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Ants and Fermat's Principle of least time

New research has uncovered that ants obey Fermat's principle of least time, so that if they have to traverse two different surfaces where they walk at different speeds, they will choose the path of least time rather than a direct path.

Ants walk at 1/12 m/s on surface 1 and 1/15 m/s on surface 2. From point A there is food at B which is 40 cm East and 40 cm South of A. The surface changes 20 m south of point A. Locate point D (i.e. calculate the length of e), if the ants are to minimise the time it takes for them to go from A to B. Method 1: Calculus

The time it takes to get from A to B is: => => Now we need to express in terms of e i.e. the distance on the boundary:

Let t = 20 – e

Using the Cosine rule we can express d1 and we can use Pythagoras to find d2 as it's in a right angled trinagle... in terms of t  So now we have: Now we must differentiate this with respect to t and set it equal to 0 in order to minimise T: Now we need to solve this, unfortunately the algebra is a little too difficult for pen and paper (This is a quartic equation) so we use Wolfram Alpha

We get an approximate value for t as 4.29373

Therefore our value for e => 20 – 4.29373 ~ 15.706 cm

Method 2: Snell's law of refraction

Light obeys Fermat's principle of least time and according to Snells law: Where i is the angle of incidence (angle made by d1 and the perpendicular to the boundary at point D)

And r is the angle of refraction (angle made by d2 and the perpendicular to the boundary at point D).

Drawing the perpendicular we can see that: We can also work out d1 and d2 in terms of e using Pythagoras.

Let e = x : So we now have: This simplifies to: Solving for x (again using Wolfram Alpha):

x =15.706 cm

Plot of distance e as a function of Time taken to get from A to B: As you can see the minimum point is about 15.7