Using polar coordinates to find the distance traveled

  • Thread starter ptolema
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  • #1
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Homework Statement



A tourist takes a tour through a city in stages. Each stage consists of 3 segments of length 100 feet, separated by right turns of 60°. Between the last segment of one stage and the first segment of the next stage, the tourist makes a left turn of 60°. At what distance will the tourist be from his initial position after 2010 stages?

Homework Equations



ε = e(π/3)i = cos(π/3) + i*sin(π/3) corresponds to a 60° turn to the left
ε6 = 1

The Attempt at a Solution



I really didn't know how to start, so my attempt might not even be relevant. I am also not very familiar with polar coordinates.

I started trying to do this algebraically, by assuming that e-(π/3)i corresponds to a 60° turn to the right. e-(π/3)i*e-(π/3)i*e(π/3)i since the tourist makes 2 right turns then a left in one stage. This equals e-(π/3)i. At this point, I don't know what to do next, or if I should even continue in this direction.
 

Answers and Replies

  • #2
ehild
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Imagine you are a tourist and draw your path on the map. Draw a straight line that corresponds to 100 feet, then turn to the right at 60°angle and draw the next segment and so on.

ehild
 
Last edited:
  • #3
HallsofIvy
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You titled this "find the distance traveled". Do you understand that that is NOT what this question asks you to find? The problem asks you to find the straight line distance between starting and ending points, NOT the total distance traveled.
 
  • #4
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You titled this "find the distance traveled". Do you understand that that is NOT what this question asks you to find? The problem asks you to find the straight line distance between starting and ending points, NOT the total distance traveled.

Yes, I understand that now. My problem is that although I can figure out the angle of the tourist after a given number of stages, I'm stuck on how to find the distance between the starting and ending points.
When I do it graphically, I get a path that looks like partial traces of hexagons. The displacement from 1 stage is 200ft, and after 2 stages, the displacement is 200√3ft. I am not sure how I would generalise this for 2010 stages.
 
  • #5
ehild
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Think in displacement vectors. You can represent the displacement also by complex numbers, as you did in the first post. What is the displacement just after the first stage?

ehild
 
  • #6
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I actually just tried it a different way. I used your suggestion, ehild, of drawing out the path.

[PLAIN]http://www.privateline.com/Cellbasics/sevencellcluster.gif [Broken]

So the tourist travels in a path that traces the outside of this hexagon cluster. At least, this is how I ended up coming up with a diagram. Working from this assumption, it takes 6 stages to return to the starting point.
From here, it get a bit trivial. 2010/6 = 335, so he returns to his starting point exactly 335 times. After 2010 stages, he is at his starting point, i.e. his distance from the starting point is 0.

Did I get anywhere with this?
 
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  • #7
ehild
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You have solved the problem, well done! The distance between the end point and starting point of the route is zero. The distance travelled is different :)
I show my picture, with the displacement during the first stage and during the second stage. They differ with the factor e-iπ/3. The sum of 6 stages is zero.

ehild
 

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