# Exponential Growth

Joe Thorley · 2019-04-03 · 2 minute read

## Exponential Growth

A variable is said to grow exponentially when its rate of change is proportional to its current value

$\frac{dV}{dt} = \mu V$

It is called exponential growth because the variable’s value at any given time ($$t$$) is an exponential function of time $V_t = V_0 {\rm e}^{\mu t}$

t <- 1:30
mu <- 0.25
V0 <- 1
Vt <- V0 * exp(mu * t)
plot(Vt ~ t, type = "l") ## Calculating $$\mu$$

The growth rate $$\mu$$ which has units of $$t^{-1}$$ is the instantaneous growth rate. It can be calculated from the value of the variable at two different time periods using the equation

$\mu = \frac{1}{t}\ln\frac{V_t}{V_0}$

(1/t * log(Vt/V0))[1:2]
##  0.25 0.25

The growth rate can be found from the doubling time ($$t_2$$; the number of time intervals required for $$V$$ to double) using

$\mu = \frac{\ln 2}{t_2}$ and from the number of doublings per time interval ($$\delta_t$$; a common expression for unicellular organisms) using $\mu = \delta_t\ln 2$

It is worth noting that as the exponential term is the product of $$\mu$$ and $$t$$ it is straightforward to switch between time units. For example to convert a $$\mu$$ in units of $$\text{day}^{-1}$$ to units of $$\text{hour}^{-1}$$ simply divide $$\mu$$ by 24.

## Negative Exponential Growth

If the growth rate is negative ($$\lambda = - \mu$$) then exponential decay occurs

plot(V0 * exp(-mu * t)  ~ t, type = "l", ylab = "Vt") and $$\frac{\ln 2}{\lambda}$$ gives the half-life ($$t_{1/2}$$).

### Mortality

In the case of a population of individuals, where $$V_0 = 1$$ and $$t = 1$$ and the growth rate is negative then the interval mortality ($$A$$) is given by

$A = 1- {\rm e}^{-Z}$ where $$Z = \lambda$$ is the instantaneous mortality rate.

Conversely,

$Z = \ln(1-A)$