# Can anyone understand anything in physics

## Understand physics - recognize relationships instead of learning by heart

Preface

1. General learning tips
1.1 Learning is an active activity
1.2 Lectures are sometimes useful
1.3 The internet is your friend

2. Mathematical foundations of physics
2.1 Introduction
2.2 notation
2.2.1 Operator priorities
2.3 arithmetic
2.3.1 Percentage calculation
2.3.2 Power calculation
2.3.2.1 Powers of ten and "prefixes"
2.3.2.2 Powers with the base e
2.3.3 Logarithms
2.4 Algebra
2.4.1 Direct and indirect proportionality
2.4.2 Exponential Laws
2.4.2.1 Half-value parameters
2.5 geometry
2.5.1 Analytical geometry
2.5.1.1 Ordinary Cartesian coordinate systems
2.5.1.2 Logarithmic and semi-logarithmic coordinate systems
2.5.1.2.1 Interpolation of a logarithmic scale
2.5.2 Trigonometry (trigonometric functions)
2.5.2.1 Trigonometry in a right triangle
2.5.2.2 Trigonometric functions and any angles
2.5.2.3 Conversion of trigonometric functions into one another
2.5.2.4 Inversion of the trigonometric functions
2.6 Analysis
2.6.1 Differential Calculus
2.6.1.1 Differentiation of linear functions
2.6.1.2 Graphical differentiation of any functions
2.6.1.3 Computational differentiation of any functions
2.6.2 Integral calculus
2.6.2.1 Integration of linear functions
2.6.2.2 Graphic integration of any functions
2.6.2.3 Computational integration of any functions
2.6.3 Differential equations
2.6.3.1 Linear differential equations of the first order
2.6.4 Approximation of functions
2.6.4.1 Taylor series
2.6.4.2 Fourier series

3. Physical quantities and units
3.1 Introduction
3.2 The International System (SI)
3.2.1 SI base quantities
3.2.2 Derived quantities
3.3 Major non-SI units
3.3.1 degrees Celsius and Fahrenheit
3.3.2 Hours, Minutes and Seconds
3.3.3 liters
3.3.4 X-ray
3.3.5 Torr and its synonyms
3.3.6 Atmospheric pressure
3.4 Vector quantities and how to convert them to scalars
3.5 Some interesting connections
3.5.1 Analogies between translation and rotation
3.5.2 Analogies between electricity, fluid dynamics and waves

4. Physical and chemical phenomena
4.1 Introduction
4.2 Some interesting connections
4.2.1 Kirchhoff's rules
4.2.2 Analogy between gas pressure and osmotic pressure
4.2.3 Analogy between osmosis and heat conduction
4.2.4 Conductors and metals
4.2.5 Analogy between the barometer formula and Boltzmann theorem
4.2.6 Analogy between gas discharge and halogenation
4.2.7 Humidity
4.2.8 Absorption coefficient and penetration depth
4.2.9 Electromagnetism and mechanical collision processes
4.2.10 total reflection

Closing word

The chapters, the headings of which are in italics, deal with areas of higher mathematics, knowledge of which is not required for a basic understanding of physics to the extent required for the partial rigor in medical physics. They are intended for particularly interested readers and can safely be skipped.

### Preface

Physics is an extremely extensive, important, but also interesting subject. However, many students with physics as an examination subject do not have the opportunity to perceive the last aspect because they have to work through the complex subject under enormous time pressure. Often the amount of required knowledge from areas that appear so different as mechanics, thermodynamics, electricity and radiation physics leads to stubborn memorization. But that is exactly the wrong way to go.

With this script, I want to show that it is much more effective to take your time and to deal thoroughly with mathematical fundamentals and relationships in order to gain a real understanding of the matter. Once you understand the science you are dealing with, it is no longer difficult to memorize laws, formulas, phenomena and numbers. And so over time you will begin to love physics.

This script is primarily aimed at students in those fields in which a physics exam has to be taken; These include, above all, scientific and medical disciplines such as biology, chemistry, human medicine, veterinary medicine, pharmacy or nutritional sciences, as well as technical fields of study such as biotechnology, computer science or electrical engineering. Secondly, it is intended for particularly interested students and laypeople.

Finally, it could possibly also be helpful for one or the other physics or mathematics students in the first semesters.

This preface would like to close with a motto:

"Don't learn by heart - use your brain!"

Claus D. Volko

### 1.1 Learning is an active activity

Especially in subjects like physics and chemistry, where understanding is the most important thing, it doesn't make much sense to just leaf through some textbooks a few times: In this way, you only notice a fraction of the material and may overlook important relationships.

You can only learn physics properly if you try to understand and actively follow the content of the teaching. Specifically, that means:

- Always have a notebook and a pen with you!
- If a passage seems important, mark it in the book and summarize it in your own words in the exercise book.
- It is advisable to make a note of the page number so that you can look it up if necessary.
- Try to understand all the derivations, even if this can be a bit tedious at times. Only when you understand how you came up with certain laws can you really understand these laws.
- In this context, it is also not bad to repeat the mathematics material of the eighth grade, especially differential and integral calculus, and look up in a mathematics book how to solve simple first-order differential equations. Although this is not required in some fields of study (e.g. medicine), it is a prerequisite for being able to understand many derivations.

### 1.2 Lectures are sometimes useful

Nowadays it has become fashionable to stay away from lectures. Unfortunately - because lectures sometimes bring something. Especially when you have the opportunity to ask the professor questions that are important to you.

To do this, however, you should have familiarized yourself with the subject in advance so that you can follow the lecture properly and ask the right questions at appropriate opportunities.

Incidentally, it is not advisable to write down the lecture like a stenographer. This only leads to the fact that one cannot follow the lecture sufficiently. You may then have written down everything down to the smallest detail in order to be able to read it through in peace at home. But that doesn't do much, because the same subject can usually be found in textbooks anyway. But you have the disadvantage that you miss the opportunity to ask the lecturer questions.

Therefore: It is better to only write down the bare essentials, i.e. only those that are not in the textbook, and try to logically understand the material presented. That brings a lot more. Only in this way does attending lectures really make sense.

### 1.3 The internet is your friend

Our generation has a decisive advantage over the previous generation: Each of us has internet access at home or at university (e.g. on the top floor of the anatomical institute). This gives us an unprecedented opportunity to search specifically for answers to questions that are still open.

If a term or context is unclear to you, you can start a search engine and type in what you want to know. For example: "Thermocouple", "Raoult's law" or "How does this damned MOS transistor work ??". Within a few seconds the search engine will spit out a list of web pages that contain the terms you are looking for.

How useful this list is, of course, depends on the search engine used. In my opinion, Google (http://www.google.com/) is particularly recommended. Google uses a special technique to order the pages according to their relevance. In fact, the pages found with Google are usually much more useful than if you use other search engines.

If Google is still not enough, you can also do it with Altavista (http://www.altavista.com/), Lycos (http://www.lycos.com/), HotBot (http: // www. hotbot.com/) or - especially for German-language sites - Fireball (http://www.fireball.de/).

If the search result is not satisfactory, you still have the option of "posting" a question in a forum.

The newsgroups de.sci.physik, de.sci.chemie and de.sci.medizin and their US counterparts sci.physics, sci.chem and sci.med are of particular interest. But be careful: Since these newsgroups are also read by experienced scientists who, unlike our professors, are not particularly interested in teaching, if you have too trivial questions you easily risk a "flame", i.e. a not particularly polite expression of displeasure. So always try to solve the problem yourself first before asking a question.

Perhaps even more interesting than these newsgroups are the online forums operated by individual institutes at the University of Vienna. In particular, the Institute for Medicinal Chemistry (http://www.univie.ac.at/Med-Chemie/) is a pioneer here. In addition to a "WWW-supported revision course" in which Prof. März and Prof. Kremser ask the students possible rigorosum questions and correct their answers, there has been a discussion forum for Prof. Goldenberg's lecture since winter semester 2001/2002. If you have specific questions of a chemical nature, you can ask them there.

Unfortunately, to date there is no comparable internet offer from the Institute for Medical Physics. On its homepage, only the collections of questions for students preparing for the partial rigorosum are currently of interest.

### 2.1 Introduction

Physics is a quantitative science: it tries to capture various phenomena of nature with numbers and to find general, quantitatively valid laws. To do this, it depends on mathematics as an auxiliary science.

Not every high school student will have exclusively positive experiences with mathematics. Some of them were probably happy about passing the school-leaving exam because they believed they had gotten through mathematics forever. That this is unfortunately not the case may initially have been a shock for many.

