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(a^x)   GROWTH,DECAY
          MOVIE

1.  How E.Coli Bacteria Grow in a Tube of Liquid Medium - Movie
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Bacteria display a characteristic four-phase pattern of growth in liquid culture.

The initial Lag phase is a period of slow growth during which the bacteria are adapting to the conditions in the fresh medium.
This is followed by a Log Phase during which growth is exponential, doubling every replication cycle. 
The Stationary Phase occurs when the nutrients become limiting, and the rate of multiplication equals the rate of death.
The Logarithmic Decline Phase occurs when cells die faster than they are replaced.(This latter occurs over a much longer period of time than the previous three.)

In this activity, you model the Log Phase. 100 Escherichia coli (E.coli) bacteria are inoculated into a fresh tube of Luria broth growth medium. They begin to grow by cell division.

We assume that every hour, 5% of all bacteria split into two. 
That is, the population increases every hour by the same growth factor a =1.05.

If the number E. coli is f[x] after x hours, then the number is f[x+1]=f[x]*a an hour later.

The following movie shows the growth of the bacteria colony (image to the left). We count the number of bacteria every hour and plot that number versus time x (image to the right).

c =100;               (* intial number of bacteria *)
a = 1.05;               (* growth factor a, 1 < a ≤ 2 *)
hours = 60;         (* hours to watch *)
mobilityOfBacteria  =1;           (* distance traveled before reproducing *)

MDBCreatePopulationGrowthMovie[c,a,hours, mobilityOfBacteria]

Questions about the Movie

Choose a very low initial number, e.g. 4, and a higher reproduction rate, e.g. 1.25. 
Observe the growth pattern of bacteria within 32 hours.

What does this bacteria pattern remind you of?

Hint: Think of milk products.

c =4;               (* intial number of bacteria *)
a = 1.25;               (* growth factor a, 1 < a ≤ 2 *)
hours = 32;         (* hours to watch *)
mobilityOfBacteria  =1;           (* distance traveled before reproducing *)

MDBCreatePopulationGrowthMovie[c,a,hours, mobilityOfBacteria]

Choose c = 4, a = 1.25 and hours = 32.  

Choose a higher mobilityOfBacteria, e.g.4 (this means that the bacteria have more time, here 4 time steps, to move freely before reproducing, and therefore fill the available region more uniformly). 

Observe the difference between this and the previous movie.

c =4;               (* intial number of bacteria *)
a = 1.25;               (* growth factor a, 1 < a ≤ 2 *)
hours = 32;         (* hours to watch *)
mobilityOfBacteria  =4;           (* distance traveled before reproducing *)

MDBCreatePopulationGrowthMovie[c,a,hours, mobilityOfBacteria]

Choose c = 4, a = 1.05 and hours = 72.  

Here the growth factor is very low, a = 1.05, so you see the discrete structure of the model in the first pictures.

mobilityOfBacteria  = 2 means that the bacteria have 2 time steps to move freely before reproducing.

c =1;               (* intial number of bacteria *)
a = 1.05;               (* growth factor a, 1 < a ≤ 2 *)
hours = 72;         (* hours to watch *)
mobilityOfBacteria  =2;           (* distance traveled before reproducing *)

MDBCreatePopulationGrowthMovie[c,a,hours, mobilityOfBacteria]

Answer the following questions either by hand, or using the computer, or both. If experimenting using the simulation, use an initial population of 100 E.coli unles otherwise indicated.

What value of the growth factor a corresponds to the statement that every hour 5% of the E.coli split into two?

Answer: a = 1.05

What value of the growth factor a would correspond to the statement that not 5% but 73% of the E.coli duplicate every hour?

Answer: a = 1.73

What value of the growth factor a would correspond to the statement that not 5% but 100% of the E.coli duplicate every hour?

Answer: a = 2

The initial population of E.coli in the tube is 100 and the growth factor a = 1.1.

What is the population of E.coli after 2 hours in the tube? 

What is the percentage increase?

What is the population after 2 hours if the initial population is 300 (same growth rate)?

What is the percentage increase?

(1.1^2)

Answer: With an initial population of 100, after two hours there are 121 E.coli in the tube. The percentage increase is 21%.

With an initial population of 300, after two hours there are 323 E. coli in the tube. The percentage increase is again 21%.

The growth factor a = 1.05.

What is the percentage increase during one entire (24 hour) day? Does this depend on the initial population?

The daily growth rate is the factor by which the initial population gets multiplied during each day. What is its value?

(1.05^24)

Answer: The percentage increase during one day is 222.5%. The initial population only affects how this number is rounded. The daily growth rate is 3.225.

The growth rate a = 1.05. How long does it take for the population to double? Does this depend on the initial population?

(Hint: either use trial and improvement, or--this is challenging--try to find a way of calculating the answer directly.)

Clear[t];
Solve[(1.05^t)==2.0,t]

The population takes between 14 and 15 hours to double; 14.2 hours to be more precise. This does not depend on the initial population (apart from issues to do with rounding).

How long does it take for the population to double if the growth rate is 1.1?

(1.1^7.3)

Clear[t];
Solve[(1.1^t)==2.0,t]

Answer: Between 7 and 8 hours (7.3, to be more precise).

Challenging question: how long does it take for the population to double if the growth rate is r?

Clear[r,t];
Solve[(r^t)==2, t]

Answer: The time taken for the population to double is  ln 2/ln r hours

1.  Atom Decay - Movie
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In this movie you see radioactive (red) atoms on a rectangular grid, decaying into non-radioactive (green) atoms. During one time step, e.g. one hour, each atom decays with a certain probability, probabilityOfDecay.

We assume that each atom decays with a probability of 10% per hour. 
That is, the population decreases every hour by the same decay factor a = 0.90.

If the number of atoms is f[x] after x hours, then the number of atoms is f[x+1]=f[x]*a an hour later.

The plot of the population is combined with a plot of the number of radioactive atom versus number of time steps. The movie stops after the number of time steps needed to reach 1/1000 of the initial population or after the radioactive population has become 0, whatever happens first. 

If the initial number is not a square, non-radioactive atoms are filled in randomly, in order to make the square complete.

initialNumber=1524;            (* intial number of atoms *)
probabilityOfDecay=0.1;            (* > 0 and < 1 *)

MDBCreateExpDecayMovie[initialNumber,probabilityOfDecay]

Questions about the Movie

Choose the same initialNumber, but a higher probability of decay: 0.2. What do you expect to see? Experiment.

initialNumber=1524;            (* intial number of atoms *)
probabilityOfDecay=0.2;            (* > 0 and < 1 *)

MDBCreateExpDecayMovie[initialNumber,probabilityOfDecay]

Answer the following questions either by hand, or using the computer, or both.

If the decay probability is 0.1, and the initial number is 1524, how many radioactive atoms remain after one hour? What percentage is this of the initial value?

With the same decay probability, but an initial number of 6096, how may radioactive atoms remain after one hour? What percentage is this of the initial value?

1524*0.9

6096*0.9

Answer: 1372 and 5486 respectively; 90% in each case.

If the decay probability is 0.2, what is the decay factor?

Answer: 0.8

If the decay probability is 0.1 and the initial number is 1524, what is the number of radioactive atoms remaining after 3 hours?

What is the percentage decrease?

Does the percentage decrease depend on the initial number?

1524*(0.9^3)

100*(1-(0.9^3))

Answer: 1111 atoms: 27.1% decrease. The percentage decrease is independent of the initial number.

If the decay factor is 0.9 and the initial number is 1524, how long does it take for the number of radioactive atoms to fall to half its initial value? 

Does this length of time depend on the initial number?

(Hint: either use trial and improvement or--and this is challenging--find a way to calculate the answer directly.)

(0.9^6.6)

Clear[t];
Solve[(0.9^t)==0.5,t]

Answer: between 6 and 7 hours; 6.6 to be more precise. This does not depend on the initial number.

