Exponential growth or decline

• Rate of increase (or decrease) is proportional to current population size

Exponential growth or decline

• Rate of increase (or decrease) is proportional to current population size

Exponential growth or decline

• Rate of increase (or decrease) is proportional to current population size

Exponential growth or decline

• Rate of increase (or decrease) is proportional to current population size

Human population growth

• Define a variable to represent the number of individuals in the population at time $t$
• $N(t)$, where $N$ is a dependent variable and $t$ is an independent variable
• is the same as $N_t$
• or just $N$ (shorthand)

Initial conditions

• Define the value of the population at time $t=0$

  $$\begin{eqnarray} N(t = 0) &=& N_0 \\ &=& 7\ 000\ 000\ 000 \end{eqnarray}$$

Parameters

• Specify the rate of increase
  • Human population increasing at approximately 1.5% each year
  • Rate of increase $\lambda = 0.015$ per year

The difference equation model

• Incrementing the human population in 1 year time steps

  $$N(t + 1) = \lambda \times N(t) + N(t)$$

• where $\lambda = 0.015$

Changing the time step

• Incrementing the human population in 1 month time steps

  $$N(t + 1) = (\lambda / 12) \times N(t) + N(t)$$

• where $\lambda = 0.015$

Return to time steps of a year

• Incrementing the human population in 1 year time steps

  $$N(t + 1) = \lambda \times N(t) + N(t)$$

• where $\lambda = 0.015$

How long until population doubles?

$$N(t + 1) = \lambda \times N(t) + N(t)$$

$$\mathit{i.e.\quad} N(t + 1) = (\lambda + 1) \times N(t)$$

How long until population doubles?

$$\begin{eqnarray} N(t + 2) &=& (\lambda + 1) \times N(t + 1) \\ &=& (\lambda + 1)\times(\lambda + 1) \times N(t) \\ &=& (\lambda + 1)^2 \times N(t) \end{eqnarray}$$

How long until population doubles?

$$N(t + n) = (\lambda + 1)^n \times N(t)$$

So, the number of years ($n$) taken to double is the $n$ that satisfies

$$(\lambda + 1)^n = 2$$

Solving the equation

$$\begin{eqnarray} (\lambda + 1)^n &=& 2 \\ \log((\lambda + 1)^n) &=& \log(2) \\ n \times \log(\lambda + 1) &=& \log(2) \\ n &=& \log(2) / \log(\lambda + 1) \\ n &=& 46.6 \end{eqnarray}$$

Doubling time

• The doubling time ($n$) does not depend on the starting population size
• This is always true of an exponentially growing population

Fixed doubling time

Model types

• Difference equation models
  • Describe updating of population in chunks of time
  • e.g. $N(t + 1) = \lambda \times N(t) + N(t)$

Model types

• Difference equation models
  • Describe updating of population in chunks of time
  • e.g. $N(t + 1) = \lambda \times N(t) + N(t)$
• Differential equation models
  • Describe continuous updating of population
  • e.g. $\frac{dN}{dt} = \lambda \times N$
  • where $\frac{dN}{dt}$ is the rate of change of $N$

Model types

• Difference equation models
  • Describe updating of population in chunks of time
  • e.g. $N(t + 1) = \lambda \times N(t) + N(t)$
• Differential equation models
  • Describe continuous updating of population
  • e.g. $\frac{dN}{dt} = \lambda \times N$
  • where $\frac{dN}{dt}$ is the rate of change of $N$
• Difference equations
  • Good approximations to continuous models when time chunks are small
  • No need to worry about differential equations today :)

Practicals

• Programming in R Practical:
  • Learn some programming skills
  • Build simple population dynamics models in R