Lensyl Urbano – Page 7 – Montessori Muddle

Linearizing an Exponential Function: Radioactive Decay

Using this data for the decay of a radioisotope, find its half life.

t (s)	A (g)
0	100
100	56.65706876
200	32.10023441
300	18.18705188
400	10.30425049
500	5.838086287
600	3.307688562

We can start with the equation for decay based on the half life:

  

 
where:

and linearize (make it so it can be plotted as a straight line) by using logarithms.

Take the log of each side (use base 2 because of the half life):



Use the rules of logarithms to simplify:

 

 



   

Finally rearrange a little:

Now, since the two variables in the last equation are A and t we can see the analogy between this equation and the equation of a straight line:

    

and,

      

where:

So if we draw a graph with log₂(A) on the y-axis, and time (t) on the x axis, the slope of the line should be:

Which we can use to find the half life (λ).

Radioactive Half Lives

Since we most commonly talk about radioactive decay in terms of half lives, we can write the equation for the amount of a radioisotope (A) as a function of time (t) as:



where:

To reverse this equation, to find the age of a sample (time) we would have to solve for t:

Take the log of each side (use base 2 because of the half life):



Use the rules of logarithms to simplify:

 

 



 

Now rearrange and solve for t:

So we end up with the equation for time (t):

Now, because this last equation is a linear equation, if we’re careful, we can use it to determine the half life of a radioisotope. As an assignment, find the half life for the decay of the radioisotope given below.

t (s)	A (g)
0	100
100	56.65706876
200	32.10023441
300	18.18705188
400	10.30425049
500	5.838086287
600	3.307688562

Comparing Covid Cases of US States

Missouri’s confirmed cases (z-score) compared to the other U.S. states from April 20th to October 3rd, 2020. The z-score is a measure of how far you are away from the average. In this case, a negative z-score is good because it indicates that you’re below the average number of cases (per 1000 people). For all the states.

Based on my students’ statistics projects, I automated the method (using R) to calculate the z-score for all the states in the U.S. We used the John Hopkins daily data.

I put graphs for all of the states on the COVID: The U.S. States Compared webpage.

The R functions (test.R) assumes all of the data is in a folder (COVID-19-master/csse_covid_19_data/csse_covid_19_daily_reports_us/), and outputs the graphs to the folder ‘images/zscore/‘ which needs to exist.

covid_data <- function(infile, state="Missouri") {
    filename <- paste(file_dir, infile, sep='')
    mydata <- read.csv(filename)
    pop <- read.csv('state_populations.txt')
    mydata <- merge(mydata, pop)
    mydata$ConfirmedPerCapita1000 <- mydata$Confirmed / mydata$Population *1000
    summary(mydata$ConfirmedPerCapita1000)
    stddev <- sd(mydata$ConfirmedPerCapita1000)
    avg <- mean(mydata$ConfirmedPerCapita1000)
    cpc1k <- mydata[mydata$Province_State == state,]$ConfirmedPerCapita1000
    zscore <- (cpc1k - avg)/stddev
    #print(infile, zscore)
    return(zscore)
}


get_zScore_history <-function(state='Missouri') {
  df <- data.frame(Date=as.Date(character()), zscore=numeric())
  for (f in datafiles){
    dateString <- as.Date(substring(f, 1, 10), format='%m-%d-%y')
    zscore <- covid_data(f, state=state)
    df[nrow(df) + 1,] = list(dateString, zscore)
  }
  df$day <- 1:nrow(df)

  plot_zScore(df, state)


