2
=
(
3w2 − xy) (
y ∂w∂x
)
− (1 + yw)(
6w ∂w∂x − y
)
(3w2 − xy)2
=
(
3w2 − xy)
(
y 1+yw3w2−xy
)
− (1 + yw)(
6w 1+yw3w2−xy
− y)
(3w2 − xy)2
=−2(
xy2 + 3w)
(wy + 1)
(3w2 − xy)3
79
CHAPTER 4. PARTIAL DIFFERENTIATION
∂2w
∂y2=
∂
∂y
(
1 + xw
3w2 − xy
)
=
(
3w2 − xy)
∂∂y (1 + xw)− (1 + xw) ∂
∂y
(
3w2 − xy)
(3w2 − xy)2
=
(
3w2 − xy)
(
x∂w∂y
)
− (1 + xw)(
6w ∂w∂y − x
)
(3w2 − xy)2
=
(
3w2 − xy)
(
x 1+xw3w2−xy
)
− (1 + xw)(
6w 1+xw3w2−xy
− x)
(3w2 − xy)2
=−2(
x2y + 3w)
(wx + 1)
(3w2 − xy)3
∂2w
∂x∂y=
∂
∂x
(
1 + xw
3w2 − xy
)
=
(
3w2 − xy)
∂∂x (1 + xw)− (1 + xw) ∂
∂x
(
3w2 − xy)
(3w2 − xy)2
=
(
3w2 − xy) (
w + x∂w∂x
)
− (1 + xw)(
6w ∂w∂x − y
)
(3w2 − xy)2
=
(
3w2 − xy)
(
w + x 1+yw3w2−xy
)
− (1 + xw)(
6w 1+yw3w2−xy
− y)
(3w2 − xy)2
=w(
3w2 − xy)2
+ x(
3w2 − xy)
(1 + yw)− 6w (1 + yw)2 − y(
3w2 − xy)
(1 + xw)
(3w2 − xy)3
∂2w
∂y∂x=
∂
∂y
(
1 + yw
3w2 − xy
)
=
(
3w2 − xy)
∂∂y (1 + yw)− (1 + yw) ∂
∂y
(
3w2 − xy)
(3w2 − xy)2
=
(
3w2 − xy)
(
w + y ∂w∂y
)
− (1 + yw)(
6w ∂w∂y − x
)
(3w2 − xy)2
=
(
3w2 − xy)
(
w + y 1+xw3w2−xy
)
− (1 + yw)(
6w 1+xw3w2−xy
− x)
(3w2 − xy)2
=w(
3w2 − xy)2
+ x(
3w2 − xy)
(1 + yw)− 6w (1 + yw)2 − y(
3w2 − xy)
(1 + xw)
(3w2 − xy)3
(79) Differentiating the equation with respect to x and y, we obtain
3w2 ∂w
∂x+ 4
∂w
∂x− 2x + y2 = 0 and 3w2 ∂w
∂y+ 4
∂w
∂y+ 2xy + 8 = 0.
Therefore, ∂w∂x = 2x−y2
3w2+4and ∂w
∂y = −2xy−83w2+4
. Solving
2x− y2 = 0
−2xy − 8 = 0, one obtains
80
CHAPTER 4. PARTIAL DIFFERENTIATION
(2,−2) as the only critical point of the function. At (2,−2), further differentiations yield
∂2w
∂x2=
2(
3w2 + 4)
−(
2x− y2)
(6w ∂w∂x )
(3w2 + 4)2,
∂2w
∂y2=
(−2x)(
3w2 + 4)
+ (2xy + 8) (6w ∂w∂x )
(3w2 + 4)2,
∂2w
∂x∂y=−2y
(
3w2 + 4)
−(
2x− y2)
(6w ∂w∂y )
(3w2 + 4)2.
Since ∂w∂x = 0 at (2,−2), conclude that ∂2w
∂x2 > 0 and ∂2w∂y2 < 0 at that point. Therefore,
∂2w∂x2
∂2w∂y2 −
(
∂2w∂x∂y
)2< 0 at (2,−2) and therefore the point (2,−2) is a saddle point.
(80) Differentiating the equation with respect to x and y, we obtain
1 + yw + xy∂w
∂x− ew ∂w
∂x= 0 and 1 + xw + xy
∂w
∂y− ew ∂w
∂y= 0.
This imply ∂w∂x = 1+yw
ew−xy and ∂w∂y = 1+xw
ew−xy . Further differentiation gives
∂2w
∂x2=
(ew − xy) y ∂w∂x − (1 + yw)
(
ew ∂w∂x − y
)
(ew − xy)2,
∂2w
∂y2=
x (ew − xy) ∂w∂y − (1 + xw)
(
ew ∂w∂y − x
)
(ew − xy)2,
∂2w
∂x∂y=
(ew − xy)(
w + y ∂w∂y
)
− (1 + yw)(
ew ∂w∂y − x
)
(ew − xy)2.
(81) Differentiating the equations with respective to x and y, we obtain
2u∂u∂x −4v ∂v
∂x = −2x− 5y
(v cos uv + 1− 6v)∂u∂x +(u cos uv − 3− 6u) ∂v
∂x = −ex cos y
2u∂u∂y −4v ∂v
∂y = 6y − 5x
(v cos uv + 1− 6v)∂u∂y +(u cos uv − 3− 6u)∂v
∂y = ex sin y.
We may now use Cramer’s rule to obtain ∂u∂x , ∂v
∂x , ∂u∂y and ∂v
∂y . For instance, we have
∂u
∂x=
1
D
∣
∣
∣
∣
∣
−2x− 5y −4v
−ex cos y u cos uv − 3− 6u
∣
∣
∣
∣
∣
81
CHAPTER 4. PARTIAL DIFFERENTIATION
and∂v
∂x=
1
D
∣
∣
∣
∣
∣
2u −2x− 5y
v cos uv + 1− 6v −ex cos y
∣
∣
∣
∣
∣
,
where
D =
∣
∣
∣
∣
∣
2u −4v
v cos uv + 1− 6v u cos uv − 3− 6u
∣
∣
∣
∣
∣
.
(82) (a) If T and Φ are defined as functions of P and V by the equations, we differentiate
the equations to obtain
∂T
∂P=
V
R,
∂Φ
∂P=
CV
P,
∂T
∂V=
P
Rand
∂Φ
∂V=
CP
V.
Therefore,∂T
∂P
∂Φ
∂V− ∂T
∂V
∂Φ
∂P=
CP − CV
R= 1.
(b) If P and V are defined as functions of T and Φ by the equations, we differentiate
the equations with respect to T and Φ to obtain
V ∂P∂T +P ∂V
∂T = RCVP
∂P∂T +CP
V∂V∂T = 0
and
V ∂P∂Φ +P ∂V
∂Φ = 0CVP
∂P∂Φ +CP
V∂V∂Φ = 1
.
Cramer’s rule then yields
∂P
∂T=
CP
V,
∂V
∂T= −CV
P,
∂P
∂Φ= −P
Rand
∂V
∂Φ=
V
R.
Therefore,∂P
∂T
∂V
∂Φ− ∂V
∂T
∂P
∂Φ= 1.
82
Chapter 5
Multiple Integrals
(1) (a)∫ 3
0
∫ 2
04xydxdy =
∫ 3
04y
[
x2
2
]2
0
dy =
∫ 3
08ydy =
[
4y2]3
0= 36
(b)∫ 4
2
∫ 3
14xydxdy =
∫ 4
24y
[
x2
2
]3
1
dy =
∫ 4
216ydy =
[
8y2]4
2= 96
(c)∫ 1
0
∫ 1
y4xydxdy =
∫ 1
04y
[
x2
2
]1
y
dy =
∫ 1
0
(
2y − 2y3)
dy =
[
y2 − 1
2y4
]1
0
=1
2
(d)
∫ 2
0
∫ x+1
x/24xydydx =
∫ 2
04x
[
y2
2
]x+1
x/2
dx =
∫ 2
04x
[
(x + 1)2
2− x2
8
]
dx
=1
2
∫ 2
0
(
3x3 + 8x2 + 4x)
dx =1
2
[
3
4x4 +
8
3x3 + 2x2
]2
0
=62
3
(e)
∫ 1
0
∫ 3√
y
y2
4xydxdy =
∫ 1
04y
[
x2
2
] 3√
y
y2
dy = 2
∫ 1
0
(
y5/3 − y5)
dy
= 2
[
3
8y8/3 − 1
6y6
]1
0
=5
12
(f)∫ 1
0
∫
√y
04xydxdy =
∫ 1
04y
[
x2
2
]
√y
0
dy =
∫ 1
02y2dy =
[
2
3y3
]1
0
=2
3
83
CHAPTER 5. MULTIPLE INTEGRALS
(2)
y = x2
y = x3=⇒ (x, y) = (0, 0) or (1, 1)
∫ 1
0
∫ x2
x3
(
x4 + y2)
dydx =
∫ 1
0
[
x4y +1
3y3
]x2
x3
dx =
∫ 1
0
(
4
3x6 − x7 − 1
3x9
)
dx
=
[
4
21x7 − 1
8x8 − 1
30x10
]1
0
=9
280= 0.032143
(3)
x = y2
y = x=⇒ (x, y) = (0, 0) or (1, 1)
∫ 1
0
∫ y
y2
√xydxdy =
∫ 1
0
√y
[
2
3x3/2
]y
y2
dy =
∫ 1
0
2
3
(
y2 − y7/2)
dx
=2
3
[
1
3y3 − 2
9y9/2
]1
0
=2
27
(4) The parabola and the line intersect each other at (−1,−3) and (4, 12).
Hence,
∫
Rx dA =
∫ 4
−1
[∫ 3x
x2−4x dy
]
dx =
∫ 4
−1x[
3x− x2 + 4]
dx
=
∫ 4
−1
[
−x3 + 3x2 + 4x]
dx =125
4.
(5) (a)∫∫
R2xydxdy = X + Y
X =
∫ 1
0
∫ x+2
02xydydx =
∫ 1
0x[
y2]x+2
0dx =
∫ 1
0
(
x3 + 4x2 + 4x)
dx =43
12
Y =
∫ 3
1
∫ 4−x
02xydydx =
∫ 3
1x[
y2]4−x
0dx =
∫ 3
1
(
x3 − 8x2 + 16x)
dx =44
3
Thus,∫ ∫
R2xydxdy =
43
12+
44
3=
73
4
(b)∫∫
R2xydxdy = X + Y
84
CHAPTER 5. MULTIPLE INTEGRALS
X =
∫ 2
1
∫ 6−2y
y2xydxdy =
∫ 2
1y[
x2]6−2y
ydy =
∫ 2
1
(
3y3 − 24y2 + 36y)
dy =37
4
Y =
∫ 1
0
∫ 4y
y2xydxdy =
∫ 1
0y[
x2]4y
ydy =
∫ 1
015y3dy =
15
4
Thus,∫ ∫
R2xydxdy =
37
4+
15
4= 13
(c)∫∫
R2xydxdy = X + Y
X =
∫ 3
1
∫ (9−y)/2
3(y−1)/22xydxdy =
∫ 3
1y[
x2](9−y)/2
3(y−1)/2dy =
1
4
∫ 3
1
(
72y − 8y3)
dy = 32
Y =
∫ 1
0
∫ 3y+1
1−y2xydxdy =
∫ 1
0y[
x2]3y+1
1−ydy =
∫ 1
0
(
8y3 + 8y2)
dy =14
3
Thus,∫ ∫
R2xydxdy = 32 +
14
3=
110
3
(d)∫∫
R2xydxdy = X + Y
X =
∫ 1
0
∫ x+2
2−2x2xydydx =
∫ 1
0x[
y2]x+2
2−2xdx =
∫ 1
0
(
12x2 − 3x3)
dx =13
4
Y =
∫ 3
1
∫ (7−x)/2
(x−1)/22xydydx =
∫ 3
1x[
y2](7−x)/2
(x−1)/2dx =
1
4
∫ 3
1
(
48x− 12x2)
dx = 22
Thus,∫∫
R2xydxdy =
13
4+ 22 =
101
4
(6)
y
x1
3
85
CHAPTER 5. MULTIPLE INTEGRALS
(a)∫ 3
0
∫ 1
0x2dxdy =
∫ 3
0
[
1
3x3
]1
0
dy =
[
1
3y
]1
0
= 1
(b)
∫ 3
0
∫ 1
0ex+ydxdy =
∫ 3
0
[
ex+y]1
0dy =
∫ 3
0
(
e1+y − ey)
dy
=[
e1+y − ey]3
0= e4 − e3 − e + 1
(c)∫ 3
0
∫ 1
0xy2dxdy =
∫ 3
0
[
1
2y2x2
]1
0
dy =
∫ 3
0
1
2y2dy =
[
1
6y3
]3
0
=9
2
(7)
y = 1
(a)
∫ 1
0
∫ 1
yx3ydxdy =
∫ 1
0
[
1
4x4y
]1
y
dy =
∫ 1
0
(
1
4y − 1
4y5
)
dy
=
[
1
8y2 − 1
24y6
]1
0
=1
12
(b)
∫ 1
0
∫ 1
yxy3dxdy =
∫ 1
0
[
1
2x2y3
]1
y
dy =
∫ 1
0
(
1
2y3 − 1
2y5
)
dy
=
[
1
8y4 − 1
12y6
]1
0
=1
24
(c)
∫ 1
0
∫ 1
yx2y2dxdy =
∫ 1
0
[
1
3x3y2
]1
y
dy =
∫ 1
0
(
1
3y2 − 1
3y5
)
dy
=
[
1
9y3 − 1
18y6
]1
0
=1
18
86
CHAPTER 5. MULTIPLE INTEGRALS
(8)
y
x
2
2
(a)
∫ π/2
0
∫ π/2
0sin (x + y) dxdy
=
∫ π/2
0[− cos (x + y)]
π/20 dy =
∫ π/2
0
(
cos y − cos(π
2+ y))
dy
=[
sin y − sin(π
2+ y)]π/2
0= 2
(b)
∫ π/2
0
∫ π/2
0cos (x + y) dxdy =
∫ π/2
0[sin (x + y)]
π/20 dy
=
∫ π/2
0
[
sin(π
2+ y)
− sin (y)]
dy =[
cos(π
2+ y)
− cos (y)]π/2
0= 0
(c)
∫ π/2
0
∫ π/2
0(1 + xy) dxdy =
∫ π/2
0
[
x +1
2x2y
]π/2
0
dy
=
∫ π/2
0
(
1
2π +
1
8π2y
)
dy =
[
1
2πy +
1
16π2y2
]π/2
0
=1
4π2 +
1
64π4
(9) R =
(x, y) | 3√
y ≤ x ≤ 2, 0 ≤ y ≤ 8
=
(x, y) |0 ≤ x ≤ 2, 0 ≤ y ≤ x3
∫ 8
0
(
∫ 2
3√
yex4
dx
)
dy =
∫ 2
0
(
∫ x3
0ex4
dy
)
dx =
∫ 2
0ex4
x3dx =1
4
(
e16 − 1)
(10) R =
(x, y) |0 ≤ x ≤ 1, 1 ≤ y ≤ x2
=
(x, y) |0 ≤ x ≤ √y, 0 ≤ y ≤ 1
∫ 1
0
(∫ 1
x2
sin y√y
dy
)
dx =
∫ 1
0
(
∫ +√
y
0
sin y√y
dx
)
dy =
∫ 1
0
sin y√y
√ydy = 1− cos 1 = 0.4597
87
CHAPTER 5. MULTIPLE INTEGRALS
(11) (a)
∫ 1
−1
∫
√1−y2
−√
1−y2
(
x2 + 3y3)
dxdy =
∫ 1
−1
[
1
3x3 + 3y3x
]
√1−y2
−√
1−y2
dy
=2
3
∫ 1
−1
√
(1− y2)dy − 2
3
∫ 1
−1
√
(1− y2)y2dy + 6
∫ 1
−1y3√
(1− y2)dy
=2
3
∫ π/2
−π/2cos2 θdθ − 2
3
∫ π/2
−π/2sin2 θ cos2 θdθ + 6
∫ π/2
−π/2sin3 θ cos2 θdθ
=2
3
∫ π/2
−π/2
1 + cos 2θ
2dθ − 2
3
∫ π/2
−π/2
sin2 2θ
4dθ − 6
∫ π/2
−π/2
(
cos2 θ − cos4 θ)
d (cos θ)
=2
3
(
1
2π
)
− 2
3
∫ π/2
−π/2
1− cos 4θ
8dθ − 6
[(
cos3 θ
3− cos5 θ
5
)]π/2
−π/2
=1
3π − 2
3
(
1
8π
)
+ 0 =1
4π
(b)
∫ 1
0
∫ y
y2
√xydxdy =
∫ 1
0
[
2
3x3/2y1/2
]y
y2
dy =
∫ 1
0
2
3
(
y2 − y7/2)
dy
=
[
2
9y3 − 4
27y9/2
]1
0
=2
27
(c)
∫ 1
0
∫ y2
0yexdxdy =
∫ 1
0[yex]y
2
0 dy =
∫ 1
0
(
yey2 − y)
dy
=
[
1
2ey2 − 1
2y2
]1
0
=1
2e− 1
(d)
88
CHAPTER 5. MULTIPLE INTEGRALS
1 4
-3
-2
-1
1
2
3
32
y2 = 2x
y2 = 8 - 2x
y
x
∫ 2
−2
∫ (8−y2)/2
y2/2
(
4− y2)
dxdy =
∫ 2
−2
[(
4− y2)
x](8−y)2/2
y2/2dy
=
∫ 2
−2
(
16− 8y2 + y4)
dy =
[
16y − 8
3y3 +
1
5y5
]2
−2
=512
15
(e)
∫ 1
0
∫ 3√
y
√y
(
x4 + y2)
dxdy =
∫ 1
0
[
1
5x5 + y2x
] 3√
y
√y
dy
=
∫ 1
0
(
1
5y5/3 + y7/3 − 6
5y5/2
)
dy =
[
3
40y8/3 +
3
10y10/3 − 12
35y7/2
]1
0
=9
280= 0.03214
(f)
y = x
y = -x
y
x
1.0
1.0
-1.0
-1.0
0.5
0.5-0.5
-0.5
∫ 1
0
∫ x
−x
(
3xy2 − y)
dydx +
∫ 0
−1
∫ −x
x
(
3xy2 − y)
dydx
=
∫ 1
0
[
3
2x2y2 − xy
]x
−x
dx =
∫ 1
02x4dx−
∫ 0
−12x4dx = 0
(g)
89
CHAPTER 5. MULTIPLE INTEGRALS
y
x
2y = x
20
∫ 2
0
∫ x/2
0ex2
dydx =
∫ 2
0
[
ex2
y]x/2
0dx =
∫ 2
0
1
2xex2
dx =
[
1
4ex2
]2
0
=1
4e4 − 1
4
(h)
1
0.5
-0.5
-1
0
0.5 1-0.5-1
y
x
∫ 1
0
∫ x3
x4
(x + y) dydx +
∫ 0
−1
∫ x4
x3
(x + y) dydx
=
∫ 1
0
[
1
2x2 + yx
]x3
x4
dx +
∫ 0
−1
[
1
2x2 + yx
]x4
x3
dx
=
∫ 1
0
(
x4 +1
2x6 − x5 − 1
2x8
)
dx +
∫ 0
−1
(
x5 +1
2x8 − x4 − 1
2x6
)
dx
=
[
1
5x5 +
1
14x7 − 1
6x6 − 1
18x9
]1
0
+
[
1
6x6 +
1
18x9 − 1
5x5 − 1
14x7
]0
−1
=31
630− 241
630= −1
3
(i)
y
x
2y = x
1
0
90
CHAPTER 5. MULTIPLE INTEGRALS
∫ 1
0
∫ 2y
0e−y2/2dxdy =
∫ 1
0
[
xe−y2/2]2y
0dy =
∫ 1
02ye−y2/2dy
=[
−2e−12y2]1
0= 2− 2e−
12
(12)
Boundaries: u = 0, u = 1, v = 0, v = 1
Jacobian:
∂ (x, y)
∂ (u, v)=
∣
∣
∣
∣
∣
∂x∂u
∂y∂u
∂x∂v
∂y∂v
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∂(u−uv)∂u
∂(uv)∂u
∂(u−uv)∂v
∂(uv)∂v
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
1− v v
−u u
∣
∣
∣
∣
∣
= u
∫ 1
0
∫ 1−x
0ey/(x+y)dydx =
∫ 1
0
∫ 1
0e
uvu
∂ (x, y)
∂ (u, v)dudv
=
∫ 1
0
∫ 1
0uevdvdu =
∫ 1
0[uev]10 du =
∫ 1
0(ue− u) du
=
[(
u2e
2− u2
2
)]1
0
=e− 1
2
(13)∫ 10 f (x) dx = A =⇒
∫ 10 f (y) dy = A
∫ 10
∫ 10 f (x) f (y) dxdy =
∫ 10 f (y)
[
∫ 10 f (x) dx
]
dy =∫ 10 Af (y) dy = A
∫ 10 f (y) dy = A2
∫ 10
∫ x0 f (x) f (y) dydx +
∫ 10
∫ 1x f (x) f (y) dydx =
∫ 10
∫ 10 f (x) f (y) dxdy = A2
Let A = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ x and B = (x, y) |0 ≤ x ≤ 1, x ≤ y ≤ 1.
Consider the following transformation that transforms the space in A to B,
v (x, y) = x
u (x, y) = y.
Then J =
∣
∣
∣
∣
∣
∂u∂x
∂v∂x
∂u∂y
∂v∂y
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
0 1
1 0
∣
∣
∣
∣
∣
= −1.
91
CHAPTER 5. MULTIPLE INTEGRALS
∫ 1
0
∫ x
0f (x) f (y) dydx =
∫ 1
0
∫ 1
uf (u) f (v) dvdu =
∫ 1
0
∫ 1
xf (x) f (y) dydx
Thus,∫ 10
∫ x0 f (x) f (y) dydx +
∫ 10
∫ 1x f (x) f (y) dydx = 2
∫ 10
∫ 1x f (x) f (y) dydx = A2.
Therefore,∫ 10
∫ 1x f (x) f (y) dydx = A2
2 .
(14) Let u = xy and v = xy3, where 4 ≤ u ≤ 8 and 5 ≤ v ≤ 15.
Jacobian:
∂ (u, v)
∂ (x, y)=
∣
∣
∣
∣
∣
∂u∂x
∂v∂x
∂u∂y
∂v∂y
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∂(xy)∂x
∂(xy3)∂x
∂(xy)∂y
∂(xy3)∂y
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
y y3
x 3xy2
∣
∣
∣
∣
∣
= 2xy3 = 2v
∫∫
Ddxdy =
∫∫
D1
∂ (x, y)
∂ (u, v)dudv =
∫∫
D1
1∂(u,v)∂(x,y)
dudv =
∫ 8
4
∫ 15
5
1
2vdvdu
=
∫ 8
4
[
ln v
2
]15
5
du =
∫ 8
4
(
ln 15
2− ln 5
2
)
du =
∫ 8
4
(
1
2ln 3
)
du = 2 ln 3
(15) (a)
∫ ∫
Ω(x + y) dxdy
=
∫ π
0
∫ 2
0(r cos θ + r sin θ) rdrdθ =
∫ π
0(cos θ + sin θ) dθ
∫ 2
0r2dr
= [sin θ − cos θ]π0
[
1
3r3
]2
0
= (2)
(
8
3
)
=16
3
(b)
∫ ∫
Ω(x + y) dxdy
=
∫ π/2
−π/2
∫ 3
2(r cos θ + r sin θ) rdrdθ =
∫ π/2
−π/2(cos θ + sin θ) dθ
∫ 3
2r2dr
= [sin θ − cos θ]π/2−π/2
[
1
3r3
]3
2
= (2)
(
27− 8
3
)
=38
3
92
CHAPTER 5. MULTIPLE INTEGRALS
(c)
∫ ∫
Ω(x + y) dxdy
=
∫ π/2
−π/2
∫ 2 cos θ
0(r cos θ + r sin θ) rdrdθ
=
∫ π/2
−π/2(cos θ + sin θ)
∫ 2 cos θ
0r2drdθ =
∫ π/2
−π/2(cos θ + sin θ)
[
1
3r3
]2 cos θ
0
dθ
=8
3
∫ π/2
−π/2
(
cos4 θ + cos3 θ sin θ)
dθ =8
3
∫ π/2
−π/2cos4 θdθ +
8
3
∫ π/2
−π/2cos3 θ sin θdθ
=8
3(2)
∫ π/2
0cos4 θdθ +
8
3
∫ π/2
−π/2cos3 θ sin θdθ =
16
3
(
3π
16
)
+8
3
[
−1
4cos4 θ
]π/2
−π/2
= π + 0 = π
(d)
∫ ∫
Ω(x + y) dxdy
=
∫ π/2
0
∫3√
θ
0(r cos θ + r sin θ) rdrdθ =
∫ π/2
0(cos θ + sin θ)
∫3√
θ
0r2drdθ
=
∫ π/2
0(cos θ + sin θ)
[
1
3r3
]3√
θ
0
dθ =1
3
∫ π/2
0θ (cos θ + sin θ) dθ3
=1
3[cos θ + θ sin θ + sin θ − θ cos θ]
π/20 =
π
6
(16) Distance from (0, 0) to(
1,√
3)
=
√
(1)2 +(√
3)2
= 2
sec θ = 2 =⇒ θ = π3
∫ ∫
R
√
x2 + y2dxdy
=
∫ π/3
0
∫ sec θ
0(r) rdrdθ =
∫ π/3
0
[
1
3r3
]sec θ
0
dθ =1
3
∫ π/3
0sec3 θdθ
=
√3
3+
ln(
2 +√
3)
6= 0.79684
(17) R =
(x, y) | − 2 ≤ x ≤ 4, x2−62 ≤ y ≤ x + 1
∫ ∫
RxydA =
∫ 4
−2
∫ x+1
x2−62
xydydx =1
8
∫ 4
−2(−x5 + 16x3 + 8x2 − 32x)dx = 36
93
CHAPTER 5. MULTIPLE INTEGRALS
(18) Let x = r cos θ and y = r sin θ. Jacobian J = ∂(x,y)∂(r,θ) = r.
R = (r, θ) |0 ≤ r ≤ 4 sin θ, 0 ≤ θ ≤ π∫ π
0
∫ 4 sin θ
0r2 sin θdrdθ =
64
3
∫ π
0sin4 θdθ = 8π
(19) Let x = r cos θ and y = r sin θ. Jacobian J = ∂(x,y)∂(r,θ) = r.
R =
(r, θ) |2 cos θ ≤ r ≤ 4 cos θ,−π2 ≤ θ ≤ π
2
∫ π2
−π2
∫ 4 cos θ
2 cos θr2 cos θdrdθ =
56
3
∫ π2
−π2
cos4 θdθ = 7π
(20) A sketch will convince you that
∫ 2
0
(
∫
√4−y2
0
√
x2 + y2dx
)
dy =
∫
R
√
x2 + y2dxdy,
where R is the region given by (x, y) : x ≥ 0, y ≥ 0 and x2 + y2 ≤ 4.
The above integral is therefore equal to
∫ π2
0
(∫ 2
0r × r dr
)
dθ =4π
3.
