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In this short note, we study an introduction to crystallography.

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Definitions of Basis, Lattice, and Crystals

Basis

A crystal is formed by the periodic repetition of a group of atoms in all directions. That group of atoms is called the basis.

The basis of a crystal can be one or more atoms.
An ideal crystal consists of infinite repetitions of the basis.

Lattice

The infinite array of mathematical points which (describes how the basis is repeated) forms a periodic spatial arrangement is called the lattice.

The lattice looks identical from whichever points you view the array.
Note that lattice is not a crystal.

$\"\"$

In three dimensions, the lattice can be identified by three independent vectors \$\\mathbf{a}_1\$, \$\\mathbf{a}_2\$ and \$\\mathbf{a}_3\$.
The position of each point \$\\mathbf{R}\$ can be written as a linear combination of these vectors, \\[\\mathbf{R}=n_1 \\mathbf{a}_1+n_2 \\mathbf{a}_2+n_3 \\mathbf{a}_3 \\tag{1}\\] where \$n_1\$, \$n_2\$, and \$n_3\$ are integers.
The vectors that can generate or span the lattice are not unique. For example, see the following figure.

Recall that the volume of a parallelepiped with axes \$\\mathbf{a}_1\$ , \$\\mathbf{a}_2\$ and \$\\mathbf{a}_3\$ is given by \$|\\mathbf{a}_1\\cdot(\\mathbf{a}_2\\times \\mathbf{a}_3)|\$ .

Primitive Cells vs. Unit Cells

Primitive Cell

Of the vectors satisfying Eq. (1), those that form a parallelepiped with the smallest volume, are called primitive translation vectors, and the parallelepiped they form is known as the primitive cell.

Unit Cell

To better demonstrate the symmetry of the entire lattice, sometimes non-primitive translation vectors are used to specify the lattice. In this case, the parallelepiped is called the unit cell.

The unit cell may or may not be identical to the primitive cell.

[caption id=\"attachment_7270\" align=\"aligncenter\" width=\"350\"] $\"\"$ Primitive cell is shown in gray[/caption]

Unit cell of a face-centered cubic structure

Primitive unit cell of a face-centered cubic structure. As drawn, the primitive translation vectors are \\[\\mathbf{a}_1=\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}})\\quad \\mathbf{a}_2=\\frac{a}{2}(\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\\quad \\mathbf{a}_3=\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{z}})\\]

Lattice Parameters and Packing Fraction

Lattice Parameters

Lattice Parameters

The lengths of axes of the unit cell (called lattice constants \$a\$, \$b\$, and \$c\$) and the angles between the axes (\$\\alpha\$, \$\\beta\$ and \$\\gamma\$) specify the unit cell. The lattice constants and the three angles between them are termed lattice parameters. See the following figure.

Packing Fraction

We try to pack \$N\$ hard spheres (cannot deform)
The total volume of the spheres is \\[V_s=N \\frac{4}{3}\\pi R^3\\]
The volume these spheres occupy \$V>V_S\$ (there are spacing)\\[{\\rm Packing\\ Fraction}=\\frac{N \\frac{4}{3}\\pi R^3}{V}\\]

Bravais Lattices

Auguste Bravais, a French physicist, identified 14 distinct lattices in three dimensions.
There are 5 Bravais lattices in two dimensions (shown below)
In crystallography, all lattices are traditionally called Bravais lattices or translation lattices.

Five Bravais Lattices in Two Dimensions

Seven Crystal Systems

Isometric (or Cubic)

$a=b=c$

$\\alpha=\\beta=\\gamma=90^\\circ$

Tetragonal

$a=b\\ne c$

$\\alpha = \\beta = \\gamma = 90^\\circ$

Orthorhombic

$a\\ne b\\ne c$

$\\alpha = \\beta = \\gamma = 90^\\circ$

Hexagonal

$a=b\\ne c$

$\\alpha = \\beta = 90^\\circ, \\gamma = 120^\\circ$

Triclinic

$a\\ne b\\ne c$

$\\alpha \\ne \\beta \\ne \\gamma$

Monoclinic

$a\\ne b\\ne c$

$\\alpha = \\beta = 90^\\circ \\ne \\gamma$

Rhombohedral (or Trigonal)

$a=b=c$

$\\alpha = \\beta = \\gamma < 120^\\circ, \\ne 90^\\circ$

Cubic Lattices


Primitive	Body-Centered	Face-Centered

Simple Cubic (SC) Crystal

Location of all lattice points \\[\\mathbf{R}=a(u\\, \\hat{\\mathbf{x}}+v\\, \\hat{\\mathbf{y}}+w\\, \\hat{\\mathbf{z}})=a[u\\ v\\ w]\\]
Polonium (Po) is the only metal that forms a simple cubic unit cell.
Each lattice point is shared by 8 neighboring units
average volume occupied by each lattice point\$=\$
average volume occupied by each atom in SC\$=\\Omega_{SC}=\\frac{a^3}{8\\times\\frac{1}{8}}=a^3\$

Body-Centered Cubic (BCC) Crystal

Location of all lattice points \\[\\begin{aligned} \\mathbf{R}&=a(u\\, \\hat{\\mathbf{x}}+v\\, \\hat{\\mathbf{y}}+w\\, \\hat{\\mathbf{z}})=a[u\\ v\\ w]\\\\ \\mathbf{R}&=a((u+0.5)\\, \\hat{\\mathbf{x}}+(v+0.5)\\, \\hat{\\mathbf{y}}+(w+0.5)\\, \\hat{\\mathbf{z}})=a[u+0.5\\ v+0.5\\ w+0.5]\\end{aligned}\\]
average volume occupied by each lattice point\$=\$
average volume occupied by each atom in BCC\$=\\Omega_{BCC}=\\frac{a^3}{8\\times\\frac{1}{8}+1}=\\frac{a^3}{2}\$
Nearest neighbor distance = \$d=2r=a\\frac{\\sqrt{3}}{2}\$
\\[\\text{Packing fraction}=\\frac{2\\times\\frac{4\\pi}{3}\\times(\\frac{a\\sqrt{3}}{4})^3}{a^3}=\\frac{\\pi\\sqrt{3}}{8}\\approx 0.68\\]

The primitive translation vectors are: \\[\\mathbf{a}_1=\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}-\\hat{\\mathbf{z}})\\] \\[\\mathbf{a}_2=\\frac{a}{2}(-\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\\] \\[\\mathbf{a}_3=\\frac{a}{2}(\\hat{\\mathbf{x}}-\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\\]

The primitive unit cell is a rhombohedron of edge \$a\\frac{\\sqrt{3}}{2}\$
The angle between adjacent edges is \$109^\\circ 28'\$

Number of nearest neighbors = 8
Number of second nearest neighbors = 6
Second nearest neighbor distance = \$a\$

Some Elements with BCC structure

Barium	Ba	Chromium	Cr
Cesium	Cs	\$\\alpha-\$Iron	\$\\alpha-\$Fe
Potassium	K	Lithium	Li
Molybdenum	Mo	Sodium	Na
Niobium	Nb	Rubidium	Rb
Tantalum	Ta	Titanium	Ti
Vanadium	V	Tungsten	W

Face-Centered Cubic (FCC) Crystal

Location of all lattice points \\[\\begin{aligned} \\mathbf{R}&=a[u\\ v\\ w]\\\\ \\mathbf{R}&=a[u+0.5\\ v\\ w]\\\\ \\mathbf{R}&=a[u\\ v+0.5\\ w]\\\\ \\mathbf{R}&=a[u\\ v\\ w+0.5]\\end{aligned}\\]
average volume occupied by each lattice point\$=\$
average volume occupied by each atom in FCC\$=\\Omega_{FCC}=\\frac{a^3}{8\\times\\frac{1}{8}+6\\times\\frac{1}{2}}=\\frac{a^3}{4}\$
Nearest neighbor distance = \$d=2r=\\frac{a}{\\sqrt{2}}\$
\\[\\text{Packing fraction}=\\frac{4\\times\\frac{4\\pi}{3}\\times(\\frac{a\\sqrt{2}}{4})^3}{a^3}=\\frac{\\pi\\sqrt{2}}{6}\\approx 0.74\\]

The primitive cell vs the unit cell

The primitive translation vectors are: \\[\\mathbf{a}_1=\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}})\\] \\[\\mathbf{a}_2=\\frac{a}{2}(\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\\] \\[\\mathbf{a}_3=\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{z}})\\]

The angle between adjacent edges is \$60^\\circ\$

The reference atom is red. Blue, green, and orange points are the 12 nearest neighbors of the reference atoms, and the purple points are its 6 next nearest neighbors atoms.

Packing fraction \$\\frac{1}{6}\\pi\\sqrt{2}\\approx 0.740\$
Some elements with fcc structure: Ar, Ag, Al, Au, Ca, Ce, \$\\beta-\$Co, Cu, Ir, Kr, La, Ne, Ni, Pb, Pd, Pr, Pt, \$\\delta-\$Pu, Rh, Sc, Sr, Th, Xe, Yb

Characteristics of Cubic structure

	simple cubic	b.c.c.	f.c.c.
volume of conventional cell	\$a^3\$	\$a^3\$	\$a^3\$
no. of lattice points per cell	1	2	4
no. of nearest neighbors (coordination number)	6	8	12
no. of 2nd nearest neighbors	12	6	6
nearest neighbor distance	\$a\$	\$\\frac{\\sqrt{3}}{2}a\\approx 0.866 a\$	\$\\frac{a}{\\sqrt{2}}\\approx 0.707a\$
2nd nearest neighbor distance	\$a\\sqrt{2}\$	\$a\$	\$a\$
packing fraction	\$\\pi/6\\approx 0.52\$	\$\\pi\\sqrt{3}/8\\approx 0.68\$	\$\\pi\\sqrt{2}/6\\approx 0.74\$

[111], [101], and [110] describe directions \$m\$, \$t\$, and \$n\$, respectively.

Crystal Structure May Change with Temperature

If the temperature changes, some materials might undergo a phase change.
For example, at atmospheric pressure (10⁵ Pa) and below 912 °C, pure iron (Fe) has a BCC structure, known as ⍺-iron. If we heat iron above 912 °C, its structure changes to an FCC structure, known as ????-iron. Above 1394 °C, the structure changes back to BCC, known as ????-iron. You may see the phase diagram of pure iron in the following figure.

$\"\"$

Hexagonal Close-Packed (HCP) Structure

30 elements crystallize in hcp form.
HCP crystal has a hexagonal lattice and a multi-atom basis.
It can be viewed as two nested simple hexagonal Bravais lattice shifted by \$\\mathbf{a}_1/3+ \\mathbf{a}_2/3+\\mathbf{a}_3/2\$.
where

\\[\\mathbf{a}_1=a\\hat{\\mathbf{x}},\\ \\mathbf{a}_2=\\frac{a}{2}\\hat{\\mathbf{x}}+\\frac{a\\sqrt{3}}{2}\\hat{\\mathbf{y}},\\quad \\mathbf{a}_3=c\\,\\hat{\\mathbf{z}}\\]

- average vol. occupied by each lattice point \\[\\Omega_{hex\\ lattice}=\\frac{\\sqrt{3}}{2}a^2 c\\]
- average vol. occupied by each atom \\[\\Omega_{HCP}=\\frac{\\Omega_{hex\\ lattice}}{2}\\]

Comparing Close-Packed Structures

[caption id=\"attachment_7293\" align=\"aligncenter\" width=\"700\"] $\"\"$ The left is the hcp structure, and the right is the fcc structure.[/caption]


hcp	fcc

In this figure, the left structure is hcp, and the right is fcc
volume fraction = 0.74
number of nearest neighbors (coordination number) is 12 for both hcp and fcc structures
Although a hexagonal close-packing of equal atoms is only obtained if \$c/a=\\sqrt{8/3}\\approx 1.63\$, the term hcp is used for any structure described previously.

Elements with hcp structures

Element	$\"\\large$	Element	$\"\\large$
Ideal	1.63
Be	1.56	Cd	1.89
Ce	1.63	\\(\\alpha\\)-Co	1.62
Dy	1.57	Er	1.57
Gd	1.59	He (2K)	1.63
Hf	1.58	Ho	1.57
La	1.62	Lu	1.59
Mg	1.62	Nd	1.61
Os	1.58	Pr	1.61
Re	1.62	Ru	1.59
Tb	1.58	Ti	1.59
Tl	1.60	Tm	1.57
Y	1.57	Zn	1.59

[adapted from Ashcroft, Mermin, Solid State Physics]

Diamond Crystal

It is the structure of carbon in a diamond crystal
It can be viewed as two interpenetrating fcc lattices displaced by \$\\frac{a}{4}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\$
Or it can be imagined as the fcc lattice with two point basis \$\\mathbf{0}\$ and \$\\frac{a}{4}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}+\\hat{\\mathbf{z}})\$.
Coordination number is 4
Packing fraction is \$\\frac{\\sqrt{3}}{16}\\pi\\approx 0.34\$

3D Diamond Structure

Elements with diamond structure: C (diamond), Si, Ge, \$\\alpha\$-Sn (gray)
average volume occupied by each atom: \$\\Omega_{DC}=\\frac{\\Omega_{FCC}}{2}=\\frac{a^3}{8}\$

Sodium Chloride Structure

Na\$^{+}\$ and Cl\$^-\$ ions are placed on alternate points of a simple cubic structure
The lattice is fcc; the basis consists of Na\$^{+}\$ and Cl\$^-\$

NaCl structure, Blue atoms represent Na atoms and Green ones represent Cl atoms.

Some compounds with sodium chloride structure

LiF	LiCl	LiBr	LiI
NaF	NaCl	NaBr	NaI
RbF	RbCl	RbBr	RbI
CsF
AgF	AgCl	AgBr
MgO	MgS	MgSe
CaO	CaS	CaSe	CaTe
SrO	SrS	SrSe	SrTe
BaO	BaS	BaSe	BaTe

[Adapted from Ashcroft, Mermin, Solid State Physics]

Cesium Chloride Structure

Cs\$^{+}\$ and Cl\$^-\$ ions are placed at \$\\mathbf{0}\$ and body center position \$\\frac{a}{2}(\\hat{\\mathbf{x}}+\\hat{\\mathbf{y}}+\\hat{\\mathbf{y}})\$, respectively.
The lattice is simple cubic; the basis consists of Cs\$^{+}\$ and Cl\$^-\$

CsCl structure. Blue spheres represent Cl atoms and the big red sphere represents the Cs atom which is much larger than the Cl atoms.

Some compounds with the cesium chloride structure:

CsCl	CsBr	CsI
TlCl	TlBr	TlI

Miller Indices

Miller Indices for Directions in Cubic Structure

It is often required to specify certain directions and planes in crystals. To this end, we use Miller indices.
\$[hkl]\$ represents the direction vector \$h\\hat{\\mathbf{x}}+k\\hat{\\mathbf{y}}+l\\hat{\\mathbf{z}}\$ of a line passing through the origin.
These integers \$h\$, \$k\$ and \$l\$ must be the smallest numbers that will give the desired direction. i.e. we write \$[111]\$ not \$[222]\$.
If a component is negative, it is conventionally specified by placing a bar over the corresponding index. For example, we write \$[1\\bar{1}1]\$ instead of \$[1 -1 1]\$.
Coordinates in angle brackets such as \$\\langle123\\rangle\$ denote a family of directions that are equivalent due to symmetry operations, such as [123], [132], [321], [\$\\bar{1}23\$], [\$\\bar{1}\\bar{2}3\$], etc.

[111], [101], and [110] describe directions \$m\$, \$t\$, and \$n\$, respectively.

Miller Indices for Planes in Cubic Structure

A crystallographic plane is denoted by the Miller indices of the direction normal to the plane, but instead of brackets we use parenthesis, i.e. (\$hkl\$)


$(100)$	$(110)$	$(111)$

“Coordinates in curly brackets or braces such as \$\\{100\\}\$ denote a family of plane normals that are equivalent due to symmetry operations, much the way angle brackets denote a family of directions.”
For a cubic system, a family \$\\{hkl\\}\$ consists of all the planes given by the permutations of the number \$h\$, \$k\$, \$l\$ and their negatives.
If the symmetry of the system is lower than that of cubic, not all the planes given by the permutations necessarily belong to a family. For example, in a rhombohedral system, we have \$\\{100\\}=\\{(100),(\\bar{1}00),(010),(0\\bar{1}0),(001),(00\\bar{1})\\}\$, but in an orthorhombic system \$\\{100\\}\$ family has only two members \$(100)\$ and \$(\\bar{1}00)\$.

Miller Indices for hcp

Miller indices contain 4 digits instead of 3 digits
\$[hkil]\$ means: \\[\\{\\alpha (h \\mathbf{a}_1+k\\mathbf{a}_2+i\\mathbf{a}_3+l \\mathbf{c}): \\alpha\\in\\mathbb{R}\\}\\] and \\[h+k+i=0\\]
Directions along axes \$\\mathbf{a}_1\$, \$\\mathbf{a}_2\$ and \$\\mathbf{a}_3\$ are of type \$\\langle \\bar{1} 2\\bar{1} 0\\rangle\$.
\$(hkil)\$ is a plane the normal direction of which is \$[hkil]\$.

CHAPTER 1: Groups

We concern ourselves with sets $G$ of objects $a, b, c, \\ldots$ called elements. The sentence \" $a$ is an element of $G$ \" will be denoted symbolically by $a \\in G$. Assume an operation called \"multiplication\" which assigns to an ordered pair of objects $a, b$ of $G$ another object $a \\cdot b$ (or simply $ab$ ) the product of $a$ and $b$. It is useful to require that $G$ be closed with respect to multiplication, namely:

(1) If $a, b \\in G$, then $a \\cdot b \\in G$.

EXAMPLES.

(a) Let $G$ be the set of positive integers. If subtraction is taken as the \"multiplication\" in $G$, then $G$ is certainly not closed, e.g., $3 \\cdot 5=3-5=-2$. If taking the greatest common divisor is our multiplication, then closure is obvious.

(b) Take $G$ to be the set of functions of one variable. If $f(x), g(x) \\in G$, define $f(x) \\cdot g(x)=f[g(x)]$, e.g., $e^{x} \\cdot \\log x=e^{\\log x}=x$.

EXERCISE 1. Write out the multiplication table and thereby show closure for the set of functions $f_{1}=x, \\quad f_{2}=\\frac{1}{x}, \\quad f_{3}=1-x, \\quad f_{4}=\\frac{1}{1-x}, \\quad f_{5}=\\frac{x}{x-1}, \\quad f_{6}=\\frac{x-1}{x}$.

SOLUTION.

	$f_{1}$	$f_{2}$	$f_{3}$	$f_{4}$	$f_{5}$	$f_{6}$
$f_{1}$	$f_{1}$	$f_{2}$	$f_{3}$	$f_{4}$	$f_{5}$	$f_{6}$
$f_{2}$	$f_{2}$	$f_{1}$	$f_{4}$	$f_{3}$	$f_{6}$	$f_{5}$
$f_{3}$	$f_{3}$	$f_{6}$	$f_{1}$	$f_{5}$	$f_{4}$	$f_{2}$
$f_{4}$	$f_{4}$	$f_{5}$	$f_{2}$	$f_{6}$	$f_{3}$	$f_{1}$
$f_{5}$	$f_{5}$	$f_{4}$	$f_{6}$	$f_{2}$	$f_{1}$	$f_{3}$
$f_{6}$	$f_{6}$	$f_{3}$	$f_{5}$	$f_{1}$	$f_{2}$	$f_{4}$

where $f_{i} \\cdot f_{j}$ is listed in the $i^{\\text {th }}$ row and $j^{\\text {th }}$ column.

We make the further requirement that multiplication obey the associative law:

(2) If $a, b, c \\in G$, then $(a b) c=a(b c)$.

This is a rather strong condition. It is not generally satisfied; consider, e.g., subtraction among the integers. For functions of one variable, as above, it is valid, however. If $f(x), g(x)$, $h(x)$ are any three functions we have $ (f g) h=f(g(h(x)))=f(g h) $

EXERCISE 2. Deduce the associative law for four elements from (2), that is, show that the five possible products of four elements written in a given sequence are all equal. Furthermore, attempt to determine the number of possible products of $n$ elements given in a linear order. For example, the elements $a_{1}, a_{2}, a_{3}, a_{4}$ in that order yield the products $\\left(a_{1} a_{2}\\right)\\left(a_{3} a_{4}\\right), a_{1}\\left(a_{2}\\left(a_{3} a_{4}\\right)\\right)$, etc. Hint: Let $\\alpha_{n}$ be the number of products of $a_{1}, a_{2}, \\ldots, a_{n}$. Find a recursion formula for $\\alpha_{n}$ and use the Lagrange generating function

$
f(x)=\\alpha_{1} x+\\alpha_{2} x^{2}+\\cdots+\\alpha_{n} x^{n}+\\cdots
$

EXERCISE 3. The associative law for $n$ elements states that all possible products of $n$ elements written in a prescribed order, e.g., $a_{1}, a_{2}, \\ldots, a_{n}$, yield the same result. Prove the associative law for any number of elements using only (2) (the associative law for three elements).

PROOF FOR EXERCISE 3: We assume the validity of the associative law for all products of $m$ factors, $m \\leq n$, and show that this implies the validity of the law for $n+1$. Consider the particular product $(n+1) \\prod_{k=1}^{n+1} a_{k}$ which is obtained from the $n+1$ elements $a_{1}, a_{2}, \\ldots, a_{n+1}$ by successively multiplying on the right, i.e.,

$
\\begin{aligned}
& \\prod_{k=1}^{1} a_{k}=a_{1} \\
& \\prod_{k=1}^{n+1} a_{k}=\\left(\\prod_{k=1}^{n} a_{k}\\right) a_{n+1}
\\end{aligned}
$

Let $P_{n+1}$ be any product of the $n+1$ elements $a_{1}, a_{2}, \\ldots, a_{n+1}$ taken in that order. Since $P_{n+1}$ is the result of at least one multiplication, we may write

$
P_{n+1}=P_{1}^{m} P_{m+1}^{n+1}, \\quad 1 \\leq m \\leq n
$

where $P_{1}^{m}$ is some product of the elements $a_{1}, a_{2}, \\ldots, a_{m}$ in that order and $P_{m+1}^{n+1}$ of the remaining elements $a_{m+1}, a_{m+2}, \\ldots, a_{n+1}$. By the induction hypothesis we have

$
P_{\\mu}^{v}=\\prod_{k=\\mu}^{v} a_{k}
$

for any $\\mu, v$ such that $v-\\mu+1 \\leq n$. Specifically, we have

$
\\begin{aligned}
P_{n+1}=\\left(\\prod_{j=1}^{m} a_{j}\\right)\\left(\\prod_{k=m+1}^{n+1} a_{k}\\right) & =\\prod_{j=1}^{m} a_{j} \\cdot\\left[\\left(\\prod_{k=m+1}^{n} a_{k}\\right) \\cdot a_{n+1}\\right] \\
& =\\left(\\prod_{j=1}^{m} a_{j} \\prod_{k=m+1}^{n} a_{k}\\right) \\cdot a_{n+1} \\
& =\\left(\\prod_{k=1}^{n} a_{k}\\right) \\cdot a_{n+1}=\\prod_{k=1}^{n+1} a_{k}
\\end{aligned}
$

each step being a simple application of (2).

1.1. The Concept of a Group

A set $G$ will be called a group if it satisfies the following conditions:

Closure. There exists an operation called multiplication which assigns to any ordered pair $a, b \\in G$ a product $a b \\in G$.
Associative Law. If $a, b, c \\in G$ then $(a b) c=a(b c)$.
Identity. There exists an $e \\in G$, called the (left) identity, such that $e a=a$ for all $a \\in G$.
Inverse. For every $a \\in G$ there is an $a^{-1} \\in G$, called the (left) inverse of $a$, such that $a^{-1} a=e$.

Let us examine the product

$
\\left(a^{-1}\\right)^{-1} a^{-1} a a^{-1}
$

On one hand,

$
\\left[\\left(a^{-1}\\right)^{-1} a^{-1}\\right]\\left[a a^{-1}\\right]=e\\left[a a^{-1}\\right]=a a^{-1}
$

and on the other

$
\\left[\\left(a^{-1}\\right)^{-1}\\right]\\left[\\left(a^{-1} a\\right) a^{-1}\\right]=\\left[\\left(a^{-1}\\right)^{-1}\\right]\\left[e a^{-1}\\right]=\\left(a^{-1}\\right)^{-1} a^{-1}=e
$

Consequently,

$
a a^{-1}=e
$

The existence of the left inverse implies the existence of a right inverse. A similar result holds for the identity; for consider the product

$
a a^{-1} a
$

First we have

$
a a^{-1} a=\\left(a a^{-1}\\right) a=e a=a
$

But also

$
a a^{-1} a=a\\left(a^{-1} a\\right)=a e
$

Consequently,

$
a e=a
$

and the existence of the right identity implies the existence of a left identity.

EXERCISE 4. Two systems of postulates are said to be equivalent if either system can be derived logically from the other. Show that the system (1), (2), (3), (4) is equivalent to the system in which (3) and (4) are replaced by:

There is a right identity $e \\in G$ such that $a e=a$ for all $a \\in G$.
To each $a \\in G$ there is a right inverse $a^{-1} \\in G$ such that $a a^{-1}=e$. Apparently the words right and left need not be included in (3), (4), (3'), or (4').

EXERCISE 5. Consider the postulate system in which (3) and (4) are replaced by:

There exists a left identity $e \\in G$; that is, $e a=a$ for all $a \\in G$.
To each $a \\in G$ there is a right inverse $a^{-1} \\in G$; that is, $a a^{-1}=e$. Determine whether this system of postulates defines a group. If not, give a counterexample.

SOLUTION. For any $a \\in G$ define multiplication by $a x=x$ for all $x \\in G$. This system satisfies the postulates (1),(2), $\\left(3^{*}\\right)$, and $\\left(4^{*}\\right)$. What group property does it not satisfy?

For ordinary numbers, the quotient $b \\div a$ of two numbers can be defined as the solution of the equation $a x=b$. Consider similar equations for elements of $G$ :
(a) $a x=b$, (b) $x a=b$, (c) $a x b=c$

If (a) is true for some $x$, then

$
a^{-1} a x=a^{-1} b=e x=x
$

Hence, if there is a solution, it is $a^{-1} b$ and it is therefore unique; $a^{-1} b$ is in fact a solution. Similar reasoning shows that (b) possesses the unique solution $x=b a^{-1}$ and (c) the unique solution $a^{-1} c b^{-1}$. The existence of a unique solution for each of the above equations demonstrates a property of the group analogous to division.

Since $a^{-1}$ is the solution of the equation $x a=e, a^{-1}$ is unique. Similarly, $e$ is the unique solution of $x a=a$. We observe that the solution of $x(a b)=e$ is $(a b)^{-1}=b^{-1} a^{-1}$. In general, the inverse of a product

$
\\left(a_{1} a_{2} \\cdots a_{n}\\right)^{-1}=a_{n}^{-1} \\cdots a_{2}^{-1} a_{1}^{-1}
$

If $x=\\left(a^{-1}\\right)^{-1}$, then $x$ satisfies the equation $x a^{-1}=e$, which has the unique solution $x=a$. Thus the inverse of the inverse of an element is the element itself.

EXERCISE 6. Show that postulates (3) and (4) may be replaced by $\\left(3^{+}\\right)$ If $a, b \\in G$, the equations

$
x a=b, \\quad a y=b
$

possess (not necessarily unique) solutions $x, y \\in G$.

A group that satisfies the commutative law,
(5) If $a, b \\in G$, then $a b=b a$,
is said to be commutative or abelian.

EXERCISE 7. Show that the six functions of Exercise 1 form a noncommutative group with respect to their rule of multiplication. Determine the identity element and the inverse to each function.

1.2. Subgroups

If $G$ is a group and $S$ is a subset of $G$ that is itself a group under the same operation as $G$, then $S$ is called a subgroup of $G$.

EXAMPLE. Take $G$ to be the set of rational numbers other than zero under ordinary multiplication. $G$ has, e.g., the subgroups

the positive rational numbers
the powers of any element
the set consisting of +1 and -1

Trivially, (d) the set $G$ itself or (e) the set consisting of the element 1.

THEOREM 1.1 Necessary and sufficient conditions for a subset $S$ of $G$ to be a subgroup are:

Closure. If $s_{1}, s_{2} \\in S$, then $s_{1} s_{2} \\in S$.
Inverse. If $s \\in S$, then $s^{-1} \\in S$.

Proof: Necessity. If $S$ is a subgroup, (i) holds by definition. The identity $e \\in S$ by the uniqueness in $G$ and existence in $S$ by the solution of the equation

$
x s=s
$

Note that (ii) is similarly established through the equation $x s=e$.
Sufficiency. If (i) and (ii) hold, then $S$ is a subgroup. From (ii) if $s \\in S$ then $s^{-1}$ is an element of $S$ and hence (i) gives $e \\in S$. The associative law holds for elements of $S$ since they are elements of $G$. The proof of the theorem is complete.

If $S$ is a subgroup of $G$ and $a \\in G$, the coset $a S$ is defined to be the set of all elements $a \\cdot s$, where $s \\in S$.

EXAMPLE. Take for $G$ the set of all rational numbers excluding zero under ordinary multiplication. Let $S$ be the set of all positive elements of $G$. There are only two cosets, $S$ and $-S=-1 S$. These have no elements in common and both sets together cover $G$. If we take instead $S=\\{+1,-1\\}$ then the cosets are $a S=\\{+a,-a\\}$. Here the same coset is given by $+a$ and $-a$. Note again that no two cosets overlap and that the cosets cover $G$. These results are valid in general.

Let $S$ be a subgroup of $G$ and take $a, b \\in G$.

LEMMA 1.2 If the cosets $aS$ and $bS$ have an element $c$ in common, then $aS = bS$.

Assume for some $s, s^{\\prime} \\in S$ we have $c = as = bs^{\\prime}$. Therefore $b = as\\left(s^{\\prime}\\right)^{-1}$. From Theorem 1.1, $s\\left(s^{\\prime}\\right)^{-1} = s^{\\prime \\prime} \\in S$ and consequently $bS = as^{\\prime \\prime}S$. Now $s^{\\prime \\prime}S = S$, since if we suppose $S$ to be any group, $s$ any element of $S$, we have $ s S \\subset S $ (read: $sS$ is a subset of $S$, or all elements of $sS$ are elements of $S$ ). Also, $ s^{-1} s \\subset S \\text { or } \\quad S \\subset s S $ Therefore $ s S=S $ In the above argument we may now write $bS = as^{\\prime \\prime}S = aS$.

LEMMA 1.3 Every $a \\in G$ is contained in some coset

$a \\in aS \\quad \\text { since } \\quad e \\in S$

and hence $ae = a \\in aS . G$ is covered by the cosets of $S$.

THEOREM 1.4 Let $G$ be a finite group of order $N$ and $S$ a subgroup of order $n$. The number $n$ of elements in the subgroup is a divisor of $N$.

Proof: The cosets $aS$ have the same number of elements as $S$. For let $S$ consist of the distinct elements $s_{1}, s_{2}, \\ldots, s_{n} . aS$ consist of $as_{1}, as_{2}, \\ldots, as_{n}$, where $ a s_{1} \\neq a s_{k}, \\quad i \\neq k $ For otherwise we would have $as_{i} = as_{k}$ and hence $s_{i} = s_{k}, i \\neq k$, contrary to the definition of the $s_{i}$.

Consequently, $aS$ consists of exactly $n$ elements. Let $j$ be the number of cosets. By Lemmas 1.2 and 1.3 the cosets cover $G$ without overlapping. It follows that
$
N=j n
$

Take $a \\in G$. We denote $aa$ by $a^{2}$ or, in general, we define all the integral powers $a^{\\mu}$ of $a$ by

$
\\begin{aligned}
& a^{\\mu}=a a \\cdots a(\\mu \\text { times }) \\text { for } \\mu>0 \\
& a^{0}=e \\
& a^{\\mu}=a^{-1} a^{-1} \\cdots a^{-1}(-\\mu \\text { times }) \\text { for } \\mu<0
\\end{aligned}
$

The set of all powers of $a$ is a group and clearly the smallest group containing $a$. The problem of determining the smallest group containing as few as two elements is already of an entirely different nature. For example, what can be said about

$
(a b)^{n}=a b \\cdot a b \\cdots a b(n \\text { times }) ?
$

If multiplication is commutative such products can be handled, but this does not apply in general.

EXERCISE 8. Show that the powers of elements obey the usual properties of exponents

$
\\begin{aligned}
& a^{\\mu} a^{\\nu}=a^{\\mu+v} \\
& \\left(a^{\\nu}\\right)^{\\mu}=a^{\\nu \\mu}
\\end{aligned}
$

The first property implies the commutative law for multiplication of powers of $a$.

The set $S$ of all powers of $a$ forms a subgroup since $S$ is closed under multiplication and inverses exist (cf. Theorem 1.1).

Case 1. The powers of $a$ are all distinct. $S$ is then called an infinite cyclic group.

Case 2. There exist integers $i$ and $k$ with, say, $i<k$ such that $a^{i} = a^{k}$. Multiplying on both sides by $a^{-i}$ we obtain $e = a^{k-i}$. Thus the set of positive integers for which $a^{\\mu} = e$ is not empty. Let $d$ be the smallest such integer

$
a^{d}=e \\Rightarrow a^{q d}=e \\quad \\text { for all integers } q
$

(read: \"implies\" for \" $\\Rightarrow$ \"). Conversely, if $a^{m} = e, m$ is a multiple of $d$, for we may write $m = q d + r$ where $0 \\leq r<d$

$
a^{r}=a^{m-q d}=a^{m} a^{-q d}=e .
$

But $d$ is the smallest positive integer for which $a^{d} = e$. Hence $r$ must be zero whence $m = q d$. The powers

$
a^{0}, a^{1}, a^{2}, \\ldots, a^{d-1}
$

are all distinct for otherwise we would have

$
a^{i}=a^{k}, \\quad 0 \\leq i<k<d, \\quad \\text { or } \\quad a^{k-i}=e
$

this equation is impossible for $0<k-1<d$. Any other power of $a$ must be equal to one of these, for example $a^{d} = e, a^{d+1} = a, \\ldots$, or, in general,

$
a^{q d+r}=a^{r}, \\quad 0 \\leq r<d
$

Thus there are only $d$ distinct powers of $a . S$ is called a cyclic subgroup of order $d$ and $d$ is called the period of $a$.

THEOREM 1.5 The period of any element of a finite group is a divisor of the order of the group.

Proof: This is an immediate consequence of Theorem 1.4. Let $G$ be a finite group of order $N$ and $a$ any element of $G$. If ${ }^{\\prime} d$ is the period of $a$, we may write $N = d j$. From $a^{d} = e$ we have

$
a^{d j}=a^{N}=e
$

This statement for prime $N$ is equivalent to Fermat's theorem in arithmetic.
COROLLARY If the order of $G$ is $p$, a prime, then $G$ must be cyclic.
PROOF: The period of any element must be a divisor of $p$ and is therefore either $p$ or 1 . The only element of period 1 is $e$. Consequently, if $a \\in G$ and $a \\neq e$ the period of $a$ must be $p$.

There is \"essentially\" one cyclic group of order $n$. Phrased differently, two cyclic groups of the same order have the \"same structure.\" The notion of \"same structure\" will be examined later in more detail.

EXAMPLES. Let us determine all possible structures of groups of order 4. The period of any element must be 1,2 , or 4 . If there is an element $a$ of period 4 then $e, a, a^{2}, a^{3}$ exhaust the group. On the other hand, if there is no element of period 4 , then all elements but $e$ must have the period 2 . Thus if $e, a, b, c$ denote the different elements of the group we have $a^{2} = b^{2} = c^{2} = e$. Consider the element $x = a b$. From $ax = aab = b$ we have clearly $x \\neq e, x \\neq a$. From the uniqueness of the solution $y = e$ of the equation $yb = b$ it follows that $x \\neq b$. Therefore $x$ must be $c$. The commutative law holds in this group, for if $x \\in G$ then $x = x^{-1}$ and consequently $ab = (ab)^{-1} = b^{-1}a^{-1} = ba$. It is a simple matter to write out the multiplication table:

	$e$	$a$	$b$	$c$
$e$	$e$	$a$	$b$	$c$
$a$	$a$	$e$	$c$	$b$
$b$	$b$	$c$	$e$	$a$
$c$	$c$	$b$	$a$	$e$

We have shown that there are essentially two groups of order 4 and both are commutative.

Groups of order 6 are essentially of two kinds, the cyclic group and the noncommutative group given in Exercise 1. This last is the simplest example of a noncommutative group. One of the unsolved problems of algebra is that of classifying all the groups of order $n$. There is, of course, always the cyclic group of order $n$ and for $n$ prime, only the cyclic group. For nonprimes there is no general theory although a classification has been achieved for special cases. The table below gives a summary for the first few cases:

$N$	4	6	8	9	10	12	14	15
$\\mu$	2	2	5	2	2	5	2	1
$\\nu$	0	1	2	0	1	3	1	0

where $\\mu$ is the total number and $\\nu$ the number of noncommutative groups of order $N$.

EXERCISE 9. The two noncommutative groups of order 8 are essentially:

The symmetries of the square, i.e., the rotations in space which take the square into itself.
The group formed by the quaternion units $\\pm 1, \\pm i, \\pm j, \\pm k$.

Construct the multiplication table for those two groups and show that they do not have the same structure.
(a) The symmetries of the square.
$\"\"$

If a rotation replaces the vertices (1234) by the vertices $\\left(a_{1} a_{2} a_{3} a_{4}\\right)$, then denote the rotation simply by $\\left(a_{1} a_{2} a_{3} a_{4}\\right)$. The identity is clearly $e = (1234)$. Denote by $a = (2341)$ the counterclockwise rotation through $90^{\\circ}$. Let $a^{2} = b = (3412)$ and $c = a^{2} = (4123)$. We have $a^{4} = e$. The powers of $a$ form a group $S$ of order 4. If $s$ denotes a rotation of $180^{\\circ}$ about the axis $1-3$ we have $s = (1432)$. The coset $sS$ is simply

$
\\begin{aligned}
& s=(1432), \\quad s a=(2143)=t, \\
& s a^{2}=(3214)=u, \\quad s a^{3}=(4321)=v ;
\\end{aligned}
$

these together with the powers of $a$ exhaust the symmetries of the square:

	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$e$	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$a$	$a$	$b$	$c$	$e$	$v$	$s$	$t$	$u$
$b$	$b$	$c$	$e$	$a$	$u$	$v$	$s$	$t$
$c$	$c$	$e$	$a$	$b$	$t$	$u$	$v$	$s$
$s$	$s$	$t$	$u$	$v$	$e$	$a$	$b$	$c$
$t$	$t$	$u$	$v$	$s$	$c$	$e$	$a$	$b$
$u$	$u$	$v$	$s$	$t$	$b$	$c$	$e$	$a$
$v$	$v$	$s$	$t$	$u$	$a$	$b$	$c$	$e$

(b) The quaternion group.