But I am of the opinion that anyone who associates bad experiences with mathematics should overcome their shyness. Mathematics is a useful tool that enables one to easily grasp, express and understand complex relationships.

What is it called so often? "We learn not for school, but for life." This is especially true for mathematics: if you can apply it correctly, it is a real help in life.

In the following, I would like to refresh your knowledge of some important chapters in mathematics and show how you can use them to better understand physics.

### 2.2 notation

#### 2.2.1 Operator priorities

I would just like to briefly remind you of a convention that elementary school students should be familiar with anyway: point calculation takes precedence over line calculation. I only mention this because in this script, for aesthetic reasons and to save space, the use of fraction lines is dispensed with. Instead, the slash is used as a division sign.

Now and then expressions of the form a / b + c appear. This specific expression could also be written like this: (a / b) + c. However, it is by no means to be understood as a / (b + c).

It should also be noted that division has higher priority than multiplication. So a / b * c means (a / b) * c and not a / (b * c).

These conventions also apply in textbooks. There, too, one rarely encounters a fraction line.

### 2.3 arithmetic

#### 2.3.1 Percentage calculation

We know this chapter from the AHS lower level; it probably doesn't pose a problem for a single student.

But I would like to point out a little something that might make life a little easier:

The percent sign% actually means "* 1/100". Percentages are therefore nothing more than ratios (proportions), in which one makes use of this percentage sign. So if you have a ratio of share to base value, you multiply it by 1 = 100/100 = 100 * 1/100 = 100% and you get the percentage that makes up the share in the base value.

This is the reason why, for example, the absorption capacity can be given as a ratio or as a percentage: Both spellings mean exactly the same thing.

Formulas from middle school like p = A * 100 / G are basically superfluous ballast and can be forgotten.

#### 2.3.2 Power calculation

The power calculation is a very fundamental area because it enables extremely small or large values ​​to be represented in a manageable manner. For this purpose, one mainly uses powers of ten, i.e. factors with the value 10n, where n is an integer.

ab is in principle a short form for [illustration not included in this excerpt].

This mathematical symbol (the big "Pi") may not be familiar to everyone: it is the product symbol. His statement in this case is that the value a b times must be multiplied by itself.

For computer freaks: In the C programming language, this would correspond to the following code loop: for (product = 1; b; b--) product * = a ;.

These calculation rules follow from this:

Figure not included in this excerpt

The following relationship exists with division:

Figure not included in this excerpt

And pull with the roots. [Figure not included in this excerpt]

##### 2.3.2.1 Powers of ten and "prefixes"

If one uses powers of ten to express extremely large or small values, it is possible to use "prefixes" such as M (mega-), m (milli-) or µ (micro-) in their place. This can make things a little clearer, but is a little harder to work with because it is necessary to know the values ​​of these prefixes - and yet it is a little easier to divide m than 1 km by 1 to divide pm.

The Greek lowercase letter µ means "my", pronounced "mü". This is why some people mistakenly say "mucrometer" instead of "micrometer".

Work with powers of ten as long as possible and only replace them with prefixes at the end.

And only if you think it looks better.

In principle, these signs are similar to the percent sign: Just as the percent sign can be replaced by "* 1/100", you could write "* [not shown in this sample] instead of the prefixes. Here x is different for each prefix, e.g. for n (nano) x = -9.

The consequence of this is that prefixes are not linked to their units, but can be "adopted" when the unit is changed.

Examples:

1 MJ / s = 1 MW

10 kPa = 10 * 101325 kbar (= 1013250000 bar ≈ 1 Gbar).

Attention: It should be noted that you always have to put a bracket around the prefix and the unit. Because [illustration not included in this excerpt] write it down; however, this notation is not common and misleading.

##### 2.3.2.2 Powers with the base e

In addition to base 10, we often find another base: Euler's number e ≈ 2.718. This constant was not determined arbitrarily, but is apparently a natural constant on which many natural processes - e.g. radioactive decay or the growth of populations - are based. These processes can be described by exponential laws (see Chapter 2.3.2.)

For those interested in mathematics: You can easily approximate the value of e by building a Taylor series. See chapter 2.5.4.1.

#### 2.3.3 Logarithms

Now we come to a chapter that is more likely to cause problems than the topics covered in the previous sections. But this is not difficult either.

The logarithmic function is one of the two inversions of the power function; the other is the root. Let us assume that it is given:

Figure not included in this excerpt

Then the following applies:

Figure not included in this excerpt

One says: b is the logarithm of c to base a. This means:

If you want to know which number you have to multiply b times to get c, you extract the bth root of c (or raise c to the power of 1 / b) and get a.

If, on the other hand, you want to know how often you have to multiply the number a to get c, calculate the logarithm of c to base a and get b.

As with the power functions, the bases 10 and e play an outstanding role.It is therefore called [Figure not included in this reading sample] is called natural logarithm (logarithmus naturalis) and is written ln.

Logarithms with different bases can easily be converted into one another. The following applies in general:

Figure not included in this excerpt

lg a = ln a / ln 10 ln a = lg a / lg e

After algebraic transformation one finally obtains ln 10 = 1 / lg e. You are welcome to use the calculator to check whether this is correct.

There are some rules for calculating logarithms that you should master because they could make your life easier in physics, among other things. The most important of them are:

log (a * b) = log a + log b log (a / b) = log a - log b

If you turn back to the section on power calculus (Chapter 2.2.1), you will find that there is a certain analogy here - which is of course not just random, as those interested in mathematics can try to prove if they have time.

In the natural sciences, one often works with logarithmic quantities in order to avoid having to work with numbers that are too large. A well-known example is the pH value (potentia hydrogenii), the negative decadic logarithm of the [figure not included in this reading sample] ion concentration in a solution. It is used in chemistry to determine whether a solution is acidic or alkaline (basic) in character.

Some logarithmic quantities can also be found in classical physics, such as the sound level measure: [Figure not included in this reading sample]. I is the current sound intensity, [illustration not included in this reading sample] a comparison value. In biophysics, the lower hearing threshold is usually used for this purpose (not to be confused with the lower hearing threshold, which indicates the frequency of the lowest sound that can just be heard); in humans this is around [figure not included in this excerpt].

The sound strength, like all other logarithmic measures, is a dimensionless quantity. Strictly speaking, it has the unit 1 - it is a pure number. For the sake of clarity, it has been given a unit, the decibel (abbreviated to dB). 1 dB, like 1 rad (radian, a measure of angle) or 1 sr (steradian, measure of solid angle) is actually a superfluous unit that you could safely leave out. But it is just the convention to state them; that's why you should do it.

### 2.4 Algebra

#### 2.4.1 Direct and indirect proportionality

Relationships exist between almost all physical quantities that can be expressed by linear equations (i.e. without exponents greater than 1 or less than 0 and also without differential quotients). One then says that these quantities are proportional to one another.

Correct handling of proportionality relationships is crucial for understanding various physical and chemical phenomena, e.g. why a lowering of the gas pressure favors reactions that proceed with the release of heat (exothermic reactions) or why the wall tension of thick vessels is greater than that of thin. Therefore I would like to dedicate myself to this topic, although it is not too difficult to see through, a little more detailed.

We already know proportionality relationships from elementary school - keyword: "final calculation". A distinction was made between two different types of inference, on the one hand the direct conclusion ("the more, the more") and on the other hand the indirect conclusion ("the more, the less"). These two types of inference correspond to the two types of proportionality relationships.

In my opinion, the most elegant definitions of the two proportionality relationships are as follows:

Quantities are directly proportional to one another if their ratio is constant; Quantities are inversely proportional to one another if their product is constant.

In a nutshell:

Direct proportionality = constant ratio; indirect proportionality = constant product.

The prerequisite is, of course, that the ambient conditions are constant. For example, the pressure of a gas (see below) only increases proportionally to the reduction in gas volume if the temperature of the gas does not change.

Algebraically, these relationships can be noted in various ways, which can be transformed into one another according to the way in which equations are solved. a and b stand for different changeable (variable) physical quantities and k for an unchangeable (constant) quantity (proportionality factor).

[Figure not included in this excerpt]

If k remains unchanged, then with direct proportionality it follows:

- The ratio between a and b remains constant.
- If you increase a, you automatically increase b.
- If you increase b, you automatically increase a.
- If you decrease a, you automatically decrease b.
- If you decrease b, you automatically decrease a as well.

If k remains unchanged, then with indirect proportionality it follows:

- The product of a and b remains constant.
- If you increase a, you automatically decrease b.
- If you increase b, you automatically decrease a.
- If you decrease a, you automatically increase b.
- If you decrease b, you automatically increase a.

That's all of the witchcraft. A lot can be done with this in practice.