If the decay factor is 0.8, how long does it take for the number of radioactive atoms to fall to half its initial value?

Does this length of time depend on the initial number?

(Hint: either use trial and improvement or--and this is challenging--find a way to calculate the answer directly.)

(0.8^3.1)

Clear[t];
Solve[(0.8^t)==0.5,t]

Answer: just over 3 hours; 3.1 to be more precise. This does not depend on the initial number.

Challenging question: how long does it take for the number of radioactive atoms to halve if the decay factor is r?

Clear[r,t];
Solve[(r^t)== 1/2,t]

Answer: The time taken for the number to halve is -ln 2/ln r hours.

Challenging question: how long does it take for the number of radioactive atoms to halve if the decay probability is (1-p)?

Answer: The time taken for the number to halve is -ln 2/ln (1-p) hours.

DEFINITION:
    Exp Functions                        f[x] = c*(a^x)

2.  Graphical Explanation  Growth, Decay
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Here the exponential growth of a population of aggressive viruses is modelled by the equation f[x] = 100*(3^x). Initially, there are 100 viruses. 

After one hour there are 300, after two hours 300 * 3 = 900,
and after three hours 900*3 = 2700 viruses.

In mathematics, a quantity f[x] that grows exponentially is one that grows at a constant rate a proportional to its size f[x]. This holds for every time x.

This implies that for any exponentially growing quantity, the larger the quantity gets, the faster it grows.

In the graph below, the exponential decay of 120 g of a radioactive isotope that decays at a rate of 50% per year is shown; this is modelled by the equation  f[x] = 120*(0.5^x).

After one year, 120*0.5 = 60 g remain, after two years 60 * 0.5 = 30 g,
and after three years, 30*0.5 = 15 g.

A quantity f[x] is said to be subject to exponential decay if it decreases at a constant rate a proportional to its value f[x].

2.    Exponential Growth Functions - Formulas
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With this growth function one can read immediately the percentage of growth i.

Very often (1+ i) is shortcutted by (1+ i) = a. a is the base of the exponential function.

i = a - 1.

This form of growth function uses the natural exponential function (e^x). The number e is the base of the natural exponential function, sometimes called Euler's number or Napier's constant; it is universal, and equal to about 2.718. 

The number k is a measure of the rate of growth, and is called the growth constant. The larger the value of k, the faster the growth.

The equation f[x] = c (a^x) and the equation f[x] = c (e^(k x)) are equivalent. Start with the equation f[x] = c (a^x) and set

Then

Exponential functions show a constant growth rate per unit time: if f[x]=c (a^x), then f[x+1] is always a * f[x], whatever the value of x.

The exponential function  f[x] = (a^x) takes a constant time to double in value. The value of h for which f[x+h] = 2 f[x] doesn't depend on x.

This value of h is called the doubling time.

We can express h in terms of logarithms:

Log[x] means Ln[x],   Log[a, 2] means Log[2]/Log[a].

2.    Exponential Decay Functions - Formulas
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With this decay function one can read immediately the percentage of decay  i.

Very often (1- i) is shortcutted by (1- i) = a. a is the base of the exponential function.

i = 1 - a.

This decay function uses the natural exponential function (e^x). e is the Euler number.
k is called the decay constant. The more negative k the stronger the decline.

The exponential functions f[x] = (a^x), 0 < a < 1 posts a constant decay factor a.

As x goes up by 1, f[x] declines constantly by the factor a - independent of x.

Given an exponential function f[x] = (a^x),  0 < a < 1.

This h is called half-life. 
This is the same as saying f[x] goes down by 50% every time x goes up by h.

h in terms of logarithm

Log[x] means Ln[x],   Log[a, 2] means Log[2]/Log[a].

2.    Average Percentage Growth/Decay Rate - Formulas
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The average percentage growth/decay rate

measures the average percentage rate of change of f[(Subscript[x, 0])] per h units.
The measurement starts at (Subscript[x, 0]) and ends with (Subscript[x, 0]) + h.

The rate is positive when f is growing and negative when f is falling.

2.   Comparing Exponential and Linear Growth
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Compare exponential growth with linear growth with an example.

Exponential growth:

Whenever you add 2 to x, f[x] gets multiplied by 100.

Linear growth:

Whenever you add 2 to x, f[x] increases by 2.

EXAMPLE:
  Step by Step

3.   Iodine 131 Decay - Treating Thyroid Problems
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Iodine 131 Decay - Example
Radioactive iodine 131 can cause thyroid problems, and help diagnose and treat thyroid problems. 
Long-term (chronic) exposure to radioactive iodine can cause nodules, or cancer of the thyroid. However, once thyroid cancer occurs, treatment with high doses of I-131 may be used to treat it. Doctors also use lower doses of I-131 to treat overactive thyroids.

Iodine 131 is a radioactive material that decays according to n[x] = (Subscript[n, 0])(e^(-0.0866x)), x in days. A scientist has a sample of 99 grams of iodine 131.

(a) Define the decay function and graph it. 

(b) How much iodine 131 is left after 9 days? After 27 days?

(c) When will 70 grams of iodine 131 be left?

(d) What is the half-life of iodine 131? What percentage decays every day?

(e) Advanced: What is the average percentage decay rate in day 1? From the beginning of day 6 to the end of day 7? 
How many grams of I-131 decay from the beginning of day 6 to the end of day 7?

Solution:

(a)  (Subscript[n, 0]) is the sample of 99 grams, the initial amount.

Clear[n,x];
n0=99;
k=-0.0866;

n[x_]=n0*(E^(k*x))

Plot n. Label the axes with "x (days)" and "Iodine 131  (grams)", the plot with "Iodine 131 Decay".

MDPlot[n[x],{x,0,80},
AxesLabel->{"x (days)","Iodine 131 (grams )"},
PlotLabel->"Iodine 131 Decay\n"]

More details▼.

(b)  First estimate with the cursor, using the above graph, how much iodine there is after 9 days and 27 days.

Then calculate the numbers.

n[9]

n[27]

Answer: The amount of Iodine after 9 days is about 45.4 grams, after 27 days about 9.55 grams.

More details▼.

(c)  To find how long it takes for the mass of Iodine-131 to fall to 70 grams, solve the equation n[x] =70.

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==70 ,x]

Answer: After approximately 4 days 70 grams of iodine is left.

More details▼.

(d)  How long does it take for the mass to fall to half the initial amount (Subscript[n, 0])? (Half-Life) 
Solve the equation 1/2 (Subscript[n, 0]) = (Subscript[n, 0])*(e^(k*x)) for x.

-> Find k;
k=-0.0866;  (* enter k < 0 *)

halfLife = - Log[2]/ k//N

Answer: The half-life is about 8 days.

Convert the 99 (e^(-0.0866 x)) form into the (a^x) form. a = 1 - i, with i the average decay rate.
Then calculate in an input cell i which is 1 - a.

Clear[f,x];
a=N[(E^(-0.0866)) ];  (* \[Placeholder] -> k *)
c =n0 = 99;

f[x_] = c*(a^x)

1-a

Answer: Each day 8.3% of the given amount decays.

More details▼.

(e)  The average percentage decay rate measures the decay of iodine 131 in percent per day. The time of measurement starts at day (Subscript[x, 0]) and ends at (Subscript[x, 0])+h.

What is the average percentage decay rate in day1? (Subscript[x, 0]) = 1 and h =1.

-> f[x];
x0=1;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

Answer: The average percentage decay rate in day equals -8.3 %.

What is the average percentage decay rate from the beginning of day 6 to the end of day 7? 
These are two days, so h = 2.
Is the average percentage decay rate for two days twice the average decay rate for 1 day?

-> f[x];
x0=6;
h=2;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

Answer: The average percentage decay rate for day 6 and day 7 equals -15.9 %.

Note that the average percentage decay rate for two days is less than twice the average percentage decay rate for 1 day.

How many grams of I-131 decay during day 6 and day 7?

n[5]-n[7]

Answer: During days 6 and 7, 10.2 grams if I-131 decay.

