# Locus of points question. (I think?)

1. Dec 13, 2011

1. The problem statement, all variables and given/known data

A straight path separates a meadow from a field. A pedestrian travels along the path at a speed of 5 km/hr, through the meadow at a speed of 4 km/hr, and through the field at a speed of 3 km/hr. Initially, the pedestrian is on the path. Draw the region which the pedestrian can cover in 1 hour.

2. Relevant equations

???

3. The attempt at a solution

I do not understand what the layout of the path, meadow and field is. Is the path supposed to be separating the field and meadow like....

|>>>>>>>>>>>>||----||***********|
|>>>>>>>>>>>>||----||***********|

I also don't get what they mean by what area can the pedestrian cover? Overall I have no clue how to do this problem.

Last edited: Dec 13, 2011
2. Dec 14, 2011

### JHamm

The question is asking you to find every point on the map that the pedestrian could make it to in an hour or less

3. Dec 14, 2011

### Staff: Mentor

I don't think I would have figured it out, without JHamm's interpretation.

Perhaps the wording could be improved. Let me try.

You allow visitors to your farm. They all walk with the same speed, and this is related to the surface underfoot. You restrict each visitor to one hour's exploration beyond the front gate, and then they get picked up by a car. Eventually, part of your field, meadow, and pathway get trampled by all these visitors. Mark the boundary of the trampled region on each side of the path.

The pathway separates the field on the left from the meadow on the right. You can assume it runs in a straight line if you wish. Each pedestrian travels along the path at a speed of 5 km/hr, through the meadow at a speed of 4 km/hr, and through the field at a speed of 3 km/hr. They start on the path at the front gate.

4. Dec 14, 2011

### JHamm

If you're unsure where to start I'd begin by plotting some points; however remember that you can travel along the path for a bit and then cut in to the meadow or the field, maybe you could get further that way.

5. Dec 14, 2011

### verty

This is a confusing question. Think about the form that the quickest path from any point will take. Then I would draw an accurate sketch to find how it should look.

6. Dec 14, 2011

EDIT: please do not respond yet; I posted too soon and there were lots of errors in my post. Please let me fix them real quick

Okay thanks to everyone who helped so far. I have tried to come up with an answer:

I made the sidewalk be the y axis and the meadow the positive x direction and the field the negative x direction. So if the pedestrian walks a certain distance up or down the sidewalk (y-axis) and then cuts into the field then the x distance he walks can be found with the equation

x = (-5/4)y +5 for if he walks in negative y direction and then cuts into the meadow, and
x = (5/4)y -5 for if he walks in positive y direction and cuts into the meadow...

these can be rearranged for y to give y = -(5/4)x + 5 and y = (5/4)x -5

The pedestrian could also cover a semicircle of area with a radius of 4 if he did not use the technique of walking on the path before cutting into the meadow. The equation we could use for the area he could cover with this technique would be x ≤ +√(16-y^2)

so the area in the meadow he could cover can be summed up with these 3 equations:

y -(5/4)x ≥-5 for walking up in the positive y direction before cutting into the meadow
y + (5/4)x ≤ 5 for walking in the negative y direction before cutting into the meadow
x ≤ +√(16-y^2) for semicircle that could be covered

I don't feel like this is right, can you guys give me another push in the right direction? Your initial posts were very helpful

Last edited: Dec 14, 2011
7. Dec 14, 2011

### Staff: Mentor

I think forget about the semicircle. Think of an escaping prisoner (in leg-irons, of course). How far away could he get in one hour, dragging himself through any combination of vegetations and pathway? Plot ALL the paths; there will be overlap of some with others. We are interested in just the locus of the end points of all of these possible routes.

8. Dec 15, 2011

K I waited too long to edit but here are the corrected equations.

http://www4d.wolframalpha.com/Calculate/MSP/MSP16919ib17h96ag5cc2300000d64f3e5e18hf591?MSPStoreType=image/gif&s=45&w=169&h=37 [Broken]
and

for the field

http://www3.wolframalpha.com/Calculate/MSP/MSP36019ib182dg3g1ef3a000023ahf29hi45d52d5?MSPStoreType=image/gif&s=2&w=180&h=36 [Broken]
and
http://www3.wolframalpha.com/Calculate/MSP/MSP192219ib14g3bi9bcbd200004518e1a7288daf20?MSPStoreType=image/gif&s=40&w=123&h=22 [Broken]

I don't see how it would be possible for the pedestrian to reach a point not covered by those inequalities. Once again this is treating the sidewalk as the y-axis, the meadow as the positive x direction, and the field as the negative x direction.

Last edited by a moderator: May 5, 2017
9. Dec 15, 2011

### Staff: Mentor

Your straight line equations look right, defining the outer limits of his location after 1 hour.

I don't know how complicated you want to make it. For example, he's allowed to spend some time in the meadow then cross over into the field, and perhaps duck back into the meadow, too?

10. Dec 15, 2011

but if you draw lines 4 or 3 km (for meadow and field) long coming from the starting point in all directions, there are some points the pedestrian could reach that are not taken care of by the straight line equations. That's why i thought a semicircle with a radius of 3 and 4 km was necessary as well.

Well since the inequalities have less than or equal to, I thought it would be sufficient to find the outermost points the pedestrian could reach, and then assume he could also reach all the points within those outer boundaries?

11. Dec 15, 2011

### Staff: Mentor

Now I see what you mean. fooplot sketch

So the extent of his reach will be farther than those straight lines. Indeed, for their entire length except their points of intersection with the axes.