# Basics

2015-07-25

Imagination will often carry us to worlds that never were. But without it we go nowhere. – Carl Sagan, popular atrophysicist

# Units

## Dimensions

A number can be used to count items or measure different concepts. For instance, $x$ might be refering to a duration, an angle, a mass, a pressure, a distance, a speed, an acceleration, a force, etc. The concept $x$ is referring to is called the dimension of $x$.

There is usually several ways to measure a similar concept. For instance, a distance might be measured in either meters, feet, stadiums, etc. To avoid confusion, we agreed on which units should be prefered; those are called SI Units (for Système International, which is French for International System).

Below is a summary of some dimensions, the corresponding SI unit, and other common units:

Dimension SI unit Other units
duration second (s) hour (h), day (d), year (y)
distance meter (m) foot (ft), mile (mi), light-year (ly)
speed meter per second (m/s) mile per hour (mph), knot (kn)
mass kilogram (kg) ton (t), pound (lb)
pressure pascal (Pa) atmosphere (atm), bar (bar)
A light year is the distance that a particle of light can travel in a year. For comparison, it takes light a little bit more than eight minutes (8 min) to get from the Sun to the Earth, meaning that the Sun is 8 light-minutes away from the Earth. Kerbin is about 45 light-seconds away from Kerbol.

## Prefixes

When considering different scales, it is practical to use different units. Using a same unit when traveling or when describing a stamp would force us to use tiny and huge numbers, making it harder to build an intuition.

Such units have been hinted above. For instance, the ångström is used for the size of atoms and molecules; the parsec is used for interstellar distances. Creating a new unit for each use case is cumberstone and makes it harder to concialiate intersecting situations.

A simpler approach is to use prefixes. The idea is to easily create new units out of a basic one. That way, a kilometer is 1,000 meters and we can simply write 355 km rather than 355,000 m.

Here are the most common prefixes:

kilo- (k-) mega- (M-) giga- (G-) tera- (T-)
10³ 10⁶ 10⁹ 10¹²

There are also prefixes to decrease the value of an unit:

milli- (m-) micro- (µ-) nano- (n-) pico- (p-)
10⁻³ 10⁻⁶ 10⁻⁹ 10⁻¹²

## Conversion

We sometimes need to switch the unit used for a measure. For example, let us convert to SI units. We know that and so:

This particular example shows how easy it is to include units in computations. As we will see below, having the units is useful when considering more complex expressions.

An addition involves two measures of the same dimension. For example, let us assume we have a distance x defined as follows:

Since we know that that , we can replace it in the expression:

x = 2 ly + 4,730 Tm
= 2 (9,461 Tm) + 4,730 Tm
= (18,922 + 4,730) Tm
= 23,652 Tm

Conversely, we could also have said that and then:

x = 2 ly + 4,730 Tm
= 2 ly + (4,730 / 9,461) ly
= (2 + 0.5) ly
= 2.5 ly

Of course, the two ways are equivalent and we can check that .

## Multiplication (and division)

We can create new units pretty easily. For example, if some object travels a distance in a duration , we can define the velocity . Conversely, assume the object has traveled at a velocity for a duration ; then, the distance it has gone through is:

By doing the operations on both the numerical values and the units, we know what our result is: in this case, it’s a distance, and it is expressed in meters.

# Functions

Let us consider a car $C$ moving along a straight road at a speed of $v$ (e.g. ). The position of the car, $C$, can be determined by the distance from $C$ to an arbitrary fix point $O$ (the origin). We will note this distance $x$ and we have thus $x=\mathit{OC}$.

We will measure time $t$ as the delay since the car was at the origin ($C=O$).

Say we want to follow the evolution of the position of the car as time passes by. In other words, we are interested in knowing $x$ as a function of $t$. We know that, at time t, we have $x=v×t$. We write it:

$x\left(t\right)=v×t$

This notation gives us a general formula to compute $x$ for any given value of $t$. For example, if we want to know the position of the car after one hour:

As another example, it is known that the intensity of the light emitted by a star decreases proportionnaly to the square of the distance to the star. This can be written as:

$L\left(r\right)=\frac{C}{{r}^{2}}$

where $C$ is some constant value (i.e. independent from $r$) which is to be determined experimentally.

# Derivatives

## Definition

Assume we know the position $x\left(t\right)$ of the car for any instant $t$ and we want to determine the velocity of the car at a given instant ${t}_{0}$.

