Inverse Trigonometric Functions
Continuity and Differentiability
UBC Calculus Online Course Notes

The Zebra Danio and its escape response

On this page we will investigate one example which illustrates the importance of the inverse trigonometric functions. This example is motivated by a problem in biology, and originates from a paper written by Larry Dill, a biologist at Simon Fraser University in Burnaby, BC.

The zebra danio, a tasty little morsel for hungry predators

The Zebra danio is a small tropical fish, which has many predators (larger fish) eager to have it for dinner. Surviving through the day means being able to sense danger quickly enough to escape from a hungry pair of jaws. However, the danio cannot spend all its time escaping. It too, must find food, mates, and carry on activities that sustain it. Thus, a finely tuned mechanism which allows it to react to danger but avoid over-reacting is definitely an advantage.


On this page we will show how derivatives of the inverse trigonometric functions enter into the discussion when we investigate the visual basis of an escape response, as suggested by Dill's model.

Before doing so, it may be useful to remember the following result about the derivative of the inverse tangent, obtained on a previous page :

\[  \frac{d}{dx} \tan^{-1}(x) = \frac 1{1+x^2} 
 \]


In particular, this fact together with the chain rule would allow us to conclude that the derivative of a composite function involving the inverse tangent would look like this:

\[  \frac{d}{dx} \tan^{-1}(~f~(~x(~t~)~)~) = \frac 1{1+f(x(t))^2}~~\frac{df}{dx}~~\frac{dx}{dt} 
 \]


This is a simple application of the chain rule which will be useful shortly.


The visual field: what the prey is seeing

In this diagram, we show the part of the world that is within the visual field of the Zebra danio. (We have shown only what one eye is seeing, and we will assume that the predator is approaching from this direction.) The eye of the danio is located at the sharp corner of the triangle and is said to subtend a visual angle of $ \theta $ . In this case, the object seen is a predator, and its apparent size, as viewed from the vantage point of the prey is $ S $ . The distance between the prey and the predator (which forms the horizontal leg) is labeled $ x $ . The quantities in this diagram are linked through trigonometric functions.

The visual field of the danio.

The top half of the triangle, forms a Pythagorean triangle, so



This implies that


\[ 
\tan (\frac{\theta}{2}) = \frac{S/2}{x} 
\]

or


\[ 
\theta = 2 \tan^{-1}( \frac{S/2}{x}) 
\]

In the last step, we have simply used the inverse tan function to express the same relationship, and then just multiplied both sides by 2.

Linking the visual angle to the escape response


What sort of visual input should the danio respond to, if it is to be efficient at avoiding the predator? In principle, we would like to consider a response that has the following features




In keeping with these reasonable expectations, the hypothesis proposed by Dill is that:


The escape response is triggered when the predator approaches so quickly, that the rate of change of the visual angle is larger than some critical value.





The hypothesis proposed above, restated in terms of symbols used in our diagram is that



The escape response is triggered when


\[ 
\frac{d \theta}{dt} > k 
\]

where $ k $ is the critical value of the derivative at which the response is triggered. We would expect that the value of the constant $ k $ would depend on the species (in this case the danio) and perhaps on "how jumpy" it is feeling on this particular encounter.


Applying the chain rule

We can now apply the chain rule to the relationship of quantities in the triangle to arrive at a connection between the speed of the approaching predator and the rate of change of the visual angle. We find that


\[ 
\theta = 2 \tan^{-1}( \frac{S}{2x}) 
\]


\[ 
\frac{d\theta }{dt} =\frac{d}{dt} ( 2 \tan^{-1} (\frac{S}{2x})) 
\]


\[ 
\frac{d \theta}{dt} = 2~ \frac{1}{1 + (\frac{S/2}{x})^2} ~ 
( \frac{-S x^{-2}}{2}) ~ \frac{dx}{dt} 
 \]

where in the last step we have applied the chain rule.

Now suppose that the predator is approaching the prey at speed $ v $ , decreasing the distance to its prey, so that


\[ 
\frac{dx}{dt} = - v 
\]
(The minus sign denotes that the distance $ x $ is decreasing.) Substituting this into the result above, rearranging, and cancelling common factors, we arrive at:


\[ 
\frac{d \theta}{dt} = \frac{-S}{x^{2}}~ \frac{v}{1 + (\frac{S/2}{x})^2} 
= \frac{S v} { \frac{S^2}{4} + x^2} 
\]



Reaction distance and the escape response.

We are now in position to determine how the size and speed of the predator influence the distance $ x $ at which the prey first reacts to the perceived threat. By hypothesis, the Zebra danio first "notices" that a hungry predator is looming dangerously nearby when its visual system detects a rate of visual angle change that is greater than a critical rate, $ k $ . This first happens when the rate of change of $ \theta $ is just equal to $ k $ , i.e.


\[ 
\frac{d \theta}{dt} = k 
\]

Using the computed expression for $ d \theta/dt $ and setting this equal to $ k $ we find that


\[ 
\frac{S v} { \frac{S^2}{4} + x^2} = k 
\]

In this equation, the quantities $  S, v,  $ which are characteristic of the predator and its size and swimming speed are assumed constant, and so is $ k $ , the prey's visual sensitivity to a "looming" threat. We are to solve for the distance $ x $ at which this critical value of $  d \theta /dt  $ is reached. We will call that distance the reaction distance, $  x_r  $ . Taking reciprocals and multiplying both sides by $ Sv $ , we find that


\[ 
 { \frac{S^2}{4} + x_r^2} = \frac {Sv}{k} 
\]


\[ 
   x_r^2 = \frac {Sv}{k}- \frac{S^2}{4} = S(\frac{v}{k}-\frac{S}{4}) 
\]

We find that for a predator of size $ S $ moving at speed $ v $ towards the prey, the escape response will be triggered at the reaction distance


\[ 
   x_r = \sqrt{ S(\frac{v}{k}-\frac{S}{4})} 
\]

This result has interesting implications which we hope you will explore below. In particular, the reaction distance $ x_r $ depends on various constants associated with the problem. We ask you to look more closely at this dependence and to see what it implies about success and failure of this type of response based on the rapidity of change of the visual angle. Is the response always successful ? How can it be made more sensitive? How might a predator try to outwit this kind of response.




For your consideration:






Other sources of information:

Lawrence M Dill (1974) The escape response of the Zebra Danio (Brachydanio rerio). I. The stimulus for escape. Animal Behaviour 22, 711-722.

Home page of Lawrence M Dill

Information about Behavioural Ecology at Simon Fraser University

Exotic Tropical Fish

Freshwater Aquarium Fish: the Zebra Danio