Examples:

General gas equation (thermodynamics): p * V = n * R * T; it follows:

- p and V are indirectly proportional to each other (p * V = k),
- p and T are directly proportional to each other (p = k * T),
- V and T are directly proportional to each other (V = k * T).

Since the general gas constant R is invariable, changing p, V, or T has no effect on the value of R; R acts as a proportionality factor in this equation. The amount of substance n cannot be changed using p, V or T either: it is a constant characteristic of the gas mixture being treated.

Exothermic chemical reactions proceed with the release of heat, i.e. the temperature decreases. Since p and T are proportional to one another, the gas pressure therefore also automatically decreases. Conversely, a reduction in gas pressure, whatever it is, favors the occurrence of exothermic reactions.

The same applies to endothermic reactions that take place with the absorption of heat: They are favored by an increase in the gas pressure.

Frequency, wavelength and speed of propagation (waves):

ν * λ = c; it follows:

- ν and λ are indirectly proportional to each other (ν * λ = k),
- ν and c are directly proportional to each other (ν = k * c),
- λ and c are directly proportional to each other (c = k * λ).

The Greek lowercase letter λ means "lambda". In physics it stands for the wavelength. The Greek lowercase letter ν means "ny"; it stands for the frequency.

Consequences:

Ultraviolet radiation (UV) has a higher frequency than visible light (the frequency is higher than that of violet light), so its wavelength is lower than that of visible light.

Infrared radiation (IR), on the other hand, has a lower frequency than visible light (the frequency is lower than that of red light), so its wavelength is higher than that of visible light.

In the case of electromagnetic waves, E = h * ν = h * c / λ also applies. That is the reason why higher-frequency, i.e. short-wave rays have greater energy than comparatively low-frequency, long-wave rays.

Laplace's law (fluid dynamics):

p = 2 * ζ / r; it follows:

- p and ζ are directly proportional to each other (p = k * ζ),

- p and r are indirectly proportional to each other (p = k / r),

- ζ and r are directly proportional to each other (ζ / r = k).

The Greek small letter ζ means "zeta" and stands for the surface density of the surface energy, also called specific surface energy or wall tension.

One consequence of these relationships is, for example, the fact that an increase in the radius r of a blood vessel by 10% also leads to an increase in the wall tension, by 10%, provided that the pressure p remains constant. This is the cause of a vascular disease called aneurysm.

Continuity condition (fluid dynamics):

Figure not included in this excerpt

The areas are therefore directly proportional to each other, as are the speeds. However, the areas are inversely proportional to the speeds. The ratios between the areas or the speeds are therefore indirectly proportional to one another.

Expressed algebraically: [Figure not included in this reading sample]

Here K is the ratio of the proportionality constants between A1 and A2 or v1 and v2. Since these two constants of proportionality are equal, K = 1.

Figure not included in this excerpt

So here is a special case of indirect proportionality: The relationships are reciprocal.

By further reshaping depending on the given and sought values, one obtains:

Figure not included in this excerpt

Admittedly, this may have been a somewhat lengthy and cumbersome explanation, since one could have obtained these relationships without further considerations by simply transforming the first equation [illustration not included in this reading sample]. But I wanted to show how different proportionality relationships can be linked with one another.

#### 2.4.2 Exponential Laws

In addition to proportionality relationships, exponential relationships are the most common way that physical quantities can be linked to one another.

The general formula of an exponential law is:

Figure not included in this excerpt

The constant k is often negative here because such laws are mostly used to describe an exponential decrease.

Such laws describe, among other things:

- the decrease in radioactive activity, a
- the altitude dependence of the air pressure (barometric altitude formula),
- the energy distribution in the flowing fluid (Boltzmann theorem),
- the charging current of a capacitor,
- the decay current of a coil.

For those who are particularly interested: In the section on differential equations, Chapter 2.5.3, I will go into how such laws can be derived from differential quotients.

In addition to the form mentioned above, the following form is also used to describe an exponential decrease:

Figure not included in this excerpt

In contrast to k, µ is always positive.

Caution: The stipulation that k can be negative or positive, while µ is always positive, was not made by an internationally recognized standardization body, but solely by myself. I will follow these rules in this script in order to

to increase the clarity. But it could be that you will come across a formula in a textbook in which µ stands for a negative value.

##### 2.4.2.1 Half-value parameters

Sometimes it is required to work with half-lives, half-value thicknesses and other "half-value parameters". Generally speaking, the half-value parameter of an exponential function is the value x for which: f (x) = f0 / 2.If we insert this into the general formula for exponential laws and de-logarithmize the equation, we get:

Figure not included in this excerpt

or.

Figure not included in this excerpt

or.

Figure not included in this excerpt

With knowledge of the half-value parameter [figure not included in this reading sample] we can calculate the constant k (or µ) with the help of the last equation. This then allows us to compute any value f (x) provided we know some value f (x).

Example:

The air pressure in the atmosphere is 101,325 Pa. An altitude difference of 5 km results in a pressure decrease of around 50%. How big is the pressure at a height of 1 km?

Figure not included in this excerpt

Caution: If someone intends to work with the approximation on which this example is based for an examination, then he should immediately put this plan aside. At least the partial rigorosum in medical physics requires that you work with the barometric altitude formula (see chapter 4.2.6), otherwise you will be penalized with points deducted.

So that I am not misunderstood: This does not mean that the derivation of exponential laws would be inexact or should not be used if the half-value was known. On the contrary: for some test examples (especially from the chapter on radioactivity) this procedure is even necessary.

### 2.5 geometry

#### 2.5.1 Analytical geometry

Analytical geometry was founded by the famous mathematician and philosopher René Descartes. He had the idea of ​​developing a coordinate system to better describe geometric figures mathematically. Subsequently, the vector calculation arose from this.

The basics should be well known from school. But we want to briefly repeat how to deal with coordinate systems.

##### 2.5.1.1 Ordinary Cartesian coordinate systems

One skill that every scientist or medical professional must possess is reading graphs. We have already dealt with this in the AHS lower level; so it shouldn't really pose a problem in and of itself.

The only thing you really have to pay attention to is the yardstick. The measured values ​​only make sense if this is known. If it is not specified, it must be calculated. To do this, measure a distance on the x-axis (abscissa), the actual value of which is known, and divide the latter by the measured stress, including the y-axis (ordinate).

Example:

Before us is a diagram that describes the decay of a radioactive isotope. We see that 25 mm on the x-axis corresponds to one second. So the x-scale is 1 s / (25 mm) = 0.04 s / mm. Now we measure an x ​​difference of 5 cm. This corresponds to a time difference of 5 cm * 0.04 s / mm = 50 mm * 0.04 s / mm = 2 s.

##### 2.5.1.2 Logarithmic and semi-logarithmic coordinate systems

Dealing with logarithmic or semi-logarithmic coordinate systems is a bit more complicated.

First of all, I would like to turn to the question of what these coordinate systems are used for. The answer is clear: they are not absolutely necessary. Like so much in mathematics, they can make life a little easier. (The opposite is of course also possible, if not more often the case.)

Logarithmic and semi-logarithmic coordinate systems allow the graphs of exponential functions (see Chapter 2.3.2) to be represented in the form of straight lines. This makes it much easier to draw the function, which should be of particular interest to all of us whose artistic talent is not as strong as their intellectual talent. Another advantage is that it is now possible to determine the exponential constant without any problems: it corresponds to the increase in the logarithmic function.

Not obvious ad hoc? Let's take a look at algebra. The exponential function reads:

Figure not included in this excerpt

If we apply it to an ordinary coordinate system, we get a curve; it is mathematically positive (i.e. counterclockwise) curved and, depending on whether the exponential constant k is positive or negative, it either increases to infinity or decreases to a limit value that is never reached (asymptotic behavior).

If we take the logarithm (and that, since the base e occurs in the function, preferably naturaliter, as the Latins would say), we get:

Figure not included in this excerpt

This equation corresponds to a linear function. We remember from school: The general formula for a linear function is:

y = k * x + d

k is called the rise of the function. This increase can easily be read from a graph. To do this, mark any two points on the functional line, measure the difference between their y-coordinates and divide them by the difference between their y-lines. Short:

Figure not included in this excerpt

Depending on whether the y-coordinate of the point further to the right is the larger or the smaller, k is positive or negative. If k is positive, the function increases linearly; if k is negative, it decreases linearly.

The exponential constant k of an exponential function is therefore equal to the increase k of the linear function that results when we log both sides of the functional equation to the base e.

This is probably also the reason why the exponential constant, which is usually abbreviated here as µ, is also called the linear absorption coefficient in the law of absorption for electromagnetic waves.

