Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt

1
2
1
2
2
4
6
8
10
1
2
Figure 1.11: Plots of f(x) = p(x)/q(x).
1.6 Hints
Hint 1.1
area = constant ×diameter
2
.
Hint 1.2
A pair (x, y) is a solution of the equation if it make the equation an identity.
Hint 1.3
The domain is the subset of R on which the function is defined.
Hint 1.4
Find the slope and x-intercept of the line.
Hint 1.5
The inverse of the function is the reflection of the function across the line y = x.
Hint 1.6
The formu la for geometric growth/decay is x(t) = x
0
r
t

, where r is the rate.
14
Hint 1.7
Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take the
leading coefficient of q(x) to be unity.
f(x) =
p(x)
q(x)
=
ax
2
+ bx + c
x
2
+ βx + χ
Use the properties of the function to solve for the unknown parameters.
Hint 1.8
Write the polynomial in factored form.
15
1.7 Solutions
Solution 1.1
area = π ×radius
2
area =
π
4
× diameter
2
The constant of proportionality is
π

4
.
Solution 1.2
1. If we multiply the equation by y
2
− 4 and divide by x + 1, we obtain the equation of a line.
y + 2 = x − 1
2. We f actor the quadratics on the right side of the equation.
x + 1
y −2
=
(x + 1)(x − 1)
(y −2)(y + 2)
.
We note that one or both sides of the equation are undefined at y = ±2 because of division by zero. There are
no solutions for these two values of y and we assume from this point that y = ±2. We multiply by (y −2)(y +2).
(x + 1)(y + 2) = (x + 1)(x − 1)
For x = −1, the equ ation becomes the identity 0 = 0. Now we consider x = −1. We divide by x + 1 to obtain
the equation of a line.
y + 2 = x − 1
y = x −3
Now we collect the solutions we have found.
{(−1, y) : y = ±2} ∪ {(x, x −3) : x = 1, 5}
The solutions are depicted in Figure /reffig not a line.
16
-6
-4
-2 2
4
6

-6
-4
-2
2
4
6
Figure 1.12: The solutions of
x+1
y−2
=
x
2
−1
y
2
−4
.
Solution 1.3
The denominator is nonzero for all x ∈ R. Since we don’t have any division by zero problem s, the domain of the
function is R. For x ∈ R,
0 <
1
x
2
+ 2
≤ 2.
Consider
y =
1
x

2
+ 2
. (1.1)
For any y ∈ (0 . . . 1/2], there is at least one value of x that satisfies Equation 1.1.
x
2
+ 2 =
1
y
x = ±

1
y
− 2
Thus the range of the function is (0 . . . 1/2]
17
Solution 1.4
Let c denote degrees Celsius and f denote degrees Fahrenheit. The line passes through the points (f, c) = (32, 0) and
(f, c) = (212, 100). The x-intercept is f = 32. We calculate the slope of the line.
slope =
100 − 0
212 − 32
=
100
180
=
5
9
The relationship between fahrenheit and celcius is
c =

5
9
(f − 32).
Solution 1.5
We p lot the various transformations of f(x).
Solution 1.6
The formula for geometric growth/decay is x(t) = x
0
r
t
, where r is the rate. Let t = 0 coincide with 6:00 pm. We
determine x
0
.
x(0) = 10
9
= x
0

11
10

0
= x
0
x
0
= 10
9
At 7:00 pm the n umber of bacteria is

10
9

11
10

60
=
11
60
10
51
≈ 3.04 × 10
11
At 3:00 pm the n umber of bacteria was
10
9

11
10

−180
=
10
189
11
180
≈ 35.4
18
Figure 1.13: Graphs of f(−x), f(x + 3), f(3 − x) + 2, and f