More details▼.

3.    E.Coli Bacterial Growth 
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E.Coli Bacterial Growth - Example: 
A colony of bacteria grows according to the law of uninhibted growth with a rate of 4% per day. The initial amount of bacteria is 83 grams.

(a) Define the growth function and graph it.

(b) What is the population after 12 days, 20 days?

(c) What is the doubling time for the population?

(d) How long will it take for the population to reach 155 grams?

(e) Advanced: What is the average percentage growth rate in day 1, from the beginning of day 5 to the end of day 7? By how much does the population grow from the beginning of day 5 to the end of day 7?

Solution:

(a)  After one day there are 1.04 times as many bacteria. The growth factor a = 1.04. The initial amount equals 83 grams, so c = 83.

Clear[f,x,a,c];
c =83;
a = 1.04;

f[x_]= c*(a^x)

Plot f. Label the axes with "x (days)" and "Bacteria  (grams)", the plot with "Growth of Bacteria".

MDPlot[f[x],{x,0,20},
AxesLabel->{"x (days)","Bacteria (gram) "},
PlotLabel->"Growth of Bacteria\n"]

More details▼.

(b)  First, estimate with the cursor, on the graph above, how many bacteria there are after 12 days and 20 days.
Then calculate the numbers.

f[12]

f[20]

Answer: The population after 12 days is about 132.8 grams of bacteria, after 20 days  about 181.8 grams of bacteria.

More details▼.

(c)  When has the initial amount c doubled? (Doubling Time) 
Solve the equation 2 c = c*(a^x) for x.

-> Find a;
a=1.04; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

Answer: The population doubles approximately after 17.6 days.

More details▼.

(d)  To find how long it takes for the mass to reach 155 grams, solve the equation f[x] =155.

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==155 ,x]

Answer: It takes approximately 15.9 days for the population to reach 155 grams

More details▼.

(e)  The average percentage growth rate measures the growth of bacteria in percent per day. The time of measurement starts at day x0 and ends at x0+h.

What is the average growth rate during day1? x0 = 1.

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

Answer: The average growth rate in day 1 equals 4%

What is the average percentage growth rate from the beginning of day 5 to the end of day 7? 
Day 5 is included, so there are 3 days. It starts with x0 = 4.

-> n[x];
x0=4;
h=3;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

Answer: The average percentage growth rate the beginning of day 5 to the end of day 7 is 12.48%

What is the absolute growth the beginning of day 5 to the end of day 7 in grams?
Day 5 is included, so it starts with day 4.

f[7]-f[4]

Answer: The absolute growth of bacteria from day 5 to day 7 equals 12.12 grams.

More details▼.

3.   Advanced: Characteristics of the Exp Functions:
Relative Growth Rates Are All the Same 
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Compare Exponential and Polynomial Growth - Example: 
Use the relative growth rate ((f[x0+h]-f[x0])/(f[x0]))to measure the growth of f[x].

(a) Determine c and a of f[x] = c*(a^x), when f[2] =16 and f[5] =1024. 
Define f[x] and graph f for x ∈ {-2.5, 7.5}.

(b) Define the polynomial function f1[x] = (x^4). Graph it with f[x] for x ∈ {-2.5, 7.5} in one plot. Calculate the relative growth rate from -2.5 to 7.5.

(c) Compare the relative growth rates of (4^x)and (x^4) for x ∈ {-2.5, 7.5}. Which one grows better?

(d) Show the rule of thumb for f[x] = (Subscript[n, 0]) (e^(k*x)):
 If |k| ≤ 0.1, the relative growth rate is a good estimate for k. 
 Use f[x] = 3.27 (e^(-0.09 x)) to show this and h = 1.

Solution:

(a)  Enter the two equations f[2] =16 and f[5] =1024 and determine c and a.

-> f[x] == value;
{x1,y1}={2,16};
{x2,y2}={5,1024};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Define f[x] and plot the exponential function f[x]. Label the plot with "Exponential Function".

Clear[f,x];
c =1;
a =4;

f[x_]= c*(a^x)

MDPlot[f[x],{x,-2.5,7.5},AxesLabel->{"x","y"},
PlotLabel->"Exponential Function\n"]

More details▼.

(b) Define f1[x] = (x^4)

Clear[f1,x];
f1[x_]= (x^4)

Plot f and f1. Label the axes with "x" and "y", the plot with "Comparing Functions".

MDPlot[ {f[x],f1[x]},
{x,-2.5,7.5},AxesLabel->{"x","y"},
PlotLabel->"Comparing  Functions\n"]

Measure the growth in f1[x] = (x^4) between -2.5 and 7.5 with the relative growth rate 
and h = 1.
For which (Subscript[x, 0]) is the relative growth rate largest resp. smallest?

Clear[f,x,f1,x0];  (* R e l a t i v e   G r o w t h   R a t e *)

f1[x_]=\[Placeholder]  ; (* enter your f *)  

start =\[Placeholder];
stop = \[Placeholder]; 
h = \[Placeholder]; 

tablevalues = Table[{x0,h,((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},{x0,start, stop, h}]//Chop;

MDShowTable[tablevalues,{"x0 value","h","Rel Growth Rate of  "<>ToString[f1[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f1[x0  +  h] -  f1[x0])/(f1[x0]))],
StandardForm]}];

Clear[f,x,f1,x0];  (* R e l a t i v e   G r o w t h   R a t e *)

f1[x_]=(x^4)  ; (* enter your f *)  

start =-2.5;
stop = 7.5; 
h = 1; 

tablevalues = Table[{x0,h,((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;

MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f1[x],StandardForm]<>"\n\n\t"<>
ToString[HoldForm[((f1[x0  +  h] -  f1[x0])/(f1[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(x\^4\)\n\n\t\!\(\*TagBox[\(\(\(\(f1[\(\(x0 + h\)\)]\)\) - \(\(f1[x0]\)\)\)\/\(f1[x0]\)\), HoldForm]\)"}, {-2.5`, 1, -87.04` "%"}, {-1.5`, 1, -98.76543209876543` "%"}, {-0.5`, 1, 0}, {0.5`, 1, 8000.` "%"}, {1.5`, 1, 671.6049382716049` "%"}, {2.5`, 1, 284.16` "%"}, {3.5`, 1, 173.26114119117037` "%"}, {4.5`, 1, 123.15195854290504` "%"}, {5.5`, 1, 95.07547298681783` "%"}, {6.5`, 1, 77.25219705192396` "%"}, {7.5`, 1, 64.97975308641976` "%"}})

Answer: At (Subscript[x, 0]) =0.5 the relative growth rate is 8000%. At (Subscript[x, 0]) = 7.5 the relative growth rate is about 65%.

More details▼.

(c) Compare the growth of f[x] and f1[x] between -2.5 and 7.5 with the relative growth rate and h = 1.
At which (Subscript[x, 0]) value is the rel. growth rate from the exponential function larger than the rel. growth rate of the polynomial function?

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=\[Placeholder] ; (* enter  f *)  
f1[x_]= \[Placeholder] ;  (* enter  f1 *)

start =\[Placeholder];
stop = \[Placeholder]; 
h = \[Placeholder]; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,{"x0 value","h","Rel Growth Rate of  "<>ToString[f[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>ToString[f1[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=(4^x) ; (* enter your f *)  
f1[x_]= (x^4) ; (* enter  f1 *)

start =-2.5;
stop = 7.5; 
h = 1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,
((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f[x],StandardForm]<>
"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>
ToString[f1[x],StandardForm]<>"\n\n\t"<>

ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(4\^x\)\n\n\t\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)", "Rel Growth Rate of  \!\(x\^4\)\n\n\t\!\(\*TagBox[\(\(\(\(f1[\(\(x0 + h\)\)]\)\) - \(\(f1[x0]\)\)\)\/\(f1[x0]\)\), HoldForm]\)"}, {-2.5`, 1, 300.` "%", -87.04` "%"}, {-1.5`, 1, 300.` "%", -98.76543209876543` "%"}, {-0.5`, 1, 300.` "%", 0}, {0.5`, 1, 300.` "%", 8000.` "%"}, {1.5`, 1, 300.` "%", 671.6049382716049` "%"}, {2.5`, 1, 300.` "%", 284.16` "%"}, {3.5`, 1, 300.` "%", 173.26114119117037` "%"}, {4.5`, 1, 300.` "%", 123.15195854290504` "%"}, {5.5`, 1, 300.` "%", 95.07547298681783` "%"}, {6.5`, 1, 300.` "%", 77.25219705192396` "%"}, {7.5`, 1, 300.` "%", 64.97975308641976` "%"}})

Answer: At (Subscript[x, 0]) = 2.5 the growth rate of the exponential function is higher.

