Exponential

Overview

Using exponential functions you may describe how the population of cell cultures grows, how the isotope C 14 decays, how the air pressure varies with height, and the like. Thus, the first chapter of this notebook is entitled "How Do Cell Cultures Grow?" and is devoted to giving an intuitive introduction to exponential growth by studying the growth of bacteria cultures.

In the second chapter, "Properties of the Exponential Function", We explain and develop the base of the exponential function and clarify the importance of the sign in an exponential function. Later in this notebook, we discuss the behavior of the function on its "run" to infinity.

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"Euler's Number and  the  Natural Exponential Function" represents the epitome of exponential functions. In the third chapter you will be able to hear Euler's Number e...if your PC has a sound card.

After the accidents at Harrisburg and Chernobyl one question arises: "How Do Isotops Decay?" which is the subjec of the fourth chapter. Additional information about radioactivity may be of interest when performing your calculations and can be obtained at some internet addresses which have been listed as a reference.

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Mathematica

These are the Mathematica commands used in this Notebook.

({{ListPlot, Plot, Table, FindRoot, ListPlay, Map, Limit, E}})

Internet

Our Homepage:

www.mathsnfun.ac.at

How comes that radioactivity was discovered? Who was involved? When?

Röntgen and Curie

The following link is a fabulous server providing informations and data on chemistry and the chemical elements

chem. elements on the internet

Examples for radioactivity and radioactive decays

isotops on the internet

PhotoDisc

www.photodisc.com

1               How Do Cell Cultures Grow?

Empirical Observations

Observing a cell culture, a bioligist realizes that the area occupied by the cell culture is increasing. The estimated increase is 45 % per hour. Calculate the occupied area after 1, 2, 3, 4, ...n hours, considering an initial area of 1000 mm².

We are going to simulate its increase using a theoretical model.

?CellGrowth

Let us start with an initial area of 1000 mm², an increase of 45% (i.e. 0.45) and an observation period of 9 hours.
However, do not use a period longer than 12 hours (first highlight the line below which contains the data, then click the CellGrowth button):

CellGrowth[\[SelectionPlaceholder]]

1000,  0.45, 9

Now observe the behavior of the same model using an increase of 8 %, 23 %, 50 %, 90 %.

All right, we end up at an area 28.3 times larger as the initial one. Wow!! 
Well, the exact value is 28....... Check the last picture of the movie!

What happens if we start with 100 mm² ? Do we end up at the factor 28.3 as well? Try it!

CellGrowth[\[SelectionPlaceholder]]

100, 0.45, 9

What we learn is that the percentage is crucial, in this case the growth rate, 45 %, not the initial value for the area.

Calculation of the Increase of Area

How many times larger than the initial area is the one occupied by the cell culture after 7 hours?

AfterOneHour = 1000 + 1000  .45

AfterTwoHours = AfterOneHour + AfterOneHour  .45

Please, continue, considering 3 and 4 hours:

AfterThreeHours  =  \[Placeholder]

AfterFourHours  =  \[Placeholder]

Let us search for a general underlying structure! Please, fill in:

AfterOneHour = 1000 (1 + .45) = 1000 1.45
AfterTwoHours = 1000 1.45 (1+ .45) = 1000 1.45 1.45 = 1000 (1.45^2)
AfterThreeHours = ...
AfterFourHours = ...

Now it's easy to calculate the area after, say, seven hours.

Area7  =  1000 (1.45^\[Placeholder])

Therefore the area has increased by a factor 13476.5 / 1000 = 13.47.

Of course, you can compute the increase in a shorter way. It is just given as

(1.45^7)

Let us view the evolving cell culture graphically, during a time period of 10 hours.

ListPlot[Points = Table[{n, 1000 (1.45^n)}, {n, 1, 22}], 
        PlotStyle -> {Red, PointSize[ .03]}, 
        AxesLabel -> {"Hours", "Area"}];

How does the evolution look like for a time period of 14, 18 or 22 hours?

Exercises

A New Type of Function

Connect the points in ListPlot ( using PlotJoined -> True ).

