Mathematical derivations

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 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 $L_0$ = 73 cm, $L_\infty$ = 265 cm and $K$ = 0.14, this gives $t_0$ = -2.3 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 $K$ = 0.14, the remaining distance halves every 5 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.