This is obtained at once by the ordinary rules of multiplication of the quaternion units

$
\\begin{array}{c|cccccccc}
& +1 & +i & +j & +k & -1 & -i & -j & -k \\
\\hline+1 & +1 & +i & +j & +k & -1 & -i & -j & -k \\
+i & +i & -1 & +k & -j & -i & +1 & -k & +j \\
+j & +j & -k & -1 & +i & -j & +k & +1 & -i \\
+k & +k & +j & -i & -1 & -k & -j & +i & +1 \\
-1 & -1 & -i & -j & -k & +1 & +i & +j & +k \\
-i & -i & +1 & -k & +j & +i & -1 & +k & -j \\
-j & -j & +k & +1 & -i & +j & -k & -1 & +i \\
-k & -k & -j & +i & +1 & +k & +j & -i & -1
\\end{array}
$

The two groups do not have the same structure since the group of symmetries has 5 elements of period 2 while the quaternion group has only one such element.

","chapter_express_content":"","public_visibility":false,"book_slug":"introduction-to-crystallography","slug":"test","regular_authors":"","express_authors":"","created_at":"2024-06-17T11:52:16.534Z","__v":0,"allow_part_see":false,"header_data":[]},{"_id":"66cf6974d17357fa85fdc0f4","book_id":"6647a03ea3fd815124d5695b","chapter_name":"test-3","sequence":3,"chapter_content":"

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","chapter_express_content":"","public_visibility":false,"book_slug":"introduction-to-crystallography","slug":"test-3","regular_authors":"","express_authors":"","created_at":"2024-08-28T16:44:40.509Z","__v":0,"header_data":[]},{"_id":"66cf6c7fd17357fa85fdc12d","book_id":"6647a03ea3fd815124d5695b","chapter_name":"test 4","sequence":4,"chapter_content":"\n Calculus Made Easy - Chapter 14: Exponential, Logarithmic, Hyperbolic Functions and their Derivatives\n\n\n\n\n

On True Compound Interest

Let there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some fixed rate; for the bigger the capital, the bigger the amount of interest on it in a given time.

Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call “simple interest,” or by what they call “compound interest.” For in the former case the capital remains fixed, while in the latter the interest is added to the capital, which therefore increases by successive additions.

(1) At simple interest. Consider a concrete case. Let the capital at start be $\$100\$, and let the rate of interest be \$10\$ percent per annum. Then the increment to the owner of the capital will be $\$10\$ every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for \$10\$ years, by the end of that time he will have received \$10\$ increments of $\$10\$ each, or $\$100\$, making, with the original $\$100\$, a total of $\$200\$ in all. His property will have doubled itself in \$10\$ years. If the rate of interest had been \$5\$ percent, he would have had to hoard for \$20\$ years to double his property. If it had been only \$2\$ percent, he would have had to hoard for \$50\$ years. It is easy to see that if the value of the yearly interest is \$\\dfrac{1}{n}\$ of the capital, he must go on hoarding for \$n\$ years in order to double his property.

Or, if \$y\$ be the original capital, and the yearly interest is \$\\dfrac{y}{n}\$, then, at the end of \$n\$ years, his property will be \\[y + n\\dfrac{y}{n} = 2y.\\]

(2) At compound interest. As before, let the owner begin with a capital of $\$100\$, earning interest at the rate of \$10\$ percent per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to $\$110\$; and in the second year (still at \$10\$%) this will earn $\$11\$ interest. He will start the third year with $\$121\$, and the interest on that will be $\$12.1\$; so that he starts the fourth year with $\$133.1\$, and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to $\$259.374\$. In fact, we see that at the end of each year, each dollar will have earned \$\\frac{1}{10}\$ of a dollar, and therefore, if this is always added on, each year multiplies the capital by \$\\frac{11}{10}\$; and if continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by \${2.59374}\$. Let us put this into symbols. Put \$y_0\$ for the original capital; \$\\dfrac{1}{n}\$ for the fraction added on at each of the \$n\$ operations; and \$y_n\$ for the value of the capital at the end of the \$n\$ operation. Then \\[y_n = y_0\\left(1 + \\frac{1}{n}\\right)^n.\\]

But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the $\$100\$ ought to have been growing. At the end of half a year it ought to have been at least $\$105\$, and it certainly would have been fairer had the interest for the second half of the year been calculated on $\$105\$. This would be equivalent to calling it \$5\$% per half-year; with \$20\$ operations, therefore, at each of which the capital is multiplied by \$\\frac{21}{20}\$. If reckoned this way, by the end of ten years the capital would have grown to $\$265.33\$.; for \\[\\left(1 + \\frac{1}{20}\\right)^{20} ={2.6533}.\\]

But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into \$10\$ parts, and reckon a one percent interest for each tenth of the year. We now have \$100\$ operations lasting over the ten years; or \\[y_n = \\$100 \\left( 1 + \\frac{1}{100} \\right)^{100};\\] which works out to $\$270.481\$.

Even this is not final. Let the ten years be divided into \$1000\$ periods, each of \$\\frac{1}{100}\$ of a year; the interest being \$\\frac{1}{10}\$ percent for each such period; then \\[y_n = \\$ 100 \\left( 1 + \\frac{1}{1000} \\right)^{1000};\\] which works out to $\$271.692\$.

Go even more minutely, and divide the ten years into \$10,000\$ parts, each \$\\frac{1}{1000}\$ of a year, with interest at \$\\frac{1}{100}\$ of \$1\$ percent. Then \\[y_n = \\$100 \\left( 1 + \\frac{1}{10,000} \\right)^{10,000};\\] which amounts to $\$271.815\$.

Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression \$\\left(1 + \\dfrac{1}{n}\\right)^n\$, which, as we see, is greater than \$2\$; and which, as we take \$n\$ larger and larger, grows closer and closer to a particular limiting value. However big you make \$n\$, the value of this expression grows nearer and nearer to the figure \\[2.71828\\ldots\\] a number never to be forgotten.

Let us take geometrical illustrations of these things. In the following figure, \$OP\$ stands for the original value. \$OT\$ is the whole time during which the value is growing. It is divided into \$10\$ periods, in each of which there is an equal step up. Here \$\\dfrac{dy}{dx}\$ is a constant; and if each step up is \$\\frac{1}{10}\$ of the original \$OP\$, then, by \$10\$ such steps, the height is doubled. If we had taken \$20\$ steps, each of half the height shown, at the end the height would still be just doubled. Or \$n\$ such steps, each of \$\\dfrac{1}{n}\$ of the original height \$OP\$, would suffice to double the height. This is the case of simple interest. Here is \$1\$ growing till it becomes \$2\$.

In the next figure, we have the corresponding illustration of the geometrical progression. Each of the successive ordinates is to be \$1 + \\dfrac{1}{n}\$, that is, \$\\dfrac{n+1}{n}\$ times as high as its predecessor. The steps up are not equal, because each step up is now \$\\dfrac{1}{n}\$ of the ordinate at that part of the curve. If we had literally \$10\$ steps, with \$\\left(1 + \\frac{1}{10} \\right)\$ for the multiplying factor, the final total would be \$\\left(1 + \\frac{1}{10}\\right)^{10}\$ or \${2.594}\$ times the original \$1\$. But if only we take \$n\$ sufficiently large (and the corresponding \$\\dfrac{1}{n}\$ sufficiently small), then the final value \$\\left(1 + \\dfrac{1}{n}\\right)^n\$ to which unity will grow will be \$2.71828\$.

The Number \$\\boldsymbol{e}\$

To this mysterious number \$2.7182818\\dots\$, the mathematicians have assigned the letter \$e\$. This number is often called Euler’s number after the Swiss mathematician Leonhard Euler. All fifth graders know that the Greek letter \$\\pi\$ (called pi) stands for \$3.141592\\dots\$; but how many of them know that \$e\$ means \$2.71828\\dots\$? Yet it is an even more important number than \$\\pi\$!

What, then, is \$e\$?

Suppose we were to let \$1\$ grow at simple interest till it became \$2\$; then, if at the same nominal rate of interest, and for the same time, we were to let \$1\$ grow at true compound interest, instead of simple, it would grow to the value of the number \$e\$.

This process of growing proportionately, at every instant, to the magnitude at that instant, some people call an exponential rate of growing. Unit exponential rate of growth is that rate which in unit time will cause \$1\$ to grow to \$2.718281\$. It might also be called the organic rate of growing because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.

If we take \$100\$ percent as the unit of rate, and any fixed period as the unit of time, then the result of letting \$1\$ grow arithmetically at unit rate, for unit time, will be \$2\$, while the result of letting \$1\$ grow exponentially at unit rate, for the same time, will be \$2.71828\\ldots\$ .

A little more about the number \$\\boldsymbol{e}\$

We have seen that we require to know what value is reached by the expression \$\\left(1 + \\dfrac{1}{n}\\right)^n\$, when \$n\$ becomes indefinitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of a calculator) got by assuming \$n = 2\$; \$n = 5\$; \$n = 10\$; and so on, up to \$n = 10,000\$. \\[\\begin{align} &\\left(1 + \\frac{1}{2}\\right)^2 &=& 2.25. \\\\ &\\left(1 + \\frac{1}{5}\\right)^5 &=& 2.488. \\\\ &\\left(1 + \\frac{1}{10}\\right)^{10} &=& 2.594. \\\\ &\\left(1 + \\frac{1}{20}\\right)^{20} &=& 2.653. \\\\ &\\left(1 + \\frac{1}{100}\\right)^{100} &=& {2.705}. \\\\ &\\left(1 + \\frac{1}{1000}\\right)^{1000} &=& {2.7169}. \\\\ &\\left(1 + \\frac{1}{10,000}\\right)^{10,000} &=& {2.7181}. \\end{align}\\]

It is, however, worth while to find another way of calculating this immensely important figure.

Accordingly, we will avail ourselves of the binomial theorem, and expand the expression \$\\left(1 + \\dfrac{1}{n}\\right)^n\$ in that well-known way.

The binomial theorem gives the rule that \\[\\begin{align} (a + b)^n &= a^n + n \\dfrac{a^{n-1} b}{1!} + n(n - 1) \\dfrac{a^{n-2} b^2}{2!} + n(n -1)(n - 2) \\dfrac{a^{n-3} b^3}{3!} + \\cdots. \\end{align}\\] Putting \$a = 1\$ and \$b = \\dfrac{1}{n}\$, we get \\[\\begin{align} \\left(1 + \\dfrac{1}{n}\\right)^n &= 1 + 1 + \\dfrac{1}{2!} \\left(\\dfrac{n - 1}{n}\\right) + \\dfrac{1}{3!} \\dfrac{(n - 1)(n - 2)}{n^2} + \\dfrac{1}{4!} \\dfrac{(n - 1)(n - 2)(n - 3)}{n^3} + \\cdots. \\end{align}\\]

Now, if we suppose \$n\$ to become indefinitely great, say a billion, or a billion billions, then \$n - 1\$, \$n - 2\$, and \$n - 3\$, etc., will all be sensibly equal to \$n\$; and then the series becomes \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\displaystyle e = 1 + 1 + \\dfrac{1}{2!} + \\dfrac{1}{3!} + \\dfrac{1}{4!} + \\cdots.}\\]

By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms:

	\$1.000000\$
dividing by \$1\$	\$1.000000\$
dividing by \$2\$	\$0.500000\$
dividing by \$3\$	\$0.166667\$
dividing by \$4\$	\$0.041667\$
dividing by \$5\$	\$0.008333\$
dividing by \$6\$	\$0.001389\$
dividing by \$7\$	\$0.000198\$
dividing by \$8\$	\$0.000025\$
dividing by \$9\$	\$0.000002\$
Total	\$2.718281\$

\$e\$ is incommensurable with \$1\$, and resembles \$\\pi\$ in being an interminable non-recurrent decimal.

The Exponential Series

We shall have need of yet another series.

Let us, again making use of the binomial theorem, expand the expression \$\\left(1 + \\dfrac{1}{n}\\right)^{nx}\$, which is the same as \$e^x\$ when we make \$n\$ indefinitely great. \\[\\begin{align} e^x &= 1^{nx} + nx \\frac{1^{nx-1} \\left(\\dfrac{1}{n}\\right)}{1!} + nx(nx - 1) \\frac{1^{nx - 2} \\left(\\dfrac{1}{n}\\right)^2}{2!} + nx(nx - 1)(nx - 2) \\frac{1^{nx-3} \\left(\\dfrac{1}{n}\\right)^3}{3!} + \\cdots\\\\ &= 1 + x + \\frac{1}{2!} \\cdot \\frac{n^2x^2 - nx}{n^2} + \\frac{1}{3!} \\cdot \\frac{n^3x^3 - 3n^2x^2 + 2nx}{n^3} + \\cdots \\\\ &= 1 + x + \\frac{x^2 -\\dfrac{x}{n}}{2!} + \\frac{x^3 - \\dfrac{3x^2}{n} + \\dfrac{2x}{n^2}}{3!} + \\cdots. \\end{align}\\]

But, when \$n\$ is made indefinitely great, this simplifies down to the following: \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\displaystyle e^x = 1 + x + \\frac{x^2}{2!} + \\frac{x^3}{3!} + \\frac{x^4}{4!} + \\cdots.}\\]

This series is called the exponential series.

Derivative of the Exponential Function

The great reason why \$e\$ is regarded of importance is that \$e^x\$ possesses a property, not possessed by any other function of \$x\$, that when you differentiate it its value remains unchanged; or, in other words, its derivative is the same as itself. This can be instantly seen by differentiating it with respect to \$x\$, thus: \\[\\begin{align} \\frac{d(e^x)}{dx} &= 0 + 1 + \\frac{2x}{1 \\cdot 2} + \\frac{3x^2}{1 \\cdot 2 \\cdot 3} + \\frac{4x^3}{1 \\cdot 2 \\cdot 3 \\cdot 4} + \\frac{5x^4}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5} + \\cdots. \\end{align}\\] or \\[\\begin{align} \\frac{d(e^x)}{dx}= 1 + x + \\frac{x^2}{1 \\cdot 2} + \\frac{x^3}{1 \\cdot 2 \\cdot 3} + \\frac{x^4}{1 \\cdot 2 \\cdot 3 \\cdot 4} + \\cdots, \\end{align}\\] which is exactly the same as the original series.

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\dfrac{d(e^x)}{dx}=e^x}\\]

Now we might have gone to work the other way, and said: Go to; let us find a function of \$x\$, such that its derivative is the same as itself. Or, is there any expression, involving only powers of \$x\$, which is unchanged by differentiation? Accordingly; let us assume as a general expression that \\[y = A + Bx + Cx^2 + Dx^3 + Ex^4 + \\cdots,\\] (in which the coefficients \$A\$, \$B\$, \$C\$, etc. will have to be determined), and differentiate it. \\[\\dfrac{dy}{dx} = B + 2Cx + 3Dx^2 + 4Ex^3 + \\cdots.\\]

Now, if this new expression is really to be the same as that from which it was derived, it is clear that \$A\$ must \$=B\$; that \$C=\\dfrac{B}{2}=\\dfrac{A}{1\\cdot 2}\$; that \$D = \\dfrac{C}{3} = \\dfrac{A}{1 \\cdot 2 \\cdot 3}\$; that \$E = \\dfrac{D}{4} = \\dfrac{A}{1 \\cdot 2 \\cdot 3 \\cdot 4}\$, etc.

The law of change is therefore that

\\[y = A\\left(1 + \\dfrac{x}{1} + \\dfrac{x^2}{1 \\cdot 2} + \\dfrac{x^3}{1 \\cdot 2 \\cdot 3} + \\dfrac{x^4}{1 \\cdot 2 \\cdot 3 \\cdot 4} + \\cdots.\\right).\\]

If, now, we take \$A = 1\$ for the sake of further simplicity, we have \\[y = 1 + \\dfrac{x}{1} + \\dfrac{x^2}{1 \\cdot 2} + \\dfrac{x^3}{1 \\cdot 2 \\cdot 3} + \\dfrac{x^4}{1 \\cdot 2 \\cdot 3 \\cdot 4} + \\cdots.\\]

Differentiating it any number of times will give always the same series over again.

If, now, we take the particular case of \$A=1\$, and evaluate the series, we shall get simply \\[\\begin{align} \\text{when } x &= 1,\\quad & y &= 2.718281\\dots; & \\text{that is, } y &= e; \\\\ \\text{when } x &= 2,\\quad & y &=(2.718281\\dots)^2; & \\text{that is, } y &= e^2; \\\\ \\text{when } x &= 3,\\quad & y &=(2.718281\\dots)^3; & \\text{that is, } y &= e^3; \\end{align}\\] and therefore \\[\\text{when } x=x,\\quad y=(2.718281\\dots.)^x;\\quad\\text{that is, } y=e^x,\\] thus finally demonstrating that \\[e^x = 1 + \\dfrac{x}{1} + \\dfrac{x^2}{1\\cdot2} + \\dfrac{x^3}{1\\cdot 2\\cdot 3} + \\dfrac{x^4}{1\\cdot 2\\cdot 3\\cdot 4} + \\cdots.\\]

Of course it follows that \$e^y\$ remains unchanged if differentiated with respect to \$y\$. Also \$e^{ax}\$, which is equal to \$(e^a)^x\$, will, when differentiated with respect to \$x\$, be \$ae^{ax}\$, because \$a\$ is a constant.

Natural or Naperian Logarithms

Another reason why \$e\$ is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If \$y\$ is the value of \$e^x\$, then \$x\$ is the logarithm, to the base \$e\$, of \$y\$. Or, if \\[y = e^x,\\] then \\[x = \\log_e y.\\]

The logarithm with base \$e\$ is called the natural logarithm. The natural logarithm is so important that it has its own shorthand: \\[\\log_e y \\qquad\\text{is often written as}\\qquad \\ln y.\\]

The two curves plotted in Figs. 14.3 and 14.4 represent these equations.

The points calculated are:

\$x\$	\$-2\$	\$-1\$	\$-0.5\$	\$0\$	\$0.5\$	\$1\$	\$1.5\$	\$2\$
\$y=e^x\$	\$0.14\$	\$0.37\$	\$0.61\$	\$1\$	\$1.65\$	\$2.71\$	\$4.50\$	\$7.39\$

For Fig. 14.3

\$y\$	\$0.1\$	\$0.5\$	\$1\$	\$2\$	\$3\$	\$4\$	\$8\$
\$x=\\ln y\$	\$-2.30\$	\$-0.69\$	\$0\$	\$0.69\$	\$1.10\$	\$1.39\$	\$2.08\$

For Fig. 14.4

It will be seen that, though the calculations yield different points for plotting, yet the result is identical. The two equations really mean the same thing.

As many persons who use ordinary logarithms, which are calculated to base \$10\$ instead of base \$e\$, are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or \\[\\ln a + \\ln b = \\ln (ab).\\] Also the rule of powers holds good; \\[n \\times \\ln a = \\ln a^n.\\] But as \$10\$ is no longer the basis, one cannot multiply by \$100\$ or \$1000\$ by merely adding \$2\$ or \$3\$ to the index. One can change the natural logarithm to the ordinary logarithm¹ simply by multiplying it by \$\\frac{1}{\\ln 10}\\approx 0.4343\$; or \\[\\begin{align} \\log_{10} x = \\frac{1}{\\ln 10}\\ln x\\approx0.4343 \\times \\ln x, \\end{align}\\] and conversely, \\[\\begin{align} \\ln x = \\frac{1}{\\log_{10} e}\\times \\log_{10} x=\\ln 10\\times \\log_{10}x\\approx2.3026 \\times \\log_{10} x. \\end{align}\\]

Using A Calculator to Find \$\\boldsymbol{e^x}\$ and \$\\boldsymbol{\\ln x}\$

Modern scientific and graphing calculators are equipped with buttons for exponentiation with the base \$e\$ and buttons for calculating natural or common logarithms. The exponential function \$e^x\$ is sometimes denoted by \$\\exp(x)\$. Therefore, when using a calculator, we may need to locate the \$\\boxed{e^x}\$ or \$\\boxed{\\text{exp}(x)}\$ button. Many calculators provide two distinct buttons for calculating the natural logarithm (\$\\ln x\$) and the common logarithm (\$\\log_{10}x\$). If your calculator has an \$\\boxed{\\text{ln}}\$ button for the natural logarithm, the \$\\boxed{\\text{log}}\$ button is likely designed to return \$\\log_{10}x\$.

Derivatives of Logarithmic and Exponential Functions

Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.

Take the equation: \\[y = \\ln x.\\] First transform this into \\[e^y = x,\\] whence, since the derivative of \$e^y\$ with regard to \$y\$ is the original function unchanged (see here), \\[\\frac{dx}{dy} = e^y,\\] and, reverting from the inverse to the original function, \\[\\frac{dy}{dx} = \\frac{1}{\\ \\dfrac{dx}{dy}\\ } = \\frac{1}{e^y} = \\frac{1}{x}.\\]

Now this is a very curious result. It may be written \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\dfrac{d(\\ln x)}{dx} = x^{-1}.}\\]

Note that \$x^{-1}\$ is a result that we could never have got by the Power Rule for differentiating powers. That rule is to multiply by the power, and reduce the power by \$1\$. Thus, differentiating \$x^3\$ gave us \$3x^2\$; and differentiating \$x^2\$ gave \$2x^1\$. But differentiating \$x^0\$ does not give us \$x^{-1}\$ or \$0 \\times x^{-1}\$, because \$x^0\$ is itself \$= 1\$, and is a constant. We shall have to come back to this curious fact that differentiating \$\\ln x\$ gives us \$\\dfrac{1}{x}\$ when we reach the chapter on integrating.

Now, try to differentiate \\[\\begin{align} y = \\ln(x+a), \\end{align}\\] that is, \\[\\begin{align} e^y = x+a; \\end{align}\\] we have \$\\dfrac{d(x+a)}{dy} = e^y\$, since the differential of \$e^y\$ remains \$e^y\$.

This gives \\[\\frac{dx}{dy} = e^y = x+a;\\] hence, reverting to the original function (see the Derivative of an Inverse Function), we get² \\[\\frac{dy}{dx} = \\frac{1}{\\;\\dfrac{dx}{dy}\\;} = \\frac{1}{x+a}.\\]

Next try \\[y = \\log_{10} x.\\]

First change to natural logarithms by multiplying by the modulus \$\\log_{10}e=\\dfrac{1}{\\ln 10}\\approx 0.4343\$. This gives us \\[\\begin{align} y = \\frac{1}{\\ln 10} \\ln x; \\end{align}\\] whence \\[\\begin{align} \\frac{dy}{dx} = \\frac{1}{x\\ln 10}. \\end{align}\\]

In general, because \\[\\log_a x=\\frac{\\log_e x}{\\log_e a}=\\frac{\\ln x}{\\ln a}\\] and \$\\dfrac{1}{\\ln a}\$ is a constant, we have \\[\\frac{d\\left(\\log_a x\\right)}{dx}=\\frac{d\\left(\\dfrac{1}{\\ln a}\\cdot \\ln x \\right)}{dx}=\\frac{1}{\\ln a}\\cdot\\frac{d(\\ln x)}{dx}=\\frac{1}{\\ln a\\cdot x}\\]

The next thing is not quite so simple. Try this: \\[y = a^x.\\]

Taking the logarithm of both sides, we get \\[\\begin{align} \\ln y &= x \\ln a, \\end{align}\\] or \\[\\begin{align} x &= \\frac{\\ln y}{\\ln a}\\\\ &= \\frac{1}{\\ln a} \\times \\ln y. \\end{align}\\]

Since \$\\dfrac{1}{\\ln a}\$ is a constant, we get \\[\\frac{dx}{dy} = \\frac{1}{\\ln a} \\times \\frac{1}{y} = \\frac{1}{a^x \\times \\ln a};\\] hence, reverting to the original function. \\[\\frac{dy}{dx} = \\frac{1}{\\;\\dfrac{dx}{dy}\\;} = a^x \\times \\ln a.\\]

We see that, since \\[\\frac{dx}{dy} \\times \\frac{dy}{dx} =1\\quad\\text{and}\\quad \\frac{dx}{dy} = \\frac{1}{y} \\times \\frac{1}{\\ln a},\\quad \\frac{1}{y} \\times \\frac{dy}{dx} = \\ln a.\\]

We shall find that whenever we have an expression such as \$\\ln y =\$ a function of \$x\$, we always have \$\\dfrac{1}{y}\\, \\dfrac{dy}{dx} =\$ the derivative of the function of \$x\$, so that we could have written at once, from \$\\ln y = x \\ln a\$, \\[\\frac{1}{y}\\, \\frac{dy}{dx} = \\ln a\\quad\\text{and}\\quad \\frac{dy}{dx} = a^x \\ln a.\\]

In summary

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ y=\\log_a x \\qquad\\Rightarrow \\qquad \\dfrac{dy}{dx}=\\dfrac{1}{x\\cdot \\ln a} }\\]

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ y=a^x \\qquad\\Rightarrow \\qquad \\dfrac{dy}{dx}=a^x\\cdot\\ln a }\\]

Let us now attempt further examples.

Examples

Example 14.1. Differentiate \$y\$ with respect to \$x\$ if \$y=e^{-ax}\$.

Solution. Let \$-ax=z\$; then \$y=e^z\$. \\[\\frac{dy}{dx} = e^z;\\quad \\frac{dz}{dx} = -a;\\quad\\text{hence}\\quad \\frac{dy}{dx} = -ae^{-ax}.\\]

Or thus: \\[\\ln y = -ax;\\quad \\frac{1}{y}\\, \\frac{dy}{dx} = -a;\\quad \\frac{dy}{dx} = -ay = -ae^{-ax}.\\]

Example 14.2. Differentiate \$y\$ with respect to \$x\$ if \$y=e^{\\frac{x^2}{3}}\$.

Solution. Let \$\\dfrac{x^2}{3}=z\$; then \$y=e^z\$. \\[\\frac{dy}{dz} = e^z;\\quad \\frac{dz}{dx} = \\frac{2x}{3};\\quad \\frac{dy}{dx} = \\frac{2x}{3}\\, e^{\\frac{x^2}{3}}.\\]

Or thus: \\[\\ln y = \\frac{x^2}{3};\\quad \\frac{1}{y}\\, \\frac{dy}{dx} = \\frac{2x}{3};\\quad \\frac{dy}{dx} = \\frac{2x}{3}\\, e^{\\frac{x^2}{3}}.\\]

Example 14.3. Given \$y = e^{\\frac{2x}{x+1}}\$, find \$\\dfrac{dy}{dx}\$.

Solution. \\[\\begin{align} \\ln y &= \\frac{2x}{x+1},\\quad \\frac{1}{y}\\, \\frac{dy}{dx} = \\frac{2(x+1)-2x}{(x+1)^2}; \\end{align}\\] hence \\[\\begin{align} \\frac{dy}{dx} &= \\frac{2}{(x+1)^2} e^{\\frac{2x}{x+1}}. \\end{align}\\]

Check by writing \$\\dfrac{2x}{x+1}=z\$.

\$y=e^{\\sqrt{x^2+a}}\$.\$\\ln y=(x^2+a)^{\\frac{1}{2}}\$. \\[\\frac{1}{y}\\, \\frac{dy}{dx} = \\frac{x}{(x^2+a)^{\\frac{1}{2}}}\\quad\\text{and}\\quad \\frac{dy}{dx} = \\frac{x \\times e^{\\sqrt{x^2+a}}}{(x^2+a)^{\\frac{1}{2}}}.\\] For if \$(x^2+a)^{\\frac{1}{2}}=u\$ and \$x^2+a=v\$, \$u=v^{\\frac{1}{2}}\$, \\[\\frac{du}{dv} = \\frac{1}{{2v}^{\\frac{1}{2}}};\\quad \\frac{dv}{dx} = 2x;\\quad \\frac{du}{dx} = \\frac{x}{{(x^2+a)}^{\\frac{1}{2}}}.\\]

Check by writing \$\\sqrt{x^2+a}=z\$.

Example 14.4. If \$y=\\log(a+x^3)\$, find \$\\dfrac{dy}{dx}\$.

Solution. Let \$(a+x^3)=z\$; then \$y=\\ln z\$. \\[\\frac{dy}{dz} = \\frac{1}{z};\\quad \\frac{dz}{dx} = 3x^2;\\quad\\text{hence}\\quad \\frac{dy}{dx} = \\frac{3x^2}{a+x^3}.\\]

Example 14.5. If \$y=\\ln\\left\\{{3x^2+\\sqrt{a+x^2}}\\right\\}\$, find \$\\dfrac{dy}{dx}\$.

Solution. Let \$3x^2 + \\sqrt{a+x^2}=z\$; then \$y=\\ln z\$. \\[\\begin{align} \\frac{dy}{dz} &= \\frac{1}{z};\\quad \\frac{dz}{dx} = 6x + \\frac{x}{\\sqrt{x^2+a}}; \\\\ \\frac{dy}{dx} &= \\frac{6x + \\dfrac{x}{\\sqrt{x^2+a}}}{3x^2 + \\sqrt{a+x^2}} = \\frac{x(1 + 6\\sqrt{x^2+a})}{(3x^2 + \\sqrt{x^2+a}) \\sqrt{x^2+a}}. \\end{align}\\]

Example 14.6. If \$y=(x+3)^2 \\sqrt{x-2}\$, find \$\\dfrac{dy}{dx}\$.

Solution. Taking the logarithm of both sides, we get \\[\\begin{align} \\ln y &=\\ln\\left[(x+3)^2\\sqrt{x-2}\\right]\\\\ &=\\ln\\left[(x+3)^2\\right]+\\ln\\left[(x-2)^{\\frac{1}{2}}\\right]\\\\ &= 2 \\ln(x+3)+ \\frac{1}{2} \\ln(x-2). \\end{align}\\] Differentiating both sides, we get \\[\\begin{align} \\frac{1}{y}\\, \\frac{dy}{dx} &= \\frac{2}{(x+3)} + \\frac{1}{2(x-2)}; \\\\ \\frac{dy}{dx} &= (x+3)^2 \\sqrt{x-2} \\left\\{\\frac{2}{x+3} + \\frac{1}{2(x-2)}\\right\\}. \\end{align}\\]

Example 14.7. If \$y=(x^2+3)^3(x^3-2)^{\\frac{2}{3}}\$, find \$\\dfrac{dy}{dx}\$.

Solution. Taking the logarithm of both sides, we get \\[\\begin{align} \\ln y &= 3 \\ln(x^2+3) + \\frac{2}{3} \\ln(x^3-2); \\\\ \\frac{1}{y}\\, \\frac{dy}{dx} &= 3 \\frac{2x}{(x^2+3)} + \\frac{2}{3} \\frac{3x^2}{x^3-2} = \\frac{6x}{x^2+3} + \\frac{2x^2}{x^3-2}. \\end{align}\\] For if \$u=\\ln(x^2+3)\$, let \$x^2+3=z\$ and \$u=\\ln z\$. \\[\\frac{du}{dz} = \\frac{1}{z};\\quad \\frac{dz}{dx} = 2x;\\quad \\frac{du}{dx} = \\frac{2x}{x^2+3}.\\] Similarly, if \$v=\\ln(x^3-2)\$, \$\\dfrac{dv}{dx} = \\dfrac{3x^2}{x^3-2}\$ and \\[\\frac{dy}{dx} = (x^2+3)^3(x^3-2)^{\\frac{2}{3}} \\left\\{ \\frac{6x}{x^2+3} + \\frac{2x^2}{x^3-2} \\right\\}.\\]

Example 14.8. If \$y=\\dfrac{\\sqrt[2]{x^2+a}}{\\sqrt[3]{x^3-a}}\$, find \$\\dfrac{dy}{dx}\$.

Solution. \\[\\begin{align} \\ln y &= \\ln\\left[\\frac{\\sqrt[2]{x^2+a}}{\\sqrt[3]{x^3-a}}\\right]\\\\ &=\\ln \\left[(x^2+a)^{\\frac{1}{2}}(x^3-a)^{-\\frac{1}{3}}\\right]\\\\ &=\\ln \\left[(x^2+a)^{\\frac{1}{2}}\\right]+\\ln \\left[(x^3-a)^{-\\frac{1}{3}}\\right]\\\\ &= \\frac{1}{2} \\ln(x^2+a) - \\frac{1}{3} \\ln(x^3-a). \\end{align}\\] Differentiating \\[\\begin{align} \\frac{1}{y}\\, \\frac{dy}{dx} &= \\frac{1}{2}\\, \\frac{2x}{x^2+a} - \\frac{1}{3}\\, \\frac{3x^2}{x^3-a} = \\frac{x}{x^2+a} - \\frac{x^2}{x^3-a} \\end{align}\\] and \\[\\begin{align} \\frac{dy}{dx} &= \\frac{\\sqrt[2]{x^2+a}}{\\sqrt[3]{x^3-a}} \\left\\{ \\frac{x}{x^2+a} - \\frac{x^2}{x^3-a} \\right\\}. \\end{align}\\]

Example 14.9. If \$y=\\dfrac{1}{\\ln x}\$, find \$\\dfrac{dy}{dx}\$.

Solution. \\[\\frac{dy}{dx} = \\frac{\\ln x \\times 0 - 1 \\times \\dfrac{1}{x}} {\\ln^2 x} = -\\frac{1}{x \\ln^2x}.\\tag{Quotient Rule}\\]

Example 14.10. If \$y=\\sqrt[3]{\\ln x} = (\\ln x)^{\\frac{1}{3}}\$, find \$\\dfrac{dy}{dx}\$.

Solution. Let \$z=\\ln x\$; \$y=z^{\\frac{1}{3}}\$. \\[\\frac{dy}{dz} = \\frac{1}{3} z^{-\\frac{2}{3}};\\quad \\frac{dz}{dx} = \\frac{1}{x};\\quad \\frac{dy}{dx} = \\frac{1}{3x \\sqrt[3]{\\ln^2 x}}.\\]

Example 14.11. If \$y=\\left(\\dfrac{1}{a^x}\\right)^{ax}\$, find \$\\dfrac{dy}{dx}\$.

Solution. \\[\\begin{align} \\ln y = ax(\\ln 1 - \\ln a^x) = -ax \\ln a^x. \\end{align}\\] Differentiating \\[\\begin{align} \\frac{1}{y}\\, \\frac{dy}{dx} = -ax \\times a^x \\ln a - a \\ln a^x. \\end{align}\\] and \\[\\begin{align} \\frac{dy}{dx} = -\\left(\\frac{1}{a^x}\\right)^{ax} (x \\times a^{x+1} \\ln a + a \\ln a^x). \\end{align}\\]

Try now the following exercises.

Exercises I

Exercise 14.1. Differentiate \$y=b(e^{ax} -e^{-ax})\$.

Answer

\$ab(e ^{ax} + e ^{-ax})\$.

Solution

(1)

\\[y=b\\left(e^{a x}-e^{-a x}\\right)\\]

\\[\\begin{align} \\frac{d y}{d x}&=b\\left(a e^{a x}+a e^{-a x}\\right)\\\\ &=a b\\left(e^{a x}+e^{-a x}\\right) \\end{align}\\]

Exercise 14.2. Find the derivative of the expression \$u=at^2+2\\ln t\$ with respect to \$t\$.

Answer

\$2at + \\dfrac{2}{t}\$.

Solution

\\[\\frac{d u}{d t}=2 a t+\\frac{2}{t}\\]

Exercise 14.3. If \$y=n^t\$, find \$\\dfrac{d(\\ln y)}{dt}\$.

Answer

\$\\ln n\$.

Solution

\\[\\frac{d\\left(\\ln n^{t}\\right)}{d t}=\\frac{d(t \\ln n)}{d t}=\\ln n\\]

Exercise 14.4. Show that if \$y=\\dfrac{1}{b}\\cdot\\dfrac{a^{bx}}{\\ln a}\$,\$\\dfrac{dy}{dx}=a^{bx}\$.

Solution

\\[\\frac{d y}{d x}=\\frac{1}{b} \\cdot \\frac{b(\\ln a) a^{b x}}{\\ln a}=a^{b x}\\]

Exercise 14.5. If \$w=pv^n\$, find \$\\dfrac{dw}{dv}\$.

Answer

\$npv^{n-1}\$.

Solution

\\[\\frac{d w}{d v}=n p v^{n-1}\\]

Differentiate

Exercise 14.6. \$y=\\ln x^n\$.

Answer

\$\\dfrac{n}{x}\$.

Solution

\\[y=\\ln x^{n} \\Rightarrow y=n \\ln x\\]

\\[\\frac{d y}{d x}=\\frac{n}{x} \\\\ \\]

Exercise 14.7. \$y=3e^{-\\frac{x}{x-1}}\$.

Answer

\$\\dfrac{3e ^{- \\frac{x}{x-1}}}{(x - 1)^2}\$.

Solution

\\[y=3 e^{-\\frac{x}{x-1}}\\] Let \$u=-\\dfrac{x}{x-1}\$. Then \$y=3e^u\$, and \\[\\frac{dy}{du}=3 e^{u}\\] \\[\\begin{align} \\frac{d u}{d x}&=\\frac{-1(x-1)-1(-x)}{(x-1)^{2}}\\\\ &=\\frac{-x+1+x}{(x-1)^{2}}\\\\ &=\\frac{1}{(x-1)^{2}} \\end{align}\\] Using the Chain Rule:

\\[\\frac{d y}{d x}=\\frac{d y}{d x} \\cdot \\frac{d x}{d x}=\\frac{3}{(x-1)^{2}} e^{\\frac{-x}{x-1}}\\]

Exercise 14.8. \$y=(3x^2+1)e^{-5x}\$.

Answer

\$6x e ^{-5x} - 5(3x^2 + 1)e ^{-5x}\$.

Solution

\\[y=(3x^2-1)e^{-5x}\\] Using the Product Rule, we get \\[\\begin{align} \\frac{d y}{d x}&=\\frac{d(3x^2-1)}{dx} e^{-5x}+(3x^2-1)\\frac{d\\left(e^{-5x}\\right)}{dx}\\\\ &=6 x e^{-5 x}+\\left(3 x^{2}-1\\right)\\left(-5 e^{-5 x}\\right) \\\\ &=6x e ^{-5x} - 5(3x^2 + 1)e ^{-5x} \\end{align}\\]

Exercise 14.9. \$y=\\ln\\left(x^a+a\\right)\$.

Answer

\$\\dfrac{ax^{a-1}}{x^a + a}\$.

Solution

\\[y=\\ln \\left(x^{a}+a\\right)\\] Let \$u=x^a+a\$. Now using the Chain Rule \\[\\begin{align} \\frac{d y}{d x}&= \\frac{d(\\ln u)}{du}\\frac{du}{dx}\\\\ &=\\frac{1}{u}\\left(a x^{a-1}\\right)\\\\ &=\\frac{a x^{a-1}}{x^a+a} \\end{align}\\]

Exercise 14.10. \$y=(3x^2-1)(\\sqrt{x}+1)\$.

Answer

\$\\left(\\dfrac{6x}{3x^2-1} + \\dfrac{1}{2\\left(\\sqrt x + x\\right)}\\right) \\left(3x^2-1\\right)\\left(\\sqrt x + 1\\right)\$.