ListPlot[Points = Table[{n, 1000 (1.45^n)}, {n, 1, 10}], 
        PlotJoined -> True, PlotRange -> {{0, 10}, {0, 45000}}, 
        Epilog -> {{Red, PointSize[ .03], 
              Point /@ Points}}];

Note that the number of cells increses most rapidly at the end of the considered time interval!

We can describe the evolution of the cell culture according to time t with the function AreaGrowth[t_] := 1000 (1.45^t).

AreaGrowth[t_] := 1000 (1.45^t);
AreaGrowth[t]

Calculate the area ofter 3.5 hours.

AreaGrowth[ 3.5 ]

As you know, the variable t is in the exponent. Thus the function is called  Exponential Function,

Considering the above example, we have c = ...
and a = ...
a is called the base of the eponential function.

Plot[AreaGrowth[t], {t, 0, 10}, 
        AxesLabel -> {"t", "Area"}];

Try to solve the following problem:
Define the function AreaGrowth1 with parameters c = 100 and a = 1.45. Plot the function.

AreaGrowth1[t_] := \[Placeholder]
AreaGrowth1[t]

Plot[ \[Placeholder] ]

For every exponential function, it is true that

Why a>0?

What happens, if a < 0?

Let us take a = -2, for example.

fn[x_] := (((-2))^x); fn[x]

Now study the table of values for whole numbers in between -10 and 10.

Table[{x, fn[x]}, {x, -10, 10, 1}]

Or look at

ListPlot[Table[{x, fn[x]}, {x, -10, 10, 1}], 
        PlotStyle -> {Red, PointSize[0.02]}];

Everything seems to be ok.

But let us now choose a different step size. Let x again range between -10 and 10. But this time the step size is 0.5.

Table[{x, fn[x]}, {x, -10, 10, 0.5}]

Oups, what is going on?

For example, what does (((-2))^0.5) mean?

Well, that is the square root of -2.

N[(Sqrt[-2])]

The result is a complex number!
Now take (((-2))^9.5), i.e. the 19th root of (2^2).

N[(((-2))^((19/2)))]

Again the result is complex.

How does this result translate to the (real) function we considered above?

Try to state your conclusion and explain why the base of some exponential function has to be positive.

Studying the plot of this (using the Plot command and specifying a plot range) allows you to immediately see what is "going wrong."

Plot[fn[x], {x, 0, 10}]

The function is not defined for all x ϵ R.

General Notation

The description of exponential growth (e.g., of some cell culture) is based on the following formula

Clear[Population, a];
Population[t_] := N0  (a^t)
Population[t]

where Population[t] gives you the number of bacteria at time t. We have

Population[0]

\(TextForm\`\(\(N\_0\)\(\ \)\(is\)\(\ \)\(the\)\(\ \)\(initial\)\(\ \)\(number\)\(\ \)\(of\)\(\ \)\(bacteria\)\(\ \)\(\((at\ time\ t = 0)\) . \ 

        Exponential\)\(\ \)\(growth\)\(\ \)\(is\)\(\ \)\(described\)\(\ \)\(by\)\(\ \)\)\)

After the Milking

It is a fact that the number of germs in cow milk increases almost exponentially. We are going to scutinize the milk of the cow "Lisa":
3 hours after milking there were 66000 germs in the milk, 2 hours later 1.1 Mio. How many germs can be observed 2 and 6 hours respectively after the milking?

There are two unknowns you have to calculate, (Subscript[N, 0]) and a. Therefore you need two equations.

Clear[a, N0];    
\(\ Solution = 

      NSolve[{66000 == N0\ a\^3, 1100000 == N0\ a\^5}, {a, N0}]\)

a must be positive, thus take the second solution.

{N0, a} = {N0, a} /. 
        Solution[[2]]

Define the function NumberOfGerms

NumberOfGerms[t_] := N0 (a^t); NumberOfGerms[t]

and study it.

\(\(Plot[\ NumberOfGerms[t], \ {t, 0, 6.5}, 
	
        AxesLabel -> {"Hours", "Germs"}, 
        Epilog -> {{Red, PointSize[ .04], 
              Point /@ Table[{t, NumberOfGerms[t]}, {t, 1, 
                    6}]}}\ \ ];\)\)

Isn' t it tremendous, how germs breed in milk?

