   SEARCH HOME Math Central Quandaries & Queries  Question from sean, a student: Two people 1.8 metres tall walk directly away from each other until they can no longer see each other (due to the curvature of the earth, which has a radius of about 6378 km). A) Find a function relating the height of two identical objects with the distance between them using the scenario above as an example. B) Sketch this function (you may use Graphmatica if you wish).Over what domain and range does the function exist? C) Describe this relation in practical terms. Sean continued

This is what I have done so far....

The distance is found when the Earths curvature interrupts their line of sight. This will occur for both individuals simultaneously (same travel speed and height (z=height, 1.8=m)) when their starting point (A) becomes the same LineOfSight height as their head (assuming eyes = height). Best way of approaching the problem is setting their start position to Top Dead Center, $x = 0, r = \mbox{max;}$ this allows the LineOfSight to follow the horizontal line $y = r.$ The disappearance occurs when the hypotenuse of the triangle OAB becomes equal to the $h = r+z = 6378001.8.$ This gives us the angle, by using arccosine =(adjacent / hypotenuse ). Noting we use the shortest distance (OA) as the adjacent, and the longer distance (OB) as the hypotenuse. $\arccos( 6378000 / 6378001.8 )= \arccos( 0.999 999 717) = 0.000 751 292$ radians. There are $2 \pi$ radians in a circle and hence 0.000 751 292 radians is $0.000 751 292 / (2 \pi ) = 0.000 119 572$ of the circle. Working out circumference $c =2 \pi \times \mbox{radius } = 2 \pi \times 6378000= 40 074 156 \mbox{ m.}$ The length of the arc from A to B is then $0.000 119 572 \times 40 074 156= 4791.742 \mbox{ m.}$ The distance between the people when they can no longer see each other is then $2 \times 4791.742= 9583 \mbox{ m.}$

Hi Sean,

I agree with what you have one so far. Part A) of the problem asks you to repeat what you have done but with the heights of the two people left as the variable $z.$ You will then have the distance between them, lets call it $D$ as a function of $z,$ that is $D = D(z).$ This expression will contain an arccosine function. The domain and range of $D(z)$ will depend on the domain and range of the arccosine function.

Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.