The velocity is the variation of position through time. Thus, to know how fast the car is going at time ${t}_{0}$, we need to look at the position of the car at two different instants. We already have ${t}_{0}$; let us also consider ${t}_{0}+h$ for some arbitrary value $h$.

The difference in position between instants ${t}_{0}$ and ${t}_{0}+h$ is thus $x\left({t}_{0}+h\right)-x\left({t}_{0}\right)$; a shorter notation for this is $\mathit{\Delta }x\left({t}_{0}\right)$. The value $h$ is not shown because it has no importance in itself. The Greek letter $\mathit{\Delta }$ (“delta”, equivalent of $d$), is generally used to denote a difference (here, the difference in position).

Notice that, the bigger $h$, the bigger we expect this difference to be: the longer the delay, the longer the car moved. To compensate for this, we will divide by how much time has passed, which is to say $h={t}_{0}+h-{t}_{0}=\mathit{\Delta }{t}_{0}$:

$\frac{\mathit{\Delta }x\left({t}_{0}\right)}{\mathit{\Delta }{t}_{0}}$

This value is the mean velocity from instant ${t}_{0}$ to instant ${t}_{0}+h$. However, the mean velocity is a value that only gives a general idea of the speed on some period of time. In this duration, the instant velocity (actual speed) can vary a lot and the mean velocity would then be far off to these values.

Since we expect the speed to not change a lot on short periods of time, a natural solution is to consider the mean velocity over shorter durations.

So, as we pick shorter and shorter durations $h$, the value $\mathit{\Delta }{t}_{0}$ becomes smaller, but so does $\mathit{\Delta }x$. Often, we will notice that the mean velocity seems to converge (becomes closer and closer) to a particular value. Instead of continuing to choose smaller and smaller values of $h$, we will pick this values and call it the limit of $\frac{\mathit{\Delta }x\left({t}_{0}\right)}{\mathit{\Delta }{t}_{0}}$ as $h$ tends to $0$ (becomes smaller and smaller). Or, for short:

$\underset{h\to 0}{lim}\frac{\mathit{\Delta }x\left({t}_{0}\right)}{\mathit{\Delta }{t}_{0}}$

Such a limit is called the derivative of $x$ at ${t}_{0}$. We have a shorter way to note this:

$\frac{dx\left({t}_{0}\right)}{dt}$

Here, the derivative of $x$ at ${t}_{0}$ corresponds to the mean velocity over an infinitely small period, that is, the instant velocity.

Finally, we can do this for any value of ${t}_{0}$. Thus, we have a new function that let us evaluate the velocity at any ${t}_{0}$:

$x\text{'}=\frac{dx}{dt}$

When the derivation is done with respect to time (i.e. $\frac{\dots }{t}$), we can simply use the dot notation: $\stackrel{\bullet }{x}$.

## Second derivative

The velocity is the derivative of the position. As a function, it can itself fluctuate and we can be interested in these variations. The derivative of the velocity is the acceleration: $a=\stackrel{\bullet }{v}$.

A shorter way of saying that the acceleration is the derivative of the derivative of the position, is to say that the acceleration is the second derivative of the position: $a=\stackrel{\bullet \bullet }{x}$.

## Formal derivation

We now have a way to compute the derivative of a function at a given point. However, it is not accurate: while we do get a better approximation by taking a smaller value for $h$, the result is still an approximation and can sometimes stay far off.

Instead, we can look at the expressions to determine the exact value for the limit. For instance, let us consider the function $f\left(x\right)=12x$ and let us search for the derivative of f at some $x$, i.e. $\frac{df\left(x\right)}{dx}$. First:

$\frac{\mathit{\Delta }f\left(x\right)}{\mathit{\Delta }x}=\frac{f\left(x+h\right)-f\left(x\right)}{\left(x+h\right)-x}=\frac{12\left(x+h\right)-12x}{h}=\frac{12\overline{)h}}{\overline{)h}}=12$

We now look at the value $\frac{\mathit{\Delta }f\left(x\right)}{\mathit{\Delta }x}$ as $\mathit{\Delta }x$ gets small; in this case, it happens to always be $12$, and does not depend on $\mathit{\Delta }x$. Thus, however small $\mathit{\Delta }x$, the value is $12$, and:

$\frac{df\left(x\right)}{dx}=\underset{h\to 0}{lim}\frac{\mathit{\Delta }f\left(x\right)}{\mathit{\Delta }x}=12$