It is more convenient to calculate NumberOfGerms[2] and NumberOfGerms[6] than to extract the numbers from the plot:

{ NumberOfGerms[ \[Placeholder] ], \[Placeholder] }

Numbers from another cow: 
There are 32000 germs 2 hours after milking; 2 hours after the first reading is taken there are 834000. How many germs are there 3 and 7 hours respectively after milking? 
Determine the exponential function, plot it and calculate the number of germs.

Clear[a, N0];
Solution  =  NSolve[ { \[Placeholder] },  {a, N0} ]

Exercises

2               Properties of Exponential Functions

The Base and the Sign of the Exponent -- the Heart of Exponential Functions

In the last chapter we have familiarized you with the idea that exponential functions can be used to model the behavior of some systems, such as the growth of cell cultures and the population increase of germs in milk. And these functions are doing a good job.
What happens, if we change the base a of the exponential function?

?BaseMovie

BaseMovie[\[SelectionPlaceholder]]

0.5, 1.5, 0.1, PointSize  ->  0.03

The base "rocks" the exponential function. Study it in detail!

Generate a movie, where a ranges between 0.01 and 0.3, and between 0.9 and 1.1, for example.

What did you see?

The function y = (a^x) is strictly monotone ............, for a > 1.

The function y = (a^x) is strictly  monotone ............, for 0 < a < 1.

A new surprise.

Describe, how the following two functions are related.

Plot[{(2^x), ((((1/2)))^x)}, {x, -3, 3}, 
        PlotStyle -> {Red, Blue}];

Pay attention:  (2^(-x))  =  ((((1/2)))^x). These curves are identical!
People often write (2^(-x)) in order to avoid fractions. Check it with a plot.

Plot[{(2^(-x)), ((((1/2)))^x)}, {x, -3, 3}, 
        PlotStyle -> {Blue, Red}];

Ok, they are identical! Of course, they are numerically identical, too.

\(\(\(2\^\(-x\) === \((1\/2)\)\^x\)\(
\)
    \)\)

Is it true, that all functions a^(-x) with a > 1 are decreasing?

Now we are going to make some experiments with the function TestExponent.

?TestExponent

Use different values for the base a.

TestExponent[\[SelectionPlaceholder]]

0.3, 0.9,  .1

Here is an example: Plot the function using a range from a1 = 1.5 to a2 = 1.9 and a step size 0.05 .

What happens to the exponential function (3^x) as the value for x increases?
Specifically, what happens as x goes to + infinity, x -> ∞.

Scale = (10^4);
Animate[
        Plot[(3^x), {x, -3, xLimit}, PlotRange -> {0, Scale}, 
          PlotStyle -> Red], {xLimit, 3, 25}];

Now try the same steps with the function 1(3^x). What changes?
Hint: Change the variable Scale.

By the way, how many millions are there in (3^25)?

3  \[Placeholder]

That is just the beginning! The first numerical impression of the exponential growth! Now calculate (3^250), (3^2500), (3^25000).

3  \[Placeholder]

Exponential growth is indeed tremendous. 
In order to handle calculations involving x -> ∞ mathematicians developed a special tool, called Limits.

Limit[(3^x), x -> ∞]

That's what we expected. Exponential functions increase at a tremendous speed and eventually become infinitely big. Infinity means greater than all possible bounds.

Infinity is a highly philosophical idea, even in mathematics. Maybe you can try to discuss the relation between this mathematical idea and the religious idea of the "infiniteness" of God with your teacher in Religion or Philosophy.
What do these ideas have in common? Was the creation of the mathematical idea of infinity inspired by religious beliefs? Do they, somehow, express the same concept?

Returning to mathematics, what happens with the function (3^(-x)), as x-> ∞?
Let view a movie first, and then perform the necessary calculations.

Scale = 14;
Animate[
        Plot[(3^(-x)), {x, -3, xLimit}, PlotRange -> {0, Scale}, 
          PlotStyle -> Red], {xLimit, 3, 25}];

Do the calculations by yourself.

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The functions y = (a^x) and y = (a^(-x)) are mirror images of each other, the mirror axis is ...

All exponential functions of the form y = (a^((+/-) x)) intersect the y-axis at ...