  # LINEAR REGRESSIONS:
  # http://r-statistics.co/Linear-Regression.html
  lmod <- lm(day ~ zscore, df)
  return(df)
}

plot_zScore <- function(df, state){
  max_z <- max( abs(max(df$zscore)), abs(min(df$zscore)))
  print(max_z)


  zplot <- plot(x=df$day, y=df$zscore, main=paste('z-score: ', state), xlab="Day since April 20th, 2020", ylab='z-score', ylim=c(-max_z,max_z))
  abline(0,0, col='firebrick')
  dev.copy(png, paste('images/zscore/', state, '-zscore.png', sep=''))
  dev.off()
}

get_states <- function(){
  lastfile <- datafiles[ length(datafiles) ]
  filename <- paste(file_dir, lastfile, sep='')
  mydata <- read.csv(filename)
  pop <- read.csv('state_populations.txt')
  mydata <- merge(mydata, pop)
  return(mydata$Province_State)
}

graph_all_states <- function(){
  states <- get_states()
  for (state in states) {
    get_zScore_history(state)
  }
}

file_dir <- 'COVID-19-master/csse_covid_19_data/csse_covid_19_daily_reports_us/'
datafiles <- list.files(file_dir, pattern="*.csv")

print("To get the historical z-score data for a state run (for example):")
print(" > get_zScore_history('New York')" )

df = get_zScore_history()

You can run the code in test.R in the R console using the commands:

> source('test.R')

which does Missouri by default, but to do other states use:

> get_zScore_history('New York')

To get all the states use:

> graph_all_states()

Basic R (using Covid data)

Once you start R you’ll need to figure out which directory you’re working in:

> getwd()

On a Windows machine your default working directory might be something like:

[1] "C:/Users/username"

On OSX or Linux you’ll get something like:

 [1] "/Users/username"

To get to the directory you want to work in use setwd(). I’ve put my files into the directory: “/Users/lurba/Documents/TFS/Stats/COVID”

To get there from the working directory above I could enter the full path above, or just the relative path like:

> setwd("TFS/Stats/COVID")

Now my data is in the file named “04-20-2020.csv” (from the John Hopkins Covid data repository on github) which I’ll read in with:

> mydata <- read.csv("04-20-2020.csv")

This creates a variable named “mydata” that has the information in it. I can see the column names by using:

> colnames(mydata)

which gives:

 [1] "Province_State"       "Country_Region"       "Last_Update"         
 [4] "Lat"                  "Long_"                "Confirmed"           
 [7] "Deaths"               "Recovered"            "Active"              
 [10] "FIPS"                 "Incident_Rate"        "People_Tested"       
 [13] "People_Hospitalized"  "Mortality_Rate"       "UID"                 
 [16] "ISO3"                 "Testing_Rate"         "Hospitalization_Rate"

Let’s take a look at the summary statistics for the number of confirmed cases, which is in the column labeled “Confirmed”:

> summary(mydata$Confirmed)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    317    1964    4499   15347   13302  253060

This shows that the mean is 15, 347 and the maximum is 253,060 confirmed cases.

I’m curious about which state has that large number of cases, so I’m going to print out the columns with the state names (“Province_State”) and the number of confirmed cases (“Confirmed”). From our colnames command above we can see that “Province_State” is column 1, and “Confirmed” is column 6, so we’ll use the command:

> mydata[ c(1,6) ]

The “c(1,6)” says that we want the columns 1 and 6. This command outputs

             Province_State Confirmed
1                   Alabama      5079
2                    Alaska       321
3            American Samoa         0
4                   Arizona      5068
5                  Arkansas      1973
6                California     33686
7                  Colorado      9730
8               Connecticut     19815
9                  Delaware      2745
10         Diamond Princess        49
11     District of Columbia      2927
12                  Florida     27059
13                  Georgia     19407
14           Grand Princess       103
15                     Guam       136
16                   Hawaii       584
17                    Idaho      1672
18                 Illinois     31513
19                  Indiana     11688
20                     Iowa      3159
21                   Kansas      2048
22                 Kentucky      3050
23                Louisiana     24523
24                    Maine       875
25                 Maryland     13684
26            Massachusetts     38077
27                 Michigan     32000
28                Minnesota      2470
29              Mississippi      4512
30                 Missouri      5890
31                  Montana       433
32                 Nebraska      1648
33                   Nevada      3830
34            New Hampshire      1447
35               New Jersey     88722
36               New Mexico      1971
37                 New York    253060
38           North Carolina      6895
39             North Dakota       627
40 Northern Mariana Islands        14
41                     Ohio     12919
42                 Oklahoma      2680
43                   Oregon      1957
44             Pennsylvania     33914
45              Puerto Rico      1252
46             Rhode Island      5090
47           South Carolina      4446
48             South Dakota      1685
49                Tennessee      7238
50                    Texas     19751
51                     Utah      3213
52                  Vermont       816
53           Virgin Islands        53
54                 Virginia      8990
55               Washington     12114
56            West Virginia       902
57                Wisconsin      4499
58                  Wyoming       317
59                Recovered         0

Looking through, we can see that New York was the state with the largest number of cases.