More details▼.

(d) Show, using the example f[x] = 3.27 (e^(-0.09 x)) that the relative growth rate is a good estimate for k if |k| ≤ 0.1.

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e   a n d  k  *)
n0=\[Placeholder];
k=\[Placeholder];

f[x_]=n0*(E^(k*x))  ; (* enter your f *)  

start = 0;
stop = 5; 
h =1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0])),k},{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,{"x0 value","h",ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"k"}];

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e   a n d  k  *)
n0=3.27;
k=-0.09;

f[x_]=n0*(E^(k*x))  ; (* enter your f *)  

start = 0;
stop = 5; 
h =1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0])),k},
{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,
{"x0 value","h",ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],

StandardForm],"k"}];

({{"x0 value", "h", "\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)", "k"}, {0, 1, -0.08606881472877177`, -0.09`}, {1, 1, -0.08606881472877184`, -0.09`}, {2, 1, -0.08606881472877183`, -0.09`}, {3, 1, -0.0860688147287717`, -0.09`}, {4, 1, -0.08606881472877186`, -0.09`}, {5, 1, -0.08606881472877188`, -0.09`}})

More details▼.

EXERCISE  SECTION 
	Test Your Knowledge     Practice

Exponential  Growth

Germs in cow milk  Germs in cow milk grow exponentially, according to the law f[x] = c*(a^x), x in hours. Two hours after milking there are about 18,000 germs in 1(cm^3) cow milk. Three hours later one will find about 1.5 million germs.

(a) What is the rate in percent per hour?

(b) Define the growth function and graph it.

(c) How many germs will one find after 7 hours, 10 hours?

(d) What is the doubling time in minutes for the number of germs?

(e) How long will it take until there are 7 million germs in the cow milk?

(f) Advanced: What is the average percentage growth rate in hour 1, from the beginning of hour 4 to the end of hour 7? What is the absolute growth from hour 4 to hour 7?

Solution: (a) 336.79%   (b) 943.4669019152035` * ((4.367902323681498`)^x)   (c) 2.8617856063833475`*^7; 2.3848213386527944`*^9   (d) 28.2 minutes   (e) 6.04 hours    (f) 336.8 %; 36299.2 %; 2.853923382200721`*^7

-> f[x] == value;
{x1,y1}={2,18000};
{x2,y2}={5,1500000};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Clear[f,x];
c =943.4669019152024`;
a =4.367902323681495`;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,6},AxesLabel->{"x (hours)","population"},PlotLabel->"Number of germs  in cow milk\n"]

f[7]

f[10]

-> Find a;
a=4.367902323681495`; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

0.470158876330054` 60

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==7000000 ,x]

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=3;
h=4;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

f[7]-f[3]

Bacterial Growth  A colony of bacteria grows according to the law of uninhibited growth with a rate of 18% per day. The initial number of bacteria is 125.

(a) Define the growth function and graph it. Label the axis and the plot.

(b) What is the population after 17 days, 31 days?

(c) What is the doubling time for the population?

(d) How long will it take for the population to reach 385?

(e) Advanced: What is the average percentage growth rate in day 1, from the beginning of day 3 to the end of day 5? What is the absolute growth from day 3 to day 5?

Solution: (a) 125 * ((1.18`)^x)   (b) 2085; 21148   (c) 4.18 days   (d) 6.79 days   (e) 18%; 64.3%; 112

Clear[f,x];
c =125;
a =1.18;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,30},AxesLabel->{"x (day)","population"},PlotLabel->"Bacterial Growth\n"]

f[17]

f[31]

-> Find a;
a=1.18; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==385 ,x]

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=2;
h=3;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

f[5]-f[2]

Cell Growth  A biologist observes that the surface area of a cell culture in a culture medium grows about 30 % per hour.The initial surface area is 1000 (mm^2).

(a) Define the growth function and graph it.

(b) What is the surface area after 8 hours, 13 hours?

(c) What is the doubling time for the surface area (hours, minutes)?

(d) How long will it take for the surface area to reach 15,000 (mm^2)?

(e) Advanced: What is the average percentage growth rate in 1 hour, from the beginning of hour 5 to the end of hour 10? What is the absolute growth from the fifth hour to the tenth hour?

Solution: (a) 1000 * (1.3^x)   (b) 8158; 30288   (c) 2 hours and 38.5 minutes   (d) 10 hours and 19.3 minutes    (e) 30%; 382.7%; 10,930 (mm^2)

Clear[f,x];
c =1000;
a =1.3;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,20},AxesLabel->{"x (hours)","area"},PlotLabel->"Cell Growth\n"]

f[8]

f[13]

-> Find a;
a=1.3; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

.6419267958111403` 60

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==15000 ,x]

.3217 60

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=4;
h=6;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

f[10]-f[4]

Population Growth  Let's assume, that the population of a town grows exponentially according to f[x] = c * (a^x), x in years. In 1990 there lived 48,000 inhabitants in that town, in 2000 there were 63,000 inhabitants.

Hint: Convert 1990 - 2100 to x = 0 to x = 110 years

(a) Define the growth function and graph it.

(b) How many inhabitants will live in that town in 2010, and in 2100 (year conversion)?
Do you really think that the exponential function models the growth for 2100 realistic?

(c) What is the doubling time for the population?

(d) How long will it take since there are 70,000 inhabitants in that town?

(e) Advanced: What is the average percentage growth rate in year 1, from the beginning of year 15 to the end of year 23? What is the absolute growth from year 15 to year 23?

Solution: (a) 48000 * ((1.0275664856754427`)^x)   (b) 82688;930087  (c) 25.49 year   (d) about 13.87 years   (e) 2.8%; 27.7%; 19477 inhabitants

-> f[x] == value;
{x1,y1}={0,48000};
{x2,y2}={10,63000};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Clear[f,x];
c =48000;
a =1.0275664856754427;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,110},AxesLabel->{"x (years)","inhabitants"},PlotLabel->"Inhabitants in a town\n"]

f[20]

f[109]

-> Find a;
a=1.0275664856754427; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==70000 ,x]

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=14;
h=9;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

f[23]-f[14]

Bacterial Growth  The number n[x] of bacteria in a colony is given by the exponential model n[x]=(Subscript[n, 0])*(e^(k*x)), x in hours. The initial amount of bacteria is 300.

(a)  If n[7] = 465,  calculate the parameter k. 
Hint: Use also n[0] = 300.

(b) Define the growth function and graph it.

(c) What is the population after 9 hours, 13 hours?

(d) What is the doubling time for the population?

(e) How long will it take for the population to reach 527?

(f) Advanced: What is the average percentage growth rate in hour 1, from the beginning of hour 3 to the end of hour 5? What is the absolute growth from hour 3 to hour 5?