Solution

\\[y=\\left(3 x^{2}-1\\right) \\sqrt{x+1}\\] Using the Product Rule: \\[\\frac{d y}{d x}=6 x \\sqrt{x+1}+\\left(3 x^{2}-1\\right) \\cdot \\frac{1}{2 \\sqrt{x+1}}\\]

Exercise 14.11. \$y=\\dfrac{\\ln(x+3)}{x+3}\$.

Answer

\$\\dfrac{1 - \\ln \\left(x + 3\\right)}{\\left(x + 3\\right)^2}\$.

Solution

\\[y=\\frac{\\ln (x+3)}{x+3}\\] Using the Quotient Rule: \\[\\begin{align} \\frac{d y}{d x}&=\\frac{\\frac{1}{x+3}(x+3)-\\ln (x+3)}{(x+3)^{2}}\\\\ &=\\frac{1-\\ln (x+3)}{(x+3)^{2}} \\end{align}\\]

Exercise 14.12. \$y=a^x \\times x^a\$.

Answer

\$a^x\\left(ax^{a-1} + x^a \\ln a\\right)\$.

Solution

\\[y=a^{x} \\cdot x^{a}\\]

\\[\\begin{align} \\frac{d y}{d x}&=\\left(a^{x} \\ln a\\right) x^{a}+a x^{a-1} \\cdot a^{x}\\\\ &=a^x\\left(a x^{a-1}+x^a \\ln a\\right) \\end{align}\\]

Exercise 14.13. It was shown by Lord Kelvin that the speed of signaling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called \$y\$, then the number of signals \$s\$ that can be sent per minute can be expressed by the formula \\[s=ay^2 \\ln \\frac{1}{y};\\] where \$a\$ is a constant depending on the length and the quality of the materials. Show that if these are given, \$s\$ will be a maximum if \$y=1 / \\sqrt{e}\$.

Solution

\\[s=ay^2\\ln\\frac{1}{y}\\qquad (y\\neq 0)\\] To differentiate \$\\ln\\frac{1}{y}=\\ln\\left(y^{-1}\\right)\$, let \$u=y^{-1}\$. Then \\[\\begin{align} \\frac{d\\left(\\ln\\frac{1}{y}\\right)}{dy}&=\\frac{d(\\ln u)}{du}\\cdot\\frac{du}{dy}\\\\ &=\\frac{1}{u}\\cdot \\left(-y^{-2}\\right)\\\\ &=-y\\cdot\\frac{1}{y^2}\\\\ &=-\\frac{1}{y} \\end{align}\\] Now using the Product Rule, we get \\[\\begin{align} \\frac{d s}{d y}&=2 a y\\, \\ln \\frac{1}{y}+a y^2\\left(-\\frac{1}{y}\\right) \\\\ &= -2 a y\\, \\ln y-a y\\\\ &=a y\\,(-2 \\ln y-1) \\end{align}\\]

\\[\\begin{align} \\frac{d s}{d y}&=0 \\Rightarrow-2 \\ln y-1=0\\\\ & \\Rightarrow \\ln y=-\\frac{1}{2} \\Rightarrow y=e^{-1 / 2} \\end{align}\\]

\\[\\begin{align} \\frac{d^{2} s}{d y^{2}}&=-2 a \\ln y-2 a y\\left(\\frac{1}{y}\\right)-a\\\\ &=-2 a \\ln y-2 a-a \\end{align}\\]

When \$y=e^{-\\frac{1}{2}}\$, \\[\\frac{d^{2} s}{d y^{2}}=-2 a \\ln e^{-\\frac{1}{2}}-3 a=a \\ln e-3 a=-2 a<0\\] Therefore, it follows from the Second Derivative Test that \$s\$ is a minimum if \$y=\\dfrac{1}{\\sqrt{e}}\$.

Exercise 14.14. Find the maximum or minimum of \\[y=x^3-\\ln x.\\]

Answer

Min.: \$y = 0.7\$ for \$x = 0.694\$.

Solution

\\[y=x^{3}-\\ln x\\]

\\[\\frac{d y}{d x}=3 x^{2}-\\frac{1}{x}=\\frac{3 x^{3}-1}{x}\\]

\\[\\frac{d y}{d x}=0 \\Rightarrow 3 x^{3}-1=0\\]

\\[\\Rightarrow x^{3}=\\frac{1}{3} \\Rightarrow x=\\frac{1}{\\sqrt[3]{3}}\\]

To show that this specific value of \$x\$ makes \$y\$ a minimum, we apply the Second Derivative Test. First we differentiate \$\\dfrac{dy}{dx}\$ to find the second derivative \\[\\frac{d^2y}{dx^2}=9x^2.\\] Since when \$x=\\dfrac{1}{\\sqrt[3]{3}}\$ \\[\\frac{d^2y}{dx^2}=9\\left(\\frac{1}{\\sqrt[3]{3}}\\right)^2=\\frac{9}{\\sqrt[3]{9}}>0\\] \$y=\\left(\\dfrac{1}{\\sqrt[3]{3}}\\right)^3-\\ln\\dfrac{1}{\\sqrt[3]{3}}=\\dfrac{1}{3}\\left(1+\\ln 3\\right)\\approx 0.7\$ is a minimum occurring when \$x=\\dfrac{1}{\\sqrt[3]{3}}\$.

Exercise 14.15. Differentiate \$y=\\ln(axe^x)\$.

Answer

\$\\dfrac{1 + x}{x}\$.

Solution

\\[y=\\ln \\left(a x e^{x}\\right)\\]

Recall that \$\\ln(AB)=\\ln A+\\ln B\$ and \$\\ln\\left(A^B\\right)=B\\ln A\$. Using these properties, we can rewrite \$y\$ as \\[y=\\ln a+\\ln x+\\ln e^x=\\ln a+\\ln x+x\\ln e\\] Since \$\\ln e=1\$, \\[y=\\ln a+\\ln x+x.\\] Now we can differentiate term by term \\[\\frac{dy}{dx}=0+\\frac{1}{x}+1=\\frac{1+x}{x}.\\]

Exercise 14.16. Differentiate \$y=(\\ln ax)^3\$.

Answer

\$\\dfrac{3}{x} (\\ln ax)^2\$.

Solution

\\[y=u^3\\qquad\\text{where}\\quad u=\\ln ax\\] Using the Chain Rule \\[\\begin{align} \\frac{dy}{dx}=&\\frac{dy}{du}\\cdot\\frac{du}{dx}\\\\ =& 3u^2\\cdot\\frac{1}{ax}\\\\ =&3\\cdot(\\ln ax)^2\\frac{1}{ax}\\\\ =&\\frac{3}{ax}\\left(\\ln ax\\right)^2 \\end{align}\\]

The Logarithmic Curve

Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation \$y=bp^x\$.

We can see, by putting \$x=0\$, that \$b\$ is the initial height of \$y\$.

Then when \\[x=1,\\quad y=bp;\\qquad x=2,\\quad y=bp^2;\\qquad x=3,\\quad y=bp^3,\\quad \\text{etc.}\\]

Also, we see that \$p\$ is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In the following figure, we have taken \$p\$ as \$\\frac{6}{5}\$; each ordinate being \$\\frac{6}{5}\$ as high as the preceding one.

If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant difference; so that, if we should plot out a new curve, the following figure, with values of \$\\ln y\$ as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that \\[\\begin{align} \\ln y = \\ln b + x \\cdot \\ln p, \\end{align}\\] whence \\[\\begin{align} \\ln y - \\ln b = x \\cdot \\ln p. \\end{align}\\] Now, since \$\\ln p\$ is a mere number, and may be written as \$\\ln p=a\$, it follows that \\[\\ln \\frac{y}{b}=ax,\\] and the equation takes the new form \\[y = be^{ax}.\\]

The Die-away Curve

If we were to take \$p\$ as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in the next figure, where each successive ordinate is \$\\frac{3}{4}\$ of the height of the preceding one.

The equation is still \\[y=bp^x;\\]

but since \$p\$ is less than one, \$\\ln p\$ will be a negative quantity, and may be written \$-a\$; so that \$p=e^{-a}\$,³ and now our equation for the curve takes the form \\[y=be^{-ax}.\\]

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton’s celebrated “law of cooling”) by the equation \\[\\theta_t=\\theta_0 e^{-at};\\] where \$\\theta_0\$ is the original excess of temperature of a hot body over that of its surroundings, \$\\theta_t\$ the excess of temperature at the end of time \$t\$, and \$a\$ is a constant—namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula, \\[Q_t=Q_0 e^{-at},\\] is used to express the charge of an electrified body, originally having a charge \$Q_0\$, which is leaking away with a constant of decrement \$a\$; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.

In fact \$e^{-at}\$ serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols, \$\\dfrac{dy}{dt}\$ is proportional at every moment to the value that \$y\$ has at that moment. For we have only to inspect the curve, the above figure, to see that, at every part of it, the slope \$\\dfrac{dy}{dx}\$ is proportional to the height \$y\$; the curve becoming flatter as \$y\$ grows smaller. In symbols, thus \\[y=be^{-ax}\\] or \\[\\ln y = \\ln b - ax \\ln e = \\ln b - ax,\\] and, differentiating, \\[\\frac{1}{y}\\, \\frac{dy}{dx} = -a;\\] hence \\[\\frac{dy}{dx} = be^{-ax} \\times (-a) = -ay;\\] or, in words, the slope of the curve is downward, and proportional to \$y\$ and to the constant \$a\$.

We should have got the same result if we had taken the equation in the form \\[\\begin{align} y &= bp^x; \\end{align}\\] for then \\[\\begin{align} \\frac{dy}{dx}= bp^x \\times \\ln p. \\end{align}\\] But \\[\\ln p = -a;\\] giving us \\[\\begin{align} \\frac{dy}{dx} = y \\times (-a) = -ay, \\end{align}\\] as before.

The Time-constant. In the expression for the “die-away factor” \$e^{-at}\$, the quantity \$a\$ is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol \$T\$. Then the die-away factor will be written \$e^{-\\frac{t}{T}}\$; and it will be seen, by making \$t = T\$ that the meaning of \$T\$ \$\\left(\\text{or of}~\\dfrac{1}{a}\\right)\$ is that this is the length of time which it takes for the original quantity (called \$\\theta_0\$ or \$Q_0\$ in the preceding instances) to die away \$\\dfrac{1}{e}\$th part—that is to \$0.3678\$—of its original value.

As an example, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when \$t = 0\$) it is \$72~^\\circ\$C hotter than the surrounding objects, and if the time-constant of its cooling is \$20\$ minutes (that is, if it takes \$20\$ minutes for its excess of temperature to fall to \$\\dfrac{1}{e}\$ part of \$72\$ degrees), then we can calculate to what it will have fallen in any given time \$t\$. For instance, let \$t\$ be \$60\$ minutes. Then \$\\dfrac{t}{T} = \\dfrac{60}{20} = 3\$, and we shall have to find the value of \$e^{-3}\$, and then multiply the original \$72\$ degrees by this. Since \$e^{-3}\$ is \$0.0498\$, at the end of \$60\$ minutes the excess of temperature will have fallen to \$72~^\\circ\\text{C} \\times 0.0498 = 3.586~^\\circ\$C.

Further Examples

Example 14.12. The strength of an electric current in a conductor at a time \$t\$ seconds after the application of the electromotive force producing it is given by the expression \$C = \\dfrac{E}{R}\\left\\{1 - e^{-\\frac{Rt}{L}}\\right\\}\$.

The time constant is \$\\dfrac{L}{R}\$.

If \$E = 10\$, \$R =1\$, \$L = 0.01\$; then when \$t\$ is very large the term \$1-e^{-\\frac{Rt}{L}}\$ becomes \$1\$, and \$C = \\dfrac{E}{R} = 10\$; also \\[\\frac{L}{R} = T = 0.01.\\]

Its value at any time may be written: \\[C = 10 - 10e^{-\\frac{t}{0.01}},\\] the time-constant being \$0.01\$. This means that it takes \$0.01\$ sec. for the variable term to fall by \$\\dfrac{1}{e} = 0.3678\$ of its initial value \$10e^{-\\frac{0}{0.01}} = 10\$.

To find the value of the current when \$t = 0.001~\$s, say, \$\\dfrac{t}{T} = 0.1\$, \$e^{-0.1} = 0.9048\$.

It follows that, after \$0.001\$ s, the variable term is \$0.9048 \\times 10 = 9.048\$, and the actual current is \$10 - 9.048 = 0.952\$.

Similarly, at the end of \$0.1\$ s, \\[\\frac{t}{T} = 10;\\quad e^{-10} = 0.000045;\\] the variable term is \$10 \\times 0.000045 = 0.00045\$, the current being \$9.9995\$.

Example 14.13. The intensity \$I\$ of a beam of light which has passed through a thickness \$l\$ cm of some transparent medium is \$I = I_0e^{-Kl}\$, where \$I_0\$ is the initial intensity of the beam and \$K\$ is a “constant of absorption.”

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through \$10\$ cm of a certain transparent medium, this means that \$(100-18) = 100 \\times e^{-K\\times10}\$ or \$e^{-10K} = 0.82\$. Now we take the natural logarithm of both sides \\[\\ln\\left(e^{-10K}\\right)=\\ln 0.82\\] or \\[-10K \\times \\ln e =\\ln 0.82\\approx -0.2;\\tag{$\\ln e=1$}\\] hence \$K\\approx 0.02\$.

To find the thickness that will reduce the intensity to half its value, one must find the value of \$l\$ which satisfies the equality \$50 = 100 \\times e^{-0.02l}\$, or \$0.5 = e^{-0.02l}\$. It is found by putting this equation in its logarithmic form, namely, \\[\\ln 0.5 = -0.02 \\times l \\times \\ln e,\\] which gives \\[l \\approx \\frac{{-0.6931}}{-0.02 \\times 1} \\approx {34.7}~\\text{centimeters}.\\]

Example 14.14. The quantity \$Q\$ of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity \$Q_0\$ of the substance by the relation \$Q = Q_0 e^{-\\lambda t}\$, where \$\\lambda\$ is a constant and \$t\$ the time in seconds elapsed since the transformation began.

For “Radium \$A\$,” if time is expressed in seconds, experiment shows that \$\\lambda = 3.85 \\times 10^{-3}\$. Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

Solution. We have \$0.5 = e^{-0.00385t}\$.\\[\\begin{align} \\ln 0.5 &= -0.00385t \\times \\ln e; \\tag{$\\ln e=1$} \\end{align}\\] and \\[\\begin{align} t \\approx 180.04~\\text{s}\\approx 3\\text{ minutes very nearly}. \\end{align}\\]

Exercises II

Exercise 14.17. Draw the curve \$y = b e^{-\\frac{t}{T}}\$; where \$b = 12\$, \$T = 8\$, and \$t\$ is given various values from \$0\$ to \$20\$.

Answer

Use your calculator to evaluate \$12 e^{-\\frac{t}{8}}\$ for different values of \$t\$ within the range \$0\\leq t\\leq 20\$. Then, connect the resulting points \$(t, 12 e^{-\\frac{t}{8}})\$ together.

Solution

To sketch the curve by hand, we can evaluate \$y=12 e^{-\\frac{t}{8}}\$ at various values of \$t\$ (\$0\\leq t\\leq 20\$). For example,

\$t\$	\$y=12 e^{-t/8}\$
0	12.000
1	10.590
5	6.423
10	3.438
15	1.840
20	0.985

Then plot each of these points \$(t, y)\$ on a set of axes, and connect these points with a smooth curve.

There are also multiple tools for plotting the curve \$y=12 e^{-\\frac{t}{8}}\$ for \$0\\leq t\\leq 20\$. For example, you may visit WolframAlpha.com and in the search bar simply type:
plot 12 e^ (-t/8) from t=0 to t=20.

Exercise 14.18. If a hot body cools so that in \$24\$ minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to \$1\$ per cent. of the original excess.

Answer

\$T = 34.625\$; \$159.45\$ minutes.

Solution

The equation of cooling is

\\[\\theta=\\theta_{0} e^{-\\frac{t}{T}}\\]

After 24 minutes

\\[\\begin{align} \\frac{\\theta}{\\theta_{0}} & =\\frac{1}{2}=e^{-\\frac{24}{T}} \\\\ \\Rightarrow \\quad \\ln \\frac{1}{2} & =\\ln \\left(e^{-\\frac{24}{T}}\\right) \\\\ -\\ln 2 & =-\\frac{24}{T} \\\\ T & =\\frac{24}{\\ln 2} \\approx 34.6247 \\end{align}\\]

time-constant is approximately \$34.6247 \\frac{1}{\\mathrm{~min}}\$.

Now we want to find \$t\$ such that

\\[\\begin{gathered} \\frac{\\theta}{\\theta_{0}}=0.01=e^{-\\frac{t}{34.6247}} \\\\ \\ln \\left(10^{-2}\\right)=\\ln \\left(e^{-\\frac{t}{34.6247}}\\right) \\\\ -2 \\ln 10=-\\frac{t}{34.6247} \\\\ \\Rightarrow \\quad t=2 \\ln 10 \\times 34.6247 \\approx 159.453 \\text { minutes } \\end{gathered}\\]

Exercise 14.19. Plot the curve \$y = 100(1-e^{-2t})\$

Solution

When \$x=0\$, \$y=0\$

When \$x\$ is a large positive number, the term \$e^{-2t}\$ becomes negligible (\$\\approx 0\$). Consequently, we have \\[y\\approx 100(1-0)=100.\\]

On the other hand, when \$x\$ is a large negative number, the terms \$e^{-2t}\$ becomes very large positive. As a result \$y\$ becomes numerically large but negative.

The curve \$y = 100(1-e^{-2t})\$ is shown below.

Exercise 14.20. The following equations give very similar curves: \\[\\begin{align} \\text{(i)}\\ y &= \\frac{ax}{x + b}; \\\\ \\text{(ii)}\\ y &= a(1 - e^{-\\frac{x}{b}}); \\\\ \\text{(iii)}\\ y &= \\frac{2a}{\\pi} \\arctan \\left(\\frac{x}{b}\\right).\\quad (\\text{it may be written as }y = \\frac{2a}{\\pi} \\tan^{-1} \\left(\\frac{x}{b}\\right)) \\end{align}\\] Draw all three curves, taking \$a= 10\$ units; \$b = 3\$ units.

Solution

Let’s assume \$b>0\$.

When \$x=0\$, then

\$y=\\frac{ax}{x + b}\$ is zero.

\$y=a(1-e^{-\\frac{x}{b}})\$ is zero because \$e^0=1\$.

\$y=\\frac{2a}{\\pi} \\arctan \\left(\\frac{x}{b}\\right)\$ is zero because \$\\arctan 0 =0\$.

When \$x\$ is very large positive

\$y=\\dfrac{ax}{x + b}\$ is \$\\approx a\$ because we can ignore \$b\$ compared to \$x\$ (for example, when \$b=3\$ and \$x=1,000,000,000\$, then \$x+b\\approx 1,000,000,000\$).

\$y=a(1-e^{-\\frac{x}{b}})\$ is \$\\approx a\$ because \$e^{-x/b}\$ is negligible for large values of \$x\$.

\$y=\\frac{2a}{\\pi} \\arctan \\left(\\frac{x}{b}\\right)\$ is \$\\approx a\$ because when \$x\$ is large, \$\\arctan(x/b)\\approx \\dfrac{\\pi}{2}\$.

For negative values of \$x\$, these curves are not comparable because \$y=\\dfrac{ax}{x + b}\$ is very large when \$x\$ is close to \$-b\$.

When \$x\$ is large negative,

\$e^{-x/b}\$ is large positive. Therefore \$y=a(1-e^{-\\frac{x}{b}})\$ is large negative when \$x\$ is large negative.

\$y=\\dfrac{ax}{x + b}\$ is \$\\approx a\$ because again \$b\$ is negligible compared to \$x\$ and thus \$x+b\\approx x\$.

\$y=\\frac{2a}{\\pi} \\arctan \\left(\\frac{x}{b}\\right)\$ is \$\\approx -a\$ because when \$x\$ is large negative, \$\\arctan(x/b)\\approx -\\dfrac{\\pi}{2}\$.

These curves are shown below.

Exercise 14.21. Find the derivative of \$y\$ with respect to \$x\$, if \\[(\\text{a})~y = x^x;\\quad (\\text{b})~y = (e^x)^x;\\quad (\\text{c})~y = e^{\\left(x^x\\right)}.\\]

Answer

a) \$x^x \\left(1 + \\ln x\\right)\$; (b) \$2x(e ^x)^x\$; (c) \$e ^{x^x} \\times x^x \\left(1 + \\ln x\\right)\$.

Solution

(a)

\\[y=x^{x}\\]

Taking the natural logarithm of both sides and recalling \$\\ln \\left(A^{B}\\right)=B \\ln A\$ :

\\[\\begin{align} \\ln y & =\\ln \\left(x^{x}\\right) \\\\ & =x \\ln x \\end{align}\\]

Now differentiating and using the Chain Rule for the left-hand side and the Product Rule for the right-hand side

\\[\\begin{align} \\frac{1}{y} \\frac{d y}{d x} & =\\ln x+x \\cdot \\frac{1}{x} \\\\ \\Rightarrow \\quad \\frac{d y}{d x} & =y(\\ln x+1) \\\\ & =x^{x}(\\ln x+1) \\end{align}\\]

(b)

\\[y=\\left(e^{x}\\right)^{x}\\]

Method 1)

Taking the natural logarithm of both sides and using the property \$\\ln \\left(A^{B}\\right)=B \\ln A\$, we have

\\[\\begin{align} \\ln y & =\\ln \\left(\\left(e^{x}\\right)^{x}\\right) \\\\ & =x \\ln \\left(e^{x}\\right) \\\\ & =x \\cdot x=x^{2} \\end{align}\\]

Now let’s differentiate both sides with respect to \$x\$:

\\[\\frac{1}{y} \\cdot \\frac{d y}{d x}=2 x\\] Simplifying, we find \\[\\frac{d y}{d x}=2 x y=2 x\\left(e^{x}\\right)^{x}\\]

Method 2) Recall that \$\\left(A^{B}\\right)^{C}=A^{B C}\$. Therefore

\\[y=\\left(e^{x}\\right)^{x}=e^{x \\cdot x}=e^{x^{2}}\\]

Let \$u=x^{2}\$ and applying the Chain Rule with \$y=e^{u}\$ where \$u=x^{2}\$, we get

\\[\\begin{align} \\frac{d y}{d x} & =\\frac{d y}{d u} \\cdot \\frac{d u}{d x} \\\\ & =e^{u} \\cdot 2 x \\\\ & =y \\cdot 2 x \\\\ & =2 x\\left(e^{x}\\right)^{x} \\end{align}\\]

(c)

\\[y=e^{\\left(x^{x}\\right)}\\]

Method 1) Let \\[y=e^{u} \\quad \\text { where } \\quad u=x^{x}\\] Then using the result of part (a) and the Chain Rule, we obtain \\[\\begin{align} \\frac{d y}{d x} & =\\frac{d y}{d u} \\cdot \\frac{d u}{d x}\\\\ &= e^{u} \\cdot\\left[x^{x}(\\ln x+1)\\right] \\\\ & =e^{\\left(x^{x}\\right)} \\cdot x^{x}(\\ln x+1) . \\end{align}\\]

Method 2) Take \$\\ln\$ of both sides

\\[\\begin{align} \\ln y & =\\ln \\left(e^{\\left(x^{x}\\right)}\\right)\\\\ & =x^{x} \\ln e \\\\ & =x^{x} \\end{align}\\] Now differentiate both sides with respect to \$x\$ to get \\[\\begin{align} \\frac{1}{y} \\frac{d y}{d x} & =\\frac{d\\left(x^{x}\\right)}{d x} \\\\ & =x^{x}(\\ln x+1) \\end{align}\\] Therefore \\[\\begin{align} \\frac{d y}{d x} & =y \\times x^{x}(\\ln x+1) \\\\ & =e^{\\left(x^{x}\\right)} \\times x^{x}(\\ln x+1) . \\end{align}\\]

Exercise 14.22. For “Thorium \$A\$,” the value of \$\\lambda\$ is \$5\$; find the “mean life,” that is, the time taken by the transformation of a quantity \$Q\$ of “Thorium \$A\$” equal to half the initial quantity \$Q_0\$ in the expression \\[Q = Q_0 e^{-\\lambda t};\\] \$t\$ being in seconds.

Answer

\$0.14\$ seconds.

Solution

\\[Q=Q_{0} e^{-5 t}\\] We need to find \$t\$ such that \$\\frac{Q}{Q_{0}}=\\frac{1}{2}\$

\\[\\frac{1}{2}=e^{-5 t}\\] Taking the natural logarithm of both sides \\[\\begin{align} \\ln (0.5) & =\\ln \\left(e^{-5 t}\\right) \\\\ & =-5 t \\ln e \\\\ & =-5 t \\\\ \\Rightarrow \\quad t & =-\\frac{\\ln (0.5)}{5} \\approx 0.1386 \\text { seconds. } \\end{align}\\]

Exercise 14.23. A condenser of capacity \$K = 4 \\times 10^{-6}\$, charged to a potential \$V_0 = 20\$, is discharging through a resistance of \$10,000\$ ohms. Find the potential \$V\$ after (a) \$0.1\$ seconds; (b) \$0.01\$ seconds; assuming that the fall of potential follows the rule \$V = V_0 e^{-\\frac{t}{KR}}\$.

Answer

(a) \$1.642\$;(b) \$15.58\$.

Solution

Here \\[V=20 e^{-\\frac{t}{4\\times 10^{-6}\\times 10,000}}\\] or \\[V=20 e^{-25t}\\]

(a) When \$t=0.1\$

\\[V=20 e^{-2.5}\\approx 1.6417\\]

(b) When \$t=0.01\$

\\[V=20 e^{-0.25}\\approx 15.576\\]

Exercise 14.24. The charge \$Q\$ of an electrified insulated metal sphere is reduced from \$20\$ to \$16\$ units in \$10\$ minutes. Find the coefficient \$\\mu\$ of leakage, if \$Q = Q_0 \\times e^{-\\mu t}\$; \$Q_0\$ being the initial charge and \$t\$ being in seconds. Hence find the time taken by half the charge to leak away.

Answer

\$\\mu = 0.00037\$, \$31\\text{ min} 4 \\text{ s}\$.

Solution

Since \\[\\frac{Q}{Q_{0}}=\\frac{16}{20}=e^{-\\mu \\times 10\\times 60}\\] we can find \$\\mu\$

\\[\\begin{align} & \\ln \\frac{16}{20}=-\\mu \\times 10 \\times 60 \\\\ & \\Rightarrow \\quad \\mu=-\\frac{1}{600} \\ln 0.8 \\approx 0.000372 \\end{align}\\]

Now we need to find \$t\$ such that

\\[\\begin{align} \\frac{1}{2} & =e^{-0.000372 t} \\\\ \\ln (0.5) & =-0.000372 t \\\\ t & \\approx 1863.77 \\mathrm{~s} \\approx 31 \\mathrm{~min}\\ 3.77 \\mathrm{~s} \\end{align}\\]

Exercise 14.25. The damping on a telephone line can be ascertained from the relation \$i = i_0 e^{-\\beta l}\$, where \$i\$ is the strength, after \$t\$ seconds, of a telephonic current of initial strength \$i_0\$; \$l\$ is the length of the line in kilometres, and \$\\beta\$ is a constant. For the Franco-English submarine cable laid in 1910, \$\\beta = 0.0114\$. Find the damping at the end of the cable (\$40\$ kilometres), and the length along which \$i\$ is still \$8\$% of the original current (limiting value of very good audition).

Answer

\$i\$ is \$63.4\$% of \$i_0\$, \$220\$ kilometres.

Solution

\\[\\frac{i}{i_{0}}=e^{-\\beta l}\\]

(a)

\\[\\frac{i}{i_{0}}=e^{-0.0114 \\times 40}\\approx 0.6338=63.38\\%.\\]

Therefore, the strength becomes \$63.38 \\%\$ of the initial strength.

(b)

\\[\\begin{align} \\frac{i}{i_{0}}=0.08 & =e^{-0.0114\\, l} \\\\ \\ln (0.08) & =-0.0114\\,l \\\\ \\ell & =-\\frac{\\ln (0.08)}{0.0114} \\approx 221.555 \\mathrm{~km} \\end{align}\\]

Exercise 14.26. The pressure \$p\$ of the atmosphere at an altitude \$h\$ kilometres is given by \$p=p_0 e^{-kh}\$; \$p_0\$ being the pressure at sea-level (\$760\$ millimetres).

The pressures at \$10\$, \$20\$ and \$50\$ kilometres being \$199.2\$, \$42.2\$, \$0.32\$ respectively, find \$k\$ in each case. Using the mean value of \$k\$, find the percentage error in each case.

Answer

\$0.1339\$, \$0.1445\$, \$0.1555\$, mean \$0.1446\$; \$-10.2\$%, practically nil, \$+71.9\$%.

Solution

\\[\\quad p=p_{0} e^{-k h}\\]

Calculating \$k\$ when \$h=10~\\mathrm{km}\$,

\\[\\frac{p}{p_{0}}=\\frac{199.2}{760}=e^{-10 k} \\Rightarrow k=-\\frac{\\ln \\left(\\frac{199.2}{760}\\right)}{10} \\approx 0.1339\\]

Calculating \$k\$ when \$h=20 \\mathrm{~km}\$

\\[\\frac{p}{p_{0}}=\\frac{42.2}{760}=e^{-20 k} \\Rightarrow k=-\\frac{\\ln \\left(\\frac{42.2}{760}\\right)}{20} \\approx 0.1445\\]

Calculating \$k\$ when \$h=50 \\mathrm{~km}\$

\\[\\frac{p}{p_0}=\\frac{0.32}{760}=e^{-50 k} \\Rightarrow k=-\\frac{\\ln \\left(\\frac{0.32}{760}\\right)}{50} \\approx 0.1555\\]

Average \$k\$ (the mean value of \$k\$): \\[k_{\\text{av}}=\\frac{0.1339+0.1445+0.1555}{3} \\approx 0.1446\\]

Calculating \$p\$ using \$k_{\\text{av}}\$ when \$h=10 \\mathrm{~km}\$:

\\[\\begin{align} & 760 \\times e^{-10 \\times 0.1446}=178.99 \\\\ & \\text { error }=\\frac{178.99-199.2}{199.2} \\approx-10.1 \\% \\end{align}\\]

Calculating \$p\$ using \$k_{\\text{av}}\$ when \$h=20 \\mathrm{~km}\$:

\\[\\begin{align} & 760 \\times e^{-20 \\times 0.1446}=42.15 \\\\ & \\text { error }=\\frac{42.2-42.15}{42.2} \\approx 0.12 \\% \\end{align}\\]

Calculating \$p\$ using \$k_{\\text{av}}\$ when \$h=50 \\mathrm{~km}\$:

\\[\\begin{array}{r} 760 \\times e^{-50 \\times 0.1446}=0.550 \\\\ \\text { error }=\\dfrac{0.55-0.32}{0.32} \\approx 72.07 \\% \\end{array}\\]

Exercise 14.27. Find the minimum or maximum of \$y = x^x\$.

Answer

Min. for \$x = \\dfrac{1}{e }\$.

Solution

In Exercise S, we showed that

\\[\\frac{d y}{d x}=\\frac{d\\left(x^{x}\\right)}{d x}=x^{x}(\\ln x+1)\\]

Setting

\\[\\frac{d y}{d x}=0 \\Leftrightarrow x^{x}=0 \\text { or } \\ln x+1=0\\]

Since \$x^{x} \\neq 0\$, at the minimum or maximum, we must have \\[\\ln x =-1\\] Then \\[\\begin{align} e^{\\ln x} & =e^{-1} \\\\ x & =e^{-1}=\\frac{1}{e} . \\end{align}\\]

To determine if this value of \$x\$ makes \$y\$ a maximum or a minimum, we apply the Second Derivative Test:

\\[\\begin{align} \\frac{d^{2} y}{d x^{2}} & =\\frac{d\\left(x^{x}\\right)}{d x}(\\ln x+1)+x^{x}\\left(\\frac{1}{x}\\right) \\\\ & =x^{x}(\\ln x+1)^{2}+\\frac{x^{x}}{x} \\end{align}\\] When \$x=\\dfrac{1}{e}\$, then \\[\\begin{align} \\frac{d^{2} y}{d x^{2}} & =\\left(\\frac{1}{e}\\right)^{\\frac{1}{e}} \\times 0+\\frac{\\left(\\frac{1}{e}\\right)^{\\frac{1}{e}}}{\\frac{1}{e}} \\\\ & =\\frac{1}{e}\\left(\\frac{1}{e}\\right)^{\\frac{1}{e}}>0 \\end{align}\\] Therefore, \$x=1/e\\approx 0.369\$ corresponds to a minimum value of \$y\$.

The graph of \$y=x^{x}\$ is shown below.

Exercise 14.28. Find the minimum or maximum of \$y = x^{\\frac{1}{x}}\$.

Answer

Max. for \$x = e\$.

Solution

\\[y=x^{\\frac{1}{x}}\\]

To differentiate, we first take the natural logarithm of both sides then use the Chain Rule:

\\[\\begin{align} \\ln y & =\\ln \\left(x^{\\frac{1}{x}}\\right) \\\\ & =\\frac{1}{x} \\ln x \\\\ \\frac{1}{y} \\frac{d y}{d x} & =-\\frac{1}{x^{2}} \\ln x+\\frac{1}{x} \\cdot \\frac{1}{x} \\\\ \\frac{d y}{d x} & =\\frac{y}{x^{2}}(1-\\ln x) \\end{align}\\]

Since \$y=x^{\\frac{1}{x}} \\neq 0\$,

\\[\\frac{d y}{d x}=0 \\quad \\Leftrightarrow \\quad 1-\\ln x=0\\] or \\[\\frac{d y}{d x}=0 \\Leftrightarrow x=e\\]

To determine whether \$x=e\$ makes \$y\$ a minimum or a maximum, we compare the value of \$y\$ at this point with the values of \$y\$ at some nearby points:

When \$x=e\$

\\[y=e^{\\frac{1}{e}} \\approx e^{\\frac{1}{2.718}} \\approx e^{0.368} \\approx 1.444\\]

When \$x=2\$ \\[y=2^{\\frac{1}{2}}=\\sqrt{2}=1.414\\]

When \$x=3\$

\\[y=3^{\\frac{1}{3}}=\\sqrt[3]{3} \\approx 1.442\\]

Therefore \$x=e\$ makes \$y=x^{\\frac{1}{x}}\$ a maximum.

We can also apply the Second Derivative Test:

\\[\\frac{d^{2} y}{d x^{2}}=\\frac{d y}{d x} \\frac{1-\\ln x}{x^{2}}+y \\frac{d}{d x}\\left(\\frac{1-\\ln x}{x^{2}}\\right)\\]

Using the Quotient Rule

\\[\\frac{d}{d x}\\left(\\frac{1-\\ln x}{x^{2}}\\right)=\\frac{-\\frac{1}{x} \\cdot x^{2}-2 x(1-\\ln x)}{x^{4}}\\] Therefore, \\[\\frac{d^{2} y}{d x^{2}}=\\frac{d y}{d x} \\frac{1-\\ln x}{x^{2}}-x^{\\frac{1}{x}} \\frac{x+2 x(1-\\ln x)}{x^{2}}\\]

When \$x=e\$, both \$\\dfrac{d y}{d x}\$ and \$1-\\ln x\$ are zero. Therefore, when \$x=e\$

\\[\\frac{d^{2} y}{d x^{2}}=0 \\times 0-e^{\\frac{1}{e}} \\frac{e+2 e(0)}{e^2}<0\\]

It follows from the Second Derivative Test that \$y=e^{\\frac{1}{e}}\\approx 1.44\$ is a maximum occurring when \$x=e\$.

The graph of \$y=x^{\\frac{1}{x}}\$ is shown below

Exercise 14.29. Find the minimum or maximum of \$y = xa^{\\frac{1}{x}}\$.

Answer

Min. for \$x = \\ln a\$.

Solution

\\[y=x a^{\\frac{1}{x}}\\]

Using the Product Rule

\\[\\frac{d y}{d x}=a^{\\frac{1}{x}}+x \\frac{d\\left(a^{\\frac{1}{x}}\\right)}{d x}\\]

To find \$\\frac{d\\left(a^{\\frac{1}{x}}\\right)}{d x}\$, let \$u=\\frac{1}{x}\$ and use the Chain Rule

\\[\\begin{align} \\frac{d a^{u}}{d x} & =\\frac{d a^{u}}{d u} \\cdot \\frac{d u}{d x} \\\\ & =\\left(a^{u} \\ln a\\right)\\left(-\\frac{1}{x^{2}}\\right) \\\\ & =-\\frac{\\ln a}{x^{2}} a^{\\frac{1}{x}} \\end{align}\\]

Therefore

\\[\\frac{d y}{d x}=a^{\\frac{1}{x}}-x \\frac{\\ln a}{x^{2}} a^{\\frac{1}{x}}=a^{\\frac{1}{x}}\\left(1-\\frac{\\ln a}{x}\\right)\\]

Since \$a^{\\frac{1}{x}}>0\$,

\\[\\frac{d y}{d x}=0 \\Leftrightarrow 1-\\frac{\\ln a}{x}=0\\]

\\[\\frac{d y}{d x}=0 \\Leftrightarrow \\boxed{x=\\ln a}\\]

Using the Second Derivative Test

\\[\\begin{gathered} \\frac{d y}{d x}=y\\left(1-\\frac{\\ln a}{x}\\right) \\\\ \\frac{d^{2} y}{d x^{2}}=\\frac{d y}{d x}\\left(1-\\frac{\\ln a}{x}\\right)+y \\cdot \\frac{\\ln a}{x^{2}} \\end{gathered}\\]

When \$x=\\ln a\$

\\[\\frac{d^{2} y}{d x^{2}}=0 \\times 0+a^{\\frac{1}{\\ln a}} \\cdot \\frac{\\ln a}{(\\ln a)^{2}}>0\\]

Therefore \$y=a^{\\frac{1}{\\ln a}}\$ is a minimum occurring when \$x=\\ln a\$.

Hyperbolic Functions

Hyperbolic functions are specific combinations of exponential functions that arise frequently in various applications, leading mathematicians to give them distinctive names and thoroughly investigate their properties. Although hyperbolic functions are combined exponential functions, in some ways they resemble the trigonometric functions. As a result, the hyperbolic functions are given individual names such as hyperbolic sine, hyperbolic cosine, hyperbolic tangent, and others. Their definitions are as follows:

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\begin{align} \\sinh x&=\\frac{e^{x}-e^{-x}}{2},\\\\ \\cosh x&=\\frac{e^{x}+e^{-x}}{2},\\\\ \\tanh x&=\\frac{\\sinh x}{\\cosh x}\\\\ \\text{coth } x&=\\frac{1}{\\tanh x}=\\frac{\\cosh x}{\\sinh x}\\\\ \\text{sech }x &=\\frac{1}{\\cosh x}\\\\ \\text{csch }x &=\\frac{1}{\\sinh x} \\end{align}}\\]

One can easily sketch the graphs of \$y = \\sinh x\$ and \$y=\\cosh x\$ by plotting the curves \$y = e^x\$ and \$y = e^{-x}\$, adding and subtracting the ordinates, and taking half of each. The following figure displays their graphs.