That way, we know the exact value of derivative of $f$ in any point. Let us take a second example with $g\left(x\right)=7{x}^{2}$:

$\frac{\mathit{\Delta }g\left(x\right)}{\mathit{\Delta }x}=\frac{7\left(x+h{\right)}^{2}-7{x}^{2}}{h}=\frac{7\left({x}^{2}+2\mathit{xh}+{h}^{2}\right)-7{x}^{2}}{h}=\frac{7x\overline{)h}+{h}^{\overline{)2}}}{\overline{)h}}=7x+h$

Here, the expression does depend on $h$. However, the smaller $h$ gets, the less influence it has on the sum: the value is becoming closer and closer to $7x$. Thus: $\frac{dg\left(x\right)}{dx}=7x$.

## Derivation rules

Now, what if we want to compute the derivative of $h\left(x\right)=12x+7{x}^{2}$? Of course, we could go through the same step as in the previous part. However keeping the same definitions of $f$ and $g$, we can notice that $h=f+g$. It means that $h\left(x\right)=f\left(x\right)+g\left(x\right)$ for all $x$’s.

It turns out that it can be shown that $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}f+\frac{d}{dx}g$ for any functions $f$ and $g$. Using this rule and knowing the derivative of $f$ and $g$, we can derive:

$\frac{d}{dx}h\left(x\right)=\frac{d}{dx}\left(f+g\right)\left(x\right)=\frac{d}{dx}f\left(x\right)+\frac{d}{dx}g\left(x\right)=12+7x$

There are a few derivation rules that can help us determine the derivative of complex functions easily.

# Integrals

## Definition

Now, let us consider the reverse situation: we know the velocity of the car $v$ at any given instant and we would like to know where it was at an arbitrary instant ${t}_{0}$. In other words, we know the derivative of the position $\stackrel{\bullet }{x}=v$ and we intend to get the position $x$ back.

The first thing to notice is that the velocity only informs us on relative motion: two cars can have the same velocity through time while being in different positions. In the example below, the positions of two cars with the same speed are shown; since the red car starts ahead, it stays ahead.

This means that we will need additional information to know where to start. Here, we will assume the car start at .

As a first approximation, we could pretend the velocity is constant, and always equal to $v\left({t}_{0}\right)$. That would make the position trivial to compute: $x\left({t}_{0}\right)\approx v\left({t}_{0}\right)×{t}_{0}$.

Now, given a constant velocity ${v}_{0}$ and a delay ${t}_{0}$, we know how to compute the distance ${x}_{0}$ as ${x}_{0}={v}_{0}×{t}_{0}$. In this situation however, the velocity changes over the time interval from to ${t}_{0}$.

Note that ${v}_{0}×{t}_{0}$ is also the area of a rectangle of height ${v}_{0}$ and width ${t}_{0}$. This maps to the green area on the graph.

For a better approximation, we will simply split this in several parts of width $h$.

For instance, we can assume that, from time to time $t=h$, velocity is constant and equal to $v\left(h\right)$. Then, the distance traveled on this duration is simply $v\left(h\right)×h$; then, from $h$ to $2h$, the car further travel $v\left(2h\right)×h$. Thus, from time to time $t=2h$, the car traveled roughly $v\left(h\right)×h+v\left(2h\right)×h$.

When we consider more steps, we will want to avoid writing the whole sum. Instead, we can use the $\mathit{\Sigma }$-notation (“sigma”, Greek equivalent of $s$) to denote a sum:

_i = 1^n v(i h) h = v(1 h) h
+ v(2 h) h
+ v(3 h) h
+ ...
+ v(n h) h

In other words, $\sum _{i=1}^{n}v\left(ih\right)×h$ means “sum the expression $v\left(ih\right)×h$ where $i$ takes each of the integer values from $1$ to $n$”.

Here, we will want to have $h×n={t}_{0}$ so that we can retrieve the distance traveled from time to time $t={t}_{0}$.

Notice that the value we are looking for is the sum of the surface areas of the green rectangles. It turns out that this tends to match the value of the surface area under the curve.