When we use a logarithmic coordinate system, we are plotting the graph of the logarithmic function in and of itself. From this k can easily be determined, namely as with any linear function (see also the section on differential calculus, Chapter 2.5.1).

The same applies to semi-logarithmic coordinate systems. These are coordinate systems with logarithmic measurements on one axis and linear measurements on the other. Here the scale is half linear (so "usually") and the other half is logarithmic.The exponential constant is then of course not equal to Δy / Δx, but in the case that the y-axis is the logarithmic and the x-axis is the linear, [figure not included in this reading sample], or in the opposite case [figure in this Excerpt not included]. Here a stands for the base that is used for the logarithmic scale. Most of the time this is the number 10.

Please note: Although the logarithms of the function values ​​are plotted on the logarithmic axis, it is not labeled with the logarithmic but with the actual values. This is the reason why the value 1 (= any number to the power of zero) is at the origin (it corresponds to the "zero point" of common coordinate systems) of every logarithmic axis. If the point is now marked at a certain distance [figure not included in this reading sample], we intuitively recognize that the base a (in our example 10) is used, with a distance of x being a power of this number (e.g. even a power of ten).

But you have to pay attention to:

If we move on the axis by a distance of length x, the value of the new point is not a greater than that of the previous one, but a times as large.

This may sound a little complicated. In fact, it is not entirely trivial. But try, following my explanations above, to draw a logarithmic (or semi-logarithmic) coordinate system yourself. Then you will surely understand what I mean.

2.5.1.2.1 Interpolation of a logarithmic scale

What can really be a headache is interpolating a logarithmic scale.

Interpolating means finding the exact value of a point that lies between two labeled points. This is not too difficult on linear scales (see Chapter 2.4.1.1): You only have to know the scale - i.e. the ratio between the actual and measured unit value - and multiply this proportion by the measured value. In principle, you are working with a direct proportionality relationship.

Since logarithmic coordinate systems are not linear, you cannot work with proportions (in this way). Perhaps you've already guessed it: the relationship between measured and actual value can be described by an exponential law.

This matter may not be entirely trivial, at least for non-mathematicians, non-technicians, and non-physicists. In principle, the exponential law is as follows:

Figure not included in this excerpt

The index t here means “actually”, g “measured”; a is the base (usually 10), [illustration not included in this reading sample] the distance on the scale that corresponds to a multiplication by a.

This law may look a little complicated, but it is the only way to get any value exactly from a logarithmic scale. So memorize it well. You could still use it. (Although the professors are satisfied with estimates, at least for the partial rigorosum in Medical Physics, it is safer to work as precisely as possible.)

Example:

A horizontal logarithmic scale is labeled with 1 Pa at the origin and 10 Pa at a distance of 2 cm. Which pressure corresponds to the point 8 mm to the right and which of those 8 mm to the left of the origin?

Solution: The base a is obviously 10, the actual value of the origin [illustration not included in this excerpt]. From this follows the formula:

Figure not included in this excerpt

It is common (and hopefully known) for coordinates on horizontal axes to increase from left to right. The coordinates (i.e. measured values) to the right of the zero point are therefore always positive. For [Figure not included in this excerpt] Pa.

8 mm to the left of the origin of the coordinate axes therefore means [Figure not included in this reading sample] Pa.

From this we can also see that on a logarithmic scale, the further we move in a negative direction, we come closer and closer to the value 0, but never quite reach it; therefore the actual value can never be negative.

And now a more difficult task.

In front of us is a semi-logarithmic coordinate system that describes the relationship between the absorber thickness (abscissa, i.e. x-axis) and the intensity of the non-absorbed radiation (ordinate, i.e. y-axis). The y-axis is logarithmic, the x-axis linear. The origin of the y-axis has the value [Figure not included in this reading sample]; at a distance of 4.5 cm it is labeled with [illustration not included in this reading sample]. From the x-axis we can immediately see that a scale length of 1 cm corresponds to an actual thickness of 5 m. Only two points are marked in the diagram; they have the coordinates (x = 2 cm, y = 10 cm) or (x = 4 cm, y = 1 cm). How much is the half-value thickness?

Medical students at the University of Vienna: If you can answer this question correctly, you don't have to worry about the pure arithmetic tasks in the Rigorosum. This one is by far more difficult than any arithmetic problem that could be asked of you. In order to answer them, you must have read carefully up to now and also understood what you have experienced.

Let's first look at the y-axis. At first glance we can see: The base a is 10, the actual value of the origin [illustration not included in this reading sample]. From this follows the formula:

[Figure not included in this excerpt]

We can now find the linear absorption coefficient µ (see Chapter 4.2.8)

calculate by dividing one equation by the other. (For this to be allowed, [Figure not included in this reading sample] must of course be different from zero. That is also the case, otherwise the y-coordinates of all points would be zero and the graph of the function would correspond to the x-axis.)

So if we divide (I) by (II), we get:

[Figure not included in this excerpt]

In order to determine the half-value thickness, we only need to insert it into the corresponding formula and get:

[Figure not included in this excerpt]

#### 2.5.2 Trigonometry (trigonometric functions)

To understand some physical relationships, it is necessary to know the meanings of the two basic trigonometric functions sine and cosine.

##### 2.5.2.1 Trigonometry in a right triangle

The term "trigonometry" is derived from the triangle (triangle, actually "three-angle"). The two angles of the right triangle, which differ from 90 degrees, can be characterized with the aid of the trigonometric functions by the ratios between the lengths of the sides of the triangle.

A right triangle consists of two cathets and the hypotenuse. The hypotenuse (usually denoted by the letter c) is the side opposite the right angle (90 °); the cathets (a and b) are the other two sides. In relation to a certain angle, the opposite side is called the opposite side and the other side is called the adjacent side.

The ratio between the opposite side and the hypotenuse is called the sine of the angle, the ratio between the adjacent side and the hypotenuse is called the cosine of the angle.

Algebraic formulation: Let α be the angle, a the opposite side, b the adjacent side and c the hypotenuse. Then the following applies:

sin (α) = a / c (I)

cos (α) = b / c (II)

The Greek lowercase letter α is - as everyone probably knows - "alpha".

If we divide (I) by (II), we get the ratio between the opposite side and the adjacent side. This is also called the tangent of the angle:

sin (α) / cos (α) = a / c / (b / c) = a / c * c / b = a / b = tan (α)

Which values ​​can trigonometric functions take? Since the hypotenuse is always the longest side of a triangle based on the Pythagorean theorem [illustration not included in this reading sample], neither the sine nor the cosine of an angle can be greater than

Be 1. In any case, negative values ​​are also impossible when it comes to an angle in a right-angled triangle, since the lengths of the sides of a triangle are always positive. (We will see later that there are indeed angles for which the trigonometric functions can take negative values.) For 0 ≤ α ≤ 90 °, the following applies:

0 ≤ sin (α) ≤ 1

0 ≤ cos (α) ≤ 1

Only the tangent function can give larger (namely up to infinitely large) values.

When do sine and cosine reach their maximum (in the area of ​​the right-angled triangle) and when do their minimum values?

Let us first look at the sine function: We have sin (α) = a / c. It follows:

- sin (α) = 0 if a = 0. This would mean that the opposite side would be infinitely small. According to the Pythagorean theorem, we get b = c. So our triangle would be a single line representing both the adjacent b and the hypotenuse c. The angle α between b and c would consequently be zero.
- sin (α) = 1 if a = c. From the Pythagoras theorem it follows that b = 0. Here we have the opposite case: The triangle would be a line which would correspond to the sides a and c. The (of course not really visible) angle α between b and c would be 90 °.

This can be justified algebraically: The sum of the angles in a triangle is always 180 °. Since the right angle γ = 90 ° and the angle between a and c = 0, it follows that α = 90 °.

Everyone probably knows that the Greek lowercase letters β and γ are called "beta" and "gamma", respectively.

Consequently, the sine function for angles occurring in a right-angled triangle has its minimum at 0 and its maximum at 90 °.

What about the cosine function? From cos (α) = b / c it follows:

- cos (α) = 0 when b = 0; In this case, α is equal to 90 ° (see above).
- cos (α) = 1 if b = c and therefore a = 0; α is then zero (as can also be seen from the explanations above).

So we see: the sine function has its maxima where the cosine function has its minima, and vice versa.

I just used the plural forms of maximum and minimum because there are actually more than two angles where these values ​​are reached - namely, an infinite number. However, these angles are not possible in a right-angled triangle. So let's say goodbye to this geometric figure and see how the two basic trigonometric functions are defined for any given angle.