(21) (a)
∫∫
x2+y2≤1cos(
x2 + y2)
dxdy =
∫ 2π
0
∫ 1
0
(
cos r2)
rdrdθ
=
∫ 2π
0
[
1
2sin r2
]1
0
dθ =
∫ 2π
0
(
1
2sin 1
)
dθ = π sin 1
(b)
∫∫
1≤x2+y2≤4cos(
x2 + y2)
dxdy =
∫ 2π
0
∫ 2
1
(
cos r2)
rdrdθ
=
∫ 2π
0
[
1
2sin r2
]2
1
dθ =
∫ 2π
0
(
1
2sin 4− 1
2sin 1
)
dθ = π sin 4− π sin 1
94
CHAPTER 5. MULTIPLE INTEGRALS
(22) (a)
∫∫
x2+y2≤1sin(
√
x2 + y2)
dxdy =
∫ 2π
0
∫ 1
0(sin r) rdrdθ
=
∫ 2π
0[sin r − r cos r]10 dθ =
∫ 2π
0(sin 1− cos 1) dθ = 2π sin 1− 2π cos 1
(b)
∫∫
1≤x2+y2≤4sin(
x2 + y2)
dxdy =
∫ 2π
0
∫ 2
1(sin r) rdrdθ
=
∫ 2π
0[sin r − r cos r]21 dθ =
∫ 2π
0(sin 2− 2 cos 2− sin 1 + cos 1) dθ
= 2π sin 2− 4π cos 2− 2π sin 1 + 2π cos 1
(23) (a)
x
y
1
1
∫∫
0≤x2+y2≤1(x + y) dxdy =
∫ π/2
0
∫ 1
0(r cos θ + r sin θ) rdrdθ
=
∫ π/2
0
∫ 1
0r2 (cos θ + sin θ) drdθ =
∫ π/2
0
[
1
3r3 (cos θ + sin θ)
]1
0
dθ
=
∫ π/2
0
1
3(cos θ + sin θ) dθ =
1
3[sin θ − cos θ]
π/20 =
2
3
(b)
2
2
1
1
y
x
95
CHAPTER 5. MULTIPLE INTEGRALS
∫∫
1≤x2+y2≤4(x + y) dxdy =
∫ π/2
0
∫ 2
1(r cos θ + r sin θ) rdrdθ
=
∫ π/2
0
∫ 2
1r2 (cos θ + sin θ) drdθ =
∫ π/2
0
[
1
3r3 (cos θ + sin θ)
]2
1
dθ
=
∫ π/2
0
(
8
3− 1
3
)
(cos θ + sin θ) dθ =7
3[sin θ − cos θ]
π/20 =
14
3
(24)
y
x1
y x= 3
∫∫
Ω
√
x2 + y2dxdy =
∫ π/3
0
∫ sec θ
0
√
(r cos θ)2 + (r sin θ)2rdrdθ
=
∫ π/3
0
∫ sec θ
0r2drdθ =
∫ π/3
0
[
1
3r3
]sec θ
0
dθ
=
∫ π/3
0
1
3sec3 θdθ =
1
3
∫ π/3
0sec θ
(
1 + tan2 θ)
dθ
=1
3
∫ π/3
0sec θdθ +
1
3
∫ π/3
0sec θ tan2 θdθ
=1
3[ln (sec θ + tan θ)]
π/30 +
1
3
∫ π/3
0tan θd (sec θ)
=1
3ln(
2 +√
3)
+1
3[tan θ sec θ]
π/30 −
∫ π/3
0sec3 θdθ
=1
6ln(
2 +√
3)
+1
3
√3
(25) (a)
1
1
1−
x
96
CHAPTER 5. MULTIPLE INTEGRALS
∫ 1
−1
∫
√1−y2
0
√
x2 + y2dxdy =
∫ π/2
−π/2
∫ 1
0r2drdθ =
∫ π/2
−π/2
[
1
3r3
]1
0
dθ
=
∫ π/2
−π/2
1
3dθ =
[
1
3θ
]π/2
−π/2
=1
3π
(b)
x
y
2
2
∫ 2
0
∫
√4−x2
0
√
x2 + y2dxdy =
∫ π/2
0
∫ 2
0r2drdθ =
∫ π/2
0
[
1
3r3
]2
0
dθ
=
∫ π/2
0
8
3dθ =
[
8
3θ
]π/2
0
=4
3π
(c)
x
y
Ω
3π
1
112
∫ 2
1/2
∫
√1−x2
0
3√
x2 + y2dxdy =
∫ π/3
0
∫ 1
12
sec θr4drdθ =
∫ π/3
0
[
1
5r5
]1
12
sec θ
dθ
=
∫ π/3
0
(
1
5− sec5 θ
160
)
dθ =1
15π − 11
640
√3− 3
1280ln(
2 +√
3)
Note: For n ≥ 2,
∫
secn θdθ =1
n− 1secn−2 θ tan θ +
n− 2
n− 1
∫
secn−2 θdθ.
97
CHAPTER 5. MULTIPLE INTEGRALS
(d)
x
y
Ω
3π
1
112
∫ 1/2
0
∫
√1−x2
0xy√
x2 + y2dxdy
=
∫ π/2
π/3
∫ 1
0r4 cos θ sin θdrdθ +
∫ π/3
0
∫ 12
sec θ
0r4 cos θ sin θdrdθ
=
∫ π/2
π/3
[
1
5r5 cos θ sin θ
]1
0
dθ +
∫ π/3
0
[
1
5r5 cos θ sin θ
] 12
sec θ
0
dθ
=1
5
∫ π/2
π/3cos θ sin θdθ +
∫ π/3
0
1
160sec5 θ cos θ sin θdθ
=1
5
∫ π/2
π/3
sin 2θ
2dθ −
∫ π/3
0
1
160sec4 θd (cos θ)
=1
5
[
−1
4cos 2θ
]π/2
π/3
− 1
160
[
−1
3cos−3 θ
]π/3
0
=1
40+
7
480=
19
480
(26)
∫∫
x2+y2≤b2(y + b) dxdy =
∫ 2π
0
∫ b
0(r sin θ + b) rdrdθ
=
∫ 2π
0
[
1
3(sin θ) r3 +
1
2br2
]b
0
dθ =
∫ 2π
0
(
1
3(sin θ) b3 +
1
2b3
)
dθ
=
[
−1
3(cos θ) b3 +
1
2b3θ
]2π
0
= πb3
(27)
∫∫
x2+y2≤1
[
1−(
x2 + y2)]
dxdy =
∫ 2π
0
∫ 1
0
(
1− r2)
rdrdθ
=
∫ 2π
0
[
−1
4r4 +
1
2r2
]1
0
dθ =
∫ 2π
0
1
4dθ =
[
1
4θ
]2π
0
=π
2
98
CHAPTER 5. MULTIPLE INTEGRALS
(28)
2
∫∫
x2+y2≤4
√
3
(
1− x2 + y2
4
)
dxdy = 2
∫ 2π
0
∫ 2
0
√
3
(
1− r2
4
)
rdrdθ
= 2
∫ 2π
0
[
− 1
18
(
12− 3r2)3/2
]2
0
dθ = 2
∫ 2π
0
(
2
3
√12
)
dθ
= 2
[(
2
3
√12
)
θ
]2π
0
=16
3π√
3
(29)
∫∫
x2+y2≤5
(
5− x2 − y2)1/6
dxdy =
∫ 2π
0
∫
√5
0
(
5− r2)1/6
rdrdθ
=
∫ 2π
0
[
−3
7
(
5− r2)7/6
]
√5
0
dθ =
∫ 2π
0
(
15
76√
5
)
dθ =
[
15
76√
5θ
]2π
0
=30
7π
6√
5 = 17.6
(30)
∫∫
x2+y2≤1
√
4− x2 − y2dA =
∫ 2π
0
∫ 1
0
√
4− r2rdrdθ
=
∫ 2π
0
[
−1
3
(
4− r2)3/2
]1
0
dθ =
∫ 2π
0
(
−√
3 +8
3
)
dθ
=
[
−√
3 +8
3
]2π
0
=16π
3− 2π
√3
(31)
x
Ω
99
CHAPTER 5. MULTIPLE INTEGRALS
∫∫
(
1− x2 − y2)
dA =
∫ π/2
−π/2
∫ cos θ
0
(
1− r2)
rdrdθ
=
∫ π/2
−π/2
[
−1
4r4 +
1
2r2
]cos θ
0
dθ =
∫ π/2
−π/2
(
−1
4cos4 θ +
1
2cos2 θ
)
dθ
=5
32π
Note:∫
cos2 θdθ =
∫
1 + cos 2θ
2dθ =
1
2θ +
1
4sin 2θ
and∫
cosn θdθ =1
n− 1cosn−1 θ sin θ +
n− 1
n
∫
cosn−2 θdθ
(32)
2
∫∫
2xdA =
∫ π/2
−π/2
∫ 2 cos θ
02r2 cos θdrdθ =
∫ π/2
−π/2
[
2
3r3 cos θ
]2 cos θ
0
dθ
=
∫ π/2
−π/2
(
16
3cos4 θ
)
dθ = 2π
(33)
∫∫
√
x2 + y2dA =
∫ π/2
−π/2
∫ 2a cos θ
0r2drdθ =
∫ π/2
−π/2
[
1
3r3
]2a cos θ
0
dθ
=
∫ π/2
−π/2
(
8
3a3 cos3 θ
)
dθ =
∫ π/2
−π/2
(
8
3a3(
1− sin2 θ)
)
d (sin θ)
=
[
8
3a3
(
sin θ − 1
3sin3 θ
)]π/2
−π/2
=32
9a3
(34)
100
CHAPTER 5. MULTIPLE INTEGRALS
∫∫
√
b2 − b2
a2(x2 + y2)dA = b
∫ π
0
∫ a sin θ
0
√
1− r2
a2rdrdθ
= b
∫ π
0
[
−1
3a2
(
1− r2
a2
)3/2]a sin θ
0
dθ =2a2b
3
∫ π/2
0
(
1− cos3 θ)
dθ
=2a2b
3
∫ π/2
0dθ − 2a2b
3
∫ π/2
0
(
1− sin2 θ)
d (sin θ)
=1
3a2bπ − 4
9a2b
(35)
x = - y
Ix =
∫∫
Ry2ρ (x, y) dA =
∫ 2
0
∫ y−y2
−yy2 (x + y) dxdy =
∫ 2
0
[(
1
2x2y2 + y3x
)]y−y2
−y
dy
=
∫ 2
0
(
−2y5 +1
2y6 + 2y4
)
dy =
[
−1
3y6 +
1
14y7 +
2
5y5
]2
0
=64
105
(36)
x = y
y = 2 - x
101
CHAPTER 5. MULTIPLE INTEGRALS
M =
∫∫
Rρ (x, y) dA =
∫ 1
0
∫ 2−y
y(6x + 3y + 3) dxdy
=
∫ 1
0
[
3x2 + 3xy + 3x]2−y
ydy
=
∫ 1
0
(
−6y2 − 12y + 18)
dx =[
−2y3 − 6y2 + 18y]1
0= 10
My =
∫∫
Rxρ (x, y) dA =
∫ 1
0
∫ 2−y
xy (6x + 3y + 3) dxdy
=
∫ 1
0
(
22− 24y + 6y2 − 4y3)
dx
=[
22y − 12y2 + 2y3 − y4]1
0= 11
Mx =
∫∫
Ryρ (x, y) dA =
∫ 1
0
∫ 2−y
yy (6x + 3y + 3) dxdy
=
∫ 1
0
(
18y − 12y2 − 6y3)
dx
=
[
9y2 − 4y3 − 3
2y4
]1
0
=7
2
Hence,
x =My
M=
11
10and y =
Mx
M=
7/2
10=
7
20
(37)
M =
∫∫
Rρ (x, y) dA =
∫ 1
0
∫ 6
0(x + y + 1) dxdy
=
∫ 1
0
[
1
2x2 + yx + x
]6
0
dy
=
∫ 1
0(24 + 6y) dy =
[
24y + 3y2]1
0= 27
Mx =
∫∫
Ryρ (x, y) dA =
∫ 1
0
∫ 6
0y (x + y + 1) dxdy
=
∫ 1
0
[
1
2x2y + y2x + xy
]6
0
dy
=
∫ 1
0
(
24y + 6y2)
dy =[
12y2 + 2y3]1
0= 14
102
CHAPTER 5. MULTIPLE INTEGRALS
My =
∫∫
Rxρ (x, y) dA =
∫ 1
0
∫ 6
0x (x + y + 1) dxdy
=
∫ 1
0
[
1
3x3 +
1
2(y + 1) x2
]6
0
dy
=
∫ 1
0(90 + 18y) dy =
[
90y + 9y2]1
0= 99
Hence,
x =My
M=
99
27=
11
3and y =
Mx
M=
14
27
Iy =
∫∫
Rx2ρ (x, y) dA =
∫ 1
0
∫ 6
0x2 (x + y + 1) dxdy
=
∫ 1
0
[
1
4x4 +
1
3(y + 1)x3
]6
0
dy =
∫ 1
0(396 + 72y) dy = 432
(38)
x = 4 - y
Let k be the constant density,
M =
∫∫
Rρ (x, y) dA =
∫ 2
0
∫ 4−y
y2/2kdxdy
= k
∫ 2
0
(
4− y − y2
2
)
dy = k
[
4y − 1
2y2 − 1
6y3
]2
0
=14k
3
My =
∫∫
Rxρ (x, y) dA =
∫ 2
0
∫ 4−y
y2/2kxdxdy
= k
∫ 2
0
[
x2
2
]4−y
y2/2
= −1
8k
∫ 2
0
(
−64 + 32y − 4y2 + y4)
dy
= −1
8k
[
−64y + 16y2 − 4
3y3 +
1
5y5
]2
0
=128k
15
103
CHAPTER 5. MULTIPLE INTEGRALS
Mx =
∫∫
Ryρ (x, y) dA =
∫ 2
0
∫ 4−y
y2/2ykdxdy = k
∫ 2
0
(
4y − y2 − y3
2
)
dy
= k
[
2y2 − 1
3y3 − 1
8y4
]2
0
=10k
3
Hence,
x =My
M=
128k/15
14k/3=
64
35and y =
Mx
M=
10k/3
14k/3=
5
7
(39)
a-a 0
a
Density = kr
M =
∫∫
Rρ (x, y) dA =
∫ π
0
∫ a
0kr · rdrdθ
=
∫ π
0dθ
∫ a
0kr2dr = π
[
kr3
3
]a
0
=1
3πka3
Mx =
∫∫
Ryρ (x, y) dA =
∫ π
0
∫ a
0kr (r sin θ) · rdrdθ
=
∫ π
0sin θdθ
∫ a
0kr3dr = [− cos θ]2π
0 ·[
kr4
4
]a
0
=1
2ka4
Hence,
y =Mx
M=
ka4
213πka3
=3a
2π
and by symmetry, x = 0.
(40) (a)
104
CHAPTER 5. MULTIPLE INTEGRALS
M =
∫∫
Rρ (x, y) dA =
∫ 1
0
∫ 1−y
0dxdy =
∫ 1
0(1− y) dy =
[
y − 1
2y2
]1
0
=1
2
Mx =
∫∫
Ryρ (x, y) dA =
∫ 1
0
∫ 1−y
0ydxdy =
∫ 1
0[yx]1−y
0 dy
=
∫ 1
0
(
y − y2)
dy =
[
1
2y2 − 1
3y3
]1
0
=1
6
My =
∫∫
Rxρ (x, y) dA =
∫ 1
0
∫ 1−y
0xdxdy
=
∫ 1
0
[
x2
2
]1−y
0
dy =
∫ 1
0
1
2(−1 + y)2 dy
=
[
1
6(−1 + y)3
]1
0
=1
6
Hence,
x =My
M=
1612
=1
3and y =
Mx
M=
1612
=1
3
(b)
a
a
0 a
M =
∫∫
Rρ (x, y) dA =
∫ π4
−π4
∫ a
0rdrdθ =
πa2
4
My =
∫∫
Rxρ (x, y) dA = 2
∫ π4
0
∫ a
0(r cos θ) rdrdθ =
1
3
√2a3
Hence,
x =My
M=
13
√2a3
πa2
4
=4√
2a
3π
105
CHAPTER 5. MULTIPLE INTEGRALS
and by symmetry, y = 0.
(41) (a)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 2
0
∫ 1
0
∫ 1
−1yz dx dy dz =
∫ 1
−1dx
∫ 1
0y dy
∫ 2
0z dz
= [x]1−1
[
1
2y2
]1
0
[
1
2z2
]2
0
= (2)
(
1
2
)
(2) = 2
(b)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 2
0
∫ 1
0
∫ 1
−1(x + y + z) dx dy dz =
∫ 2
0
∫ 1
0
[
1
2x2 + xy + xz
]1
−1
dy dz
=
∫ 2
0
∫ 1
0(2y + 2z) dy dz = 2
∫ 2
0
[(
1
2y2 + yz
)]1
0
dz = 2
∫ 2
0
(
1
2+ z
)
dz = 6
(42) (a)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0dz dy dx =
∫ 1
0
∫ 1−x
0(1− x− y) dy dx
=
∫ 1
0
[
y − xy − 1
2y2
]1−x
0
dx =
∫ 1
0
(
1
2− x +
1
2x2
)
dx =1
6
(b)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0xy dz dy dx =
∫ 1
0x
∫ 1−x
0
(
y − xy − y2)
dy dx
=
∫ 1
0x
[
1
2y2 − 1
2xy2 − 1
3y3
]1−x
0
dx =1
6
∫ 1
0
(
x− 3x2 + 3x3 − x4)
dx =1
120
106
CHAPTER 5. MULTIPLE INTEGRALS
(43) (a)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 1
−1
∫ 2
0
∫ 1−x2
0(x + y) dz dy dx =
∫ 1
−1
∫ 2
0(x + y)
(
1− x2)
dy dx
=
∫ 1
−1
∫ 2
0
(
x− x3 + y − x2y)
dy dx =
∫ 1
−1
(
2x− 2x3 + 2− 2x2)
dx =8
3
(b)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ 1
−1
∫ 2
0
∫ 1−x2
0y2 dz dy dx =
∫ 2
0y2 dy
∫ 1
−1
∫ 1−x2
0dz dx
=
[
1
3y3
]2
0
∫ 1
−1
(
1− x2)
dx =
(
8
3
)[
x− 1
3x3
]1
−1
=32
9
(44) (a)
∫∫∫
Ryz2dxdydz =
∫ 1
0
∫ 1
0
∫ 1
0yz2dzdydx =
∫ 1
0
∫ 1
0
[
1
3yz3
]1
0
dydx
=
∫ 1
0
∫ 1
0
1
3ydydx =
∫ 1
0
[
1
6y2
]1
0
dx
=
∫ 1
0
1
6dx =
1
6
(b)
107
CHAPTER 5. MULTIPLE INTEGRALS
∫∫∫
Rydxdydz =
∫ 1
0
∫ 1−y
0
∫ 1−x−y
0ydzdxdy =
∫ 1
0
∫ 1−y
0[yz]1−x−y
0 dxdy
=
∫ 1
0
∫ 1−y
0(y (1− y − x)) dxdy =
∫ 1
0
[
y
(
x− yx− 1
2x2
)]1−y
0
dy
=
∫ 1
0
(
1
2y − y2 +
1
2y3
)
dy
=
[
1
4y2 − 1
3y3 +
1
8y4
]1
0
=1
24
(c)
10.75
0.50.25
0
1
0.75
0.5
0.25
0
1
0.75
0.5
0.25
0
x y
z
x y
z
∫∫∫
Rxydxdydz =
∫ 1
0
∫ 1
0
∫ 1−y
0xydzdydx =
∫ 1
0xdx ·
∫ 1
0
∫ 1−y
0ydzdy
=
[
x2
2
]1
0
·∫ 1
0[yz]1−y
0 dy =1
2
∫ 1
0
(
y − y2)
dy
=1
2
[
1
2y2 − 1
3y3
]1
0
=1
12
(45) (a)
∫∫∫
D
(
x2 + y2)
dx dy dz =
∫ 2π
0
∫ 1
0
∫ r cos θ+r sin θ+2
0r3 dz dr dθ
=
∫ 2π
0
∫ 1
0
(
r4 cos θ + r4 sin θ + 2r3)
dr dθ
=
∫ 2π
0
[
1
5r5 cos θ +
1
5r5 sin θ +
1
2r4
]1
0
dθ
=
∫ 2π
0
(
1
5cos θ +
1
5sin θ +
1
2
)
dθ
=
[
1
5sin θ − 1
5cos θ +
1
2θ
]2π
0
= π
108
CHAPTER 5. MULTIPLE INTEGRALS
(b)
∫∫∫
Df (x, y, z) dx dy dz
=
∫ π
−π
∫ 1+cos θ
0
∫ r cos θ+r sin θ+2
0r3 dz dr dθ
=
∫ π
−π
∫ 1+cos θ
0
(
r4 cos θ + r4 sin θ + 2r3)
dr dθ
=
∫ π
−π
[
1
5r5 cos θ +
1
5r5 sin θ +
1
2r4
]1+cos θ
0
dθ
=
∫ π
−π
[
1
5(1 + cos θ)5 cos θ +
1
5(1 + cos θ)5 sin θ +
1
2(1 + cos θ)4
]
dθ
=1
5
(
105π
8
)
+1
5(0) +
1
2
(
35π
4
)
= 7π
(46) D =
(x, y, z) | 0 ≤ x2 + y2 + z2 ≤ 1
R = (ρ, φ, θ) | 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π∫∫∫
Dρ (x, y, z) dx dy dz
=
∫ ∫ ∫
Rρ (ρ sinφ cos θ, ρ sinφ sin θ, ρ cos φ) ρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π
0
∫ 1
0ρ4 sinφdρ dφ dθ =
∫ 2π
0dθ
∫ π
0sinφdφ
∫ 1
0ρ4 dρ
= [θ]2π0 [− cos φ]π0
[
1
5ρ5
]1
0
= (2π) (2)
(
1
5
)
=4
5π
(47) D =
(x, y, z) | 1 ≤ x2 + y2 + z2 ≤ 4
R = (ρ, φ, θ) | 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π∫∫∫
Dρ (x, y, z) dx dy dz
=
∫ 2π
0
∫ π
0
∫ 2
1ρ3 sinφdρ dφ dθ =
∫ 2π
0dθ
∫ π
0sinφdφ
∫ 2
1ρ3 dρ
= [θ]2π0 [− cos φ]π0
[
1
4ρ4
]2
1
= (2π) (2)
(
15
4
)
= 15π
(48)
z =√
x2 + y2
x2 + y2 + z2 = 1=⇒ x2 + y2 = 1
2
(The intersection of the surface is a cycle centred at the origin and radius is√
22 .)
109
CHAPTER 5. MULTIPLE INTEGRALS
D =
(x, y, z) | 0 ≤ x2 + y2 ≤ 12 ,√
x2 + y2 ≤ z ≤√
1− x2 − y2
R =
(ρ, φ, θ) | 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π4 , 0 ≤ θ ≤ 2π
∫∫∫
Dρ (x, y, z) dx dy dz
=
∫ 2π
0
∫ π/4
0
∫ 1
0ρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π/4
0sinφ
[
1
3ρ3
]1
0
dφ dθ =1
3
∫ 2π
0dθ
∫ π/4
0sinφdφ =
1
3[θ]2π
0 [− cos φ]π/40
=1
3(2π)
(
1−√
2
2
)
=
(
2−√
2)
π
3
(49) S = (r, θ, z) |0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4− r,
V =
∫ 2π
0
∫ 4
0
∫ 4−r
0r dz dr dθ =
∫ 2π
0
∫ 4
0
(
4r − r2)
dr dθ = (2π)
(
32− 64
3
)
=64
3π
(50)
R
h
y
z
x
(a) Let σ = kz.
Mass = M =
∫∫∫
TσdV =
∫ 2π
0
∫ R
0
∫ h
0rkzdzdrdθ
=
∫ 2π
0
∫ R
0
[
1
2rkz2
]h
0
drdθ =
∫ 2π
0
∫ R
0
∫
1
2rkh2drdθ
=
∫ 2π
0
[
1
4r2kh2
]R
0
dθ =
∫ 2π
0
1
4R2kh2dθ =
[
1
4R2kh2θ
]2π
0
=1
2kπR2h2
110
CHAPTER 5. MULTIPLE INTEGRALS
(b) By symmetry, x = y = 0
Mz =
∫∫∫
TzσdV =
∫ 2π
0
∫ R
0
∫ h
0rkz2dzdrdθ =
∫ 2π
0dθ
∫ R
0rdr
∫ h
0kz2dz
= (2π)×[
1
2r2
]R
0
×[
1
3kz3
]h
0
= (2π)× 1
2R2 × 1
3kh3 =
1
3πkR2h3
z =13πkR2h3
12kπR2h2
=2
3h
(c)
Iz =
∫∫∫
Tσ(
x2 + y2)
dV =
∫ 2π
0dθ
∫ R
0r3dr
∫ h
0kzdz = (2π)×
[
1
4r4
]R
0
×[
1
2kz2
]h
0
= (2π)× 1
4R4 × 1
2kh2 =
1
4πR4kh2 =
1
2MR2
(51) (a) Since σ = constant
M = σ
∫∫∫
TdV = σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
0dz = σ × (2π)×
[
1
2r2
]R
0
× [z]h0 = πσR2h
and
Iz = σ
∫∫∫
T
(
x2 + y2)
dV = σ
∫ 2π
0dθ
∫ R
0r3dr
∫ h
0dz = σ × (2π)×
[
1
4r4
]R
0
× [z]h0
= σ × (2π)× 1
4R4 × h =
1
2πσR4h =
M
2R2
(b)
Ix = σ
∫∫∫
T
(
y2 + z2)
dV = σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
0
(
r2 sin2 θ + z2)
dz
= σ
∫ 2π
0sin2 θdθ
∫ R
0r3dr
∫ h
0dz + σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
0z2dz
= σ
∫ 2π
0sin2 θdθ
∫ R
0r3dr
∫ h
0dz + σ × (2π)×
[
1
2r2
]R
0
×[
1
3z3
]h
0
= σ ×[
1
2θ − sin 2θ
4
]2π
0
×[
1
4r4
]R
0
× [z]h0 + σ × (2π)× 1
4R4 × h
=1
4σπR4h +
1
3σπR2h3 = M
(
R2
4+
h2
3
)
111
CHAPTER 5. MULTIPLE INTEGRALS
(c) By part (b), the moment of inertia is
M
(
R2
4+
h2
3
)
−M
(
h
2
)2
= M
(
R2
4+
h2
12
)
(52)
h
y
x
z
Rh
rz
=
(a)
V =
∫∫∫
TdV =
∫ 2π
0dθ
∫ R
0rdr
∫ h
hrR
dz =
∫ 2π
0dθ
∫ R
0r
(
h− hr
R
)
dr
= (2π)×[
1
2hr2 − 1
3hr3
]R
0
= πhR2 − 2
3πhR2 =
1
3πhR2
(b)
M = σV =1
3σπhR2
zM = σ
∫∫∫
TzdV = σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
hrR
zdz = σ
∫ 2π
0dθ
∫ R
0r
(
h2
2− h2r2
2R2
)
dr
= σ
∫ 2π
0
1
8h2R2dθ =
1
4σπh2R2
Hence,
z =3
4h
By symmetry, x = y = 0.
112
CHAPTER 5. MULTIPLE INTEGRALS
(c)
Iz = σ
∫∫∫
T
(
x2 + y2)
dV = σ
∫ 2π
0dθ
∫ R
0r3dr
∫ h
hrR
dz
= σ
∫ 2π
0dθ
∫ R
0r3
(
h− hr
R
)
dr
= σ
∫ 2π
0
1
20hR4dθ =
1
10σπhR4 =
3
10MR2
(d)
Ix = σ
∫∫∫
T
(
y2 + z2)
dV = σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
hrR
(
r2 sin2 θ + z2)
dz
= σ
∫ 2π
0sin2 θdθ
∫ R
0r3dr
∫ h
hrR
dz + σ
∫ 2π
0dθ
∫ R
0rdr
∫ h
hrR
z2dz
= σ
∫ 2π
0sin2 θdθ
∫ R
0r3
(
h− hr
R
)
dr + σ
∫ 2π
0dθ
∫ R
0
1
3r
(
h3 − h3r3
R3
)
dr
= σ
∫ 2π
0
1
20hR4 sin2 θdθ + σ
∫ 2π
0
1
10h3R2dθ =
1
20σπhR4 +
1
5σπR2h3
(53)
(a)
V =
∫∫∫
TdV =
∫ 2π
0dθ
∫ 1
0rdr
∫ 1−r2
0dz =
∫ 2π
0dθ
∫ 1
0r(
1− r2)
dr
=
∫ 2π
0
1
4dθ =
π
2
(b) Note that σ = kz
M =
∫∫∫
TσdV = k
∫∫∫
TzdV = k
∫ 2π
0dθ
∫ 1
0rdr
∫ 1−r2
0zdz
= k
∫ 2π
0dθ
∫ 1
0
1
2r(
1− r2)2
dr = k
∫ 2π
0
1
12dθ =
πk
6
113
CHAPTER 5. MULTIPLE INTEGRALS
(c) Note that σ = k(
r2 + z2)
M =
∫∫∫
TσdV = k
∫∫∫
T
(
r2 + z2)
dV
= k
∫ 2π
0dθ
∫ 1
0rdr
∫ 1−r2
0
(
r2 + z2)
dz
= k
∫ 2π
0dθ
∫ 1
0r
(
r2(
1− r2)
+1
3
(
1− r2)3)
dr
= k
∫ 2π
0
1
8dθ =
πk
4
(54)
V =
∫∫∫
dV =
∫ π/2
−π/2dθ
∫ 2a cos θ
0rdr
∫ r2/a
0dz =
∫ π/2
−π/2dθ
∫ 2a cos θ
0
r3
adr
=
∫ π/2
−π/24a3 cos4 θdθ =
3
2πa3
(55)
V =
∫∫∫
dV =
∫ π/2
−π/2dθ
∫ 2a cos θ
0rdr
∫ r
0dz =
∫ π/2
−π/2dθ
∫ 2a cos θ
0r2dr
=
∫ π/2
−π/2
8
3a3 cos3 θdθ =
32
9a3
(56)
V =
∫∫∫
dV =
∫ π/2
−π/2dθ
∫ 2 cos θ
0rdr
∫ 12(4+r cos θ)
0dz
=
∫ π/2
−π/2dθ
∫ 2 cos θ
0
1
2r (4 + r cos θ) dr
=
∫ π/2
−π/2
(
4
3cos4 θ + 4 cos2 θ
)
dθ =5
2π
(57)
V =
∫∫∫
dV =
∫ π/2
−π/2dθ
∫ a cos θ
0rdr
∫ a−r
0dz =
∫ π/2
−π/2dθ
∫ a cos θ
0r (a− r) dr
=
∫ π/2
−π/2
(
−1
3a3 cos3 θ +
1
2a3 cos2 θ
)
dθ =1
4πa3 − 4
9a3
114
CHAPTER 5. MULTIPLE INTEGRALS
(58)
z =
z = r + 1
52 - r2
Where the sphere x2 + y2 + z2 = 25 intersects the cone z =√
x2 + y2 + 1, we have
r2 + (r + 1)2 = 25, thus, r = 3.
V =
∫ 2π
0dθ
∫ 3
0rdr
∫
√25−r2
r+1dz
=
∫ 2π
0dθ
∫ 3
0r(√
25− r2 − (r + 1))
dr
=
∫ 2π
0
[
−1
3
(
25− r2)3/2 − 1
3r3 − 1
2r2
]3
0
dθ
=
∫ 2π
0
41
6dθ =
41
3π
(59)
x
x
y
1
z = x2 + y2
z = x
The surfaces intersect at x2 + y2 = x, thus r = cos θ.
