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We have a toilet paper roll with the diameter $R=8\;\mathrm{cm}$ with an inside hollow tube of diameter $r=2\;\mathrm{cm}$. Every layer of the paper has the thickness $d=200µm$ and the layers lies perfectly on top of each other. By how many does the number of pieces of the paper differ had we used a piece of the length $l_{1}=9\;\mathrm{cm}$ instead of $l_{2}=13\;\mathrm{cm}?$ A part of the solution has to be an estimate of the approximation error (if you use one).

Bonus: Calculate the precise length of the spiral the toilet paper makes.

Measure the tensile strenght of office paper. Use a common office paper with the density 80 g\cdot m^{−2}.

How tall could be a tower built from aluminium cans of diet soft drink?

Mage Greyhald celebrated his 100th birthday a long time ago and has begun to fear that Death will pay him a long delayed visit. He decided thus that he will ecase himself into a magic chest, where Death can't reach him. Unfortuntely he forgot to tell the craftsmen to add breathing holes. Air in the chest takes up a volume;$V_{0}=400l$, the percentage of the volume that is oxygen is $φ_{0}=0,21$. With every breath he uses up only $k=20%$ of the volume's oxygen $V_{d}=0,5l$. The frequency of breaths of the mage after the closing of the chest rises according to the relation

$$\\f(t)=f_0 \cdot \frac{\varphi_0}{\varphi (t)}\,,$ wheref_{0}=15breaths\cdot min^{−1}is$ the initial brath frequency is $φ(t)$ and the percetage of the volume that is oxygen at time case $t$. Determine how long until Death will come for Greyhald if the minimum volume of oxygen in the air required for survival is $φ_{s}=0,06$.

On Discworld it is not unusual to be an alchemist. So we have decided that you should try it. Imagine that you are sitting an exam to enter the guild of alchemists. Together with the brochure of the series you got three wrapped pieces of metal. They are thin plates of metal so be careful with them so that yu won't destroy them and ideally don't touch them. It is your task to find out which (precious?) metals we sent you. We don't require you return the metals and so you can use whatever procedures to determine that, even destructive processes but we shall acknowledge only the sufficiently scientific ones. Your solution will be the description of the procedure required and to determine as precisely as possible the cmposition of the individual specimens and you should menntion the label that was on their packaging. Don't forget that it is even good to determine what metals they aren't.

Note If someone wouldlike to become a new participant in this seminar and they would like to solve this task then they should write an email to alchymie@fykos.cz and they will recieve the package from a week later up to 10 days.

Copy the function $iterace_stanMap$ from the series and using the following commands choose ten very close initial conditions for some $K$.

K=…;

X01=…;

Y01=…;

Iter1 = iterace_stanMap(X01,Y01,1000,K);

…

X10=…;

Y10=…;

Iter10 = iterace_stanMap(X10,Y10,1000,K);

</pre> Between $Iter1$ and $Iter10$ there are hidden a thousand iterations of given initial conditions using the Standard map. As to see how the ten points look after the $nth$ iteration, you have to write

n=…;

plot(Iterace1(n,1),Iterace1(n,2),„o“,…,Iterace10(n,1),Iterace10(n,2),„o“)

xlabel („x“);

ylabel („y“);

axis([0,2*pi,-pi,pi],„square“);

refresh;

</pre> we write $"o"$ into $plot$ so that the points will draw themselves as circles. The rest of the commands is then included so that the graph will include the whole square and that it would have the correct labels.

• Set some strong kicks, $Kat$ least approx. -0,6, and place the 10 initial conditions very close to each other somewhere in the middle of the chaotic region (ie for example „on the tip of a pen“). How do the ten iteration's distances with respect to each other change? Document on graphs. How do the ten initially very close initial conditions change after 1 000 iterations? What can we learn from this about the „willingness to mix“of the given area?
• Take again a large kick and set your ten initial conditions along the horizontal equilibrium of the rotor ie $x=0$, $y=0$. How will these ten initial conditions change in time with respect to each other? What can we say about their distance after a large amount of kickso?

• *Bonus:** Try to code and plot the behaviour of some other map. (For inspiration you can look at the sample solution of the last series.)

• Show that for arbitrary values of parameters $K$ and $T$ you can express the Standard map from the series express as

$$x_{n} = x_{n-1} y_{n-1},$$

$$\\ y_n = y_{n-1} K \sin(x),$$

where $x$, y$ are somehow scaled d$φ⁄dt,φ$. Show that the physical parameter $K$, x, y$$.