##### 2.5.2.2 Trigonometric functions and any angles

For a general definition of the trigonometric functions sine and cosine that is valid for all angles, we use the unit circle. This is a circle with radius 1 (which is actually a dimensionless number). Its center point should be at the origin (zero point) of an ordinary Cartesian coordinate system.

Anyone who cannot get used to the idea that the radius of the unit circle is dimensionless, although it is a specification of a length and the length is a dimension, should imagine that the coordinate axes are not the distance between one lying on the circle Point is plotted from the origin, but its ratio to the radius of the circle.

To determine the sine or cosine of a certain angle, we draw a straight line that goes through the origin of the coordinate axes in such a way that it encloses the desired angle with the positive (i.e. to the right of the origin) part of the x-axis. If we now consider the coordinates of the point at which the straight line intersects the circle, the following applies:

The x-coordinate of the intersection is equal to the cosine of the angle; the y-coordinate of the intersection is equal to the sine of the angle.

This agrees with the explanations for the special case of the right-angled triangle. Because: If we look at an angle between 0 and 90 °, we get a right-angled triangle if we draw a straight line parallel to the y-axis (i.e. vertical) through the intersection. Since the radius of the unit circle is equal to 1, the ratio sin (α) = a / c between the y-coordinate of the point and the radius is equal to the y-coordinate of the point. The same applies to the ratio cos (α) = b / c.

So we see: The minimum of the trigonometric functions sine and cosine is actually not zero, but -1. This value is reached by the sine at an angle of 270 ° and the cosine at 180 °. Zero points are found for sine at 0, 180 ° and 360 °, for cosine at 90 ° and 270 °.

The full circle is known to have an angle of 360 °. Here sine and cosine have the same values ​​as with 360 ° - 360 ° = 0. If we go beyond 360 °, we basically go through the circle a second time. We notice that the same values ​​result as in the first run, only the angles are shifted by 360 °. Formulated algebraically: sin (ϕ) = sin (ϕ - 360 °). The same also applies to cosines.

The Greek lowercase letter ϕ means "phi". In mathematics it is mainly used to indicate any angle. In physics, the electrical potential is also referred to with this letter.

If we go through the circle a third time, the same values ​​result again - which is clear: because sin (ϕ) = sin (ϕ - 360 °), for example sin (800 °) = sin (800 ° - 360 °) = sin (440 °) = sin (440 ° - 360 °) = sin (80 °).

One can therefore formulate generally for positive angles ϕ: sin (ϕ) = sin (ϕ mod 360 °)

cos (ϕ) = cos (ϕ mod 360 °)

The operator mod is called "modulo" and specifies the remainder of the division.

What about negative (i.e. clockwise) angles? Now, since sin (ϕ) = sin (ϕ - 360 °), sin (ϕ) = sin (ϕ + 360 °) must also apply. If we add 360 ° sufficiently often to a negative angle, we finally arrive at a positive angle that has exactly the same sine and cosine values. So the following applies:

sin (ϕ) = sin (ϕ + n * 360 °)

cos (ϕ) = cos (ϕ + n * 360 °)

It should be noted that n must be an integer, i.e. only zero after the decimal point.

The last two equations best illustrate what sine and cosine are: namely, periodic functions. Their values ​​are repeated every 360 °. One could therefore also say that the period of sin (ϕ) and cos (ϕ) is 360 °.

That is the reason why these trigonometric functions are so important for physics: They can be used to describe periodic processes in a mathematically exact manner. Examples of this are the oscillations of a pendulum, the propagation of a wave or the time course of an alternating voltage.

But also some physical quantities (namely those scalar quantities which are derived from vectors) are defined with the help of such functions. (See Chapter 3.4.1, where we will deal with this in detail.)

The trigonometric functions also play a major role in optical phenomena such as refraction and reflection, because here all calculations work with angles.

Finally, we need the trigonometric functions in mechanics, e.g. to determine the thrust and pressure components of a force acting at an angle on a body. To do this, we draw a right triangle and use sine or cosine to calculate the relationships between the components.

Example:

Someone is pressing with a force of 10 N on a body that is stuck on an inclined plane. The print is in the vertical direction; on the other hand, the plane is inclined by 60 ° (i.e. the angle between the inclined plane and the ground is 60 °). How big is the thrust and how big is the pressure component of this force?

By definition, the shear component is the component of the force vector that is parallel to the attack surface, and the pressure component is the component perpendicular to it. (It is also sometimes said: The thrust component is normal to the body axis, i.e. it is perpendicular to it, and the pressure component is parallel to it. That means exactly the same thing.) All three together form a right-angled triangle. Since we know the amount of the force vector, we only need one more angle of this triangle to be able to calculate the two force components.

With the help of a sketch it can be determined that the angle between the force vector and the pressure component is equal to the angle of inclination of the inclined plane. To do this, let's draw the ground and the inclined plane, viewed from the front, as straight lines that intersect but are not perpendicular to each other (so that their intersection angle is different from 90 °). In order to keep the following explanations concise, we denote the straight line that represents the floor as b and the straight line that represents the plane as e.

Let us now consider the smaller of the two angles they include: Let it be the angle of inclination α (in the specific example 60 °).

Now we choose any point K on e: it indicates the position of the body on which the force is being exerted.

If we draw a straight line f through point K, which is perpendicular to b, we get the straight carrier line of the force vector. At the same time, a right-angled triangle results, which is enclosed by the straight lines b, e and f. In addition to the right angle between f and b, we know of another angle, namely that between b and e: It is α. Since the sum of the angles in each triangle is 180 °, the result for the angle β between e and f is 90 ° - α (in our example 30 °).

Let us now draw a straight line d through point K, which is perpendicular to e.This is the carrier line of the pressure component of the force. Since the angle between d and e is 90 ° and a straight line encloses an angle of 180 ° with itself, the value 180 ° - (90 ° + β) = 90 ° - β = 90 results for the angle between d and f ° - (90 ° - α) = α.

With this knowledge, the rest of the example should not cause any problem: Since the amount of the force vector f is equal to the length of the hypotenuse, the thrust component e is equal to the length of the opposite side of α and the pressure component d is equal to the length of the adjacent side, we get:

Figure not included in this excerpt

Sample: According to the Pythagorean Theorem, the condition [illustration not included in this reading sample] must be fulfilled. This is actually the case, as you are welcome to check with the calculator (or roughly in your head).

##### 2.5.2.3 Conversion of trigonometric functions into one another

If we draw the graphs of sin (ϕ) and cos (ϕ), we see that they would be congruent in themselves if one were not shifted by 90 ° compared to the other. It is said that the phase distance between sine and cosine is 90 °. Hence:

sin (ϕ) = cos (ϕ + 90 °) cos (ϕ) = sin (ϕ + 90 °)

Sine functions can therefore be converted into cosine functions and vice versa.

##### 2.5.2.4 Inversion of the trigonometric functions

The inverse of the sine function is the arc sine (arcsin), that of the cosine is the arc cosine (arccos). There is of course also an arctan. Both functions are supported by every standard pocket calculator. They are used to determine an angle whose sine or cosine (or tangent) corresponds to a certain value. The emphasis here is of course on the word "one", because there are an infinite number of angles for which a certain trigonometric function delivers the same value. But if you know an angle, then, as explained in the previous chapter, you can easily determine any other with the same sine, cosine and tangent:

sin (arcsin (ϕ)) = sin (n * 360 ° + arcsin (ϕ)) cos (arcsin (ϕ)) = cos (n * 360 ° + arcsin (ϕ)) tan (arcsin (ϕ)) = tan ( n * 360 ° + arcsin (ϕ))

The same goes for arccos and arctan. Here, too, n is an integer.

So far we have given the angles in degrees as we are used to from everyday life. In physics, however, another measure is mostly used, the radiant measure.

An angle can be specified by the length of the arc of the unit circle, which is measured between the x-axis and the point of intersection of the straight line with the circle.

Since the circumference of the unit circle is 2 * π, the following applies:

2 * π = 360 °

It follows:

1 ° = π / 180

(Certainly everyone knows the Greek lowercase letter π: It is called "pi" and in mathematics stands for the circle number, the value of which is approx. 3.141592654.)

Although the arc length of the unit circle is dimensionless (it corresponds to the ratio of the arc length of an actual circle to its radius), it is given a unit: 1 radian, or 1 rad for short. It is thus clearly recognizable that it is a question of an angle specification.

For the value of 1 rad, transforming the last equation gives:

1 rad = 180 ° / π

### 2.6 Analysis

#### 2.6.1 Differential Calculus

The differential calculus is used to find a function that indicates the rise of a given antiderivative or its course.