V =
∫∫∫
dV =
∫ π/2
−π/2dθ
∫ cos θ
0rdr
∫ r cos θ
r2
dz
=
∫ π/2
−π/2dθ
∫ cos θ
0r(
r cos θ − r2)
dr
=
∫ π/2
−π/2
(
−1
4cos4 θ +
1
3cos4 θ
)
dθ =1
32π
115
CHAPTER 5. MULTIPLE INTEGRALS
(60)
cone
hyperbola
z
Where the hyperboloid intersects the cone, we have r2 + a2 = 2r2, thus r = a.
V =
∫ 2π
0dθ
∫ a
0rdr
∫
√a2+r2
√2r
dz =
∫ 2π
0dθ
∫ a
0r(√
a2 + r2 −√
2r)
dr
=
∫ 2π
0
[
1
3
(
a2 + r2)3/2 −
√2
3r3
]a
0
dθ =
∫ 2π
0
(
1
3a3√
2− 1
3a3
)
dθ
=2
3πa3√
2− 2
3πa3 =
2
3πa3
(√2− 1
)
(61)
y
x
z
21
Sphere: x2 + y2 + z2 = 1
Cone: z = 3(x2 + y2)
21
The cone z =√
3 (x2 + y2) has equation z =√
3r. It intersects the sphere z2 + r2 = 1 at
3r2 + r2 = 1 or r = 12 .
V =
∫ 2π
0dθ
∫ 12
0rdr
∫
√1−r2
√3r
dz =
∫ 2π
0dθ
∫ 12
0r(√
1− r2 −√
3r)
dr
=
∫ 2π
0
[
−1
3
(
1− r2)3/2 − 1
3r3√
3
] 12
0
dθ =
∫ 2π
0
(
−1
6
√3 +
1
3
)
dθ =2
3π − 1
3π√
3
116
CHAPTER 5. MULTIPLE INTEGRALS
(62)
V =
∫ 2π
0dθ
∫ 2
1rdr
∫
√
9− r2
4
0dz
=
∫ 2π
0dθ
∫ 2
1
1
2r√
(36− r2)dr =
∫ 2π
0
[
−1
6
(
36− r2)3/2
]2
1
dθ
=
∫ 2π
0
(
−16
3
√32 +
35
6
√35
)
dθ = −128
3π√
2 +35
3π√
35 = 27.273
(63) (a) r = 1. Cylinder of radius 1, with the z-axis as its axis.
φ = π2 . The xy-plane.
φ = 3π4 . The cone shown in diagram.
=34
y
x
z
θ = π4 . A plane prependicular to the xy-plane with the angle between this plane and
the xz-plane is π4 .
z = 1. The plane parallel to the xy-plane through (0, 0, 1) .
ρ = cos φ. The sphere with centre at (x, y, z) =(
0, 0, 12
)
and radius 12 .
(64)
V =
∫ 2π
0
∫ π
0
∫ R
0ρ2 sin φdρdφdθ =
∫ 2π
0
∫ π
0
1
3R3 sinφdφdθ =
∫ 2π
0
2
3R3dθ =
4
3πR3
(65) r = ρ sinφ, θ = θ, z = ρ cos φ.
(66)
R =
(x, y, z) | − 2 ≤ x ≤ 2,−√
4− x2 ≤ y ≤√
4− x2, 0 ≤ z ≤√
4− (x2 + y2)
=
(ρ, φ, θ) |0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π
2, 0 ≤ θ ≤ 2π
117
CHAPTER 5. MULTIPLE INTEGRALS
∫ 2
−2
∫
√4−x2
−√
4−x2
∫
√4−(x2+y2)
0z2√
x2 + y2 + z2 dzdydx
=
∫ 2π
0
∫ π/2
0
∫ 2
0
(
ρ3 cos2 φ) (
ρ2 sinφ)
dρdφdθ
=
∫ 2π
0
∫ π/2
0
[
1
6ρ6
]2
0
cos2 φ sinφdφdθ
=32
3
∫ 2π
0
∫ π/2
0cos2 φ sinφdφdθ
=32
3
∫ 2π
0
[
−1
3cos3 φ
]π/2
0
dθ
=32
9
∫ 2π
0dθ =
64π
9
(67) Substitute x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cos φ into z =√
3x2 + 3y2, then
ρ cos φ =√
3 (ρ sinφ cos θ)2 + 3 (ρ sinφ sin θ)2 =√
3ρ sinφ =⇒ tan φ = 1√3
=⇒ φ = π6 .
S =
(ρ, φ, θ) |0 ≤ ρ ≤ 4, 0 ≤ φ ≤ π6 , 0 ≤ θ ≤ 2π
V =∫ 2π0
∫ π/60
∫ 40 ρ2 sinφdρ dφ dθ = (2π)
(
1−√
32
)
(
643
)
= 643 π(
2−√
3)
= 17.958
(68)
∫ 3
0
∫
√9−z2
0
∫ x
0xy dydxdz =
∫ 3
0
∫
√9−z2
0x
[
y2
2
]x
0
dxdz =1
2
∫ 3
0
∫
√9−z2
0x3 dxdz
=1
2
∫ 3
0
[
x4
4
]
√9−z2
0
dz =1
8
∫ 3
0
(
9− z2)2
dz
=1
8
∫ 3
0
(
81− 18z2 + z4)
dz =81
5
(69)
V =
∫ α
0
∫ π
0
∫ R
0ρ2 sinφdρdφdθ =
∫ α
0
∫ π
0
1
3R3 sin φdφdθ =
∫ α
0
2
3R3dθ =
2
3αR3
118
CHAPTER 5. MULTIPLE INTEGRALS
(70) Now, σ = k (R− ρ) , the mass is
M =
∫ ∫ ∫
σdV =
∫ 2π
0
∫ π
0
∫ R
0k (R− ρ) ρ2 sinφdρdφdθ
=
∫ 2π
0
∫ π
0
[
k (sinφ)
(
−1
4ρ4 +
1
3Rρ3
)]R
0
dφdθ
=
∫ 2π
0
∫ π
0
1
12R4k sinφdφdθ
=
∫ 2π
0
1
6R4kdθ =
1
3πR4k
(71)
h
y
RBase
Now, σ = kρ, tan α = Rh . On the base, ρ cos φ = h. The mass is
M =
∫ ∫ ∫
kρdV =
∫ 2π
0
∫ α
0
∫ h sec φ
0kρ3 sin φdρdφdθ
=
∫ 2π
0
∫ α
0
[
1
4kρ4 sinφ
]h sec φ
0
dφdθ =
∫ 2π
0
∫ α
0
1
4
h4
cos4 φk sin φdφdθ
=
∫ 2π
0
1
12h4k
(
sec3 α− 1)
dθ =1
6πh4k
(
sec3 α− 1)
=1
6πh4k
(
(
R2 + h2
h2
)3/2
− 1
)
=1
6πhk
(
(
R2 + h2)3/2 − h3
)
(72)
V =
∫∫∫
conedV =
∫ 2π
0
∫ α
0
∫ h sec φ
0ρ2 sinφdρdφdθ =
∫ 2π
0
∫ α
0
[
1
3ρ3 sinφ
]h sec φ
0
dφdθ
=
∫ 2π
0
∫ α
0
1
3
h3
cos3 φsinφdφdθ =
∫ 2π
0
1
6h3(
sec3 α− 1)
dθ
=1
3πh3
(
sec2 α− 1)
=1
3πh3 tan2 α =
1
3πR2h
119
CHAPTER 5. MULTIPLE INTEGRALS
(73) (a) i.
Iz =
∫∫∫
ballσ(
x2 + y2)
dV = σ
∫ 2π
0
∫ π
0
∫ R
0ρ2 sin2 φ · ρ2 sinφdρdφdθ
= σ
∫ 2π
0
∫ π
0
∫ R
0ρ4 sin3 φdρdφdθ = σ
∫ 2π
0
∫ π
0
R5
5sin3 φdφdθ
= σ
∫ 2π
0
4
15R5dθ =
8
15σπR5 =
2
5MR2
By parallel axis theorem,
I =
(
2
5MR2 + MR2
)
=7
5MR2
(b)
Mxy =
∫∫∫
TzσdV = σ
∫ 2π
0
∫ π/2
0
∫ R
0ρ2 sinφ · ρ cos φdρdφdθ
= σ
∫ 2π
0
∫ π/2
0
∫ R
0ρ3 sinφ cos φdρdφdθ = σ
∫ 2π
0
∫ π/2
0
R4
4sinφ cos φdφdθ
= σ
∫ 2π
0
∫ π/2
0
R4
4
sin 2φ
2dφdθ = σ
∫ 2π
0
1
8R4dθ =
1
4σπR4
Thus
z =Mxy
M/2=
14σπR4
23σπR3
=3
8R
and by symmetry x = y = 0.
(74) (a)
Iz =
∫∫∫
Tσ(
x2 + y2)
dV = σ
∫ 2π
0
∫ π
0
∫ R2
R1
ρ2 sin2 φ · ρ2 sinφdρdφdθ
= σ
∫ 2π
0
∫ π
0
∫ R2
R1
ρ4 sin3 φdρdφdθ = σ
∫ 2π
0
∫ π
0
(
R52
5− R5
1
5
)
sin3 φdφdθ
= σ
∫ 2π
0
4
15
(
R52 −R5
1
)
dθ =8
15σπ(
R52 −R5
1
)
=2
5M
R52 −R5
1
R32 −R3
1
(b) Setting R2 = R and R1 → R in part (a), we get
Iz = limR1→R
2
5M
R5 −R51
R3 −R31
=2
5M lim
R1→R
R5 −R51
R3 −R31
=2
5M lim
R1→R
−5R41
−3R21
=2
3MR2
120
CHAPTER 5. MULTIPLE INTEGRALS
(c) By parallel axis theorem,
I =2
3MR2 + MR2 =
5
3R2M
(d)
Mxy =
∫∫∫
TzσdV = σ
∫ 2π
0
∫ π/2
0
∫ R2
R1
ρ2 sinφ · ρ cos φdρdφdθ
= σ
∫ 2π
0
∫ π/2
0
∫ R2
R1
ρ3 sinφ cos φdρdφdθ
= σ
∫ 2π
0
∫ π/2
0
(
R42
4− R4
1
4
)
sinφ cos φdφdθ
= σ
∫ 2π
0
∫ π/2
0
(
R42
4− R4
1
4
)
sin 2φ
2dφdθ
= σ
∫ 2π
0
(
R42
8− R4
1
8
)
dθ =1
4πσ(
R42 −R4
1
)
and
M =
∫∫∫
TσdV = σ
∫ 2π
0
∫ π/2
0
∫ R2
R1
ρ2 sinφdρdφdθ =2
3πσ(
R32 −R3
1
)
Thus
z =Mxy
M=
14πσ
(
R42 −R4
1
)
23πσ
(
R32 −R3
1
) =3
8
R42 −R4
1
R32 −R3
1
and by symmetry x = y = 0.
(e) Setting R2 = R, R1 → R in part (a), we get
z = limR1→R
3
8
R4 −R41
R3 −R31
=1
2R
(75) (a) R sec φ ≤ ρ ≤ 2R cos φ, 0 ≤ φ ≤ π4 , 0 ≤ θ ≤ 2π.
(b) Sphere: ρ2 sin2 φ + (ρ cos φ−R)2 = R2, therefore ρ = 2R cos φ
121
CHAPTER 5. MULTIPLE INTEGRALS
(76) (a) σ = kρ
M =
∫∫∫
σdV =
∫ 2π
0
∫ π/2
0
∫ 2R cos φ
0ρ2 sinφ · kρdρdφdθ
=
∫ 2π
0
∫ π/2
0
[
1
4kρ4 sinφ
]2R cos φ
0
dφdθ
= k
∫ 2π
0
∫ π/2
04R4 cos4 φ sin φdφdθ = k
∫ 2π
0
4
5R4dθ =
8
5kπR4
(b) σ = kρ sinφ
M =
∫∫∫
σdV =
∫ 2π
0
∫ π/2
0
∫ 2R cos φ
0ρ2 sin φ · kρ sinφdρdφdθ
=
∫ 2π
0
∫ π/2
0
[
1
4kρ4 sin2 φ
]2R cos φ
0
dφdθ
= k
∫ 2π
0
∫ π/2
04R4 cos4 φ sin2 φdφdθ
= k
∫ 2π
0
1
8πR4dθ =
1
4π2kR4
(c) σ = kρ cos2 θ sinφ
M =
∫∫∫
σdV =
∫ 2π
0
∫ π/2
0
∫ 2R cos φ
0ρ2 sinφ · kρ cos2 θ sinφdρdφdθ
= k
∫ 2π
0
∫ π/2
0
[
1
4kρ4 sin2 φ cos2 θ
]2R cos φ
0
dφdθ
= k
∫ 2π
0
∫ π/2
04R4 cos4 φ sin2 φ cos2 θdφdθ
= k
∫ 2π
0
1
8πR4 cos2 θdθ =
1
8π2kR4
(77)
x
z
4
122
CHAPTER 5. MULTIPLE INTEGRALS
Where the sphere ρ = 2√
2 cos φ intersects the sphere ρ = 2, we have 2 = 2√
2 cos φ giving
φ = π4 .
V =
∫ 2π
0
∫ π/4
0
∫ 2
0ρ2 sin φdρdφdθ +
∫ 2π
0
∫ π/2
π/4
∫ 2√
2 cos φ
0ρ2 sinφdρdφdθ
=
∫ 2π
0
∫ π/4
0
8
3sinφdφdθ +
∫ 2π
0
∫ π/2
π/4
16
3
√2 cos3 φ sin φdφdθ
=
∫ 2π
0
(
8
3− 4
3
√2
)
dθ +
∫ 2π
0
1
3
√2dθ
=16
3π − 8
3π√
2 +2
3π√
2 =16
3π − 2π
√2
(78)
x
z
V =
∫ 2π
0
∫ π
0
∫ 1−cos φ
0ρ2 sinφdρdφdθ =
∫ 2π
0
∫ π
0
1
3(1− cos φ)3 sinφdφdθ
=
∫ 2π
0
1
12
[
(1− cos φ)4]π
0dθ =
∫ 2π
0
4
3dθ =
8
3π
123
Chapter 6
Vector Calculus
(1) Let θ be the angle the two vectors a and b. Since (a • b) = ‖a‖ ‖b‖ cos θ,
‖a× b‖2 = (‖a‖ ‖b‖ sin θ)2 = ‖a‖2 ‖b‖2 sin2 θ = ‖a‖2 ‖b‖2(
1− cos2 θ)
= ‖a‖2 ‖b‖2 − ‖a‖2 ‖b‖2 cos2 θ = ‖a‖2 ‖b‖2 − (a • b)2
A
B
C
(2)−−→AB = b1i + b2j− (a1i + a2j) = (b1 − a1) i + (b2 − a2) j
−→AC = c1i + c2j− (a1i + a2j) = (c1 − a1) i + (c2 − a2) j
The area of the triangle is
1
2
∥
∥
∥
−−→AB∥
∥
∥
∥
∥
∥
−→AC∥
∥
∥ sin θ
=1
2
∥
∥
∥
−−→AB ×−→AC
∥
∥
∥=
1
2
∥
∥
∥
∥
∥
∥
∥
∥
det
i j k
(b1 − a1) (b2 − a2) 0
(c1 − a1) (c2 − a2) 0
∥
∥
∥
∥
∥
∥
∥
∥
=1
2
∥
∥
∥
∥
∥
det
(
(b1 − a1) (b2 − a2)
(c1 − a1) (c2 − a2)
)
k
∥
∥
∥
∥
∥
=1
2
∣
∣
∣
∣
∣
det
(
(b1 − a1) (b2 − a2)
(c1 − a1) (c2 − a2)
)∣
∣
∣
∣
∣
where θ is the angle between−−→AB and
−→AC.
(3) Let x be the position vector of the point (x, y, z) .
125
CHAPTER 6. VECTOR CALCULUS
B
0
x
(XB)2 = (OX)2 − (OB)2 = ‖x‖2 −∥
∥
∥
−−→OB
∥
∥
∥
2= ‖x‖2 − (‖x‖ cos θ)2
= ‖x‖2 −(
‖x‖ x • a‖x‖ ‖a‖
)2
= ‖x‖2 − (x • a)2
‖a‖2.
(4) (a)
∇f |(2,1) =∂ (y − 3x)
∂xi +
∂ (y − 3x)
∂yj
∣
∣
∣
∣
(2,1)
= −3i + j
(b)
∇f |(2,1,−1) =2x2 + z
xi + 2yj + (−4z + lnx)k
∣
∣
∣
∣
(2,1,−1)
=7
2i + 2j + (4 + ln 2)k
(5) (a)
∇ • F =
(
∂
∂xi +
∂
∂yj +
∂
∂zk
)
•((
x2 − y)
i + (4z) j +(
x2)
k)
=∂
∂x
(
x2 − y)
+∂
∂y(4z) +
∂
∂z
(
x2)
= 2x
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
(
x2 − y)
(4z)(
x2)
∣
∣
∣
∣
∣
∣
∣
∣
= −4i−2xj + k
(b)
∇ • F =
(
∂
∂xi +
∂
∂yj +
∂
∂zk
)
• ((yz) i + (xz) j + (xy)k) = 0
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
(yz) (xz) (xy)
∣
∣
∣
∣
∣
∣
∣
∣
= 0
126
CHAPTER 6. VECTOR CALCULUS
(c)
∇ • F =
(
∂
∂xi +
∂
∂zj
)
• ((−y) i + (x) j) =∂
∂x(−y) +
∂
∂y(x) = 0
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
(−y) (x) 0
∣
∣
∣
∣
∣
∣
∣
∣
= 2k
(d)
∇ • F =
(
∂
∂xi +
∂
∂yj +
∂
∂zk
)
•(
(2xz) i + (−xy) j +(
−z2)
k)
= −x
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
(2xz) (−xy) 0
∣
∣
∣
∣
∣
∣
∣
∣
= 2xj− yk
(6) (a) div F (x, y, z) = ∇ • F (x, y, z) = 2x + 6y − 4z
div F (x0, y0, z0) = 2x0 + 6y0 − 4z0
curlF (x, y, z) = ∇× F (x, y, z) =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
x2 3y2 −2z2
∣
∣
∣
∣
∣
∣
∣
∣
= 0
(b) div F (x, y, z) = 0
curlF (x, y, z) =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
eyz ezx exy
∣
∣
∣
∣
∣
∣
∣
∣
= 〈xexy − xezx, yeyz − yexy, zezx − zeyz〉
curlF (0, 1, 2) =⟨
0, e2 − 1, 2− 2e2⟩
(c) div F (x, y, z) = 0
curlF (x, y, z) =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
sin yz cos zx xy
∣
∣
∣
∣
∣
∣
∣
∣
= 〈x− x sin zx, y cos yz − y,−z sin zx− z cos yz〉
curlF(
0, 1,π
2
)
= 〈0,−1, 0〉
(d) div F (x, y, z) = 2xy− 6xy2z2 + 1
div F (2, 1, 1) = 4− 12 + 1 = −7
127
CHAPTER 6. VECTOR CALCULUS
curlF (x, y, z) =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
x2y − z2 −2xy3z2 xy + z
∣
∣
∣
∣
∣
∣
∣
∣
=⟨
x + 4xy3z,−2z − y,−2y3z2 − x2⟩
curlF (2, 1, 1) = 〈10,−3,−6〉
(7) (a)
∇2f = ∇ •∇f
= ∇ •(
∂(
ln(
x2 + y2))
∂xi +
∂(
ln(
x2 + y2))
∂yj
)
=
(
∂
∂xi +
∂
∂yj
)
•(
2x
x2 + y2i +
2y
x2 + y2j
)
=∂
∂x
(
2x
x2 + y2
)
+∂
∂y
(
2y
x2 + y2
)
= 2−x2 + y2
(x2 + y2)2− 2−x2 + y2
(x2 + y2)2= 0
(b)
∇2f = ∇ •∇f
=
(
∂
∂xi +
∂
∂yj
)
•((
3Ax2 + 2Bxy + Cy2)
i +(
Bx2 + 2Cxy + 3Dy2)
j)
=∂
∂x
(
3Ax2 + 2Bxy + Cy2)
+∂
∂y
(
Bx2 + 2Cxy + 3Dy2)
= 6Ax + 2By + 2Cx + 6Dy
(c)
∇2f = ∇ •∇f
= ∇ •(
∂r−1
∂xi +
∂r−1
∂yj +
∂r−1
∂zj
)
=
(
∂
∂xi +
∂
∂yj +
∂
∂zk
)
•(
−r−2 x
ri− r−2 y
rj−−r−2 z
rk)
= − ∂
∂x
(
r−3x)
− ∂
∂y
(
r−3y)
− ∂
∂z
(
r−3z)
= −r−3 + 3r−4 x2
r− r−3 + 3r−4 y2
r− r−3 + 3r−4 z2
r
= −3r−3 + 3r−5(
x2 + y2 + z2)
= −3r−3 + 3r−3 = 0
128
CHAPTER 6. VECTOR CALCULUS
(8)
∇ • (f (r) r)
=
(
∂
∂xi+
∂
∂yj+
∂
∂zk
)
• (f (r)xi+f (r) yj+f (r) zk)
=
(
f (r)∂x
∂x+ x
∂f (r)
∂x
)
+
(
f (r)∂y
∂y+ y
∂f (r)
∂y
)
+
(
f (r)∂z
∂z+ z
∂f (r)
∂z
)
=
(
f (r) + xf ′ (r)∂r
∂x
)
+
(
f (r) + yf ′ (r)∂r
∂y
)
+
(
f (r) + zf ′ (r)∂r
∂z
)
=
(
f (r) + f ′ (r)x2
r
)
+
(
f (r) + f ′ (r)y2
r
)
+
(
f (r) + f ′ (r)z2
r
)
= 3f (r) + f ′ (r)x2 + y2 + z2
r= 3f (r) + rf ′ (r)
(9) (a) To find a φ such that ∇φ = A,
φx = ex cos y + yz ⇒ φ =
∫
(ex cos y + yz) dx = ex cos y + xyz + f (y, z)
φy = xz − ex sin y ⇒ φ =
∫
(xz − ex sin y) dy = ex cos y + xyz + g (x, z)
φz = xy + z + 2⇒ φ =
∫
(xy + z + 2) dz = xyz +1
2z2 + 2z + h (x, y)
Thus, φ = ex cos y + xyz + 12z2 + 2z + C.
(b) To find a φ such that ∇φ = A,
φx = y + z ⇒ φ =
∫
(y + z) dx = xy + xz + f (y, z)
φy = x + z ⇒ φ =
∫
(x + z) dy = xy + yz + g (x, z)
φz = x + y ⇒ φ =
∫
(x + y) dz = xz + yz + h (x, y)
Thus, φ = xy + yz + xz + C.
(c) To find a φ such that ∇φ = A,
φx = y sin z ⇒ φ =
∫
y sin zdx = xy sin z + f (y, z)
φy = x sin z ⇒ φ =
∫
x sin zdy = xy sin z + g (x, z)
φz = xy cos z ⇒ φ =
∫
xy cos zdz = xy sin z + h (x, z)
Thus, φ = xy sin z + C.
129
CHAPTER 6. VECTOR CALCULUS
(10)
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
(
2x2 − 3)
(−4z) cos z
∣
∣
∣
∣
∣
∣
∣
∣
=
[
∂
∂y(cos z)− ∂
∂z(−4z)
]
i−[
∂
∂x(cos z)− ∂
∂z
(
2x2 − 3)
]
j
+
[
∂
∂x(−4z)− ∂
∂y
(
2x2 − 3)
]
k
= 4i 6= 0
Hence, F is not a conservative vector field.
(11) Let r (t) =[
4− t2, t]
, −3 ≤ t ≤ 2, then r′ (t) = [−2t, 1].
∫
Γy2 dx + x dy =
∫ 2
−3
[
(t)2 (−2t) +(
4− t2)
(1)]
dt =
∫ 2
−3
(
4− t2 − 2t3)
dt =245
6
(12) Let r (t) = [cos t, sin t, t], 0 ≤ t ≤ 2π, then r′ (t) = [− sin t, cos t, 1].
∫
Γy sin z ds =
∫ 2π
0(sin t) sin (t)
√
(− sin t)2 + (cos t)2 + (1)2 dt =√
2
∫ 2π
0sin2 t dt
=
√2
2
∫ 2π
0(1− cos 2t) dt =
√2π
(13) Let x (t) = 1 + 2t, y (t) = 2, z (t) = 3− 2t and 0 ≤ t ≤ 1.
f (x (t) , y (t) , z (t)) = (1 + 2t) + 2 (3− 2t) = 7− 2t
x′ (t) = 2, y′ (t) = 0, z′ (t) = −2 and
√
(
dxdt
)2+(
dydt
)2+(
dzdt
)2=√
8
∫
C (x + yz) ds =∫ 10 (7− 2t)
(√8)
dt = 12√
2
(14) F =[
x4, xy]
R = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ x∫
Γx4 dx + xy dy =
∫
ΓF • dr =
∫ 1
0
∫ x
0
[
∂
∂x(xy)− ∂
∂y
(
x4)
]
dydx =
∫ 1
0
∫ x
0y dydx
=
∫ 1
0
[
y2
2
]x
0
dx =1
2
∫ 1
0x2 dx =
1
6
(15) F =[
y2, 3xy]
130
CHAPTER 6. VECTOR CALCULUS
R = (r, θ) |1 ≤ r ≤ 2, 0 ≤ θ ≤ π∫
Γy2 dx + 3xy dy =
∫
ΓF • dr =
∫ π
0
∫ 2
1
[
∂
∂x(3xy)− ∂
∂y
(
y2)
]
rdrdθ =
∫ π
0
∫ 2
1y rdrdθ
=
∫ π
0
∫ 2
1r2 sin θ drdθ =
7
3
∫ π
0sin θ dθ =
14
3
(16) (a) Let x = t, y = t, z = 0, 0 ≤ t ≤ 1, then dxdt = 1, dy
dt = 1, dzdt = 0.
∫
ABf (x, y, z) ds =
∫ 1
0(t) (t)2
√
12 + 12 + 02 dt =√
2
∫ 1
0t3 dt =
√2
4
(b) Let x = 2t, y = 3t, z = 6t, 0 ≤ t ≤ 1, then dxdt = 2, dy
dt = 3, dzdt = 6.
∫
AB f (x, y, z) ds =∫ 10
[
(2t) (3t) + (6t)2]√
22 + 32 + 62 dt = 294∫ 10 t2 dt = 98
(17) (a) Let x = cos t, y = sin t, where 0 ≤ t ≤ π2 .
∫
Cxyds =
∫ π2
0cos t sin t
√
(
d (cos t)
dt
)2
+
(
d (sin t)
dt
)2
dt
=
∫ π2
0cos t sin tdt =
1
2
∫ π2
0sin 2tdt
=
[
−1
4cos 2t
]π/2
0
=1
2
(b) Let x = t, y = 2t, z = −t where 0 ≤ t ≤ 1.
∫
C(xy + y + z) ds
=
∫ 1
0(t× 2t + 2t− t)
√
(
d (t)
dt
)2
+
(
d (2t)
dt
)2
+
(
d (−t)
dt
)2
dt
=
∫ 1
0
(
2t2 + t)√
6dt
=√
6
∫ 1
0
(
2t2 + t)
dt =7
6
√6
(c) Let x = cos t, y = sin t, z = t where 0 ≤ t ≤ 2π.
∫
Cyds =
∫ 2π
0sin t
√
(
d (cos t)
dt
)2
+
(
d (sin t)
dt
)2
+
(
d (t)
dt
)2
dt
=
∫ 2π
0
√2 sin tdt = 0
131
CHAPTER 6. VECTOR CALCULUS
(18) (a) Let x = cos θ, y = sin θ, 0 ≤ θ ≤ π, then dxdθ = − sin θ, dy
dθ = cos θ.
∫
Cf (x, y) ds =
∫ π
0(cos θ + sin θ)
√
(− sin θ)2 + (cos θ)2 dθ
=
∫ π
0(cos θ + sin θ) dθ = [sin θ − cos θ]π0 = 2
(b)
∫
Cf (x, y) ds =
∫ π
0(cos θ)2 sin θ
√
(− sin θ)2 + (cos θ)2 dθ =
∫ π
0cos2 θ sin θ dθ
=
[
−1
3cos3 θ
]π
0
=2
3
(19) x = t cos t, y = t sin t, z = t, 0 ≤ t ≤ 1, then dxdt = cos t− t sin t, dy
dt = sin t + t cos t, dzdt = 1
and
√
(
dxdt
)2+(
dydt
)2+(
dzdt
)2=√
(cos t− t sin t)2 + (sin t + t cos t)2 + 12 =√
t2 + 2.
(a)∫
C f (x, y, z) ds =∫ 10 (t)
√t2 + 2 dt =
[
13
(
t2 + 2)3/2
]1
0=√
3− 23
√2
(b)
∫
Cf (x, y, z) ds =
∫ 1
0
√
t2 + 2 dt =
[
t
2
√
t2 + 2 + ln∣
∣
∣t +√
t2 + 2∣
∣
∣
]1
0
= ln(√
3 + 1)
− ln√
2 +1
2
√3
= 1.5245 =1
2
[
ln(√
3 + 2)
+√
3]
(20) (a) x = cos t + t sin t, y = sin t − t cos t, 0 ≤ t ≤ 2π, then dxdt = t cos t, dy
dt = t sin t and√
(
dxdt
)2+(
dydt
)2= t.
