Problem Solving and Symbolic Mathematics Algebra

Chapter 3
Problem Solving and Symbolic
Mathematics: Algebra
EXERCISES
Exercise 3.1. Write the following expression in a simpler
form:
(x 2 + 2x)2 − x 2 (x − 2)2 + 12x 4
B=
.
6x 3 + 12x 4
x 2 (x 2 + 4x + 4) − x 2 (x 2 − 4x + 4) + 12x 4
B =
6x 3 + 12x 4
=
x 4 + 4x 3 + 4x 2 − x 4 + 4x 3 − 4x 2 + 12x 4
6x 3 + 12x 4
=
6x + 4
12x 4 + 8x 3
12x + 8
=
=
6x 3 + 12x 4
12x + 6
6x + 3
Exercise 3.2. Manipulate the van der Waals equation into
an expression for P in terms of T and Vm . Since the pressure
is independent of the size of the system (it is an intensive
variable), thermodynamic theory implies that it can depend
on only two independent intensive variables.
a
P + 2 (Vm − b) = RT
Vm
a
RT
P+ 2 =
Vm
Vm − b
a
RT
− 2
P =
Vm − b
Vm
Exercise 3.3. a. Find x and y if ρ = 6.00 and φ = π/6
radians
x = (6.00) cos (π/6) = (6.00)(0.866025) = 5.20
y = ρ sin (φ) = (6.00)(0.500) = 3.00
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00049-5
© 2013 Elsevier Inc. All rights reserved.
b. Find ρ and φ if x = 5.00 and y = 10.00.
√
ρ = x 2 + y 2 = 125.0 = 11.18
φ = arctan (y/x) = arctan (2.00) = 1.107 rad
= 63.43◦
Exercise 3.4. Find the spherical polar coordinates of the
point whose Cartesian coordinates are (2.00, 3.00, 4.00).
√
r = (2.00)2 + (3.00)2 − (4.00)2 = 29.00 = 5.39
3.00
φ = arctan
= 0.98279 rad = 56.3◦
2.00
4.00
= 0.733 rad = 42.0◦
θ = arccos
5.39
Exercise 3.5. Find the Cartesian coordinates of the point
whose cylindrical polar coordinates are ρ = 25.00,φ =
60.0◦ , z = 17.50
x = ρ cos (φ) = 25.00 cos (60.0◦ )
= 25.00 × 0.500 = 12.50
y = ρ sin (φ) = 25.00 sin (60.0◦ )
= 25.00 × 0.86603 = 21.65
z = 17.50
Exercise 3.6. Find the cylindrical polar coordinates of the
point whose Cartesian coordinates are (−2.000,−2.000,
3.000).
ρ = ( − 2.00)2 + ( − 2.00)2 = 2.828
−2.00
φ = arctan
= 0.7854 rad = 45.0◦
−2.00
z = 3.000
e11
e12
Mathematics for Physical Chemistry
Exercise 3.7. Find the cylindrical polar coordinates of
the point whose spherical polar coordinates are r = 3.00,
θ = 30.00◦ , φ = 45.00◦ .
z = r cos (θ ) = (3.00) cos (30.00◦ ) = 3.00 × 0.86603
= 2.60
ρ = r sin (θ ) = (3.00) sin (30.00◦ ) = (3.00)(0.500)
b. B = (3 + 7i)3 − (7i)2 .
B ∗ = (3 − 7i)3 − (−7i)2 = (3 − 7i)3 − (7)2
Exercise 3.12. Write a complex number in the form x +i y
and show that the product of the number with its complex
conjugate is real and nonnegative
(x + i y)(x − i y) = x 2 + i x y − i x y + y 2 = x 2 + y 2
= 1.50
φ = 45.00◦
The square of a real number is real and nonnegative, and x
and y are real.
