Remember that straight lines can be represented by the formula \[\text{n_marbles}=A+B\cdot \text{length}\] The coefficient \(A\) is the value where the line intercepts the vertical axis

The coefficient \(B\) is how much length goes up when n_marbles increases. This is called slope

In our case \(A\) and \(B\) are

(Intercept)      length 
-14.5353768   0.1914521 

The formula from Hooke’s Law is \[\text{force}=K\cdot(L-\text{length})\] Since force is the weight of the marbles, we can write \[-m g\cdot\text{n_marbles}=K\cdot(L-\text{length})\] which can be re-written as \[\text{n_marbles}=\frac{K}{m g}\cdot\text{length} - \frac{K}{m g}\cdot L\]

We also see that \[\texttt{coef(model)[1]}=-\frac{K}{m g}L = -\texttt{coef(model)[1]}\cdot L \] Therefore \[L=-\frac{\texttt{coef(model)[1]}}{\texttt{coef(model)[2]}}\]

When there are no balls, the natural length of the coil is \(L\)

This value is hard to measure directly

But, using the formula from the regression, we have

-coef(model)[1]/coef(model)[2]

(Intercept) 
   75.92175 

I cleaned up all the files

There are two that I could not recover

The rest are either “coins” or “marbles”

We want to know the number of cells every day, which we represent with the vector ncell

Each element of the vector is the number of cells in day t

We start with an initial number of cells, that we call initial

Each day, the number of cells increases by a factor rate

The number of cells on day t is ncell[t]

Each day the number of cell is multiplied by rate

Therefore ncell[t] = rate * ncell[t-1]

This is a recursive formula

Can you write an explicit formula? (non-recursive)

The solution of the recurrence is

ncell[t] = initial * rate^t

In R we can do this easily when t is a vector

t <- seq(from=start, to=end, by=step)
ncell <- initial * rate ^ t

We have many cases when the values increase (or decrease) with a factor

For example, the cost of DNA sequencing

In general if the relation is \[y = I\cdot R^x\] we can get a better picture applying logarithms \[\log(y) = \log(I\cdot R^x) = \log(I) + x \log(R)\]

Sometimes the formula is different. For example the area of a circle is \[a=\pi r^2\] and the volume of a sphere is \[v=\frac{4}{3}\pi r^3\]

In general you can have \[y= A x^B\] then, using logarithms, you have \[\log(y)=\log(A x^B)=\log(A) + B\log(x)\] Now we need also \(\log(x)\)

When the logarithmic scale shows a straight line, we can use a linear model

We have to be careful to use log and exp in the correct place

Let’s consider this case

example <- data.frame(t, vol=2*t^3)
head(example)

  t vol
1 1   2
2 2  16
3 3  54
4 4 128
5 5 250
6 6 432

model <- lm(log(vol)~log(t), data=example)
model

Call:
lm(formula = log(vol) ~ log(t), data = example)

Coefficients:
(Intercept)       log(t)  
     0.6931       3.0000  

exp(coef(model)[1])

(Intercept) 
          2