Arc length =∫ 2π0
√
(
dxdt
)2+(
dydt
)2dt =
∫ 2π0 t dt =
[
12 t2]2π
0= 2π2
(b) x = θ − sin θ, y = 1 − cos θ, 0 ≤ θ ≤ π, then dxdθ = 1 − cos θ, dy
dt = sin θ and√
(
dxdθ
)2+(
dydθ
)2=√
2− 2 cos θ.
Arc length =∫ π0
√2− 2 cos θ dθ =
∫ π0 2 sin θ
2 dθ =[
−4 cos θ2
]π
0= 4
132
CHAPTER 6. VECTOR CALCULUS
(21) (a)
curlF
=
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
3 + 2xy x2 − 3y2 0
∣
∣
∣
∣
∣
∣
∣
∣
=
[
0− ∂
∂z
(
x2 − 3y2)
,∂
∂z(3 + 2xy)− 0,
∂
∂x
(
x2 − 3y2)
− ∂
∂y(3 + 2xy)
]
= [0, 0, 0] = 0
Since curlF = 0, the F is conservative.
(b) Since F is conservative, then the potential function f exists.
fx = 3 + 2xy =⇒ f (x, y) =∫
(3 + 2xy) dx = 3x + x2y + g (y)
fy = ∂∂y
[
3x + x2y + g (y)]
= x2 + g′ (y) = x2 − 3y2
So, g′ (y) = −3y2 =⇒ g (y) =∫ (
−3y2)
dy = −y3 + C.
Thus, f (x, y) = 3x + x2y − y3 + C.
r (π) = [eπ sinπ, eπ cos π] = [0,−eπ]
r (0) =[
e0 sin 0, e0 cos 0]
= [0, 1]
By the Foundamental Theorem of Line Integrals,
∫
ΓF • dr =f (0,−eπ)− f (0, 1) =
[
− (−eπ)3]
−[
− (1)3]
= e3π + 1.
(22) Let f (x, y, z) = 3x− 6yz, g (x, y, z) = 2y + 3xz and h (x, y, z) = 1− 4xz2.
(a) Let x = t, y = t2, z = t3, 0 ≤ t ≤ 1, then dxdt = 1, dy
dt = 2t, dzdt = 3t2.
fdx
dt+ g
dy
dt+ h
dz
dt= (3x− 6yz) + 2t (2y + 3xz) + 3t2
(
1− 4xz2)
=(
3t− 6t5)
+ 2t(
2t2 + 3t4)
+ 3t2(
1− 4t7)
= −12t9 + 4t3 + 3t2 + 3t
∫
y
ABF • dr =
∫ 10
(
−12t9 + 4t3 + 3t2 + 3t)
dt = 2310
(b) Let x = t, y = t, z = t, 0 ≤ t ≤ 1, then dxdt = dy
dt = dzdt = 1.
f dxdt + g dy
dt + hdzdt =
(
3t− 6t2)
+(
2t + 3t2)
+(
1− 4t3)
= −4t3 − 3t2 + 5t + 1∫
y
ABF • dr =
∫ 10
(
−4t3 − 3t2 + 5t + 1)
dt = 32
(c) From A to C, let x = 0, y = 0, z = t, 0 ≤ t ≤ 1, then dxdt = dy
dt = 0, dzdt = 1.
f dxdt + g dy
dt + hdzdt = 0 + 0 + (1) = 1
133
CHAPTER 6. VECTOR CALCULUS
∫
y
ACF • dr =
∫ 10 dt = 1
From C to B, let x = s, y = s, z = 0, 0 ≤ s ≤ 1, then dxds = dy
ds = 1, dzds = 0.
f dxds + g dy
ds + hdzds = (3s− 6s) + (2s + 3s) + 0 = 2s
∫
y
CBF • dr =
∫ 10 2s ds = 1
∫
y
ABF • dr =
∫
y
ACF • dr +
∫
y
CBF • dr = 1 + 1 = 2
(23) Let x (t) = t and y (t) = t3, −2 ≤ t ≤ 3. Then r (t) =[
t, t3]
, F (r (t)) =[
(t)(
t3)
, t3 − t]
=[
t4, t3 − t]
and r′ (t) =[
1, 3t2]
.
Work done =∫
C F • dr =∫ 3−2
[
t4, t3 − t]
•[
1, 3t2]
dt =∫ 3−2
(
t4 + 3t5 − 3t3)
dt = 13554
(24) r (t) = [cos t, sin t], 0 ≤ t ≤ π, then r′ (t) = [− sin t, cos t].
F (r (t)) =[
− (sin t)2 , (cos t) (sin t)]
=[
− sin2 t, sin t cos t]
Work done =
∫ π
0
[
− sin2 t, sin t cos t]
• [− sin t, cos t] dt =
∫ π
0
(
sin3 t + sin t cos2 t)
dt
=
∫ π
0sin t dt = 2
(25) (a) Let x = 1 + t, y = 2− t and 0 ≤ t ≤ 1, then dxdt = 1 and dy
dt = −1.
F (r (t)) = 〈2 (1 + t)− (2− t) + 4, 3 (1 + t) + 5 (2− t)− 6〉 = 〈3t + 4, 7− 2t〉r′ (t) = 〈1,−1〉∫
Γ F • dr = F (r (t)) • r′ (t) dt =∫ 10 [(3t + 4)− (7− 2t)] dt =
∫ 10 (5t− 3) dt = −1
2
(b) F (r (t)) =⟨
(
t2) (
t3)
,(
t2)2
, t3⟩
=⟨
t5, t4, t3⟩
r′ (t) =⟨
1, 2t, 3t2⟩
∫
Γ F • dr =∫ 10
[(
t5)
(1) +(
t4)
(2t) +(
t3) (
3t2)]
dt =∫ 10 6t5 dt = 1
(c) From (0, 0) to (1, 0), let x = t, y = 0 and 0 ≤ t ≤ 1, then dxdt = 1and dy
dt = 0.
Then F (r (t)) = 〈0, 0〉 and∫
Γ1F•dr = 0.
From (1, 0) to (1, 1), let x = 1, y = s and 0 ≤ s ≤ 1, then dxds = 0 and dy
ds = 1.
Then F (r (s)) = 〈6s, 2s〉, r′ (s) = 〈0, 1〉 and∫
Γ2F • dr =
∫ 10 2s ds = 1.
From (1, 1) to (0, 1), let x = 1− u, y = 1 and 0 ≤ u ≤ 1, then dxdu = −1, dy
du = 0.
F (r (u)) =⟨
6 (1− u)2 , 2⟩
, r′ (u) = 〈−1, 0〉 and∫
Γ3F • dr =
∫ 10 −6u2 du = −2.
From (0, 1) to (0, 0), let x = 0, y = 1− v and 0 ≤ v ≤ 1, then dxdv = 0 and dy
dv = −1.
F (r (v)) = 〈0, 2 (1− v)〉, r′ (v) = 〈0,−1〉 and∫
Γ4F • dr =
∫ 10 2 (v − 1) dv = −1.
Therefore,∫
Γ F • dr =∫
Γ1+∫
Γ2+∫
Γ3+∫
Γ4= 0 + 1− 2− 1 = −2.
134
CHAPTER 6. VECTOR CALCULUS
(26) (a) Let x = t, y = t and z = t, where 0 ≤ t ≤ 1.
∫
CF•dr =
∫
C
[(
x2 − y)
i+(
y2 − z)
j+(
z2 − x)
k]
• [dxi+dyj+dzk]
=
∫
C
(
x2 − y)
dx+(
y2 − z)
dy+(
z2 − x)
dz
=
∫ 1
0
(
t2 − t)
dt+(
t2 − t)
dt+(
t2 − t)
dt
= 3
∫ 1
0
(
t2 − t)
dt = −1
2
(b) Let x = t, y = t2 and z = t3, where 0 ≤ t ≤ 1.
∫
CF•dr =
∫
C
[(
x2 − y)
i+(
y2 − z)
j+(
z2 − x)
k]
• [dxi+dyj+dzk]
=
∫
C
(
x2 − y)
dx+(
y2 − z)
dy+(
z2 − x)
dz
=
∫ 1
0
(
t2 − t2)
dt+(
(
t2)2 − t3
)
dt2+(
(
t3)2 − t
)
dt3
=
∫ 1
0
(
t2 − t2)
dt+2t(
t4 − t3)
dt+3t2(
t6 − t)
dt
=
∫ 1
0
(
2t5 − 2t4 + 3t8 − 3t3)
dt = −29
60
(27) (a) Let x = cos t and y = sin t, where 0 ≤ t ≤ 2π,
∮
C−ydx + xdy =
∫ 1
0− sin td (cos t) + cos td (sin t)
=
∫ 2π
0
(
sin2 t + cos2 t)
dt = 2π.
(b) Let x = 1− t and y = t, where 0 ≤ t ≤ 1
∫
C(x + y) dx +
(
x2 + y2)
dy =
∫ 1
0(1) d (1− t) +
∫ 1
0
(
(1− t)2 + t2)
dt
= −1 +
∫ 1
0
(
2t2 + 1− 2t)
dt = −1
3.
135
CHAPTER 6. VECTOR CALCULUS
(28) Since
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
y sin z x sin z xy cos z
∣
∣
∣
∣
∣
∣
∣
∣
=
(
∂
∂y(xy cos z)− ∂
∂z(x sin z)
)
i+
(
∂
∂x(xy cos z)− ∂
∂z(y sin z)
)
j
+
(
∂
∂x(x sin z)− ∂
∂y(y sin z)
)
k
= 0
Therefore, F is conservative and there exists a function φ such that ∇φ = F. In this case,
φ = xy sin z+ constant and
∫ (1,2,3)
(0,0,0)[(xy sin z) i + (x sin z) j + (xy cos z)k] • dr
= xy sin z|(1,2,3)(0,0,0) = (1) (2) sin (3)− (0) (0) sin 0 = 2 sin 3
(29) Since cos2 t + sin2 t = 1, so, let x− 1 = cos t and y = sin t, then x = cos t− 1, y = sin t.
(30) (a) F = 〈3y, 5x〉, P = 3y, Q = 5x and R = (r, θ) |0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
∫
∂R F • dr =∫ ∫
R
(
∂Q∂x − ∂P
∂y
)
dA =∫ 2π0
∫ 10 (5− 3) r dr dθ = 2 [θ]2π
0
[
12r2]1
0= 2π
(b) F =⟨
y2, x2⟩
, P = y2, Q = x2 and R = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
∫
∂RF • dr =
∫ 1
0
∫ 1
0(2x− 2y) dx dy =
∫ 1
0
[
x2 − 2xy]1
0dy
=
∫ 1
0(1− 2y) dy =
[
y − y2]1
0= 0
(c) F = 〈xy, x− y〉, P = xy, Q = x− y, and R = (x, y) |0 ≤ x ≤ 1, 1− x ≤ y ≤ 1.
∫
∂R F • dr =∫ 10
∫ 11−x (1− x) dy dx =
∫ 10
(
x− x2)
dx =[
12x2 − 1
3x3]1
0= 1
6
136
CHAPTER 6. VECTOR CALCULUS
(31) (a) Let x = a cos t and y = a sin t, where 0 ≤ t ≤ 2π.
∮
C
(
−x2y)
dx +(
xy2)
dy
=
∫ 2π
0
(
−a2 cos2 t× a sin t)
d (a cos t) +(
a2 cos t× a sin2 t)
d (a sin t)
=
∫ 2π
0
(
a4 cos2 t sin2 t + a4 cos2 t sin2 t)
dt
= 2a4
∫ 2π
0
(
cos2 t sin2 t)
dt = 2a4
∫ 2π
0
(
sin 2t
2
)2
dt
=1
2a4
∫ 2π
0sin2 2tdt
=1
2a4
∫ 2π
0
(
1− cos 4t
2
)
dt =1
2πa4
and
∫∫
x2+y2≤a2
(
∂(
xy2)
∂x− ∂
(
−x2y)
∂y
)
dxdy =
∫∫
x2+y2≤a2
(
y2 + x2)
dxdy
Use polar coordinates, x = r cos θ and y = r sin θ, we have
∫∫
x2+y2≤a2
(
∂(
xy2)
∂x− ∂
(
−x2y)
∂y
)
dxdy
=
∫ 2π
0
∫ a
0
(
r2 cos2 θ + r2 sin2 θ)
rdrdθ
=
∫ 2π
0
∫ a
0r3drdθ =
∫ 2π
0
1
4a4dθ =
1
2πa4
(b)
x
y(0,1)
(1,0)
(-1,0)
(0,-1)
137
CHAPTER 6. VECTOR CALCULUS
On C1– the line x + y = 1, x = 1− t and y = t, where 0 ≤ t ≤ 1.
∫
C1
(−y) dx + (x) dy =
∫ 1
0− (t) d (1− t) + (1− t) dt
=
∫ 1
0tdt + (1− t) dt =
∫ 1
0dt = 1
On C2– the line −x + y = 1, x = −t and y = 1− t, where 0 ≤ t ≤ 1.
∫
C2
(−y) dx + (x) dy =
∫ 1
0− (1− t) d (−t) + (−t) d (1− t)
=
∫ 1
0(1− t) dt + tdt =
∫ 1
0dt = 1
On C3– the line −x− y = 1, x = t− 1 and y = −t, where 0 ≤ t ≤ 1.
∫
C3
(−y) dx + (x) dy =
∫ 1
0− (−t) d (t− 1) + (t− 1) d (−t)
=
∫ 1
0tdt− (t− 1) dt =
∫ 1
0dt = 1
On C4– the line x− y = 1, x = t and y = t− 1, where 0 ≤ t ≤ 1.
∫
C4
(−y) dx + (x) dy =
∫ 1
0− (t− 1) dt + td (t− 1)
=
∫ 1
0− (t− 1) dt + tdt =
∫ 1
0dt = 1
Hence,
∮
C(−y) dx + (x) dy =
∫
C1
+
∫
C2
+
∫
C3
+
∫
C4
(−y) dx + (x) dy = 1 + 1 + 1 + 1 = 4
and
∫∫
square
(
∂ (x)
∂x− ∂ (−y)
∂y
)
dxdy =
∫∫
square2dxdy = 2× area of square
= 2×√
2×√
2 = 4
Therefore,
∮
C(−y) dx + (x) dy =
∫∫
square
(
∂ (x)
∂x− ∂ (−y)
∂y
)
dxdy.
138
CHAPTER 6. VECTOR CALCULUS
and the Green’s Theorem is true in this case.
(32) By Green’s Theorem,
∮
xy2dx +(
x2y + 2x)
dy =
∫∫
Region
(
∂(
x2y + 2x)
∂x− ∂
(
xy2)
∂y
)
dxdy
=
∫∫
Region2dxdy
= 2× area of the region
(33) (a)
X = cos u sin v, Y = sinu sin v, Z = cos v
Xu = − sinu sin v, Yu = cos u sin v, Zu = 0
Xv = cos u cos v, Yv = sinu cos v, Zv = − sin v
and
E = (Xu)2 + (Yu)2 + (Zu)2 = sin2 v
G = (Xv)2 + (Yv)
2 + (Zv)2 = 1
F = XuXv + YuYv + ZuZv = 0
∫
Sf (x, y, z) dxdydz =
∫ π/2
0
∫ π/2
0cos u sin v sinu sin v cos v
√
sin2 vdudv
=
∫ π/2
0cos u sinudu
∫ π/2
0cos v sin3 vdv
=1
8
(b)
O
x
y
z
139
CHAPTER 6. VECTOR CALCULUS
On S1, the plane x = 0
∫
S1
f (x, y, z) dS =
∫ 1
0
∫ 1−z
0(y + z) dydz =
1
3
On S2, the plane y = 0
∫
S2
f (x, y, z) dS =
∫ 1
0
∫ 2
0(z) dxdz = 1
On S3, the plane x = 2
∫
S3
f (x, y, z) dS =
∫ 1
0
∫ 1−z
0(y + z) dydz =
1
3
On S4, the plane z = 0
∫
S4
f (x, y, z) dS =
∫ 1
0
∫ 2
0ydxdy = 1
On S5, the plane y + z = 1
∫
S5
f (x, y, z) dS =
∫ 1
0
∫ 2
0(1)
√
1 + (0)2 + (−1)2dxdy = 2√
2
Hence,∫
Sf (x, y, z) dS = 1 +
1
3+ 1 +
1
3+ 2√
2 =8
3+ 2√
2
(34)
z =√
2− x2 − y2, zx =−x
√
2− x2 − y2, zy =
−y√
2− x2 − y2.
140
CHAPTER 6. VECTOR CALCULUS
Let x = r cos θ, y = r sin θ, 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.
∫∫
x2+y2≤1
√
1 + z2x + z2
ydxdy
=
∫∫
x2+y2≤1
√
1 +x2
2− x2 − y2+
y2
2− x2 − y2dxdy
=
∫ 2π
0
∫ 1
0r
√
1 +r2
2− r2drdθ
=
∫ 2π
0
∫ 1
0r
√
2
2− r2drdθ
= −∫ 2π
0
[√2√
2− r2]1
0dθ
=
∫ 2π
0
(
2−√
2)
dθ = 2π(
2−√
2)
(35)
Let X = a cos u, Y = a sinu, Z = v, where 0 ≤ u ≤ 2π and 0 ≤ v ≤ 10− a cos u− 2a sinu.
Then Xu = −a sinu, Yu = a cos u, Zu = 0 and Xv = 0, Yv = 0, Zv = 1.
E = (−a sinu)2 + (a cos u)2 + (0)2 = a2
G = (0)2 + (0)2 + (1)2 = 1
F = (−a sinu) (0) + (a cos u) (0) + (0) (1) = 0
and√
EG− F 2 =√
a2 cos2 u + a2 sin2 u = a
A =
∫ 2π
0
∫ 10−a cos u−2a sin u
0advdu = a
∫ 2π
0(10− a cos u− 2a sinu) du = 20πa
141
CHAPTER 6. VECTOR CALCULUS
(36) z = 2x, zx = 2, zy = 0.
∫∫
x2+y2≤1
√
1 + z2x + z2
ydxdy =
∫∫
x2+y2≤1
√
1 + (2)2 + (0)2dxdy
=√
5
∫∫
x2+y2≤1dxdy =
√5π
(37)
z = 4−x2−y2, zx = −2x, zy = −2y. Let x = r cos θ, y = r sin θ, 1 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π
∫∫
1≤x2+y2≤4
√
1 + z2x + z2
ydxdy =
∫∫
1≤x2+y2≤4
√
1 + 4x2 + 4y2dxdy
=
∫ 2π
0
∫ 2
1r√
1 + 4r2drdθ
=
∫ 2π
0
[
1
12
(
1 + 4r2)3/2
]2
1
dθ
=
∫ 2π
0
(
17
12
√17− 5
12
√5
)
dθ =1
6π(
17√
17− 5√
5)
(38) (a) Let f (x, y, z) = 1, z = φ (x, y) = 1− x− y, and ∂z∂x = ∂z
∂y = −1.
R is the region inside x2 + y2 = a2 or R = (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π.
S =
∫ ∫
Sf (x, y, z) dS =
∫ ∫
Rf (x, y, φ (x, y))
√
(
∂z
∂x
)2
+
(
∂z
∂y
)2
+ 1 dx dy
=
∫ 2π
0
∫ a
0
(√3)
r dr dθ =√
3πa2
(b) Let f (x, y, z) = 1, z = φ (x, y) = 23
(
x3/2 + y3/2)
, and ∂z∂x =
√x, ∂z
∂y =√
y.
142
CHAPTER 6. VECTOR CALCULUS
R = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
S =
∫ ∫
Sf (x, y, z) dS =
∫ 1
0
∫ 1
0
√
(√x)2
+ (√
y)2 + 1 dx dy
=
∫ 1
0
[
2
3(x + y + 1)3/2
]1
0
dy =2
3
∫ 1
0
[
(y + 2)3/2 − (y + 1)3/2]
dy
=2
3
[
2
5(y + 2)5/2 − 2
5(y + 1)5/2
]1
0
=4
15
[(
35/2 − 25/2)
−(
25/2 − 15/2)]
=4
15
(
35/2 − 27/2 + 1)
= 1.4066
(39) (a) σ (x, y, z) = z, let z = φ (x, y) =√
4− x2 − y2, and ∂z∂x = −x
z , ∂z∂y = −y
z .
R = (r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
S =
∫ ∫
Sσ (x, y, z) dS =
∫ 2π
0
∫ 2
0(z)
√
(
−x
z
)2+(
−y
z
)2+ 1r dr dθ
=
∫ 2π
0
∫ 2
0
√
x2 + y2 + z2r dr dθ =
∫ 2π
0dθ
∫ 2
02r dr = [θ]2π
0
[
r2]2
0= 8π
(b) Find the equation for the area, a plane, within ∆ABC:−−→AB =
−−→OB −−→OA = 〈0, 1, 0〉 − 〈1, 0, 0〉 = 〈−1, 1, 0〉 and
−→AC = 〈−1, 0, 1〉.
normal vector = n =−−→AB ×−→AC =
∣
∣
∣
∣
∣
∣
∣
∣
i j k
−1 1 0
−1 0 1
∣
∣
∣
∣
∣
∣
∣
∣
= 〈1, 1, 1〉
Equation for the plane with normal vector n and passing through A (1, 0, 0):
(x− 1) + (y − 0) + (z − 0) = 0 =⇒ x + y + z = 1
σ (x, y, z) = xyz, let z = φ (x, y) = 1− x− y, and ∂z∂x = ∂z
∂y = −1.
R = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x
S =
∫ ∫
Sσ (x, y, z) dS =
∫ 1
0
∫ 1−x
0(xyz)
√
(−1)2 + (−1)2 + 1 dy dx
=√
3
∫ 1
0
∫ 1−x
0xy (1− x− y) dy dx =
√3
∫ 1
0
[
1
2xy2 − 1
2x2y2 − 1
3xy3
]1−x
0
dx
=√
3
∫ 1
0
(
−1
6x4 +
1
2x3 − 1
2x2 +
1
6x
)
dx =
√3
120
(c) σ (x, y, z) = z2
On the surface S1, z = 0 and σ (x, y, z) = 0, thus∫ ∫
S1σ (x, y, z) dS1 = 0.
On the surface S2, z = 1 and σ (x, y, z) = 1 and S2 = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1,thus
∫ ∫
S2σ (x, y, z) dS2 =
∫ ∫
S2dS2 =
∫ 10
∫ 10 dx dy = 1.
143
CHAPTER 6. VECTOR CALCULUS
On the surface S3 and S4, x = 0 and 1, and S3 = S4 = (y, z) |0 ≤ y ≤ 1, 0 ≤ z ≤ 1,thus
∫ ∫
S3σ (x, y, z) dS3 =
∫ ∫
S4σ (x, y, z) dS4 =
∫ 10
∫ 10 z2 dy dz = 1
3 .
On the surface S5 and S6, y = 0 and 1, and S5 = S6 = (x, z) |0 ≤ x ≤ 1, 0 ≤ z ≤ 1,thus
∫ ∫
S5σ (x, y, z) dS5 =
∫ ∫
S6σ (x, y, z) dS6 =
∫ 10
∫ 10 z2 dx dz = 1
3 .
Therefore,∫ ∫
S σ (x, y, z) dS =∫ ∫
S1+∫ ∫
S2+... +
∫ ∫
S6= 7
3 .
(40) (a) The equation for the plane that including ∆ABC:
unit normal vector = 〈6,3,2〉√62+32+22
=⟨
67 , 3
7 , 27
⟩
6 (x− 1) + 3 (y − 0) + 2 (z − 0) = 0 =⇒ 6x + 3y + 2z = 6
Let z = φ (x, y) = 3− 3x− 32y, then F = 〈xy, y, zx〉 =
⟨
xy, y, 3x− 3x2 − 32xy
⟩
Using the formula
∫ ∫
SF • n dS =
∫ ∫
R
(
−P∂z
∂x−Q
∂z
∂y+ R
)
dx dy
F • n =⟨
xy, y, 3x− 3x2 − 32xy
⟩
•⟨
67 , 3
7 , 27
⟩
= 67x + 3
7y + 37xy − 6
7x2
R = (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 2− 2x
dS = 1|n • k| dx dy = 7
2 dx dy
∫ ∫
SF • n dS =
∫ 1
0
∫ 2−2x
0
(
6
7x +
3
7y +
3
7xy − 6
7x2
)
7
2dy dx =
7
4
=
∫ 1
0
[
3xy − 3x2y − 3
4xy2
]2−2x
0
dx
or using the formula
∫∫
SF • n dS =
∫∫
DF (r (u, v)) • (ru × rv) dudv
Let r (x, y) =⟨
x, y, 3− 3x− 32y⟩
and D = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 2− 2x, then
rx = 〈1, 0,−3〉 and ry =⟨
0, 1,−32
⟩
.
rx × ry =
∣
∣
∣
∣
∣
∣
∣
∣
i j k
1 0 −3
0 1 −32
∣
∣
∣
∣
∣
∣
∣
∣
=⟨
3, 32 , 1⟩
144
CHAPTER 6. VECTOR CALCULUS
F (r (x, y)) =⟨
xy, y, 3x− 3x2 − 32xy
⟩
∫∫
DF (r (x, y)) • (rx × ry) dx dy
=
∫ 1
0
∫ 2−2x
0
⟨
xy, y, 3x− 3x2 − 3
2xy
⟩
•⟨
3,3
2, 1
⟩
dy dx
=
∫ 1
0
∫ 2−2x
0
(
3
2xy +
3
2y + 3x− 3x2
)
dy dx
=
∫ 1
0
(
9x3 − 15x2 + 3x + 3)
dx =7
4
(b) Let f (x, y, z) = x2 + y2 + z2 − a2 and z = φ (x, y) =√
a2 − x2 − y2.
F = 〈x, y, z〉 =⟨
x, y,√
a2 − x2 − y2⟩
Using the formula
∫ ∫
SF • n dS =
∫ ∫
R
(
−P∂z
∂x−Q
∂z
∂y+ R
)
dx dy
n = ∇f‖∇f‖ = 〈2x,2y,2z〉
2√
x2+y2+z2=
⟨
2x,2y,2√
a2−x2−y2⟩
2a =⟨
xa , y
a , 1a
√
a2 − x2 − y2⟩
F • n = x2
a + y2
a + 1a
(
a2 − x2 − y2)
= a
R = (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π
dS = 1|n • k|r dr dθ =
a√
a2 − x2 − y2r dr dθ =
ar√a2 − r2
dr dθ
∫ ∫
S F • n dS =∫ 2π0
∫ a0 (a)
ar√a2 − r2
dr dθ = a2∫ 2π0
[
−√
a2 − r2]a
0dθ = a3
∫ 2π0 dθ =
2πa3
or using the formula
∫ ∫
SF • n dS =
∫ ∫
DF (r (u, v)) • (ru × rv) du dv
Let
r (φ, θ) = 〈a sinφ cos θ, a sinφ sin θ, a cos φ〉 and D =
(φ, θ) |0 ≤ φ ≤ π
2, 0 ≤ θ ≤ 2π
then rφ = 〈a cos φ cos θ, a cos φ sin θ,−a sinφ〉 and rθ = 〈−a sinφ sin θ, a sinφ cos θ, 0〉.
145
CHAPTER 6. VECTOR CALCULUS
rφ × rθ =
∣
∣
∣
∣
∣
∣
∣
∣
i j k
a cos φ cos θ a cos φ sin θ −a sinφ
−a sinφ sin θ a sinφ cos θ 0
∣
∣
∣
∣
∣
∣
∣
∣
=⟨
a2 sin2 φ cos θ, a2 sin2 φ sin θ, a2 sinφ cos φ⟩
F (r (φ, θ)) = 〈a sinφ cos θ, a sinφ sin θ, a cos φ〉F (r (φ, θ)) • (rφ × rθ) = a3 sinφ
Therefore,
∫∫
DF (r (φ, θ)) • (rφ × rθ) dφ dθ =
∫ 2π
0
∫ π/2
0a3 sinφdφ dθ = 2πa3
(c) On the surface S1, z = 0, F = 〈x, 2y, 0〉 and n = 〈0, 0,−1〉, thus
∫∫
S1
F • ndS =
∫∫
S1
(0) dS = 0.
On the surface S2, z = 1, F = 〈x, 2y, 3〉, n = 〈0, 0, 1〉, dS = 1|n • k| dx dy = dx dy,
and S2 = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1, thus
∫∫
S2
F • ndS =
∫∫
S2
(3) dS = 3
∫ 1
0
∫ 1
0dx dy = 3.
On the surface S3, x = 0, F = 〈0, 2y, 3z〉, n = 〈−1, 0, 0〉, thus
∫∫
S3
F • ndS =
∫∫
S3
(0) dS = 0.
On the surface S4, x = 1, F = 〈1, 2y, 3z〉, n = 〈1, 0, 0〉, dS4 = 1|n • i| dx dy = dy dz,
and S4 = (y, z) |0 ≤ y ≤ 1, 0 ≤ z ≤ 1, thus
∫∫
S4
F • ndS =
∫∫
S4
(1) dS =
∫ 1
0
∫ 1
0dy dz = 1.
