An ant starts at the origin of the coordinate plane. The ant only has enough energy to walk for 10 units along either the x- or y-axis before stopping. These lines are the two ant highways.
However, if it goes off the highway, more energy is needed for the same distance. That is, the further the ant deviates, the greater the energy consumption, as represented by the factor 1+0.4d, where d is the distance to the nearest highway. For example, if the ant is 1 unit away from the nearest highway, it uses 1.4 times the energy on the highway to walk 1 meter.
What is the area of the set of points the ant can reach before getting tired and stopping?
To simplify, consider how far Ant gets taking off from the origin at an angle theta from the x-axis, where theta is less than 45 deg. This angle is chosen so that the x-axis is always the closest "highway". It is d = y = s sin(theta) away, where s is how far Ant has traveled in a straight radial direction off-road. (All 45 degree wedges will have the same shape reflectively). If we call X the distance Ant gets to at angle theta, we have the line integral:
Int{0 to X} [1+0.4 y] ds = 10
Int{0 to X} [1 + 0.4 s sin(theta)] ds = 10
Int{0 to X} ds + 0.4 sin(theta) Int{0 to X} [ s ] ds = 10
X + 0.4 sin(theta) 0.5 X^2 = 10
0.2 sin(theta) X^2 + X - 10 = 0, a quadratic in X. Using the quadratic formula: [-b (+/-) sqrt (b^2 -4ac)]/2a
X(theta) = [ -1 (+/-) sqrt( 1 + 8 sin(theta)) ] / [0.4 sin(theta)]
=2.5 csc(theta) (-1 ± sqrt(8 sin(theta) + 1))
(The positive root is the right one.)
This is only valid 0 to 45 deg. and goes to a minimum near 5.58 at 45 deg.
I didn't try to get an area.... Maybe I will consult Wolfram Alpha...
WA gives about 19 cm^2 in each of the 8 wedges, for 156.29 total cm^2. These numbers pass a sanity test since since the wedges have two sides 5 and 10 cm.
Edited on August 24, 2018, 7:52 pm
attempt....