All exponential functions of the form y = c.(a^((+/-) x)) , where c ϵ R, intersect the y-axis at ...

The function y = (a^x), a > 1, converges to ... , for x -> ∞.

The function y = (a^(-x)), a > 1, converges to ..., for x-> ∞.

Exercises

3               Euler's Number e and the Natural Exponential Function

Euler's Number

In science one chooses a special base in order to avoid chaos. This base is called Euler's number, e.
In Mathematica we denote it by E, or can access this constant by using the palette: Palette / BasicInput e .

e is an irrational number, i.e., you cannot write e as a fraction of whole numbers. But you can calculate e to "arbitrarily" high precision, e.g. to 100 significant digits.

N[E,100]

The Natural Exponential Function

Let us start with a little game:

Insert arbitrary integers into the expression (((1 + (1/n)))^(n ))and observe what happens. Probably, you take integers ranging from n = 1 to n = 10.

N[Table[(((1 + (1/n)))^n), {n, 1, 10}]]

Vary n up to 100.
What happens to the value of the expression (((1 + (1/n)))^n)?

What happens, if n -> ∞?

Limit[(((1 + (1/n)))^n), n -> ∞]

Therefore the limit of the sequence (((1 + (1/n)))^(n ))is e.

How does the graph of  (e^x) look like?

Plot[(E^x),{x, -5, 2}];

Plot (e^x)and (e^(-x)) together.

Plot[ { \[Placeholder]

Listen to e

How does the number e sound? That is quite a peculiar question, right? But, if your PC has a sound card, you can now listen to the "sound" generated by Euler's number.

irratDigits=RealDigits[N[E,4000]][[1]];
ListPlay[irratDigits];

It may be very hard to believe it, but that's how irrational numbers sound.
Why?
That's not too tricky to see. Let's look at the first 200 digits of e.

N[E,200]

You cannot find any periodicity in these numbers. Therefore nothing can oscillate and the sound is just noise.
You can compare it to the sound generated by some rational number, e.g., 1/17.

ratDigits = RealDigits[
            N[(1/17), 
              4000]][[1]];
ListPlay[ratDigits];

Looking at the first 200 digits, you will immediately see a periodicity, which is characteristic for rational numbers. Periodical sequence of numbers oscillate, thus you can hear a note.

N[(1/17), 200]

If you want to know more about sounds and music within Mathematica, see the Notebook Angular Functions.

Exercises

4               How Does an Isotope Decay?

100 Years ago, Scientists Discovered Fascinating Facts

The below picture shows the yellowed photograph of a hand, taken by Wilhelm Röntgen in 1895. He was the first to roentgenize a hand with high energetic radiation, so called X-rays.

The photograph in the upper left corner is the first radiograph which shows the hand of Mrs Röntgen. Notice how you can see his wedding ring.

Wilhelm Röntgen was awarded the Noble Prize in Physics.

Noble Prize Laureate Röntgen

Nowadays, radiophotographs are an essential tool in surgery.

Röntgen was an excellent experimental physicist who, aside from the Noble Prize award money, never profitted from his discovery.

Afterwards this discovery, devolopment proceeded at a tremendous speed.

Antoine Henri Becquerel, radioactive radiation of uranium, ( 1903 Noble Prize Award )
Marie and Pierre Curie, discovery of radium and polonium
                                             ( 1903, 1910  Noble Prize Awards )
Ernest Rutherford, discovery of the structure of atoms and alpha- and beta-radiation
                                             ( 1908 Noble Prize Award )

By the way: Marie and Pierre Currie were a nice couple. At the WWW server

Madame Curie

you can find some nice photos showing both together and madame Curie's favourite photo of her husband!

Madame Curie

Madame Curie was the first to introduce the idea of radioactivity, although the phenomenon was discovered by Henri Becquerel. After extracting uranium from ore, she realized that the rest material was more "active" than the pure uranium.

She concluded that new radioactive elements were contained in the ore. That was the day of the discovery of radium and polonium (Po)!

Let's help madame Curie conduct her experiments with polonium. Polonium is still in practical use, for irradiation of cancer cells, which was performed even at the time of madame Curie.
Data on polonium can be found on the WWW server

Chemical Elements on the Internet

Radioactive nuclei decay into nuclei of another chemical elements which in in this case is thallium. Thallium itself is radioactive. It decays into lead very fast. This process is called radioactive decay.

