Find the optimum distance B travels ..

Rajeev

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I have been stuck with this problem , I think it may require a combination of Trigonometry , Geometry and Calculus . Guys please try to solve , thank you ..


A and B are 12 feet apart at points x and y respectively.

A and B have to swap places. A walks directly towards point y. B is covid conscious and wants to maintain a 6 ft distance from A.

Both A and B walk at constant and equal speed.

Find the optimum distance B travels maintaining a distance of 6 feet from A ?
 
I have been stuck with this problem , I think it may require a combination of Trigonometry , Geometry and Calculus . Guys please try to solve , thank you ..


A and B are 12 feet apart at points x and y respectively.

A and B have to swap places. A walks directly towards point y. B is covid conscious and wants to maintain a 6 ft distance from A.

Both A and B walk at constant and equal speed.

Find the optimum distance B travels maintaining a distance of 6 feet from A ?
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment
 
I have tried couple of things but it didn’t work out .

1. B walks through the semi circle between the point y and x but then using coordinate geometry of distance between two points I was able to see that distance between them becomes less that 6 feet at some point

2. Second option I tried was B taking an angular position and at its 6 feet height A is directly perpendicular to him. at center . but again distance is not maintained consistently .

3,. Another option was B taking an angle of 60 degree and making equilateral triangle with A when A has moved 6 feet . This also didn’t yield result .

Like I said this may require a different approach and I have run out of option .
 
Let's put A and B on the x axis, A is at 0, B is at 12. We know that at some point B must be at y=6 to avoid getting close to A. Which point is this? My guess is when A's and B's x coordinates are the same. So, I would let B travel on a straight line towards that point (you need to calculate its coordinates). Then it should travel on a straight line towards (0,0).
 
Let's put A and B on the x axis, A is at 0, B is at 12. We know that at some point B must be at y=6 to avoid getting close to A. Which point is this? My guess is when A's and B's x coordinates are the same. So, I would let B travel on a straight line towards that point (you need to calculate its coordinates). Then it should travel on a straight line towards (0,0).

I have tried all such options . They end up at some point distance less than 6 .

One option is both A and B walk 3 feet towards each other and then B must take some sort of curve and then straight line .. like this image .. but I am not sure how to calculate the exact path
 

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I have tried all such options . They end up at some point distance less than 6 .

One option is both A and B walk 3 feet towards each other and then B must take some sort of curve and then straight line .. like this image .. but I am not sure how to calculate the exact path
First of all, we are optimizing distance, not time. So the fewer curved segments there are, the better. B can move to a point with y=6, wait there until B passes, then continue - both straight segments.
 
First of all, we are optimizing distance, not time. So the fewer curved segments there are, the better. B can move to a point with y=6, wait there until B passes, then continue - both straight segments.
Stopping and waiting not allowed - they are moving at constant speed.
 
Stopping and waiting not allowed - they are moving at constant speed.

yes , they keep moving with constant speed . no stopping is allowed . Thats why it will need a curve and possibly calculus to solve .

My calculus is very rusty so I am hoping some experts here help :)
 
I would use coordinate geometry, and put A initially at \((12,0)\) and B at the origin. Without loss of generality we can let their speeds be 1 unit of distance per 1 unit of time.

For the first leg of B's path, let's let that path be described parametrically by the line:

[MATH]\left\langle \frac{1}{\sqrt{a^2+1}}t,\frac{a}{\sqrt{a^2+1}}t \right\rangle[/MATH]
A's location is on the line:

[MATH]\langle 12-t,0\rangle[/MATH]
The square of distance \(D\) between A and B is then:

[MATH]D^2=\left(\frac{1}{\sqrt{a^2+1}}t-(12-t)\right)^2+\left(\frac{a}{\sqrt{a^2+1}}t\right)^2[/MATH]
Use of differential calculus reveals that the minimum distance is

[MATH]D_{\min}=72\left(1-\frac{1}{\sqrt{a^2+1}}\right)[/MATH]
We want this to be 36 units, hence:

[MATH]72\left(1-\frac{1}{\sqrt{a^2+1}}\right)=36[/MATH]
And taking the positive root, we obtain:

[MATH]a=\sqrt{3}[/MATH]
This minimum occurs then for:

[MATH]6^2=\left(\frac{3}{2}t-12\right)^2+\left(\frac{\sqrt{3}}{2}t\right)^2[/MATH]
[MATH]48=3(t-8)^2+t^2[/MATH]
[MATH]t=6[/MATH]
And so B has traveled 6 units of distance. Here's a live graph of this first leg of the journey:


Beyond this point, I am thinking B should cannot walk along the moving circumference of A's position without simply being pushed back along a horizontal line. It seems to me B should continue the linear trajectory until reaching the line \(y=6\), or for:

[MATH]0\le t\le\frac{12}{\sqrt{3}}[/MATH]
Then B should travel some distance long this horizontal line until the moment it can begin a linear trajectory towards the final destination being tangent to the moving circle as it was during the first leg of the journey.