On the surface S5, y = 0, F = 〈x, 0, 3z〉, n = 〈0,−1, 0〉, thus
∫∫
S5
F • ndS =
∫∫
S5
(0) dS = 0.
On the surface S6, y = 1, F = 〈x, 2, 3z〉, n = 〈0, 1, 0〉, dS6 = 1|n•j| dx dy = dx dy, and
146
CHAPTER 6. VECTOR CALCULUS
S6 = (x, z) |0 ≤ x ≤ 1, 0 ≤ z ≤ 1, thus
∫∫
S6
F • ndS =
∫∫
S6
(2) dS = 2
∫ 1
0
∫ 1
0dx dz = 2.
Therefore,∫∫
SF • n dS =
∫∫
S1
+
∫∫
S2
+... +
∫∫
S6
= 6.
(41) For the top end-face of the cylinder x2 + y2 = 4, 0 ≤ z ≤ 3, n = 〈0, 0, 1〉. Using the
formula∫∫
SF • n dS =
∫∫
R
(
−P∂z
∂x−Q
∂z
∂y+ R
)
dx dy
F =⟨
4x,−2y2, z2x2⟩
=⟨
4x,−2y2, 9x2⟩
F • n = 9x2
R = (r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
dS = 1|n • k|r dr dθ = r dr dθ
∫∫
SF • n dS =
∫ 2π
0
∫ 2
09 (r cos θ)2 r dr dθ = 9
∫ 2π
0
∫ 2
0r3 cos2 θ dr dθ
= 9
∫ 2π
0
1 + cos 2θ
2dθ
∫ 2
0r3 dr
= 9
[
θ
2+
1
4sin 2θ
]2π
0
[
1
4r4
]2
0
= 36π
or using the formula
∫∫
SF • n dS =
∫ ∫
DF (r (u, v)) • (ru × rv) du dv
Let
r (r, θ) = 〈r cos θ, r sin θ, 3〉 and D = (r, θ) |0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
then rφ = 〈cos θ, sin θ, 0〉 and rθ = 〈−r sin θ, r cos θ, 0〉.
rr × rθ =
∣
∣
∣
∣
∣
∣
∣
∣
i j k
cos θ sin θ 0
−r sin θ r cos θ 0
∣
∣
∣
∣
∣
∣
∣
∣
= 〈0, 0, r〉
F (r (r, θ)) =⟨
4r cos θ,−2r2 sin2 θ, 9r2 cos2 θ⟩
F (r (r, θ)) • (rr × rθ) = 9r3 cos2 θ∫∫
D F (r (r, θ)) • (rr × rθ) dr dθ =∫ 2π0
∫ 20 9r3 cos2 θ dr dθ = 36π
147
CHAPTER 6. VECTOR CALCULUS
(42) (a)
O
x
y
z
Let f = x + 2y + z − 3 = 0, then n = ∇f =i + 2j + k
√
(1)2 + (2)2 + (1)2= 1√
6(i + 2j + k)
∫
SF • ndS =
∫
S(−i + 2j + 3k) •
(
1√6
(i + 2j + k)
)
dS =
∫
S
√6dS
=√
6
∫ 3
0
∫ 3−x2
0
√
1 + (−1)2 + (−2)2dydx
=27
2
(b)
O
x
y
z
Let f = ex − y = 0, then n = ∇f =exi− j
√
(ex)2 + (1)2=
exi− j√e2x + 1
∫
SF • ndS =
∫
S(−2i + 2yj + zk) •
(
exi− j√e2x + 1
)
dS =
∫
S
(−2ex − 2y)√e2x + 1
dS
Let Z = v, X = u, Y = eu, where 0 ≤ v ≤ 1 and 0 ≤ u ≤ ln 2.
Note that Xu = 1, Yu = eu, Zu = 0 and Xv = 0, Yv = 0, Zv = 1
E = (Xu)2 + (Yu)2 + (Zu)2 = (1)2 + (eu)2 + 02 = 1 + e2u
G = (Xv)2 + (Yv)
2 + (Zv)2 = (0)2 + (0)2 + (1)2 = 1
F = XuXv + YuYv + ZuZv = (1) (0) + (eu) (0) + (0) (1) = 0
∫
SF • ndS =
∫ 1
0
∫ ln 2
0
(−2eu − 2eu)√e2u + 1
√
(1 + e2u) (1)− 02dudv
= −4
∫ 1
0
∫ ln 2
0eududv = −4
∫ 1
0dv = −4
148
CHAPTER 6. VECTOR CALCULUS
(43) (a)
∫
D∇ • Fdxdydz =
∫
D
(
∂
∂xi+
∂
∂yj+
∂
∂zk
)
•(
x3i+y3j+z3k)
dxdydz
=
∫
D
(
∂x3
∂x+
∂y3
∂y+
∂z3
∂z
)
dxdydz = 3
∫
D
(
x2 + y2 + z2)
dxdydz
= 3
∫ π/2
0
∫ 2π
0
∫ 1
0r4 sinφdrdθdφ =
6
5π
On S1– the surface of the hemi-sphere x2 + y2 + z2 = 1 and n = r.
∫
S1
F • ndS =
∫
S1
F • rdS =
∫
S1
(
x3i+y3j+z3k)
• (xi+yj+zk) dS
=
∫
S1
(
x4+y4+z4)
dS
Let X = sin v cos u, Y = sin v sinu, Z = cos v, then√
EG− F 2 =√
sin2 v = sin v.
∫
S1
F • ndS
=
∫ 2π
0
∫ π/2
0
(
(sin v cos u)4 +(sin v sinu)4 + (cos v)4)
sin vdvdu
=
∫ 2π
0
∫ π/2
0
(
sin4 v cos4 u + sin4 v sin4 u + cos4 v)
sin vdvdu
=
∫ 2π
0
∫ π/2
0
(
sin4 v(
cos4 u + sin4 u)
+ cos4 v)
sin vdvdu
= −∫ 2π
0
[
(
cos4 u + sin4 u)
(
1
5cos5 v − 2
3cos3 v + cos v
)
+1
5cos5 v
]π/2
0
du
=
∫ 2π
0
(
8
15
(
cos4 u + sin4 u)
+1
5
)
du
=8
15
∫ 2π
0
(
(
1 + cos 2u
2
)2
+
(
1− cos 2u
2
)2)
du +2
5π
=8
15
∫ 2π
0
(
1 + 2 cos 2u + cos2 2u
4+
1− 2 cos 2u + cos2 2u
4
)
du +2
5π
=4
15
∫ 2π
0
(
1 + cos2 2u)
du +2
5π
=4
15
∫ 2π
0
(
1 +
(
1 + cos 4u
2
))
du +2
5π
=4
15
[
3u
2+
sin 4u
8
]2π
0
+2
5π =
6π
5
149
CHAPTER 6. VECTOR CALCULUS
On S2– the bottom surface (z = 0) and n = −k
∫
S2
F • ndS =
∫
S2
(
x3i+y3j+z3k)
• (−k) dS = −∫
S2
z3dS = 0
(b)
∫
D∇ • Fdxdydz =
∫
D
(
∂
∂xi+
∂
∂yj+
∂
∂zk
)
•(
2xzi+yj−z2k)
dxdydz
=
∫
D
(
∂2xz
∂x+
∂y
∂y+
∂(
−z2)
∂z
)
dxdydz
=
∫
D(2z + 1− 2z) dxdydz
=
∫∫
x2+4y2≤16
∫ y
0dzdxdy =
∫ ∫
x2+4y2≤16
ydxdy
Let x = 4r cos θ, y = 2r sin θ and |J | = 8r
∫
D∇ • Fdxdydz =
∫ π/2
0
∫ 1
016r2 sin θdrdθ =
16
3
On the plane z = y : n =−j + k√
2
∫
S1
F • ndS =
∫
S1
(
2xzi+yj−z2k)
• −j + k√2
dS =1√2
∫
S1
(
−y − z2)
dS
= − 1√2
∫∫
R1
(
y + y2)
√
1 + z2x + z2
ydxdy
= − 1√2
∫∫
R1
(
y + y2)√
1 + 1dxdy
= −∫∫
R1
(
y + y2)
dxdy
= −8
∫ π/2
0
∫ 1
0
(
2r2 sin θ + 4r3 sin2 θ)
drdθ
= −2π − 16
3
On the plane z = 0 : n = −k
∫
S2
F • ndS =
∫
S2
(
2xzi+yj−z2k)
• (−k) dS = −∫
S2
z2dS = 0
150
CHAPTER 6. VECTOR CALCULUS
On the plane x2 + 4y2 = 16 : n =xi + 4yj
√
x2 + 16y2
∫
S3
F • ndS =
∫
S3
(
2xzi+yj−z2k)
•(
xi + 4yj√
x2 + 16y2
)
dS
=
∫
S4
2x2z + 4y2
√
x2 + 16y2dS
Let X = 4 cos u, Y = 2 sinu and Z = v Then Xu = −4 sinu, Yu = 2 cos u, Zu = 0,
and Xv = 0, Yv = 0, Zv = 1.
E = X2u + Y 2
u + Z2u = 16 sin2 u + 4 cos2 u, G = 1, F = 0
∫
S3
F • ndS
=
∫ π/2
0
∫ 2 sin u
0
2 (4 cos u)2 (v) + 4 (2 sinu)2√
(4 cos u)2 + 16 (2 sinu)2
√
16 sin2 u + 4 cos2 udvdu
=
∫ π/2
0
∫ 2 sin u
0
(
16v cos2 u + 8 sin2 u)
dvdu
=
∫ π/2
0
(
32 sin2 u cos2 u + 16 sin3 u)
du
= 2π +32
3
On the plane x = 0 : n = −i
∫
S4
F • ndS =
∫
S2
(
2xzi+yj−z2k)
• (−i) dS = −∫
S4
2xzdS = 0
Hence,
∫
SF • ndS =
∫
S1
F • ndS +
∫
S2
F • ndS +
∫
S3
F • ndS +
∫
S4
F • ndS =16
3
(44) (a) F = 〈x, y, z〉
∇• F =∂
∂x(x) +
∂
∂y(y) +
∂
∂z(z) = 3
151
CHAPTER 6. VECTOR CALCULUS
D = (ρ, φ, θ) |0 ≤ ρ ≤ a, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π∫ ∫
DF • n dS =
∫ ∫ ∫
D(∇ • F) dV =
∫ 2π
0
∫ π
0
∫ a
0(3) ρ2 sinφdρ dφ dθ
= 3 (Volume of a sphere with radius a) = 4πa3
(b) F = 〈x, y, z〉
∇• F =∂
∂x(x) +
∂
∂y(y) +
∂
∂z(z) = 3
D = (x, y, z) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1∫ ∫
DF • n dS =
∫ ∫ ∫
D(∇ • F) dV =
∫ 1
0
∫ 1
0
∫ 1
0(3) dx dy dz
= 3 (Volume of a unit cube) = 3
(c) F =⟨
x2, y2, z2⟩
∇• F =∂
∂x
(
x2)
+∂
∂y
(
y2)
+∂
∂z
(
z2)
= 2x + 2y + 2z
D = (ρ, φ, θ) |0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π∫∫
DF • n dS
=
∫ ∫ ∫
D(∇ • F) dV =
∫ 2π
0
∫ π
0
∫ 1
0(2x + 2y + 2z) ρ2 sinφdρ dφ dθ
=
∫ 2π
0
∫ π
0
∫ 1
02 (ρ sinφ cos θ + ρ sinφ sin θ + ρ cos φ) ρ2 sinφdρ dφ dθ
=
∫ 1
0ρ3 dρ
∫ π
0
∫ 2π
02(
sin2 φ cos θ + sin2 φ sin θ + sin φ cos φ)
dθ dφ
=
[
1
4ρ4
]1
0
∫ π
02[
sin2 φ sin θ − sin2 φ cos θ + θ sinφ cos φ]2π
0dθ dφ
=1
4
∫ π
04π sinφ cos φdφ = π
[
1
2sin2 φ
]π
0
= 0
152
CHAPTER 6. VECTOR CALCULUS
(45) The total force acting on D is
−∫
SρgzndS • k = −
∫
Sρgzk • ndS = −
∫
D∇ • ρgzkdxdydz
= −ρg
∫
D
(
∂
∂xi+
∂
∂yj+
∂
∂zk
)
• zkdxdydz
= −ρg
∫
D
∂z
∂zdxdydz
= −ρg
∫
Ddxdydz
= −ρg × volume of D
(46) (a) F = 〈2z,−y, x〉
∇ × F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
2z −y x
∣
∣
∣
∣
∣
∣
∣
∣
= 〈0, 1, 0〉
The equation for the area within ∆ABC:
unit normal vector = n = 〈1,1,1〉√12+12+12
=⟨
1√3, 1√
3, 1√
3
⟩
(x− 2) + (y − 0) + (z − 0) = 0 =⇒ x + y + z = 2
find the integral by using surface integral
S = (x, y) |0 ≤ x ≤ 2, 0 ≤ y ≤ 2− xdS = 1
|n • k| dx dy =√
3 dx dy
∫
∂SF • dr =
∫
S(∇× F) • n dS =
∫ 2
0
∫ 2−x
0
(
1√3
)√3 dy dx =
∫ 2
0[y]2−x
0 dx
=
∫ 2
0(2− x) dx =
[
2x− 1
2x2
]2
0
= 2
Find the integral by using line integral
Γ1 (AB) : r (t) = 〈2− 2t, 2t, 0〉, 0 ≤ t ≤ 1
F (r (t)) = 〈0,−2t, 2− 2t〉r′ (t) = 〈−2, 2, 0〉F (r (t)) • r′ (t) = 〈0,−2t, 2− 2t〉 • 〈−2, 2, 0〉 = −4t∫
Γ1F (r (t)) • r′ (t) dt =
∫ 10 (−4t) dt = −2
Γ2 (BC) : r (s) = 〈0, 2− 2s, 2s〉, 0 ≤ s ≤ 1
F (r (s)) = 〈4s, 2s− 2, 0〉r′ (s) = 〈0,−2, 2〉
153
CHAPTER 6. VECTOR CALCULUS
F (r (s)) • r′ (s) = 〈4s, 2s− 2, 0〉 • 〈0,−2, 2〉 = 4− 4s∫
Γ2F (r (s)) • r′ (s) ds =
∫ 10 (4− 4s) ds = 2
Γ3 (CA) : r (u) = 〈2u, 0, 2− 2u〉, 0 ≤ u ≤ 1
F (r (u)) = 〈4− 4u, 0, 2u〉r′ (u) = 〈2, 0,−2〉F (r (u)) • r′ (u) = 〈4− 4u, 0, 2u〉 • 〈2, 0,−2〉 = 8− 12u∫
Γ3F (r (u)) • r′ (u) du =
∫ 10 (8− 12u) du = 2
Therefore,∫
∂S F • dr =∫
Γ1+∫
Γ2+∫
Γ3= −2 + 2 + 2 = 2.
(b) F = 〈xz,−y, xy〉
∇ × F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
xz −y xy
∣
∣
∣
∣
∣
∣
∣
∣
= 〈x, x− y, 0〉
The equation for the plane that including ∆ABC:
unit normal vector = n = 〈1,1,1〉√12+12+12
=⟨
1√3, 1√
3, 1√
3
⟩
(x− 1) + (y − 0) + (z − 0) = 0 =⇒ x + y + z = 1
Find the integral by using surface integral.
S = (x, y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1− xdS = 1
|n • k| dx dy =√
3 dx dy
∫
∂SF • dr =
∫
S(∇× F) • n dS =
∫ 1
0
∫ 1−x
0
[
1√3
(2x− y)
]√3 dy dx
=
∫ 1
0
[
2xy − 1
2y2
]1−x
0
dx =
∫ 1
0
(
3x− 1
2− 5
2x2
)
dx =1
6
Find the integral by using line integral
Γ1 (AB) : r (t) = 〈1− t, t, 0〉, 0 ≤ t ≤ 1
F (r (t)) =⟨
0,−t, t− t2⟩
r′ (t) = 〈−1, 1, 0〉F (r (t)) • r′ (t) =
⟨
0,−t, t− t2⟩
• 〈−1, 1, 0〉 = −t∫
Γ1F (r (t)) • r′ (t) dt =
∫ 10 (−t) dt = −1
2
Γ2 (BC) : r (s) = 〈0, 1− s, s〉, 0 ≤ s ≤ 1
F (r (s)) = 〈0, s− 1, 0〉r′ (s) = 〈0,−1, 1〉F (r (s)) • r′ (s) = 〈0, s− 1, 0〉 • 〈0,−1, 1〉 = 1− s∫
Γ2F (r (s)) • r′ (s) ds =
∫ 10 (1− s) ds = 1
2
Γ3 (CA) : r (u) = 〈u, 0, 1− u〉, 0 ≤ u ≤ 1
F (r (u)) =⟨
u− u2, 0, 0⟩
154
CHAPTER 6. VECTOR CALCULUS
r′ (u) = 〈1, 0,−1〉F (r (u)) • r′ (u) =
⟨
u− u2, 0, 0⟩
• 〈1, 0,−1〉 = u− u2
∫
Γ3F (r (u)) • r′ (u) du =
∫ 10
(
u− u2)
du = 16
Therefore,∫
∂S F • dr =∫
Γ1+∫
Γ2+∫
Γ3= −1
2 + 12 + 1
6 = 16 .
(47) (a)
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
x2 2x z2
∣
∣
∣
∣
∣
∣
∣
∣
= 2k
By Stokes’ Theorem,
∮
CF • dr =
∫∫
S∇× F • ndS =
∫∫
4x2+y2≤42k • kdS
= 2
∫∫
4x2+y2≤4dS = 2× 2π = 4π
(b)
∇× F =
∣
∣
∣
∣
∣
∣
∣
∣
i j k∂∂x
∂∂y
∂∂z
y zx x2
∣
∣
∣
∣
∣
∣
∣
∣
= −xi− 2xj+(z − 1)k
By Stokes’ Theorem,
∮
CF • dr =
∫∫
S∇× F • ndS
=
∫∫
S(−xi− 2xj+ (z − 1)k) •
(
1√3i +
1√3j+
1√3k
)
dS
=1√3
∫∫
S(−3x + z − 1) dS
=1√3
∫ 1
0
∫ 1−x
0(−3x− x− y)
√
1 + (zx)2 + (zy)2dydx
=1√3
∫ 1
0
∫ 1−x
0(−4x− y)
√
1 + (−1)2 + (−1)2dydx
=
∫ 1
0
∫ 1−x
0(−4x− y) dydx =
[
−4xy − 1
2y2
]1−x
0
=
∫ 1
0
(
−3x +7
2x2 − 1
2
)
dx
= −5
6
155
CHAPTER 6. VECTOR CALCULUS
(48) By Stokes’ Theorem,
∫∫
S∇× (yi) • ndS =
∮
C(yi) • dr =
∮
C(yi) • (dxi + dyj + dzk) =
∮
Cydx
=
∫ 2π
0sin td (cos t) = −
∫ 2π
0sin2 tdt = −π
156
Chapter 7
Differential Equations
(1) (a) dydx = 4y
x(y−3) =⇒ (y−3)dyy = 4dx
x =⇒(
1− 3y
)
dy = 4dxx
∫
(
1− 3y
)
dy = 4∫
dxx =⇒ y − 3 ln |y| = 4 ln |x|+ C
(b) dydx = y3+4y
x(y2+2y+4)=⇒ (y2+2y+4)dy
y3+4y= dx
x =⇒(
1y + 2
y2+4
)
dy = dxx
∫
(
1y + 2
y2+4
)
dy =∫
dxx =⇒ ln |y|+ tan−1 y
2 = ln |x|+ C
(c) dydx − 2
xy = − 1x2 p (x) = − 2
x µ (x) = e∫
p(x)dx = e−∫
2x
dx = e−2 ln|x| = 1x2
1x2
(
dydx − 2
xy)
= − 1x4 =⇒ d
dx
(
1x2 y)
= − 1x4 =⇒ 1
x2 y = −∫
1x4 dx = 1
3x3 + C
y = 13x + Cx2
(d) (x− 1) dydx = y + 2 (x− 2)3 =⇒ dy
dx = yx−1 + 2(x−2)3
x−1 =⇒ dydx −
yx−1 = 2(x−2)3
x−1
p (x) = − 1x−1
µ (x) = e−∫
1x−1
dx = 1x−1
1x−1
(
dydx −
yx−1
)
= 2(x−2)3
(x−1)2=⇒ d
dx
(
yx−1
)
= 2x− 8 + 6x−1 − 2
(x−1)2
yx−1 =
∫
[
2x− 8 + 6x−1 − 2
(x−1)2
]
dx = x2 − 8x + 6 ln |x− 1|+ 2x−1 + C
y =(
x2 − 8x + 6 ln |x− 1|+ 2x−1 + C
)
(x− 1)
(e) x3 dydx +
(
2− 3x2)
y = x3 =⇒ dydx +
(2−3x2)x3 y = 1
p (x) =(2−3x2)
x3
µ (x) = e∫ (2−3x2)
x3 dx = e
(
−3 ln x− 1
x2
)
= 1x3 e−
1
x2
1x3 e−
1
x2
[
dydx +
(2−3x2)x3 y
]
= 1x3 e−
1
x2 =⇒ ddx
(
1x3 e−
1
x2 y)
= 1x3 e−
1
x2
1x3 e−
1
x2 y =∫
1x3 e−
1
x2 dx = 12e−
1
x2 + C =⇒ y = 12x3 + Cx3e
1
x2
(f) dydx + (2 cos x) y = sin2 x cos x
157
CHAPTER 7. DIFFERENTIAL EQUATIONS
p (x) = 2 cos x
µ (x) = e∫
2 cos xdx = e2 sin x
e2 sin x[
dydx + (2 cos x) y
]
= e2 sin x sin2 x cos x =⇒ ddx
(
e2 sin xy)
= e2 sin x sin2 x cos x
e2 sin xy =∫
e2 sin x sin2 x cos xdx =∫
e2 sin x sin2 xd (sinx)
Let u = sin2 x and dv = e2 sin xd (sinx), then du = 2 sinx cos xdx and v = 12e2 sin x.
e2 sin xy =1
2e2 sin x sin2 x−
∫ (
1
2e2 sin x
)
(2 sinx cos xdx)
=1
2e2 sin x sin2 x−
∫
e2 sin x sinx cos xdx
=1
2e2 sin x sin2 x−
∫
e2 sin x sinxd (sinx)
=1
2e2 sin x sin2 x− 1
2e2 sin x sinx +
1
2
∫
e2 sin x cos xdx
=1
2e2 sin x sin2 x− 1
2e2 sin x sinx +
1
4e2 sin x + C
= e2 sin x
(
1
2sin2 x− 1
2sinx +
1
4
)
+ C
y = 12 sin2 x− 1
2 sin x + 14 + Ce−2 sin x or y = Ce−2 sin x − 1
4 cos 2x− 12 sinx + 1
2
(2) (a) The substitution y = xv(x) reduces the differential equation to
v + xdv
dx=
2 + v
1 + 3v, or x
dv
dx=
2− 3v2
1 + 3v.
Therefore, one has∫ 1 + 3v
2− 3v2dv =
∫ dx
x. Integration yields
√6− 6
12ln
(
y
x+
√6
3
)
−√
6 + 6
12ln
(
y
x−√
6
3
)
= lnCx,
where C is an arbitrary constant.
(b) Re-writing the equation as y′ +
(
3
x
)
y = 4 +2
x2and multiplying both sides of this
equation by µ = e∫
3x
dx = x3, we obtain
d
dx
(
x3y)
= 4x3 + 2x.
Integration yields
x3y = x4 + x2 + C, or y = x +1
x+
C
x3.
158
CHAPTER 7. DIFFERENTIAL EQUATIONS
(c) Let v = yx , then dv
dx = 1x
dydx −
yx2 =⇒ dy
dx = x(
dvdx + y
x2
)
.
x2(
dvdx + y
x2
)
− y + 3x3y − x4 = 0 =⇒ x2 dvdx + 3x3y − x4 = 0 =⇒ dv
dx + 3x2v = x2
p (x) = 3x2
µ (x) = e∫
3x2dx = ex3
ex3 ( dvdx + 3x2v
)
= x2ex3
=⇒ ddv
(
ex3
v)
= x2ex3
=⇒ ex3
v =∫
x2ex3
dx = 13ex3
+ C
v = 13 + Ce−x3
=⇒ yx = 1
3 + Ce−x3
=⇒ y = 13x + Cxe−x3
(d) Let v = yx , then dv
dx = 1x
dydx −
yx2 =⇒ dy
dx = x(
dvdx + y
x2
)
.
x(
dvdx + y
x2
)
= 2 yx +
(
xy
)3= 2v + 1
v3 =⇒ x dvdx + v = 2v + 1
v3 =⇒ x dvdx = v + 1
v3
=⇒ x dvdx = v4+1
v3 =⇒ v3dvv4+1
= dxx
∫
v3dvv4+1
=∫
dxx =⇒ 1
4 ln∣
∣v4 + 1∣
∣ = ln |x|+ C =⇒ 14 ln
∣
∣
∣
( yx
)4+ 1∣
∣
∣ = ln |x|+ C
(e) Let v = 1y , then dv = − 1
y2 dy =⇒ dy = −y2dv = − 1v2 dv.
− dvv2dx
+ 1v = 1
v2 =⇒ dvv2dx
= 1v − 1
v2 =⇒ dvdx = v − 1 =⇒ dv
v−1 = dx∫
dvv−1 =
∫
dx =⇒ ln |v − 1| = x + C =⇒ ln∣
∣
∣
1y − 1
∣
∣
∣= x + C
(f) Let v = 1y4 , then dv = − 4
y5 dy =⇒ dy = −14y5dv.
−14y5 dv
dx − 12xy = −y5 =⇒ dv
dx + 2xy4 = 4 =⇒ dv
dx + 2vx = 4
p (x) = 2x
µ (x) = e∫
2x
dx = e2 ln x = x2
x2(
dvdx + 2v
x
)
= 4x2 =⇒ ddx
(
x2v)
= 4x2 =⇒ x2v =∫
4x2dx = 43x3 + C
v = 43x + C 1
x2 =⇒ 1y4 = 4
3x + C 1x2
(g) Let v = y − 4x, then dvdx = dy
dx − 4 =⇒ dydx = dv
dx + 4.
dvdx + 4 = v2 =⇒ dv
dx = v2 − 4 =⇒ dvv2−4
= dx∫
dvv2−4
=∫
dx =⇒ −∫
dv4−v2 =
∫
dx =⇒ −14 ln
∣
∣
∣
v+2v−2
∣
∣
∣= x + C
Use the formula∫
dua2−u2 = 1
2a ln∣
∣
∣
u+au−a
∣
∣
∣+ C.
(h) Putting v =1
y, the equation is then transformed into
− 1
v2
dv
dx+
1
xv=
x
v2, or
dv
dx− 1
xv = −x,
which is a linear equation with unknown v. Multiplying the equation by µ =
e∫
− 1x
dx =1
x, one obtains
d
dx
(
1
xv
)
= −1, which implies1
xv = −x + C, or
1
xy=
−x + C.
(i) Substituting y = w− 13 into the differential equation, we obtain
x3
3w− 4
3dw
dx+ x2w− 1
3 = w− 43 cos x,
159
CHAPTER 7. DIFFERENTIAL EQUATIONS
ordw
dx+
(
3
x
)
w =3 cos x
x3, which is a linear equation with unknown w. Multiplying
the equation by µ = x3, we haved
dx(x3w) = 3 cos x. Integration then gives w =
3 sinx + C
x3, or y =
x3√
3 sinx + C.
(3) (a) The equation is separable and may be written as y dy = ex dx, which may be inte-
grated to give y2 = 2ex + C, where C is a constant of integration. Applying the
initial condition y(0) = 1, we conclude that C = −1 and that the solution to the
initial value problem satisfies y2 = 2ex − 1.
(b) The change of variable y = xv(x) reduces the equation to
dx
x=
dv
cot v= tan v dv,
or∫ dx
x=∫
tan v dv, which implies lnx = − ln cos v + lnC. Therefore, the solution
of the differential equation satisfies
x cosy
x= C. By the initial condition y(π) = 0, we have C = π. Therefore,
x cosy
x= π.
(c) The differential equation is separable, which may be integrated to give
∫
(1− y5)dy =
∫
x ex2
dx,
or y − y6
6= 1
2ex2
+ C. Since y(0) = 0, we conclude that C = −12 . As such, the
solution of the initial value problem satisfies
y6 − 6y + 3ex2
= 3.