Studying information on radioactive elements, you will often come across the half-life period. This is the period of time needed for exactly half of the radioactive material to decay.

For example, the radioactive element polonium 210 has a half-life period of 138 days. If you have a probe of 3.2 mg, after 138 days only 1.6 mg polonium 210 will be left. The rest of the probe is not disappeared, but has decayed via thallium 206 (half-life period 4.2 min) into the stable lead 206.

Nevertheless we are going to study the decay of 3.2 mg polonium into thallium. The initial amount of polonium will be called (Subscript[N, 0]).

From Measured Data to a Law of Radioactive Decay

After 138 days there is 1.6 mg polonium,
after 276 days there is  ....... 0.8 mg,
...  left
continue until ......................0.05 mg.

The rest of the probe, 3.15 mg, is, more or less, lead and thallium!
Study the decrease of polonium graphically!

Measured1 = {{0, 3.2}, {138, 
            1.6}, {276,  .8}, {414,  .4}, {552,  .2}, {690,  .1}, {828,  .05}};
ListPlot[Measured1, AxesLabel -> {"Time", "Number"}, 
        PlotStyle -> {Red, PointSize[ .03]}, 
        PlotRange -> {{0, 900}, {0, 5}}];

There are some similarities to the earlier example using the function TestExponential[■]. Which law governs this behavior?

Mathematicians choose the following formula to describe this decrease.

This formula has proven a success in practice. Scientists call it law of the radioactive decay.

Number[t] is the remaining number of nuclei of some radioactive element at time t. In order to avoid huge numbers, Number[t] and (Subscript[N, 0]) are often given in mg.

There is a minus sign in the exponent, because the number of radioactive nuclei decreases. λ.....lambda is called the decay constant.

What is the value of λ in this case? Find it out experimentally.

?DecayExperiment

DecayExperiment[\[SelectionPlaceholder]]

3.2, Measured1, 0.003, 0.008,  .001,  PointSize  ->  0.03

Do the experiments, first using a coarse step size and then a finer one, adjusting the limits for λ. Which λ do you think fits?

The dimension of λ is decays / day. Therefore the exponent -λ*t is a pure number, since the unit "days" cancel.

Of course, λ can be calculated. Use the command FindRoot to do so.

Clear[λ];N0 = 3.2;
Sol=FindRoot[N0(E^(-λ 138))==1.6,{λ, 0.007}]

The Mathematica command FindRoot needs a starting value for λ. This value is best chosen close to what you feel a solution may be. In our case it can be identified by looking at the movie. It is 0.007.

λ = λ /. Sol

What is the law of radioactive decay for (Subscript[N, 0]) = 3.2 mg polonium?

Number[t_]:=N0(E^(-λ t));Number[t]

View it graphically (using the Plot command).Note: you must finish the command below yourself!

Plot[ Num\[Placeholder]

Now to something completely different ...
A new example: The isotope phosphor 32 has a half-life period of 14.3 days. There is 24.3 g in the lab. Do the following ...

Generate a ListPlot for 4 different half-life periods, use the function DecayExperiment[■] and calculate the decay constant. Formulate the law of radioactive decay of phosphor 32 and plot the behavior.

\(\(MeasuredData = {{0, 24.3}, {14.3, 
            12.15}, 
		{\ \ \[Placeholder]\ }, {\ \[Placeholder]\ }};\)\)
ListPlot[MeasuredData, AxesLabel -> {"Time", "Number"}, 
        PlotStyle -> {Red, PointSize[ .02]}, 
        PlotRange -> {{0, 100}, {0, 25}}];

Exercises

Are Half-Life Periods and Decay Constants Related?

Usually, half-life periods are denoted by  τ = "tau". It is not too tricky to see the relation between the half-life period λ and the decay constant τ. Try to find the relation yourself. We will guide you.

Take the values for λ and τ from the polonium example of the last chapter:

PolLambda = 0.0050228056
PolTau = 138

Here is some other data on the barium isotope to help with the calculation:

BarLambda = 0.0533190076
BarTau = 13

Now multiply in both cases λ and τ:

PolTau   * PolLambda

BarTau \[Placeholder]

But that means that you can calculate τ, if you know λ, and vice versa.