There are two methods for this: graphical and computational differentiation. Their specific advantages and disadvantages are quickly explained: While the graphical method can be applied to any function graph, the computational method requires knowledge of the function equation. But it is much more precise.

In some fields of study, e.g. medicine, only graphic differentiation is required. However, it is worthwhile to recall the rules of the computational method, because this is the only way to understand some of the derivations of physical laws.

##### 2.6.1.1 Differentiation of linear functions

As said above, differentiating means finding the slope of a function. In the case of linear functions, i.e. functions of the form y = k * x + d, the increase is the same at every point: It is the quotient of the difference between the y coordinates of two points broken by that of the x coordinates of the same two points: Δy / Δx. Now Δy is nothing else than y1 - y2. We obtain:

y1 - y2 = k * x1 + d - (k * x2 + d)

y1 - y2 = k * x1 + d - k * x2 - d y1 - y2 = k * (x1 - x2)

Δy = k * Δx

Δy / Δx = k

Thus the coefficient k is equal to the slope of the linear function. So the slope function is:

Figure not included in this excerpt

The expression dy / dx is called the differential quotient. See also chapter 2.5.1.3.

If we plot its graph, we will find that it is a straight line that runs parallel to the x-axis.

If the equation of the antiderivative is given, the associated increase function can be read off directly from it.

If we only have one graph, we choose any two points, measure the differences between their y-coordinates Δy and their x-coordinates Δx and calculate the quotient Δy / Δx.

It wasn't too difficult.

##### 2.6.1.2 Graphical differentiation of any functions

Not all functions are differentiable. Functions that cannot be defined by an equation but only by a set of points, for example, cannot be differentiated. Strictly speaking, this chapter should therefore be called "Graphical Differentiation of Any Differentiable Functions", but that would sound too bulky.

The rise of functions of higher order, in which the variable does not only occur in the first power, is not constant, but changes continuously; it depends on the function variable itself. This is the reason why you actually have to plot or calculate a slope function here. In this chapter we would first like to devote ourselves to the graphical method.

The rise of the function at a certain point X, the x-coordinate of which is x, is equal to the rise of the tangents through this point X. That is the straight line which the function intersects at this single point X only.

Graphical differentiation is about drawing a straight line that is as similar as possible to this tangent - with a little practice, it will soon no longer be as difficult as it might sound. In addition, graphical differentiation is never as precise as the computational method.

The rise of the tangents is then determined in the same way as the rise of any other linear function.

When drawing the rise function, a suitable scale must be selected so that there is space for both the rise maximum and the rise minimum in the diagram.

Since you are under time pressure during the exam, it makes no sense to determine and draw the rise of as many points as possible. The following procedure is recommended:

- First you determine the points at which the slope is zero. These are the so-called local maxima and minima - perhaps they are also known to you as "extreme points". An extreme point is characterized by the fact that the direction of the rise changes in it. If the function falls to this point and rises after it, it is called the low point; the opposite case is called the high point. At one extreme point itself, the function neither increases nor decreases: the increase function has a zero point here.
- Then you determine the points at which the rise or fall is maximum. These are the turning points, i.e. those points at which the curvature of the function changes. If it was previously mathematically curved positively (counterclockwise), it is later mathematically negatively (clockwise) curved or vice versa. The second step is to measure the local maximum increases or decreases in the functions and record the largest value as close as possible to the edge of the diagram. This results in the scale that must be observed from now on for all further intermediate points.

From a mathematical point of view, the curvature corresponds to the increase in the slope function, i.e. the second derivative of the antiderivative. At the turning point the curvature is neither positive nor negative, but exactly zero.

- Finally, in order to increase the accuracy, the increase in points between the turning points and extreme points can also be calculated. If you are practiced to a certain extent, this is often no longer necessary: ​​You can also estimate very well how the increase function must run.
- However, if a function has neither extreme nor turning points, you have to take several points that are fairly far apart, determine the rise prevailing there and try to draw a curve that looks as good as possible, on which with a high degree of probability not only the ascertained rise values, but also as many as possible between the points used for the measurement.

It might be useful to also note the following special cases:

- A second order function of the form [illustration not included in this reading sample] has a linear derivative function of the form y = k * x + d, where (as we will see in the next chapter) k = 2 * a and d = b. The graph of the derivative function is a straight line. The task can therefore be solved perfectly by determining the increase in any two points, plotting the increase values ​​on a suitable scale and connecting them with one another with a straight line.
- The rise function of the sine function y = sin (x) is equal to the cosine function dy / dx = cos (x).
- The rise function of the cosine function y = cos (x) is equal to the sine function mirrored around the x-axis, i.e. dy / dx = -sin (x).
- The increase function of the logarithm function y = ln x is dy / dx = 1 / x.

##### 2.6.1.3 Computational differentiation of any functions

The mean rise of a function is Δy / Δx. In order to calculate the instantaneous increase at a certain point x, one lets Δx approach zero and thus obtains the differential quotient dy / dx.

Mathematically exactly formulated:

Figure not included in this excerpt

If the differential quotient is formulated in general, i.e. for any value x, it results in the derivative function.

Many mathematicians have already dealt with the question of what the differential quotient looks like for different functions. In doing so, you came across some useful rules. The three most basic are:

Figure not included in this excerpt

That means: The exponents become coefficients in the differential quotient; the exponent itself is reduced by 1. (II) says that the differential quotient of the sum of two functions corresponds to the sum of the differential quotients of the two functions. Effectively this means that you can add any number of expressions of the type k * xn to one another; In order to calculate the differential quotient of the sum, the differential quotients of the individual functions are calculated and then added together.

Example:

Figure not included in this excerpt

There is also the product rule (III), the quotient rule (IV) and the chain rule

(V):

Figure not included in this excerpt

It may not be immediately clear how to understand the chain rule.

Therefore an example:

Figure not included in this excerpt

The chain rule is used to resolve function nesting in product chains.

#### 2.6.2 Integral calculus

Integration is the reverse of differentiation: The rise function is given to find out the antiderivative to it. Mathematicians of many generations have racked their brains about how to do this until it was finally discovered that the antiderivative (in principle!) Describes the area between the graph of the slope function and the x-axis.

##### 2.6.2.1 Integration of linear functions

The integration of a linear function is particularly easy: Since your graph is a straight line, the area between it and the x-axis is equal to the area of ​​a rectangular triangle. As is known, this is x * y / 2 if x and y stand for the two sides.

If the graph goes through the zero point (i.e. the function has the form y = k * x), the area we are looking for is:

Figure not included in this excerpt

The expression ∫ (y (x) * dx) stands for the indefinite integral from y to x. You always write "* dx" because y (x) stands for the differential quotient of the antiderivative dY / dx. From this it follows: ∫ (y (x) * dx) = ∫ (dY / dx * dx) = ∫dY = Y.

However, this is only one of an infinite number of antiderivatives. We can add any constant value that is independent of the form variable x to the antiderivative found. The result is then also a valid antiderivative.

This can easily be justified by the first rule of computational differentiation: [Figure not included in this reading sample].

But that doesn't mean that every arbitrary function can be taken as an antiderivative! (Otherwise the integration would be too easy - or pointless.) No, the emphasis is on "independent of x". Because only under this condition does the factor zero occur when differentiating, which means that the added constant is omitted. [Figure not included in this excerpt].

Such an antiderivative is also called an indefinite integral. The definite integral is the area in a certain range x1 to x2. This is the difference between the area up to the point with the x coordinate x2 minus that up to x1. This means:

[Figure not included in this excerpt]

For a function of the form [illustration not included in this excerpt]. If we want to calculate the area from the origin to a certain point with the x-coordinate x2, x1 = 0, and it is sufficient to simply insert the x-coordinate of this point into the antiderivative - that's it.

But what does the matter look like if the graph of the function to be integrated does not go through the origin of the coordinate axes?

In this case the function has the form y = k * x + d, where d is different from zero. Now we can no longer simply multiply y by x for integration, because there is no longer a triangle between the zero point and a certain point. The corner of the triangle, which was at the origin earlier, is shifted either to the right or to the left, depending on whether d is negative or positive; the graph begins to "grow" elsewhere. But we can easily calculate its x-coordinate. It is the point of intersection of the function with the x-axis (also called "zero point"). That is, the y-coordinate of this point is zero. We obtain:

0 = k * x + d

x = -d / k

If we subtract this x-value from the x-coordinate of the point up to which we want to calculate the integral, we get Δx. This is one leg of the triangle; the other is still the y-coordinate of the point. The area is thus:

Figure not included in this excerpt

The indefinite integral is therefore:

Figure not included in this excerpt

This is the general formula of the indefinite integral of a linear function. The determination of a certain integral works analogously to my explanations regarding the special case y = k * x.