(d) dydx = x2y−y
y+1 =y(x2−1)
y+1 =⇒ y+1y dy =
(
x2 − 1)
dx =⇒(
1 + 1y
)
dy =(
x2 − 1)
dx
∫
(
1 + 1y
)
dy =∫ (
x2 − 1)
dx =⇒ y + ln |y| = x3
3 − x + C
x = 3, y = −1 =⇒ −1 + ln |−1| = 9− 3 + C =⇒ C = −7
y + ln |y| = x3
3 − x− 7
(e) dydx + 2xy = 0 =⇒ dy
dx = −2xy =⇒ dyy = −2xdx =⇒ ln |y| = −x2 + C
x = 0, y = 2 =⇒ ln |2| = C
ln |y| = −x2 + ln 2 =⇒ y = e−x2+ln 2 = 2e−x2
(f) Let v = yx , then dv
dx = 1x
dydx −
yx2 =⇒ dy
dx = x(
dvdx + y
x2
)
.
x(
dvdx + y
x2
)
= 1+(y/x)2
2(xy/x2)= 1+v2
2v =⇒ x dvdx + v = 1+v2
2v =⇒ x dvdx = 1−v2
2v =⇒ 2v1−v2 dv = dx
x
160
CHAPTER 7. DIFFERENTIAL EQUATIONS
∫
2v1−v2 dv =
∫
dxx =⇒ − ln
∣
∣1− v2∣
∣ = ln |x|+ C =⇒ − ln∣
∣
∣1−
( yx
)2∣
∣
∣= ln |x|+ C
x = 1, y = −2 =⇒ − ln∣
∣
∣1−
(−21
)2∣
∣
∣= ln |1|+ C =⇒ C = − ln 3
− ln∣
∣
∣1−
( yx
)2∣
∣
∣= ln |x| − ln 3
(g) dydx + xy = xy2 =⇒ dy
dx = xy (y − 1) =⇒ dyy(y−1) = xdx =⇒
(
1y−1 − 1
y
)
dy = xdx
∫
(
1y−1 − 1
y
)
dy =∫
xdx =⇒ ln |y − 1| − ln |y| = x2
2 + C =⇒ ln∣
∣
∣
y−1y
∣
∣
∣= x2
2 + C
x = 0, y = 2 =⇒ ln∣
∣
2−12
∣
∣ = C =⇒ C = − ln 2
ln∣
∣
∣
y−1y
∣
∣
∣= x2
2 − ln 2
(4) (a) With M = 2xy+cosx and N = x2+sin y, a simple calculation gives∂M
∂y=
∂N
∂x= 2x.
So the equation is exact. Integrating the equation∂f
∂x= M with respect to x results
in
f(x, y) =
∫
(2xy + cos x) dx = x2y + sinx + g(y),
where g(y) is an arbitrary function of f . If we substitute this into the second equation∂f
∂y= N , we obtain
dg
dy= sin y, or g(y) = − cos y. We therefore conclude that the
solution of the equation satisfies f(x, y) = C, or x2y + sinx− cos y = C.
(b) Since∂
∂y
[
y + 2xy3]
=∂
∂x
[
1 + 3x2y2 + x]
= 1 + 6xy2, the equation is exact. We
consider the equations
∂f
∂x= y + 2xy3
∂f
∂y= 1 + 3x2y2 + x
Integrating the first equation with respect to x, we obtain
f(x, y) =
∫
(y + 2xy3)dx = xy + x2y3 + g(y).
Substituting this into the second equation, we conclude that
x + 3x2y2 +dg
dy= 1 + 3x2y2 + x.
Thusdg
dy= 1 and g(y) = y. Therefore, f(x, y) = xy + x2y3 + y, which implies that
the solution of the equation satisfies
xy + x2y3 + y = C.
161
CHAPTER 7. DIFFERENTIAL EQUATIONS
(5) Let y (t) be the number of bacteria at time t (hour) and dydt be the growth rate of the
bacteria. So dydt = ay, where a is a constant.
dydt = ay =⇒ dy
y = adt =⇒ ln |y| = at + C =⇒ y (t) = eateC = A0eat, where A0 is a
constant.
Let y (0) = A0, then y (6) = A0e6a = 2A0.
=⇒ e6a = 2 =⇒ a = 16 ln 2
Thus, y (t) = A0e(t ln 2)/6.
y (24) = A0e(24 ln 2)/6 = 16A0
So the number of bacteria is 16 times as much as the number at the beginning.
(6) Let x (t) be the weight of the radioactive material at time t (year) and dxdt be the rate of
change of the weight. So dxdt = −bx, where b is a constant.
dxdt = −bx =⇒ x (t) = B0e
−bt, where B0 is a constant.
Let x (0) = 100 = B0, then x (2) = B0e−2b = 0.95× 100.
=⇒ e−2b = 0.95 =⇒ b = −12 ln 0.95
Thus, x (t) = 100e(t ln 0.95)/2.
x (t) = 100e(t ln 0.95)/2 = 0.9× 100 =⇒ e(t ln 0.95)/2 = 0.9 =⇒ t = 2 ln 0.9ln 0.95 = 4.1082
So it takes 4.11 years for 10% of the original mass to decay.
(7) Let z (t) be the amount of the deposit at time t (year) and dzdt be the rate of change of
the deposit. So dzdt = 0.07z and z (t) = C0e
0.07t, where C0 is a constant.
Let z (0) = C0 = 10000, then z (t) = 10000e0.07t.
z (2) = 10000e0.07(2) = 11503
So there will be $11503 in the account after 2 years.
(8) Let w (t) be the temperature of the bar at time t (hour) and dwdt be the rate of change of
the temperature. So dwdt = d (w (t)− S), where d is a constant and S is the surrounding
temperature.
dww(t)−S = (d) dt =⇒ ln |w (t)− S| = dt + c =⇒ w (t) = S + Doe
dt, where D0 is a constant.
Let w (0) = 100 and S = 20, then 20 + D0 = 100 =⇒ D0 = 80.
w(
13
)
= 20 + 80ed/3 = 50 =⇒ ed/3 = 38 =⇒ d = 3 ln 3
8
Thus, w (t) = 20 + 80e(3 ln 38)t.
162
CHAPTER 7. DIFFERENTIAL EQUATIONS
(9) Let T (t) be the temperature of the copper ball at time t. By Newton’s Law of Cooling,
we havedT
dt= k(T − 30),
where k is a physical constant, with initial condition T (0) = 100. The equation is
separable, and an integration gives T (t)−30 = C ekt for any t > 0. T (0) = 100⇒ C = 70,
and T (3) = 70⇒ e3k =40
70⇒
k = −13 ln
(
74
) ∼= −0.187. Thus T (t) = 30 + 70 e−13
ln( 74)t for any t > 0. If T (t) = 40, a
simple calculation shows that t = 3× ln 7
ln(7/4)∼= 10.4 minutes.
(10) Dividing the equation by L and then multiplying both sides by
µ = eRL
t, we obtaind
dt
[
eRL
t × i]
=E0
Le
RL
t cos ωt.
Integration yields eRL
t × i =E0
L
∫
eRL
t cos ωt dt, which implies
i(t) =E0
R2 + ω2L2(R cos ωt + ωL sin ωt) + C e−
RL
t,
where C is an arbitrary constant. Using the initial condition i(0) = 0, we conclude that
C =−E0R
R2 + ω2L2. Therefore,
i(t) =E0
R2 + ω2L2
[
R cos ωt + ωL sin ωt−R e−RL
t]
.
(11) Let S(t) be the amount of pollutants at any instant t, S(0) = S0 be the initial amount
of pollutants, V be the volume of water in the tank and r be the rate of influx / outflux.
Since the influx has no pollutants at all, we have
dS
dt=
(
dS
dt
)
in
−(
dS
dt
)
out
= −rS
V
Integration gives S(t) = S0e−( r
V )t for any t > 0.
(a) If S(t50%) =1
2S0, then e−( r
V )t50% =1
2, or
( r
V
)
t50% = ln 2. Thus t50% =V
r× ln 2.
(b) If S(t10%) =1
10S0, then e−( r
V )t10% =1
10, or
( r
V
)
t10% = ln 10. Thus t10% =
V
r× ln 10.
As V = 5000 and r = 4000, t50% =5000
4000× ln 2 ∼= 0.866 hours (or 52 minutes) and
163
CHAPTER 7. DIFFERENTIAL EQUATIONS
t10% =5000
4000× ln 10 ∼= 2.88 hours (or 2 hours 53 minutes).
(12) The equation of motion is my = mg, or y = g, where g ∼= 9.8 m/sec2 is the acceleration
due to gravity. The general solution is y(t) = 12gt2+c1t+c2, where c1 and c2 are arbitrary
constants. Using the initial conditions y(0) = 0 and y(0) = −5, we obtain c2 = 0 and
c1 = −5 and therefore y(t) = 12gt2 − 5t. Since y(t) = 0 if and only if t = 5
g∼= 0.51 sec,
the minimum value of y(t) is equal to −252g∼= −1.276. As such, the maximum height
reached by the stone is 1.276 meters above the point from which the stone was released.
When y(t) = 650, we solve 12gt2 − 5t − 650 = 0 to obtain t =
5 +√
25 + 1300g
g∼= 12.04
sec, meaning that the stone will strike the ground approximately 12.04 seconds after
release. Since y = gt − 5, the velocity when the stone strikes the ground is given by√
25 + 1300g ∼= 112.98 m/sec.
(13) (a) Auxiliary equation: m2 + 4m + 4 = 0. ∴ m1 = m2 = −2.
The general solution is: y = e−2x(c1 + c2x)
(b) Auxiliary equation: m2 − 7m + 12 = 0. ∴ m1 = 3, m2 = 4.
The general solution is: y = c1e3x + c2e
4x.
(c) Auxiliary equation: m2 + 4m + 9 = 0. ∴ m = −2±√
5i.
The general solution is: y = e−2x(c1 cos√
5x + c2 sin√
5x)
(d) Auxiliary equation: m2 − 2m + 10 = 0. ∴ m = 1± 3i.
The general solution is: y = ex(A cos 3x + B sin 3x).
(e) Auxiliary equation: m2 − 2m + 1 = 0. ∴ m = 1, 1.
The general solution is: y = c1ex + c2xex.
(f) Auxiliary equation: m2 + 2m + 2 = 0. ∴ m = −1± i.
The general solution is: y = e−x (c1 cos x + c2 sinx).
(14) (a) Let yp (x) = Ke3x, then y′p (x) = 3Ke3x.
y′ − 5y = e3x =⇒ 3Ke3x − 5(
Ke3x)
= e3x =⇒ −2K = 1 =⇒ K = −12
Therefore, yp (x) = −12e3x
(b) Let yp (x) = Ke−5x, then y′p (x) = −5Ke−5x.
y′ + 6y = 4e−5x =⇒ −5Ke−5x + 6(
Ke−5x)
= 4e−5x =⇒ K = 4
Therefore, yp (x) = 4e−5x
(c) Let yp (x) = (Ax + B) e2x, then
y′p (x) = Ae2x + 2 (Ax + B) e2x = [2Ax + (A + 2B)] e2x.
164
CHAPTER 7. DIFFERENTIAL EQUATIONS
y′ − 5y = xe2x
[2Ax + (A + 2B)] e2x − 5[
(Ax + B) e2x]
= xe2x
[−3Ax + (A− 3B)] = x
Comparing the coefficients of the polynomial,
−3A = 1
A− 3B = 0=⇒
A = −13
B = −19
Therefore, yp (x) =(
−13x− 1
9
)
e2x.
(d) Let yp (x) = (Ax + B) e2x, then y′p (x) = [2Ax + (A + 2B)] e2x.
y′ + 6y = (2x− 1) e2x =⇒ [2Ax + (A + 2B)] e2x + 6 (Ax + B) e2x = (2x− 1) e2x
=⇒ [8Ax + (A + 8B)] = 2x− 1
Comparing the coefficients of the polynormial,
8A = 2
A + 8B = −1=⇒
A = 14
B = − 532
Therefore, yp (x) =(
14x− 5
32
)
e2x.
(e) Let yp (x) =(
Ax2 + Bx + C)
e2x, then
y′p (x) = (2Ax + B) e2x + 2e2x(
Ax2 + Bx + C)
=[
2Ax2 + (2A + 2B)x + (B + 2C)]
e2x.
y′ − 5y =(
−9x2 + 6x)
e2x
=⇒[
2Ax2 + (2A + 2B)x + (B + 2C)]
e2x−5(
Ax2 + Bx + C)
e2x =(
−9x2 + 6x)
e2x
=⇒ y′p (x) =[
−3Ax2 + (2A− 3B)x + (B − 3C)]
e2x =(
−9x2 + 6x)
e2x
Comparing the coefficients of the polynormial,
− 3A = −9
2A− 3B = 6
B − 3C = 0
=⇒
A = 3
B = 0
C = 0
Therefore, yp (x) = 3x2e2x.
(15) (a) Putting yp = Ax2 +Bx+C, where A, B and C are coefficients to be determined, we
obtain y′p = 2Ax + B and y′′p = 2A. Substituting into the equation and comparing
coefficients, we have
A =1
2, B =
3
2, C =
7
4
Therefore, yp = 14(2x2 + 6x + 7).
165
CHAPTER 7. DIFFERENTIAL EQUATIONS
(b) Putting yp = A cos 4x + B sin 4x into the equation, we have
y′p = −A sin 4x + 4B cos 4x and y′′p = −16A cos 4x − 16B sin 4x. Comparing coeffi-
cients, we obtain A = 225 , B = − 1
25 . Therefore, yp = 125(2 cos 4x− sin 4x).
(c) Solve for yh: m− 5 = 0 =⇒ m = 5 =⇒ yh = c1e5x.
Let yp = Ax2 + Bx + C, then y′p = 2Ax + B.
(2Ax + B)− 5(
Ax2 + Bx + C)
= −5Ax2 + (2A− 5B)x + (B − 5C) = 8x2 − 2
Comparing the coefficients of the polynormial,
−5A = 8
2A− 5B = 0
B − 5C = −2
=⇒
A = −85
B = −1625
C = 34125
Therefore, yp = −85x2 − 16
25x + 34125 and y = c1e
5x − 85x2 +−16
25x + 34125 .
(d) yh = c1e5x
Let yp = Ke3x, then y′p = 3Ke3x.
3Ke3x − 5Ke3x = e3x =⇒ K = −12
Therefore, yp = −12e3x and y = c1e
5x − 12e3x.
(e) yh = c1e5x
Let yp = A sinx + B cos x, then y′p = A cos x−B sinx.
(A cos x−B sinx)−5 (A sinx + B cos x) = (A− 5B) cos x+(−B − 5A) sinx = cos x
Comparing the coefficients of sine and cosine,
A− 5B = 1
−B − 5A = 0=⇒
A = 126
B = − 526
Therefore, yp = 126 sinx− 5
26 cos x and y = c1e5x + 1
26 sinx− 526 cos x.
(f) Let yp = Ae3x + Bx2 + Cx + D, where A, B, C and D are constants to be deter-
mined. Substituting yp into the equation y′′ +3y′ +2y = 10e3x +4x2 and comparing
coefficients, we obtain
A =1
2, B = 2, C = −6 and D = 7.
Hence yp = 12e3x + 2x2 − 6x + 7.
(g) Putting yp = A cos 2x + B sin 2x + Cex, we obtain y′p = 2B cos 2x− 2A sin 2x + Cex
and y′′p = −4A cos 2x− 4B sin 2x + Cex. By comparing coefficients, we have
−3A + 4B = 2, − 4A− 3B = 0, 4C = 3.
Solving the equations, we conclude that A = − 625 , B = 8
25 and C = 34 . Therefore,
yp = − 625 cos 2x + 8
25 sin 2x + 34ex.
166
CHAPTER 7. DIFFERENTIAL EQUATIONS
(h) yh = c1e5x
Let yp = A sinx + B cos x + Ce3x, then y′p = A cos x−B sinx + 3Cex.
(A cos x−B sinx + 3Cex)− 5(
A sin x + B cos x + Ce3x)
= e3x + cos x
=⇒ (A− 5B) cos x− (5A + B) sinx− 2Ce3x = e3x + cos x
Comparing the coefficients of each function term,
A− 5B = 1
5A + B = 0
−2C = 1
=⇒
A = 126
B = − 526
C = −12
Therefore, yp = 126 sinx− 5
26 cos x− 12e3x and y = c1e
5x+ 126 sinx− 5
26 cos x− 12e3x.
(i) Solve for yh: m2 + 4m + 8 = 0 =⇒ m = −2± 2i.
With α = −2 and β = 2, yh = e−2x (c1 sin 2x + c2 cos 2x).
Let yp = Ax2 + Bx + C, then y′p = 2Ax + B and y′′p = 2A.
2A + 4 (2Ax + B) + 8(
Ax2 + Bx + C)
= 8Ax2 + (8A + 8B) x + (2A + 4B + 8C) =
8x2 + 8x + 18
Comparing the coefficients of the polynormial,
8A = 8
8A + 8B = 8
2A + 4B + 8C = 18
=⇒
A = 1
B = 0
C = 2
Therefore, yp = x2 + 2 and y = e−2x (c1 sin 2x + c2 cos 2x) + x2 + 2.
(j) Solve for yh: m2 + 4m + 5 = 0 =⇒ m = −2± i.
With α = −2 and β = 1, yh = e−2x (c1 sinx + c2 cos x).
Let yp = A sinx + B cos x, then y′p = A cos x−B sinx and y′′p = −A sinx−B cos x.
4 (A−B) sinx + 4 (A + B) cos x
= (−A sinx−B cos x) + 4 (A cos x−B sinx) + 5 (A sinx + B cos x)
Comparing the coefficients of the sine and cosine,
4 (A−B) = −2
4 (A + B) = 2=⇒
A = 0
B = 12
Therefore, yp = 12 cos x and y = e−2x (c1 sinx + c2 cos x) + 1
2 cos x.
(k) Solve for yh: m2 − 2m = 0 =⇒ m = 0, 2. Then yh = c1 + c2e2x.
167
CHAPTER 7. DIFFERENTIAL EQUATIONS
Let yp = Aex sinx + Bex cos x, then
y′p = Aex cos x + Aex sinx + Bex cos x−Bex sinx
= [(A−B) sinx + (A + B) cos x] ex
y′′p = [(A−B) sinx + (A + B) cos x] ex + ex [(A−B) cos x− (A + B) sin x]
= ex (2A cos x− 2B sinx) .
and
ex (2A cos x− 2B sinx)− 2 [(A−B) sinx + (A + B) cos x] ex = ex sin x
ex (−2A sinx− 2B cos x) = ex sin x
Comparing the coefficients of the sine and cosine,
−2A = 1
−2B = 0=⇒
A = −12
B = 0
Therefore, yp = −12ex sin x and y = c1 + c2e
2x − 12ex sin x.
(l) Solve for yh: m2 −m + 2 = 0 =⇒ m = 12 ±
√7
2 i.
With α = 12 and β =
√7
2 , yh = ex2
(
c1 sin√
72 x + c2 cos
√7
2 x)
.
Let yp =(
Ax2 + Bx + C)
ex, then
y′p =[
Ax2 + (2A + B)x + (B + C)]
ex
y′′p = ex[
Ax2 + (4A + B) x + (2A + 2B + C)]
.
Substituting into the differential equation and simplifying yield
ex[
2Ax2 + (2A + 2B) x + (2A + B + C)]
=(
6x2 + 8x + 7)
ex
Comparing the coefficients of polynomial,
2A = 6
2A + 2B = 8
2A + B + 2C = 7
=⇒
A = 3
B = 1
C = 0
Therefore, yp =(
3x2 + x)
ex and y = ex2
(
c1 sin√
72 x + c2 cos
√7
2 x)
+(
3x2 + x)
ex.
(m) Solve for yh: m2 + 2m + 3 = 0 =⇒ m = −1±√
2i.
With α = −1 and β =√
2, yh = e−x(
c1 sin√
2x + c2 cos√
2x)
.
168
CHAPTER 7. DIFFERENTIAL EQUATIONS
Let yp = Ax2 + Bx + C + D sin x + E cos x, then
y′p = 2Ax + B + D cos x− E sinx
y′′p = 2A−D sinx− E cos x.
Substituting into the differential equation and simplifying yield
3Ax2+(4A + 3B) x+(2A + 2B + 3C)+(2D − 2E) sinx+(2D + 2E) cos x = x2+sinx
Comparing the coefficients of each function term,
3A = 1
4A + 3B = 0
2A + 2B + 3C = 0
2D − 2E = 1
2D + 2E = 0
=⇒
A = 13
B = −49
C = 227
D = 14
E = −14
Therefore, yp = 13x2− 4
9x+ 227 + 1
4 sin x− 14 cos x and y = e−x
(
c1 sin√
2x + c2 cos√
2x)
+13x2 − 4
9x + 227 + 1
4 sinx− 14 cos x.
(16) Substitution of yp = K(x) ekx, where K(x) is a function to be determined, into the
differential equation yields
K ′′(x) + (2k + p)K ′(x) + (k2 + pk + q)K(x) = A.
Since k2 + pk + q = 0, one has K ′′(x) + (2k + p)K ′(x) = A.
(i) If 2k + p 6= 0 (distinct roots), then K(x) =Ax
2k + p.
(ii) If 2k + p = 0 (repeated root), we have K(x) =Ax2
2.
(17) Putting yp = w(x) ekx,where w(x) is a function to be determined, we obtain
w′′ + (2k + p)w′ + (k2 + pk + q)w = f(x).
(a) Substitution of y = e−xw into the equation y′′ + 2y′ + y = 3x2e−x yields w′′ = 3x2.
Therefore, w =x4
4and thus yp =
x4
4e−x is a particular solution.
(b) Substitution of y = e4xw into the equation y′′− 8y′ = 2e4x sin 3x yields w′′− 16w =
2 sin 3x. Putting w = A cos 3x + B sin 3x and using the method of undetermined
coefficients, we conclude that w = − 225 sin 3x. Therefore yp = − 2
25 sin 3xe4x is a
particular solution of the equation.
169
CHAPTER 7. DIFFERENTIAL EQUATIONS
(18) (a)
W(
et sin t, et cos t)
= det
[
et sin t et cos t
et (sin t + cos t) et (cos t− sin t)
]
=(
et sin t) [
et (cos t− sin t)]
−(
et cos t) [
et (sin t + cos t)]
= −e2t
(b)
W (y1, y2, y3) = det
1 sin 2t cos 2t
(1)′ (sin 2t)′ (cos 2t)′
(1)′′ (sin 2t)′′ (cos 2t)′′
= det
1 sin 2t cos 2t
0 2 cos 2t −2 sin 2t
0 −4 sin 2t −4 cos 2t
= det
[
2 cos 2t −2 sin 2t
−4 sin 2t −4 cos 2t
]
= −8 cos2 2t− 8 sin2 2t = −8
(c)
W (y1, y2, y3, y4) = det
1 x x2 x3
0 1 2x 3x2
0 0 2 6x
0 0 0 6
= 12
(19) (a) The homogeneous equation has two independent solutions
y1 = e−2x cos 2x and y2 = e−2x sin 2x. Therefore,
W (x) =
∣
∣
∣
∣
∣
e−2x cos 2x e−2x sin 2x
−2e−2x (cos 2x + 2 sin 2x) 2e−2x (− sin 2x + cos 2x)
∣
∣
∣
∣
∣
= 2e−4x.
Using variation of parameters, one has a particular solution of the form yp =
v1e−2x cos 2x + v2e
−2x sin 2x, where
v1 =
∫ −y2R(x)
W (x)dx = −
∫
e−2x sin 2xe−2x
2e−4xdx =
cos 2x
4,
v2 =
∫
y1R(x)
W (x)dx =
∫
e−2x cos 2xe−2x
2e−4xdx =
sin 2x
4.
Therefore,
yp =cos 2x
4e−2x cos 2x +
sin 2x
4e−2x sin 2x.
170
CHAPTER 7. DIFFERENTIAL EQUATIONS
(b) The homogeneous equations has independent solutions y1 = e−x and y2 = e2x.
Therefore, W (x) = 3ex. The equation has a particular solution of the form yp =
v1e−x + v2e
2x, where
v1 =
∫ −e2x sin 2x
3exdx = −1
3
∫
ex sin 2x dx =1
15(−ex sin 2x + 2ex cos 2x)
and
v2 =
∫
e−x sin 2x
3exdx =
1
3
∫
e−2x sin 2x dx =e−2x
12(− cos 2x− sin 2x) .
Therefore,
yp =1
15(− sin 2x + 2 cos 2x) +
1
12(− sin 2x− cos 2x) = − 3
20sin 2x +
1
20cos 2x.
(c) We have y1 = ex, y2 = e3x and W (x) = 2e4x. The equation has a particular solution
of the form yp = v1ex + v2e
3x, where
v1 =
∫ −e3x
2e4x
1
1 + e−xdx =
1
2ln(1 + ex)− x
2
and
v2 =
∫
ex
2e4x
1
1 + e−xdx = −1
4e−2x +
1
2e−x − 1
2ln(1 + e−x).
Then yp = ex(
12 ln(1 + ex)− x
2
)
+ e3x(
−14e−2x + 1
2e−x − 12 ln(1 + e−x)
)
.
(20) (a) For y′ + 1xy = 0, dy
dx = − yx =⇒ dy
y = −dxx =⇒ ln |y| = − ln |x|+ C
=⇒ yh = eCe− ln|x| = c11x .
Let yp = v (x) yh = v (x) 1x , then y′p = 1
xv′ (x)− 1x2 v (x) and v′ (x) = ln x
x1yh
= lnx.
Thus, v (x) =∫
lnxdx = x (lnx− 1) and yp = lnx−1. Therefore, y = c11x +lnx−1.
(b) yh = c1e5x
v′ (x) = e−5x(
8x2 − 2)
v (x) =∫
e−5x(
8x2 − 2)
dx = e−5x(
−85x2 − 16
25x + 34125
)
yp = −85x2 − 16
25x + 34125
y = c1e5x − 8
5x2 − 1625x + 34
125
(c) yh = c1e5x
v′ (x) = e−5x(
e3x)
= e−2x
v (x) =∫
e−2xdx = −12e−2x
yp = −12e3x
y = c1e5x − 1
2e3x
171
CHAPTER 7. DIFFERENTIAL EQUATIONS
(d) yh = c1e5x
v′ (x) = e−5x (cos x)
v (x) =∫
e−5x (cos x) dx = 126e−5x (sin x− 5 cos x)
yp = 126 (sin x− 5 cos x)
y = c1e5x + 1
26 sinx− 526 cos x
(e) Solve for yh: m2 −m− 2 = 0 =⇒ m = −1, 2.
Thus, yh = c1e−x + c2e
2x.
Let y1 = e−x and y2 = e2x, then W (y1, y2) = 3ex.
v1 =∫
− e2x(4x2)3ex dx = −4
3ex(
x2 − 2x + 2)
v2 =∫ e−x(4x2)
3ex dx = −13e−2x
(
2x2 + 2x + 1)
yp = −43ex
(
x2 − 2x + 2)
(e−x)− 13e−2x
(
2x2 + 2x + 1) (
e2x)
= −2x2 + 2x− 3
y = c1e−x + c2e
2x − 2x2 + 2x− 3
(f) yh = c1e−x + c2e
2x
Let y1 = e−x and y2 = e2x, then W (y1, y2) = 3ex.
v1 =∫
− e2x sin 2x3ex dx = 2
15ex cos 2x− 115ex sin 2x
v2 =∫
e−x sin 2x3ex dx = − 1
12e−2x (cos 2x + sin 2x)
yp = 115 (2 cos 2x− sin 2x)− 1
12 (cos 2x + sin 2x) = 120 cos 2x− 3
20 sin 2x
y = c1e−x + c2e
2x + 120 cos 2x− 3
20 sin 2x
(g) yh = c1e−x + c2e
2x
Let y1 = e−x and y2 = e2x, then W (y1, y2) = 3ex.
v1 =∫
− e2x(e2x)3ex dx = −1
9e3x
v2 =∫ e−x(e2x)
3ex dx = 13x
yp = −19e2x + 1
3xe2x
y = c1e−x + c2e
2x − 19e2x + 1
3xe2x
(h) Solve for yh: m2 − 2m + 1 = 0 =⇒ m = 1, 1. Then yh = c1ex + c2xex.
Let y1 = ex and y2 = xex. Then W (y1, y2) = e2x.
v1 =∫
−xex
(
ex
x
)
e2x dx = −x
v2 =∫ ex
(
ex
x
)
e2x dx = lnx
yp = −xex + xex lnx
y = c1ex + c2xex − xex + xex lnx
(i) Solve for yh: m2 + 4 = 0 =⇒ m = ±2i. Then with α = 0 and β = 2, yh =
c1 cos 2x + c2 sin 2x.
172
CHAPTER 7. DIFFERENTIAL EQUATIONS
Let y1 = cos 2x and y2 = sin 2x, then W (y1, y2) = 2.
v1 =∫
− sin 2x(4 sec2 2x)2 dx = − sec 2x
v2 =∫ cos 2x(4 sec2 2x)
2 dx = ln |sec 2x + tan 2x|yp = 2 sin 2x ln |sec 2x + tan 2x| − 2
y = c1 cos 2x + c2 sin 2x + sin 2x ln |sec 2x + tan 2x| − 1
(21) By Hooke’s Law, the spring constant k satisfies the equation −98 = k(0.7), so that
k = 140N/m Then with m = 10 kg, y0 = −0.05 m, and v0 = −0.1 m/s, the motion of
the mass is given by
y = −0.05 cos√
14x +−0.0267 sin√
14x
(22) Kirchoff’s loop law gives i′′(t)+20i′(t)+200i(t) = 0, and the solution is i = e−10t(c1 cos 10t+
c2 sin 10t). The initial conditions are i(0) = 0 and i′(0) = E(0)L − R
L i(0) − q(0)LC =
120.5 − 10
0.5(0)− 0.510−2 (0) = 24. This yields c1 = 0 and c2 = 12
5 ; thus i(t) = 125 e−10t sin 10t for
any t > 0.
(23) (a) Let A =
[
0 1
8 −2
]
, then λ1 = 2, v1 =
[
1
2
]
, λ2 = −4 and v2 =
[
−1
4
]
.
y = c1e2xv1 + c2e
−4xv2
Therefore,
y1 = c1e2x − c2e
−4x
y2 = 2c1e2x + 4c2e
−4x.