−1
(x).
Solution 1.7
We write p(x) and q(x) as general quadratic polynomials.
f(x) =
p(x)
q(x)
=
ax
2
+ bx + c
αx
2
+ βx + χ
We wil l u se the properties of the function to solve for the unknown parameters.
19
Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take
the leading coefficient of q(x) to be unity.
f(x) =
p(x)
q(x)
=
ax
2
+ bx + c
x
2
+ βx + χ
f(x) has a second order zero at x = 0. This means that p(x) has a second order zero there and that χ = 0.
f(x) =

ax
2
x
2
+ βx + χ
We n ote that f (x) → 2 as x → ∞. This d etermine s the parameter a.
lim
x→∞
f(x) = lim
x→∞
ax
2
x
2
+ βx + χ
= lim
x→∞
2ax
2x + β
= lim
x→∞
2a
2
= a
f(x) =
2x
2
x
2
+ βx + χ

Now we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0.
f(x) =
2x
2
x
2
+ χ
Finally, we use that f(1) = 1 to determine χ.
f(x) =
2x
2
x
2
+ 1
20
Solution 1.8
Consider the polynomial
p(x) = (x + 2)
40
(x − 1)
30
(x − π)
30
.
It is of degree 100. Since the factors only vanish at x = −2, 1, π, p(x) only vanishes there. Since factors are non-
negative, the polynomial is non-negative.
21
Chapter 2
Vectors
2.1 Vectors

2.1.1 Scalars and Vectors
A vector is a quantity having both a magnitude and a direction. Examples of vector quantities are velocity, force
and position. One can represent a vector in n-dimensi onal space with an arrow whose initial point is at the origin,
(Figure 2.1). The magnitude is the length of the vector. Typographically, variables representing vectors are often
written in capital letters, bold face or with a vector over-line, A, a,a. The magnitude of a vector is denoted |a|.
A scalar has only a magnitude. Examples of scalar quantities are mass, time and speed.
Vector Algebra. Two vectors are equal if they have the same magnitude and direction. The negative of a vector,
denoted −a, is a vector of the same magnitude as a but in the opposite direction. We add two vectors a and b by
placing the tail of b at the head of a and defining a + b to be the vector with tail at the origin and head at the head
of b. (See Figure 2.2.)
The difference, a − b, is defined as the sum of a and the negative of b, a + (−b). The result of multiplying a by
a scalar α is a vector of magnitude |α||a| with the same/opposite direction if α is positive/negative. (See Figure 2.2.)
22
x
z
y
Figure 2.1: Graphical representation of a vector in three dimensions.
a+b
a
b
-a
a
2a
Figure 2.2: Vector arithmetic.
Here are the properties of adding vectors and multip lyin g them by a scalar. They are evident from geometric
23
considerations.
a + b = b + a αa = aα commutative laws
(a + b) + c = a + (b + c) α(βa) = (αβ)a associative laws
α(a + b) = αa + αb (α + β)a = αa + βa distributive laws

Zero and Unit Vectors. The additive identity element for vectors is the zero vector or null vector. This is a vector
of magnitud e zero which is den oted as 0. A un it vector is a vector of magnitude one. If a is nonzero then a/|a| is a
unit vector i n the direction of a. Unit vectors are often denoted with a caret over-line,
ˆ
n.
Rectangular Unit Vectors. In n dimensional Cartesian space, R
n
, the unit vectors in the directions of the
coordinates axes are e
1
, . . . e
n
. These are called the rectangular unit vectors. To cut down on subscripts, the unit
vectors in three dimensional space are often denoted with i, j and k. (Figure 2.3).
x
z
y
j
k
i
Figure 2.3: Rectangular unit vectors.
24
Components of a Vector. Consider a vector a with tail at the origin and head having the Cartesian coordinates
(a
1
, . . . , a
n
). We can represent this vector as the sum of n rectangular component vectors, a = a
1
e