What if we calculate the specific integral for a range in which the graph is below the x-axis, e.g. for y = x the range from -1 to 0? Let's try it out: [Figure not included in this excerpt]. So we get a negative value. Now it is the amount of this value which corresponds to the area between the graph and the x-axis.

Now the interesting thing is what happens if we take an area in which the graph is partly above and partly below the x-axis. Maybe you already know or can guess it. To provide even more clarity, an example: We integrate y = x in the range from -1 to +1 and get: [Figure not included in this reading sample]. The area below the x-axis is therefore subtracted from that above the x-axis. Since both areas had the same amount in this example, the total area was 0.

Perhaps this is a good time to explain the real nature of the antiderivative: it specifies the rule by which the sum of all function values ​​in a given section can be calculated.

This is again expressed somewhat imprecisely. After all: What is the sum of all function values ​​of a continuous, i.e. uninterrupted function in a certain range? If it doesn't just consist of a single point, there are an infinite number of points in between. For each of these infinitely many points there is a function value. So the sum of the function values ​​would have to be infinitely large in such a range. In fact, a certain "resolution" is used, which depends on the degree of the function. You can imagine it like this: Instead of cutting a pretzel stick into an infinite number of tiny pieces, measuring their lengths in thousands of years of work and then adding them up (although you would have cut your fingers very soon anyway, which is not exactly pleasant ), choose larger pieces and calculate the average length of each; these mean lengths are then added.But if you want to know more, you should consult a book on higher mathematics: This topic clearly goes beyond the material of this script.

That is, for example, the reason why one calculates the integral over this range and then divides it by the range span to calculate the mean value of any function in a certain range: In principle, this procedure is the calculation of the arithmetic mean, i.e. adding all values ​​and Dividing by their number, very similar.

##### 2.6.2.2 Graphic integration of any functions

Just as not all functions can be differentiated, not all functions can be integrated either. The same restriction applies to the following section as to Chapter 2.5.1.2.

The graphical integration of any functions is even easier than the graphical differentiation of the same: Since the antiderivative essentially describes the area between the function graph and the x-axis, it is sufficient to determine the area up to the given point. The best way to do this is to draw a square grid, count the number of whole boxes contained in the area and estimate how many boxes total the remaining part of the area. The finer you draw the grid, the more accurate the result. In tests, however, the grid is often already drawn so that you only need to read it off.

The number of boxes must then be multiplied by the value of one box. To do this, you have to know the scale on both the x and y axes.

When drawing the antiderivative, a suitable scale has to be chosen so that there is space for both maxima and minima in the diagram.

It should be noted that areas between the x-axis and underlying parts of the graph of the function to be integrated (for which y has a negative value) must be subtracted from the total area.

The following special cases should perhaps also be noted, even if they can be derived using the method described (as well as from the explanations in Chapter 2.5.1.2):

- The antiderivative of the sine function y = sin (x) is equal to the cosine function mirrored around the x-axis, i.e. ∫ (y (x) * dx) = -cos (x) + c.
- The antiderivative of the cosine function y = cos (x) is equal to the sine function ∫ (y (x) * dx) = sin (x) + c.
- y = 1 / x results in the integrated logarithm function ∫ (y (x) * dx) = ln | x | + c.

| x | stands for the amount of x. Without the absolute dashes, the antiderivative for negative x-values ​​would not be defined because there is no value u for which applies [figure not included in this reading sample].

##### 2.6.2.3 Mathematical integration of any functions

This chapter is anything but trivial. In some cases the computational determination of antiderivatives is extremely complicated; in the case of non-polynomial functions, one often does not even bother to integrate them exactly, but converts them into Taylor series and then integrates them (see Chapter 2.5.4.1). An example of a function that cannot be integrated in elementary terms is [Figure not included in this reading sample].

As I said, integrating means finding the antiderivative associated with a function. If one differentiates this antiderivative, the given function is obtained as a derivative.

Figure not included in this excerpt

There is also an analogue to the second rule of differentiation:

Figure not included in this excerpt

On the other hand, the counterpart to the chain rule, the so-called substitution rule, is somewhat problematic:

Figure not included in this excerpt

Although this conversion is always valid, it does not help in every case. It only fulfills its purpose (namely to bring us closer to the simplest solution) if dv / dx itself does not depend on x.

Examples:

Figure not included in this excerpt

That looks quite complicated, but I warned you: Integral calculus is part of higher mathematics, so you need a little more skills than working at a cash register.

The application of this rule (also called "partial integration") only makes sense if one of the two factors can be differentiated in such a way that after a finite number of steps at some point a derivative results in a constant. We use this factor for u.

Example:

y (x) = x * cos (x). We take here u (x) = x; then du / dx = 1. v (x) is then cos (x); from this follows ∫ (v (x) * dx) = sin (x). Inserted in (IV), this results in ∫ (y (x) * dx) = x * sin (x) - ∫ (sin (x) * dx) = x * sin (x) + cos (x) + c. By differentiating the result, it is easy to verify that we have calculated correctly

On the other hand, if we had with [Figure not included in this excerpt]. This form again contains an expression that we could only integrate with formula (IV). Let us now set for u (x) = x2and for v (x) = sin (x), after a further step following rule (IV) we get back to the initial form ∫ (x * cos (x) * dx). On the other hand, let's do it the other way around, i.e. we set u (x) = sin (x) and v (x) = x2a, we move further away from the solution, i.e. the simplest form of the antiderivative, which manages without an integral sign.

For functions of the form y (x) = u (x) / v (x), the determination of the antiderivative is differently difficult. One can differentiate between the following cases:

- The form variable x occurs in the numerator u (x) to the same degree or to a higher degree than in the denominator v (x): We first divide the numerator by the denominator. If the result still contains a fraction in which the form variable x occurs both in the numerator and in the denominator (although in the numerator, if we have calculated correctly, it must be in a lower maximum degree than in the denominator), we proceed afterwards as described below.

Example:

y (x) = (x + 1) / (x + 2). By dividing we get: y (x) = 1 - 1 / (x - 2). This then results through elementary integration: ∫ (y (x) * dx) = ∫ ((1 - 1 / (x + 2)) * dx) = ∫ (1 * dx) - ∫ (1 / (x + 2) * dx) = x - ln | x + 2 | + c

- Special case: If the form variable x occurs both in u (x) and in v (x) exclusively in the first degree, the antiderivative can also be determined without division. You only have to eliminate x from the numerator. You can do this by writing the fraction as the sum of two or more partial fractions. Partial fractions are those fractions in which x does not appear in the numerator and only in the first degree in the denominator.

Example:

y (x) = (x + 1) / (x + 2). This fraction can also be reformed without division: (x + 1 + 1 - 1) / (x + 2) = (x + 2 - 1) / (x + 2) = (x + 2) / (x + 2) - 1 / (x + 2) = 1 - 1 / (x + 2). The integration then takes place as described above.

- The form variable x occurs in the numerator u (x) to a lower degree than in the denominator v (x): Here we have to work with partial fraction decomposition. The hardest part is writing the denominator as the product of several linear functions of x. There is no recipe for this. Often, however, one gets quite far if one emphasizes x as far as possible and then tries to find out whether the remainder of the term can be determined by a formula like a2- b2= (a + b) * (a - b) can be decomposed. If that succeeds, you only have to calculate the denominator of the partial fractions. To do this, a system has to be solved from several equations, the number of which corresponds to the number of partial fractions.

Example:

Figure not included in this excerpt

Example:

y (x) = tan (x). From chapter 2.4.2.1 we know: tan (x) = sin (x) / cos (x). We also know that d (cos (x)) / dx = -sin (x). The following transformation is also permitted:

y (x) = - (- sin (x)) / cos (x) = -d (cos (x)) / dx / cos (x)

There is thus a function of the form y (x) = k * u (x) / v (x), where u (x) = dv / dx. The function v (x) is here cos (x), k = -1. The integral is therefore:

∫ (y (x) * dx) = -ln | cos (x) | + c

In principle there are no more rules. What cannot be solved with these rules can only be calculated by conversion or approximation using a polynomial function. One method for this is the already mentioned Taylor series formation, see Chapter 2.5.4.1.

#### 2.6.3 Differential equations

##### 2.6.3.1 Linear differential equations of the first order

Solving differential equations is not always easy; Sometimes it is also a headache for mathematicians (or at least math students). In the following section we will only deal with the simplest kind of these equations. If that is too much for you, you can probably skip this chapter - at least medical professionals are guaranteed not to need it. However, knowing how to approach differential equations helps us to gain a deeper understanding of certain scientific relationships.