Solution: (a) 0.0626   (b) 300 (e^(0.06260784727587933` x))   (c) 527; 677 bacteria   (d) 11.07 hours   (e) 9 hours    (f) 6.4%, 20.6%, rounded 70 bacteria

-> n[x] == value;
{x1,y1}={7,465};
{x2,y2}={0,300};

k= ((Log[y1/y2])/(x1-x2));n0=y1 (E^(-k x1));
{n0,k}//N

Clear[n,x];
n0=300;
k=0.06260784727587933`;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,15},AxesLabel->{"x (hours)","population"},PlotLabel->"Bacteria Growth\n"]

n[9]

n[13]

-> Find k;
k=0.06260784727587933`; (* enter k > 0 *)

doublingTime =  Log[2]/ k//N

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==527 ,x]

-> f[x];
x0=0;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> f[x];
x0=2;
h=3;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> n[x] == value;
{x1,y1}={0,300};
{x2,y2}={7,465};

k= ((Log[y1/y2])/(x1-x2));n0=y1 (E^(-k x1));
{n0,k}//N

n[5]-n[2]

Fruit flies  A population of fruit flies is increaing exponentially, according to the law n[x]=n0*(e^(k*x)), x in days. After three days there are 170 flies and two days later there are 520 flies. 

(a) What are the parameter (Subscript[n, 0]), k?

(b) Define the growth function and graph it.

(c) How many flies will there be after 7 days, 15 days?

(d) What is the doubling time for the number of flies?

(e) How long will it take until there are 1000 flies?

(f) Advanced: What is the average percentage growth rate in day 1, from the beginning of day 4 to the end of day 6? What is the absolute growth from day 4 to day 6?

Solution: (a) 31.7773, 0.559015  (b) 31.777321508691188` (e^(0.5590151872626056` x))   (c) rounded 1591; 139244 flies   (d) 1.24 days  (e) 6.17 days  (f) 74.9%, 434.97%, 739 flies

-> n[x] == value;
{x1,y1}={3,170};
{x2,y2}={5,520};

k= ((Log[y1/y2])/(x1-x2));n0=y1 (E^(-k x1));
{n0,k}//N

Clear[n,x];
n0=31.777321508691188`;
k=0.5590151872626056`;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,15},AxesLabel->{"x (hours)","population"},PlotLabel->"Fruit Flies\n"]

n[7]

n[15]

-> Find k;
k=0.5590151872626056`; (* enter k > 0 *)

doublingTime =  Log[2]/ k//N

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==1000 ,x]

-> f[x];
x0=0;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> f[x];
x0=3;
h=3;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> n[x] == value;
{x1,y1}={0,300};
{x2,y2}={7,465};

k= ((Log[y1/y2])/(x1-x2));n0=y1 (E^(-k x1));
{n0,k}//N

n[6]-n[3]

GROUP PROJECT   World Population Growth

Download the World Population Data Sheet  
http://www.prb.org/pdf05/05WorldDataSheet_Eng.pdf

The most important data are provided below in the tables.

Then answer the following questions, according to the function n[x] = (Subscript[n, 0])(e^(k x)), x in years:

(a) Find the population of the world in millions in Mid-2005. What is the percentage of  population living in the more developed and less developed regions?

(b) Find the rate of natural increase of the world. Define the growth function for the world population. Let x = 0 correspond to 2005 and graph it til 2100.
Hint: (Subscript[n, 0])should be in millions. Use an appropriate PlotRange.

(c) What will be the population of the world in 2050 and 2100? 

(d) What is the doubling time for the population?

(e) How long will it take for the population to reach 8000 millions?

(f) Find out which African nation has the highest population growth rates in the world. In Austria is the population growth only 0.1 % per year. Define the growth function for each country. Let x = 0 correspond to 2005 and graph it til 2100.

Why does the Austrian growth curve seems to be linear?

Hint: (Subscript[n, 0])should be in millions. Use an appropriate PlotRange.

(g) When would the two countries have populations of the same size?

(h) Compare the two countries Libya and Nicaragua. Define the growth function for each country, letting x = 0 correspond to 2005 and graph it.

(i) Make projections for each population in the year 2025.

(j) When reach Nicaragua 10 million? What would the population be at that time in Libya?
             
Solution: (a) more: 18.7%; less: 81.3%   (b) 6477 * (e^(0.011928570865273812` x))  (c) 11078.9; 20115.1  (d) 58 years 1.2 months  (e) 17 years 8.4 months  (f)  14 * (e^(0.03343477608623742` x)); 8.2` * (e^(0.0009995003330834232` x))  (g) Never  (h) Libya: 5.8`  *(e^(0.023716526617316065` x)); Nicaragua: 5.8` * (e^(0.026641930946421092` x))  (i) Libya: 9.32024 millions; Nicaragua: 9.88182 millions  (j) Nicaragua: after 20.44 years; Libya: 9.42 millions

({{Region, "Mid-2005\n(million)""", "Rate of Natural Increase\n(%)"}, {World Population, 6477, 1.2}, {More Developed , 1211, 0.1}, {Less Developed, 5266, 1.5}})

pMore=1211/6477 *100 //N
pLess=5266/6477 *100 //N

Clear[n,x];
k=N[  Log[1.012] ];  (* \[Placeholder] -> a *)
n0 =c = 6477;

n[x_] =n0*(E^(k x))

MDPlot[n[x],{x,0,95},AxesLabel->{"x (years)","population\n (millions)"},PlotLabel->"Population Growth in the World\n",PlotRange->{ 0, 20000 }]

n[45]

n[95]

-> Find k;
k=0.011928570865273812`; (* enter k > 0 *)

doublingTime =  Log[2]/ k//N

12 0.108

Clear[x];
NSolve[ n[x]==8000 ,x]

12* 0.704

({{Region, "Mid-2005\n(million)""", "Rate of Natural Increase\n(%)"}, {Niger, 14.0, 3.4}, {Austria , 8.2, 0.1}})

Clear[f,x];
k=N[  Log[1.034] ];  (* \[Placeholder] -> a *)
n0 =c = 14;

fNiger[x_] =n0*(E^(k x))

Clear[f,x];
k=N[  Log[1.001] ];  (* \[Placeholder] -> a *)
n0 =c = 8.2;

fAustria[x_] =n0*(E^(k x))

MDPlot[fNiger[x],{x,0,95},AxesLabel->{"x (years)","population\n (millions)"},PlotLabel->"Population Growth of Niger\n"]

MDPlot[fAustria[x],{x,0,95},AxesLabel->{"x (years)","population\n (millions)"},PlotLabel->"Population Growth of Austria\n",PlotRange->{ 8.2, 10 }]

({{Region, "Mid-2005\n(million)""", "Rate of Natural Increase\n(%)"}, {Libya, 5.8, 2.4}, {Nicaragua , 5.8, 2.7}})

Clear[f,x];
k=N[  Log[1.024] ];  (* \[Placeholder] -> a *)
n0 =c = 5.8;

fLibya[x_] =n0*(E^(k x))

Clear[f,x];
k=N[  Log[1.027] ];  (* \[Placeholder] -> a *)
n0 =c = c;

fNicaragua[x_] =n0*(E^(k x))

MDPlot[fLibya[x],{x,0,100},AxesLabel->{"x (years)","population\n (millions)"},PlotLabel->"Population Growth of Libya\n"]

MDPlot[fNicaragua[x],{x,0,100},AxesLabel->{"x (years)","population\n (millions)"},PlotLabel->"Population Growth of Nicaragua\n"]

fLibya[20]

fNicaragua[20]

Clear[x];
NSolve[ fNicaragua[x]==10 ,x]

fLibya[20.446234791958553]

Exponential Decay

Radioactive Decay  Strontium 90 is a radioactive material that decays according to the function n[x] = (Subscript[n, 0])(e^(-0.0244 x)), x in years. A scientist has a sample of 200 grams of strontium 90.

(a) Define the decay function and graph it. 

(b) How much strontium 90 is left after 8 years and 21 years?

(c) When will 50 grams of strontium 90 be left?

(d) What is the half-life of strontium 90? What percentage decays every year?

(e) Advanced: What is the average percentage decay rate in year 1, from the beginning of year 11 to the end of year 15? What is the absolute decay (that is, by how much does the mass fall) from the beginning of year 11 to the end of year 15?