1
+ ··· + a
n
e
n
.
(See Figure 2.4.) Another notation for the vector a is a
1
, . . . , a
n
. By the Pythagorean theorem, the magnitude of
the vector a is |a| =

a
2
1
+ ··· + a
2
n
.
x
z
y
a
a
a
1
3
i
k

j
a
2
Figure 2.4: Components of a vector.
2.1.2 The Kronecker Delta and Einstein Summation Convention
The Kronecker Delta tensor is defined
δ
ij
=

1 if i = j,
0 if i = j.
This notation will be useful in our work with vectors.
Consider writing a vector in terms of its rectangular components. Instead of using ellipses: a = a
1
e
1
+···+ a
n
e
n
, we
could write the expression as a sum: a =

n
i=1
a
i
e
i

. We can shorten this notation by leaving out the sum: a = a
i
e
i
,
where it is understood that whenever an index is repeated in a term we sum over that index from 1 to n. This is the
25
Einstein summation convention. A repeated index is called a summation index or a dummy index. Other indices can
take any valu e from 1 to n and are called free indices.
Example 2.1.1 Consider the matrix equation: A · x = b. We can write out the matrix and vectors explicitly.



a
11
··· a
1n
.
.
.
.
.
.
.
.
.
a
n1
··· a
nn







x
1
.
.
.
x
n



=



b
1
.
.
.
b
n




This take s m uch less space when we use the summation convention.
a
ij
x
j
= b
i
Here j is a summation index and i is a free index.
2.1.3 The Dot and Cross Product
Dot Product. The dot product or scalar product of two vectors is defined,
a · b ≡ |a||b|cos θ,
where θ is the angle from a to b. From this definition one can derive the following properties:
• a · b = b · a, commutative.
• α(a · b) = (αa) · b = a · (αb), associativity of scalar multiplication.
• a · (b + c) = a · b + a · c, distributive. (See Exercise 2.1.)
• e
i
e
j
= δ
ij
. In three dimensions, this is
i · i = j · j = k · k = 1, i · j = j · k = k · i = 0.
• a · b = a
i
b
i
≡ a
1
b

1
+ ··· + a
n
b
n
, dot product in terms of rectangular components.
• If a · b = 0 then either a and b are orthogonal, (perpendicular), or one of a and b are zero.
26
The Angle Between Two Vectors. We can use the dot product to find the angle between two vectors, a and
b. From the definition of the dot product,
a · b = |a||b|cos θ.
If the vectors are nonzero, then
θ = arccos

a · b
|a||b|

.
Example 2.1.2 What is the angle between i and i + j?
θ = arccos

i · (i + j)
|i||i + j|

= arccos

1
√
2


=
π
4
.
Parametric Equation of a Line. Consider a line in R
n
that passes through the point a and is parallel to the
vector t, (tangent). A parametric equation of the line is
x = a + ut, u ∈ R.
Implicit Equation of a Line In 2D. Consider a line in R
2
that passes through the point a and is normal,
(orthogonal, perpendicular), to the vector n. All the lines that are normal to n have the property that x · n is a
constant, where x is any point on the line. (See Figure 2.5.) x · n = 0 is the line that is normal to n and passes
through the origin. The line that is normal to n and passes through the poin t a is
x · n = a · n.
The normal to a line determines an orie ntation of the line. The normal points in the direction that is above the
line. A point b is (above/on/below) the line if (b −a) ·n is (positive/zero/negative). The signed distance of a point
27
=0
=1
=
a n
n
a
=-1
x n
x n
x n
x n