As for derivation, when $h$ tends to zero, our rough approximation will become more precise. Since we stil want $h×n={t}_{0}$ we will instead make $n$ grow instead, and set $h$ to $\frac{{t}_{0}}{n}$:

$\sum _{i=1}^{n}v\left(i\frac{{t}_{0}}{n}\right)×\frac{{t}_{0}}{n}$

As $n$ tends to infinity (grows larger and larger), we expect the sum to converge (come closer and closer) to some fixed value. Again, there is a short notation for this:

An ever shorter notation when there is no ambiguity is simply:

Notations with $dx$ and $dt$ make it easy to reason with derivatives and integrals. Since $v$ is the derivative of $x$:

${\int }_{{t}_{1}}^{{t}_{2}}v={\int }_{{t}_{1}}^{{t}_{2}}\frac{dx}{\overline{)dt}}\overline{)dt}=\sum _{{t}_{1}}^{{t}_{2}}dx=x\left({t}_{2}\right)-x\left({t}_{1}\right)$

This highlights the fact that integrals are just summing up all the variations between two points (here, from $t={t}_{1}$ to $t={t}_{2}$). In particular, to know the exact value of $x\left({t}_{2}\right)$, we need to know the initial value $x\left({t}_{1}\right)$.

There is also a shorter notation for the difference of a value between two points:

$x\left({t}_{2}\right)-x\left({t}_{1}\right)=\left[x{\right]}_{{t}_{1}}^{{t}_{2}}$

## Illustration

For example, if , then:

x(30 s) - x(0 s) = _0^30 s t 2 m/s^2 t t
= (_0^30 s 2 t t) m/s^2
= [t^2]_0^30 s m/s^2
= (900 s^2 - 0 s^2) m/s^2
= 900 m

In particular, with we are given the additional information that $C$ did start at $O$, i.e. $x\left(0\right)=0$, then:

This means that an object starting at rest getting a constant push of will travel in .

## Antiderivative

More generally, in the previous example, we could write:

Another way to write it is:

$x\left(t\right)={t}^{2}m/{s}^{2}+C$

where $C=x\left(0\right)$ is a value independent of $t$ which depends on the initial conditions (e.g. the position of the car at the initial instant). Each of the possible expression of $x$ (depending on $C$) is a primitive of $\stackrel{\bullet }{x}$.

## Geometric integrals

There should be more explanations here, but these formulas shows how we compute the areas and volumes of common shapes.

Circle circumference:

d = _C s
= _0^2 R
= R _0^2
= 2 R

Disk area:

A = _D A
= _r=0^R _ =0^2 r r
= 2 _0^R r r
= 2 _0^R
= R^2

Sphere area:

A = _S A
= _ =0^ _ =0^2 (R ) (R )
= 2 R^2 _0^
= 2 R^2 _0^
= 4 R^2

Sphere enclosed volume:

V = _B V
= _r=0^R _ =0^ _ =0^2 (r ) (r ) r
= 2 _r=0^R _ =0^ r^2 r
= 2 _0^R _0^
= 4 R^3

# Differential equations

## Definition

A differential equation is an equation whose unknown is a function and involving a derivative of this function. For example:

$\frac{df}{dx}=3{x}^{2}$

This is a differential equation and we already know how to solve it (find the expression of $f$). Here, $f\left(x\right)={x}^{3}+C$ for any constant value $C$ (there are several possible solutions).

## Exponential

The exponential function is defined as $f\left(x\right)={e}^{x}$ where $e$ is a mathematical constant whose value is about $2.718$. It was picked so that:

$\frac{df}{dx}=f$

In other words:

$\frac{d}{dx}\left({e}^{x}\right)={e}^{x}$

If we define $f\left(x\right)={e}^{g\left(x\right)}$ instead, derivation rules gives us:

$\frac{df\left(x\right)}{dx}=g\text{'}\left(x\right){e}^{g\left(x\right)}$

so that $\frac{df}{dx}=g\text{'}f$.

## First order

Now, consider the following equation:

$\frac{df}{dx}=f$

We already know that $f\left(x\right)={e}^{x}$ is a solution; however, so is $f\left(x\right)=2{e}^{x}$. Actually, the set of solutions to this equation is the functions $f\left(x\right)=C{e}^{x}$ where $C$ is any constant value.

The remark we made before tell us how to solve a differential equation of the form:

$\frac{df}{dx}=hf$

where $h$ is also a function. We just need to find $g$ such that $g\text{'}=h$, i.e. the solutions are:

$f\left(x\right)=C{e}^{{\int }_{0}^{x}h\left(x\right)dx}$