First-order linear differential equations are equations in which at least one differential quotient occurs. However, it is always only in the first derivative (therefore 1st order), and it is not present in any other than the first power (therefore linear).

Such differential equations can be solved by separating the differential quotient. ∫ (y (x) * dx) is, as we have already read and written enough, the integral from y to x. If we have a term of the form y (x) * dx on both sides of the equation, we can integrate both sides and thus eliminate the differential quotient.

Example:

We now want to derive the general form of the exponential law (see Chapter 2.3.2) from the differential quotient df (x) / dx = k * f (x). We reform the equation as follows:

1 / f (x) * df (x) = k * dx

Now we integrate both sides and get: ln | f (x) | = k * x + c

It doesn't matter which side we write the c on. In this case c is the difference between the two constants that the antiderivatives can contain as a summand.

By de-logarithmizing we finally get:

Figure not included in this excerpt

As simple as that.

#### 2.6.4 Approximation of functions

##### 2.6.4.1 Taylor series

Again, this is a topic that is only marginally interesting (if at all) for non-mathematicians, non-physicists and non-technicians.

The Taylor series formation is a very good method to approximate and to discuss non-polynomial functions (e.g. trigonometric functions or any functions) with the help of polynomial functions (i.e. functions of the form [illustration not included in this reading sample]).

Taylor polynomials are characterized by the fact that for a certain point their first n derivatives give exactly the same values ​​as the original non-polynomial function. That means: If f (x) is the original function and [figure not included in this reading sample].

A Taylor polynomial with n = ∞ is called a Taylor series.

With the help of Taylor polynomials it is not only possible to differentiate and integrate the most complicated functions approximately, but also to bring them into a form that can be processed by the computer. Therefore, Taylor's series formation is one of the most important chapters of the mathematics lecture in any technical field of study.

The so-called Taylor Theorem (which is actually not from Brook Taylor himself, just as the Pythagorean Theorem was not an invention of Pythagoras) reads:

If the polynomial function Tn (x) and its derivatives [figure not included in this reading sample] are defined, the following applies:

Figure not included in this excerpt

The expression ([Figure not included in this reading sample] means: i-th derivative of the function Tn (x), whereby the value x0 is used for the parameter x.

The "zeroth derivative" [figure not included in this reading sample] stands for the original function Tn (x).

The exclamation mark is the symbol for the faculty. The factorial of a number n is defined as follows:

Figure not included in this excerpt

In practice, one usually works with x0 = 0 and uses the identifier x instead of h. Then the formula simplifies to:

Figure not included in this excerpt

This formula is also called the MacLaurin Formula.

Instead of [Figure not included in this reading sample] one can now insert the i-th derivative of a non-polynomial function at the point x = 0. In this way the formula of the Taylor polynomial of the nth degree belonging to this function is obtained.

Example:

With the help of Taylor's series formation, the value of Euler's number e, on which all natural exponential laws are based, can easily be calculated approximately.

If [Figure not included in this excerpt]. The first derivative is therefore equal to the antiderivative, which in turn implies that all derivatives must be equal to the antiderivative. Since [Figure not included in this reading sample], after inserting it into the Taylor formula:

Figure not included in this excerpt

In order to calculate e ourselves, we have to calculate the value of the function at the point x = 1, because [the figure is not included in this reading sample]. We obtain:

Figure not included in this excerpt

The higher we set the value of n, the closer we get to the actual value of e.

For n = 0 we get: e = 1/0! = 1,000

For n = 1 we get: e = 1 + 1/1! = 2,000 For n = 2 we get: e = 2 + 1/2! = 2.500 For n = 3 we get: e = 2.5 + 1/3! ≈ 2.667 For n = 4 we get: e ≈ 2.667 + 1/4! ≈ 2.708 For n = 5 we get: e ≈ 2.708 + 1/5! ≈ 2.717 For n = 6 we get: e ≈ 2.717 + 1/6! ≈ 2.718 For n = 7 we get: e ≈ 2.718 + 1/7! ≈ 2.718

With n = 6, we have already calculated Euler's number to three decimal places.

For all those who, after much back and forth, finally decided to study the natural sciences or medicine because they thought that Nobel Prizes would be awarded for these subjects in contrast to mathematics, I will quickly explain how the Taylor formula for a polynomial 2 . Degree can be derived:

According to the condition formulated in Taylor's theorem, at the point x0:

[Figure not included in this excerpt]

This corresponds to the Taylor formula for a 2nd degree polynomial.

The derivation of the formula for polynomials of higher degree is carried out in the same way. Perhaps there will be someone who wants to try to derive the Taylor formula for polynomials of any degree.

A C program follows as a technical application example, with which the cosine of angles can be calculated relatively precisely. Attention: If you are using a 32-bit compiler, do not enter a higher "precision" (i.e. maximum exponent) than 14, otherwise overflow errors could occur. With 16-bit compilers (e.g. those for MS-DOS) an overflow error could occur even with a lower precision.

Figure not included in this excerpt

##### 2.6.4.2 Fourier series

Just as the Taylor series allow non-polynomial functions to be represented as a sum of an infinite number of polynomial functions, the Fourier series allow non-harmonic (i.e. non-trigonometric) functions to be represented as a sum of an infinite number of harmonic (trigonometric) functions.

An important area of ​​application is synthetic sound generation, e.g. by a synthesizer or a computer. Likewise, the recognition and processing of speech by artificial intelligence systems is based on Fourier transformation: this is the only way to prepare complex acoustic signals for further analysis and to break down sounds into their fundamental and overtones.

The general formula of a Fourier series is:

Figure not included in this excerpt

The coefficients can be calculated according to the Euler-Fourier formulas as follows:

Figure not included in this excerpt

### 3.1 Introduction

Pure algebra deals exclusively with numbers, variables (which stand for unknown numbers) and their relationships. However, if a unit of measurement such as the meter appears in an equation, then we have already exceeded the limit of pure mathematics: Welcome to physics!

In fact, the previous chapter was by no means exclusively concerned with mathematics. But now we want to finally concentrate on the actual physics.

### 3.2 The International System (SI)

The international system of units SI was established in the middle of the 20th century. The practical thing about this system is that all other quantities can be derived from seven basic quantities. In addition, it is very easy to convert SI values ​​into one another because the units are coordinated with one another.

An example should make this clear:

The mechanical performance is defined as the quotient of work broken down by time: P = W / t. The electrical power is defined as the product of voltage and current strength: P = U * I. In the first case the SI unit is [P] = [W] / [t] = 1 J / s, in the second case [P] = [U] * [I] = 1 V * A; both units correspond to one watt: [P] = 1 W. That is, x J / s = x V * A = x W.

If, on the other hand, one were not to use the SI unit Joule (J), for example, for the voltage U, but a different unit, for example electron volts (eV), one would have to work with a conversion factor.

An electron volt is the work that is done when the charge of an electron moves in an electric field with a voltage of one volt per second. By definition, the following applies: [Figure not included in this excerpt]

If a quantity is written in square brackets, e.g. [P], it means its unit.

#### 3.2.1 SI base quantities

There are seven SI basic quantities: time, length, mass, amount of substance, temperature, light intensity, current intensity. The SI units belonging to them are called base units. The following table summarizes the basic quantities together with their units.

Figure not included in this excerpt

For the units I always wrote 1 s, 1 m etc. Why not just s, m etc.? In principle, this is a matter of agreement; The Système Internationale stipulates that the units must be written in this way. But there is also a rational explanation: x meters is actually x by 1 meter. This makes it clear that the unit of measurement actually represents a multiplication factor.

A few more personal comments on two of these basic sizes:

The name for the SI unit of mass, the kilogram, is unfortunately a little unfortunate. The prefix "kilo" suggests that the SI base unit would be the gram. In fact, instead of powers of ten, prefixes are not added to the kilogram, but to the gram; for example, for 10-6kg is not about 1 µkg (one micro kilogram), but 1 mg (one milligram).

From a purely formal point of view, it would be wise to replace the term kilogram with a new one. On the other hand, like many other units, the kilogram has historical roots; this unit already existed before the introduction of the Système Internationale.

It is also a bit strange that the luminous intensity and not the luminous flux was chosen as the SI base quantity. In the relevant literature, the luminous flux is usually defined first and then the luminous intensity (luminous flux per solid angle). Another variable can be derived from the luminous flux, the illuminance. It is defined as the luminous flux per illuminated area.

If one were to start from the luminous intensity, one would have to define the luminous flux as the product of luminous intensity and solid angle and the illuminance as the quotient of the product of luminous intensity and solid angle broken by the illuminated surface; I think this is a bit cumbersome.