Each marble has mass \(m\). The force points down, so \[-m g\cdot\text{n_marbles}=K\cdot(L-\text{length})\] which can be re-written as \[\text{n_marbles}=\underbrace{-\frac{K}{m g}\cdot L}_{A} + \underbrace{\frac{K}{m g}}_{B}\cdot\text{length} \]

In this case it is easier to use centimeters, grams and seconds

Thus, force is measured in dyne, and length in cm

Looking at Hooke’s law \[\text{force}=K\cdot(L-\text{length})\] we can see that \(K\) is measured in dyne/cm

Our model is \[\text{n_marbles}=\underbrace{-\frac{K}{m g}\cdot L}_{\texttt{coef(model)[1]}} + \underbrace{\frac{K}{m g}}_{\texttt{coef(model)[2]}}\cdot\text{length}\] therefore \[K=\texttt{coef(model)[2]}\cdot m\cdot g\]

Taking the mass of marbles as 20gr, we calculate the elasticity constant as

coef(model)[2] * 20 * 9.8

length 
  37.5 

The units are dyne/cm.

The label length comes from coef(model)[2]

The number is the same as before. The units are correct this time

This is the same as last class. Since

\[\texttt{coef(model)[1]}=-\frac{K}{m g}L = -\texttt{coef(model)[1]}\cdot L \] we can calculate \(L\) as \[L=-\frac{\texttt{coef(model)[1]}}{\texttt{coef(model)[2]}} = 75.922\]

We need very little math for our course: arithmetic, algebra, and logarithms

If \(x=p^m\) then \[\log_p(x) = m\]

If we use another base, for example \(q\), then \[\log_q(x) = m\cdot\log_q(p)\]

So if we use different bases, there is only a scale factor

The easiest one is natural logarithm

• They only work with positive numbers. Not with 0
• If \(x=p\cdot q\) then \[\log(x)=\log(p)+\log(q)\]
• If \(\log(x)=m\) then \(x=\exp(m)\)

Basic linear model \[y=A+B\cdot x\] Exponential \[y=I\cdot R^x\qquad\log(y)=log(I)+log(R)\cdot x\] Power of \(x\) \[y=C\cdot x^E\qquad\log(y)=log(C)+E\cdot\log(x)\]

The easiest way to decide is to draw several plots, placing log() in different places, and looking which one seems more like a straight line

For example, let’s analyze data from Kleiber’s Law (Physiological Reviews 1947 27:4, 511-541)

The following data shows a summary. The complete table has 26 animals

plot(kcal ~ kg, data=kleiber)

plot(log(kcal) ~ kg, data=kleiber)

plot(log(kcal) ~ log(kg), data=kleiber)

Depending on the context, we may want to use different versions of semi-log and log-log plots

For understanding the data, we do

plot(log(kcal) ~ kg, data=kleiber)

For publishing in a paper, we do

plot(kcal ~ kg, data=kleiber, log="y")

The plot that seems more straight line is the log-log plot

Therefore we need a log-log model.

model <- lm(log(kcal) ~ log(kg), data=kleiber)
coef(model)

(Intercept)     log(kg) 
      4.206       0.756 

If \[\log(kcal)=4.206 + 0.756\cdot \log(kg)\] then \[kcal=\exp(4.206) \cdot kg^{0.756}\]

Therefore:

• The average energy consumption of a 1kg animal is \(\exp(4.206) = 67.072\) kcal
• The energy consumption increases at a rate of \(0.756\) kcal/kg.

Models are the essence of scientific research

They provide us with two important things

• An explanation for the observed patterns of nature
• A method to predict what will happen in the future

predict(model, newdata)

where newdata is a data frame with column names corresponding to the independent variables

If we omit newdata, the prediction uses the original data as newdata

predict(model) == predict(model, newdata=data)

We want to predict the metabolic rate, depending on the weight

The independent variable is \(kg\), the dependent variable is \(kcal\)

But our model uses only \(\log(kg)\) and \(\log(kcal)\)

So we have to undo the logarithm, using \(\exp()\)

plot(log(kcal) ~ log(kg), data=kleiber)
lines(predict(model) ~ log(kg), data=kleiber)

## Visually

plot(kcal ~ kg, data=kleiber, log="xy")
lines(exp(predict(model)) ~ kg, data=kleiber)

plot(count~Date, data=trans)