A catenary, which is the graph of the hyperbolic cosine, is a curve that describes the shape of a homogeneous chain or cord hanging under its own weight.

When \$x\$ is positive large, \$e^{x}\$ is very large, while \$e^{-x}\$ is extremely small; hence \$e^{x}\\pm e^{-x}\\approx e^{x}\$ leading to: \\[\\tanh x=\\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\approx\\frac{e^{x}}{e^{x}}=1.\\] Similarly, when \$x\$ is negative large, \$e^x\$ is negligible, making \$e^x\\pm e^{-x}\\approx \\pm e^{-x}\$, and hence: \\[\\tanh x=\\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\approx\\frac{-e^{x}}{e^{x}}=-1.\\] The graph of \$y=\\tanh x\$ is shown below.

Identities Between Hyperbolic Functions

The properties of the hyperbolic functions closely resemble the corresponding properties of the trigonometric functions.

It follows directly from the definitions that \\[\\begin{align} \\cosh x-\\sinh x & =e^{-x}\\\\ \\cosh x+\\sinh x & =e^{x} \\end{align}\\] Now using \$A^2-B^2=(A-B)(A+B)\$, we get \\[\\begin{align} \\cosh^2 x-\\sinh^2 x&=(\\cosh x-\\sinh x)(\\cosh x+\\sinh x)\\\\ &= e^{-x}\\cdot e^x=e^{-x+x}=e^0=1 \\end{align}\\] As such we have proved one of the fundamental identities: \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ \\cosh^{2}x-\\sinh^{2}x=1}\\] Now if we divide each term of the above equation by \$\\cosh^2 x\$, we obtain \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{1-\\tanh^{2}x=\\text{sech}^{2}x=\\dfrac{1}{\\cosh^{2}x}}\\] Similarly, dividing each term of \$\\cosh^2 x-\\sinh^2 x=1\$ by \$\\sinh^2 x\$ gives us \\[\\text{coth}^2 x-\\text{csch}^{2}x=1\\]

Example 14.15. Prove \$\\sinh(x+y)=\\sinh x\\cosh y+\\cosh x\\sinh y\$.

Solution. Let’s simplify the right-hand side \\[\\begin{align} \\sinh x\\cosh y+\\cosh x\\sinh y= & \\frac{e^{x}-e^{-x}}{2}\\frac{e^{y}+e^{-y}}{2}+\\frac{e^{x}+e^{-x}}{2}\\frac{e^{y}-e^{-y}}{2}\\\\[6pt] = & \\frac{1}{4}\\left[e^{x+y}+\\cancel{e^{x-y}}-\\bcancel{e^{-x+y}}-e^{-x-y}\\right]\\\\[6pt] & \\quad+\\frac{1}{4}\\left[e^{x+y}-\\cancel{e^{x-y}}+\\bcancel{e^{-x+y}}-e^{-x-y}\\right]\\\\[6pt] = & \\frac{1}{2}\\left[e^{x+y}-e^{-(x+y)}\\right]\\\\[6pt] = & \\sinh(x+y) \\end{align}\\]

Therefore, \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\sinh(x+ y)=\\sinh x\\cosh y+\\cosh x\\sinh y.}\\] In a similar way, we can prove \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\cosh(x+ y)=\\cosh x\\cosh y+\\sinh x\\sinh y.}\\] It follows from the above identity that \\[\\cosh 2x=\\cosh^2 x+\\sinh^2 x=(1+\\sinh^2 x)+\\sinh^2 x=1+2\\sinh^2 x\\] Hence, \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\sinh^2 x=\\dfrac{1}{2}\\left(\\cosh 2x-1\\right).}\\] Similarly, \\[\\cosh 2x=\\cosh^2x+\\sinh^2 x=\\cosh^2 x+(\\cosh^2 x-1)=2\\cosh^2 x-1\\] Therefore, \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\cosh^2 x=\\frac{1}{2}\\left(\\cosh 2x+1\\right).}\\]

Another important set of identities, which follow at once from the definitions, is:⁴ \\[\\sinh(-x)=-\\sinh x\\] \\[\\cosh(-x)=\\cosh x\\] \\[\\tanh(-x)=-\\tanh x\\]

Derivatives of the Hyperbolic Functions

As the hyperbolic functions are combinations of exponential functions, and we have just learned how to differentiate exponential functions, we can apply the differentiation rules learned in Chapters 5 and 6 to determine the derivatives of the hyperbolic functions.

For example, since \$\\sinh x=\\frac{1}{2}(e^x-e^{-x})\$, we have \\[\\begin{align} \\frac{d(\\sinh x)}{dx}&=\\frac{d}{dx}\\left(\\frac{e^x}{2}-\\frac{e^{-x}}{2}\\right)\\\\ &=\\frac{1}{2}\\frac{d(e^x)}{dx}-\\frac{1}{2}\\frac{d(e^{-x})}{dx}\\\\ &=\\frac{e^x}{2}-\\frac{-e^{-x}}{2}=\\frac{e^x+e^{-x}}{2}=\\cosh x. \\end{align}\\] Similarly, we can show that \$\\dfrac{d(\\cosh x)}{dx}=\\sinh x\$.

Example 14.16. Show \$\\dfrac{d}{dx}\\tanh x=\\text{sech}^{2}x\$.

Solution. Since \$\\tanh x=\\sinh x/\\cosh x\$, we can use the Quotient Rule: \\[\\begin{align} \\frac{d}{dx}\\tanh x= & \\frac{d}{dx}\\frac{\\sinh x}{\\cosh x}\\\\[6pt] = & \\frac{\\cosh x-\\frac{d(\\sinh x)}{dx}-\\sinh x\\frac{d(\\cosh x)}{dx}}{\\cos^2 x}\\\\[6pt] = & \\frac{\\cosh^2 x-\\sinh^2 x}{\\cosh^2 x}=\\frac{1}{\\cosh^{2}x} && {\\small(\\cosh^{2}x-\\sinh^{2}x=1)}\\\\[6pt] = & \\text{ sech}^{2}x \\end{align}\\]

In summary, \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\begin{array}{rl} \\dfrac{d(\\sinh x)}{dx} & =\\cosh x\\\\[9pt] \\dfrac{d(\\cosh x)}{dx} & =\\sinh x\\\\[9pt] \\dfrac{d(\\tanh x)}{dx} & =\\text{sech}^{2}x=1-\\tanh^{2}x \\end{array}}\\]

Inverse Hyperbolic Functions

The inverse hyperbolic sine function is denoted by \$\\text{arcsinh }x\$ or \$\\sinh^{-1} x\$, and is defined by \\[y=\\text{arcsinh }x=\\sinh^{-1} x\\qquad \\text{if}\\qquad x=\\sinh y.\\] We may also define the inverse hyperbolic cosine function, denoted by \$\\text{arccosh }x\$ or \$\\cosh^{-1} x\$, as \\[y=\\text{arccosh }x=\\cosh^{-1} x\\qquad \\text{if}\\qquad x=\\cosh y.\\] However, note that if \$x=\\cosh y\$, then \$\\cosh(-y)\$ is also equal to \$x\$. This means that the above definition assigns two values of \$y\$ to each value of \$x\$. To make \$\\text{arccosh }x\$ a single-valued function, we agree that it produces only non-negative values of \$y\$. Furthermore, because \$\\cosh y\\geq 1\$, the inverse hyperbolic cosine function does not take any values of \$x\$ less than \$1\$.

Similarly, the inverse hyperbolic tangent function is defined by \\[y=\\text{arctanh }x=\\tanh^{-1} x\\qquad \\text{if}\\qquad x=\\tanh y.\\] Since \$-1<\\tanh y<1\$, the inverse hyperbolic tangent function does not take any values of \$x\$ greater than or equal to \$1\$ or less than or equal to \$-1\$.

In summary,

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\begin{align}y=\\text{ arcsinh } x & \\Leftrightarrow y=\\sinh x \\\\ y=\\text{ arccosh } x & \\Leftrightarrow x=\\cosh y & & (x\\geq 1,y\\geq 0)\\\\ y=\\text{ arctanh } x & \\Leftrightarrow x=\\tanh y & & (-1<x<1) \\end{align} }\\]

We can find explicit formulas for the inverse hyperbolic functions.

Example 14.17. Show that \$\\text{arcsinh } x=\\sinh^{-1}x=\\ln\\left(x+\\sqrt{x^{2}+1}\\right)\$.

Solution. If \$y=\\text{ arcsinh }x\$, then \\[x=\\sinh y=\\frac{e^{y}-e^{-y}}{2}.\\] Multiplying by \$y\$ by \$2e^{y}\$, we obtain \\[2xe^{y}=\\left(e^{y}\\right)^{2}-1\\] or \\[\\left(e^{y}\\right)^{2}-2xe^{y}-1=0\\] This is a quadratic equation in terms of \$e^{x}\$. Therefore, by the quadratic formula, \\[e^{y}=\\frac{2x\\pm\\sqrt{4x^{2}+4}}{2}=x\\pm\\sqrt{x^{2}+1}\\] Since \$e^{y}>0\$ and \$\\sqrt{x^2+1}\$ is greater than \$x\$, only the positive sign is acceptable. Hence, \\[e^{y}=x+\\sqrt{x^{2}+1}.\\] Taking the natural logarithm of each side produces \\[y=\\ln\\left(x+\\sqrt{x^{2}+1}\\right).\\] Therefore, \\[\\sinh^{-1}(x)=\\ln\\left(x+\\sqrt{x^{2}+1}\\right).\\]

In a similar fashion, we can derive explicit formulas for the other inverse hyperbolic functions. \\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\begin{align} & \\text{ arcsinh }x=\\ln\\left(x+\\sqrt{x^{2}+1}\\right) & & (x \\text{ can be any number})\\\\ & \\text{ arccosh }x=\\ln\\left(x+\\sqrt{x^{2}-1}\\right) & & (x\\geq1)\\\\ & \\text{ arctanh }x=\\frac{1}{2}\\ln\\frac{1+x}{1-x} & & (-1<x<1) \\end{align} }\\]

Derivatives of the Inverse Hyperbolic Functions

With what we have learned so far, we can find the derivatives of the inverse hyperbolic functions.

Example 14.18. If \$y=\\text{ arcsinh } x\$ (also denoted by \$\\sinh^{-1} x\$), find \$\\dfrac{dy}{dx}\$.

Solution. If \$y=\\text{ arcsinh } x\$, then \$x=\\sinh y\$, and \\[\\dfrac{dx}{dy}=\\cosh y\\] By the derivative of an inverse function (see here) \\[\\begin{align} \\frac{dy}{dx}&=\\frac{1}{\\dfrac{dx}{dy}}=\\frac{1}{\\cosh y}=\\frac{1}{\\sqrt{1+\\sinh^2 y}}=\\frac{1}{\\sqrt{1+x^2}}. \\end{align}\\]

We can also find the derivative of \$\\text{arcsinh }x\$ by differentiating \$\\ln(x+\\sqrt{x^2+1})\$. Using the chain rule and the fact that \$\\dfrac{d\\left(\\sqrt{x^2+1}\\right)}{dx} = \\dfrac{x}{\\sqrt{x^2+1}}\$, we get \\[\\begin{align} \\frac{d}{dx} \\ln(x+\\sqrt{x^2+1}) &= \\frac{1}{x+\\sqrt{x^2+1}}\\cdot\\left(1+\\frac{x}{\\sqrt{x^2+1}}\\right)= \\frac{1}{\\sqrt{x^2+1}} . \\end{align}\\] This confirms the result we obtained earlier.

Example 14.19. If \$y=\\text{ arccosh }x\$ (also denoted by \$\\cosh^{-1} x\$), find \$\\dfrac{dy}{dx}\$.

Solution. If \$y=\\text{ arccosh } x\$, then \$x=\\cosh y\$, and \\[\\frac{dx}{dy}=\\sinh y;\\] Hence, \\[\\frac{dy}{dx}=\\frac{1}{\\dfrac{dx}{dy}}=\\frac{1}{\\sinh y}=\\frac{1}{\\sqrt{\\cosh^2 y-1}}=\\frac{1}{\\sqrt{x^2-1}}.\\]

Example 14.20. If \$y=\\text{ arctanh } x\$ (also denoted by \$\\tanh^{-1} x\$), find \$\\dfrac{dy}{dx}\$. Show that \$\\displaystyle{\\frac{d}{dx}\\text{arctanh }x=\\frac{d}{dx}\\tanh^{-1}x=\\frac{1}{1-x^{2}}\\quad(-1<x<1).}\$

Solution. If \$y=\\text{ arctanh } x\$, then \$x=\\tanh y\$, and \\[\\frac{dx}{dy}=\\text{ sech}^2 y=1-\\tanh^2 y.\\] Thus \\[\\frac{dy}{dx}=\\frac{1}{\\dfrac{dx}{dy}}=\\frac{1}{1-\\tanh^2 y}=\\frac{1}{1-x^2}.\\]

In summary,

\\[\\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\\begin{align} & \\frac{d\\left(\\text{arcsinh }x\\right)}{dx}=\\frac{d\\left(\\sinh^{-1}x\\right)}{dx}=\\frac{1}{\\sqrt{x^{2}+1}} & & (x\\in\\mathbb{R})\\\\ & \\frac{d\\left(\\text{arccosh }x\\right)}{dx}=\\frac{d\\left(\\cosh^{-1}x\\right)}{dx}=\\frac{1}{\\sqrt{x^{2}-1}} & & (x>1)\\\\ & \\frac{d\\left(\\text{arctanh }x\\right)}{dx}=\\frac{d\\left(\\tanh^{-1}x\\right)}{dx}=\\frac{1}{1-x^{2}} & & (-1<x<1) \\end{align} }\\]

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Group Theory Study Guide

\n\n

I. Definition of a Group

\n\n

A group is a set G with an operation called multiplication that satisfies the following four axioms:

\n\n

Closure: For all a, b in G, the product ab is also in G.
Associative Law: For all a, b, c in G, (ab)c = a(bc).
Identity: There exists an element e in G (called the identity) such that ea = a for all a in G. It is shown that this implies the existence of a right identity: ae = a.
Inverse: For every a in G, there exists an element a^-1 in G (called the inverse of a) such that a^-1a = e. It is shown that this implies the existence of a right inverse: aa^-1 = e.

\n\n

II. Theorems and Lemmas

\n\n

Theorem 1.1: A subset S of a group G is a subgroup if and only if:
1. Closure: For all s₁, s₂ in S, the product s₁s₂ is also in S.
2. Inverse: For all s in S, s^-1 is also in S.
Lemma 1.2: If two cosets aS and bS of a subgroup S have an element c in common, then aS = bS.
Lemma 1.3: Every element a of a group G is contained in some coset of a subgroup S.
Theorem 1.4 (Lagrange's Theorem): If G is a finite group of order N and S is a subgroup of order n, then n divides N.
Theorem 1.5: The period of any element in a finite group is a divisor of the order of the group.

\n\n

III. Corollaries

\n\n

Corollary 1.6: If the order of a group G is a prime number p, then G is cyclic.

\n\n

IV. Definitions (continued)

\n\n

Abelian Group: A group G is called abelian or commutative if for all a, b in G, ab = ba.
Subgroup: A subset S of a group G that is itself a group under the same operation as G.
Coset: If S is a subgroup of G and a is in G, the coset aS is the set of all elements of the form as where s is in S.
Order of a Group: The number of elements in a finite group.
Period of an Element: The smallest positive integer d such that a^d = e.
Cyclic Group: A group generated by a single element. A cyclic group can be finite or infinite.

\n\n

V. Examples

\n\n

Example 1 (Subgroups): The set of non-zero rational numbers under multiplication forms a group. Subgroups include: positive rational numbers, powers of any element, {+1, -1}, the group itself, and {1}.
Example 2 (Cosets): Considering the group of non-zero rational numbers under multiplication:
- If S is the set of positive rational numbers, the cosets are S and -S.
- If S is {+1, -1}, the cosets are aS = {a, -a}.
Example 3 (Groups of Order 4): There are two groups of order 4, both abelian: the cyclic group of order 4 and the Klein four-group. The multiplication table for the Klein four-group is provided.

\n\n

Several exercises are presented in the excerpt, focusing on proving alternative group axioms, exploring counterexamples, and classifying small groups (order 6, 8). Pay special attention to the exercises involving:

\n\n

VI. Exercises

\n\n

Equivalent postulate systems.
Counterexample to a modified postulate system.
Properties of exponents for powers of elements in a group.
Classification of groups of order 6 and 8. The symmetries of a square and the quaternion units are specifically mentioned as examples of non-commutative groups of order 8.

\n\n

VII. Unsolved Problems

\n\n

The classification of all groups of order n is mentioned as an unsolved problem in algebra.

\n\n

This study guide provides a comprehensive overview of the concepts and results presented in the excerpt. Working through the examples and exercises will further enhance your understanding.Group Theory Study Guide

\n\n

I. Definition of a Group

\n\n

A group is a set G with an operation called multiplication that satisfies the following four axioms:

\n\n

Closure: For all a, b in G, the product ab is also in G.
Associative Law: For all a, b, c in G, (ab)c = a(bc).
Identity: There exists an element e in G (called the identity) such that ea = a for all a in G. It is shown that this implies the existence of a right identity: ae = a.
Inverse: For every a in G, there exists an element a^-1 in G (called the inverse of a) such that a^-1a = e. It is shown that this implies the existence of a right inverse: aa^-1 = e.

\n\n

II. Theorems and Lemmas

\n\n

Theorem 1.1: A subset S of a group G is a subgroup if and only if:
1. Closure: For all s₁, s₂ in S, the product s₁s₂ is also in S.
2. Inverse: For all s in S, s^-1 is also in S.
Lemma 1.2: If two cosets aS and bS of a subgroup S have an element c in common, then aS = bS.
Lemma 1.3: Every element a of a group G is contained in some coset of a subgroup S.
Theorem 1.4 (Lagrange's Theorem): If G is a finite group of order N and S is a subgroup of order n, then n divides N.
Theorem 1.5: The period of any element in a finite group is a divisor of the order of the group.

\n\n

III. Corollaries

\n\n

Corollary 1.6: If the order of a group G is a prime number p, then G is cyclic.

\n\n

IV. Definitions (continued)

\n\n

Abelian Group: A group G is called abelian or commutative if for all a, b in G, ab = ba.
Subgroup: A subset S of a group G that is itself a group under the same operation as G.
Coset: If S is a subgroup of G and a is in G, the coset aS is the set of all elements of the form as where s is in S.
Order of a Group: The number of elements in a finite group.
Period of an Element: The smallest positive integer d such that a^d = e.
Cyclic Group: A group generated by a single element. A cyclic group can be finite or infinite.

\n\n

V. Examples

\n\n

Example 1 (Subgroups): The set of non-zero rational numbers under multiplication forms a group. Subgroups include: positive rational numbers, powers of any element, {+1, -1}, the group itself, and {1}.
Example 2 (Cosets): Considering the group of non-zero rational numbers under multiplication:
- If S is the set of positive rational numbers, the cosets are S and -S.
- If S is {+1, -1}, the cosets are aS = {a, -a}.
Example 3 (Groups of Order 4): There are two groups of order 4, both abelian: the cyclic group of order 4 and the Klein four-group. The multiplication table for the Klein four-group is provided.

\n\n

VI. Exercises

\n\n

Equivalent postulate systems.
Counterexample to a modified postulate system.
Properties of exponents for powers of elements in a group.
Classification of groups of order 6 and 8. The symmetries of a square and the quaternion units are specifically mentioned as examples of non-commutative groups of order 8.

\n\n

VII. Unsolved Problems

\n\n

The classification of all groups of order n is mentioned as an unsolved problem in algebra.

\n\n

This study guide provides a comprehensive overview of the concepts and results presented in the excerpt. Working through the examples and exercises will further enhance your understanding.

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Definition. The partial derivative, with respect to $x$, of a function $z = f(x, y)$ is the limit of the ratio of the partial increment $\\Delta_x z$, with respect to $x$, to the increment $\\Delta x$ as $\\Delta x$ approaches zero. The partial derivative, with respect to $x$, of the function $z = f(x, y)$ is denoted by one of the symbols
$
z_x;\\quad f_x(x, y);\\quad \\frac{\\partial z}{\\partial x};\\quad \\frac{\\partial f}{\\partial x};\\quad D_1 f.
$

Thus, by definition,
$
\\frac{\\partial z}{\\partial x} = \\lim_{\\Delta x \\to 0} \\frac{\\Delta_x z}{\\Delta x} = \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x, y) - f(x, y)}{\\Delta x}.
$

Similarly, the partial derivative, with respect to $y$, of a function $z = f(x, y)$ is defined as the limit of the ratio of the partial increment of the function $\\Delta_y z$ with respect to $y$ to the increment of $\\Delta y$ as $\\Delta y$ approaches zero. The partial derivative with respect to $y$ is denoted by one of the following symbols:
$
z_y;\\quad f_y;\\quad \\frac{\\partial z}{\\partial y};\\quad \\frac{\\partial f}{\\partial y}; \\quad D_2 f.
$
Thus,
$
\\frac{\\partial z}{\\partial y} = \\lim_{\\Delta y \\to 0} \\frac{\\Delta_y z}{\\Delta y} = \\lim_{\\Delta y \\to 0} \\frac{f(x, y + \\Delta y) - f(x, y)}{\\Delta y}.
$

Noting that $\\Delta_x z$ is calculated with $y$ held constant, and $\\Delta_y z$ with $x$ held constant, we can formulate the definitions of partial derivatives as follows:

The partial derivative of the function $z = f(x, y)$ with respect to $x$ is the derivative with respect to $x$ calculated on the assumption that $y$ is constant.

The partial derivative of the function $z = f(x, y)$ with respect to $y$ is the derivative with respect to $y$ calculated on the assumption that $x$ is constant.

It is clear from this definition that the rules for computing partial derivatives coincide with the rules given for functions of one variable, and the only thing to remember is with respect to which variable the derivative is sought.

Example 1. Given the function $z = x^2 \\sin y$; find the partial derivatives $\\dfrac{\\partial z}{\\partial x}$ and $\\dfrac{\\partial z}{\\partial y}$.

Solution

$
\\frac{\\partial z}{\\partial x} = 2x \\sin y; \\quad \\frac{\\partial z}{\\partial y} = x^2 \\cos y.
$

Example 2. Given $z = x^y$, find its partial derivatives.

Solution.

Here $ \\frac{\\partial z}{\\partial x} = y x^{y-1}, \\quad \\frac{\\partial z}{\\partial y} = x^y \\ln x. $

The partial derivatives of a function of any number of variables are determined similarly. Thus, if we have a function $u$ of four variables $x, y, z, t$:

$
u = f(x, y, z, t)
$

then

$
\\frac{\\partial u}{\\partial x} = \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x, y, z, t) - f(x, y, z, t)}{\\Delta x},
$

$
\\frac{\\partial u}{\\partial y} = \\lim_{\\Delta y \\to 0} \\frac{f(x, y + \\Delta y, z, t) - f(x, y, z, t)}{\\Delta y},
$

and so forth.

Example 3. Given
$
u = x^2 + y^2 + xtz^3,
$
find its partial derivatives.

Solution.

$ \\begin{aligned} &\\frac{\\partial u}{\\partial x} = 2x + t z^3; &&\\frac{\\partial u}{\\partial y} = 2y;\\\\ &\\frac{\\partial u}{\\partial z} = 3xtz^2; &&\\frac{\\partial u}{\\partial t} = xz^3. \\end{aligned} $

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در سراسر حساب دیفرانسیل و انتگرال، ما با کمیت‌هایی که در حال رشد هستند و با نرخ‌های رشد سروکار داریم. ما همه کمیت‌ها را به دو دسته تقسیم می‌کنیم: ثابت‌ها و متغیرها. آن‌هایی که به عنوان مقادیر ثابت در نظر می‌گیریم و آن‌ها را ثابت می‌نامیم، معمولاً به صورت جبری با حروفی از ابتدای الفبا مانند \$a\$، \$b\$ یا \$c\$ نشان می‌دهیم؛ در حالی که آن‌هایی را که قابلیت رشد دارند یا (همان‌طور که ریاضیدانان می‌گویند) \"تغییر می‌کنند\"، با حروفی از انتهای الفبا مانند \$x\$، \$y\$، \$z\$، \$u\$، \$v\$، \$w\$ یا گاهی \$t\$ نشان می‌دهیم.

علاوه بر این، ما معمولاً با بیش از یک متغیر به طور همزمان سروکار داریم و به نحوه‌ای که یک متغیر به دیگری وابسته است فکر می‌کنیم: برای مثال، ما به نحوه‌ای که ارتفاعی که یک پرتابه به آن می‌رسد به زمان رسیدن به آن ارتفاع وابسته است فکر می‌کنیم. یا از ما خواسته می‌شود یک مستطیل با مساحت داده شده را در نظر بگیریم و بررسی کنیم که چگونه هر افزایشی در طول آن، کاهش متناسبی در عرض آن ایجاد می‌کند. یا به نحوه‌ای فکر می‌کنیم که هر تغییری در شیب یک نردبان باعث می‌شود ارتفاعی که به آن می‌رسد تغییر کند.

فرض کنید دو چنین متغیری داریم که به یکدیگر وابسته‌اند. تغییری در یکی باعث تغییری در دیگری می‌شود، زیرا این وابستگی وجود دارد. بیایید یکی از متغیرها را \$x\$ و دیگری را که به آن وابسته است \$y\$ بنامیم.

فرض کنید \$x\$ را تغییر دهیم، به این معنا که آن را تغییر دهیم یا تصور کنیم که با افزودن مقداری که آن را \$dx\$ می‌نامیم، تغییر کرده است. بنابراین \$x\$ به \$x + dx\$ تبدیل می‌شود. سپس، به دلیل اینکه \$x\$ تغییر کرده است، \$y\$ نیز تغییر می‌کند و به \$y + dy\$ تبدیل می‌شود. در اینجا مقدار \$dy\$ ممکن است در برخی موارد مثبت و در موارد دیگر منفی باشد؛ و (مگر به ندرت) اندازه آن با \$dx\$ برابر نخواهد بود.

دو مثال بگیرید

مثال ۳.۱. بگذارید \$x\$ و \$y\$ به ترتیب قاعده و ارتفاع یک مثلث قائم‌الزاویه باشند (شکل زیر)، که شیب ضلع دیگر آن ثابت و برابر با \$30^\\circ\$ است. اگر فرض کنیم این مثلث گسترش یابد و زوایای خود را همانند ابتدا حفظ کند، آنگاه وقتی قاعده رشد می‌کند تا به \$x + dx\$ برسد، ارتفاع به \$y + dy\$ تبدیل می‌شود. در اینجا، افزایش \$x\$ منجر به افزایش \$y\$ می‌شود. مثلث کوچکی که ارتفاع آن \$dy\$ و قاعده آن \$dx\$ است، مشابه مثلث اصلی است؛ و واضح است که مقدار نسبت \$\\dfrac{dy}{dx}\$ با نسبت \$\\dfrac{y}{x}\$ برابر است. از آنجا که زاویه \$30^\\circ\$ است، دیده می‌شود که در اینجا \\[\\frac{dy}{dx} = \\tan 30^\\circ \\approx \\frac{1}{1.73}.\\]

شکل ۳.۱

مثال ۳.۲. بگذارید \$x\$ نمایانگر فاصله افقی پایۀ نردبان \$AB\$ با طول ثابت از دیوار در شکل بعدی باشد؛ و بگذارید \$y\$ ارتفاعی باشد که به دیوار می‌رسد. اکنون \$y\$ به وضوح به \$x\$ وابسته است. به راحتی می‌توان دید که اگر پایۀ نردبان \$A\$ را کمی از دیوار دور کنیم، سر نردبان \$B\$ کمی پایین‌تر می‌آید. بیایید این را به زبان علمی بیان کنیم. اگر \$x\$ را به \$x + dx\$ افزایش دهیم، آنگاه \$y\$ به \$y + dy\$ تبدیل می‌شود. اگر \$dx > 0\$ باشد، آنگاه \$dy < 0\$ است؛ یعنی وقتی \$x\$ یک افزایش مثبت دریافت می‌کند، افزایشی که در \$y\$ حاصل می‌شود منفی است.

شکل ۳.۲

بله، اما چقدر؟ فرض کنید نردبان به قدری بلند بود که وقتی پایۀ آن \$A\$ در فاصله \$19\$ اینچی از دیوار قرار داشت، سر آن \$B\$ دقیقاً \$15\$ فوت از زمین فاصله داشت. اکنون، اگر پایۀ نردبان را \$1\$ اینچ بیشتر بیرون بکشید، سر نردبان چقدر پایین می‌آید؟ همه را به اینچ تبدیل کنید: \$x = 19\$ اینچ، \$y = 180\$ اینچ. اکنون افزایش \$x\$ که آن را \$dx\$ می‌نامیم، \$1\$ اینچ است؛ یعنی \$x + dx = 20\$ اینچ.

\$y\$ چقدر کاهش می‌یابد؟ ارتفاع جدید \$y + dy\$ خواهد بود. اگر ارتفاع را با استفاده از قضیه فیثاغورس محاسبه کنیم،1 آنگاه می‌توانیم مقدار \$dy\$ را بیابیم. طول نردبان برابر است با \\[\\sqrt{ (180)^2 + (19)^2 } = 181 \\text{ اینچ}.\\] بنابراین، ارتفاع جدید که \$y + dy\$ است، باید به گونه‌ای باشد که \\[\\begin{align} (y + dy)^2 &= (181)^2 - (20)^2 = 32761 - 400 = 32361, \\\\ y + dy &= \\sqrt{32361} \\approx 179.89 \\text{ اینچ}. \\end{align}\\] اکنون \$y\$ برابر \$180\$ است، بنابراین \$dy\$ تقریباً \$179.89 - 180 = -0.11\$ اینچ است.

بنابراین می‌بینیم که افزایش \$dx\$ به مقدار \$1\$ اینچ منجر به کاهش \$dy\$ به مقدار تقریباً \$0.11\$ اینچ شده است.

و نسبت \$dy\$ به \$dx\$ را می‌توان به صورت زیر بیان کرد: \\[\\frac{dy}{dx} \\approx - \\frac{0.11}{1}.\\]

همچنین به راحتی می‌توان دید که (به جز در یک موقعیت خاص) \$dy\$ اندازه‌ای متفاوت از \$dx\$ خواهد داشت.

اکنون در سراسر حساب دیفرانسیل، ما به دنبال یک چیز عجیب هستیم، یک نسبت صرف، یعنی نسبتی که \$dy\$ به \$dx\$ دارد وقتی هر دوی آن‌ها بی‌نهایت کوچک هستند.

باید در اینجا توجه داشت که ما فقط می‌توانیم این نسبت \$\\dfrac{dy}{dx}\$ را پیدا کنیم وقتی \$y\$ و \$x\$ به نوعی به یکدیگر مرتبط باشند، به طوری که هرگاه \$x\$ تغییر کند \$y\$ نیز تغییر کند. برای مثال، در اولین مثالی که گرفتیم، اگر قاعده \$x\$ مثلث بلندتر شود، ارتفاع \$y\$ مثلث نیز بیشتر می‌شود، و در مثال دوم، اگر فاصله \$x\$ پایۀ نردبان از دیوار افزایش یابد، ارتفاع \$y\$ که نردبان به آن می‌رسد به صورت متناسبی کاهش می‌یابد، ابتدا به آرامی و سپس با سرعت بیشتری هنگامی که \$x\$ بزرگتر می‌شود. در این موارد، رابطه بین \$x\$ و \$y\$ کاملاً مشخص است، می‌توان آن را به صورت ریاضی بیان کرد، به صورت \$\\dfrac{y}{x} = \\tan 30^\\circ\$ و \$x^2 + y^2 = l^2\$ (که در آن \$l\$ طول نردبان است)، و \$\\dfrac{dy}{dx}\$ معنایی را دارد که در هر مورد یافتیم.

اگر در حالی که \$x\$ همانند قبل فاصله پایۀ نردبان از دیوار است، \$y\$ به جای ارتفاعی که به آن می‌رسد، طول افقی دیوار، یا تعداد آجرهای آن، یا تعداد سال‌هایی که از ساخت آن می‌گذرد باشد، هر تغییری در \$x\$ به طور طبیعی هیچ تغییری در \$y\$ ایجاد نمی‌کند؛ در این مورد \$\\dfrac{dy}{dx}\$ هیچ معنایی ندارد و امکان یافتن عبارتی برای آن وجود ندارد. هرگاه از دیفرانسیل‌های \$dx\$، \$dy\$، \$dz\$ و غیره استفاده می‌کنیم، وجود نوعی رابطه بین \$x\$، \$y\$، \$z\$ و غیره ضمنی است، و این رابطه را \"تابع\" در \$x\$، \$y\$، \$z\$ و غیره می‌نامند؛ دو عبارت داده شده در بالا، یعنی \$\\dfrac{y}{x} = \\tan 30^\\circ\$ و \$x^2 + y^2 = l^2\$، به ترتیب توابع \$x\$ و \$y\$ هستند. چنین عبارات به صورت ضمنی (یعنی بدون نشان دادن صریح) شامل روش‌هایی برای بیان \$x\$ بر حسب \$y\$ یا \$y\$ بر حسب \$x\$ هستند، و به این دلیل آن‌ها را توابع ضمنی در \$x\$ و \$y\$ می‌نامند؛ آن‌ها را می‌توان به صورت زیر نوشت: \\[y = x \\tan 30^\\circ \\quad\\text{یا}\\quad x = \\frac{y}{\\tan 30^\\circ}\\] و \\[y = \\sqrt{ l^2 - x^2} \\quad\\text{یا}\\quad x = \\sqrt{ l^2 - y^2}.\\]

این عبارات اخیر به صورت صریح (یعنی به وضوح) مقدار \$x\$ بر حسب \$y\$ یا \$y\$ بر حسب \$x\$ را بیان می‌کنند، و به این دلیل آن‌ها را توابع صریح \$x\$ یا \$y\$ می‌نامند. برای مثال، \$x^2 + 3 = 2y - 7\$ یک تابع ضمنی در \$x\$ و \$y\$ است؛ می‌توان آن را به صورت \$y = \\dfrac{x^2 + 10}{2}\$ (تابع صریح \$x\$) یا \$x = \\sqrt{2y - 10}\$ (تابع صریح \$y\$) نوشت.2 می‌بینیم که یک تابع صریح در \$x\$، \$y\$، \$z\$ و غیره صرفاً چیزی است که مقدار آن هنگام تغییر \$x\$، \$y\$، \$z\$ و غیره تغییر می‌کند، چه یکی در یک زمان و چه چند تا با هم. به همین دلیل، مقدار تابع صریح را متغیر وابسته می‌نامند، زیرا به مقدار سایر کمیت‌های متغیر در تابع بستگی دارد؛ این متغیرهای دیگر را متغیرهای مستقل می‌نامند زیرا مقدار آن‌ها از مقدار فرضی تابع تعیین نمی‌شود. برای مثال، اگر \$u = x^2 \\sin \\theta\$، \$x\$ و \$\\theta\$ متغیرهای مستقل هستند و \$u\$ متغیر وابسته است.

گاهی اوقات رابطه دقیق بین چند کمیت \$x\$، \$y\$، \$z\$ یا معلوم نیست یا بیان آن راحت نیست؛ فقط می‌دانیم یا بیان آن راحت است که نوعی رابطه بین این متغیرها وجود دارد، به طوری که نمی‌توان \$x\$ یا \$y\$ یا \$z\$ را به تنهایی تغییر داد بدون اینکه بر سایر کمیت‌ها تأثیر بگذارد؛ وجود یک تابع در \$x\$، \$y\$، \$z\$ سپس با نماد \$F(x, y, z)\$ (تابع ضمنی) یا با \$x = F(y, z)\$، \$y = F(x, z)\$ یا \$z = F(x, y)\$ (تابع صریح) نشان داده می‌شود. گاهی اوقات از حرف \$f\$ یا \$\\phi\$ به جای \$F\$ استفاده می‌شود، بنابراین \$y = F(x)\$، \$y = f(x)\$ و \$y = \\phi(x)\$ همگی یک معنا دارند، یعنی اینکه مقدار \$y\$ به مقداری که \$x\$ به خود می‌گیرد بستگی دارد به روشی که بیان نشده است.

ما نسبت \$\\dfrac{dy}{dx}\$ را \"مشتق \$y\$ نسبت به \$x\$\" می‌نامیم. این یک نام علمی جدی برای این چیز بسیار ساده است. اما ما قرار نیست از نام‌های جدی بترسیم، وقتی خود چیزها اینقدر ساده هستند. به جای ترسیدن، به سادگی یک نفرین کوتاه بر حماقت دادن نام‌های طولانی می‌فرستیم؛ و پس از تخلیه ذهن خود، به خود چیز ساده، یعنی نسبت \$\\dfrac{dy}{dx}\$ می‌پردازیم.

در جبر معمولی که در مدرسه آموخته‌اید، همیشه به دنبال یک کمیت مجهول بودید که آن را \$x\$ یا \$y\$ می‌نامیدید؛ یا گاهی دو کمیت مجهول را به طور همزمان جستجو می‌کردید. اکنون باید یاد بگیرید که به روشی جدید شکار کنید؛ شکار اکنون نه \$x\$ است و نه \$y\$. به جای این، باید به دنبال این توله عجیب به نام \$\\dfrac{dy}{dx}\$ بروید. فرایند یافتن مقدار \$\\dfrac{dy}{dx}\$ را \"دیفرانسیل گرفتن\" می‌نامند. اما به یاد داشته باشید، آنچه مورد نیاز است، مقدار این نسبت است وقتی که هر دوی \$dy\$ و \$dx\$ خود بی‌نهایت کوچک در نظر گرفته می‌شوند. مقدار واقعی مشتق آن است که در حالت حدی به آن نزدیک می‌شود وقتی هر دوی آن‌ها به صورت بی‌نهایت کوچک در نظر گرفته می‌شوند.

اکنون بیایید یاد بگیریم چگونه به دنبال \$\\dfrac{dy}{dx}\$ برویم.

چگونه مشتقات را بخوانیم

هرگز نباید به اشتباه مبتدیانه فکر کنیم که \$dx\$ به معنای \$d\$ ضربدر \$x\$ است، زیرا \$d\$ یک عامل نیست—بلکه به معنای \"یک عنصر از\" یا \"مقداری از\" چیزی است که پس از آن می‌آید. \$dx\$ این‌گونه خوانده می‌شود: \"دی-ایکس\".

در صورتی که خواننده کسی را برای راهنمایی در این موارد نداشته باشد، می‌توان در اینجا به سادگی گفت که مشتقات به روش زیر خوانده می‌شوند. مشتق

\$\\dfrac{dy}{dx}\$ این‌گونه خوانده می‌شود: \"دی-وای بر دی-ایکس\" یا \"دی-وای روی دی-ایکس\".

\$\\dfrac{du}{dt}\$ این‌گونه خوانده می‌شود: \"دی-یو بر دی-تی\".