Note that we could have searched for the row with the maximum number of Confirmed cases using the command:

> d2[which.max(d2$Confirmed),]

Merging Datasets

In class, we’ve been editing the original data file to add a column with the state populations (called “Population”). I have this in a separate file called “state_populations.txt” (which is also a comma separated variable file, .csv, even if not so labeled). So I’m going to import the population data:

> pop <- read.csv("state_population.txt")

Now I’ll merge the two datasets to add the population data to “mydata”.

> mydata <- merge(mydata, pop)

Graphing (Histograms and Boxplots)

With the datasets together we can try doing a histogram of the confirmed cases. Note that there is a column labeled “Confirmed” in the mydata dataset, which we’ll address as “mydata$Confirmed”:

> hist(mydata$Confirmed)

Histogram of confirmed Covid-19 cases as of 04-20-2020.

Note that on April 20th, most states had very few cases, but there were a couple with a lot of cases. It would be nice to see the data that’s clumped in the 0-50000 range broken into more bins, so we’ll add an optional argument to the hist command. The option is called breaks and we’ll request 20 breaks.

> hist(mydata$Confirmed, breaks=20)

A more discretized version of the confirmed cases histogram.

Calculations (cases per 1000 population)

Of course, simply looking at the number of cases in not very informative because you’d expect, with all things being even, that states with the highest populations would have the highest number of cases. So let’s calculate the number of cases per capita. We’ll multiply that number by 1000 to make it more human readable:

> mydata$ConfirmedPerCapita1000 <- mydata$Confirmed / mydata$Population * 1000

Now our histogram would look like:

> hist(mydata$ConfirmedPerCapita1000, breaks=20)

The dataset still has a long tail, but we can see the beginnings of a normal distribution.

The next thing we can do is make a boxplot of our cases per 1000 people. I’m going to set the range option to zero so that the plot has the long tails:

> boxplot(mydata$ConfirmedPerCapita1000, range=0)

Boxplot of US states’ confirmed cases per 1000 people.

The boxplot shows, more or less, the same information in the histogram.

Finding Specific Data in the Dataset

We’d like to figure out how Missouri is doing compared to the rest of the states, so we’ll calculate the z-score, which tells how many standard deviations you are away from the mean. While there is a built in z-score function in R, we’ll first see how we can use the search and statistics methods to find the relevant information.

First, finding Missouri’s number of confirmed cases. To find all of the data in the row for Missouri we can use:

> mydata[mydata$Province_State == "Missouri",]

which gives something like this. It has all of the data but is not easy to read.

   Province_State Population Country_Region         Last_Update     Lat
26       Missouri    5988927             US 2020-04-20 23:36:47 38.4561
      Long_ Confirmed Deaths Recovered Active FIPS Incident_Rate People_Tested
26 -92.2884      5890    200        NA   5690   29      100.5213         56013
   People_Hospitalized Mortality_Rate      UID ISO3 Testing_Rate
26                 873       3.395586 84000029  USA      955.942
   Hospitalization_Rate ConfirmedPerCapita1000
26             14.82173              0.9834817

To extract just the “Confirmed” cases, we’ll add that to our command like so:

> mydata[mydata$Province_State == "Missouri",]$Confirmed

[1] 5890

Which just gives the number 5890. Or the “ConfirmedPerCapita1000”:

> mydata[mydata$Province_State == "Missouri",]$ConfirmedPerCapita1000
[1] 0.9834817

This method would also be useful later on if we want to automate things.

z-score

We have the mean from when we did the summary command, but there’s also a mean command.

> mean(mydata$ConfirmedPerCapita1000)
[1] 2.006805

Similarly you can get the standard deviation with the sd function.