Solution: (a) 200 (e^(-0.0244` x))   (b) 164.5; 119.8 grams   (c) 56.8 years   (d) 28.4 years; 2.4 % per year   (e) -2.4 %; -11.5 %;  17.99 grams

Clear[n,x];
n0=200;
k=-0.0244;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,100},AxesLabel->{"x (years)","amount (gram)"},PlotLabel->"Radioactive Decay of Strontium-90\n"]

n[8]

n[21]

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==50 ,x]

-> Find k;
k=-0.0244;  (* enter k < 0 *)

halfLife = - Log[2]/ k//N

Clear[f,x];
a=N[(E^(-0.0244)) ];  (* \[Placeholder] -> k *)
c =n0 = 200;

f[x_] = c*(a^x)

percent=(1-0.9758952735664502)*100

-> f[x];
x0=1;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> f[x];
x0=10;
h=5;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

n[10]-n[15]

Intensity of Laser Beams  Laser beams are very intense rays of light, which can be focused on a tiny spot. They are commonly used in medicine. The intensity decreases exponentially with the penetration of tissue exponentially. At a depth of 6 mm, the intensity decreases to 10 % of its initial value.The initial intensity is 800 Watt/(m^2).

(a) Define the decay function f[x] = c*(a^x) (where x is the depth in mm and c is the intensity) and graph it. 

(b) What is the intensity at a depth of 4 mm? At a depth of 10 mm?

(c) At what depth is the intensity 500 Watt/(m^2)?

(d) What is the initial intensity if the intensity at a depth of 8 mm is 130 Watt/(m^2)?
Hint: a remains the same. Use the Alg button, to solve the equation for c. Change x to c, so Clear[c] and solve for c.

Solution: (a)800*(0.6812^x)  (b) 172; 17 W/(m^2)   (c) 1.2 mm   (d) 2800 W/(m^2)

-> f[x] == value;
{x1,y1}={0,800};
{x2,y2}={6,0.1*800};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Clear[f,x];
c =800;
a =0.6812920690579612;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,20},AxesLabel->{"x (mm)","intensity (Watt/(m^2)"},PlotLabel->"Intensity of Laser Beams\n"]

f[4]

f[10]

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==500 ,x]

Clear[c];
Solve[{ c *(a^8) == 130, \[Placeholder] == \[Placeholder] },{ c } ]

Estimating the Age of a Fossil  Living organisms absorb carbon and release it. In the atmosphere - and also in living organisms - there is the same, constant amount of radioactive carbon-14, as a proportion of the total amount of carbon. After an organism dies, the amount of carbon-14 decays exponentially.

A fossilised leaf contains 67 % of its normal amount of carbon-14. The half-life of carbon-14 may be taken to be 5760 years.

(a) Define the decay function, according to n[x] = (Subscript[n, 0])(e^(k* x)), x in years, and graph it for an initial amount of 6 grams. 
Hint: Plot til 10000 years. Use the PlotRange -> {0,6}.

(b) How old is the fossil?

(c) How long will it take for 6 grams of carbon-14 to decay to 1 gram?

Solution: (a) 6 * (e^(-0.00012033805218054606` x))   (b) 3328 years   (c) 14889.4 years

-> Find  half-life;
halfLife=5760; 

k = - Log[2]/ halfLife//N

Clear[n,x];
n0=6;
k=-0.00012033805218054606;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,10000},AxesLabel->{"x (years)","amount (g)"},PlotLabel->"Estimating the Age of a Fossile\n",PlotRange->{ 0, 6 } ]

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==n0*0.67 ,x]

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==1 ,x]

Height of Mount Everest  Atmospheric pressure decreases exponentially with altitude (height) above sea level. (This model assumes a constant temperature of the atmosphere in order to simplify the formula for atmospheric pressure.)

At an altitude of 5500 m, the pressure is half its sea level value. Atmospheric pressure at sea level is 1013 hPa.

(a) Define the decay function, according to n[x] = (Subscript[n, 0])(e^(k*x)), for x in meters, and graph it. 
Hint: Plot to 10000 m. Use the PlotRange from 0 to 1040hPa.

(b) An airliner flies at an altitude of 11000 m. What is atmospheric pressure at this altitude?
What is the decay rate per 1000 m?
Hint: use (Subscript[x, 0])=0 and h = 1000.

(c) At the summit of Mount Everest, the barometer indicates a pressure of 332 hPa.
What is the height of Mount Everest?

(d) Advanced: What is the average percentage decay rate per meter, from 300 m to 1000 m? What is the absolute drop in pressure from 300 m to 1000 m?
Hint: Switch to n[x].

Solution: (a) 1013 * (e^(-0.00012602676010180824` x));   (b) 253.25 hPa; -11.8%   (c) 8851.5 m  (d) -0.0126 %; -8.44 %;  82.4 hPa

-> f[x];
x0=1013;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> Find  half-life;
halfLife=5500; 

k = - Log[2]/ halfLife//N

Clear[n,x];
n0=1013;
k=-0.00012602676010180824;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,10000},AxesLabel->{"x (meter)","pressure (hPa)"},PlotLabel->"Atmospheric Pressure\n",PlotRange->{ 0, 1040 }]

n[11000]

-> f[x];
x0=1013;
h=1000;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==332.0 ,x]

-> f[x];
x0=1013;
h=1;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

-> f[x];
x0=1013;
h=700;

((n[x0+h]-n[x0])/(n[x0]))*100"%"

n[1000] -n[300]

GROUP PROJECT 1   Iceman Frozen Fritz

Over 5000 years ago, a man climbed up to the icy heights of the Schnalstal glacier (border Austria - Italy) and died. He was found by chance in 1991, with his clothes and equipment, mummified and frozen: an archaeological sensation and an unique snapshot of a Copper Age man.
To find out, when the Iceman lived the scientists used the Radio Carbon Dating Method. They measure the amount of carbon-14 in the dead organism.

Source: http://www.cbv.ns.ca/marigold/history/iceman/iceman.html

(a) The half-life of carbon-14 is between 5690 years and 5770 years. Define the decay function for both half-lives (according to n[x] = (Subscript[n, 0]) * (e^(k * x)),n1[x], n2[x]) and graph them.
Hint: Use (Subscript[n, 0])=1 at the time when the man died, choose the PlotRange from 0 to 1.

(b) It is said that the Iceman lived 5300 years ago. What is the percentage of carbon-14 that the scientists must have measured? Calculate with both decay functions. 
Hint: (Subscript[n, 0])=1.

(c) With a probability of 56% the iceman lived between 3350 and 3300 BC. Calculate the percentage of carbon-14 for both half-lives. Is there a difference in the percentage of carbon-14 to (b)?

Hint: Use x = 3350 + 1991 = 5341  for 3350 BC and x = 5291 for 3300 BC
             
Solution: (a) n0 * (e^(-0.0001218184851599201` x));  n0 * (e^(-0.00012012949403118636` x))   (b) 52.43 %; 52.90 %   (c) 52.17 % - 52.49 %; 52.64 % - 52.96 %

-> Find  half-life;
halfLife=5690; 

k = - Log[2]/ halfLife//N

Clear[n,x];
n0=1;
k=-0.0001218184851599201;

n1[x_]=n0*(E^(k*x))

-> Find  half-life;
halfLife=5770; 

k = - Log[2]/ halfLife//N

Clear[n,x];
n0=1;
k=-0.00012012949403118636;

n2[x_]=n0*(E^(k*x))

MDPlot[n1[x],{x,0,5500},AxesLabel->{"x (years)","carbon-14 (amount)"},PlotLabel->"Iceman's Age\n (half-life: 5690 years)\n",PlotRange->{ 0, 1 }]

MDPlot[n2[x],{x,0,5500},AxesLabel->{"x (years)","carbon-14 (amount)"},PlotLabel->"Iceman's Age\n (half-life: 5770 years)\n",PlotRange->{ 0, 1 }]

n1[5300]

n2[5300]

n1[5341]

n1[5291]

n2[5341]

n2[5291]

GROUP PROJECT 2   Demonstration of Radioactive Decay (using coins)

In your class each student take 1 penny and stands up. Then everybody flips the coin. 
If a student has heads, the student is radioactive and decays; so the student has to sit down.
If a student has tails, the student is not radioactive and remains standing.