• Look at the model of the kicked rotor from the series and take this time the passed impuls$I(φ)=I_{0}$, after the period $T$ then $I(φ)=-I_{0}$, after another one $I_{0}$ and this way keep on kicking the rotor on and on.
• Make a map $φ_{n},dφ⁄dt_{n}$ on the basis of values $φ_{n-1},dφ⁄dt_{n-1}$ before the doublekick ± $I$ Why not?
• Solve $φ_{n},dφ⁄dt_{n}$ on the basis of some initial conditions $φ_{0},dφ⁄dt_{0}$ for an arbitrary $n$.
• *Bonus:** Try using the ingeredients from this series to design kicking which $will$ result in chaotic dynamics. Take care though because $φ$ is periodic with a period 2π and shouldn't d$φ⁄dt$ unscrew forever through kicking.

• Assume a pen of length 10 cm with a center of mass precisely in the middle and $g=9.81\;\mathrm{m}\cdot \mathrm{s}^{-2}.Now$ imagine that you put the pen on the table with a null deviation $δx$ with an accuracy of $ndecimal$ places and with a null velocity. How long after making the pen stand can you be sure with just $n-decimal$ places of the nullness of the displacement?

• Consider a model of weather with the biggest Ljapun's exponent $λ=1.16\cdot 10^{-5}s^{-1}$. The weather forecast stops being useful if its error becomes bigger than 20 %. If you had determined the state of the weather with an accuracy of 1 %, how long do you estimate that your forecast would be good for? Give the answer in days and hours.

• Take Lorenz's model of convection from the last part, copy the function $f(xi,t)$ amd simulate and draw the values of the parameters $X(t)$ for two different trajectories using the commands X01=1;

Y01=2;

Z01=5;

X02=…;

Y02=…;

Z02=…;

nastaveni = odeset('InitialStep', 0.01,'MaxStep',0.1);

pocPodminka1=[X01,Y01,Z01];

reseni1=ode45(@f,[0,45],pocPodminka1,nastaveni);

pocPodminka2=[X02,Y02,Z02];

reseni2=ode45(@f,[0,45],pocPodminka2,nastaveni);

plot(reseni1.x,reseni1.y(:,1),reseni2.x,reseni2.y(:,1));

pause()

</pre> Instead of three dots $X02,Y02,Z02you$ have to give the initial conditions for the second trajectory. Run the code for at least five different orders of magnitude that are all still small and note the time, in which the second trajectory shall differ qualitatively from the first(ie will go in the opposite direaction). Don't decrease the deviation under cca 10^{$-8}$, because then the imprecision's of numerical integration start to show. Chart the dependency of the ungluing time on the order of magnitude of the deviation.

Bonus: Attempt to use the gained dependency of the ungluing time on the size of the deviation estimate Ljapun's exponent. You will need more than five runs and you can assume that at the moment of ungluing it will always overcome some constant $Δ_{c}$.

• Look at the equations of the Lorenz model and write a script to simulate them in Octave (maybe even refresh your knowledge of the second part of series). Together with the sketching command your script should have the following form: …

function xidot = f(t,xi)

…

xdot=…;

ydot=…;

zdot= …;

xidot = [xdot;ydot;zdot];

endfunction

config = odeset('InitialStep', 0.01,'MaxStep',0.1);

initialCondition=[0.2,0.3,0.4];

solution=ode45(@f,[0,300],initialCondition,config);

plot3(solution.y(:,1),solution.y(:,2),solution.y(:,3)); </pre> Just instead of three dots fill in the rest of the code (just as in the second part of the series) and use $σ=9,5$, $b=8⁄3.Then$ figure out with a precision of at least units for what positive $r$ the system goes from asymptomatic stopping to chaotic oscillation(it is independent of the initial conditions).

• Here is the full text of the Octave script for simulating and visualising the movement of a particle in a gravitational field of a massive object in the plane $xy$, where all the constants and parameters are equal to one: clear all

pkg load odepkg

function xidot = f(t,xi)

alfa=0.1;

vx=xi(3);

vy=xi(4);

r=sqrt(xi(1)^2+xi(2)^2);

ax=-xi(1)/r^3;

ay=-xi(2)/r^3;

xidot = [vx;vy;ax;ay];

endfunction

config = odeset('InitialStep', 0.01,'MaxStep',0.1);

x0=0;

y0=1;

vx0=…;

vy0=0;

initialCondition=[x0,y0,vx0,vy0];

solution=ode45(@f,[0,100],initialCondition,config)

plot(solution.y(:,1),solution.y(:,2));

pause()</pre>

• Choose initial conditions $x0=0,y0=1,vy0=0$ and and a nonzero initial velocity in the direction $x$ such that the particle will be bound (ie. it won't escape the center.)
• Add to the gravitational force the following force $-α\textbf{r}⁄r^{4}$, where $αis$ a small positive number. Choose gradually increasing $α$ beginning with $α=10^{-3}$ and and show that they cause quasiperiodic movement.

Estimate on the basis of macroscopically measureable quantities the number of cells in the human body and the number of particles in one mole , how many molecules of oxygen„are used“ daily by a human body cell. Find the relevant information needed for the calculation and don't forget to cite your sources properly.