Figure 2.5: Equation for a line.
b from the line x · n = a · n is
(b − a) ·
n
|n|
.
Implicit Equation of a Hyperplane. A hyperplane in R
n
is an n −1 dimensional “sheet” which passes through
a given point and is normal to a given direction. In R
3
we call this a plane. Consider a hyperplane that passes through
the point a and is normal to the vector n. All the hyperplanes that are normal to n have the property that x ·n is a
constant, where x is any point in the hyperplane. x · n = 0 is the hyperplane that is n ormal to n and passes through
the origin. The hyperplane that is normal to n and passes through the point a is
x · n = a · n.
The normal determines an orientation of the hyperplane. The normal points in the direction that is above the
hyperplane. A point b is (above/on/below) the hyperplane if (b − a) · n is (positive/zero/negative). The signed
28
distance of a point b from the hyperplane x ·n = a ·n is
(b − a) ·
n
|n|
.
Right and Left-Handed Coordinate Systems. Consider a rectangular co ordinate system in two dimensions.
Angles are measured from the positive x axis in the direction of the positive y axis. The re are two ways of labeling the
axes. (See Figure 2.6.) In one the angle increases in the counterclockwise direction and in the other the angle increases
in the clockwise direction. The former is the familiar Cartesian coordinate system.
x y
xy

θ
θ
Figure 2.6: There are two ways of labeling the axes in two dimensions.
There are also two ways of labeling the axes i n a three-dimensional rectangular coordinate system. These are called
right-handed and left-handed coordinate systems. See Figure 2.7. Any other labelling of the axes could be rotated into
one of these configurations. The right-handed system is the one that is used by default. If you put your right thumb in
the direction of the z axis in a right-handed coordinate system, then your fingers curl in the direction from the x axis
to the y axis.
Cross Product. The cross product or vector product is defined,
a × b = |a||b|sin θ n,
where θ is the angle from a to b and n is a unit vector that is orthogonal to a and b and in the direction s uch that
the ordered triple of vectors a, b and n form a right-handed system.
29
x
z
yj
i
k
z
k
j
i
y
x
Figure 2.7: Right and left handed coordinate systems.
You can visualize the direction of a × b by applying the right hand rule. Curl the fingers of your right hand in the
direction from a to b. Your thumb points in the direction of a × b. Warning: Unless you are a lefty, get in the habit
of putting down your p enci l before applying the right hand rule.
The dot and cross products behave a little differently. First note that unlike the dot product, the cross product is not
commutative. The magnitudes of a × b and b × a are the same, but their directions are opposite. (See Figure 2.8.)

Let
a × b = |a||b|sin θ n and b × a = |b||a|sin φ m.
The angle from a to b is the same as the angle from b to a. Since {a, b, n} and {b, a, m} are right-handed systems,
m points in the opposite direction as n. Since a ×b = −b ×a we say that the cross product is anti-commutative.
Next we n ote that since
|a × b| = |a||b|sin θ,
the magnitude of a × b is the area of the parallelogram defined by the two vectors. (See Figure 2.9.) The area of the
triangle defined by two vectors is then
1
2
|a × b|.
From the d efini tion of the cross product, one can derive the following properties:
30
a
b
b a
a b
Figure 2.8: The cross product is anti-commutative.
b
sin
b
b
a
θ
a
Figure 2.9: The parallelogram and the triangle defined by two vectors.
• a × b = −b × a, anti-commutative.
• α(a × b) = (αa) × b = a × (αb), associativity of scalar multiplication.
• a × (b + c) = a × b + a × c, distributive.
• (a × b) × c = a × (b × c). The cross product is not associative.

• i × i = j × j = k × k = 0.
31
• i × j = k, j × k = i, k × i = j.
•
a × b = (a
2
b
3
− a
3
b
2
)i + (a
3
b
1
− a
1
b
3
)j + (a
1
b
2
− a
2
b
1
)k =







i j k
a
1
a
2
a
3
b
1
b
2
b
3






,
cross product in terms of rectangular components.
• If a · b = 0 then either a and b are parallel or one of a or b is zero.
Scalar Triple Product. Consider the volume of the parallelopiped defined by three vectors. (See Figure 2.10.)
The area of the base is ||b||c|sin θ|, where θ is the angle between b and c. The height is |a|cos φ, where φ is the angle
between b ×c and a. Thus the volume of the parallelopiped is |a||b||c|sin θ cos φ.