Good results are for 75 and more students.

(a) Count the students who remain standing after each "half-life", print it into a table and graph it.

(b) After two or three "half-lives" answer the following question: 
What is happening to the number of students standing? Do the same number of students sit down each time the coins are flipped?

(c) About how many students would have had to sit down if we started with twice as many? What about if there are only half as many in this class? What does that tell you about how the quantity of "radioactive isotopes" affects the number that decay?

(d) Can you predict which of you is going to be the first to sit down? Why or why not?

(e) How is radioactive decay sort of like gambling or playing the lottery?
             
Solution: None

Rel Growth Rate Problems

Determine c and a of f[x] = c*(a^x), when f[3] = 0.216 and f[7] = 0.0279936. 

(a) Define f[x] and graph f for x ∈ {-1.5, 8.5}.

(b) What is the average percentage decay rate for h = 1, 2, 3? Compare the three solutions. Does the decay rate increase linearly with h (that is, is its graph against h a straight line)?

Solution: (a) ((0.6`)^x)  (b) -40%, -64%, -78.4%, no

-> f[x] == value;
{x1,y1}={3,0.216};
{x2,y2}={7,0.0279936};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Clear[f,x];
c =1;
a =0.6;

f[x_]= c*(a^x)

MDPlot[f[x],{x,-1.5,8.5},AxesLabel->{"x","y"},PlotLabel->"Exponential Function\n"]

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=1;
h=2;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> n[x];
x0=1;
h=3;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

Polynomial ↔ Exponential Function
(a) Define the polynomial function f1[x] =(x^(-0.6))  and the exponential function
  f[x]=(0.6^x).

(b) Graph f1[x] and f[x]  in the same plot for x∈{0.1,8.1}. f1[x] is the green curve.

(c) Calculate the percentage decay rate of f1[x] from 0.1 to 8.1  with h = 1.

(d) Calculate the percentage decay rate of f[x] from 0.1 to 8.1  with h = 1.
Hint: Copy the program below.

Solution: None provided

Clear[f1,x];
f1[x_]= (x^(-0.6))

Clear[f,x];
c =1;
a =0.6;

f[x_]= c*(a^x)

MDPlot[{f1[x],f[x]},{x,0.1,8.1},AxesLabel->{"x","y"},PlotLabel->"Comparing Functions \n"]

Clear[f,x,x0];  (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=\[Placeholder]  ; (* enter your f *)  

start =\[Placeholder];
stop = \[Placeholder]; 
h = \[Placeholder]; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N},{x0,start, stop, h}]//Chop;

MDShowTable[tablevalues,{"x0 value","h","Rel Growth Rate of  "<>ToString[f[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f[x0  +  h] -  f[x0])/(f[x0]))],
StandardForm]}];

Clear[f,x,x0];  (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=(x^(-0.6))  ; (* enter your f *)  

start =0.1;
stop = 8.1; 
h = 1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;

MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f[x],StandardForm]<>"\n\n\t"<>
ToString[HoldForm[((f[x0  +  h] -  f[x0])/(f[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(1\/x\^0.6`\)\n\n\t\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)"}, {0.1`, 1, -76.27728513382841` "%"}, {1.1`, 1, -32.157157746389395` "%"}, {2.1`, 1, -20.83840054047118` "%"}, {3.1`, 1, -15.443557610039571` "%"}, {4.1`, 1, -12.274024062122212` "%"}, {5.1`, 1, -10.185965694858385` "%"}, {6.1`, 1, -8.705862123957218` "%"}, {7.1`, 1, -7.601696378039493` "%"}, {8.1`, 1, -6.746277725371944` "%"}})

Clear[f,x,x0];  (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=(0.6^x)  ; (* enter your f *)  

start =0.1;
stop = 8.1; 
h = 1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;

MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f[x],StandardForm]<>"\n\n\t"<>
ToString[HoldForm[((f[x0  +  h] -  f[x0])/(f[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(0.6`\^x\)\n\n\t\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)"}, {0.1`, 1, -40.00000000000001` "%"}, {1.1`, 1, -40.` "%"}, {2.1`, 1, -40.00000000000001` "%"}, {3.1`, 1, -39.999999999999986` "%"}, {4.1`, 1, -40.` "%"}, {5.1`, 1, -40.` "%"}, {6.1`, 1, -40.00000000000001` "%"}, {7.1`, 1, -40.` "%"}, {8.1`, 1, -40.00000000000001` "%"}})

Compare the relative growth rates of (7^x)and (x^7) for x ∈ {-2.5, 7.5}. 

(a) Which one grows better?

(b) At which (Subscript[x, 0]) value is the relative growth rate from the exponential function higher than the relative growth rate of the polynomial function?

Solution: (b) 3.5

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=\[Placeholder] ; (* enter  f *)  
f1[x_]= \[Placeholder] ;  (* enter  f1 *)

start =\[Placeholder];
stop = \[Placeholder]; 
h = \[Placeholder]; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,{"x0 value","h","Rel Growth Rate of  "<>ToString[f[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>ToString[f1[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=(7^x) ; (* enter your f *)  
f1[x_]= (x^7) ; (* enter  f1 *)

start =-2.5;
stop = 7.5; 
h = 1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,
((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f[x],StandardForm]<>
"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>
ToString[f1[x],StandardForm]<>"\n\n\t"<>

ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(7\^x\)\n\n\t\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)", "Rel Growth Rate of  \!\(x\^7\)\n\n\t\!\(\*TagBox[\(\(\(\(f1[\(\(x0 + h\)\)]\)\) - \(\(f1[x0]\)\)\)\/\(f1[x0]\)\), HoldForm]\)"}, {-2.5`, 1, 600.` "%", -97.20063999999999` "%"}, {-1.5`, 1, 600.0000000000001` "%", -99.95427526291724` "%"}, {-0.5`, 1, 599.9999999999999` "%", -200.` "%"}, {0.5`, 1, 599.9999999999999` "%", 218600.` "%"}, {1.5`, 1, 600.0000000000001` "%", 3472.245084590764` "%"}, {2.5`, 1, 600.` "%", 954.1350399999999` "%"}, {3.5`, 1, 600.` "%", 480.7795099952279` "%"}, {4.5`, 1, 600.` "%", 307.4283358307361` "%"}, {5.5`, 1, 600.` "%", 221.9991090548751` "%"}, {6.5`, 1, 600.` "%", 172.292291784362` "%"}, {7.5`, 1, 600.0000000000001` "%", 140.16163760402375` "%"}})

Show the rule of thumb for f[x] = (Subscript[n, 0]) (e^(k*x)):
 If |k| ≤ 1, the relative growth rate is a good estimation for k. 
 
 Use f[x] = 2.58 (e^(0.07 x)) to show this from -5 to 5 and h = 1.
 
 Solution: None provided

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e   a n d  k  *)
n0=\[Placeholder];
k=\[Placeholder];

f[x_]=n0*(E^(k*x))  ; (* enter your f *)  

start = \[Placeholder];
stop = \[Placeholder]; 
h =\[Placeholder]; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0])),k},{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,{"x0 value","h",ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"k"}];

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e   a n d  k  *)
n0=2.58;
k=0.07;

f[x_]=n0*(E^(k*x))  ; (* enter your f *)  

start = -5;
stop = 5; 
h =1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0])),k},
{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,
{"x0 value","h",ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],

StandardForm],"k"}];

({{"x0 value", "h", "\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)", "k"}, {-5, 1, 0.07250818125421649`, 0.07`}, {-4, 1, 0.07250818125421658`, 0.07`}, {-3, 1, 0.07250818125421624`, 0.07`}, {-2, 1, 0.07250818125421671`, 0.07`}, {-1, 1, 0.07250818125421642`, 0.07`}, {0, 1, 0.07250818125421661`, 0.07`}, {1, 1, 0.07250818125421646`, 0.07`}, {2, 1, 0.07250818125421628`, 0.07`}, {3, 1, 0.07250818125421667`, 0.07`}, {4, 1, 0.07250818125421649`, 0.07`}, {5, 1, 0.0725081812542164`, 0.07`}})

Test Your Knowledge

Growth of an Insect Population  The size of a certain insect population at time x (in days) obeys the function n[x] = 400 * (e^(0.028 x)).