> sd(mydata$ConfirmedPerCapita1000)
[1] 2.400277

We can now calculate the z-score for Missouri:

$\text{z-score} = \frac{(X - \mu)}{ \sigma}$

which gives a results of:

$\text{z-score} = -0.43$

So it looks like Missouri was doing reasonable well back in April, at least compared to the rest of the country.

Logic Gates Programming Assignment

To follow up on the introduction to Logic Gates post, this assignment is intended to help students practice using functions and logic statements.

Write a set of function that act as logic gates. That is, they take in one or two inputs, and gives a single true or false output based on the truth tables. Write functions for all 8 logic gates in the link. An example python program with a function for an AND gate (the function is named myAND) is given in the glowscript link below.
Write a function that uses these functions to simulate an half-adder circuit. Create a truth table for the input and output.
Write a function that uses the gate functions to simulate a full-adder circuit. Create a truth table for the input and output.

Here’s a program with an example function for an AND gate (called myAND), and a simple program to test it.

The half-adder circuit is shown below.

The full adder circuit:

Full adder circuit by en:User:Cburnett on wikimedia.org

Logic Gates

Logic gates are the building blocks of computers. The gates in the figure above take one or two inputs (A and B) and give different results based on the type of gate. Note that the last row of gates are just the opposite of the gates in the row above (NAND gives the opposite output to AND).

As an example, two gates, an AND and an XOR, can be used to make a half-adder circuit

By feeding in the four different combinations of inputs for A and B ([0, 0], [1, 0], [0, 1], and [1, 1]) you can see how these two gates add the two numbers in binary.

Creating a truth table for the half adder.

I find this to be an excellent introduction to how computers work and why they’re in binary.

Atom Size and Element Properties?

Are the elements of larger atoms harder to melt than those of smaller atoms?

We can investigate this type of question if we assume that bigger atoms have more protons (larger atomic number), and compare the atomic number to the properties of the elements.

Data: Properties of the First 20 Elements

Question 1.

Your job is to use the data linked above to draw a graph to show the relationship between Atomic Number of the element and the property you are assigned.

Question 2.

What is the relationship between the number of valence electrons of the elements in the data table and the property you were assigned.

Bonus Question

Bonus 1: The atomic number can be used as a proxy for the size of the element because it gives the number of protons, but it’s not a perfect proxy. What is the relationship between the atomic number and the atomic mass of the elements?

Atom Board: Montessori Work

These atom boards worked very well for practicing how to interpret atomic symbols. The protons (blue) and electrons (red) are magnetic so they snap into place quite satisfyingly. Their poles are oriented so that the electrons will only attach properly to the slots in the electron shells and the protons only attach the right way up to the nucleus. The neutrons are wooden and non-magnetic.

Procedure for Building an Atom

Nucleus

Step 1: Number of protons (+ charge).

The number of protons is given by the element name. Carbon will always have six protons, Hydrogen will have one proton. I have students memorize the first twenty elements in the correct order, so they can quickly determine the atomic (proton) number.
¹⁴C: Protons = 6⁺

Step 2: Number of neutrons.

Neutrons = atomic mass – number of protons
The atomic mass is given at the top left corner of the atomic symbol: 14 in the example above for ¹⁴C.
¹⁴C: Neutrons = 14 – 6 = 8

Electron Shells

Step 3: Number of electrons (- charge).

Electrons = number of protons – charge
The charge is given to the top right of the atomic symbol. In this case, there is no charge
¹⁴C: Electrons = 6 + 0 = 6

Step 4: Electron Shells

Electrons go in shells around the nucleus.
Start with the smallest shell, fill it, and then add the next shell until you’ve placed all of the electrons.
The first shell can hold only 2 electrons, the second shell can hold 8, and the third 8. The electron configuration tells how many electrons are in each shell.
¹⁴C: Electron configuration: 2-4

They’ve also turned out to be useful when explaining ionic bonding. Since it’s easy to add or remove electron shells, you can clearly show how many electrons can be donated or received to figure out how many atoms are involved in the reactions.