# Bug spiral problem

1. Jun 26, 2005

### anisotropic

Hello,

Can anyone please help me with this question? I need to get it done VERY soon. If someone could post a full solution I would be extremely grateful. It's probably very easy for someone who understands parametric equations and polar coordinates and such. I don't at the moment!

I scanned the problem in from the textbook. >http://img.photobucket.com/albums/v493/photohub/spiralthumb.jpg [Broken]<

note: I already posted this in the general Calc section only after I realized there is a more appropriate section, namely this one. However, it doesn't seem like this forum allows users to delete their own threads. I don't mean to SPAM.

Last edited by a moderator: May 2, 2017
2. Jun 26, 2005

### OlderDan

3. Jun 26, 2005

### anisotropic

Let's just say I'm studying for an exam (ETA 24 hours + a bit) worth 50%. I honestly can't afford to look at this question too long, and as it stands, I've thought about it, and can't figure out where to start. I'm not gonna waste anymore time.

If you want to help, I'd be grateful. If not, so be it I guess.

4. Jun 26, 2005

### OlderDan

Are you required to do this in polar coordinates? At first glance it looks like it might be easier in rectangular coordintes.

5. Jun 26, 2005

### anisotropic

It doesn't specify at all, but I'm pretty sure polar coordinates are in order. The question is pulled from the parametric equations/polar coordinates chapter review...

6. Jun 26, 2005

### OlderDan

The connection to parametric equations is clear. The speed is constant, so there will be a time parameter involved. I think cartesian coordinates would be easiest. Look at the bug that starts at (a/2, a/2) and recognize that the direction of its motion is determined by its own position. It starts off headed for the point (-a/2, a/2), but as it progresses along its path (x, y) it must always be headed toward the point (-y, x). The distance it travels in any time dt is always the same. I'll have to look a the implications of that, but I believe it's the correct starting point. Of course that can be transformed into polar coordintes at some point. I'm not sure the easiest place to do it yet.

7. Jun 27, 2005

### Gokul43201

Staff Emeritus
Unfortunately anisotropic, we are bound by forum guidelines, to not provide solutions unless the poster shows some effort first. However, seeing that this is not a homework problem, I suggest that you search within this site for a solution. There was a similar thread, a little over a year ago, involving Bill, Mike and someone else.
https://www.physicsforums.com/search.php? [Broken]

Last edited by a moderator: May 2, 2017
8. Jun 27, 2005

### PhilG

Here's one way to start out, but I wouldn't know if it is the best way. You can use a complex number to represent the position of each bug. For example, let z1 represent the position of the bug that starts out in the upper right hand corner, and z2 represent the position of the bug that starts out in the upper left hand corner. The path z2(t) is just the path z1(t) rotated 90 degrees counterclockwise. In complex numbers, this corresponds to multiplying by i. Since bug 1 is always heading toward bug 2, it must be that d(z1)/dt is in the same direction as z2-z1. The distance between bugs 1 and 2 is s = a - vt, where v is the constant speed of the bug. Since z2 = i * z1, you can get a differential equation for z1, the position of bug 1. After solving for the complex "position" z1(t), you can pick off r(t) and theta(t), and hopefully find r as a function of theta from these. I'm sure there is a perfectly good way to do it without using complex numbers, but I got stuck when I tried.

Also note that part (b) can be done without doing part (a). Actually, it can be done without doing any calculation at all.

9. Jun 27, 2005

### OlderDan

Here's the approach that looks right to me

$$\frac{{dy}}{{dx}} = \frac{{y - x}}{{y + x}}$$

$$r^2 = x^2 + y^2$$

$$rdr = xdx + ydy = \left( {x + y\frac{{y - x}}{{y + x}}} \right)dx = \frac{{x^2 + y^2 }}{{y + x}}dx = \frac{{r^2 }}{{y + x}}dx$$

Express everything in terms of polar coordinates and do the simple integral that results to show

$$r = r_{\rm{o}} e^{\left( {\pi /4 - \theta } \right)}$$

10. Jun 27, 2005

### PhilG

Ah, nice work! You were right about cartesian coordinates being the way to go.

11. Jun 27, 2005

### quasar987

How do you get thge expression for dy/dx ?!

12. Jun 27, 2005

### PhilG

If (x(t), y(t)) is the position of the bug starting in the upper right hand corner, (-y(t), x(t)) is the position of the bug starting in the upper left hand corner.

13. Jun 27, 2005

### anisotropic

Ok well all I came up with is, for bugs numbered 1, 2, 3, and 4...

1: (r, theta)
2: (r, theta + pi/2)
3: (r, theta + pi)
4: (r, theta + 3pi/2)

I don't even know if this makes sense, but they are the polar coordinates of the four bugs at any given time (I think). I have no clue what to do now though. I've been doing polar stuff for only a few days.

This assignment gives us a bonus on the final course mark, which is why I really need it. I'm worried I'm gonna get sub 70% tomorrow on the exam.

I don't quite understand that, but I'll try to. I think I found another solution on the Net also. http://rec-puzzles.org/new/sol.pl/analysis/bugs [Broken]

Last edited by a moderator: May 2, 2017
14. Jun 27, 2005

### OlderDan

Your equations correctly relate the relative positions of the bugs, but they tell you nothing about the relationship between r and $$\theta$$ . That relationship comes from the slope of the path the bug follows.

The problem at that link is basically the same problem except that the bugs are walking clockwise instead of counter clockwise, and the solution is formulated a bit differently. The argument made about the bugs always walking at angle pi/4 to the radial is equivalent to my expression for the slope in cartesian coordinates. I think it is easiear to visualize in cartesian coordinates, but both views are correct. What it boils down to is (in your counter clockwise problem)

$$dr = -rd\theta$$

You will see the equivalent expression at the link written as

$$\frac{dr}{d\theta} = r$$

The sign difference is because of the direction difference. This relationship is important for part b) as well as part a). It tells you that for any change in r, dr, the bug moves a distance

$$dl = -\sqrt{dr^2 + (rd\theta)^2} = -\sqrt{2} dr$$

(dr is negative as the bug moves) so the total distance moved will be

$$l = \sqrt{2}\ \ r_o$$

If you write x and y in polar form in the equation I posted, work out the differential dx and simplify you will get the equation above. Separate variables, integrate, and impose the initial conditions of the starting point in the upper right corner and you have the answer I posted.

Last edited by a moderator: May 2, 2017
15. Jun 27, 2005

### quasar987

Can you explain this part plz? It is not evident.

16. Jun 28, 2005

### PhilG

Sure thing. The path of bug 2 (upper left hand corner) is the same shape as the path of bug 1 (upper right hand corner), but rotated 90 degrees counterclockwise about the origin. This rotates the y axis into the negative x-axis, and the x-axis into the y axis. So if the y coordinate of bug 1 is y_1, then the x-coordinate of bug 2 is -y_1. If the x-coordinate of bug 1 is x_1, then the y-coordinate of bug 2 is x_1.

17. Jun 28, 2005

Neat!