(b) Let A =
[
0 1
8 −2
]
, then λ1 = 2, v1 =
[
1
2
]
, λ2 = −4 and v2 =
[
−1
4
]
.
P =
[
12 −1
4
1 1
]
and P−1 =
[
43
13
−43
23
]
.
f (x) =
[
0
ex
]
and g (x) = P−1f (x) =
[
13ex
23ex
]
.
w1 = e2x(∫
e−2x 13exdx + c1
)
= e2x(
13
∫
e−xdx + c1
)
= −13ex + c1e
2x
w2 = e−4x(∫
e4x 23exdx + c2
)
= e−4x(
23
∫
e5xdx + c2
)
= 215ex + c2e
−4x
y = Pw =
[
12 −1
4
1 1
][
−13ex + c1e
2x
215ex + c2e
−4x
]
=
[
12c1e
2x − 15ex − 1
4c2e−4x
c1e2x − 1
5ex + c2e−4x
]
.
(c) With P =
1 1 2
1 3 4
2 4 5
and D =
−2 0 0
0 4 0
0 0 1
, we have
173
CHAPTER 7. DIFFERENTIAL EQUATIONS
A = PDP−1, where A =
−5 −15 −21
3 1 3
0 6 7
. Putting
y1
y2
y3
= P
u1
u2
u3
into the equation, one concludes that
d
dt
u1
u2
u3
= P−1AP
u1
u2
u3
= D
u1
u2
u3
.
This implies u1 = −2u1, u2 = 4u2 and u3 = u3. Integration yields u1 = c1e−2t,
u2 = c2e4t and u3 = c3e
t, where c1, c2 and c3 are arbitrary constants. We thus
conclude that
y = Pu = P
c1e−2t
c2e4t
c3et
=
c1e−2t + c2e
4t + c3et
c1e−2t + 3c2e
4t + 4c3et
2c1e−2t + 4c2e
4t + 5c3et
.
(d) Let A =
1 −3 3
3 −5 3
6 −6 4
, then λ1 = −2, v1 =
1
1
0
, λ2 = −2, v2 =
−1
0
1
, λ3 = 4 and
v3 =
1212
1
.
P =
1 −1 12
1 0 12
0 1 1
and P−1 =
−12
32 −1
2
−1 1 0
1 −1 1
.
f (x) =
0
0
x2 + 5
and g (x) = P−1f (x) =
−12
32 −1
2
−1 1 0
1 −1 1
0
0
x2 + 5
=
−12x2 − 5
2
0
x2 + 5
.
w1 = e−2x[∫
e2x(
−12x2 − 5
2
)
dx + c1
]
= −18
(
2x2 − 2x + 11)
+ c1e−2x
w2 = c2e−2x
w3 = e4x(∫
e−4x(
x2 + 5)
dx)
= − 132
(
8x2 + 4x + 41)
+ c3e4x
174
CHAPTER 7. DIFFERENTIAL EQUATIONS
y = Pw =
316x + c1e
−2x − c2e−2x + 1
2c3e4x − 3
8x2 − 12964
316x + c1e
−2x + 12c3e
4x − 38x2 − 129
64
c2e−2x − 1
8x + c3e4x − 1
4x2 − 4132
(e) Diagonalize A =
[
1 −2
−2 4
]
to obtain AP = PD, where
P =
[
2 −12
1 1
]
and D =
[
0 0
0 5
]
.
Substituting y = Pz into the system y′ = Ay +
[
1
t
]
, one obtains
z′ = Dz + P−1
[
1
t
]
, or
[
z′1z′2
]
=
[
0 0
0 5
][
z1
z2
]
+
[
15(t + 2)15(4t− 2)
]
We thus have z′1 = 15(t + 2) and z′2 = 5z2 + 1
5(4t− 2). Solving these equations, we
conclude that z1 = t2
10 + 2t5 + C1 and
e−5t z2 =
∫
1
5e−5t(4t− 2)dt + C2,
or z2 = C2e5t − 4
25t +
6
125. Finally, using y = Pz, one has
y1 = −1
2C2e
5t +t2
5+
22
25t + 2C1 −
3
125,
y2 = C2e5t +
t2
10+
6
25t + C1 +
6
125.
175
Chapter 8
Laplace and Fourier Transformations
(1) (a)
L[
e3t cos2 2t]
= L
[
e3t
(
1
2+
1
2cos 4t
)]
= L
[
e3t 1
2+
1
2e3t cos 4t
]
=1
2L[
e3t]
+1
2L[
e3t cos 4t]
=1
2
1
s− 3+
1
2
s− 3
(s− 3)2 + 16.
(b)
L[
t2 sin 3t]
=d2
ds2
[
3
s2 + 9
]
=d
ds
[
− 6s
(s2 + 9)2
]
=18 (s2 − 3)
(s2 + 9)3.
(2) (a) L−1[
1s2−2s+9
]
= L−1[
1(s−1)2+8
]
= 1√8L−1[ √
8(s−1)2+8
]
= 1√8et sin
√8t
(b) L−1[
3s2+4s+6
]
= L−1[
3(s+2)2+2
]
= 3√2L−1[ √
2(s+2)2+2
]
= 3√2e−2t sin
√2t
(c)
L−1
[
s + 4
s2 + 4s + 8
]
= L−1
[
s + 4
(s + 2)2 + 4
]
= L−1
[
s
(s + 2)2 + 4+
4
(s + 2)2 + 4
]
= L−1
[
s
(s + 2)2 + 4
]
+ 2L−1
[
2
(s + 2)2 + 4
]
= e−2t (cos 2t− sin 2t) + 2e−2t sin 2t = e−2t (sin 2t + cos 2t)
(d) L−1[
3s+7s2−2s−3
]
= L−1[
3s+7(s−3)(s+1)
]
= 4L−1[
1s−3
]
− L−1[
1s+1
]
= 4e3t − e−t
177
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(e)
L−1
[
3s + 1
(9s− 1) (s2 + 1)
]
= L−1
[
54
41 (9s− 1)
]
− L−1
[
6s− 13
41 (s2 + 1)
]
=6
41L−1
[
1
s− 1/9
]
− 6
41L−1
[
s
s2 + 1
]
+13
41L−1
[
1
s2 + 1
]
=6
41et/9 − 6
41cos t +
13
41sin t
(f) L−1[
s+1s2+s
]
= L−1[
1s
]
= 1
(3) (a)
L−1
[
1
s (s2 − 3s + 2)
]
= L−1
[
1
2s− 1
s− 1+
1
2 (s− 2)
]
=1
2− et +
1
2e2t.
(b) If a 6= b, then
L−1
[
αs + β
(s2 + a2) (s2 + b2)
]
= αL−1
[
s
(s2 + a2) (s2 + b2)
]
+ βL−1
[
1
(s2 + a2) (s2 + b2)
]
=α
b2 − a2(cos at− cos bt) +
β
b2 − a2
(
sin at
a− sin bt
b
)
.
If a = b, then
L−1
[
αs + β
(s2 + a2) (s2 + b2)
]
= αL−1
[
s
(s2 + a2)2
]
+ βL−1
[
1
(s2 + a2)2
]
=αt sin at
2a+
β(sin at− at cos at)
2a3.
(c) Since q − p2
4 > 0, we may put ω2 = q − p2
4 and complete square to obtain
L−1
[
αs + β
(s2 + ps + q)2
]
= L−1
α (s + p/2) + (β − αp/2)(
(s + p/2)2 + ω2)2
= e−pt/2
[
α cos ωt +β − αp/2
2ω3/2(sinωt− ωt cos ωt)
]
.
178
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(4) (a)
L[
y′ − 5y]
= L[
8t2 − 2]
L[
y′]
− 5L [y] = 8L[
t2]
− 2L [1]
[sY (s)− y (0)]− 5Y (s) = 82!
s3− 2
1
s
(s− 5) Y (s)− 1 =16− 2s2
s3
Y (s) =s3 − 2s2 + 16
s3 (s− 5)=
91
125 (s− 5)+
34
125s− 16
25s2− 16
5s3
y (t) =91
125L−1
[
1
s− 5
]
+34
125L−1
[
1
s
]
− 16
25L−1
[
1
s2
]
− 8
5L−1
[
2
s3
]
=91
125e5t +
34
125− 16
25t− 8
5t2
(b)
L[
y′ − 5y]
= L[
e3t]
[sY (s)− y (0)]− 5Y (s) =1
s− 3
Y (s) =s− 2
(s− 3) (s− 5)=
3
2 (s− 5)− 1
2 (s− 3)
y (t) =3
2L−1
[
1
s− 5
]
− 1
2L−1
[
1
s− 3
]
=3
2e5t − 1
2e3t
(c)
L[
y′ − 5y]
= L [cos t]
[sY (s)− y (0)]− 5Y (s) =s
s2 + 1
Y (s) =s2 + s + 1
(s2 + 1) (s− 5)=
31
26 (s− 5)− 5s− 1
26 (s2 + 1)
y (t) =31
26L−1
[
1
s− 5
]
− 5
26L−1
[
s
s2 + 1
]
+1
26L−1
[
1
s2 + 1
]
=31
26e5t − 5
26cos t +
1
26sin t
179
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(d)
L[
y′ − 5y]
= L[
e3t + cos t]
[sY (s)− y (0)]− 5Y (s) =1
s− 3+
s
s2 + 1
Y (s) =22
13 (s− 5)− 1
2 (s− 3)− 5s− 1
26 (s2 + 1)
y (t) =22
13L−1
[
1
s− 5
]
− 1
2L−1
[
1
s− 3
]
− 5
26L−1
[
s
s2 + 1
]
+1
26L−1
[
1
s2 + 1
]
=22
13e5t − 1
2e3t − 5
26cos t +
1
26sin t
(e)
L[
y′′ − y′ − 2y]
= L[
4t2]
[
s2Y (s)− sy (0)− y′ (0)]
− [sY (s)− y (0)]− 2Y (s) =8
s3
(
s2 − s− 2)
Y (s) =8
s3+ 1
Y (s) =7
3 (s + 1)+
2
3 (s− 2)− 3
s+
2
s2− 4
s3
y (t) =7
3L−1
[
1
s + 1
]
+2
3L−1
[
1
s− 2
]
− 3L−1
[
1
s
]
+ 2L−1
[
1
s2
]
− 2L−1
[
2
s3
]
=7
3e−t +
2
3e2t − 3 + 2t− 2t2
(f)
L[
y′′ − y′ − 2y]
= L [sin 2t]
[
s2Y (s)− sy (0)− y′ (0)]
− [sY (s)− y (0)]− 2Y (s) =2
s2 + 4(
s2 − s− 2)
Y (s) =2
s2 + 4+ 1
Y (s) =s− 6
20 (s2 + 4)− 7
15 (s + 1)+
5
12 (s− 2)
y (t) =1
20L−1
[
s
s2 + 4
]
− 3
20L−1
[
2
s2 + 4
]
− 7
15L−1
[
1
s + 1
]
+5
12L−1
[
1
s− 2
]
=1
20cos 2t− 3
20sin 2t− 7
15e−t +
5
12e2t
180
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(g)
L[
y′′ − y′ − 2y]
= L[
e2t]
[
s2Y (s)− sy (0)− y′ (0)]
− [sY (s)− y (0)]− 2Y (s) =1
s− 2(
s2 − s− 2)
Y (s) =1
s− 2+ 1
Y (s) =2
9 (s− 2)− 2
9 (s + 1)+
1
3 (s− 2)2
y (t) =2
9L−1
[
1
s− 2
]
− 2
9L−1
[
1
s + 1
]
+1
3L−1
[
1
(s− 2)2
]
=2
9e2t − 2
9e−t +
1
3te2t
(h)
L[
y′′ + 4y′ + 8y]
= L[
8t2 + 8t + 18]
[
s2Y (s)− sy (0)− y′ (0)]
+ 4 [sY (s)− y (0)] + 8Y (s) =16
s3+
8
s2+
18
s(
s2 + 4s + 8)
Y (s) =16
s3+
8
s2+
18
s+ 1
Y (s) = − 2s + 7
s2 + 4s + 8+
2
s+
2
s3
y (t) = −2L−1
[
s
(s + 2)2 + 4
]
− 7
2L−1
[
2
(s + 2)2 + 4
]
+ 2L−1
[
1
s
]
+ L−1
[
2
s3
]
= −2e−2t (cos 2t− sin 2t)− 7
2e−2t sin 2t + 2 + t2
= t2 + 2− 3
2e−2t sin 2t− 2e−2t cos 2t
(i)
L[
y′′ + 4y′ + 5y]
= L [2 cos t− 2 sin t]
[
s2Y (s)− sy (0)− y′ (0)]
+ 4 [sY (s)− y (0)] + 5Y (s) =2s
s2 + 1− 2
s2 + 1(
s2 + 4s + 5)
Y (s) =2s
s2 + 1− 2
s2 + 1+ 1
Y (s) =1
2
s
s2 + 1− 1
2
s
s2 + 4s + 5− 1
s2 + 4s + 5
181
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
y (t) =1
2L−1
[
s
s2 + 1
]
− 1
2L−1
[
s
(s + 2)2 + 1
]
− L−1
[
1
(s + 2)2 + 1
]
=1
2cos t− 1
2e−2t (cos t− 2 sin t)− e−2t sin t =
1
2cos t− 1
2e−2t cos t
(j)
L[
y′′ + 2y′ + 3y]
= L[
t2 + sin t]
[
s2Y (s)− sy (0)− y′ (0)]
+ 2 [sY (s)− y (0)] + 3Y (s) =2
s3+
1
s2 + 1(
s2 + 2s + 3)
Y (s) =2
s3+
1
s2 + 1+ 1
Y (s) =19s + 167
108 (s2 + 2s + 3)− s− 1
4 (s2 + 1)+
2
27s− 4
9s2+
2
3s3
y (t) =19
108L−1
[
s
(s + 1)2 + 2
]
+167
108L−1
[
1
(s + 1)2 + 2
]
− 1
4L−1
[
s
s2 + 1
]
+1
4L−1
[
1
s2 + 1
]
+2
27L−1
[
1
s
]
− 4
9L−1
[
1
s2
]
+1
3L−1
[
2
s3
]
=19
108e−t
(
cos√
2t−√
2
2sin√
2t
)
+167
108
√2
2e−t sin
√2t− 1
4cos t
+1
4sin t +
2
27− 4
9t +
1
3t2
=19
108e−t cos
√2t +
37
27
√2
2e−t sin
√2t− 1
4cos t +
1
4sin t +
2
27− 4
9t +
1
3t2
(k)
L[
y′′ − 2y′]
= L[
et sin t]
[
s2Y (s)− sy (0)− y′ (0)]
− 2 [sY (s)− y (0)] =1
(s− 1)2 + 1(
s2 − 2s)
Y (s) =1
(s− 1)2 + 1+ 1
Y (s) =3
4 (s− 2)− 1
2 (s2 − 2s + 2)− 3
4s
y (t) =3
4L−1
[
1
s− 2
]
− 1
2L−1
[
1
(s− 1)2 + 1
]
− 3
4L−1
[
1
s
]
=3
4e2t − 1
2et sin t− 3
4
182
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(l) Apply Laplace transform to the equation, i.e. L [y′′ − y′ + 2y] = L[(
6t2 + 8t + 7)
et]
.
Then
[
s2Y (s)− sy (0)− y′ (0)]
− [sY (s)− y (0)] + 2Y (s)
= 62
(s− 1)3+ 8
1
(s− 1)2+ 7
1
s− 1
Thus
(
s2 − s + 2)
Y (s) = 62
(s− 1)3+ 8
1
(s− 1)2+ 7
1
s− 1+ 1
Y (s) =1
(s− 1)2+
6
(s− 1)3
y (t) = L−1
[
1
(s− 1)2
]
+ 3L−1
[
2
(s− 1)3
]
= tet + 3t2et.
(5) (a) Taking Laplace Transform on both sides of the differential equations and making use
of the initial conditions, we have
Y (s) =As
(s2 + ω2)(s2 + k2)+
Bω
(s2 + ω2)(s2 + k2).
i. If ω = k, then
y (t) = L−1
[
As
(s2 + ω2)2+
Bω
(s2 + ω2)2
]
= Atsinωt
2ω+ B
sin ωt− ωt cos ωt
2ω2.
ii. If ω 6= k, then
y (t) = L−1
[
As
(s2 + ω2) (s2 + k2)+
Bω
(s2 + ω2) (s2 + k2)
]
=A
k2 − ω2(cos ωt− cos kt) +
Bω
k2 − ω2
(
sin kt
k− sinωt
ω
)
.
(b) Since L[
5e−2t · sin 2t]
=10
(s + 2)2 + 4, we obtain
s2Y (s)− s + sY (s)− 1 + Y (s) =10
(s + 2)2 + 4.
183
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
Therefore,
Y (s) =s + 1
s2 + s + 1+
1
s2 + s + 1× 10
(s + 2)2 + 4
=s + 1
s2 + s + 1− 10
37
3s− 4
s2 + s + 1+
10
37
3s + 5
(s + 2)2 + 4
=7
37
s
(s + 1/2)2 + 3/4+
77
37
1
(s + 1/2)2 + 3/4
+30
37
s
(s + 2)2 + 4+
50
37
1
(s + 2)2 + 4,
from which it follows that
y (t) = e−t/2
(
7
37cos
√3
2t− 70
37√
3sin
√3
2t
)
+ e−2t
(
30
37cos 2t− 5
37sin 2t
)
.
(c) Take Laplace Transform to obtain sY (s) + 4Y (s) =5s
s2 + 4. Thus
Y (s) =5s
(s + 4)(s2 + 4).
Since5s
(s + 4)(s2 + 4)= − 1
s + 4+
1 + s
s2 + 4, we conclude that
y(t) = −e−4t +sin 2t
2+ cos 2t.
(6) The circuit equation is given by
2di
dt+ 20i +
1
0.08
∫ t
0i (τ) dτ = 200 e−t.
Denoting L [i (t)] by I (s), we obtain 2sI (s) + 20I (s) +25
2
I (s)
s=
200
s + 1, which implies
I (s) =400s
(s + 1) (4s2 + 40s + 25)
=100s
(s + 1)(
s + 5 + 52
√3) (
s + 5− 52
√3)
=400
11 (s + 1)+
100
33
( √3− 6
s + 5 + 52
√3
)
− 100
33
( √3 + 6
s + 5− 52
√3
)
.
184
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
Taking inverse Laplace Transform, we obtain
i (t) =400
11e−t +
100(√
3− 6)
33e−(5+ 5
2
√3)t − 100
(√3 + 6
)
33e−(5− 5
2
√3)t.
(7) (a)
(f ∗ g) (t) =
∫ t
0eaτdτ =
[
1
aeaτ
]t
0
=eat − 1
a.
(b)
(f ∗ g) (t) =
∫ t
0(t− τ) sinωτdτ = t
∫ t
0sinωτdτ −
∫ t
0τ sin ωτdτ
= t
[
−cos ωt− 1
ω
]t
0
−[
τcos ωτ
ω
]t
0−∫ t
0
(
−cos ωτ
ω
)
dτ =− sinωt + ωt
ω2.
(c)
(f ∗ g) (t) =
∫ t
0τ2ea(t−τ)dτ =
[
−τ2
aea(t−τ)
]t
0
−∫ t
0
(
−2τ
aea(t−τ)
)
dτ
=
[
−τ2
aea(t−τ)
]t
0
−[
2τ
a2ea(t−τ)
]t
0
+
∫ t
0
2
a2ea(t−τ) dτ
= −2at + 2 + a2t2 − 2eat
a3.
(8) (a) f (t) =
2t if 0 < t ≤ 1
t if t > 1and f ′ (t) =
2 if 0 < t ≤ 1
1 if t > 1.
F (s) = L [f (t)] =
∫ ∞
0f (t) e−stdt
=
∫ 1
02te−stdt +
∫ ∞
1te−stdt =
∫ ∞
0te−stdt +
∫ 1
0te−stdt
=
[
−1
ste−st
]∞
0
+1
slim
x→∞
∫ x
0e−stdt +
[
−1
ste−st
]1
0
+1
s
∫ 1
0e−stdt
=1
slim
x→∞
[
−1
se−st
]x
0
− 1
se−s +
1
s
[
−1
se−st
]1
0
=1
s2− 1
se−s − 1
s2e−s +
1
s2
=2
s2− 1
se−s − 1
s2e−s
185
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
L[
f ′ (t)]
=
∫ ∞
0f ′ (t) e−stdt
=
∫ 1
02e−stdt +
∫ ∞
1e−stdt =
∫ ∞
0e−stdt +
∫ 1
0e−stdt
= limx→∞
[
−1
se−st
]∞
0
+
[
−1
se−st
]1
0
=2
s− 1
se−s
(b) f (t) =
t2 if 0 < t ≤ 1
0 if t > 1, f ′ (t) =
2t if 0 < t < 1
0 if t > 1,
f ′′ (t) =
2 if 0 < t < 1
0 if t > 1.
L[
f ′′ (t)]
=
∫ ∞
0f ′′ (t) e−stdt =
∫ 1
02e−stdt
=
[
−2
se−st
]1
0
= −2
se−s +
2
s=
2
s
(
1− e−s)
(9) (a)
L−1
[
s2
(s2 + 4)2
]
= L−1
[
s
(s2 + 4)× s
(s2 + 4)
]
=
∫ t
0cos 2x cos 2 (t− x) dx
=1
2
∫ t
0[cos (2t) + cos (4x− 2t)] dx
=1
2cos (2t)
∫ t
0dx +
1
2
∫ t
0cos (4x− 2t) dx
=1
4sin 2t +
1
2t cos 2t
(b)
L−1
[
1
(s2 + 1)3
]
= L−1
[
1
(s2 + 1)2× 1
(s2 + 1)
]
=
∫ t
0
sinx− x cos x
2sin (t− x) dx
=1
2
∫ t
0(sin x− x cos x) (sin t cos x− cos t sinx) dx
=1
2sin t
∫ t
0
(
sin x cos x− x cos2 x)
dx +1
2cos t
∫ t
0
(
x sinx cos x− sin2 x)
dx
=3
8sin t− 1
8t2 sin t− 3
8t cos t
186
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(c)
L−1
[
s
(s2 + 4)3
]
= L−1
[
1
(s2 + 4)2× s
s2 + 4
]
=
∫ t
0
sin 2x− 2x cos 2x
16cos 2 (t− x) dx
=1
16
∫ t
0(sin 2x− 2x cos 2x) (cos 2t cos 2x + sin 2t sin 2x) dx
=1
16cos 2t
∫ t
0
(
sin 2x cos 2x− 2x cos2 2x)
dx
+1
16sin 2t
∫ t
0
(
sin2 2x− 2x sin 2x cos 2x)
dx
=1
64t sin 2t− 1
32t2 cos 2t
(10) (a) We take Laplace Transform of the equation of motion to obtain
Y (s) = αsG(s) + (β +cα
M)G(s) +
1
MF (s)G(s),
where F (s) = L [f(t)] and
G(s) =1
s2 + cM s + k
M
=1
(s + c2M )2 + 4kM−c2
4M2
.
Suppose g(t) = L−1 [G(s)].
i. If c2 < 4kM , define 4kM−c2
4M2 = ω2. Then g(t) = e−c
2Mt · sinωt
ωand
y(t) = αe−c
2Mt(cos ωt− c
2Mωsinωt) + (β +
cα
M)g(t) +
1
M(f ∗ g)(t).
ii. If c2 = 4kM , then g(t) = t · e− c2M
t and therefore
y(t) =(
α− cα
2Mt)
e−c
2Mt + (β +
cα
M)g(t) +
1
M(f ∗ g)(t).
iii. If c2 > 4kM , put 4kM−c2
4M2 = −Ω2. Then g(t) = e−c
2Mt · e
Ωt − e−Ωt
2Ωand therefore
y(t) = αe−c
2Mt
(
eΩt + e−Ωt
2− ck
4M· e
Ωt − e−Ωt
Ω
)
+ (β +cα
M)g(t) +
1
M(f ∗ g)(t).
187
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(11) Taking Laplace Transform, we obtain
2sI1 + 2sI2 =E0
s
4sI1 +
(
5s + 20 +20
s
)
I2 = 0.
A simple calculation shows that
I2 = −2E0
s(
s + 20 + 201s
) = −2E0
(s2 + 20s + 20)
=−2E0
(
s + 10 + 4√
5) (
s + 10− 4√
5) =
√5E0
20
(
1
s + 10 + 4√
5− 1
s + 10− 4√
5
)
.
Therefore,
i2(t) =
√5 E0
20
(
e−(10+4√
5)t − e−(10−4√
5)t)
.
Similarly, I1 =E0
2s2− I2 implies
i1(t) = L−1
[
E0
2s2
]
− i2(t) =E0 t
2−√
5E0
20
(
e−(10+4√
5)t − e−(10−4√
5)t)
.
(12) (a) L [3 sin 2t] = L
[
y (t) +∫ t0 (t− τ) y (τ) dτ
]
.
Let G (s) = L [g (t− τ)] and g (t) = t, so G (s) = 1s2
3
(
2
s2 + 22
)
= Y (s) + Y (s) G (s)
6
s2 + 4=
(
1 +1
s2
)
Y (s)
Y (s) =6s2
(s2 + 4) (s2 + 1)=
8
s2 + 4− 2
s2 + 1
y (t) = L−1 [Y (s)] = L
−1
[
8
s2 + 4− 2
s2 + 1
]
= 4 sin 2t− 2 sin t
(b) L[
e−t]
= L
[
y (t) + 2∫ t0 cos (t− τ) y (τ) dτ
]
.
188
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
Let G (s) = L [g (t− τ)] and g (t) = cos t, so G (s) = ss2+1
.
1
s + 1= Y (s) + 2Y (s) G (s)
1
s + 1=
(
1 +2s
s2 + 1
)
Y (s)
Y (s) =s2 + 1
(s + 1)3=
1
s + 1− 2
(s + 1)2+
2
(s + 1)3
y (t) = L−1 [Y (s)] = L
−1
[
1
s + 1− 2
(s + 1)2+
2
(s + 1)3
]
= e−t − 2te−t + t2e−t
(13) Since m = 1, c = 2 and k = 15, the spring mass-system after the blow is given by
y′′(t) + 2y′(t) + 5y(t) = δ(t)
with initial conditions y(0) = y′(0) = 0.
Apply Laplace transform to the system, we have s2Y (s) + 2sY (s) + 5Y (s) = 1 or
Y (s) =1
s2 + 2s + 5=
1
(s + 1)2 + 22
Therefore, y(t) = 12e−t sin(2t).
(14) Apply Laplace transform to the equation y(t) = 14e−3t sin(2t) to obtain
Y (s) =1
4
2
(s + 3)2 + 22=
1
2s2 + 12s + 26.
Therefore, m = 2, c = 12, k = 26.
(15)
G(ω) =
∫ ∞
−∞g(x)e−iωxdx =
∫ ∞
−∞f(kx)e−iωxdx
Let kx = t, then kdx = dt
G(ω) =
∫ ∞
−∞f(t)e−iω( t
k) dt
k=
1
k
∫ ∞
−∞f(t)e−i(ω
k)tdt =
1
kF (
w
k)
189
CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS
(16) Determine the Fourier transform of the function f(x) = e−x2
F (w) =
∫ ∞
−∞f(x)e−iωxdx =
∫ ∞
−∞e−x2
e−iωxdx
By table, if f = e−t2
2σ2 , then F (w) = σ√
2πe−σ2ω2
2 . Set σ2 = 12 , then
F (w) =
∫ ∞
−∞e−x2
e−iωxdx =
√
1
2
√2πe
−12
ω2
2 =√
πe−14
ω2
(17) Considering delta function, δ(t), with continuous g(t) and t0 ∈ (a, b), then
∫ b
ag(t)δ(t− t0)dt = g(t0)
where δ(t) = limǫ→0 fǫ(t), fǫ(t) =
1ǫ , 0 ≤ t ≤ ǫ
0, otherwise. So,
∫ ba δ(t)dt = 1, δ(t) =
limǫ→0 fǫ(t) = 0 for t 6= 0 and δ(0) = limǫ→0 fǫ(0) =∞. So,
∫ b
ag(t)δ(t− t0)dt =
∫ b
ag(t) lim
ǫ→0fǫ(t− t0)dt = g(t0)
⇒ limǫ→0
∫ b
ag(t)fǫ(t− t0)dt = g(t0)
190
Chapter 9
Partial Differential Equations
(1) Differentiation yields∂2u1
∂t2= cos x ·
[
−c2 cos ct]
= c2 ∂2u1
∂x2.
Similar calculations for u2. Since u = cos(x± ct) = cos ct cos x∓ sin ct sinx, we conclude
by the previous calculations and by the Principle of superposition that u satisfies the wave
equation.