I will leave you to work with that for now, and revisit this topic later. :)
 
The more I think about it, the more I have doubt about whether what I posted above is on the right track. Once point B moves beyond the tangent point, it rises above the circle, which means there is some part of the upward component of the velocity function that should be utilized to move the point forward instead. The slope of the trajectory should match that of the moving circle.

I'm going to have to ponder this some more...
 
Here is my quick take at this.
Person A: Walks from x=0 to x=3. Then walks along y= sqrt(9-x^2). Then walks from x=9 to x=12.
Person B: Walks from x=12 to x=9. Then walks along y = -sqrt(9-x^2). Then walks along x=3 to x=0.
 
Here is my quick take at this.
Person A: Walks from x=0 to x=3. Then walks along y= sqrt(9-x^2). Then walks from x=9 to x=12.
Person B: Walks from x=12 to x=9. Then walks along y = -sqrt(9-x^2). Then walks along x=3 to x=0.
A walks directly towards point y
 
I would use coordinate geometry, and put A initially at \((12,0)\) and B at the origin. Without loss of generality we can let their speeds be 1 unit of distance per 1 unit of time.

For the first leg of B's path, let's let that path be described parametrically by the line:

[MATH]\left\langle \frac{1}{\sqrt{a^2+1}}t,\frac{a}{\sqrt{a^2+1}}t \right\rangle[/MATH]
A's location is on the line:

[MATH]\langle 12-t,0\rangle[/MATH]
The square of distance \(D\) between A and B is then:

[MATH]D^2=\left(\frac{1}{\sqrt{a^2+1}}t-(12-t)\right)^2+\left(\frac{a}{\sqrt{a^2+1}}t\right)^2[/MATH]
Use of differential calculus reveals that the minimum distance is

[MATH]D_{\min}=72\left(1-\frac{1}{\sqrt{a^2+1}}\right)[/MATH]
We want this to be 36 units, hence:

[MATH]72\left(1-\frac{1}{\sqrt{a^2+1}}\right)=36[/MATH]
And taking the positive root, we obtain:

[MATH]a=\sqrt{3}[/MATH]
This minimum occurs then for:

[MATH]6^2=\left(\frac{3}{2}t-12\right)^2+\left(\frac{\sqrt{3}}{2}t\right)^2[/MATH]
[MATH]48=3(t-8)^2+t^2[/MATH]
[MATH]t=6[/MATH]
And so B has traveled 6 units of distance. Here's a live graph of this first leg of the journey:


Beyond this point, I am thinking B should cannot walk along the moving circumference of A's position without simply being pushed back along a horizontal line. It seems to me B should continue the linear trajectory until reaching the line \(y=6\), or for:

[MATH]0\le t\le\frac{12}{\sqrt{3}}[/MATH]
Then B should travel some distance long this horizontal line until the moment it can begin a linear trajectory towards the final destination being tangent to the moving circle as it was during the first leg of the journey.

I will leave you to work with that for now, and revisit this topic later. :)

From the graph , I assume you want B to take a semicircle to the destination of radius 6 I believe ?

I have thoroughly tested this condition and at some point distance is less than 6 between A and B .
 
The more I think about it, the more I have doubt about whether what I posted above is on the right track. Once point B moves beyond the tangent point, it rises above the circle, which means there is some part of the upward component of the velocity function that should be utilized to move the point forward instead. The slope of the trajectory should match that of the moving circle.

I'm going to have to ponder this some more...

Looking forward to it ???
 
From the graph , I assume you want B to take a semicircle to the destination of radius 6 I believe ?

I have thoroughly tested this condition and at some point distance is less than 6 between A and B .

Initially, I had B continuing along the straight line trajectory until reaching the line \(y=6\). Last night I was able to mathematically confirm my suspicion that if B tries to walk along the perimeter of the moving circle he will simply be "pushed back" along a horizontal line.

I'm still struggling to find a better way to approach this. :)
 
An attempt by a friend , remains incomplete though

solution1.jpg

Another attempt. Total 17.7 feets. From point of view of B. Go straight at 60 deg to 6 ft. Then make a curve to the point of 7 ft from y, which is 6 ft from horizontal 7ft point. Then another curve to 8 ft point which is 6 ft from the horizontal 8ft point. Then B can go straight to x. Total 17.7 feet.
 
I'm still working on this...I have approximated what I think the path up to \(y=6\) should look like here:


I'm continuing to work with some rather difficult differential equations in the meantime.
 
I'm still working on this...I have approximated what I think the path up to \(y=6\) should look like here:


I'm continuing to work with some rather difficult differential equations in the meantime.


Thank you so much . As I expected it will require getting into calculus and differential equations to sort it out . Look forward to it when you have completed it .

regards .
 
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