φ
θ
b c
a
b
c
Figure 2.10: The parallelopiped defined by three vectors.
Note that
|a · (b × c)| = |a · (|b||c|sin θ n)|
= ||a||b||c|sin θ cos φ|.
32
Thus |a · (b × c)| is the volume of the parallelopiped. a ·(b ×c) is the volume or the negative of the volume depending
on whether {a, b, c} is a right or left-handed system.
Note that parentheses are unnecessary in a ·b ×c. There is only one way to interpret the expression. If you did the
dot product first then you would be left with the cross product of a scalar and a vector which is meani ngless . a ·b ×c
is called the scalar triple product.
Plane Defined by T hre e Points. Three points which are not collinear define a plane. Consider a plane that
passes through the three points a, b and c. One way of expressing that the point x lies in the plane is that the vectors
x −a, b −a and c −a are coplanar. (See Figure 2.11.) If the vectors are coplanar, then the parallelopiped defined by
these three vectors will have zero volume. We can express this in an equation using the scalar triple product,
(x − a) · (b − a) × (c − a) = 0.
b
c
x
a
Figure 2.11: Three points define a plane.
2.2 Sets of Vectors in n Dimensions
Orthogonality. Consider two n-dimensional vectors
x = (x
1

, x
2
, . . . , x
n
), y = (y
1
, y
2
, . . . , y
n
).
33
The inner product of these vectors can be defined
x|y ≡ x · y =
n

i=1
x
i
y
i
.
The vectors are orthogonal if x · y = 0. The norm of a vector is the length of the vector generalized to n dimensions.
x =
√
x · x
Consider a set of vectors
{x
1
, x

2
, . . . , x
m
}.
If each pair of vectors in the set is orthogonal, then the set is orthogonal.
x
i
· x
j
= 0 if i = j
If in addition each vector in the set has norm 1, then the set is orthonormal.
x
i
· x
j
= δ
ij
=

1 if i = j
0 if i = j
Here δ
ij
is known as the Kronecker d elta fu nction.
Completeness. A set of n, n-dimensional vectors
{x
1
, x
2
, . . . , x

n
}
is compl ete if any n-dimensional vector can be written as a linear combination of the vectors in the set. That is, any
vector y can be written
y =
n

i=1
c
i
x
i
.
34
Taking the inner product of each side of this equation with x
m
,
y · x
m
=

n

i=1
c
i
x
i

· x

m
=
n

i=1
c
i
x
i
· x
m
= c
m
x
m
· x
m
c
m
=
y · x
m
x
m

2
Thus y has the expansion
y =
n


i=1
y · x
i
x
i

2
x
i
.
If in addition the set is orthonormal, then
y =
n

i=1
(y · x
i
)x
i
.
35
2.3 Exercises
The Dot and Cross Product
Exercise 2.1
Prove the distributive law for the dot product,
a · (b + c) = a · b + a · c.
Hint, Solution
Exercise 2.2
Prove that
a · b = a

i
b
i
≡ a
1
b
1
+ ··· + a
n
b
n
.
Hint, Solution
Exercise 2.3
What is the angle between the vectors i + j and i + 3j?
Hint, Solution
Exercise 2.4
Prove the distributive law for the cross produ ct,
a × (b + c) = a × b + a × b.
Hint, Solution
Exercise 2.5
Show that
a × b =






i j k

a
1
a
2
a
3
b
1
b
2
b
3






Hint, Solution
36
Exercise 2.6
What is the area of the quadrilateral with vertices at (1, 1), (4, 2), (3, 7) and (2, 3)?
Hint, Solution
Exercise 2.7
What is the volume of the tetrahedron with vertices at (1, 1, 0), (3, 2, 1), (2, 4, 1) and (1, 2, 5)?
Hint, Solution
Exercise 2.8
What is the equation of the plane that passes through the points (1, 2, 3), (2, 3, 1) and (3, 1, 2)? What is the distance
from the point (2, 3, 5) to the plane?
Hint, Solution

37