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(Figure 1.3a). The question is different: the choice that the forager faces is how long to stay in the patch. We will call this the patch residence time, and denote it by t. The energetic value of food removed by the forager when the residence time is t is denoted by G(t). Clearly G(0) = 0 (since nothing can be gained when no time is spent in the patch). Since the patch is exhaustible, G(t) must plateau as t increases. Time for a pause.

One of the biggest difficulties in this kind of work is getting intuition about functional forms of equations for use in models and learning how to pick them appropriately. Colin Clark and I talk about this a bit in our book (Clark and Mangel 2000). Two possible forms for the gain function are G(t) = at/(b +1) and G(t) = at 1 Kb +12). Take some time before reading on and either sketch these functions or pick values for a and b and graph them. Think about what the differences in the shapes mean. Also note that I used the same constants (a and b) in the expressions, but they clearly must have different meanings. Think about this and remember that we will be measuring gain in energy units (e.g. kilocalories) and time in some natural unit (e.g. minutes). What does this imply for the units of a and b, in each expression?

Back to work. Suppose that the travel time between the patches is t. The problem that the forager faces is the choice of residence in the patch - how long to stay (alternatively, should I stay or should I go now?). To predict the patch residence time, we proceed as follows.

Envision a foraging cycle that consists of arrival at a patch, residence (and foraging) for time t and then travel to the next patch, after which the process begins again. The total time associated with one feeding cycle is thus t + t and the gain from that cycle is G(t), so that the rate of gain is R(t) = G(t)/(t + t). In Figure 1.3, I also show an example of a gain function (panel b) and the rate of gain function (panel c). Because the gain function reaches a plateau, the rate of gain has a peak. For residence times to the left of the peak, the forager is leaving too soon and for residence times to the right of the peak the forager is remaining too long to optimize the rate of gain of energy.

The question is then: how do we find the location of the peak, given the gain function and a travel time? One could, of course, recognize that R(t) is a function of time, depending upon the constant t and use calculus to find the residence time that maximizes R(t), but I promised plane geometry in this warm-up. We now proceed to repeat a remarkable construction done by Eric Charnov (Charnov 1976). We begin by recognizing that R(t) can be written as

and that the right hand side can be interpreted as the slope of the line that joins the point (t, G(t)) on the gain curve with the point (—t, 0) on the abscissa (x-axis). In general (Figure 1.3d), the line between (—t, 0) and the curve will intersect the curve twice, but as the slope of the line increases the points of intersection come closer together, until they meld when the line is tangent to the curve. From this point of tangency, we can read down the optimal residence time. Charnov called this the marginal value theorem, because of analogies in economics. It allows us to predict residence times in a wide variety of situations (see the Connections at the end of this chapter for more details).

Egg size in Atlantic salmon and parent-offspring conflict (calculus)

We now come to an example of great generality - predicting the size of propagules of reproducing individuals - done in the context of a specific system, the Atlantic salmon Salmo salar L. (Einum and Fleming 2000). As with most but not all fish, female Atlantic salmon lay eggs and the resources they deposit in an egg will support the offspring in the initial period after hatching, as it develops the skills needed for feeding itself (Figure 1.4). In general, larger eggs will improve the chances of offspring survival, but at a somewhat decreasing effect. We will let x denote the mass of a single egg and S(x) the survival of an offspring through the critical period of time (Einum and Fleming used both 28 and 107 days with similar results) when egg mass is x. Einum and Fleming chose to model S(x) by

where xmin = 0.0676 g and a = 1.5066 are parameters fit to the data. We will define c = (xmin)a so that S(x) = 1 — cx—a, understanding that S(x) = 0 for values of x less than the minimum size. This function is shown in Figure 1.5a; it is an increasing function of egg mass, but has a decreasing slope. Even so, from the offspring perspective, larger eggs are better.

However, the perspective of the mother is different because she has a finite amount of gonads to convert into eggs (in the experiments of Einum and Fleming, the average female gonadal mass was 450 g). Given gonadal mass g, a mother who produces eggs of mass x will make g/x eggs, so that her reproductive success (defined as the expected number of eggs surviving the critical period) will be

and we can find the optimal egg size by setting the derivative of R(g, x) with respect to x equal to 0 and solving for x.

Show that the optimal egg size based on Eq. (1.8) is xopt = {c(a + 1)}l=a and for the values from Einum and Fleming that this is 0.1244 g. For comparison, the observed egg size in their experiments was about 0.12 g.

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