با مشتقات مرتبه دوم بعداً مواجه خواهیم شد. آن‌ها به این صورت هستند:

\$\\dfrac{d^2 y}{dx^2}\$؛ که این‌گونه خوانده می‌شود: \"دی-دو-وای روی دی-ایکس-به-توان-دو\"

و به این معناست که عملیات دیفرانسیل گرفتن از \$y\$ نسبت به \$x\$ دو بار انجام شده است (یا باید انجام شود).

روش دیگری برای نشان دادن اینکه یک تابع دیفرانسیل گرفته شده است، قرار دادن یک علامت پریم بر روی نماد تابع است. بنابراین اگر \$y=F(x)\$ باشد، که به این معناست که \$y\$ یک تابع نامشخص از \$x\$ است، می‌توانیم به جای \$\\dfrac{d\\bigl(F(x)\\bigr)}{dx}\$ بنویسیم \$F^\\prime(x)\$. به طور مشابه، \$F^{\\prime\\prime}(x)\$ به این معناست که تابع اصلی \$F(x)\$ دو بار نسبت به \$x\$ دیفرانسیل گرفته شده است.

قضیه فیثاغورس بیان می‌کند که در یک مثلث قائم‌الزاویه، مربع وتر برابر است با مجموع مربع‌های دو ضلع دیگر.↩︎

در اینجا فرض کرده‌ایم که \$x\$ مثبت است. با این حال، اگر به ما داده شود که \$x\$ منفی است، باید آن را به صورت \$x=-\\sqrt{2y - 10}\$ بیان کنیم.↩︎

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در سراسر حساب دیفرانسیل و انتگرال، ما با کمیت‌هایی که در حال رشد هستند و با نرخ‌های رشد سروکار داریم. همه کمیت‌ها را به دو دسته طبقه‌بندی می‌کنیم: ثابت‌ها و متغیرها. آنهایی که به عنوان مقادیر ثابت در نظر می‌گیریم و ثابت می‌نامیم، معمولاً آنها را با حروف ابتدای الفبا مانند \$a\$، \$b\$ یا \$c\$ به صورت جبری نشان می‌دهیم؛ در حالی که آنهایی را که قابلیت رشد یا (به قول ریاضیدانان) «تغییر» دارند، با حروف انتهایی الفبا مانند \$x\$، \$y\$، \$z\$، \$u\$، \$v\$، \$w\$ یا گاهی اوقات \$t\$ نشان می‌دهیم.

علاوه بر این، معمولاً همزمان با بیش از یک متغیر سروکار داریم و به نحوه وابستگی یک متغیر به متغیر دیگر فکر می‌کنیم: برای مثال، به نحوه وابستگی ارتفاعی که یک پرتابه به آن می‌رسد به زمان رسیدن به آن ارتفاع فکر می‌کنیم. یا از ما خواسته می‌شود مستطیلی با مساحت معین را در نظر بگیریم و بررسی کنیم که چگونه هر افزایش در طول آن، کاهش متناظر در عرض آن را الزامی می‌کند. یا به نحوه تغییر ارتفاع نردبان با تغییر شیب آن فکر می‌کنیم.

فرض کنید دو متغیر داریم که به یکدیگر وابسته‌اند. تغییر در یکی به دلیل این وابستگی، باعث تغییر در دیگری می‌شود. یکی از متغیرها را \$x\$ و دیگری را که به آن وابسته است \$y\$ می‌نامیم.

فرض کنید \$x\$ را تغییر می‌دهیم، یعنی آن را تغییر می‌دهیم یا تصور می‌کنیم که با اضافه کردن مقداری که آن را \$dx\$ می‌نامیم، تغییر می‌کند. بنابراین باعث می‌شویم \$x\$ به \$x + dx\$ تبدیل شود. سپس، به دلیل تغییر \$x\$، \$y\$ نیز تغییر کرده و به \$y + dy\$ تبدیل می‌شود. در اینجا \$dy\$ ممکن است در برخی موارد مثبت و در برخی دیگر منفی باشد؛ و (به جز در موارد معجزه‌آسا) اندازه آن با \$dx\$ یکسان نخواهد بود.

دو مثال را در نظر بگیرید

مثال ۳.۱. فرض کنید \$x\$ و \$y\$ به ترتیب قاعده و ارتفاع یک مثلث قائم‌الزاویه (شکل زیر) باشند که شیب ضلع دیگر آن \$30^\\circ\$ ثابت است. اگر فرض کنیم این مثلث بزرگ شود و همچنان زوایای آن مانند قبل باقی بماند، آنگاه وقتی قاعده به اندازه‌ی \$x + dx\$ رشد کند، ارتفاع به \$y + dy\$ تبدیل می‌شود. در اینجا، افزایش \$x\$ منجر به افزایش \$y\$ می‌شود. مثلث کوچک که ارتفاع آن \$dy\$ و قاعده آن \$dx\$ است، شبیه مثلث اصلی است؛ و واضح است که مقدار نسبت \$\\dfrac{dy}{dx}\$ با مقدار نسبت \$\\dfrac{y}{x}\$ یکسان است. از آنجایی که زاویه \$30^\\circ\$ است، مشاهده می‌شود که در اینجا \\[\\frac{dy}{dx} =\\tan 30^\\circ\\approx \\frac{1}{1.73}.\\]

شکل ۳.۱

مثال ۳.۲. در شکل بعدی، فرض کنید \$x\$ نشان‌دهنده فاصله افقی انتهای پایینی نردبان \$AB\$ با طول ثابت از دیوار باشد؛ و \$y\$ ارتفاعی باشد که به دیوار می‌رسد. اکنون \$y\$ به وضوح به \$x\$ بستگی دارد. به راحتی می‌توان دید که اگر انتهای پایینی \$A\$ را کمی دورتر از دیوار بکشیم، انتهای بالایی \$B\$ کمی پایین‌تر می‌آید. بیایید این را به زبان علمی بیان کنیم. اگر \$x\$ را به \$x + dx\$ افزایش دهیم، آنگاه \$y\$ به \$y + dy\$ تبدیل می‌شود. اگر \$dx>0\$ باشد، آنگاه \$dy<0\$ است. یعنی وقتی \$x\$ افزایش مثبت دریافت می‌کند، افزایشی که به \$y\$ می‌رسد منفی است.

شکل ۳.۲

بله، اما چقدر؟ فرض کنید نردبان آنقدر بلند بود که وقتی انتهای پایینی A، ۱۹ اینچ از دیوار فاصله داشت، انتهای بالایی B دقیقاً ۱۵ فوت از زمین فاصله داشت. حال، اگر انتهای پایینی را ۱ اینچ بیشتر بکشید، انتهای بالایی چقدر پایین می‌آید؟ همه چیز را به اینچ تبدیل کنید: x = 19 اینچ، y = 180 اینچ. اکنون افزایش x که آن را dx می‌نامیم، 1 اینچ است: یا x + dx = 20 اینچ.

چقدر y کاهش می‌یابد؟ ارتفاع جدید y + dy خواهد بود. اگر ارتفاع را با استفاده از قضیه فیثاغورث محاسبه کنیم،۱ می‌توانیم بفهمیم dy چقدر خواهد بود. طول نردبان \\[\\sqrt{ (180)^2 + (19)^2 } = 181 \\text{ اینچ}.\\] است. واضح است که ارتفاع جدید، که y + dy است، به گونه‌ای خواهد بود که \\[\\begin{align} (y + dy)^2 &= (181)^2 - (20)^2 = 32761 - 400 = 32361, \\\\ y + dy &= \\sqrt{32361} \\approx 179.89 \\text{ اینچ}. \\end{align}\\] اکنون y برابر با ۱۸۰ است، بنابراین dy تقریباً برابر با ۱۷۹.۸۹-۱۸۰ = -۰.۱۱ اینچ است.

بنابراین می‌بینیم که افزایش dx به میزان 1 اینچ، باعث کاهش dy به میزان تقریباً 0.11 اینچ شده است.

و نسبت dy به dx را می‌توان به صورت زیر بیان کرد: \\[\\frac{dy}{dx} \\approx - \\frac{0.11}{1}.\\]

همچنین به راحتی می‌توان دید که (به جز در یک موقعیت خاص) dy اندازه‌ای متفاوت از dx خواهد داشت.

اکنون در سراسر حساب دیفرانسیل، ما به دنبال چیزی عجیب، یک نسبت صرف، یعنی نسبتی که \$dy\$ به \$dx\$ دارد، زمانی که هر دوی آنها بی‌نهایت کوچک هستند، می‌گردیم.

