---
title: "Mathematical derivations"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Mathematical derivations}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
editor_options:
  markdown:
    wrap: 72
---

```{r setup, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)

# Values similar to those estimated for female blacktip sharks, as used in
# the growth vignette
L0 <- 73
Linf <- 265
K <- 0.14
```

This vignette sets out the mathematics behind the models used in
`mustelus`, step by step. It is intended as a reference for the other
vignettes, which state results without deriving them.

## The von Bertalanffy growth function

The [growth vignette](growth.html) introduces the von Bertalanffy growth
function as a differential equation in which growth rate declines
linearly as length increases:

$$\frac{dL}{dt} = K(L_\infty - L)$$

For an individual animal, age and time advance together, so the rate of
change in length with time is the same as its rate of change with age,
$a$. Written in terms of age, the equation is

$$\frac{dL}{da} = K(L_\infty - L)$$

The aim is to find length as a function of age, $L(a)$.

### The general solution

The equation is separable. Dividing both sides by $L_\infty - L$ and
multiplying by $da$ puts the terms in $L$ on one side and the terms in
$a$ on the other:

$$\frac{dL}{L_\infty - L} = K \, da$$

Integrating both sides,

$$\int \frac{dL}{L_\infty - L} = \int K \, da$$

The left-hand side follows from $\int dx/(b - x) = -\ln(b - x)$, which
holds while length is below the asymptote, $L < L_\infty$:

$$-\ln(L_\infty - L) = Ka + c$$

where $c$ is a constant of integration. Multiplying both sides by $-1$
and exponentiating,

$$L_\infty - L = e^{-c} e^{-Ka}$$

Writing $A = e^{-c}$ for the constant and rearranging gives the general
solution:

$$L(a) = L_\infty - A e^{-Ka}$$

The distance remaining to the asymptote, $L_\infty - L$, decays
exponentially with age at rate $K$. The constant $A$ is fixed by any one
known point on the curve, and the choice of that point is what
distinguishes the two forms of the equation.

### In terms of $t_0$

Beverton and Holt (1957) fixed the curve by the hypothetical age at
which length would be zero, $t_0$. Setting $L(t_0) = 0$ in the general
solution,

$$0 = L_\infty - A e^{-K t_0}$$

$$A = L_\infty e^{K t_0}$$

Substituting this back into the general solution,

$$L(a) = L_\infty - L_\infty e^{K t_0} e^{-Ka}$$

and collecting terms gives the familiar form:

$$L(a) = L_\infty\left(1 - e^{-K(a - t_0)}\right)$$

### In terms of $L_0$

The curve can instead be fixed by length at age zero, $L_0$, which for
a chondrichthyan is length at birth. Setting $L(0) = L_0$ in the general
solution, and noting that $e^{0} = 1$,

$$L_0 = L_\infty - A$$

$$A = L_\infty - L_0$$

Substituting this back into the general solution,

$$L(a) = L_\infty - (L_\infty - L_0)e^{-Ka}$$

Writing $L_\infty$ as $L_0 + (L_\infty - L_0)$ and collecting the terms
in $L_\infty - L_0$ gives the form used by `growth()`:

$$L(a) = L_0 + (L_\infty - L_0)\left(1 - e^{-Ka}\right)$$

### Converting between the two forms

Both forms are the general solution with a different expression for
$A$, and they describe the same curve when the two expressions are
equal:

$$L_\infty e^{K t_0} = L_\infty - L_0$$

Solving for $L_0$,

$$L_0 = L_\infty\left(1 - e^{K t_0}\right)$$

and dividing by $L_\infty$ and taking logarithms to solve for $t_0$,

$$t_0 = \frac{1}{K}\ln\left(1 - \frac{L_0}{L_\infty}\right)$$

Because length at birth lies between zero and the asymptote, the term
inside the logarithm lies between zero and one, and $t_0$ is negative.
For the example curve in the growth vignette, with $L_0$ = `r L0` cm,
$L_\infty$ = `r Linf` cm and $K$ = `r K`, this gives $t_0$ =
`r round(log(1 - L0 / Linf) / K, 1)` years.

### Checking the solution

Differentiating the $L_0$ form with respect to age,

$$\frac{dL}{da} = K(L_\infty - L_0)e^{-Ka}$$

From the general solution with $A = L_\infty - L_0$, the distance
remaining to the asymptote is $L_\infty - L(a) = (L_\infty - L_0)e^{-Ka}$.
Substituting this in recovers the original equation:

$$\frac{dL}{da} = K\left(L_\infty - L(a)\right)$$

### Time to grow half of the remaining distance

The growth vignette notes that an animal takes $\ln(2)/K$ years to grow
half of the remaining distance to $L_\infty$, whatever its current
length. From the general solution, the distance remaining at age $a$ is
$L_\infty - L(a) = A e^{-Ka}$. After a further interval $\Delta$ it is

$$L_\infty - L(a + \Delta) = A e^{-K(a + \Delta)} = \left(L_\infty - L(a)\right)e^{-K\Delta}$$

so in any interval of length $\Delta$ the remaining distance shrinks by
the same factor, $e^{-K\Delta}$. Setting that factor to one half,

$$e^{-K\Delta} = \frac{1}{2}$$

$$\Delta = \frac{\ln 2}{K}$$

which depends on neither age nor current length. With $K$ = `r K`, the
remaining distance halves every `r round(log(2) / K, 1)` years.

## References

Beverton, R.J.H. and Holt, S.J. (1957) *On the Dynamics of Exploited
Fish Populations*. Fishery Investigations Series II, Volume 19. Ministry
of Agriculture, Fisheries and Food, London.

von Bertalanffy, L. (1938) A quantitative theory of organic growth
(inquiries on growth laws. II). *Human Biology* **10**, 181–213.