Therefore we have :

This constant can be derived, c.f. the Notebook Logarithmic Functions.

Consider the following relationship:
A big λ, i.e., the element decays fast, results in τ being very ...

There is another point to mention. The formula τ  =  (0.6931471/λ), connecting λ and τ is independent of (Subscript[N, 0]). That means that the percentage of radioactive isotopes remaining after a certain time period is independent of the initial number of isotops. 
E.g., take 10 kg and 3.2 mg of polonium respectively. After 139 days there will be 5 kg and 1.6 mg respectively remaining!

Further Examples.

How many days does the decay of 3.2 mg of a barium isotope ( half-life period τ=13 days ) take, such that
a)  0.2 mg
b)  41.2 %
remain?
c) How many mg are left after 1 month ( = 30 days )?
First of all give an estimate without doing any calculation, considering the half-life period τ=13 days!

solution to a)

A Plot may give you a feeling for the decay of barium.

N0 = 3.2;
Number[t_]:=N0 (E^((1/13) (-0.6931471) t));
Plot[Number[t], {t, 0, 70}, 
        AxesLabel -> {"Time", "Number"}];

Since FindRoot needs a starting value, try to "measure" with your cursor. After approximately ... days the mass is 0.2 mg.

FindRoot[Number[t] == 0.2, {t, \[Placeholder] }]

solution to b)

41.2 % of (Subscript[N, 0])  is 0.412 (Subscript[N, 0]).
Choose 13 days for the starting value, since by then 50 % of the isotope will have decayed, which is rather close to 41.3 %.

FindRoot[ Number[t] == 0.412 N0, {t, 13 }]

FindRoot[ 
      Number[t]  ==  \[Placeholder],  {t,   \[Placeholder] }]

solution to c)

Estimate: After 13 days the mass of the remaining isotopes is 1.8 mg, after further 13 days  ...  mg,
after one month approximately  ... mg.

Number[30]

Now it is your turn! The isotope cobalt 60 has a half-life period of 5.3 years. There are 24.3 g stored in the lab.

Calculate the decay constant λ and formulate the law of radioactive decay. 
(λ = 0.130782)
How many years does it take to reduce the amount of cobalt to 11.2g?
 ( 5.92 years )

The last example is yours, as well. You can find the data for the decay of sulfur 35 on the WWW server

Isotopes on the Internet

in a table displaying the properties of some isotopes.

We have 67.3 g in a test tube. Find the decay constant and formulate the law of this decay. Plot the behavior. For many days do you have to wait until only 63.4 % of the sulfur remains?
(57.46 days)

Exercises

Mathematica Initialization Cells

Off[General::spell];
Off[General::spell1];
Off[Plot::plnr];
\(\(Off[Solve::ifun];\)
\)
<<  ProgrammingInMathematica`AutoAnimation`
\(\(\(<< Graphics`\)\(
\)
    \)\)
\(\($TextStyle = {FontFamily\  -> \ "Arial", \ 
          FontSize\  -> \ 11};\)
\)
\(\(SetOptions[Plot, Background -> RGBColor[1, \ 1, \ 0.878401], 
        PlotStyle -> RGBColor[0, \ 0, \ 0.996109], 
        DefaultColor -> RGBColor[0.502, 0.000, 0.502], 
        AxesStyle -> RGBColor[0.502, 0.000, 0.502], 
	
        AxesLabel\  -> \ {"x", "y"}];\)
\)
\(\(SetOptions[ListPlot, Background -> RGBColor[1, \ 1, \ 0.878401], 
        PlotStyle -> {RGBColor[0, \ 0, \ 0.996109], PointSize[ .03]}, 
        DefaultColor -> RGBColor[0.502, 0.000, 0.502], 
        AxesStyle -> RGBColor[0.502, 0.000, 0.502], 
	
        AxesLabel\  -> \ {"x", "y"}];\)
\)
\(\(SetOptions[ImplicitPlot, 
        Background -> RGBColor[1, \ 1, \ 0.878401], 
	