لازم به ذکر است که ما فقط می‌توانیم این نسبت \$\\dfrac{dy}{dx}\$ را زمانی پیدا کنیم که \$y\$ و \$x\$ به نحوی با یکدیگر مرتبط باشند، به طوری که هر زمان که \$x\$ تغییر کند، \$y\$ نیز تغییر کند. به عنوان مثال، در مثال اول، اگر قاعده \$x\$ مثلث بلندتر شود، ارتفاع \$y\$ مثلث نیز بیشتر می‌شود، و در مثال دوم، اگر فاصله \$x\$ پای نردبان از دیوار افزایش یابد، ارتفاع \$y\$ که نردبان به آن می‌رسد به طور متناظر کاهش می‌یابد، در ابتدا به آرامی، اما هرچه \$x\$ بزرگتر می‌شود، سریعتر و سریعتر. در این موارد، رابطه بین \$x\$ و \$y\$ کاملاً مشخص است، می‌توان آن را به صورت ریاضی بیان کرد، به ترتیب \$\\dfrac{y}{x} = \\tan 30^\\circ\$ و \$x^2 + y^2 = l^2\$ (که در آن \$l\$ طول نردبان است) و \$\\dfrac{dy}{dx}\$ معنایی را دارد که در هر مورد یافتیم.

```html

گاهی اوقات رابطه دقیق بین چندین کمیت \$x\$، \$y\$، \$z\$ یا شناخته شده نیست یا بیان آن مناسب نیست؛ فقط می‌دانیم، یا بیان آن راحت است، که نوعی رابطه بین این متغیرها وجود دارد، به طوری که نمی‌توان \$x\$ یا \$y\$ یا \$z\$ را به تنهایی بدون تأثیر بر مقادیر دیگر تغییر داد؛ وجود یک تابع در \$x\$، \$y\$، \$z\$ با نماد \$F(x, y, z)\$ (تابع ضمنی) یا با \$x = F(y, z)\$، \$y = F(x, z)\$ یا \$z = F(x, y)\$ (تابع صریح) نشان داده می‌شود. گاهی اوقات به جای \$F\$ از حرف \$f\$ یا \$\\phi\$ استفاده می‌شود، به طوری که \$y = F(x)\$، \$y = f(x)\$ و \$y = \\phi(x)\$ همگی به یک معنی هستند، یعنی اینکه مقدار \$y\$ به مقدار \$x\$ به نحوی بستگی دارد که بیان نشده است.

ما نسبت \$\\dfrac{dy}{dx}\$ را «مشتق \$\\boldsymbol{y}\$ نسبت به \$\\boldsymbol{x}\$» می‌نامیم. این یک نام علمی جدی برای این چیز بسیار ساده است. اما ما از نام‌های جدی نمی‌ترسیم، وقتی خود چیزها خیلی آسان هستند. به جای ترسیدن، ما فقط یک نفرین کوتاه بر حماقت دادن نام‌های طولانی و سخت تلفظ می‌کنیم؛ و پس از رهایی ذهنمان، به سراغ خود چیز ساده، یعنی نسبت \$\\dfrac{dy}{dx}\$ می‌رویم.

در جبر معمولی که در مدرسه یاد گرفتید، همیشه به دنبال یک کمیت مجهول بودید که آن را \$x\$ یا \$y\$ می‌نامیدید؛ یا گاهی اوقات دو کمیت مجهول وجود داشت که باید همزمان به دنبال آنها می‌گشتید. اکنون باید یاد بگیرید که به روش جدیدی به شکار بروید؛ روباه اکنون نه \$x\$ است و نه \$y\$. به جای این، باید به دنبال این توله عجیب به نام \$\\dfrac{dy}{dx}\$ بگردید. فرآیند یافتن مقدار \$\\dfrac{dy}{dx}\$ «مشتق‌گیری» نامیده می‌شود. اما به یاد داشته باشید، چیزی که می‌خواهیم مقدار این نسبت است وقتی که هر دو \$dy\$ و \$dx\$ خود بی‌نهایت کوچک هستند. مقدار واقعی مشتق مقداری است که در حالت حدی به آن نزدیک می‌شود که هر یک از آنها بی‌نهایت کوچک در نظر گرفته شوند.

حالا بیایید یاد بگیریم که چگونه به دنبال \$\\dfrac{dy}{dx}\$ برویم.

نحوه خواندن مشتقات

هرگز نباید در اشتباه تازه‌کارها بیفتیم که فکر کنیم \$dx\$ به معنی \$d\$ ضربدر \$x\$ است، زیرا \$d\$ یک عامل نیست - به معنی «یک جزء از» یا «کمی از» هر چیزی است که بعد از آن می‌آید. \$dx\$ را اینگونه می‌خوانیم: «دی-اکس».

در صورتی که خواننده کسی برای راهنمایی او در چنین مواردی نداشته باشد، می‌توان به سادگی گفت که مشتقات را به روش زیر می‌خوانیم. مشتق

\$\\dfrac{dy}{dx}\$ به صورت «دی-وای به روی دی-اکس» یا «دی-وای تقسیم بر دی-اکس» خوانده می‌شود.

\$\\dfrac{du}{dt}\$ به صورت «دی-یو به روی دی-تی» خوانده می‌شود.

بعداً با مشتق دوم آشنا می‌شویم. آنها به این شکل هستند:

\$\\dfrac{d^2 y}{dx^2}\$؛ که به صورت «دی-دو-وای به روی دی-اکس-به توان دو» خوانده می‌شود،

و به این معنی است که عمل مشتق‌گیری \$y\$ نسبت به \$x\$ دو بار انجام شده است (یا باید انجام شود).

روش دیگر برای نشان دادن اینکه یک تابع مشتق‌گیری شده است، قرار دادن یک علامت پریم (') روی نماد تابع است. بنابراین اگر \$y=F(x)\$، که به این معنی است که \$y\$ یک تابع نامشخص از \$x\$ است، می‌توانیم به جای \$\\dfrac{d\\bigl(F(x)\\bigr)}{dx}\$ بنویسیم \$F^\\prime(x)\$. به طور مشابه، \$F^{\\prime\\prime}(x)\$ به این معنی است که تابع اصلی \$F(x)\$ دو بار نسبت به \$x\$ مشتق‌گیری شده است.

قضیه فیثاغورث بیان می‌کند که در یک مثلث قائم الزاویه، مربع وتر برابر است با مجموع مربعات دو ضلع دیگر.↩︎

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:::{.problem} \n1.15. An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s^2; 2.0 s after the start a bolt begins falling from the ceiling of the car. Find:\n(a) the bolt's free fall time;\n(b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

Ans.

:::

Can we add more lines here? llllhjjkjk hii i 2025-03-31

","chapter_express_content":"

This is the express version

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1. Find the dimensions of the gravitational constant $G$ in SI units. Using its value and the acceleration due to gravity, determine Earth's mass M. Earth's radius $R \\approx 6400$ km.

Answer

[$G$] = m$^3$/(kg$\\cdot$s$^2$); $M \\approx 6 \\times 10^{24}$ kg.

Hint

Near the Earth's surface, the force of gravity imparts acceleration $g$ to bodies; $mg = G mM/R^2$, whence $M = gR^2/G$, where $R$ is the radius of the Earth.

2. Calculate the gravitational force between two touching lead spheres (diameter 1 m each). Lead density $11.3 \\, \\text{g/cm}^3$.

Answer

$F \\approx 2.34 \\times 10^{-3}$ N.

3. A lead sphere (radius $R$) contains a spherical cavity that touches its surface and passes through its center. Total mass $M$. Using the law of gravitation, find the force on a small mass $m$ placed at distance $d > R$ from the sphere's center along the line connecting centers of the sphere and cavity (on the cavity side).

Answer

$F = \\frac{1}{7} G Mm \\left[\\dfrac{8}{d^2} - \\dfrac{1}{(d-R)^2}\\right]$.

Hint

The sought force is found as the difference between the forces that would act on the small sphere from the entire lead sphere and from the part that was removed from the sphere.

4. At what distance from Earth's surface is gravitational acceleration $1 \\, \\text{m/s}^2$?

Answer

$h \\approx 13600$ km.

5. Determine gravitational acceleration $g$ at height $h = 20$ km above Earth, given surface gravity $g_0 = 981 \\, \\text{cm/s}^2$ and Earth radius $R = 6400$ km.

Answer

$g \\approx 975$ cm/s$^2$.

Solution

The acceleration due to gravity is found from the formulas $g_0 = G \\frac{M}{R^2}$ and $g = G \\frac{M}{(R+h)^2}$. Hence $\\frac{g_0}{g} = (\\frac{R+h}{R})^2 = 1 + \\frac{2h}{R} + \\frac{h^2}{R^2} \\approx 1 + \\frac{2h}{R},$ since $h/R \\ll 1$, and therefore $h^2/R^2 \\ll 2h/R$. Thus, $g = g_0 \\times (\\frac{R}{R+h})^2 \\approx \\frac{g_0 R}{R+2h}.$

6. Prove that gravitational force on a mass $m$ inside Earth is $F = mgr/R_0$, where $r$ is distance from center and $R_0$ is Earth's radius. Assume constant Earth density.

Answer

Hint. Let's show that the gravitational force acting from an arbitrary spherical shell on a material point inside it is zero. For this, let's divide the spherical shell into thin spherical layers. A cone with a small solid angle with its vertex at this point cuts out elements from the layer, and it can be shown that $\\Delta m_1 / (\\Delta m_2) = r_1^2 / r_2^2.$ Therefore, the resultant gravitational forces towards these elements are equal to zero. Dividing the entire layer into such elements, we obtain the initial statement. Consequently, only the mass of that part of the Earth enclosed within the sphere of radius $r$ acts on the point.

7. A body falls through a straight shaft drilled along Earth's rotation axis. Find its maximum velocity (ignore air resistance).

Answer

$v = \\sqrt{gR}$.

Solution

If the mass of the body falling through the shaft is $m$, then the acceleration $a$ of the body depends on the distance $r$ to the center of the Earth according to the law $a =\\frac{F}{m} = \\frac{mgr/R}{m} = g \\frac{r}{R}$ where $R$ is the radius of the Earth (see problem 6).

The acceleration is maximum at $r = R$ and becomes zero at $r = 0$. This means that the speed of the body is maximum at $r = 0$. After the body passes the center of the Earth, its acceleration is directed opposite to the velocity, and the speed of the body will decrease. To find the speed of the body at $r = 0$, let's use the law of conservation of energy. Assuming that the potential energy of the body is zero at the center of the Earth, we get $mv^2/2 = \\Pi$, where $\\Pi$ is the potential energy of the body on the surface of the Earth. It is equal to the work that must be done to very slowly move the body from the center to the surface of the Earth. Let's plot the graph of the dependence of the force that needs to be applied to the body on the distance $r$ (see figure). The work done in moving the body is numerically equal to the area of the shaded triangle (see problems 4.5 and 4.17), therefore
$\\Pi = mgR/2\\ \\text{ and }\\ mv^2/2 = mgR/2.$ Hence
$v = \\sqrt{gR}.$
$\"Description\"$

8. What direction and horizontal speed must an airplane fly along the equator to compensate for reduced weight due to Earth's rotation?

Answer

In the direction from east to west with a speed $v = 464$ m/s.

Hint

The aircraft must be stationary in the coordinate system associated with the fixed stars. For this, the angular velocity of the aircraft must be equal to the angular velocity of the Earth and directed in the opposite direction. When flying over the equator $v = \\omega R_E = 2\\pi R_E / T,$ where $R_E$ is the radius of the Earth; $T$ is its period of rotation.

9. Why are space rockets typically launched west-to-east? Why are equatorial launches most advantageous?

Answer

The rotational speed of the Earth is used.

10. At a planet's equator, bodies weigh half as much as at poles. Planet density $ρ = 3 \\times 10^3 \\, \\text{kg/m}^3$. Determine its rotation period.

Answer

$T = \\sqrt{6\\pi/(\\rhoG)} \\approx 2$ h 41.6 min.

Solution

The weight $P$ of a body is the force with which the body acts on a suspension or support that is stationary relative to the body. According to Newton's third law, the support acts on the body with a force $P'$, equal and opposite to $P$. Thus, two forces are applied to bodies at rest on the planet: the force of gravity $F_T$ and the reaction force of the support $P'$. These forces impart the necessary centripetal acceleration to the body during the rotation of the planet around its axis. At the pole and the equator, these forces are directed along the same line. According to Newton's second law $mv^2/R = F_T - P',$ where $v$ is the speed of the rotational motion of the body together with the Earth around its axis. At the pole $v = 0$ and the centripetal acceleration is absent. The reaction force of the support and, consequently, the weight of the body at the pole are equal to the force of gravity: $P_0 = F_T.$ At the equator $v = 2\\pi R / T$, where $T$ is the period of revolution. The weight of the body at the equator $P_e = F_T - m \\cdot 4\\pi^2 R / T^2.$ According to the condition, $P_e = 0.5 P_0 = 0.5 F_T,$ then $4\\pi^2 m R / T^2 = 0.5 F_T.\\tag{1}$ According to the law of universal gravitation $F_T = G \\frac{mM}{R^2}.$ Here $m$ is the mass of the body; $M$ is the mass of the planet; $R$ is the radius of the planet, but $M = \\frac{4}{3} \\pi R^3 \\rho,$ where $\\rho$ is the density of the planet. Substituting the expressions for $F_T$ and $M$ into formula (1), we get $0.5 G \\frac{m \\cdot 4/3 \\pi R^3 \\rho}{R^2} = \\frac{4\\pi^2 m R}{T^2}.$ After simplification, we find $T = \\sqrt{6\\pi/(\\rho G)}.$

11. Find the average density of a planet where spring scales show 10% less weight at the equator than at poles. Day length $T = 24$ hours.

Answer

$\\rho = 30\\pi/(T^2G) \\approx 180$ kg/m$^3$.

12. How long should an Earth day be for equatorial weightlessness?

Answer

$t \\approx 1$ h 25 min.

Hint

See problem 10.

13. Derive the dependence of body weight on geographic latitude.

Solution

Two forces act on the weighed body: the force of gravity $F_T$, directed towards the center of the Earth, and the reaction force of the support $P'$ (see figure); $P'$ is numerically equal to the weight of the body. The resultant $Q$ of the forces $F_T$ and $P'$ imparts centripetal acceleration $a = \\omega^2 r$ to the body. Since $r = R \\cos \\phi$ ($\\phi$ is the geographical latitude), then $a = \\omega^2 R \\cos \\phi\\quad\\text{and}\\quad Q = m\\omega^2 R \\cos \\phi$ where $m$ is the mass of the body. In turn, for the force of gravity $F_T$, we can write $F_T = G mM/R^2 = mg_0,$ where $g_0 = G M/R^2$ ($M$ is the mass of the Earth) is a coefficient with the dimension of acceleration, numerically close to the acceleration of free fall at the pole. According to the cosine theorem from the triangle of forces, we get $P' = \\sqrt{F_T^2 + Q^2 - 2F_T Q \\cos \\phi} = F_T \\sqrt{1 + (\\frac{Q}{F_T})^2 - \\frac{2Q}{F_T} \\cos \\phi}.$ Considering that $Q/F_T \\ll 1$, we can neglect the square of this quantity in the radicand. Using the known approximation $\\sqrt{1 - \\alpha} \\approx 1 - \\alpha/2$ for $\\alpha \\ll 1$, we obtain the final expression: $P' \\approx F_T (1 - \\frac{Q}{F_T} \\cos \\phi) = m(g_0 - \\omega^2 R \\cos^2 \\phi)$. $\"Description\"$

14. Calculate the Sun-Earth mass ratio given: Moon completes 13 orbits/year; average Sun-Earth distance is 390× greater than Moon-Earth distance.

Answer

336,000 times.

Solution

The equations of motion of the Earth around the Sun and the Moon around the Earth will be: $G \\frac{m_E M_S}{R_{ES}^2} = m_E \\omega_E^2 R_{ES};$ $G \\frac{m_M m_E}{R_{ME}^2} = m_M \\omega_M^2 R_{ME},$ or $G \\frac{M_S}{R_{ES}^3} = \\omega_E^2;$ $G \\frac{m_E}{R_{ME}^3} = \\omega_M^2.$ Dividing the first equation by the second and considering that $\\omega = 2\\pi/T$, we find $\\frac{M_S}{m_E} = \\frac{R_{ES}^3}{R_{ME}^3} \\frac{T_M^2}{T_E^2}.$

15. Determine the Sun's mass using gravitational constant γ, Earth's orbital period $T$, and Earth-Sun distance $L = 1.5 \\times 10^{11}$ m.

Answer

$m_S \\approx 2 \\times 10^{30}$ kg.

Hint

See problem 7.14.

16. Can a satellite orbit in a plane not passing through Earth's center?

Answer

No. Because in this case, the force of gravity would have a component not lying in the plane of the orbit.

17. A satellite orbits Earth at height $h$ (Earth radius $R$). For circular orbit, express orbital velocity and period in terms of $h$, $R$, and surface gravity $g$.

Answer

$v = R\\sqrt{g/(R+h)}$; $T = 2\\pi \\frac{R+h}{R} \\sqrt{(R+h)/g}$.

Solution

Assume that the satellite moves in a circular orbit, the center of which is at the center of the Earth. In this case, the satellite moves with centripetal acceleration $a_{\\text{cp}} = \\frac{v^2}{R+h},$ where $v$ is the speed of the satellite in orbit; $R$ is the average radius of the Earth; $h$ is the average altitude of the satellite. This centripetal acceleration is provided to the satellite by the force of gravity $F = G \\frac{mM}{(R+h)^2},$ where $M$ is the mass of the Earth; $m$ is the mass of the satellite; $G$ is the gravitational constant. According to Newton's second law, we have $\\frac{m\\,v^2}{R+h} = G frac{mM}{(R+h)^2},$ whence $v^2 = G \\frac{M}{R+h}.$ The value $G M$ can be found from the following relation: the weight of the body on the surface of the Earth $mg = G \\frac{mM}{R^2},$ whence $G M = gR^2.$ Substituting the value $G M$ into the formula for the speed of the satellite, we get $v = R\\sqrt{\\frac{g}{R+h}}.$ The period of revolution of the satellite $T = 2\\pi/\\omega,$ where $\\omega = v/(R+h)$ is the angular velocity of the satellite in orbit. Consequently, $T = \\frac{2\\pi(R+h)}{v} = 2\\pi \\frac{R+h}{R} \\sqrt{\\frac{R+h}{g}}.$

18. Find average angular $\\omega$ and linear $v$ velocities of a satellite with 105-minute orbital period.

Answer

$\\omega = 2\\pi/T \\approx 0.001$ rad/s; $v = \\sqrt[3]{2\\pi g R_E^2 / T} \\approx 7.4 \\times 10^3$ m/s.

Solution

The radius of the orbit $R$ can be found from Newton's second law equation $mv^2/R = G \\frac{Mm}{R^2}.$ Expressing the linear velocity $v$ in terms of the angular velocity $\\omega$, we get $v = \\omega R$. Using the relation $G M = gR_E^2$ (see problem 17), we obtain $R = \\sqrt[3]{\\frac{gR_E^2}{\\omega^2}},$ and further, $v = \\omega R = \\sqrt[3]{gR_E^2 \\omega} = \\sqrt[3]{2\\pi g R_E^2 / T}.$

19. What orbital radius and velocity would give a satellite Earth's rotation period? What trajectory would it appear to have from Earth? What orbital plane would make it appear stationary?

Answer

$R = \\sqrt[3]{gR_E^2/\\omega^2} \\approx 42000$ km; $v = \\sqrt[3]{gR_E^2 \\omega} \\approx 3.1$ km/h.

Hint

See the previous problem. The angular velocity of rotation of the satellite must be equal to the angular velocity $\\omega$ of the Earth's rotation. From the Earth, it will seem that the satellite moves forward and backward along the meridian. It will appear stationary if its orbit lies in the plane of the equator.

20. What is the first cosmic velocity for a planet with twice Earth's mass and radius?

Answer

$v_1 \\approx 8$ km/s, the same as Earth's.

Hint

The first cosmic velocity is the speed of a satellite moving in a circular orbit around a planet near its surface. Therefore $mv_1^2/R = G mM/R^2,$ whence $v_1 = \\sqrt{\\frac{G M}{R}}.$ Since $M/R = M_E/R_E$, and $G M_E = gR_E^2$, we get $v_1 = \\sqrt{gR_E}.$

21. What is the first cosmic velocity for a planet with Earth's density but half its radius?

Answer

$v_1 \\approx 4$ km/s, i.e., half that of Earth.

22. A satellite required $A = 3.2 \\times 10^{10}$ J work to reach low circular orbit. Find its mass (Earth radius $R_3 = 6400$ km).

Answer

1000 kg.

Solution

Neglecting the change in the potential energy of the satellite, we can write $A = mv^2/2$. To move in an orbit with radius $R \\approx R_E$, the satellite must have a speed $v = \\sqrt{gR_E}$ (see problem 17), then $A = mgR_E/2.$ Hence $m = 2A/(gR_E).$

23. Compare energy costs for: a) lifting a satellite to $h_1 = 3200$ km vs. orbital insertion at that height; b) same for $h_2 = 6400$ km.

Answer

In the first case 1:1, in the second 2:1.

Solution

If the potential energy of a body of mass $m$ in the gravitational field is considered zero infinitely far from the Earth (mass $M$), then at a distance $R$ from the center of the Earth it should be equal to $-G Mm/R$ (this formula can be easily obtained based on the analogy of gravitational and electrostatic fields or by considering the work done when moving the body over a small segment $\\Delta R$). To lift the satellite to a height $h$ above the Earth's surface, work must be done equal to the change in the potential energy of the satellite: $A = \\Pi_{R_E+h} - \\Pi_{R_E} = -G \\frac{Mm}{R_E+h} - (-G \\frac{Mm}{R_E}) = mg \\frac{hR_E}{R_E+h}$ (since $G M/R_E^2 = g$). To launch the satellite at this altitude, it needs to be given kinetic energy $mv^2/2$, i.e., perform work $A_1 = mv^2/2.$ For the satellite to rotate in orbit at height $h$ from the Earth, it needs to be given a speed $v = R_E \\sqrt{g/(R_E+h)}$ (see problem 17), then $A_1 = \\frac{1}{2} \\frac{mgR_E^2}{R_E+h}.$ The ratio of the works is $A/A_1 = 2h/R_E.$ For $h = 3200$ km $A/A_1 = 1$, for $h = 6400$ km $A/A_1 = 2$.

24. When would an astronaut experience weightlessness during interplanetary flight?

Answer

The astronaut will be in a state of weightlessness from the moment the spacecraft engines are turned off in outer space.

Solution

Let us consider the forces acting on the spacecraft and on the bodies located within it. The spacecraft and the bodies within it always move with the same acceleration. This acceleration for the spacecraft is determined by the thrust force $\\vec{F}_T$, the drag force $\\vec{F}_d$, and the resultant gravitational force $\\vec{F}_g$, which is proportional to its mass $M$. For any body inside the spacecraft, the acceleration is determined by the normal force $\\vec{N}$ (equal in magnitude to the force exerted on the support) and the resultant gravitational force $\\vec{F}'_g$, which is proportional to the body's mass $m$: $ \\vec{a} = \\frac{\\vec{F}_T + \\vec{F}_d + \\vec{F}_g}{M} = \\frac{\\vec{N} + \\vec{F}'_g}{m} $ However, the gravitational force per unit mass (the gravitational field) is the same for both the spacecraft and the body: $ \\frac{\\vec{F}_g}{M} = \\frac{\\vec{F}'_g}{m} $ From this, it follows that: $ \\vec{N} = \\frac{m}{M}(\\vec{F}_T + \\vec{F}_d) $ Consequently, the force exerted by the body on the support (the \"apparent weight\" of the body) is: $ \\vec{P} = -\\vec{N} = -\\frac{m}{M}(\\vec{F}_T + \\vec{F}_d) $ This relationship is valid for any stage of the spacecraft's motion. It can also be used to determine the g-forces experienced by the cosmonaut during takeoff and during entry into the dense layers of the atmosphere. When the spacecraft enters outer space ($\\vec{F}_d = 0$) and its engines are shut down ($\\vec{F}_T = 0$), then $\\vec{P} = 0$. That is, all bodies on the spacecraft are in a state of weightlessness and do not experience the interactions and internal stresses associated with weight.

25. How would the operation of pendulum clocks ('walkers') and spring (wrist) watches change inside an interplanetary spacecraft?

Answer

Pendulum clocks will stop after the rocket engine is turned off. The movement of the spacecraft will not affect the running of wristwatches.

26. How can body mass be measured under weightlessness conditions?

Answer

Many methods are possible, based on the application of Newton's second and third laws. For example:

measure the ratio of the velocities acquired by bodies with known and unknown masses after burning the thread compressing a spring between the bodies;
act on the body with a known force $F$ and measure the acceleration $a$ obtained by the body in this case, then $m = F/a$;
measure the tension force of a pre-calibrated spring when rotating the body attached to this spring with a known constant angular velocity, etc.

27. Is it possible to create artificial gravity inside a spaceship?

Answer

Yes, by rotating the spacecraft or turning on the engines.

28. Does the potential energy of objects relative to Earth change when they move inside an orbiting satellite?

Answer

Yes. The force of gravity acts on bodies in a state of weightlessness (see problem 24). The change in potential energy is still given by the formula $\\Delta \\Pi = mg_h \\Delta h,$ where $g_h$ is the acceleration due to gravity at the given altitude.

29. Do Pascal's and Archimedes' laws remain valid under weightlessness conditions?

Answer

Formally, both laws are applicable, but Archimedes' law loses its meaning, since both the body and the displaced liquid are weightless.

30. How would the linear and angular velocities of a satellite change under conditions of weak atmospheric drag? Assume a circular orbit.

Answer

The linear and angular velocities of the satellite will increase.

Hint

The total energy of the satellite is $E = \\Pi + K$. The potential energy of the satellite $\\Pi = -G mM/r$ (see problem 23). We find the kinetic energy $K$ from the relation $mv^2/r = G mM/r^2$, $K = G mM/(2r)$. Thus, the total energy $E = -G mM/(2r)$. Under conditions of motion with weak friction, the work done against friction forces is produced at the expense of the total energy $E$ of the satellite; in this case, as seen from the last relation, $|E|$ must increase, which is possible if $r$ decreases, and consequently $v$ and $\\omega$ (see problem 17) increase.

31. In which case does a spacecraft experience greater heating due to atmospheric friction: during launch or during re-entry? Explain why.

Answer

When falling to Earth. When launched, its speed in the denser layers of the atmosphere near the Earth's surface is low, while during descent it is high.

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Recall that a function $f$ from a set $U$ to a set $V$ is a rule that assigns, to each $x\\in U$, one and only one element $y\\in V$. We call $x$ the independent variable and $y$ the dependent variable. We express it by writing $y=f(x)$. The sets $U$ and $V$ are called the domain and co-domain of $f$, respectively. To mention that $f$ is a function with the domain $U$ and the co-domain $V$, we write $f:U\\to V$. In the calculus of a single variable, $U$ and $V$ are subsets of $\\mathbb{R}$. In this chapter, we deal with functions where $U$ is a subset of $\\mathbb{R}^n$.

\n\n

\nDefinition 1.\n

Definition: A function $f:U\\to\\mathbb{R}$ where $U\\subseteq \\mathbb{R}^n$ is a rule that assigns one and only one real number to each point ${\\mathbf x}=(x_1,\\cdots,x_n)$ of $U$.

\n\n

In a concise fashion, we may also write $f:U\\subseteq\\mathbb{R}^n\\to\\mathbb{R}$. When $n>1$, the function $f$ is called a real-valued function of a vector variable or simply a scalar field. In most of the examples in this chapter, $n$ is 2 or 3. When $n=2$, the independent variables are often denoted by $x$ and $y$ (or sometimes by $x_1$ and $x_2$), the dependent variable by $z$, and we write $z=f(x,y)$. When $n=3$, the independent variables are denoted by $x$, $y$ and $z$ (or by $x_1$, $x_2$ and $x_3$) and the dependent variable by another letter from the end of the alphabets like $u$ or $w$. In the most general case, we write $y=f(x_1,x_2,\\cdots,x_n)$ or $y=f({\\mathbf x})$; here $y$ is the dependent variable.

\n\n

Elementary examples of multivariable functions

\n\n

Polynomials are the simplest type of functions. A polynomial function of two variables $x$ and $y$ is the sum of a finite number of terms $c x^m y^n$ (called monomials), where $m$ and $n$ are nonnegative integers and $c$ is a real number. The degree of the monomial $c x^m y^n$ is $m+n$ providing $c\\neq 0$. The degree of the monomial of three variables $cx^m y^n z^p$ is $m+n+p$ provided $c\\neq 0$. The degree of a polynomial is the highest degree of its constituting monomials. Hence, the function defined by
\\[z=-4 x^3 y^2+4 x^4+3 x y^2+7,\\]
is a polynomial of degree 5.

\n\n

Rational functions are the second simplest type of functions. A rational function is the quotient of one polynomial by another. Therefore, the general form of a rational function of two variables is
\\[R(x,y)=\\frac{P(x,y)}{Q(x,y)}\\]
where $P(x,y)$ and $Q(x,y)$ are polynomials.

\n\n

The functions that are generated by a finite number of operations addition, subtraction, multiplication, division, and raising to a fractional power are called algebraic functions, for example,
\\[z=\\sqrt[5]{\\frac{x^3-y}{x^2 y^2+y}}+\\sqrt[3]{\\frac{x+\\sqrt{xy}}{x+y^3}}.\\]

\n\n

[/expand]

\n\n

When the domain of a function $y=f(x_1,x_2,\\cdots,x_n)$ is not specified explicitly, we assume its domain is the set of all possible points in $\\mathbb{R}^n$ at which $f$ produces real values. This set is called the natural domain or simply the domain of the function. The set of all possible outputs of the function is called the range of the function.

\n\n

\nDefinition 2.\nnoitnifed wen a si ereh\n

\n\n

[su_spoiler title=\"Example 1\" open=\"yes\" style=\"fancy\"]

\n\n

Let $f(x,y)=\\frac{1}{\\sqrt{x^2-y^2}}$. Find the natural domain of $f$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]

\n\n

The expression under the square root cannot be negative, and because it appears in the denominator it cannot be zero. Thus the domain consists of all points $(x,y)$ such that $x^2-y^2>0$; that is,
\\[y^2<x^2,\\quad \\text{or}\\quad -|x|<y<|x|.\\]
The domain of $f$ is shaded in the following figure. The dashed lines in this figure are not included in the domain.

\n\n

$\"\"$

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Example 2\" open=\"yes\" style=\"fancy\"]

\n\n

Let $f(x,y)=\\ln(xy-1)$. Find the natural domain of $f$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]
The function $f$ produces real values if the argument of the logarithmic function, which is here $xy-1$, is positive. Hence, the domain of $f$ consists of all points $(x,y)$ such that $xy>1$, which is equivalent to
\\[\\begin{cases}
y>\\frac{1}{x} & \\text{if } x>0\\\\
\\\\
y<\\frac{1}{x} & \\text{ if } x<0
\\end{cases}\\]
The domain of $f$ is shaded in the following figure.

\n\n

$\"\"$

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Example 3\" open=\"yes\" style=\"fancy\"]

\n\n

Let $f(x,y)=\\arcsin(x+y)$. Find the natural domain of $f$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]

\n\n

Because $-1\\leq \\sin x\\leq 1$, $\\arcsin u$ is defined when $-1\\leq u\\leq 1$. Therefore, we must have
\\[-1\\leq x+y\\leq 1\\quad\\text{or}\\quad -1-x\\leq y\\leq 1-x;\\]
that is, $y$ must be between the two lines $y=-x+1$ and $y=-x-1$. The domain of $f$ is the shaded area between the two lines in the following figure.

\n\n

$\"\"$

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Example 4\" open=\"yes\" style=\"fancy\"]

\n\n

Let $f(x,y,z)=\\sqrt{9-x^2-y^2-z^2}$. Find the natural domain of $f$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]

\n\n

The expression under the square root must be non-negative; that is, $9-x^2-y^2-z^2\\geq 0$ or $x^2+y^2+z^2\\leq 9$. Notice that $x^2+y^2+z^2=9$ is the equation of a sphere of radius 3 centered at the origin. Therefore, the domain of $f$ consists of all points on or within this sphere (See the following figure).

\n\n

$\"\"$

\n\n

[/su_spoiler]

\n\n

[wp-svg-icons icon=\"stop-2\" wrap=\"i\"] Composition of functions

\n\n

If $g$ is a function of one variable and $f$ is a function of three variables, then the composition of $g$ and $f$, $g\\circ f$, is the function of three variables defined by
\\[g\\circ f(x,y,z)=g(f(x,y,z)).\\]
The domain of $g\\circ f$ consists of all points $(x,y,z)$ in the domain of $f$ such that $f(x,y,z)$ is in the domain of $g$. The extension of this to functions of several variables is easy. Let $f$ be a function of $n$ variables and $g$ be a function of a single variable, then
\\[g\\circ f (x_1,\\cdots,x_n)=g(f(x_1,\\cdots,x_n))\\]
and the domain of $g\\circ f$ is the set of all points $(x_1,\\cdots,x_n)$ in the domain of $f$ such that $f(x_1,\\cdots,x_n)$ is in the domain of $g$.

\n\n

[su_spoiler title=\"Example 5\" open=\"yes\" style=\"fancy\"]

\n\n

Given $g(t)=\\ln(3-t)$ and $f(x,y,z)=\\sqrt{x^2+y^2+z^2-16}$, evaluate $g\\circ f(2,-2,3)$ and find the domain of $g\\circ f$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]
\\begin{align*}
g\\circ f(2,-2,3)&=g(f(2,-2,3))\\\\
&=g\\left(\\sqrt{2^2+(-2)^2+3^2-16}\\right)\\\\
&=g(\\sqrt{1})\\\\
&=\\ln(3-1)=\\ln 2
\\end{align*}

\n\n

The domain of $f$ is the set of all points $(x,y,z)$ in $\\mathbb{R}^3$ such that $x^2+y^2+z^2-16\\geq 0$, and the domain of $g$ is the set of all $t$ in $\\mathbb{R}$ such that $3-t>0$; that is the domain of $g$ is the interval $(-\\infty, 3)$. Therefore, the domain of $g\\circ f$ is the set of all points $(x,y,z)$ in $\\mathbb{R}^3$ such that
\\[x^2+y^2+z^2\\geq 16,\\tag{i}\\]
\\[\\text{and}\\quad \\sqrt{x^2+y^2+z^2-16}<3.\\tag{ii}\\]
We can simplify (ii)
\\[\\sqrt{x^2+y^2+z^2-16}<3\\Rightarrow x^2+y^2+z^2-16<9\\]
or
\\[x^2+y^2+z^2<25.\\tag{iii}\\]
As (i) and (ii) must both hold, the domain of $g\\circ f$ is the set of all $(x,y,z)$ such that
\\[16\\leq x^2+y^2+z^2 < 25,\\]
which is the set of all points between a sphere of radius 5 and a sphere of radius 4 both centered at the origin plus the points on the latter sphere.

\n\n

[/su_spoiler]

\n\n

In a similar way, if ${\\mathbf r}:I\\subseteq \\mathbb{R}\\to \\mathbb{R}^n$ is a vector-valued function and $f:U\\subseteq \\mathbb{R}^n\\to \\mathbb{R}$, the composite function $f\\circ{\\mathbf r}$ is a function from $I\\cap \\{t|\\ {\\mathbf r}(t)\\in U \\}$ to $\\mathbb{R}$, and is defined by $f\\circ{\\mathbf r}(t)=f({\\mathbf r}(t))$.

\n\n

[caption id=\"attachment_1351\" align=\"aligncenter\" width=\"600\"] $\"\"$ Composite function $ f\\circ{\\mathbf r}(t)=f({\\mathbf r}(t))$[/caption]

\n\n

[su_spoiler title=\"Example 6\" open=\"yes\" style=\"fancy\"]

\n\n

Noting that ${\\mathbf r}(\\theta)=(\\cos \\theta,\\sin\\theta)$ for $0\\leq \\theta<2\\pi$ gives a circle of radius 1 and centered at the origin,
(a) find the distance from the points on that circle to the point $(2,2\\sqrt{3})$
(b) find the closest point on that circle to the point $(2,2\\sqrt{3})$.

\n\n

[/su_spoiler]

\n\n

[su_spoiler title=\"Solution\" style=\"fancy\"]

\n\n

(a) Distance from the point $(x,y)$ to the point $(2,2\\sqrt{3})$ is given by
\\[f(x,y)=\\sqrt{(x-2)^2+(y-2\\sqrt{3})^2};\\]
therefore, the distance of the points on the circle of radius 1 centered at the origin is
\\[f({\\mathbf r}(\\theta))=\\sqrt{(\\cos \\theta-2)^2+(\\sin \\theta -2\\sqrt{3})^2}\\]

\n\n

(b) We want to minimize the function $h=f\\circ{\\mathbf r}$, which is a function of a single variable $\\theta$. Thus, we can use the method of optimization for functions of one variable. In other words, $h$ has minimum either at its critical points, where $h'(\\theta)=0$ or at the end points where $\\theta=0$ and $\\theta=2\\pi$.

\n\n

\\[h'(\\theta)=\\frac{-2\\sin\\theta(\\cos\\theta-2)+2\\cos\\theta(\\sin\\theta-2\\sqrt{3})}{2\\sqrt{(\\cos \\theta-2)^2+(\\sin \\theta -2\\sqrt{3})^2}}\\]

\n\n

\\[h'(\\theta)=0 \\Rightarrow -2\\sin\\theta(\\cos\\theta-2)+2\\cos\\theta(\\sin\\theta-2\\sqrt{3})=0\\]

\n\n

\\[\\text{or}\\quad 4\\sin \\theta-4\\sqrt{3} \\cos\\theta=0\\]

\n\n

[Note that it follows from $-1\\leq\\cos\\theta\\leq 1$ that $ (\\cos\\theta-2)^2>0$, so the denominator of $h'(\\theta)$ is never zero and hence $h'(\\theta)$ always exists. Similarly we can argue that $h$ is always differentiable because $(\\sin\\theta-2\\sqrt{3})^2>0$.]

\n\n

Obviously when $\\theta=0$ or $\\theta=\\pi$ at which $\\cos\\theta=0$, we have $h'(\\theta)\\neq 0$. So we can assume $\\cos\\theta\\neq 0$, and divide both sides of the above equation by $\\cos\\theta$:
\\[ 4\\sin \\theta-4\\sqrt{3} \\cos\\theta=0 \\Rightarrow \\tan\\theta=\\frac{1}{\\sqrt{3}} \\Rightarrow \\theta=\\frac{\\pi}{6} \\text{ or } \\pi+\\frac{\\pi}{6}.\\]

\n\n

Therefore, we should check the points $\\theta=\\frac{\\pi}{6},\\ \\pi+\\frac{\\pi}{6},\\ 0,\\ 2\\pi$. You can verify that when $\\theta=\\frac{\\pi}{6}$, the distance is minimum and the closest point on the circle is $\\mathbf{r}(\\frac{\\pi}{6})=(\\cos\\frac{\\pi}{6},\\sin\\frac{\\pi}{6})=\\left(\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)$.

\n\n

[/su_spoiler]

\n\n

","chapter_express_content":"","public_visibility":false,"allow_part_see":false,"book_slug":"introduction-to-crystallography","slug":"functions-of-several-variables","regular_authors":"","express_authors":"","created_at":"2025-04-26T12:42:36.598Z","__v":0,"header_data":[]},{"_id":"68300ba632fc992bf36d6e10","book_id":"6647a03ea3fd815124d5695b","chapter_name":"The Integrating Factor","sequence":15,"chapter_content":"

Let \\[ Pdx + Qdy = 0 \\] be a differential equation which is not exact. The theoretical method of integrating such an equation is to find a function \$\\mu(x, y)\$ such that the expression \\[ \\mu(Pdx + Qdy) \\] is a total differential \$du\$. When \$\\mu\$ has been found the problem reduces to a mere quadrature.

The main question which arises is as to whether or not integrating factors exist. It will be proved that on the assumption that the equation itself has one and only one solution,* which depends upon one arbitrary constant, there exists an infinity of integrating factors.

Let the general solution be written in the form \\[ \\phi(x, y) = c, \\] where c is the arbitrary constant. Then, taking the differential, \\[ \\frac{\\partial\\phi}{\\partial x}dx + \\frac{\\partial\\phi}{\\partial y}dy = 0 \\] or, as it may be written, \\[ \\phi_xdx + \\phi_ydy = 0. \\] Since, therefore, \\[\\phi(x, y) = c\\] is the general solution of \\[Pdx + Qdy = 0,\\] the relation \\[\\frac{\\phi_x}{P} = \\frac{\\phi_y}{Q}\\] must hold identically, whence it follows that a function \$\\mu\$ exists such that \\[\\phi_x = \\mu P, \\quad \\phi_y = \\mu Q.\\] Consequently \\[\\mu(Pdx + Qdy) = d\\phi,\\] that is to say an integrating factor \$\\mu\$ exists.

Let \$F(\\phi)\$ be any function of \$\\phi\$, then the expression \\[\\mu F(\\phi)\\{Pdx + Qdy\\} = F(\\phi)d\\phi\\] is exact. If, therefore, \$\\mu\$ is any integrating factor, giving rise to the solution \$\\phi = c\$, then \$\\mu F(\\phi)\$ is an integrating factor. Since \$F(\\phi)\$ is an arbitrary-function of \$\\phi\$, there exists an infinity of integrating factors.

Since the equation \\[\\mu(Pdx + Qdy) = 0\\] is exact, the integrating factor satisfies the relation \\[\\frac{\\partial(\\mu P)}{\\partial y} = \\frac{\\partial(\\mu Q)}{\\partial x}\\] or \\[P\\frac{\\partial\\mu}{\\partial y} - Q\\frac{\\partial\\mu}{\\partial x} + \\mu\\left(\\frac{\\partial P}{\\partial y} - \\frac{\\partial Q}{\\partial x}\\right) = 0.\\] Thus \$\\mu\$ satisfies a partial differential equation of the first order. In general, therefore, the direct evaluation of \$\\mu\$ depends upon an equation of a more advanced character than the ordinary linear equation under consideration. It is, however, to be noted that any particular solution, and not necessarily the general solution of the partial differential equation is sufficient to furnish an integrating factor. Moreover, in many particular cases, the partial differential equation has an obvious solution which gives the required integrating factor.

As an instance, suppose \$\\mu\$ that is a function of x alone, then \\[\\frac{1}{\\mu}\\frac{d\\mu}{dx} = \\frac{1}{Q}\\left(\\frac{\\partial P}{\\partial y} - \\frac{\\partial Q}{\\partial x}\\right).\\] It is therefore necessary that the right-hand member of this equation should be independent of y. When this is the case, \$\\mu\$ is at once obtainable by a quadrature. Now suppose also that Q is unity, then P must be a linear function of y. The equation is therefore of the form \\[dy + (py - q)dx = 0,\\] where p and q are functions of x alone. The equation is therefore linear; the integrating factor, determined by the equation \\[\\frac{d\\mu}{dx} = p\\mu\\] is \\[\\mu = e^{\\int pdx}.\\] (cf. § 2-13).

An example of an equation in which an integrating factor can readily be obtained is \\[axdy + \\beta ydx + x^m y^n(axdy + bydx) = 0.\\] Consider first of all the expression \$axdy + \\beta ydx\$; an integrating factor is \$x^{\\beta-1}y^{\\alpha-1}\$ and since \\[x^{\\beta-1}y^{\\alpha-1}(axdy + \\beta ydx) = d(x^\\beta y^\\alpha),\\] the more general expression \\[x^{\\beta-1}y^{\\alpha-1}\\Phi(x^\\beta y^\\alpha)\\] is also an integrating factor. In the same way \\[x^{b-m-1}y^{a-n-1}F(x^b y^a)\\] is an integrating factor for \$x^m y^n(axdy + bydx)\$. If, therefore, \$\\Phi\$ and \$F\$ can be so determined that \\[x^{\\beta-1}y^{\\alpha-1}\\Phi(x^\\beta y^\\alpha) = x^{b-m-1}y^{a-n-1}F(x^b y^a).\\] an integrating factor for the original equation will have been obtained. Let \\[\\Phi(z) = z^\\rho, F(z) = z^r,\\] then \$x^\\lambda y^\\mu\$ will be an integrating factor if \\[\\lambda = (\\rho+1)\\beta - 1 = (r+1)b - m - 1,\\] \\[\\mu = (\\rho+1)\\alpha - 1 = (r+1)a - n - 1.\\] These equations determine \$\\rho\$ and \$r\$, and consequently \$\\lambda\$ and \$\\mu\$ if only \$a\\beta - ba \\neq 0\$. If, on the other hand, \$a = k\\alpha, b = k\\beta\$, the original equation is \\[(1 + kx^m y^n)(axdy + \\beta ydx) = 0.\\] The integrating factor is now \\[\\frac{x^{\\beta-1}y^{\\alpha-1}}{1+kx^m y^n}\\] 2.21. The Darboux Equation.—A type of equation which was investigated by Darboux is the following: * \\[-Ldy + Mdx + N(xdy - ydx) = 0.\\] where L, M, N are polynomials in x and y of maximum degree m. It will be shown that when a certain number of particular solutions of the form \\[f(x, y) = 0,\\] in which \$f(x, y)\$ is an irreducible polynomial, are known, the equation may be integrated.

Let the general solution be \\[u(x, y) = \\text{const.}\\] then the given equation is equivalent to \\[\\frac{\\partial u}{\\partial x}dx + \\frac{\\partial u}{\\partial y}dy = 0,\\] and therefore \\[L\\frac{\\partial u}{\\partial x} + M\\frac{\\partial u}{\\partial y} - N\\left(x\\frac{\\partial u}{\\partial x} + y\\frac{\\partial u}{\\partial y}\\right) = 0\\] Replace x by \$\\frac{x}{z}\$, y by \$\\frac{y}{z}\$, where z is a third independent variable, then \$u\\left(\\frac{x}{z}, \\frac{y}{z}\\right)\$ is a homogeneous rational function of x, y, z of degree zero, and by Euler’s Theorem (§ 1.232) \\[x\\frac{\\partial u}{\\partial x} + y\\frac{\\partial u}{\\partial y} + z\\frac{\\partial u}{\\partial z} = 0.\\] Moreover \$u\\left(\\frac{x}{z}, \\frac{y}{z}\\right)\$ satisfies the relation \\[A(u) \\equiv L\\frac{\\partial u}{\\partial x} + M\\frac{\\partial u}{\\partial y} + N\\frac{\\partial u}{\\partial z} = 0,\\] in which L, M, N are homogeneous polynomials in x, y, z of degree m.

The theory depends on the fact that if \\[u(x, y) = \\text{const.}\\] is a solution of the given equation, \$u\\left(\\frac{x}{z}, \\frac{y}{z}\\right)\$ is homogeneous and of degree zero, and satisfies the relation \$A(u) = 0\$. The converse is clearly also true.

Now let \\[f(x, y) = 0\\] be any particular solution, where \$f(x, y)\$ is an irreducible polynomial of degree \$h\$, and let \\[g(x, y, z) = z^h f\\left(\\frac{x}{z}, \\frac{y}{z}\\right).\\] Then, since g is homogeneous and of degree h, \\[x\\frac{\\partial g}{\\partial x} + y\\frac{\\partial g}{\\partial y} + z\\frac{\\partial g}{\\partial z} = hg.\\] Also \\[A(g) \\equiv L\\frac{\\partial g}{\\partial x} + M\\frac{\\partial g}{\\partial y} + N\\frac{\\partial g}{\\partial z}\\] \\[= z^h\\left(L\\frac{\\partial f}{\\partial x} + M\\frac{\\partial f}{\\partial y} + N\\frac{\\partial f}{\\partial z}\\right) + hz^{h-1}fN\\] \\[= hz^{-1}Ng,\\] since \$f=0\$ is a solution. This relation may be written in the form \\[A(g) = Kg,\\] since \$A(g)\$ is a polynomial of degree \$m+h-1\$ and \$g\$ is a polynomial of degree \$h\$, K is a polynomial of degree \$m-1\$.

