Find the optimum distance B travels ..

[Rajeev]

Meanwhile still trying to understand the statement you made :

“ B continuing along the straight line trajectory until reaching the line y=6.”



I guess you mean B traveling in a slope till a distance of 6 , probably with an angle of 60 degree making an equilateral triangle with its initial position and A . I tried that but kind of stuck there .

 

[MarkFL]

I have thrown in the towel for now. This has consumed me and spit me out, I don't mind admitting.

But, I have posted this problem at another site to get feedback from some rather brilliant mathematicians.

 

[Rajeev]

I would love to note the address of other site to see those brilliant mathematicians at work, if it is a public forum ?

 

[MarkFL]

I would love to note the address of other site to see those brilliant mathematicians at work, if it is a public forum ?

I don't like posting links to one math help site from another because it can be said they are competing sites. Also, I didn't mean to imply the folks there are any more brilliant than anyone here. It's just that I have known them a long time and they will be likely to respond to my bleatings for help.

The feedback I've gotten from two folks there who's abilities are easily greater than mine have come to the same conclusion that I did essentially. They agree, without my having posted anything to bias their perceptions beforehand, that B moving at an angle of \(60^{\circ}\) from the line separating the two initial positions is optimal. Once they have both moved 6 ft., this is where things become difficult, and it is agreed that the resulting analysis leads to differential equations that likely cannot be reduced to closed form and would need to be approximated numerically.

If indeed a solution is found, I will keep you updated.

 

[Rajeev]

I don't like posting links to one math help site from another because it can be said they are competing sites. Also, I didn't mean to imply the folks there are any more brilliant than anyone here. It's just that I have known them a long time and they will be likely to respond to my bleatings for help.

The feedback I've gotten from two folks there who's abilities are easily greater than mine have come to the same conclusion that I did essentially. They agree, without my having posted anything to bias their perceptions beforehand, that B moving at an angle of \(60^{\circ}\) from the line separating the two initial positions is optimal. Once they have both moved 6 ft., this is where things become difficult, and it is agreed that the resulting analysis leads to differential equations that likely cannot be reduced to closed form and would need to be approximated numerically.

If indeed a solution is found, I will keep you updated.

Thank you so much . Much appreciated . ?

 

[MarkFL]

Thank you so much . Much appreciated . ?

I happened to mention this problem on yet another site (not a math help site) that I help admin because several people there are into math, and the owner of the site, who uses AutoCAD at work came up with this:

His exact distance was 17' 5". He used 2 inch line segments for the curved part of the path.

 

[MarkFL]

One of the mathematicians at the second site determine an approximate value of 17.5 ft.

Here is the second leg of the journey by his calculations:

[MATH]x(t)=12 - vt - 6\cos(\phi(t))[/MATH]
[MATH]y(t) = 6\sin(\phi(t))[/MATH]
where:

[MATH]\phi(t)=2\arctan\left(\frac 1{\sqrt 3} e^{2vt/6}\right)[/MATH]
Here's a live graph.

Desmos | Graphing Calculator

www.desmos.com

 

[MarkFL]

I also wanted to ask, where did this problem originate?

 

[Rajeev]

I happened to mention this problem on yet another site (not a math help site) that I help admin because several people there are into math, and the owner of the site, who uses AutoCAD at work came up with this:

His exact distance was 17' 5". He used 2 inch line segments for the curved part of the path.

One of my friends did exactly like this , with an approximated answer of 17.5 feet . I posted his handwritten solution in my earlier comment

 

[Rajeev]

I also wanted to ask, where did this problem originate?

A friend of mine , math enthusiast shared it with me . This probably came from maths discussion group of his IIT Delhi alumni , one of the premier institutions of India .

 

[Rajeev]

One of the mathematicians at the second site determine an approximate value of 17.5 ft.

Here is the second leg of the journey by his calculations:

[MATH]x(t)=12 - vt - 6\cos(\phi(t))[/MATH]
[MATH]y(t) = 6\sin(\phi(t))[/MATH]
where:

[MATH]\phi(t)=2\arctan\left(\frac 1{\sqrt 3} e^{2vt/6}\right)[/MATH]
Here's a live graph.

Desmos | Graphing Calculator

www.desmos.com

I was hoping to understand how he arrived at these equations, look forward to hear when you get a chance .

Much obliged . ??

 

[MarkFL]

He didn't post his derivations. My own were very close, but I must have made an error somewhere.