        PlotStyle -> RGBColor[0, \ 0, \ 0.996109], 
	
        AxesLabel -> {"x", "y"}, 
        DefaultColor -> RGBColor[0.502, 0.000, 0.502], 
        AxesStyle -> RGBColor[0.502, 0.000, 0.502]];\)\)

\(\(\(
\)\(\(BeginPackage["PlotTitle`"];\)\(\ \ \ \ \ \ \ \ \ \ \ \ \)\(
\)
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.FontSize, FontColor and FontWeight may be chosen seperately.";\)\(
\)
    \(Begin["`Private`"];\)\(
\)
    \(\ \)\(SetAttributes[PlotTitle, \ HoldAllComplete];\)\(
\)
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FontColor\  -> \ 
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\)\(
\)\(\ \)\(
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", \ r];\)\(
\)
    \(PlotTitle[te_, \ r___Rule]\  := \ 
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\ \ \ TraditionalForm[
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FontColor\  -> \ \((\(FontColor /. {r}\) /. 
                  Options[
                    PlotTitle])\), 
					FontWeight\  -> \ \((\(FontWeight /. {r}\) /. 
                  Options[
                    PlotTitle])\)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\ ];\)\(
\)
    \(End[];\)\(
\)
    \(EndPackage[];\)\(
\)
    \(		\)\( (*\ 
      Typical\ Call

        Plot[\ x^3\  - \ 5, \ {x, \ \(-4\), 4}, 
          PlotLabel\  -> \ PlotTitle[\ "That is:", \ x^3\  - \ 5]\ ]; 

      Plot[\ x^2, \ {x, \ \(-4\), 4}, 
        PlotLabel\  -> \ PlotTitle[\ "A quadratic function"\ ]\ ]; 

      Plot[\ Sin[\ x^2\  - 8\ x\  + \ 5], \ {x, \ \(-4\), 4}, \ 
        PlotLabel\  -> \ 
          PlotTitle[\ 
            Sin[\ x^2\  - 8\ x\  + \ 5]\ ]\ ];
	*) \)\)\)\)

ClearAll[CellGrowth];
\(\(CellGrowth::usage = \*""CellGrowth[ 1000, .45, 9 ] animates the growing of a cell culture. In this case: initial occupied area = 1000(mm^2) , 0.45 is the rate of increase. n is the number of hours you want to consider."";\)
\)
\(\(CellGrowth[Area_, Rate_, n_] := 
	
        Module[{sol, Radius, str1, str2, cha, list, 
			aus, 
            round}, 
			
			\(Animate[
              sol = Solve[Area\ \((1 + Rate)\)\^Hours == r\^2\ π, r]; 
              Radius = r /. sol[[2]]; 
              str1 = " The cell culture "; str2 = "grows and grows."; 
              cha = Characters[str1 <> StringReverse[str2]]; 
				list = 
                Table[{Radius\ \((\(-Cos[\(m\ π\)\/\(Length[cha]\/2\)]\))\), Radius\ Sin[\(m\ π\)\/\(Length[cha]\/2\)], 
                    cha[[m]]}, {m, 
                    Length[cha]}]; 
              aus = N[\@\(\(Area\ \((1 + Rate)\)\^n\)\/π\)]; 
              round = TextListPlot[list, 
							Axes -> False, 
                  DisplayFunction -> Identity]; 
              Show[Graphics[{RGBColor[0.996109, \ 0.781262, \ 0.566415], 
                    Disk[{0, 0}, Radius]}], round, 
                PlotLabel -> "Initial area = " <> 
                    ToString[Area] <> " mm²", 
                PlotRange -> {{\(-aus\), 1.1\ aus}, {\(-1.1\)\ aus, 
                      aus}}, 
			AspectRatio -> Automatic, 
                Epilog -> {{Red, 
                      Text["after " <> 
                          ToString[
                            Hours] <> " hours", { .5\ aus, \(-aus\)}, {\(-1\), \(-1\)}], Text[
                        ToString[
                            N[Area\ \((1 + Rate)\)\^Hours, 
                              5]] <> "mm²", { .7\ aus, \(-aus\) +  .1\ aus}, {\(-1\), \(-1\)}]}}, DisplayFunction -> $DisplayFunction], {Hours, 
                1, n, 1}];\)];\)\)