Exercise 3.8. Simplify the expression
(4 + 6i)(3 + 2i) + 4i
(4 + 6i)(3 + 2i) + 4i = 12 + 18i + 8i − 12 + 4i = 30i
Exercise 3.13. If z = (3.00 + 2.00i)2 , find R(z), I (z), r ,
and φ.
z = 9.00 + 6.00i − 4.00 = 5.00 + 6.00i
Exercise 3.9. Express the following complex numbers in
the form r eiφ :
a. z = 4.00 + 4.00i
√
r = 32.00 = 5.66
4.00
φ = arctan
= 0.785
4.00
z = 4.00 + 4.00i = 5.66e
R(z) = 5.00
I (z) = 6.00
√
r = 25.00 + 36.00 = 7.781
φ = arctan (6.00/5.00) = 0.876 rad = 50.2◦
0.785i
b. z = −1.00
Exercise 3.14. Find the square roots of z = 4.00 + 4.00i.
Sketch an Argand diagram and locate the roots on it.
z = −1 = eπi
z = r eiφ
√
r = 32.00 = 5.657
Exercise 3.10. Express the following complex numbers in
the form x + i y
φ = arctan (1.00) = 0.785398 rad = 45.00◦
a. z = 3eπi/2
x = r cos (φ) = 3 cos (π/2) = 3 × 0 = 0
y = r sin (φ) = 3 sin (π/2) = 3 × 1 = 3
z = 3i
b. z = e3πi/2
x = r cos (φ) = cos (3π/2) = 0
y = r sin (φ) = sin (3π/2) = −1
z = −i
Exercise 3.11. Find the complex conjugates of
a. A = (x + i y)2 − 4ei x y
A = x 2 + 2i x y + y 2 − 4ei x y
A∗ = x 2 − 2i x y + y 2 − 4e−i x y
= (x − i y)2 − 4e−i x y
Otherwise by changing the sign in front of every i:
A∗ = (x − i y)2 − 4e−i x y
√
z=
√
5.657 exp
√
5.657e0.3927i
0.785398+2π
2
i =
√
5.657e3.534i
To sketch the Argand diagram, we require the real and
imaginary parts. For the first possibility
√
√
z = 5.657( cos (22.50◦ )) + i sin (22.50◦ )
√
= 5.657(0.92388 + i(0.38268)
= 2.1973 + 0.91019i
For the second possibility
√
√
z = 5.657( cos (202.50◦ )) + i sin (202.50◦ )
√
= 5.657(−0.92388 + i(−0.38268)
= −2.1973 − 0.91019i
Exercise 3.15. Find the four fourth roots of −1.
4
−1 = eπi , e3πi , e5πi , e7πi
eπi = eπi/4 , e3πi/4 , e5πi/4 , e7πi/4
CHAPTER | 3 Problem Solving and Symbolic Mathematics: Algebra
Exercise 3.16. Estimate the number of house painters in
Chicago.
The 2010 census lists a population of 2,695,598 for the
city of Chicago, excluding surrounding areas. Assume that
about 20% of Chicagoans live in single-family houses or
duplexes that would need exterior painting. With an average
family size of four persons for house-dwellers, this would
give about 135,000 houses. Each house would be painted
about once in six or eight years, giving roughly 20,000
house-painting jobs per year. A crew of two painters might
paint a house in one week, so that a crew of two painters
could paint about 50 houses in a year. This gives about 400
two-painter crews, or 800 house painters in Chicago.
e13
9. A surface is represented in cylindrical polar
coordinates by the equation z = ρ 2 . Describe the
shape of the surface. This equation represents a
paraboloid of revolution, produced by revolving a
parabola around the z axis.