(a) Define the function and graph it. 

(b) Determine the initial number of insects (when x = 0).

(c) What is the daily growth rate of the insect population?
Hint: Use the Perc Growth/Decay button for x0 = 1 and h = 1.

(d) What will  the population be after 8 days?

(e) When will the insect population reach 700?

(f) When will the insect population double?

Clear[n,x];
n0=400;
k=0.028;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,50},AxesLabel->{"x (days)","population"},PlotLabel->"Growth of an Insect Population\n"]

n[0]

Clear[f,x];
a=N[(E^0.028) ];  (* \[Placeholder] -> k *)
c =n0 = 400;

f[x_] = c*(a^x)

-> n[x];
x0=1;
h=1;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

n[8]

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==700 ,x]

-> Find k;
k=0.028; (* enter k > 0 *)

doublingTime =  Log[2]/ k//N

Bacterial Growth  A culture of bacteria grows exponentially, according to f[x] = c * (a^x) (x in hours). After 3 hours there are 800 and two hours later there are 873.

(a) Define the function and graph it. What is the daily growth rate?
Hint: A day has 24 hours.

(b) How long will it take until there are 10,000 bacteria?

(c) Calculate the doubling time.

-> f[x] == value;
{x1,y1}={3,800};
{x2,y2}={5,873};

a=Exp[ ((Log[y1/y2])/(x1-x2))];c=y1 /(a^( x1));
{c,a}//N

Clear[f,x];
c =702;
a =1.044629120788809;

f[x_]= c*(a^x)

MDPlot[f[x],{x,0,100},AxesLabel->{"x (hours)","population"},PlotLabel->"Bacterial Growth\n"]

-> n[x];
x0=1;
h=24;

((f[x0+h]-f[x0])/(f[x0]))*100"%"

-> Calc  c,a;
Clear[x];
NSolve[ f[x]==10000 ,x]

-> Find a;
a=1.044629120788809; (* enter a > 1 *)

doublingTime =  Log[a,2]//N

C-14 Decay Estimating the Age of a Tree  A piece of charocoal is found to contain 35 % of the amount of carbon-14 that it originally had.

(a)  Define the function n[x] = (Subscript[n, 0])(e^(-k*x)), using 5730 years as half-life of carbon-14, and graph it.

Hint: Use (Subscript[n, 0]) = 1 to plot the function. Choose an appropriate PlotRange.

(b) When did the tree from which the charcoal came die?

-> Find  half-life;
halfLife=5730; 

k = - Log[2]/ halfLife//N

Clear[n,x];
n0=1;
k=-0.0001209680943385594;

n[x_]=n0*(E^(k*x))

MDPlot[n[x],{x,0,10000},AxesLabel->{"x","y"},PlotLabel->"C-14 Decay\n",PlotRange->{ 0, 1 }]

-> Calc  n0,k;
Clear[x];
NSolve[ n[x]==0.35 ,x]

Compare the relative growth rates of (3.5^x)and (x^3.5) for x ∈ {0.1, 15.1}. 

(a) Which one grows better?

(b) At which (Subscript[x, 0]) value is the relative growth rate from the exponential function higher than the relative growth rate of the polynomial function?

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=\[Placeholder] ; (* enter  f *)  
f1[x_]= \[Placeholder] ;  (* enter  f1 *)

start =\[Placeholder];
stop = \[Placeholder]; 
h = \[Placeholder]; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,{"x0 value","h","Rel Growth Rate of  "<>ToString[f[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>ToString[f1[x],StandardForm]<>"\n\n\t"<>ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

Clear[f,x,x0]; (* R e l a t i v e   G r o w t h   R a t e *)

f[x_]=(3.5^x) ; (* enter your f *)  
f1[x_]= (x^3.5) ; (* enter  f1 *)

start = 0.1;
stop = 15.1; 
h = 1; 

tablevalues = Table[{x0,h,((f[x0+h]-f[x0])/(f[x0]))100"%"//N,
((f1[x0+h]-f1[x0])/(f1[x0]))100"%"//N},
{x0,start, stop, h}]//Chop;
MDShowTable[tablevalues,
{"x0 value","h","Rel Growth Rate of  "<>
ToString[f[x],StandardForm]<>
"\n\n\t"<>ToString[HoldForm[((f[x0+h]-f[x0])/(f[x0]))],
StandardForm],"Rel Growth Rate of  "<>
ToString[f1[x],StandardForm]<>"\n\n\t"<>

ToString[HoldForm[((f1[x0+h]-f1[x0])/(f1[x0]))],
StandardForm]}];

({{"x0 value", "h", "Rel Growth Rate of  \!\(3.5`\^x\)\n\n\t\!\(\*TagBox[\(\(\(\(f[\(\(x0 + h\)\)]\)\) - \(\(f[x0]\)\)\)\/\(f[x0]\)\), HoldForm]\)", "Rel Growth Rate of  \!\(x\^3.5`\)\n\n\t\!\(\*TagBox[\(\(\(\(f1[\(\(x0 + h\)\)]\)\) - \(\(f1[x0]\)\)\)\/\(f1[x0]\)\), HoldForm]\)"}, {0.1`, 1, 249.99999999999994` "%", 441342.75959630375` "%"}, {1.1`, 1, 250.00000000000006` "%", 861.37568435365` "%"}, {2.1`, 1, 249.99999999999994` "%", 290.83945305028055` "%"}, {3.1`, 1, 249.99999999999983` "%", 166.0587645530979` "%"}, {4.1`, 1, 250.00000000000006` "%", 114.6604907143461` "%"}, {5.1`, 1, 249.99999999999994` "%", 87.13663080140056` "%"}, {6.1`, 1, 250.00000000000009` "%", 70.11781578221333` "%"}, {7.1`, 1, 249.99999999999994` "%", 58.59640751219707` "%"}, {8.1`, 1, 250.` "%", 50.29596009319089` "%"}, {9.1`, 1, 250.00000000000006` "%", 44.038946112559564` "%"}, {10.1`, 1, 249.99999999999994` "%", 39.157179443683546` "%"}, {11.1`, 1, 250.` "%", 35.24411313611802` "%"}, {12.1`, 1, 250.00000000000006` "%", 32.03854764965893` "%"}, {13.1`, 1, 250.` "%", 29.36516879056797` "%"}, {14.1`, 1, 250.` "%", 27.1020107787465` "%"}, {15.1`, 1, 250.00000000000085` "%", 25.161641581889683` "%"}})

Here are some Internet resources:

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This project has been funded with support from the European Commission.

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Duration:1.10.2005-30.9.2008

This document reflects the views only of the authors, and the Commission cannot be held responsible for any use which may be made of the information contained therein.

Authors: Susanne Kissling,  Reinhard Simonovits, Bernd Thaller

Tube:  http://biology.clc.uc.edu/fankhauser/Labs/Microbiology/Growth_Curve/     Growth_Curve.htm

Nuclear Medicine:  http://www.siraonline.com

Cheese:  http://homepage.mac.com/adrober/.cv/adrober/         Sites/.Pictures/Photo%20Album%20Pictures/           2004-04-16%2019.38.46%20-0700/Image-229DE884901811D8.jpg-       thumb_269_202.jpg

Ötzi:  www.ethlife.ethz.ch/articles/oetziorigin.html