ClearAll[BaseMovie];
\(\(BaseMovie::usage = \*""BaseMovie[.5, 1.5, .025, PointSize -> .07];
   draws the exponential function (a^x). a ranges between 0.5 and 1.5, step size 0.025.
		 PointSize determines the size of points in the plot."";\)
\)
Options[BaseMovie] = PointSize ->  .02;
\(\(BaseMovie[a1_, a2_, Step_, opts___Rule] := 
        Module[{xbeg = \(-20\), xend = 20, PtSize}, 
          PtSize = \(PointSize /. {opts}\) /. Options[BaseMovie]; 
          Animate[Plot[Base\^x, {x, xbeg, xend}, 
              PlotRange -> {{xbeg, xend}, {0, 10}}, 
              Epilog -> {Red, 
                  Text["Base = " <> ToString[Base], {\(3\ xend\)\/4, 
                      1}, {\(-1\), \(-1\)}], 
	
						Green, 
                  PointSize[PtSize], Point[{0, 1}]}], {Base, a1, a2, 
              Step}];];\)
\)

ClearAll[TestExponent];
\(\(TestExponent::usage = \*""TestExponent[ a1, a2, StepSize] shows the graphs of the exponential function (a^((+/-) x)), where the base is varied from a1 to a2 with step size StepSize."";\)

\)
\(\(TestExponent[a1_, a2_, StepSize_] := 
        Animate[Which[bas > 1, 
            Plot[{\((bas)\)\^t, \((bas)\)\^\(-t\)}, {t, \(-10\), 10}, 
              PlotStyle -> {Blue, Red}, 
              PlotRange -> {{\(-10\), 10}, {0, 20}}, 
              Epilog -> {{Red, 
                    Text[StringForm["y = `` ", \((bas)\)\^\(\(\ \)\(-\ \ x\)\)], Scaled[{0, 0.95}]], {Blue, 
                      Text[StringForm["y = `` ", \((bas)\)\^x], 
                        Scaled[{0.99, 0.95}]]}}}], 0 < bas < 1, 
            Plot[{\((bas)\)\^t, \((bas)\)\^\(-t\)}, {t, \(-10\), 10}, 
              PlotStyle -> {Blue, Red}, 
              PlotRange -> {{\(-10\), 10}, {0, 20}}, 
              Epilog -> {{Blue, 
                    Text[StringForm["y = ``", \((bas)\)\^\(\(\ \ \)\(-\ \ x\)\)], Scaled[{0, 0.95}]], {Red, 
                      Text[StringForm["y =`` ", \((bas)\)\^x], 
                        Scaled[{0.99, 0.95}]]}}}]], {bas, a1, a2, 
            StepSize}];\)
\)
Values1 = {{0, 3.2}, {138, 
            1.6}, {276,  .8}, {414,  .4}, {552,  .2}, {690,  .1}, {828,  .05}};

ClearAll[DecayExperiment];
DecayExperiment::usage="DecayExperiment[3.2, MeasuredData, 0.003, 0.008, 0.001,  PointSize -> 0.02] shows λ -  in this case values of exponential function 3.2 \!\(TraditionalForm\`E\^\(\(-λ\)\\ t\)\) from 0.003 until 0.008 with step size 0.001  .";

Options[DecayExperiment] = PointSize ->  .02;
DecayExperiment[N0_, Ppoints_, lambda1_, lambda2_, Step_, 
            opts___Rule] := 
          Module[{PtSize, xend}, 
            PtSize = PointSize /. {opts} /. Options[BaseMovie]; 
            xend = Max[
                Table[Ppoints[[i, 
                    1]], {i, 1, Length[Ppoints]}]]; 
            Animate[Plot[N0 (E^(-lambda t)), {t, 0, xend}, 
                AxesLabel -> {"Time", "Number"}, 
                Epilog -> {{Red, PointSize[PtSize], 
                      Point /@ Ppoints, {Blue, 
                        Text["λ = " <> 
                            ToString[lambda], {((3 xend)/4), 
                            (N0/4)}, {-1, -1}]}}}, 
                PlotRange -> {{0, xend}, {0, 1.6 N0}}], {lambda, 
                lambda1, lambda2, Step}];];