The operator \$A\$ has the property that if \$F\$ is any function of \$u\$, \$v\$, \$w\$, …, where \$u\$, \$v\$, \$w\$, … are themselves functions of \$x, y, z\$, \\[A(F) = \\frac{\\partial F}{\\partial u}A(u) + \\frac{\\partial F}{\\partial v}A(v) + \\frac{\\partial F}{\\partial w}A(w) + \\dots\\] Let \\[f_1(x, y) = 0, f_2(x, y) = 0, \\dots, f_p(x, y) = 0\\] be particular solutions of the given equation, where \$f_r(x, y)\$ is an irreducible polynomial of degree \$h_r\$. Let \\[g_r(x, y, z) = z^{h_r}f_r\\left(\\frac{x}{z}, \\frac{y}{z}\\right) \\quad (r=1, 2, \\dots, p),\\] and consider the function \\[u(x, y, z) = \\prod_{r=1}^{p} (g_r)^{a_r}\\] where \$a_1, a_2, \\dots, a_p\$ are constants to be determined. Now \\[A(u) = \\sum_{r=1}^{p} \\frac{\\partial u}{\\partial g_r} A(g_r)\\] \\[= \\sum_{r=1}^{p} a_r g_1^{a_1} \\dots g_r^{a_r-1} \\dots g_p^{a_p} \\cdot K_r g_r\\] \\[= u \\sum_{r=1}^{p} a_r K_r,\\] where \$K_r\$ is, for every value of r, a polynomial of degree \$m-1\$. Also \$u(x, y, z)\$ is a polynomial in x, y, z of degree \$h_1a_1 + h_2a_2 + \\dots + h_pa_p\$. If \$u(x, y, z)\$ is to furnish the required solution when \$z=1\$, it must be a polynomial in x, y, z of degree zero, and must satisfy the relation \$A(u)=0\$, whence \\[h_1a_1 + h_2a_2 + \\dots + h_pa_p = 0,\\] \\[K_1a_1 + K_2a_2 + \\dots + K_pa_p = 0.\\] Each polynomial \$K_r\$ contains at most \$\\frac{1}{2}m(m+1)\$ terms, so that the last equation, being an identity in x, y, z, is equivalent to not more than \$\\frac{1}{2}m(m+1)\$ relations between the constants \$a_1, a_2, \\dots, a_p\$. There are, therefore, in all, at most \\[\\frac{1}{2}m(m+1)+1\\] equations between the p unknown constants a. Suitable values can therefore be given to these constants if the number p exceeds the number of equations, that is if \\[p > \\frac{1}{2}m(m+1)+2.\\] If, therefore, \$\\frac{1}{2}m(m+1)+2\$ particular solutions are known, the general solution can be obtained without quadratures. If \$p = \\frac{1}{2}m(m+1)+1\$ and the discriminant of the equations is zero, the same result holds. Let \$p = \\frac{1}{2}m(m+1)+1\$ and let the discriminant be not zero. In this case, let the constants be determined by the equations \\[h_1a_1 + h_2a_2 + \\dots + h_pa_p = -m-2,\\] \\[K_1a_1 + K_2a_2 + \\dots + K_pa_p = -\\left(\\frac{\\partial L}{\\partial x} + \\frac{\\partial M}{\\partial y} + \\frac{\\partial N}{\\partial z}\\right)\\] There are now \$\\frac{1}{2}m(m+1)+1\$ non-homogeneous equations which determine the constants a. This determination of the constants gives rise to a function \$u(x, y, z)\$ such that \\[x\\frac{\\partial u}{\\partial x} + y\\frac{\\partial u}{\\partial y} + z\\frac{\\partial u}{\\partial z} = -(m+2)u,\\] \\[A(u) \\equiv L\\frac{\\partial u}{\\partial x} + M\\frac{\\partial u}{\\partial y} + N\\frac{\\partial u}{\\partial z} = -\\left(\\frac{\\partial L}{\\partial x} + \\frac{\\partial M}{\\partial y} + \\frac{\\partial N}{\\partial z}\\right)u.\\] Eliminate \$\\frac{\\partial u}{\\partial z}\$ between these equations, then \\[\\left\\{L\\frac{\\partial u}{\\partial x} + M\\frac{\\partial u}{\\partial y} + \\left(\\frac{\\partial L}{\\partial x} + \\frac{\\partial M}{\\partial y} + \\frac{\\partial N}{\\partial z}\\right)u\\right\\} - \\frac{N}{z}\\left\\{x\\frac{\\partial u}{\\partial x} + y\\frac{\\partial u}{\\partial y} + (m+2)u\\right\\} = 0.\\] But since N is homogeneous and of degree m, \\[x\\frac{\\partial N}{\\partial x} + y\\frac{\\partial N}{\\partial y} + z\\frac{\\partial N}{\\partial z} = mN,\\] and therefore, eliminating \$\\frac{\\partial N}{\\partial z}\$, \\[(Lz-Nx)\\frac{\\partial u}{\\partial x} + (Mz-Ny)\\frac{\\partial u}{\\partial y} + \\left(z\\frac{\\partial L}{\\partial x} + z\\frac{\\partial M}{\\partial y} - x\\frac{\\partial N}{\\partial x} - y\\frac{\\partial N}{\\partial y} - 2N\\right)u = 0.\\] Let \$z=1\$, then \$u(x, y)\$ satisfies the equation \\[(L-Nx)\\frac{\\partial u}{\\partial x} + (M-Ny)\\frac{\\partial u}{\\partial y} + \\left\\{\\frac{\\partial(L-Nx)}{\\partial x} + \\frac{\\partial(M-Ny)}{\\partial y}\\right\\}u = 0.\\] But this is precisely the condition that \$u(x, y)\$ should be an integrating factor for the equation \\[-Ldy + Mdx + N(xdy - ydx) = 0.\\] If, therefore, \$\\frac{1}{2}m(m+1)+1\$ particular solutions are known, an integrating factor can be obtained.

To return to the Jacobi equation (§ 2-14), \\[(a_1 + b_1x + c_1y)(xdy - ydx) - (a_2 + b_2x + c_2y)dy + (a_3 + b_3x + c_3y)dx = 0.\\] In this case \$m=1\$. The equation will have a solution of the linear form \\[\\alpha x + \\beta y + \\gamma = \\text{const.}\\] \\[A(f) = (a_2z+b_2x+c_2y)\\frac{\\partial f}{\\partial x} + (a_3z+b_3x+c_3y)\\frac{\\partial f}{\\partial y} + (a_1z+b_1x+c_1y)\\frac{\\partial f}{\\partial z} = \\lambda f,\\] where \$\\lambda\$ is a constant and \$f = \\alpha x + \\beta y + \\gamma z\$. This leads to three equations between \$\\alpha, \\beta, \\gamma, \\lambda\$, namely, \\[\\gamma(a_1 - \\lambda) + \\alpha a_2 + \\beta a_3 = 0, \\quad \\gamma b_1 + \\alpha(b_2 - \\lambda) + \\beta b_3 = 0, \\quad \\gamma c_1 + \\alpha c_2 + \\beta(c_3 - \\lambda) = 0,\\] whence \\[\\begin{vmatrix} a_1-\\lambda & a_2 & a_3 \\\\ b_1 & b_2-\\lambda & b_3 \\\\ c_1 & c_2 & c_3-\\lambda \\end{vmatrix} = 0.\\] It will be assumed that this equation has three distinct roots, \$\\lambda_1, \\lambda_2, \\lambda_3\$, to which correspond three values of f, namely, U, V, W. Then \\[U^i V^j W^k = \\text{const.}\\] will be the general solution, when z is made equal to unity, if \\[i+j+k=0,\\] \\[\\lambda_1 i + \\lambda_2 j + \\lambda_3 k = 0.\\] It is sufficient to take \$i = \\lambda_2 - \\lambda_3, j = \\lambda_3 - \\lambda_1, k = \\lambda_1 - \\lambda_2\$. The general solution is therefore \\[U^{\\lambda_2-\\lambda_3} V^{\\lambda_3-\\lambda_1} W^{\\lambda_1-\\lambda_2} = \\text{const.}\\]

","chapter_express_content":"","public_visibility":false,"allow_part_see":false,"book_slug":"introduction-to-crystallography","slug":"the-integrating-factor","regular_authors":"Kamyar Davoudi, Ehsan","express_authors":"","created_at":"2025-05-14T17:04:06.907Z","__v":0,"header_data":[]},{"_id":"68470ee332fc992bf36d83e4","book_id":"6647a03ea3fd815124d5695b","chapter_name":"Irodov","sequence":16,"chapter_content":"

PART ONE

PHYSICAL FUNDAMENTALS OF MECHANICS

1.1. KINEMATICS

Average vectors of velocity and acceleration of a point:
$
\\langle \\vec{v} \\rangle = \\frac{\\Delta \\vec{r}}{\\Delta t}, \\quad \\langle \\vec{a} \\rangle = \\frac{\\Delta \\vec{v}}{\\Delta t}, \\tag{1.1a}
$
where $\\Delta \\vec{r}$ is the displacement vector (an increment of a position vector).
Velocity and acceleration of a point:
$
\\vec{v} = \\frac{d\\vec{r}}{dt}, \\quad \\vec{a} = \\frac{d\\vec{v}}{dt}. \\tag{1.1b}
$
Acceleration of a point expressed in projections on the tangent and the normal to a trajectory:
$
a_t = \\frac{dv_t}{dt}, \\quad a_n = \\frac{v^2}{R}, \\tag{1.1c}
$
where $R$ is the radius of curvature of the trajectory at the given point.
Distance covered by a point:
$
s = \\int v , dt, \\tag{1.1d}
$
where $v$ is the magnitude of the velocity vector of a point.
Angular velocity and angular acceleration of a rigid body:
$
\\vec{\\omega} = \\frac{d\\vec{\\theta}}{dt}, \\quad \\vec{\\alpha} = \\frac{d\\vec{\\omega}}{dt}. \\tag{1.1e}
$
Relation between linear and angular quantities for a rotating rigid body:
$
\\vec{v} = \\vec{\\omega}\\times \\vec{r}, \\quad a_n = \\omega^2 R, \\quad |a_t| = \\alpha R, \\tag{1.1f}
$
where $\\vec{r}$ is the position vector of the considered point relative to an arbitrary point on the rotation axis, and $R$ is the distance from the rotation axis.

:::{.problem}
1.1. A motorboat going downstream overtakes a raft at a point $A$; $ t = 60 $ min later it turns back and after some time passes the raft at a distance $ l = 6.0 $ km from the point $A$. Find the flow velocity assuming the duty of the engine to be constant.

Ans.

$ v = l/2t = 3.0 $ km per hour.

:::

:::{.problem}
1.2. A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.

Ans.

$\\langle v \\rangle = 2v_0 (v_1 + v_2)/(2v_0 + v_1 + v_2) $.

:::

:::{.problem}
1.3. A car starts moving rectilinearly, first with acceleration $ a = 5.0 , \\text{m/s}^2 $ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $ a $, comes to a stop. The total time of motion equals $ t = 25 , \\text{s} $. The average velocity during that time is equal to $ \\langle v \\rangle = 72 , \\text{km} $ per hour. How long does the car move uniformly?

Ans.

$\\Delta t = t \\sqrt{1 - 4\\langle v \\rangle / a t} = 15 \\, \\text{s} $.

:::

:::{.problem}
1.4. A point moves rectilinearly in one direction. Fig. 1.1 shows the distance $ s $ traversed by the point as a function of the time $ t $. Using the plot find: (a) the average velocity of the point during the time of motion; (b) the maximum velocity; (c) the time moment $ t_0 $ at which the instantaneous velocity is equal to the mean velocity averaged over the first $ t_0 $ seconds.

Ans.

(a) 10 cm/s; (b) 25 cm/s; (c) $ t_0 = 16 \\, \\text{s} $.

:::

:::{.problem}
1.5. Two particles, $ 1 $ and $ 2 $, move with constant velocities $ \\vec{v}_1 $ and $ \\vec{v}_2 $. At the initial moment their position vectors are equal to $ \\vec{r}_1 $ and $ \\vec{r}_2 $. How must these four vectors be interrelated for the particles to collide?

Ans.

$(\\vec{r}_1 - \\vec{r}_2)/|\\vec{r}_1 - \\vec{r}_2| = (\\vec{v}_2 - \\vec{v}_1)/|\\vec{v}_2 - \\vec{v}_1| $.

:::

:::{.problem}
1.6. A ship moves along the equator to the east with velocity $ v_0 = 30 , \\text{km/hour} $. The southeastern wind blows at an angle $ \\theta = 60^\\circ $ to the equator with velocity $ v = 45 , \\text{km/hour} $. Find the wind velocity $ v' $ relative to the ship and the angle $ \\theta' $ between the equator and the wind direction in the reference frame fixed to the ship.

Ans.

$ v' = \\sqrt{v_0^2 + v^2 + 2v_0v \\cos \\theta} \\approx 40 $ km per hour, $ \\theta' = 19^\\circ $.

:::

:::{.problem}
1.7. Two swimmers leave point $ A $ on one bank of the river to reach point $ B $ lying right across on the other bank. One of them crosses the river along the straight line $ AB $ while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point $ B $. What was the velocity $ u $ of his walking if both swimmers reached the destination simultaneously? The stream velocity $ v_0 = 2.0 , \\text{km/hour} $ and the velocity $ v' $ of each swimmer with respect to water equals 2.5 km per hour.

Ans.

$ u = \\frac{v_0}{(1 - v_0^2/v'^2)^{1/2} - 1} = 3.0 $ km per hour.

:::

:::{.problem}
1.8. Two boats, $ A $ and $ B $, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat $ A $ along the river, and the boat $ B $ across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats $ t_A / t_B $ if the velocity of each boat with respect to water is $ \\eta = 1.2 $ times greater than the stream velocity.

Ans.

$ t_A/t_B = \\eta / \\sqrt{\\eta^2 - 1} = 1.8 $.

:::

:::{.problem}
1.9. A boat moves relative to water with a velocity which is $ n = 2.0 $ times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?

Ans.

$ \\theta = \\arcsin (1/n) + \\pi / 2 = 120^\\circ $.

:::

:::{.problem}
1.10. Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of $ \\theta = 60^\\circ $ to the horizontal. The initial velocity of each body is equal to $ v_0 = 25 , \\text{m/s} $. Neglecting the air drag, find the distance between the bodies $ t = 1.70 , \\text{s} $ later.

Ans.

$ l = v_0 t \\sqrt{2(1 - \\sin \\theta)} = 22 \\, \\text{m} $.

:::

:::{.problem}
1.11. Two particles move in a uniform gravitational field with an acceleration $ g $. At the initial moment the particles were located at one point and moved with velocities $ v_1 = 3.0 , \\text{m/s} $ and $ v_2 = 4.0 , \\text{m/s} $ horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Ans.

$ l = (v_1 + v_2) \\sqrt{v_1 v_2 / g} = 2.5 \\, \\text{m} $.

:::

:::{.problem}
1.12. Three points are located at the vertices of an equilateral triangle whose side equals $ a $. They all start moving simultaneously with velocity $ v $ constant in magnitude, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

Ans.

$ t = 2a/3v $.

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:::{.problem}
1.13. Point $ A $ moves uniformly with velocity $ \\vec{v} $ so that the vector $ \\vec{v} $ is continually \"aimed\" at point $ B $ which in its turn moves rectilinearly and uniformly with velocity $ u < v $. At the initial moment of time $ \\vec{v} \\perp \\vec{u} $ and the points are separated by a distance $ l $. How soon will the points converge?

Ans.

It is seen from Fig. 1a that the points A and B converge with velocity $ v - u \\cos \\alpha $, where the angle $ \\alpha $ varies with time. The points merge provided the following two conditions are met: $ \\begin{cases} \\int_{0}^{t} (v - u \\cos \\alpha) dt = l, \\\\ \\int_{0}^{t} v \\cos \\alpha dt = u t, \\end{cases} $ where $ t $ is the sought time. It follows from these two equations that: $ t = v l / (v^2 - u^2). $

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:::{.problem}
1.14. A train of length $ l = 350 , \\text{m} $ starts moving rectilinearly with constant acceleration $ a = 3.0 \\cdot 10^{-2} , \\text{m/s}^2 $; $ t = 30 , \\text{s} $ after the start the locomotive headlight is switched on (event $ 1 $), and $ \\Delta t = 60 , \\text{s} $ after that event the tail signal light is switched on (event $ 2 $). Find the distance between these events in the reference frames fixed to the train and to the Earth. How and at what constant velocity $ V $ relative to the Earth must a certain reference frame $ K $ move for the two events to occur in it at the same point?

Ans.

$ x_1 - x_2 = l - a \\Delta t (t + \\Delta t / 2) = 0.24 $ km. Toward the train with velocity $ V = 4.0 $ m/s.

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:::{.problem}
1.15. An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s$^2$; 2.0 s after the start a bolt begins falling from the ceiling of the car. Find:
(a) the bolt's free fall time;
(b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

Ans.

(a) 0.7 s; (b) 0.7 and 1.3 m respectively.

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:::{.problem}
1.16. Two particles, $ 1$ and $ 2$, move with constant velocities $ v_1$ and $ v_2$ along two mutually perpendicular straight lines toward the intersection point $ O$. At the moment $ t = 0$ the particles were located at the distances $ l_1$ and $ l_2$ from the point $ O$. How soon will the distance between the particles become the smallest? What is it equal to?

Ans.

$ t_m = \\frac{v_1 l_1 + v_2 l_2}{v_1^2 + v_2^2}, \\quad l_{min} = \\frac{|l_1 v_2 - l_2 v_1|}{\\sqrt{v_1^2 + v_2^2}}$.

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:::{.problem}
1.17. From point $ A$ located on a highway (Fig. 1.2) one has to get by car as soon as possible to point $ B$ located in the field at a distance $ l$ from the highway. It is known that the car moves in the field $ \\eta$ times slower than on the highway. At what distance from point $ D$ one must turn off the highway?

Ans.

$ CD = l / \\sqrt{\\eta^2 - 1}$.

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:::{.problem}
1.18. A point travels along the $ x$ axis with a velocity whose projection $ v_x$ is presented as a function of time by the plot in Fig. 1.3. Assuming the coordinate of the point $ x = 0$ at the moment $ t = 0$, draw the approximate time dependence plots for the acceleration $ a_x$, the $ x$ coordinate, and the distance covered $ s$.

Ans.

See Fig. 1b.

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:::{.problem}
1.19. A point traversed half a circle of radius $ R = 160$ cm during time interval $ t = 10.0 , \\text{s}$. Calculate the following quantities averaged over that time:
(a) the mean velocity $ \\langle v \\rangle$;
(b) the magnitude of the mean velocity vector $ |\\langle v \\rangle|$;
(c) the magnitude of the mean vector of the total acceleration $ |\\langle a \\rangle|$ if the point moved with constant tangential acceleration.

Ans.

(a) $\\langle v \\rangle = \\pi R / t = 50 \\, \\text{cm/s};$ (b) $|\\langle v \\rangle| = 2R / t = 32 \\, \\text{cm/s};$ (c) $|\\langle a \\rangle| = 2\\pi R / t^2 = 10 \\, \\text{cm/s}^2.$

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:::{.problem}
1.20. A position vector of a particle varies with time $ t$ as $ \\vec{r} = -\\vec{u}t(1 - ct)$, where $ \\vec{u}$ is a constant vector and $ c$ is a positive factor. Find:
(a) the velocity $ \\vec{v}$ and the acceleration $ \\vec{a}$ of the particle as functions of time;
(b) the time interval $ \\Delta t$ taken by the particle to return to the initial points, and the distance $ s$ covered during that time.

Ans.

(a) $\\vec{v} = \\vec{u}(1 - 2ct), \\quad \\vec{a} = -2c\\vec{u} = \\text{const};$ (b) $\\Delta t = 1/c,\\quad s = u/2c.$

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:::{.problem}
1.21. At the moment $ t = 0$ a particle leaves the origin and moves in the positive direction of the $ x$ axis. Its velocity varies with time as $ v = v_0(1 - t/\\tau)$, where $ v_0$ is the initial velocity vector whose magnitude equals $ v_0 = 10.0 , \\text{cm/s}; , \\tau = 5.0 , \\text{s}$. Find:
(a) the $ x$ coordinate of the particle at the moments of time 6.0, 10, and 20 s;
(b) the moments of time when the particle is at the distance 10.0 cm from the origin;
(c) the distance $ s$ covered by the particle during the first 4.0 and 8.0 s; draw the approximate plot $ s(t)$.

Ans.

(a) $ x = v_0 t (1 - t/2\\tau), x = 0.24, 0. and -4.0 \\, \\text{m};$ (b) 1.1, 9 and 11 s; (c) $ s = \\left\\{ \\begin{array}{ll} (1 - t/2\\tau) v_0 t & \\text{for } t \\leq \\tau, \\\\ [1 + (1 - t/\\tau)^2] v_0 t/2 & \\text{for } t \\geq \\tau. \\end{array} \\right.$ 24 and 34 cm respectively.

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:::{.problem}
1.22. The velocity of a particle moving in the positive direction of the $ x$ axis varies as $ v = \\alpha \\sqrt{x}$, where $ \\alpha$ is a positive constant. Assuming that at the moment $ t = 0$ the particle was located at the point $ x = 0$, find:
(a) the time dependence of the velocity and the acceleration of the particle;
(b) the mean velocity of the particle averaged over the time that the particle takes to cover the first $ s$ meters of the path.

Ans.

(a) $ v = \\alpha^2 t/2, \\quad a = \\alpha^2/2;$ (b) $\\langle v \\rangle = \\alpha \\sqrt{s}/2.$

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:::{.problem}
1.23. A point moves rectilinearly with deceleration whose magnitude depends on the velocity $ v$ of the particle as $ a = a\\sqrt{v}$, where $ a$ is a positive constant. At the initial moment the velocity of the point is equal to $ v_0$. What distance will it traverse before it stops? What time will it take to cover that distance?

Ans.

(a) $ s = (2/a) v_0^{3/2};$ (b) $ t = 2 \\sqrt{v_0/a}.$

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:::{.problem}
1.24. (continued) (c) the time dependence of the angle $ \\alpha$ between the vectors $ \\vec{a}$ and $ \\vec{v}$;
(d) the mean velocity vector averaged over the first $ t$ seconds of motion, and the magnitude of this vector.

Ans.

(a) $ y = -\\frac{b}{a^2}x^2;$ (b) $ \\vec{v} = a\\vec{i} - 2bt\\vec{j}, \\quad \\vec{a} = -2b\\vec{j}, \\quad v = \\sqrt{a^2 + 4b^2t^2}, \\quad a = 2b;$ (c) $ \\tan \\alpha = \\frac{a}{2bt};$ (d) $ \\langle \\vec{v} \\rangle = a\\vec{i} - bt\\vec{j}, \\quad |\\langle \\vec{v} \\rangle| = \\sqrt{a^2 + b^2t^2}.$

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:::{.problem}
1.25. A point moves in the plane $ xy$ according to the law $ x = at$, $ y = at (1 - \\alpha t)$, where $ a$ and $ \\alpha$ are positive constants, and $ t$ is time. Find:
(a) the equation of the point's trajectory $ y(x)$; plot this function;
(b) the velocity $ \\vec{v}$ and the acceleration $ \\vec{a}$ of the point as functions of time;
(c) the moment $ t_0$ at which the velocity vector forms an angle $ \\pi/4$ with the acceleration vector.

Ans.

(a) $ y = x - \\frac{\\alpha}{a}x^2;$ (b) $ \\vec{v} = a\\vec{i} + a(1 - 2\\alpha t)\\vec{j}, \\quad \\vec{a} = -2\\alpha a\\vec{j};$ (c) $ t_0 = \\frac{1}{\\alpha}.$

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:::{.problem}
1.26. A point moves in the plane $ xy$ according to the law $ x = a \\sin \\omega t$, $ y = a (1 - \\cos \\omega t)$, where $ a$ and $ \\omega$ are positive constants. Find:
(a) the distance $ s$ traversed by the point during the time $ \\tau$;
(b) the angle between the point's velocity and acceleration vectors.

Ans.

(a) $ s = a\\omega \\tau;$ (b) $ \\pi/2.$

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:::{.problem}
1.27. A particle moves in the plane $ xy$ with constant acceleration $ \\vec{a}$ directed along the negative $ y$ axis. The equation of motion of the particle has the form $ y = ax - bx^2$, where $ a$ and $ b$ are positive constants. Find the velocity of the particle at the origin of coordinates.

Ans.

$ v_0 = \\sqrt{\\frac{(1 + a^2)g}{2b}}.$

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:::{.problem}
1.28. A small body is thrown at an angle to the horizontal with the initial velocity $ \\vec{v}_0$. Neglecting the air drag, find:
(a) the displacement of the body as a function of time $ \\vec{r}(t)$;
(b) the mean velocity vector $ \\langle \\vec{v} \\rangle$ averaged over the first $ t$ seconds and over the total time of motion.

Ans.

(a) $ \\vec{r} = \\vec{v}_0 t + \\frac{1}{2}\\vec{g}t^2;$ (b) $ \\langle \\vec{v} \\rangle_t = \\vec{v}_0 + \\frac{1}{2}\\vec{g}t, \\quad \\langle \\vec{v} \\rangle = \\vec{v}_0 - \\frac{\\vec{g}(\\vec{v}_0 \\cdot \\vec{g})}{g^2}.$

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:::{.problem}
1.29. A body is thrown from the surface of the Earth at an angle $ \\alpha$ to the horizontal with the initial velocity $ \\vec{v}_0$. Assuming the air drag to be negligible, find:
(a) the time of motion;
(b) the maximum height of ascent and the horizontal range; at what value of the angle $ \\alpha$ they will be equal to each other;
(c) the equation of trajectory $ y(x)$, where $ y$ and $ x$ are displacements of the body along the vertical and the horizontal respectively;
(d) the curvature radii of trajectory at its initial point and at its peak.

Ans.

(a) $ \\tau = \\frac{2v_0 \\sin \\alpha}{g};$ (b) $ h = \\frac{v_0^2 \\sin^2 \\alpha}{2g}, \\quad l = \\frac{v_0^2 \\sin 2\\alpha}{g}, \\quad \\alpha = 76^\\circ;$ (c) $ y = x \\tan \\alpha - \\frac{g}{2v_0^2 \\cos^2 \\alpha}x^2;$ (d) $ R_1 = \\frac{v_0^2}{g \\cos \\alpha}, \\quad R_2 = \\frac{v_0^2 \\cos^2 \\alpha}{g}.$

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:::{.problem}
1.30. Using the conditions of the foregoing problem, draw the approximate time dependence of magnitudes of the normal $ a_n$ and tangential $ a_t$ acceleration vectors, as well as of the projection of the total acceleration vector $ a_v$ on the velocity vector direction.

Ans.

See Fig. 2.

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:::{.problem}
1.31. A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle $ \\alpha$ with the horizontal. Having fallen the distance $ h$, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time?

Ans.

$ l = 8h \\sin \\alpha.$

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:::{.problem}
1.32. A cannon and a target are 5.10 km apart and located at the same level. How soon will the shell launched with the initial velocity 240 m/s reach the target in the absence of air drag?

Ans.

0.41 or 0.71 min later, depending on the initial angle.

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:::{.problem}
1.33. A cannon fires successively two shells with velocity $ v_0 = 250 , \\text{m/s}$; the first at the angle $ \\theta_1 = 60^\\circ$ and the second at the angle $ \\theta_2 = 45^\\circ$ to the horizontal, the azimuth being the same. Neglecting the air drag, find the time interval between firings leading to the collision of the shells.

Ans.

$ \\Delta t = \\frac{2v_0}{g} \\frac{\\sin (\\theta_1 - \\theta_2)}{\\cos \\theta_1 + \\cos \\theta_2} = 11 \\, \\text{s}.$

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:::{.problem}
1.34. A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to $ v_0$. Due to the wind the balloon gathers the horizontal velocity component $ v_x = ay$, where $ a$ is a constant and $ y$ is the height of ascent. Find how the following quantities depend on the height of ascent:
(a) the horizontal drift of the balloon $ x(y)$;
(b) the total, tangential, and normal accelerations of the balloon.

Ans.

(a) $ x = \\frac{a}{2v_0} y^2;$ (b) $ a = a v_0, \\quad a_t = \\frac{a^2 y}{\\sqrt{1 + (a y / v_0)^2}}, \\quad a_n = \\frac{a v_0}{\\sqrt{1 + (a y / v_0)^2}}.$

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:::{.problem}
1.35. A particle moves in the plane $ xy$ with velocity $ \\vec{v} = a\\vec{i} + b x \\vec{j}$, where $ \\vec{i}$ and $ \\vec{j}$ are the unit vectors of the $ x$ and $ y$ axes, and $ a$ and $ b$ are constants. At the initial moment of time the particle was located at the point $ x = y = 0$. Find:
(a) the equation of the particle's trajectory $ y(x)$;
(b) the curvature radius of trajectory as a function of $ x$.

Ans.

(a) $ y = \\frac{b}{2a} x^2;$ (b) $ R = \\frac{v^2}{a_n} = \\frac{v^2}{\\sqrt{a^2 - a_t^2}} = \\frac{a}{b} \\left[1 + \\left(\\frac{b x}{a}\\right)^2\\right]^{3/2}.$

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:::{.problem}
1.36. A particle $ A$ moves in one direction along a given trajectory with a tangential acceleration $ a_t = \\vec{a} \\cdot \\vec{\\tau}$, where $ \\vec{a}$ is a constant vector coinciding in direction with the $ x$ axis (Fig. 1.4), and $ \\vec{\\tau}$ is a unit vector coinciding in direction with the velocity vector at a given point. Find how the velocity of the particle depends on $ x$ provided that its velocity is negligible at the point $ x = 0$.

Ans.

$ v = \\sqrt{2a x}.$

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:::{.problem}
1.37. A point moves along a circle with a velocity $ v = a t$, where $ a = 0.50 , \\text{m/s}^2$. Find the total acceleration of the point at the moment when it covered the $ n$-th ($ n = 0.10$) fraction of the circle after the beginning of motion.

Ans.

$ a = a \\sqrt{1 + (4\\pi n)^2} = 0.8 \\, \\text{m/s}^2.$

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:::{.problem}
1.38. A point moves with deceleration along the circle of radius $ R$ so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment $ t = 0$ the velocity of the point equals $ v_0$. Find:
(a) the velocity of the point as a function of time and as a function of the distance covered $ s$;
(b) the total acceleration of the point as a function of velocity and the distance covered.

Ans.

(a) $ v = \\frac{v_0}{1 + v_0 t / R} = v_0 e^{-s / R};$ (b) $ a = \\frac{\\sqrt{2} v_0^2}{R e^{2s / R}} = \\frac{\\sqrt{2} v^2}{R}.$

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:::{.problem}
1.39. A point moves along an arc of a circle of radius $ R$. Its velocity depends on the distance covered $ s$ as $ v = a \\sqrt{s}$, where $ a$ is a constant. Find the angle $ \\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $ s$.

Ans.

$ \\tan \\alpha = \\frac{2s}{R}.$

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:::{.problem}
1.40. A particle moves along an arc of a circle of radius $ R$ according to the law $ l = a \\sin \\omega t$, where $ l$ is the displacement from the initial position measured along the arc, and $ a$ and $ \\omega$ are constants. Assuming $ R = 1.00 , \\text{m}, , a = 0.80 , \\text{m}, , \\text{and} , \\omega = 2.00 , \\text{rad/s}$, find:
(a) the magnitude of the total acceleration of the particle at the points $ l = 0$ and $ l = \\pm a$;
(b) the minimum value of the total acceleration $ a_{min}$ and the corresponding displacement $ l_m$.

Ans.

(a) $ a_0 = \\frac{a^2 \\omega^2}{R} = 2.6 \\, \\text{m/s}^2, \\quad a_a = a \\omega^2 = 3.2 \\, \\text{m/s}^2;$ (b) $ a_{min} = a \\omega^2 \\sqrt{1 - \\left(\\frac{R}{2a}\\right)^2} = 2.5 \\, \\text{m/s}^2, \\quad l_m = \\pm a \\sqrt{1 - \\frac{R^2}{2a^2}} = \\pm 0.37 \\, \\text{m}.$

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:::{.problem}
1.41. A point moves in the plane so that its tangential acceleration $ a_t = a$, and its normal acceleration $ a_n = b t^4$, where $ a$ and $ b$ are positive constants, and $ t$ is time. At the moment $ t = 0$ the point was at rest. Find how the curvature radius $ R$ of the point's trajectory and the total acceleration $ a$ depend on the distance covered $ s$.

Ans.

$ R = \\frac{a^3}{2b s}, \\quad a = a \\sqrt{1 + \\left(\\frac{4b s^2}{a^3}\\right)^2}.$

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:::{.problem}
1.42. A particle moves along the plane trajectory $ y(x)$ with velocity $ v$ whose magnitude is constant. Find the acceleration of the particle at the point $ x = 0$ and the curvature radius of the trajectory at that point if the trajectory has the form:
(a) of a parabola $ y = a x^2$;
(b) of an ellipse $ \\left(\\frac{x}{a}\\right)^2 + \\left(\\frac{y}{b}\\right)^2 = 1$; $ a$ and $ b$ are constants here.

Ans.

(a) $ a = 2a v^2, \\quad R = \\frac{1}{a};$ (b) $ a = \\frac{b v^2}{a^2}, \\quad R = \\frac{a^2}{b}.$

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:::{.problem}
1.43. A particle $ A$ moves along a circle of radius $ R = 50 , \\text{cm}$ so that its position vector $ \\vec{r}$ relative to the point $ O$ (Fig. 1.5) rotates with the constant angular velocity $ \\omega = 0.40 , \\text{rad/s}$. Find the magnitude of the velocity of the particle, and the magnitude and direction of its total acceleration.

Ans.

$ v = 2R \\omega = 0.40 \\, \\text{m/s}, \\quad a = 4R \\omega^2 = 0.32 \\, \\text{m/s}^2.$

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:::{.problem}
1.44. A wheel rotates around a stationary axis so that the rotation angle $ \\phi$ varies with time as $ \\phi = k t^2$, where $ k = 0.20 , \\text{rad/s}^2$. Find the total acceleration $ a$ of the point $ A$ at the rim at the moment $ t = 2.5 , \\text{s}$ if the linear velocity of the point $ A$ at this moment $ v = 0.65 , \\text{m/s}$.

Ans.

$ a = \\frac{v}{t} \\sqrt{1 + 4a^2 t^4} = 0.7 \\, \\text{m/s}^2.$

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:::{.problem}
1.45. A shell acquires the initial velocity $ v = 320 , \\text{m/s}$, having made $ n = 2.0$ turns inside the barrel whose length is equal to $ l = 2.0 , \\text{m}$. Assuming that the shell moves inside the barrel with a uniform acceleration, find the angular velocity of its axial rotation at the moment when the shell escapes the barrel.

Ans.

$ \\omega = \\frac{2\\pi n v}{l} = 2.0 \\cdot 10^3 \\, \\text{rad/s}.$

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:::{.problem}
1.46. A solid body rotates about a stationary axis according to the law $ \\phi = a t - b t^3$, where $ a = 6.0 , \\text{rad/s}$ and $ b = 2.0 , \\text{rad/s}^3$. Find:
(a) the mean values of the angular velocity and angular acceleration averaged over the time interval between $ t = 0$ and the complete stop;
(b) the angular acceleration at the moment when the body stops.

Ans.

(a) $ \\langle \\omega \\rangle = \\frac{2a}{3} = 4 \\, \\text{rad/s}, \\quad \\langle \\beta \\rangle = \\sqrt{3ab} = 6 \\, \\text{rad/s}^2;$ (b) $ \\beta = 2\\sqrt{3ab} = 12 \\, \\text{rad/s}^2.$

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:::{.problem}
1.47. A solid body starts rotating about a stationary axis with an angular acceleration $ \\beta = a t$, where $ a = 2.0 \\cdot 10^{-2} , \\text{rad/s}^3$. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle $ \\alpha = 60^\\circ$ with its velocity vector?

Ans.

$ t = \\sqrt[3]{\\frac{4}{a} \\tan \\alpha} = 7 \\, \\text{s}.$

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:::{.problem}
1.48. A solid body rotates with deceleration about a stationary axis with an angular deceleration $ \\alpha \\propto \\sqrt{\\omega}$, where $ \\omega$ is its angular velocity. Find the mean angular velocity of the body averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to $ \\omega_0$.

Ans.

$ \\langle \\omega \\rangle = \\frac{\\omega_0}{3}.$

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:::{.problem}
1.49. A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle $ \\phi$ as $ \\omega = \\omega_0 - k \\phi$, where $ \\omega_0$ and $k$ are positive constants. At the moment $ t = 0$ the angle $ \\phi = 0$. Find the time dependence of:
(a) the rotation angle;
(b) the angular velocity.

Ans.

(a) $ \\phi = \\frac{\\omega_0}{k} (1 - e^{-kt});$ (b) $ \\omega = \\omega_0 e^{-k t}.$

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:::{.problem}
1.50. A solid body starts rotating about a stationary axis with an angular acceleration $ \\beta = \\beta_0 \\cos \\phi$, where $ \\beta_0$ is a constant vector and $ \\phi$ is an angle of rotation from the initial position. Find the angular velocity of the body as a function of the angle $ \\phi$. Draw the plot of this dependence.

Ans.

$ \\omega_z = \\pm \\sqrt{2 \\beta_0} \\sin \\phi,$ see Fig. 3.

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:::{.problem}
1.51. A rotating disc (Fig. 1.6) moves in the positive direction of the $ x$ axis. Find the equation $ y(x)$ describing the position of the instantaneous axis of rotation, if at the initial moment the axis $ C$ of the disc was located at the point $ O$ after which it moved:
(a) with a constant velocity $ v$, while the disc started rotating counterclockwise with a constant angular acceleration $\\alpha$ (the initial angular velocity is equal to zero);
(b) with a constant acceleration $ a$, while the disc started rotating counterclockwise with a constant angular velocity $ \\omega$.

Ans.

(a) $ y = \\frac{v^2}{\\alpha x}$ (hyperbola); (b) $ y = \\frac{\\sqrt{2a x}}{\\omega}$ (parabola).

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:::{.problem}
1.15. An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration $a = 1.2 , \\text{m/s}^2$; 2.0 s after the start a bolt begins falling from the ceiling of the car. Find:
(a) the bolt's free fall time;
(b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

Ans.

(a) 0.7 s; (b) 0.7 and 1.3 m respectively.

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:::{.problem}
1.16. Two particles, $1$ and $2$, move with constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines toward the intersection point $O$. At the moment $t = 0$ the particles were located at the distances $l_1$ and $l_2$ from the point $O$. How soon will the distance between the particles become the smallest? What is it equal to?

Ans.

$t_m = \\frac{v_1 l_1 + v_2 l_2}{v_1^2 + v_2^2}, \\quad l_{min} = \\frac{|l_1 v_2 - l_2 v_1|}{\\sqrt{v_1^2 + v_2^2}}$.

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:::{.problem}
1.17. From point $A$ located on a highway (Fig. 1.2) one has to get by car as soon as possible to point $B$ located in the field at a distance $l$ from the highway. It is known that the car moves in the field $\\eta$ times slower than on the highway. At what distance from point $D$ one must turn off the highway?

Ans.

$CD = l / \\sqrt{\\eta^2 - 1}$.

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:::{.problem}
1.18. A point travels along the $x$ axis with a velocity whose projection $v_x$ is presented as a function of time by the plot in Fig. 1.3. Assuming the coordinate of the point $x = 0$ at the moment $t = 0$, draw the approximate time dependence plots for the acceleration $a_x$, the $x$ coordinate, and the distance covered $s$.

Ans.

See Fig. 1b.

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:::{.problem}
1.19. A point traversed half a circle of radius $R = 160$ cm during time interval $t = 10.0 , \\text{s}$. Calculate the following quantities averaged over that time:
(a) the mean velocity $\\langle v \\rangle$;
(b) the magnitude of the mean velocity vector $|\\langle v \\rangle|$;
(c) the magnitude of the mean vector of the total acceleration $|\\langle a \\rangle|$ if the point moved with constant tangential acceleration.

Ans.

(a) $\\langle v \\rangle = \\pi R / t = 50 \\, \\text{cm/s};$ (b) $|\\langle v \\rangle| = 2R / t = 32 \\, \\text{cm/s};$ (c) $|\\langle a \\rangle| = 2\\pi R / t^2 = 10 \\, \\text{cm/s}^2.$

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:::{.problem}
1.20. A position vector of a particle varies with time $t$ as $\\vec{r} = -\\vec{k}t(1 - ct)$, where $\\vec{k}$ is a constant vector and $c$ is a positive factor. Find:
(a) the velocity $\\vec{v}$ and the acceleration $\\vec{a}$ of the particle as functions of time;
(b) the time interval $\\Delta t$ taken by the particle to return to the initial points, and the distance $s$ covered during that time.

Ans.

(a) $\\vec{v} = \\vec{k}(1 - 2ct), \\quad \\vec{a} = -2c\\vec{k} = \\text{const};$ (b) $\\Delta t = 1/c, s = k/2c.$

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:::{.problem}
1.21. At the moment $t = 0$ a particle leaves the origin and moves in the positive direction of the $x$ axis. Its velocity varies with time as $v = v_0(1 - t/\\tau)$, where $v_0$ is the initial velocity vector whose magnitude equals $v_0 = 10.0 , \\text{cm/s}; , \\tau = 5.0 , \\text{s}$. Find:
(a) the $x$ coordinate of the particle at the moments of time 6.0, 10, and 20 s;
(b) the moments of time when the particle is at the distance 10.0 cm from the origin;
(c) the distance $s$ covered by the particle during the first 4.0 and 8.0 s; draw the approximate plot $s(t)$.

Ans.

(a) $x = v_0 t (1 - t/2\\tau), x = 0.24, 0. and -4.0 \\, \\text{m};$ (b) 1.1, 9 and 11 s; (c) $s = \\left\\{ \\begin{array}{ll} (1 - t/2\\tau) v_0 t & \\text{for } t \\leq \\tau, \\\\ [1 + (1 - t/\\tau)^2] v_0 t/2 & \\text{for } t \\geq \\tau. \\end{array} \\right.$ 24 and 34 cm respectively.

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:::{.problem}
1.22. The velocity of a particle moving in the positive direction of the $x$ axis varies as $v = k \\sqrt{x}$, where $k$ is a positive constant. Assuming that at the moment $t = 0$ the particle was located at the point $x = 0$, find:
(a) the time dependence of the velocity and the acceleration of the particle;
(b) the mean velocity of the particle averaged over the time that the particle takes to cover the first $s$ meters of the path.

Ans.

(a) $v = k^2 t/2, \\quad a = k^2/2;$ (b) $\\langle v \\rangle = k \\sqrt{s}/2.$

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:::{.problem}
1.23. A point moves rectilinearly with deceleration whose magnitude depends on the velocity $v$ of the particle as $a = k \\sqrt{v}$, where $k$ is a positive constant. At the initial moment the velocity of the point is equal to $v_0$. What distance will it traverse before it stops? What time will it take to cover that distance?

Ans.

(a) $s = (2/k) v_0^{3/2};$ (b) $t = 2 \\sqrt{v_0/k}.$

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:::{.problem}
1.24. A position vector of a point $A$ relative to the origin varies with time $t$ as $\\vec{r} = k t \\vec{i} - b t^2 \\vec{j}$, where $k$ and $b$ are positive constants, and $\\vec{i}$ and $\\vec{j}$ are the unit vectors of the $x$ and $y$ axes. Find:
(a) the equation of the point's trajectory $y(x)$; plot this function;
(b) the time dependence of the velocity $\\vec{v}$ and acceleration $\\vec{a}$ vectors, as well as of the magnitudes of these quantities;
(c) the time dependence of the angle $\\theta$ between the vectors $\\vec{a}$ and $\\vec{v}$;
(d) the mean velocity vector averaged over the first $t$ seconds of motion, and the magnitude of this vector.

Ans.

(a) $y = -\\frac{b}{k^2}x^2;$ (b) $\\vec{v} = k \\vec{i} - 2b t \\vec{j}, \\quad \\vec{a} = -2b \\vec{j}, \\quad v = \\sqrt{k^2 + 4b^2 t^2}, \\quad a = 2b;$ (c) $\\tan \\theta = \\frac{k}{2b t};$ (d) $\\langle \\vec{v} \\rangle = k \\vec{i} - b t \\vec{j}, \\quad |\\langle \\vec{v} \\rangle| = \\sqrt{k^2 + b^2 t^2}.$

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:::{.problem}
1.25. A point moves in the plane $xy$ according to the law $x = k t$, $y = k t (1 - \\alpha t)$, where $k$ and $\\alpha$ are positive constants, and $t$ is time. Find:
(a) the equation of the point's trajectory $y(x)$; plot this function;
(b) the velocity $\\vec{v}$ and the acceleration $\\vec{a}$ of the point as functions of time;
(c) the moment $t_0$ at which the velocity vector forms an angle $\\pi/4$ with the acceleration vector.

Ans.

(a) $y = x - \\frac{\\alpha}{k}x^2;$ (b) $\\vec{v} = k \\vec{i} + k(1 - 2\\alpha t) \\vec{j}, \\quad \\vec{a} = -2\\alpha k \\vec{j};$ (c) $t_0 = \\frac{1}{\\alpha}.$

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:::{.problem}
1.26. A point moves in the plane $xy$ according to the law $x = k \\sin \\omega t$, $y = k (1 - \\cos \\omega t)$, where $k$ and $\\omega$ are positive constants. Find:
(a) the distance $s$ traversed by the point during the time $\\tau$;
(b) the angle between the point's velocity and acceleration vectors.

Ans.

(a) $s = k \\omega \\tau;$ (b) $\\pi/2.$

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:::{.problem}
1.27. A particle moves in the plane $xy$ with constant acceleration $\\vec{a}$ directed along the negative $y$ axis. The equation of motion of the particle has the form $y = k x - b x^2$, where $k$ and $b$ are positive constants. Find the velocity of the particle at the origin of coordinates.

Ans.

$v_0 = \\sqrt{\\frac{(1 + k^2)g}{2b}}.$

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:::{.problem}
1.28. A small body is thrown at an angle to the horizontal with the initial velocity $\\vec{v}_0$. Neglecting the air drag, find:
(a) the displacement of the body as a function of time $\\vec{r}(t)$;
(b) the mean velocity vector $\\langle \\vec{v} \\rangle$ averaged over the first $t$ seconds and over the total time of motion.

Ans.

(a) $\\vec{r} = \\vec{v}_0 t + \\frac{1}{2}\\vec{g}t^2;$ (b) $\\langle \\vec{v} \\rangle_t = \\vec{v}_0 + \\frac{1}{2}\\vec{g}t, \\quad \\langle \\vec{v} \\rangle = \\vec{v}_0 - \\frac{\\vec{g}(\\vec{v}_0 \\cdot \\vec{g})}{g^2}.$

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:::{.problem}
1.29. A body is thrown from the surface of the Earth at an angle $\\alpha$ to the horizontal with the initial velocity $\\vec{v}_0$. Assuming the air drag to be negligible, find:
(a) the time of motion;
(b) the maximum height of ascent and the horizontal range; at what value of the angle $\\alpha$ they will be equal to each other;
(c) the equation of trajectory $y(x)$, where $y$ and $x$ are displacements of the body along the vertical and the horizontal respectively;