(2) Write u (x, t) = X (x) T (t) and substitute into the partial differential equation:
X (x) T ′ (t) = c2X ′′ (x)T (t)
Hence, the two differential equations are:
X ′′ (x) + λX (x) = 0, X (0) = 0, X (L) = 0 (SLP)
T ′ (t) = −λc2T (t) (DET)
Follow the same way as in Example 9.2.1,
n0 = 1, λn =(nπ
L
)2, Xn (x) = βn sin
(nπ
Lx)
Solving the DET,
T (t) = ae−λc2t, and hence Tn (t) = ane−(nπc/L)2t
191
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
The solution u (x, t) now becomes:
u (x, t) =∞∑
n=1
Bne−(nπc/L)2t sin(nπ
Lx)
By the initial condition,
u (x, 0) = f (x) =∞∑
n=1
Bne−(nπc/L)2(0) sin(nπ
Lx)
=∞∑
n=1
Bn sin(nπ
Lx)
∫ L
0sin2
(nπ
Lx)
dx =L
2and
Bn =
∫ L0 f (x) sin
(
nπL x)
dx∫ L0 sin2
(
nπL x)
dx=
2
L
∫ L
0f (x) sin
(nπ
Lx)
dx
For different function f (x) in part (a) to (f), find the corresponding Bn:
(a) f (x) = 1
Bn =2
L
∫ L
0(1) sin
(nπ
Lx)
dx =2
L
∫ L
0sin(nπ
Lx)
dx =2
L
[−L
nπcos(nπ
Lx)
]L
0
=−2
nπ[cos (nπ)− 1] =
−2
nπ[(−1)n − 1] =
4nπ , n is odd,
0, n is even.
Therefore, the solution is u (x, t) =∑∞
n odd4
nπe−(nπc/L)2t sin(
nπL x)
.
(b) f (x) = x
Bn =2
L
∫ L
0x sin
(nπ
Lx)
dx =2
L
[−L
nπx cos
(nπ
Lx)
]L
0
− 2
L
∫ L
0
−L
nπcos(nπ
Lx)
dx
=−2
nπL cos (nπ) +
2
nπ
[
L
nπsin(nπ
Lx)
]L
0
=2L
nπ(−1)n+1 +
2
nπ(0)
Therefore,
u (x, t) =∞∑
n=1
2L (−1)n+1
nπe−(nπc/L)2t sin
(nπ
Lx)
.
192
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
(c) f (x) = A + B−AL x
Bn =2
L
∫ L
0
(
A +B −A
Lx
)
sin(nπ
Lx)
dx
= A
[
2
L
∫ L
0sin(nπ
Lx)
dx
]
+B −A
L
[
2
L
∫ L
0x sin
(nπ
Lx)
dx
]
= A2
nπ
[
(−1)n+1 + 1]
+B −A
L
2L
nπ(−1)n+1
=2
nπ
[
A + B (−1)n+1]
(by (a) and (b))
Therefore, u (x, t) =∑∞
n=1
2[A+B(−1)n+1]nπ e−(nπc/L)2t sin
(
nπL x)
.
(d) f (x) = sin(
3πL x)
Bn =2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,2L
∫ L0 sin2
(
3πL x)
dx n = 3.
B3 =2
L
∫ L
0sin2
(
3π
Lx
)
dx =1
L
∫ L
0
[
1− cos
(
6π
Lx
)]
dx
=1
L
[
x− L
6πsin
(
6π
Lx
)]L
0
= 1
Therefore,
u (x, t) = e−(3πc/L)2t sin
(
3π
Lx
)
.
(e) f (x) = cos(
3πL x)
Bn =2
L
∫ L
0cos
(
3π
Lx
)
sin(nπ
Lx)
dx
=1
L
∫ L
0
[
sinπ
L(3 + n)x− sin
π
L(3− n)x
]
dx
=1
L
[ −L
π (3 + n)cos
π
L(3 + n)x− −L
π (3− n)cos
π
L(3− n)x
]L
0
=−1
π (3 + n)[cos π (3 + n)− 1]− −1
π (3− n)[cos π (3− n)− 1] (n 6= 3)
=(−1)n+4 + 1
π (3 + n)− (−1)n−2 + 1
π (3− n)
=[(−1)n + 1] (3− n)− [(−1)n + 1] (3 + n)
π (9− n2)
=2n [(−1)n + 1]
π (n2 − 9)=
0, n is odd,4n
π(n2−9), n is even.
193
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Therefore,
u (x, t) =∞∑
n even
4n
π (n2 − 9)e−(nπc/L)2t sin
(nπ
Lx)
.
(f) f (x) =
x, 0 ≤ x ≤ L2 ,
L− x, L2 ≤ x ≤ L.
Bn
=2
L
∫ L2
0x sin
(nπ
Lx)
dx +2
L
∫ L
L2
(L− x) sin(nπ
Lx)
dx
=2
L
∫ L
0x sin
(nπ
Lx)
dx +2
L
∫ L
L2
(L− 2x) sin(nπ
Lx)
dx
=2L
nπ(−1)n+1 − 2
nπ
[
(L− 2x) cos(nπ
Lx)]L
L2
+2
nπ
∫ L
L2
(−2) cos(nπ
Lx)
dx
=2L
nπ(−1)n+1 +
2L
nπ(−1)n − 4
nπ
[
L
nπsin(nπ
Lx)
]L
L2
=4L
n2π2sin(nπ
2
)
=
4L(−1)k
(2k+1)2π2, where n = 2k + 1, k = 0, 1, 2, ...,
0, otherwise.
Therefore,
u (x, t) =∞∑
k=0
4L
(2k + 1)2 π2(−1)k e−[(2k+1)πc/L]2t sin
[
(2k + 1)π
Lx
]
.
(3) Write u (x, t) = v (x, t) + l (x) and substitute into the partial differential equation:
vt (x, t) = c2[
vxx (x, t) + l′′ (x)]
The boundary conditions imply that
u (0, t) = v (0, t) + l (0) = A, u (L, t) = v (L, t) + l (L) = B, 0 ≤ t ≤ ∞.
Furthermore, the initial condition also implies that
u (x, 0) = v (x, 0) + l (x) = f (x) , 0 ≤ x ≤ L.
Note that l (x) = A + B−AL x, l′′ (x) = 0, l (0) = A and l (L) = B.
194
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Then v (x, t) should satisfy the following initial boundary value problem
∂v∂t = c2 ∂2v
∂x2 ,
v (0, t) = 0, v (L, t) = 0, 0 ≤ t <∞,
v (x, 0) = f (x)− l (x) , 0 ≤ x ≤ L.
Solve the initial boundary value problem in question 2:
Write v (x, t) = X (x)T (t) and substitute into the partial differential equation:
X (x) T ′ (t) = c2X ′′ (x)T (t)
Hence, the two differential equations are:
X ′′ (x) + λX (x) = 0, X (0) = 0, X (L) = 0 (SLP)
T ′ (t) = −λc2T (t) (DET)
Follow the same way as in Example 9.2.1,
n0 = 1, λn =(nπ
L
)2, Xn (x) = βn sin
(nπ
Lx)
Solving the DET,T (t) = Ae−λc2t, and hence
Tn (t) = Ane−(nπc/L)2t
The solution v (x, t) now becomes:
v (x, t) =∞∑
n=1
Bne−(nπc/L)2t sin(nπ
Lx)
By the initial condition,
v (x, 0) = f (x)− l (x) =∞∑
n=1
Bn sin(nπ
Lx)
Since∫ L0 sin2
(
nπL x)
dx = L2 ,
Bn =
∫ L0 [f (x)− l (x)] sin
(
nπL x)
dx∫ L0 sin2
(
nπL x)
dx=
2
L
∫ L
0[f (x)− l (x)] sin
(nπ
Lx)
dx
195
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
For different function f (x) in part (a) and (b), find the corresponding Bn:
(a) f (x) = sin(
3πL x)
Bn =2
L
∫ L
0
[
sin
(
3π
Lx
)
−A− B −A
Lx
]
sin(nπ
Lx)
dx
=2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx− 2
L
∫ L
0
(
A +B −A
Lx
)
sin(nπ
Lx)
dx
By (2) (d),
2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx
=
0, n 6= 3,
1, n = 3.
By (2) (c)
2
L
∫ L
0
(
A +B −A
Lx
)
sin(nπ
Lx)
dx =2
nπ
[
A + (−1)n+1 B]
Therefore,
v (x, t) = e−( 3πcL )
2t sin
(
3π
Lx
)
−∞∑
n=1
2
nπ
[
A + (−1)n+1 B]
e−(nπcL )
2t sin
(nπ
Lx)
and
u (x, t) = e−( 3πcL )
2t sin
(
3π
Lx
)
−∞∑
n=1
2
nπ
[
A + (−1)n+1 B]
e−(nπcL )
2t sin
(nπ
Lx)
+
(
A +B −A
Lx
)
.
(b) f (x) = Q (x) =
x, 0 ≤ x ≤ L2 ,
L− x, L2 ≤ x ≤ L.
Bn =2
L
∫ L
0
[
Q (x)−A− B −A
Lx
]
sin(nπ
Lx)
dx
=2
L
∫ L
0Q (x) sin
(nπ
Lx)
dx− 2
L
∫ L
0
[
A +B −A
Lx
]
sin(nπ
Lx)
dx
196
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
By (2) (f)
2
L
∫ L
0Q (x) sin
(nπ
Lx)
dx =
4L(−1)k
(2k+1)2π2, n = 2k + 1, k = 0, 1, 2, ...,
0, otherwise.
By (2) (c)
2
L
∫ L
0
(
A +B −A
Lx
)
sin(nπ
Lx)
dx =2
nπ
[
A + (−1)n+1 B]
Therefore,
v (x, t)
=
∞∑
k=0
4L (−1)k
(2k + 1)2 π2−
∞∑
k=0
2[
A + (−1)2k+3 B]
(2k + 1)π
e−[(2k+1)πc/L]2t sin
(
(2k + 1) π
Lx
)
and
u (x, t)
=
∞∑
k=0
4L (−1)k
(2k + 1)2 π2−
∞∑
k=0
2[
A + (−1)2k+3 B]
(2k + 1)π
e−[(2k+1)πc/L]2t sin
(
(2k + 1) π
Lx
)
+
(
A +B −A
Lx
)
.
(4) (a) Separation of variables yields
u(x, t) =∞∑
n=1
bn · e−4n2t sin nx,
with
bn =2
π
∫ π
0x(π − x) sinnx dx =
4[1− (−1)n]
πn3.
Thus
u(x, t) =8
π
∑
n odd
e−4n2t
n3sinnx.
(b) The function w(x) = 30 + 70x satisfies w(0) = 30, w(1) = 100 and w′′(x) = 0. It is
now clear that v(x, t) = u(x, t) − w(x) would satisfy the equation vt = c2 vxx, with
boundary conditions v(0, t) = v(1, t) = 0 and initial condition v(x, 0) = 20 − 70x.
197
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
We may now solve for v(x, t) to obtain
u(x, t) = (30 + 70x) +∞∑
n=1
(
100× (−1)n + 40
nπ
)
e−n2π2c2t sinnπx.
(5) We substitute u(x, t) = φ(t) · w(x, t) into ut = c2uxx − ku to obtain
φ′(t)w + φ(t)wt = φ(t)[
c2wxx − kw]
If w(x, t) satisfies the standard heat equation wt = c2 wxx, then φ′(t) = −k φ(t), or
φ(t) = e−kt. For the given IBVP, we let u(x, t) = e−t · w(x, t) to obtain
wt = wxx for 0 < x < 1, t > 0;
w(0, t) = 0 and w(1, t) = 0 for t ≥ 0;
w(x, 0) = sinπx + 12 sin 3πx for 0 ≤ x ≤ 1
It follows that
w(x, t) = e−π2t · sinπx +1
2e−9π2t · sin 3πx
and therefore
u(x, t) = e−t · w(x, t) = e−(π2+1)t · sinπx +1
2e−(9π2+1)t · sin 3πx.
(6)
2
∫ 1
0sin pnx sin pmx dx =
∫ 1
0[cos(pn − pm)x− cos(pn + pm)x] dx.
For n 6= m, we have
2
∫ 1
0sin pnx · sin pmx dx =
sin(pn − pm)
pn − pm− sin(pn + pm)
pn + pm.
Since pn is a positive root of the equation tan p = − pσ , it follows that sin pn = −
(pn
σ
)
cos pn.
Therefore,
2
∫ 1
0sin pnx · sin pmx dx =
sin pn cos pm − cos pn sin pm
pn − pm− sin pn cos pm + cos pn sin pm
pn + pm= 0.
Similarly, we have
∫ 1
0sin2 pnx dx =
1
2
∫ 1
0[1− cos 2pnx ] dx =
1
2− sin 2pn
4pn=
1
2− sin pn · cos pn
2pn=
1
2+
cos2 pn
2σ.
198
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
(7) (a) Solving the S-L Problem, one has λn = (n− 12)2 as eigenvalues and Xn(x) = sin(n−
12)x as the corresponding eigenfunctions . Therefore,
u(x, t) =∞∑
n=1
bn e−c2(n− 12)2 t sin(n− 1
2)x.
Using the initial condition x(π − x) = u(x, 0), one has
x(π − x) =∞∑
n=1
bn sin(n− 1
2)x,
where
bn =
∫ π0 x(π − x) sin(n− 1
2)x dx∫ π0 sin2(n− 1
2)x dx=
8
π
[
4 + (−1)n (2n− 1)π
(2n− 1)3
]
Therefore,
u(x, t) =8
π
∞∑
n=1
[
4 + (−1)n (2n− 1)π
(2n− 1)3
]
e−c2(n− 12)2 t sin(n− 1
2)x.
(b) We have u(x, t) = a0+∑∞
n=1 an e−16n2t cos nx, where a0+∑∞
n=1 an cos nx = u(x, 0) =
2xπ if 0 ≤ x ≤ π
22π (π − x) if π
2 ≤ x ≤ π
The arbitrary constants a0 and an’s may now be calculated by Euler’s Formula. The
result is
u(x, t) =1
2+
8
π2
∑
n even
[
cos nπ2 − 1
n2
]
e−16n2t cos nx.
(c) We substitute
u(x, t) =∞∑
n=1
Tn(t) sin nπx,
where Tn(t)’s are to be determined, into the p.d.e. to obtain T ′′n (t)+n2π2c2 Tn(t) = 0
for n 6= 3 and T ′′3 (t) = −9π2c2 T3(t)+1. Initial conditions on u(x, t) imply T ′
n(0) = 0
for every n = 1, 2, 3, . . ., T1(0) = 3, T4(0) = −2, and Tn(0) = 0 whenever n 6= 1, 4.
Solving the ODE, we obtain Tn(t) = 0 when n 6= 1, 3, 4; T1(t) = 3 cos πct, T3(t) =1
9π2c2[1− cos 3πct] and T4(t) = −2 cos 4πct. Therefore,
u(x, t) = 3 cos πct · sinπx + 1
[
1− cos 3πct
9π2c2
]
sin 3πx− 2 cos 4πct · sin 4πx.
199
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
(d) We substitute
u(x, t) =∞∑
n=1
Tn(t) sin pnx
into the equation to obtain T ′1(t) = −p2
1 T1(t) + 1 and T ′n(t) = −p2
n Tn(t) for n > 1,
with Tn(0) = 0 for n = 1, 2, 3, . . .. Here, pn’s are positive roots of the equation
tan p = −p. Thus Tn(t) = 0 whenever n > 1 and T1(t) =1− e−p2
1 t
p21
. Hence
u(x, t) =
(
1− e−p21 t
p21
)
sin p1x.
(e) We put u(x, t) = v(x, t) + x · cos t into the equation to obtain
∂v
∂t= c2 ∂2v
∂x2+ x · sin t,
with boundary conditions v(0, t) = 0 and v(1, t) = 0 and initial condition v(x, 0) = 0
for 0 ≤ x ≤ 1.
Substituting v(x, t) =∑∞
n=1 Tn(t) sinnπx into the equation and bearing in mind
that
x · sin t =∞∑
n=1
2(−1)n+1 sin t
nπsin nπx,
we obtain
T ′n(t) + c2n2π2Tn(t) =
2(−1)n+1 sin t
nπ.
Solving for the equation, with initial condition Tn(0) = 0, we have
Tn(t) =2 (−1)n+1
nπ
(
e−c2n2π2t
c4n4π4 + 1− cos t
c4n4π4 + 1+
c2n2π2 sin t
c4n4π4 + 1
)
Hence
u(x, t) = x · cos t +2
π
∞∑
n=1
(−1)n+1
(
e−c2n2π2t − cos t + c2n2π2 sin t
n(c4n4π4 + 1)
)
sinnπx.
(8) Write u (x, t) = X (x) T (t) and substitute into the partial differential equation:
X (x)T ′′ (t) = c2X ′′ (x)T (t)
200
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Hence, the two differential equations are:
T ′′ (t) + λc2T (t) = 0
X ′′ (x) + λX (x) = 0, X (0) = 0, X (L) = 0 (SLP)
Follow the same way as in Example 9.2.1,
n0 = 1, λn =(nπ
L
)2, Xn (x) = βn sin
(nπ
Lx)
Solving T ′′ (t) + λc2T (t) = 0 with λn =(
nπL
)2,
Tn (t) = an cos(nπc
Lt)
+ bn sin(nπc
Lt)
The solution u (x, t) now becomes:
u (x, t) =∞∑
n=1
[
An cos(nπc
Lt)
+ Bn sin(nπc
Lt)]
sin(nπ
Lx)
By the initial condition,
u (x, 0) = f (x) =∞∑
n=1
An sin(nπ
Lx)
ut (x, t) = g (x) =nπc
L
∞∑
n=1
[
−An sin(nπc
Lt)
+ Bn cos(nπc
Lt)]
sin(nπ
Lx)
ut (x, 0) = g (x) =nπc
L
∞∑
n=1
Bn sin(nπ
Lx)
Since∫ L0 sin2
(
nπL x)
dx = L2 ,
An =
∫ L0 f (x) sin
(
nπL x)
dx∫ L0 sin2
(
nπL x)
dx=
2
L
∫ L
0f (x) sin
(nπ
Lx)
dx and
Bn =L
nπc
∫ L0 g (x) sin
(
nπL x)
dx∫ L0 sin2
(
nπL x)
dx=
2
nπc
∫ L
0g (x) sin
(nπ
Lx)
dx
For different function f (x) in part (a) to (e), find the corresponding An and Bn:
201
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
(a) f (x) ≡ sin(
3πL x)
and g (x) ≡ 0. By (2) (d)
An =2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,
1, n = 3.
Bn =2
nπc
∫ L
0(0) sin
(nπ
Lx)
dx = 0
Therefore, the solution is u (x, t) = cos(
3πcL t)
sin(
3πL x)
.
(b) f (x) ≡ 0 and g (x) ≡ sin(
3πL x)
. By (2) (d)
An =2
L
∫ L
0(0) sin
(nπ
Lx)
dx = 0
Bn =2
nπc
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,L
3πc , n = 3.
Therefore, u (x, t) = L3πc sin
(
3πcL t)
sin(
3πL x)
.
(c) f (x) ≡ sin(
3πL x)
and g (x) ≡ sin(
3πL x)
. By (2) (d)
An =2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,
1, n = 3.
Bn =2
nπc
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,L
3πc , n = 3.
Therefore, u (x, t) =[
cos(
3πcL t)
+ L3πc sin
(
3πcL t)]
sin(
3πL x)
.
(d) f (x) ≡ sin(
3πL x)
and g (x) ≡ sin(
4πL x)
. By (2) (d)
An =2
L
∫ L
0sin
(
3π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 3,
1, n = 3.
Bn =2
nπc
∫ L
0sin
(
4π
Lx
)
sin(nπ
Lx)
dx =
0, n 6= 4,2
4πc
∫ L0 sin2
(
4πL x)
dx, n = 4.
B4 =1
4πc
∫ L
0
[
1− cos
(
8π
Lx
)]
dx =1
4πc
[
x− L
8πsin
(
8π
Lx
)]L
0
=L
4πc
Therefore, u (x, t) = cos(
3πcL t)
sin(
3πL x)
+ L4πc sin
(
4πcL t)
sin(
4πL x)
.
202
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
(e) f (x) = ǫQ (x) =
ǫx, 0 ≤ x ≤ L2 ,
ǫ (L− x) , L2 ≤ x ≤ L.
and g (x) = 0. By (2) (f)
An =2
L
∫ L
0ǫQ (x) sin
(nπ
Lx)
dx =
4ǫL(−1)k
(2k+1)2π2, where n = 2k + 1, k = 0, 1, 2, ...,
0, otherwise.
Bn =2
nπc
∫ L
0(0) sin
(nπ
Lx)
dx = 0
Therefore, u (x, t) =∑∞
k=0
4ǫL(−1)k
(2k+1)2π2cos[
(2k+1)πcL t
]
sin[
(2k+1)πL x
]
.
(9) Write u (x, t) = X (x) T (t) and substitute into the partial differential equation:
X (x)T ′′ (t) = c2X ′′ (x)T (t)
Hence, the two differential equations are:
T ′′ (t) + λc2T (t) = 0
X ′′ (x) + λX (x) = 0, X ′ (0) = 0, X ′ (L) = 0 (SLP)
Follow the same way as in Example 9.2.1,
n0 = 0, λn =(nπ
L
)2, Xn (x) = αn cos
(nπ
Lx)
Solving T ′′ (t) + λc2T (t) = 0 with λn =(
nπL
)2,
Tn (t) = an cos(nπc
Lt)
+ bn sin(nπc
Lt)
The solution u (x, t) now becomes:
u (x, t) =
∞∑
n=0
[
An cos(nπc
Lt)
+ Bn sin(nπc
Lt)]
cos(nπ
Lx)
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CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
By the initial condition,
u (x, 0) = A0 +∞∑
n=1
An cos(nπ
Lx)
ut (x, t) = B0 +nπc
L
∞∑
n=1
[
−An sin(nπc
Lt)
+ Bn cos(nπc
Lt)]
cos(nπ
Lx)
ut (x, 0) = B0 +nπc
L
∞∑
n=1
Bn cos(nπ
Lx)
Since∫ L
0cos2
(nπ
Lx)
dx =
L, n = 0,L2 , n ≥ 1.
We have
A0 =
∫ L0 f (x) dx
L=
1
L
∫ L
0f (x) dx and
An =
∫ L0 f (x) cos
(
nπL x)
dx∫ L0 cos2
(
nπL x)
dx=
2
L
∫ L
0f (x) cos
(nπ
Lx)
dx and
B0 =
∫ L0 g (x) dx
L=
1
L
∫ L
0g (x) dx and
Bn =L
nπc
∫ L0 g (x) cos
(
nπL x)
dx∫ L0 cos2
(
nπL x)
dx=
2
nπc
∫ L
0g (x) cos
(nπ
Lx)
dx.
Thus,
ut (x, t) = B0 +nπc
L
∞∑
n=1
[
−An sin(nπc
Lt)
+ Bn cos(nπc
Lt)]
cos(nπ
Lx)
and
u (x, t) = A0 + B0t +∞∑
n=1
[
An cos(nπc
Lt)
+ Bn sin(nπc
Lt)]
cos(nπ
Lx)
.
(10) Separation of variables and the superposition principle give
u(x, t) =∞∑
n=1
[αn cos nπt + βn sinnπt] sinnπx
Initial conditions are
u(x, 0) =
0.02x if 0 ≤ x ≤ 12
0.02(1− x) if 12 ≤ x ≤ 1
and ut(x, 0) = 0
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CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Therefore, we have
βn = 0 and αn = 2
[
∫ 1/2
00.02x · sinnπx dx +
∫ 1
1/20.02(1− x) · sinnπx dx
]
.
This implies (see Ex. 3 of Revision with L = 1 and K = 0.01)
u(x, t) =0.08
π2
∑
n odd
(−1)n−1
2
n2cos nπt · sin nπx.
(11) We substitute u(x, t) =∑∞
n=1 Tn(t) sinnx into the wave equation
utt = uxx + sin 100x · sin 100t
to obtain ordinary differential equations
T ′′n (t) + n2Tn(t) = 0 when n 6= 100 and T ′′
100(t) + 10000 · T100(t) = sin 100t,
with initial conditions
T1(0) = 2, T ′1(0) = 0, T4(0) = −3, T ′
4(0) = 0 and Tn(0) = T ′n(0) = 0 for n 6= 1, 4
Solving these ODE, we obtain
T1(t) = 2 cos t,
T4(t) = −3 cos 4t,
T100(t) = L−1
100
(s2 + 1002)2
=sin 100t− 100t · cos 100t
20000.
Therefore,
u(x, t) = 2 cos t · sinx− 3 cos 4t · sin 4x +
[
sin 100t− 100t · cos 100t
20000
]
sin 100x
(12) Differentiating E(t) with respect to t, one obtains
E′(t) =
∫ L
0[ρ ut(x, t) · utt(x, t) + T ux(x, t) · uxx(x, t)] dx
=
∫ L
0
[
ρ ut(x, t) · c2uxx(x, t) + T ux(x, t) · uxt(x, t)]
dx
= T
[∫ L
0ut(x, t) · uxx(x, t) dx +
∫ L
0ux(x, t) · uxt(x, t) dx
]
205
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Integrating the second integral by parts, we have
∫ L
0ux(x, t) · uxt(x, t) dx
= [ux(L, t) · ut(L, t)− ux(0, t) · ut(0, t)]−∫ L
0uxx(x, t) · ut(x, t) dx
= −∫ L
0uxx(x, t) · ut(x, t) dx because ut(0, t) = ut(L, t) = 0
Therefore, E′(t) = 0, and the total energy is conserved.
(13) Let u(x, y) = X(x)Y (y).
Solving the following two ODEs:
1) X ′′(x) + λX(x) = 0 where 0 < x < 2;
2) Y ′′(y)− λY (y) = 0 where 0 < y < 1.
So, with the given boundary conditions,
λn =n2π2
4, Xn = sin
nπx
2, Yn = β[e
nπ2
y − e−nπ2
y], n = 1, 2, 3 . . . .
Therefore,
un(x, y) = βn[enπ2
y − e−nπ2
y] sinnπx
2, n = 1, 2, 3....
By Principle of Superposition,
u(x, y) =∞∑
n=1
βn[enπ2
y − e−nπ2
y] sinnπx
2
For boundary condition: u(x, 1) = x(2− x), 0 < x < 2,
βn =2
2[enπ2 − e−
nπ2 ]
∫ 2
0x(2− x) sin
nπx
2dx, n = 1, 2, 3....
=1
[enπ2 − e−
nπ2 ]
−8(−2 + 2 cos nπ + nπ sinnπ)
n3π3
=1
[enπ2 − e−
nπ2 ]
16(1− (−1)n)
n3π3
206
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Finally,
u(x, y) =∞∑
n=1
[
1
[enπ2 − e−
nπ2 ]
16(1− (−1)n)
n3π3
]
[enπ2
y − e−nπ2
y] sinnπx
2
=32
π3
∑
n=odd
1
n3
(
enπ2
y − e−nπ2
y
enπ2 − e−
nπ2
)
sinnπx
2
(14) Set Solution of uxx + uyy = 0 as u(x, y) = X(x)Y (y). Then
Y ′′X + X ′′Y = 0Y ′′
Y= −X ′′
X= −λ
Solving the following two ODEs:
(1) Y ′′(x) + λY (x) = 0 where 0 < y < 1;
(2) X ′′(y)− λX(y) = 0 where 0 < x < 1.
So, with the given boundary conditions,
λn = n2π2, Yn = sinnπy, X(x) = αenπx + βe−nπx
Since u(1, y) = 0,
X(1) = αenπ + βe−nπ = 0⇒ β = −αe2nπ
Therefore,
Xn = αn[enπx − e2nπe−nπx] = αn[enπx − enπ(2−x)], n = 1, 2, 3, . . .
and
un(x, y) = αn[enπx − enπ(2−x)] sinnπy, n = 1, 2, 3 . . . .
By Principle of Superposition,
u(x, y) =∞∑
n=1
αn[enπx − enπ(2−x)] sinnπy, n = 1, 2, 3....
From boundary condition, u(0, y) = 3 sin πy − 4 sin 6πy for 0 ≤ y ≤ 1, we have
α1 =3
(1− e2π), α6 = − 4
(1− e12π)and αn = 0, n 6= 1, 6
207
CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS
Hence
u(x, y) =3
(1− e2π)[eπx − eπ(2−x)] sinπy − 4
(1− e12π)[e6πx − e6π(2−x)] sin 6πy.
(15) Using polar coordinates, we substitute u(r, θ) = R(r)·S(θ) into the equations and proceed
as in the lecture notes. Since S(θ + 2π) = S(θ), we have
S(θ) = an cos nθ + bn sin nθ when n ≥ 1 and S(θ) = a0 when n = 0,
R(r) = cn rn + dn r−n when n ≥ 1 and R(r) = c0 ln r + d0 when n = 0
Consequently, the Principle of Superposition implies that
u(r, θ) = (A0 ln r + B0) +∞∑
n=1
rn [An cos nθ + Bn sinnθ] + r−n [Cn cos nθ + Dn sinnθ] ,
where A0, B0, An, Bn, Cn and Dn’s are arbitrary constants.
(a) To ensure that u(r, θ) is finite as r → 0, we must choose A0 = Cn = Dn = 0.
Furthermore, the boundary condition implies that
B0 +∞∑
n=1
[An cos nθ + Bn sinnθ] =
1 for 0 ≤ θ ≤ π;
0 for π < θ < 2π
Therefore,
u(r, θ) =1
2+∑
n odd
[
2
nπ
]
rn sinnθ.
(b) The requirement for limr→∞ u(r, θ) to be finite implies that
A0 = An = Bn = 0
Therefore,
u(r, θ) = B0 +∞∑
n=1
r−n[Cn cos nθ + Dn sin nθ].
Since u(1, θ) = 1 + cos 2θ − 3 sin 4θ, we may compare coefficients to obtain B0 = 1,
C2 = 1, D4 = −3 and all other coefficients = 0. Therefore,
u(r, θ) = 1 +cos 2θ
r2− 3 sin 4θ
r4.
208