11. Find the complex conjugate of the quantity
e2.00i + 3eiπ
e2.00i + 3eiπ = e2.00i − 3 = cos (2.00)
+ i sin (2.00) − 3
(e
2.00i
iπ ∗
+ 3e ) = cos (2.00) − i sin (2.00) − 3
= e−2.00i − 3
13. Find the difference 3.00eπi − 2.00e2πi .
3.00eπi − 2.00e2πi = −3.00 − 2.00 = 5.00
PROBLEMS
1. Manipulate the van der Waals equation into a cubic
equation in Vm . That is, make a polynomial with terms
proportional to powers of Vm up to Vm3 on one side of
the equation.
a
P + 2 (Vm − b) = RT
Vm
Multiply by Vm2
(P Vm2 + a)(Vm − b) = RT Vm2
P Vm3 + aVm − b P Vm2 − ab = RT Vm2
P Vm3 − (b + RT )Vm2 + aVm − ab = 0
15. Find the four fourth roots of 3.000i.
⎧
πi/2
⎪
⎪ 3.000e
⎪
⎨
5πi/2
3.000e
3.000i =
⎪ 3.000e9πi/2
⎪
⎪
⎩
3.000e13πi/2
⎧ √
4
⎪
3.000eπi/8 = 1.316eπi/8
⎪
⎪
√
⎨
4
√
3.000e5πi/8 = 1.316e5πi/8
4
√
3.000i =
4
⎪
3.000e9πi/8 = 1.316e9πi/8
⎪
⎪
⎩√
4
3.000e13πi/8 = 1.316e13πi/8
3 + 2i 2
17. If z =
, find R(z), I (z), r, and φ.
4 + 5i
3. A Boy Scout finds a tall tree while hiking and wants to
estimate its height. He walks away from the tree and
finds that when he is 45 m from the tree, he must look
upward at an angle of 32◦ to look at the top of the tree.
His eye is 1.40 m from the ground, which is perfectly
level. How tall is the tree?
(4 + 5i)−1 =
3 + 2i
=
4 + 5i
=
◦
h = (45 m) tan (32 ) + 1.40 m = 28.1 m + 1.40 m
= 29.5 m ≈ 30 m
The zero in 30 m is significant, which we indicated
with a bar over it.
5. Express the equation y = b, where b is a constant, in
plane polar coordinates.
=
22
7i
−
41 41
=
=
y = ρ sin (φ) = b
b
= b csc (φ)
ρ =
sin (φ)
7. Find the values of the plane polar coordinates that
correspond to x = 3.00, y = 4.00.
√
ρ = 9.00 + 16.00 = 5.00
4.00
= 53.1◦ = 0.927 rad
φ = arctan
3.00
2
4 − 5i
4 − 5i
=
16 + 25
41
4 − 5i
(3 + 2i)
41
−15 + 8
10
12
+
i+
41
41
41
22
7i
−
41 41
2
22
41
2
2 × 22 × 7
5
−
i+
(41)2
41
0.43378 − 0.18322i
R(z) = 0.43378
I (z) = −0.18322
(0.43378)2 + (0.18322)2 = 0.47079
−0.18322
φ = arctan
= arctan ( − 0.42238)
0.43378
= −0.39965
r =
e14
Mathematics for Physical Chemistry
The principal value of the arctangent is in the fourth
quadrant, equal to −0.39965 rad. Since φ ranges from
0 to 2π , we subtract 0.39965 from 2π to get
φ = 5.8835 rad
19. Estimate the number of grains of sand on the beaches
of the major continents of the earth. Exclude islands
and inland bodies of water.
Assume that the earth has seven continents with an
average radius of 2000 km. Since the coastlines are
somewhat irregular, assume that each continent has a
coastline of roughly 10000 km = 1 × 107 m for a
total coastline of 7 × 107 m. Assume that the average
stretch of coastline has sand roughly 5 m deep and
50 m wide. This gives a total volume of beach sand
of 1.75 × 1010 m3 . Assume that the average grain of
sane is roughly 0.3 mm in diameter, so that each cubic
millimeter contains roughly 30 grains of sand. This is
equivalent to 3 × 1010 grains per cubic meter, so that
we have roughly 5 × 1020 grains of sand. If we were to
include islands and inland bodies of water, we would
likely have a number of grains of sand nearly equal to
Avogadro’s constant.
21. Estimate the number of blades of grass in a lawn with
an area of 1000 square meters.
Assume approximately 10 blades per square
centimeter.
100 cm 2
(1000 m2 )
number = (10 cm−2 )
1m
= 1 × 108