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**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 28

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*Last edited by engrymbiff (2010-06-18 07:25:26)*

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Let the point A be (0, 0), i.e. origin, and lets say that they meet at a point C (b, k)

AC=2BC

√(b²+k²)=2k

b²+k²=4k²

k=b/√3

they meet at C (b, b/√3) and AC is 30 degrees inclined from the X-axis.

If two or more thoughts intersect with each other, then there has to be a point.

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**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 28

ZHero wrote:

Let the point A be (0, 0), i.e. origin, and lets say that they meet at a point C (b, k)

AC=2BC

√(b²+k²)=2k

b²+k²=4k²

k=b/√3they meet at C (b, b/√3) and AC is 30 degrees inclined from the X-axis.

Sorry, maybe I did not explain the problem well enough. "**A's direction of movement is always in the nearest-distance-direction**". So, A's direction will change according to B's movement. This will result in some sort of parable for A's path. NOT just pythagoras

Thanks anyway for your try

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,751

Hi engrymbiff;

Your problem is a pursuit curve.

http://mathworld.wolfram.com/PursuitCurve.html

http://curvebank.calstatela.edu/pursuit2/pursuit2.htm

google for more about pursuit curves.

Your problem has 2 difficult features about it:

1) It is a reflection of the one on the top of the page.

2) The ratio of the velocities is 2:1

If you can make the change that A starts at the origin and B the pursuer starts at (b,0) then the equation of B's pursuit is y = c x^2 - log(x), with c to be determined.

Your assumption that the path is a parabola is close but not true.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 28

Thanks bobbym!! That does help a lot!

If someone has anything else to fill in with it still will be much appreciated!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,751

Hi engrymbiff;

I am working on a solution that does not use differential equations. Unfortunately the distance formula involves an integral that may not be integrable. If so that would indicate their may not be an analytical answer, only a numerical one.

Please come back and tell me what you learn new about the problem. Better yet, if you solve it please post your solution.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 28

Perfect, I'm also working on it right now actually. I'll update you if I manage to solve it!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,751

Hi engrymbiff;

The solution for the conditions I have given in post #4: Leads to this DE.

Where c is the ratio of B to A in this case 1:2, So c = 1 / 2, c is always less than 1.

I am having trouble with setting the initial conditions. Okay , got it. If A starts at y=5 and B starts at ( 0 , 0 ) we set the initial conditions as y ' (5) = 0 and y(5) = 0

Solving the DE we get:

Plugging in x = 0 we get y = 10 / 3. So A will catch B at ( 0 , 10 / 3 ). This can be easily checked as being correct.

http://www.mathsisfun.com/graph/functio … 3333333333

Notice that A comes from the right and doesn't start at the origin, but to make it do so should be a simple translation.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

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