(d) the curvature radii of trajectory at its initial point and at its peak.

Ans.

(a) $\\tau = \\frac{2v_0 \\sin \\alpha}{g};$ (b) $h = \\frac{v_0^2 \\sin^2 \\alpha}{2g}, \\quad l = \\frac{v_0^2 \\sin 2\\alpha}{g}, \\quad \\alpha = 76^\\circ;$ (c) $y = x \\tan \\alpha - \\frac{g}{2v_0^2 \\cos^2 \\alpha}x^2;$ (d) $R_1 = \\frac{v_0^2}{g \\cos \\alpha}, \\quad R_2 = \\frac{v_0^2 \\cos^2 \\alpha}{g}.$

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:::{.problem}
1.30. Using the conditions of the foregoing problem, draw the approximate time dependence of magnitudes of the normal $a_n$ and tangential $a_t$ acceleration vectors, as well as of the projection of the total acceleration vector $a_v$ on the velocity vector direction.

Ans.

See Fig. 2.

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:::{.problem}
1.31. A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle $\\alpha$ with the horizontal. Having fallen the distance $h$, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time?

Ans.

$l = 8h \\sin \\alpha.$

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Ans.

0.41 or 0.71 min later, depending on the initial angle.

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:::{.problem}
1.33. A cannon fires successively two shells with velocity $v_0 = 250 , \\text{m/s}$; the first at the angle $\\theta_1 = 60^\\circ$ and the second at the angle $\\theta_2 = 45^\\circ$ to the horizontal, the azimuth being the same. Neglecting the air drag, find the time interval between firings leading to the collision of the shells.

Ans.

$\\Delta t = \\frac{2v_0}{g} \\frac{\\sin (\\theta_1 - \\theta_2)}{\\cos \\theta_1 + \\cos \\theta_2} = 11 \\, \\text{s}.$

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:::{.problem}
1.34. A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to $v_0$. Due to the wind the balloon gathers the horizontal velocity component $v_x = k y$, where $k$ is a constant and $y$ is the height of ascent. Find how the following quantities depend on the height of ascent:
(a) the horizontal drift of the balloon $x(y)$;
(b) the total, tangential, and normal accelerations of the balloon.

Ans.

(a) $x = \\frac{k}{2v_0} y^2;$ (b) $a = k v_0, \\quad a_t = \\frac{k^2 y}{\\sqrt{1 + (k y / v_0)^2}}, \\quad a_n = \\frac{k v_0}{\\sqrt{1 + (k y / v_0)^2}}.$

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:::{.problem}
1.35. A particle moves in the plane $xy$ with velocity $\\vec{v} = k \\vec{i} + b x \\vec{j}$, where $\\vec{i}$ and $\\vec{j}$ are the unit vectors of the $x$ and $y$ axes, and $k$ and $b$ are constants. At the initial moment of time the particle was located at the point $x = y = 0$. Find:
(a) the equation of the particle's trajectory $y(x)$;
(b) the curvature radius of trajectory as a function of $x$.

Ans.

(a) $y = \\frac{b}{2k} x^2;$ (b) $R = \\frac{v^2}{a_n} = \\frac{v^2}{\\sqrt{a^2 - a_t^2}} = \\frac{k}{b} \\left[1 + \\left(\\frac{b x}{k}\\right)^2\\right]^{3/2}.$

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:::{.problem}
1.36. A particle $A$ moves in one direction along a given trajectory with a tangential acceleration $a_t = \\vec{k} \\cdot \\vec{\\tau}$, where $\\vec{k}$ is a constant vector coinciding in direction with the $x$ axis (Fig. 1.4), and $\\vec{\\tau}$ is a unit vector coinciding in direction with the velocity vector at a given point. Find how the velocity of the particle depends on $x$ provided that its velocity is negligible at the point $x = 0$.

Ans.

$v = \\sqrt{2k x}.$

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:::{.problem}
1.37. A point moves along a circle with a velocity $v = k t$, where $k = 0.50 , \\text{m/s}^2$. Find the total acceleration of the point at the moment when it covered the $n$-th ($n = 0.10$) fraction of the circle after the beginning of motion.

Ans.

$a = k \\sqrt{1 + (4\\pi n)^2} = 0.8 \\, \\text{m/s}^2.$

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:::{.problem}
1.38. A point moves with deceleration along the circle of radius $R$ so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment $t = 0$ the velocity of the point equals $v_0$. Find:
(a) the velocity of the point as a function of time and as a function of the distance covered $s$;
(b) the total acceleration of the point as a function of velocity and the distance covered.

Ans.

(a) $v = \\frac{v_0}{1 + v_0 t / R} = v_0 e^{-s / R};$ (b) $a = \\frac{\\sqrt{2} v_0^2}{R e^{2s / R}} = \\frac{\\sqrt{2} v^2}{R}.$

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:::{.problem}
1.39. A point moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered $s$ as $v = k \\sqrt{s}$, where $k$ is a constant. Find the angle $\\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$.

Ans.

$\\tan \\alpha = \\frac{2s}{R}.$

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:::{.problem}
1.40. A particle moves along an arc of a circle of radius $R$ according to the law $l = k \\sin \\omega t$, where $l$ is the displacement from the initial position measured along the arc, and $k$ and $\\omega$ are constants. Assuming $R = 1.00 , \\text{m}, , k = 0.80 , \\text{m}, , \\text{and} , \\omega = 2.00 , \\text{rad/s}$, find:
(a) the magnitude of the total acceleration of the particle at the points $l = 0$ and $l = \\pm k$;
(b) the minimum value of the total acceleration $a_{min}$ and the corresponding displacement $l_m$.

Ans.

(a) $a_0 = \\frac{k^2 \\omega^2}{R} = 2.6 \\, \\text{m/s}^2, \\quad a_k = k \\omega^2 = 3.2 \\, \\text{m/s}^2;$ (b) $a_{min} = k \\omega^2 \\sqrt{1 - \\left(\\frac{R}{2k}\\right)^2} = 2.5 \\, \\text{m/s}^2, \\quad l_m = \\pm k \\sqrt{1 - \\frac{R^2}{2k^2}} = \\pm 0.37 \\, \\text{m}.$

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:::{.problem}
1.41. A point moves in the plane so that its tangential acceleration $a_t = k$, and its normal acceleration $a_n = b t^4$, where $k$ and $b$ are positive constants, and $t$ is time. At the moment $t = 0$ the point was at rest. Find how the curvature radius $R$ of the point's trajectory and the total acceleration $a$ depend on the distance covered $s$.

Ans.

$R = \\frac{k^3}{2b s}, \\quad a = k \\sqrt{1 + \\left(\\frac{4b s^2}{k^3}\\right)^2}.$

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:::{.problem}
1.42. A particle moves along the plane trajectory $y(x)$ with velocity $v$ whose magnitude is constant. Find the acceleration of the particle at the point $x = 0$ and the curvature radius of the trajectory at that point if the trajectory has the form:
(a) of a parabola $y = k x^2$;
(b) of an ellipse $\\left(\\frac{x}{k}\\right)^2 + \\left(\\frac{y}{b}\\right)^2 = 1$; $k$ and $b$ are constants here.

Ans.

(a) $a = 2k v^2, \\quad R = \\frac{1}{2}k;$ (b) $a = \\frac{b v^2}{k^2}, \\quad R = \\frac{k^2}{b}.$

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:::{.problem}
1.43. A particle $A$ moves along a circle of radius $R = 50 , \\text{cm}$ so that its position vector $\\vec{r}$ relative to the point $O$ (Fig. 1.5) rotates with the constant angular velocity $\\omega = 0.40 , \\text{rad/s}$. Find the magnitude of the velocity of the particle, and the magnitude and direction of its total acceleration.

Ans.

$v = 2R \\omega = 0.40 \\, \\text{m/s}, \\quad a = 4R \\omega^2 = 0.32 \\, \\text{m/s}^2.$

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:::{.problem}
1.44. A wheel rotates around a stationary axis so that the rotation angle $\\phi$ varies with time as $\\phi = k t^2$, where $k = 0.20 , \\text{rad/s}^2$. Find the total acceleration $a$ of the point $A$ at the rim at the moment $t = 2.5 , \\text{s}$ if the linear velocity of the point $A$ at this moment $v = 0.65 , \\text{m/s}$.

Ans.

$a = \\frac{v}{t} \\sqrt{1 + 4k^2 t^4} = 0.7 \\, \\text{m/s}^2.$

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:::{.problem}
1.45. A shell acquires the initial velocity $v = 320 , \\text{m/s}$, having made $n = 2.0$ turns inside the barrel whose length is equal to $l = 2.0 , \\text{m}$. Assuming that the shell moves inside the barrel with a uniform acceleration, find the angular velocity of its axial rotation at the moment when the shell escapes the barrel.

Ans.

$\\omega = \\frac{2\\pi n v}{l} = 2.0 \\cdot 10^3 \\, \\text{rad/s}.$

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:::{.problem}
1.46. A solid body rotates about a stationary axis according to the law $\\phi = k t - b t^3$, where $k = 6.0 , \\text{rad/s}$ and $b = 2.0 , \\text{rad/s}^3$. Find:
(a) the mean values of the angular velocity and angular acceleration averaged over the time interval between $t = 0$ and the complete stop;
(b) the angular acceleration at the moment when the body stops.

Ans.

(a) $\\langle \\omega \\rangle = \\frac{2k}{3} = 4 \\, \\text{rad/s}, \\quad \\langle \\alpha \\rangle = \\sqrt{3k b} = 6 \\, \\text{rad/s}^2;$ (b) $\\alpha = 2\\sqrt{3k b} = 12 \\, \\text{rad/s}^2.$

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:::{.problem}
1.47. A solid body starts rotating about a stationary axis with an angular acceleration $\\alpha = k t$, where $k = 2.0 \\cdot 10^{-2} , \\text{rad/s}^3$. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle $\\theta = 60^\\circ$ with its velocity vector?

Ans.

$t = \\sqrt[3]{\\frac{4}{k} \\tan \\theta} = 7 \\, \\text{s}.$

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:::{.problem}
1.48. A solid body rotates with deceleration about a stationary axis with an angular deceleration $\\alpha \\propto \\sqrt{\\omega}$, where $\\omega$ is its angular velocity. Find the mean angular velocity of the body averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to $\\omega_0$.

Ans.

$\\langle \\omega \\rangle = \\frac{\\omega_0}{3}.$

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:::{.problem}
1.49. A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle $\\phi$ as $\\omega = \\omega_0 - k \\phi$, where $\\omega_0$ and $k$ are positive constants. At the moment $t = 0$ the angle $\\phi = 0$. Find the time dependence of:
(a) the rotation angle;
(b) the angular velocity.

Ans.

(a) $\\phi = \\frac{\\omega_0}{k} (1 - e^{-k t});$ (b) $\\omega = \\omega_0 e^{-k t}.$

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:::{.problem}
1.50. A solid body starts rotating about a stationary axis with an angular acceleration $\\alpha = \\alpha_0 \\cos \\phi$, where $\\alpha_0$ is a constant vector and $\\phi$ is an angle of rotation from the initial position. Find the angular velocity of the body as a function of the angle $\\phi$. Draw the plot of this dependence.

Ans.

$\\omega_z = \\pm \\sqrt{2 \\alpha_0} \\sin \\phi,$ see Fig. 3.

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:::{.problem}
1.51. A rotating disc (Fig. 1.6) moves in the positive direction of the $x$ axis. Find the equation $y(x)$ describing the position of the instantaneous axis of rotation, if at the initial moment the axis $C$ of the disc was located at the point $O$ after which it moved:
(a) with a constant velocity $v$, while the disc started rotating counterclockwise with a constant angular acceleration $\\alpha$ (the initial angular velocity is equal to zero);
(b) with a constant acceleration $a$ (and the zero initial velocity), while the disc started rotates counterclockwise with a constant angular velocity $\\omega$.

Ans.

(a) $y = \\frac{v^2}{\\alpha x}$ (hyperbola); (b) $y = \\frac{\\sqrt{2a x}}{\\omega}$ (parabola).

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:::{.problem}
1.52. A point $A$ is located on the rim of a wheel of radius $R = 0.50 , \\text{m}$ which rolls without slipping along a horizontal surface with velocity $v = 1.00 , \\text{m/s}$. Find:
(a) the magnitude and the direction of the acceleration vector of the point $A$;
(b) the total distance $s$ traversed by the point $A$ between the two successive moments at which it touches the surface.

Ans.

(a) $a_A = \\frac{v^2}{R} = 2.0 \\, \\text{m/s}^2$, directed toward the center of the wheel; (b) $s = 8R = 4.0 \\, \\text{m}$.

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:::{.problem}
1.53. A ball of radius $R = 10.0 , \\text{cm}$ rolls without slipping down an inclined plane so that its center moves with constant acceleration $a = 2.50 , \\text{cm/s}^2$; $t = 2.00 , \\text{s}$ after the beginning of motion its position corresponds to that shown in Fig. 1.7. Find:
(a) the velocities of the points $A, B,$ and $O$;
(b) the accelerations of these points.

Ans.

(a) $v_A = 2a t = 10.0 \\, \\text{cm/s}, \\quad v_B = \\sqrt{2}a t = 7.1 \\, \\text{cm/s}, \\quad v_O = 0;$ (b) $a_A = 2a \\sqrt{1 + \\left(\\frac{a t^2}{2R}\\right)^2} = 5.6 \\, \\text{cm/s}^2, \\quad a_B = a \\sqrt{1 + \\left(1 - \\frac{a t^2}{R}\\right)^2} = 2.5 \\, \\text{cm/s}^2, \\quad a_O = \\frac{a^2 t^2}{R} = 2.5 \\, \\text{cm/s}^2$.

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:::{.problem}
1.54. A cylinder rolls without slipping over a horizontal plane. The radius of the cylinder is equal to $r$. Find the curvature radii of trajectories traced out by the points $A$ and $B$ (see Fig. 1.7).

Ans.

$R_A = 4r, \\quad R_B = 2\\sqrt{2}r$.

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:::{.problem}
1.55. Two solid bodies rotate about stationary mutually perpendicular intersecting axes with constant angular velocities $\\omega_1 = 3.0 , \\text{rad/s}$ and $\\omega_2 = 4.0 , \\text{rad/s}$. Find the angular velocity and angular acceleration of one body relative to the other.

Ans.

$\\omega = \\sqrt{\\omega_1^2 + \\omega_2^2} = 5 \\, \\text{rad/s}, \\quad \\alpha = \\omega_1 \\omega_2 = 12 \\, \\text{rad/s}^2$.

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:::{.problem}
1.56. A solid body rotates with angular velocity $\\vec{\\omega} = a t \\vec{i} + b t^3 \\vec{j}$, where $a = 0.50 , \\text{rad/s}^2, , b = 0.060 , \\text{rad/s}^3$, and $\\vec{i}$ and $\\vec{j}$ are the unit vectors of the $x$ and $y$ axes. Find:
(a) the magnitudes of the angular velocity and the angular acceleration at the moment $t = 10.0 , \\text{s}$;
(b) the angle between the vectors of the angular velocity and the angular acceleration at that moment.

Ans.

(a) $\\omega = a t \\sqrt{1 + \\left(\\frac{b t^2}{a}\\right)^2} = 8 \\, \\text{rad/s}, \\quad \\alpha = a \\sqrt{1 + \\left(\\frac{2b t^2}{a}\\right)^2} = 1.3 \\, \\text{rad/s}^2;$ (b) $17^\\circ$.

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:::{.problem}
1.57. A round cone with half-angle $\\theta = 30^\\circ$ and the radius of the base $R = 5.0 , \\text{cm}$ rolls uniformly and without slipping over a horizontal plane as shown in Fig. 1.8. The cone apex is hinged at the point $O$ which is on the same level with the point $C$, the cone base center. The velocity of point $C$ is $v = 10.0 , \\text{cm/s}$. Find the magnitudes of:
(a) the vector of the angular velocity of the cone and the angle it forms with the vertical;
(b) the vector of the angular acceleration of the cone.

Ans.

(a) $\\omega = \\frac{v}{R \\cos \\theta} = 2.3 \\, \\text{rad/s}, \\quad 60^\\circ;$ (b) $\\alpha = \\left(\\frac{v}{R}\\right)^2 \\tan \\theta = 2.3 \\, \\text{rad/s}^2$.

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:::{.problem}
1.58. A solid body rotates with a constant angular velocity $\\omega_0 = 0.50 , \\text{rad/s}$ about a horizontal axis $AB$. At the moment $t = 0$, the axis $AB$ starts turning about the vertical with a constant angular acceleration $\\alpha_0 = 0.10 , \\text{rad/s}^2$. Find the angular velocity and angular acceleration of the body after $t = 3.5 , \\text{s}$.

Ans.

$\\omega = \\omega_0 \\sqrt{1 + \\left(\\frac{\\alpha_0 t}{\\omega_0}\\right)^2} = 0.6 \\, \\text{rad/s}, \\quad \\alpha = \\alpha_0 \\sqrt{1 + \\left(\\frac{\\omega_0 t}{\\alpha_0}\\right)^2} = 0.2 \\, \\text{rad/s}^2$.

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The Fundamental Equation of Dynamics

:::{.problem}
1.59. An aerostat of mass $m$ starts coming down with a constant acceleration $a$. Determine the ballast mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected.

Ans.

$\\Delta m = \\frac{2m a}{g + a}$.

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:::{.problem}
1.60. In the arrangement of Fig. 1.9, the masses $m_0, m_1,$ and $m_2$ of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration $a$ with which the body $m_0$ comes down, and the tension of the thread binding together the bodies $m_1$ and $m_2$, if the coefficient of friction between these bodies and the horizontal surface is equal to $k$. Consider possible cases.

Ans.

$a = \\frac{m_0 - k(m_1 + m_2)}{m_0 + m_1 + m_2} g, \\quad T = \\frac{(1 + k)m_0}{m_0 + m_1 + m_2} m_2 g$.

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:::{.problem}
1.61. Two touching bars $1$ and $2$ are placed on an inclined plane forming an angle $\\alpha$ with the horizontal (Fig. 1.10). The masses of the bars are equal to $m_1$ and $m_2$, and the coefficients of friction between the inclined plane and these bars are equal to $k_1$ and $k_2$ respectively, with $k_1 > k_2$. Find:
(a) the force of interaction of the bars in the process of motion;
(b) the minimum value of the angle $\\alpha$ at which the bars start sliding down.

Ans.

(a) $F = \\frac{(k_1 - k_2)m_1 m_2 g \\cos \\alpha}{m_1 + m_2};$ (b) $\\tan \\alpha_{\\text{min}} = \\frac{k_1 m_1 + k_2 m_2}{m_1 + m_2}$.

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:::{.problem}
1.62. A small body was launched up an inclined plane set at an angle $\\alpha = 15^\\circ$ against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is $\\eta = 2.0$ times less than the time of its descent.

Ans.

$k = \\left[\\frac{(\\eta^2 - 1)}{(\\eta^2 + 1)}\\right] \\tan \\alpha = 0.16$.

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:::{.problem}
1.63. The following parameters of the arrangement of Fig. 1.11 are available: the angle $\\alpha$ which the inclined plane forms with the horizontal, and the coefficient of friction $k$ between the body $m_1$ and the inclined plane. The masses of the pulley and the threads, as well as the friction in the pulley, are negligible. Assuming both bodies to be motionless at the initial moment, find the mass ratio $m_2/m_1$ at which the body $m_2$:
(a) starts coming down;
(b) starts going up;
(c) is at rest.

Ans.

(a) $\\frac{m_2}{m_1} > \\sin \\alpha + k \\cos \\alpha;$ (b) $\\frac{m_2}{m_1} < \\sin \\alpha - k \\cos \\alpha;$ (c) $\\sin \\alpha - k \\cos \\alpha < \\frac{m_2}{m_1} < \\sin \\alpha + k \\cos \\alpha$.

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:::{.problem}
1.64. The inclined plane of Fig. 1.11 forms an angle $\\alpha = 30^\\circ$ with the horizontal. The mass ratio $m_2/m_1 = \\eta = 2/3$. The coefficient of friction between the body $m_1$ and the inclined plane is equal to $k = 0.10$. The masses of the pulley and the threads are negligible. Find the magnitude and the direction of acceleration of the body $m_2$ when the formerly stationary system of masses starts moving.

Ans.

$a = \\frac{g(\\eta - \\sin \\alpha - k \\cos \\alpha)}{\\eta + 1} = 0.05 \\, g$.

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:::{.problem}
1.65. A plank of mass $m_1$ with a bar of mass $m_2$ placed on it lies on a smooth horizontal plane. A horizontal force growing with time $t$ as $F = a t$ ($a$ is constant) is applied to the bar. Find how the accelerations of the plank $a_1$ and of the bar $a_2$ depend on $t$, if the coefficient of friction between the plank and the bar is equal to $k$. Draw the approximate plots of these dependences.

Ans.

When $t \\leq t_0$, the accelerations $a_1 = a_2 = \\frac{a t}{m_1 + m_2};$ when $t \\geq t_0$, $a_1 = \\frac{k g m_2}{m_1}, \\quad a_2 = \\frac{a t - k m_2 g}{m_2}$. Here $t_0 = \\frac{k g m_2 (m_1 + m_2)}{a m_1}$. See Fig. 4.

:::

:::{.problem}
1.66. A small body $A$ starts sliding down from the top of a wedge (Fig. 1.12) whose base is equal to $l = 2.10 , \\text{m}$. The coefficient of friction between the body and the wedge surface is $k = 0.140$. At what value of the angle $\\alpha$ will the time of sliding be the least? What will it be equal to?

Ans.

$\\tan 2\\alpha = -\\frac{1}{k}, \\quad \\alpha = 49^\\circ; \\quad t_{\\text{min}} = 1.0 \\, \\text{s}$.

:::

:::{.problem}
1.67. A bar of mass $m$ is pulled by means of a thread up an inclined plane forming an angle $\\alpha$ with the horizontal (Fig. 1.13). The coefficient of friction is equal to $k$. Find the angle $\\beta$ which the thread must form with the inclined plane for the tension of the thread to be minimum. What is it equal to?

Ans.

$\\tan \\beta = k; \\quad T_{\\text{min}} = \\frac{m g (\\sin \\alpha + k \\cos \\alpha)}{1 + k^2}$.

:::

:::{.problem}
1.68. At the moment $t = 0$, the force $F = a t$ is applied to a small body of mass $m$ resting on a smooth horizontal plane ($a$ is a constant). The permanent direction of this force forms an angle $\\alpha$ with the horizontal (Fig. 1.14). Find:
(a) the velocity of the body at the moment of its breaking off the plane;
(b) the distance traversed by the body up to this moment.

Ans.

(a) $v = \\frac{m g^2 \\cos \\alpha}{2 a \\sin^2 \\alpha};$ (b) $s = \\frac{m^2 g^3 \\cos \\alpha}{6 a^2 \\sin^3 \\alpha}$.

:::

:::{.problem}
1.69. A bar of mass $m$ resting on a smooth horizontal plane starts moving due to the force $F = \\frac{m g}{3}$ of constant magnitude. In the process of its rectilinear motion, the angle $\\alpha$ between the direction of this force and the horizontal varies as $\\alpha = k s$, where $k$ is a constant, and $s$ is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle $\\alpha$.

Ans.

$v = \\sqrt{\\frac{2 g}{3 k} \\sin \\alpha}$.

:::

:::{.problem}
1.70. A horizontal plane with the coefficient of friction $\\mu$ supports two bodies: a bar and an electric motor with a battery on a block. A thread attached to the bar is wound on the shaft of the electric motor. The distance between the bar and the electric motor is equal to $l$. When the motor is switched on, the bar, whose mass is twice as great as that of the other body, starts moving with a constant acceleration $a$. How soon will the bodies collide?

Ans.

$t = \\sqrt{2l/(3a+\\mu g)}$.

:::

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جابجایی

هنگامی که نیروهایی به یک جسم وارد می‌شوند، تمام اجسام تا حدی تغییر شکل می‌دهند. این تغییر شکل را می‌توان بر اساس حرکت ذرات مادی توصیف کرد که مفهومی است به نام جابجایی.

در مکانیک کلاسیک، یک ذره به عنوان یک جرم نقطه‌ای در نظر گرفته می‌شود که می‌تواند در راستای محورهای x، y و z حرکت کند و فاقد درجات آزادی چرخشی است.

اگر موقعیت یک ذره در پیکربندی اولیه¹ با X و در پیکربندی جاری با x نمایش داده شود، بردار جابجایی به صورت زیر تعریف می‌شود:
\\[ \\mathbf{u} = \\mathbf{x} - \\mathbf{X}. \\] این میدان برداری سه مؤلفه دارد که در راستای محورهای x₁، x₂ و x₃ (یا x، y و z) به ترتیب با u₁، u₂ و u₃ (یا u، v و w) نشان داده می‌شوند.

به طور کلی، میدان جابجایی یکنواخت نیست؛ نقاط مختلف جسم ممکن است جابجایی‌های مختلفی را تجربه کنند. بنابراین، به طور کلی تابعی از مکان و زمان t است. دو روش رایج برای بیان این وابستگی وجود دارد:

به عنوان تابعی از مکان اولیه (مرجع). در این حالت، معمولاً جابجایی را با حرف بزرگ U نشان می‌دهیم:
\$\\mathbf{U} = \\mathbf{U}(\\mathbf{X}, t)\$
به عنوان تابعی از مکان جاری (تغییر شکل یافته):
\$\\mathbf{u} = \\mathbf{u}(\\mathbf{x}, t)\$

در فلزات با تغییر شکل‌های کشسان کوچک، هنگام ارزیابی توابع مختصات، اغلب می‌توان تمایز بین دستگاه‌های مختصات X و x را نادیده گرفت.

در این فصل، تمرکز ما بر نظریه کرنش کوچک است، که در آن هم جابجایی و هم گرادیان آن کوچک فرض می‌شوند. برای رفتار کشسان متداول در فلزات — که کرنش کمتر از ‪0.2‬٪ است — این فرض مناسب است.

چرخش و کرنش

جابجایی می‌تواند بدون تغییر شکل رخ دهد. به عنوان مثال، یک جسم ممکن است تحت انتقال جسم صلب یا چرخش جسم صلب قرار گیرد که در آن موقعیت نسبی ذرات بدون تغییر باقی می‌ماند. برای مثال، به تیر نشان داده شده در شکل زیر توجه کنید. قسمت‌های انتهایی آن چرخش محض را بدون اینکه تحت نیروهای یا گشتاورهای داخلی قرار گیرند، تجربه می‌کنند. این قسمت‌ها جابجایی نشان می‌دهند، اما تغییر شکل ندارند. در این موارد، اگرچه جابجایی وجود دارد، هیچ تغییر شکل داخلی ایجاد نمی‌شود و در نتیجه هیچ نیروی داخلی (یا تنش)ای بوجود نمی‌آید.

\"\" — هنگام اعمال نیرو، دو قسمت انتهایی صرفاً می‌چرخند

آنچه باعث ایجاد تنش‌های داخلی می‌شود جابجایی‌های نسبی بین ذرات مجاور است. جابجایی نسبی بین دو نقطه با افزایش فاصله اولیه آن‌ها بیشتر می‌شود. بنابراین، برای توصیف تغییر شکل، باید حرکت نسبی دو نقطه را بر فاصله اولیه آن‌ها نرمال‌سازی کنیم.

به عنوان مثال، دو نقطه A و B را در نظر بگیرید که خط متصل‌کننده آن‌ها موازی با محور x و فاصله آن‌ها Δx است. فرض کنید نقطه A دارای جابجایی \$(u, v)\$ باشد، در حالی‌که نقطه B دارای جابجایی \$(u + \\Delta u, v + \\Delta v)\$ است.

مؤلفه‌های جابجایی نسبی بر واحد فاصله اولیه بین دو نقطه به صورت زیر خواهند بود:
\\[ \\begin{aligned} &\\frac{u_B-u_A}{|AB|}=\\frac{(u + \\Delta u) - u}{\\Delta x} = \\frac{\\Delta u}{\\Delta x},\\\\ &\\frac{v_B-v_A}{|AB|}=\\frac{(v + \\Delta v) - v}{\\Delta x} = \\frac{\\Delta v}{\\Delta x} \\end{aligned} \\]
هنگامی که \$\\Delta x \\to 0\$، این عبارت‌ها به مشتقات جزئی نزدیک می‌شوند:
\\[ \\frac{\\partial u}{\\partial x} \\quad \\text{و} \\quad \\frac{\\partial v}{\\partial x}. \\] این مطلب نشان می‌دهد که برای توصیف تغییر شکل، نیاز است مشتقات مکانی (یا گرادیان) مؤلفه‌های جابجایی را محاسبه کنیم.

تانسور گرادیان جابجایی به صورت زیر داده می‌شود:
\\[ \\begin{bmatrix} u\\\\ v\\\\ w \\end{bmatrix} \\begin{bmatrix} \\dfrac{\\partial }{\\partial x} & \\dfrac{\\partial }{\\partial y}& \\dfrac{\\partial }{\\partial z} \\end{bmatrix} = \\begin{bmatrix} \\dfrac{\\partial u}{\\partial x} & \\dfrac{\\partial u}{\\partial y} & \\dfrac{\\partial u}{\\partial z} \\\\ \\dfrac{\\partial v}{\\partial x} & \\dfrac{\\partial v}{\\partial y} & \\dfrac{\\partial v}{\\partial z} \\\\ \\dfrac{\\partial w}{\\partial x} & \\dfrac{\\partial w}{\\partial y} & \\dfrac{\\partial w}{\\partial z} \\end{bmatrix} \\]

گرادیان جابجایی به تنهایی برای توصیف تغییر شکل کافی نیست، زیرا حرکت نسبی بین نقاط یک جسم می‌تواند ناشی از چرخش جسم صلب نیز باشد که منجر به ایجاد تنش نمی‌شود. اگرچه در برخی مسائل چرخش ممکن است مهم باشد، اما وقتی هدف ما تمرکز بر تغییر شکل ناشی از تنش باشد، باید اثرات چرخش را کم کنیم.

برای حذف چرخش‌ها، عنصری را در نظر می‌گیریم که اضلاع آن موازی با محورهای x و y هستند (شکل زیر را ببینید). برای چرخش‌های کوچک، چرخش یک خط که در ابتدا موازی با محور x است به صورت زیر است:
\\[ \\omega_1 \\approx \\tan \\omega_1 \\approx \\frac{\\Delta v}{\\Delta x}, \\] و چرخش یک خط که در ابتدا موازی با محور y است به صورت زیر است:
\\[ \\omega_2 \\approx \\tan \\omega_2 \\approx -\\frac{\\Delta u}{\\Delta y}. \\]

چرخش متوسط این عنصر در صفحه ‪xy‬ به صورت زیر تعریف می‌شود:
\\[ \\omega_{yx} = \\frac{1}{2} \\left( \\frac{\\partial v}{\\partial x} - \\frac{\\partial u}{\\partial y} \\right). \\]

به طور کلی، چرخش توسط عبارت زیر توصیف می‌شود:
\\[ \\begin{bmatrix} 0 & \\omega_{xy} & \\omega_{xz}\\\\ \\omega_{yx} & 0 & \\omega_{yz}\\\\ \\omega_{zx} & \\omega_{zy} & 0 \\end{bmatrix}= \\begin{bmatrix} 0 & \\frac{1}{2}\\left(\\frac{\\partial u}{\\partial y}-\\frac{\\partial v}{\\partial x}\\right) & \\frac{1}{2}\\left(\\frac{\\partial u}{\\partial z}-\\frac{\\partial w}{\\partial x}\\right)\\\\ \\frac{1}{2}\\left(\\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial x}\\right) & 0 & \\frac{1}{2}\\left(\\frac{\\partial v}{\\partial z}-\\frac{\\partial w}{\\partial y}\\right)\\\\ \\frac{1}{2}\\left(\\frac{\\partial w}{\\partial x}-\\frac{\\partial u}{\\partial z}\\right) & \\frac{1}{2}\\left(\\frac{\\partial w}{\\partial y}-\\frac{\\partial v}{\\partial z}\\right) & 0 \\end{bmatrix} \\]
\\[ \\omega_{ij}=\\frac{1}{2}\\left(\\frac{\\partial u_i}{\\partial x_j}-\\frac{\\partial u_j}{\\partial x_i}\\right) \\] توجه کنید که این یک ماتریس پادمتقارن است.

تفاضل چرخش از گرادیان جابجایی منجر به کرنش می‌شود که تغییرات محلی در شکل جسم را توصیف می‌کند.

\\[ \\bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\\begin{bmatrix} \\epsilon_{xx} & \\epsilon_{xy} & \\epsilon_{xz}\\\\ \\epsilon_{yx} & \\epsilon_{yy} & \\epsilon_{yz}\\\\ \\epsilon_{zx} & \\epsilon_{zy} & \\epsilon_{zz} \\end{bmatrix}= \\begin{bmatrix} \\dfrac{\\partial u}{\\partial x} & \\dfrac{1}{2}\\left(\\dfrac{\\partial u}{\\partial y}+\\dfrac{\\partial v}{\\partial x}\\right) & \\dfrac{1}{2}\\left(\\dfrac{\\partial u}{\\partial z}+\\dfrac{\\partial w}{\\partial x}\\right)\\\\ \\dfrac{1}{2}\\left(\\dfrac{\\partial v}{\\partial x}-\\dfrac{\\partial u}{\\partial x}\\right) & \\dfrac{\\partial v}{\\partial y} & \\dfrac{1}{2}\\left(\\dfrac{\\partial v}{\\partial z}+\\dfrac{\\partial w}{\\partial y}\\right)\\\\ \\dfrac{1}{2}\\left(\\dfrac{\\partial w}{\\partial x}+\\dfrac{\\partial u}{\\partial z}\\right) & \\dfrac{1}{2}\\left(\\dfrac{\\partial w}{\\partial y}+\\dfrac{\\partial v}{\\partial z}\\right) & \\dfrac{\\partial w}{\\partial z} \\end{bmatrix}}. \\]
بنابراین، مؤلفه‌های کرنش به صورت زیر داده می‌شوند:
\\[ \\bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\\epsilon_{ij}=\\frac{1}{2}\\left(\\frac{\\partial u_i}{\\partial x_j}+\\frac{\\partial u_j}{\\partial x_i}\\right)}. \\]
مؤلفه‌های روی قطر اصلی کرنش‌های نرمال نامیده می‌شوند و سایر مؤلفه‌ها کرنش برشی نامیده می‌شوند.

توجه کنید که ماتریس متناظر با کرنش متقارن است.
توجه کنید که \\[ \\frac{\\partial u_i}{\\partial x_j}=\\epsilon_{ij}+\\omega_{ij}. \\]
در کاربردهای مهندسی، معمولاً از نماد گذاری ساده شده برای مؤلفه‌های کرنش نرمال استفاده می‌شود: وقتی هر دو اندیس یکسان باشند، از فقط یک زیرنویس استفاده می‌شود. به عنوان مثال، ϵ_‪xx‬ اغلب به صورت ϵ_x و ϵ_‪yy‬ به صورت ϵ_y نوشته می‌شود. \\[ \\epsilon_x=\\frac{\\partial u}{\\partial x},\\quad \\epsilon_y=\\frac{\\partial v}{\\partial y}, \\quad \\epsilon_z=\\frac{\\partial w}{\\partial z}. \\]
در بسیاری از مسائل مهندسی، اغلب دو برابر کرنش برشی ϵ_‪ij‬ را به عنوان Ɣ_‪ij‬ نشان می‌دهند: \\[ \\begin{aligned} \\gamma_{xy}&=2\\epsilon_{xy}=\\frac{\\partial u}{\\partial y}+\\frac{\\partial v}{\\partial x},\\\\ \\gamma_{xz}&=2\\epsilon_{xz}=\\frac{\\partial u}{\\partial z}+\\frac{\\partial w}{\\partial x},\\\\ \\gamma_{yz}&=2\\epsilon_{yz}=\\frac{\\partial v}{\\partial z}+\\frac{\\partial w}{\\partial y}. \\end{aligned} \\]

تفسیر هندسی

عنصری دو بعدی را در نظر بگیرید که تغییر شکل داده است. برای سادگی، گوشه عنصر در حالت تغییر شکل یافته را به گوشه عنصر قبل از تغییر شکل منتقل می‌کنیم. کرنش‌های نرمال تغییر طول بر واحد طول اولیه هستند، در حالی که کرنش برشی نیمی از کاهش زاویه نشان داده شده در شکل زیر است.

علامت مؤلفه‌های کرنش

کرنش نرمال زمانی مثبت در نظر گرفته می‌شود که طول در جهت مربوطه افزایش یابد. به عنوان مثال، ϵ_‪xx‬ زمانی مثبت است که عنصر در راستای محور x کشیده شود.

کرنش برشی ϵ_‪xy‬ زمانی مثبت در نظر گرفته می‌شود که زاویه بین محورهای +x و +y (یا بین –x و –y) کاهش یابد و زاویه بین +x و –y (یا –x و +y) افزایش یابد (شکل زیر را ببینید). به عبارت دیگر، تغییر زاویه‌های ناشی از یک تنش برشی مثبت به عنوان کرنش برشی مثبت تعریف می‌شود و بالعکس.

\"\" — چپ: عنصر تحت کرنش برشی مثبت قرار گرفته است. راست: عنصر تحت کرنش برشی منفی قرار گرفته است

واحدهای کرنش

توجه داشته باشید که کرنش بی‌بعد است، بنابراین واحدی ندارد.

منابع

‪Malvern, L. E. (1969‬). ‪Introduction to the mechanics of a continuous medium‬. ‪Prentice-Hall‬.
‪McClintock, F. A‬., & ‪Argon, A. S. (1966‬). ‪Mechanical behavior of materials‬. ‪Addison-Wesley‬.

پیکربندی، آرایش مکانی تمام ذرات یک جسم در لحظه‌ای مشخص از زمان است.↩︎

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Theorem 1

\n Theorem 2. Given any \$m\$ -dimensional subspace \$\\mathcal{M}\$ in an \$n\$ -dimensional vector space \$\\mathcal{V}\$ , we can find a basis \$\\{x_{1}, \\ldots, x_{m}, x_{m+1}, \\ldots, x_{n}\\}\$ in \$\\mathcal{V}\$ so that \$x_{1}, \\ldots, x_{m}\$ are in \$\\mathcal{M}\$ and form, therefore, a basis of \$\\mathcal{M}\$ .\n

\n\n

Theorem 2

\n Theorem 1. A subspace \$\\mathcal{M}\$ in an \$n\$ -dimensional vector space \$\\mathcal{V}\$ is a vector space of dimension \$\\leq n\$ .\n

\n\n

\nProof.\n

\n Proof. It is possible to give a deceptively short proof of this theorem that runs as follows. Every set of \$n+1\$ vectors in \$\\mathcal{V}\$ is linearly dependent, hence the same is true of \$\\mathcal{M}\$ ; hence, in particular, the number of elements in each basis of \$\\mathcal{M}\$ is \$\\leq n\$ , Q.E.D.\n

\n The trouble with this argument is that we defined dimension \$n\$ by requiring in the first place that there exist a finite basis, and then demanding that this basis contain exactly \$n\$ elements. The proof above shows only that no basis can contain more than \$n\$ elements; it does not show that any basis exists. Once the difficulty is observed, however, it is easy to fill the gap. If \$\\mathcal{M}=\\mathcal{O}\$ , then \$\\mathcal{M}\$ is \$0\$ -dimensional, and we are done. If \$\\mathcal{M}\$ contains a non-zero vector \$x_{1}\$ , let \$\\mathcal{M}_{1}\$ ( \$\\subset \\mathcal{M}\$ ) be the subspace spanned by \$x_{1}\$ . If \$\\mathcal{M}=\\mathcal{M}_{1}\$ , then \$\\mathcal{M}\$ is \$1\$ -dimensional, and we are done. If \$\\mathcal{M} \\neq \\mathcal{M}_{1}\$ , let \$x_2\$ be an element of \$\\mathcal{M}\$ not contained in \$\\mathcal{M}_1\$ , and let \$\\mathcal{M}_2\$ be the subspace spanned by \$x_{1}\$ and \$x_{2}\$ ; and so on. Now we may legitimately employ the argument given above; after no more than \$n\$ steps of this sort, the process reaches an end, since (by Section: Dimension , Theorem 2) we cannot find \$n+1\$ linearly independent vectors. ◻\n

\n∎\n

\n\n

The following result is an important consequence of this second and correct proof of Theorem 1.

\n\n

We shall denote the dimension of a vector space \$\\mathcal{V}\$ by the symbol \$\\operatorname{dim} \\mathcal{V}\$ . In this notation Theorem 1 asserts that if \$\\mathcal{M}\$ is a subspace of a finite-dimensional vector space \$\\mathcal{V}\$ , then \$\\operatorname{dim} \\mathcal{M} \\leq \\operatorname{dim} \\mathcal{V}\$ .

\n\n

EXERCISES

\n\n

\nExercise 1.\n

\n Exercise 1. If \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are finite-dimensional subspaces with the same dimension, and if \$\\mathcal{M} \\subset \\mathcal{N}\$ , then \$\\mathcal{M}=\\mathcal{N}\$ .\n

\n\n

\nExercise 2.\n

\n Exercise 2. If \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are subspaces of a vector space \$\\mathcal{V}\$ , and if every vector in \$\\mathcal{V}\$ belongs either to \$\\mathcal{M}\$ or to \$\\mathcal{N}\$ (or both), then either \$\\mathcal{M}=\\mathcal{V}\$ or \$\\mathcal{N}=\\mathcal{V}\$ (or both).\n

\n\n

\nExercise 3.\n

\n Exercise 3. If \$x\$ , \$y\$ , and \$z\$ are vectors such that \$x+y+z=0\$ , then \$x\$ and \$y\$ span the same subspace as \$y\$ and \$z\$ .\n

\n\n

\nExercise 4.\n

\n Exercise 4. Suppose that \$x\$ and \$y\$ are vectors and \$\\mathcal{M}\$ is a subspace in a vector space \$\\mathcal{V}\$ ; let \$\\mathcal{H}\$ be the subspace spanned by \$\\mathcal{M}\$ and \$x\$ , and let \$\\mathcal{K}\$ be the subspace spanned by \$\\mathcal{M}\$ and \$y\$ . Prove that if \$y\$ is in \$\\mathcal{H}\$ but not in \$\\mathcal{M}\$ , then \$x\$ is in \$\\mathcal{K}\$ .\n

\n\n

\nExercise 5.\n

\n Exercise 5. Suppose that \$\\mathcal{L}\$ , \$\\mathcal{M}\$ , and \$\\mathcal{N}\$ are subspaces of a vector space.\n

\n Show that the equation \\[\\mathcal{L} \\cap(\\mathcal{M}+\\mathcal{N})=(\\mathcal{L} \\cap \\mathcal{M})+(\\mathcal{L} \\cap \\mathcal{N})\\] is not necessarily true.\n
\n Prove that \\[\\mathcal{L} \\cap\\big(\\mathcal{M}+(\\mathcal{L} \\cap \\mathcal{N})\\big)=(\\mathcal{L} \\cap \\mathcal{M})+(\\mathcal{L} \\cap \\mathcal{N})\\] \n

\n\n

\nExercise 6.\n

\n Exercise 6. \n

\n Can it happen that a non-trivial subspace of a vector space \$\\mathcal{V}\$ (i.e., a subspace different from both \$\\mathcal{O}\$ and \$\\mathcal{V}\$ ) has a unique complement?\n
\n If \$\\mathcal{M}\$ is an \$m\$ -dimensional subspace in an \$n\$ -dimensional vector space, then every complement of \$\\mathcal{M}\$ has dimension \$n-m\$ .\n

\n\n

\nExercise 7.\n

\n Exercise 7. \n

\n Show that if both \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are three-dimensional subspaces of a five-dimensional vector space, then \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are not disjoint.\n
\n If \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are finite-dimensional subspaces of a vector space, then \\[\\operatorname{dim} \\mathcal{M}+\\operatorname{dim} \\mathcal{N}=\\operatorname{dim}(\\mathcal{M}+\\mathcal{N})+\\operatorname{dim}(\\mathcal{M} \\cap \\mathcal{N}).\\] \n

\n\n

\nExercise 8.\n

\n Exercise 8. A polynomial \$x\$ is called even if \$x(-t)=x(t)\$ identically in \$t\$ (see Section: Subspaces , (3)), and it is called odd if \$x(-t)=-x(t)\$ .\n

\n (a) Both the class \$\\mathcal{M}\$ of even polynomials and the class \$\\mathcal{N}\$ of odd polynomials are subspaces of the space \$\\mathcal{P}\$ of all (complex) polynomials.\n

\n (b) Prove that \$\\mathcal{M}\$ and \$\\mathcal{N}\$ are each other’s complements.\n

","chapter_express_content":"

\nExample 1.\nsvs\n

\n\n

Let me start $y=x^2$ and \$u=t^{3/2}\$.

\n\n

This is a test

\n\n

","public_visibility":false,"allow_part_see":false,"book_slug":"introduction-to-crystallography","slug":"test-import-html","regular_authors":"","express_authors":"","created_at":"2025-12-15T00:44:16.915Z","__v":0,"header_data":[]}]},"seoMetadata":{"meta_title":"","meta_description":"","meta_keywords":"","meta_author":"","og_image":""},"error":null}},"actionData":null,"errors":null}

	\\(1.000000\\)
dividing by \\(1\\)	\\(1.000000\\)
dividing by \\(2\\)	\\(0.500000\\)
dividing by \\(3\\)	\\(0.166667\\)
dividing by \\(4\\)	\\(0.041667\\)
dividing by \\(5\\)	\\(0.008333\\)
dividing by \\(6\\)	\\(0.001389\\)
dividing by \\(7\\)	\\(0.000198\\)
dividing by \\(8\\)	\\(0.000025\\)
dividing by \\(9\\)	\\(0.000002\\)
Total	\\(2.718281\\)

	simple cubic	b.c.c.	f.c.c.
volume of conventional cell	\\(a^3\\)	\\(a^3\\)	\\(a^3\\)
no. of lattice points per cell	1	2	4
no. of nearest neighbors (coordination number)	6	8	12
no. of 2nd nearest neighbors	12	6	6
nearest neighbor distance	\\(a\\)	\\(\\frac{\\sqrt{3}}{2}a\\approx 0.866 a\\)	\\(\\frac{a}{\\sqrt{2}}\\approx 0.707a\\)
2nd nearest neighbor distance	\\(a\\sqrt{2}\\)	\\(a\\)	\\(a\\)
packing fraction	\\(\\pi/6\\approx 0.52\\)	\\(\\pi\\sqrt{3}/8\\approx 0.68\\)	\\(\\pi\\sqrt{2}/6\\approx 0.74\\)

	$\"\"$
(a) Tetrahedral bond in a diamond structure	(b) Diamond structure projected on a cube face. Fractions denote the height above the base in units of \\(a\\)

$\"\"$	$\"\"$
Determination of indices for a digonal axis if Type I - $[2\\overline{11}0]$	Determination of indices for a digonal axis if Type II - $[10\\overline{1}0]$

	$e$	$a$	$b$	$c$
$e$	$e$	$a$	$b$	$c$
$a$	$a$	$e$	$c$	$b$
$b$	$b$	$c$	$e$	$a$
$c$	$c$	$b$	$a$	$e$

	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$e$	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$a$	$a$	$b$	$c$	$e$	$v$	$s$	$t$	$u$
$b$	$b$	$c$	$e$	$a$	$u$	$v$	$s$	$t$
$c$	$c$	$e$	$a$	$b$	$t$	$u$	$v$	$s$
$s$	$s$	$t$	$u$	$v$	$e$	$a$	$b$	$c$
$t$	$t$	$u$	$v$	$s$	$c$	$e$	$a$	$b$
$u$	$u$	$v$	$s$	$t$	$b$	$c$	$e$	$a$
$v$	$v$	$s$	$t$	$u$	$a$	$b$	$c$	$e$

$\"\"$	$\"\"$	$\"\"$
Oblique	Rectangular	Centerd Rectangular

$\"\"$	$\"\"$
Square	Hexgonal (Rhombic)

	$e$	$a$	$b$	$c$
$e$	$e$	$a$	$b$	$c$
$a$	$a$	$e$	$c$	$b$
$b$	$b$	$c$	$e$	$a$
$c$	$c$	$b$	$a$	$e$

	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$e$	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$a$	$a$	$b$	$c$	$e$	$v$	$s$	$t$	$u$
$b$	$b$	$c$	$e$	$a$	$u$	$v$	$s$	$t$
$c$	$c$	$e$	$a$	$b$	$t$	$u$	$v$	$s$
$s$	$s$	$t$	$u$	$v$	$e$	$a$	$b$	$c$
$t$	$t$	$u$	$v$	$s$	$c$	$e$	$a$	$b$
$u$	$u$	$v$	$s$	$t$	$b$	$c$	$e$	$a$
$v$	$v$	$s$	$t$	$u$	$a$	$b$	$c$	$e$

Definitions of Basis, Lattice, and Crystals

Primitive Cells vs. Unit Cells

Lattice Parameters and Packing Fraction

Lattice Parameters

Packing Fraction

Bravais Lattices

Five Bravais Lattices in Two Dimensions

Seven Crystal Systems

Cubic Lattices

Simple Cubic (SC) Crystal

Body-Centered Cubic (BCC) Crystal

Some Elements with BCC structure

Face-Centered Cubic (FCC) Crystal

Characteristics of Cubic structure

Crystal Structure May Change with Temperature

Hexagonal Close-Packed (HCP) Structure

Comparing Close-Packed Structures

Elements with hcp structures

Diamond Crystal

Sodium Chloride Structure

Cesium Chloride Structure

Miller Indices

Miller Indices for Directions in Cubic Structure

Miller Indices for Planes in Cubic Structure

Miller Indices for hcp

Further Reading

CHAPTER 1: Groups

EXAMPLES.

1.1. The Concept of a Group

1.2. Subgroups

Productivity Pack Minutes Saved by the User. How about $\\int f(x) dx$?

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On True Compound Interest

The Number \\(\\boldsymbol{e}\\)

A little more about the number \\(\\boldsymbol{e}\\)

The Exponential Series

Derivative of the Exponential Function

Natural or Naperian Logarithms

Using A Calculator to Find \\(\\boldsymbol{e^x}\\) and \\(\\boldsymbol{\\ln x}\\)

Derivatives of Logarithmic and Exponential Functions

Examples

Exercises I

The Logarithmic Curve

The Die-away Curve

Further Examples

Exercises II

Hyperbolic Functions

Identities Between Hyperbolic Functions

Derivatives of the Hyperbolic Functions

Inverse Hyperbolic Functions

Derivatives of the Inverse Hyperbolic Functions

Group Theory Study Guide

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Elementary examples of multivariable functions

[wp-svg-icons icon=\"stop-2\" wrap=\"i\"] Composition of functions

PART ONE

PHYSICAL FUNDAMENTALS OF MECHANICS

1.1. KINEMATICS

The Fundamental Equation of Dynamics

جابجایی

چرخش و کرنش

تفسیر هندسی

علامت مؤلفه‌های کرنش

واحدهای کرنش

منابع

EXERCISES

This is a test

	$e$	$a$	$b$	$c$
$e$	$e$	$a$	$b$	$c$
$a$	$a$	$e$	$c$	$b$
$b$	$b$	$c$	$e$	$a$
$c$	$c$	$b$	$a$	$e$

	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$e$	$e$	$a$	$b$	$c$	$s$	$t$	$u$	$v$
$a$	$a$	$b$	$c$	$e$	$v$	$s$	$t$	$u$
$b$	$b$	$c$	$e$	$a$	$u$	$v$	$s$	$t$
$c$	$c$	$e$	$a$	$b$	$t$	$u$	$v$	$s$
$s$	$s$	$t$	$u$	$v$	$e$	$a$	$b$	$c$
$t$	$t$	$u$	$v$	$s$	$c$	$e$	$a$	$b$
$u$	$u$	$v$	$s$	$t$	$b$	$c$	$e$	$a$
$v$	$v$	$s$	$t$	$u$	$a$	$b$	$c$	$e$

Introduction to Crystallography

By Unknown Author

Table of Contents

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