Find the Equation of a Line (not simple)

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In summary, the conversation discusses a puzzle involving two points on a Cartesian plane and their movements. Point 1 moves along the y-axis at a constant velocity and period T, while point 2 moves at a constant velocity and always directly towards point 1. The velocities and period are such that the two points will eventually meet at the origin, creating a curve. The conversation also touches on the difficulty of solving the problem and provides some hints to approach it, such as using vectors and derivatives.
  • #1
Samky
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TL;DR Summary
Old puzzle -- I need help on how to approach
An old puzzle that was given to me long ago. I don't know how to approach this problem, any help is appreciated. I'm removing some specifics because I don't want it solved for me, I want to know how to approach it. Feel free to solve a simplified version if that helps you explain it to me.

Imagine two points on a standard Cartesian plane. Point one at the origin, and point 2 somewhere on the x axis. Now:

1) Point 1 moves along the y-axis at constant velocity and with period T

2) Point 2 moves at constant velocity and always directly at point 1

3) The velocities and period are such that the two points will meet at the origin

By tracing the path of point 2, we have a line.

Find the equation of this line.

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Simplified version example:

Point 2 starts at (10, 0). Point 1 moves at 10 meters per second. Point 1 moves up 10 meters then down 10 meters. After two seconds they meet at (0, 0)
 
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  • #2
You get a tractrix, or two combined tractrices in this case. It has analytic expressions for the arc length, so maybe you can find an analytic solution for the full problem. If not, numerical methods will find an approximation.
 
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  • #3
Thanks a lot, I'll look into that!

I wrote some MATLAB code to get a good approximation, so I have some graphs and numbers, but that's sort of cheating. I'm sure a mathematician could solve the original problem in closed form but I have no clue how to begin... but this should help me out, thanks again.
 
  • #4
Samky said:
By tracing the path of point 2, we have a line.

It's probably a good idea to call this a curve and not a line.
 
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  • #5
Samky said:
1) Point 1 moves along the y-axis at constant velocity and with period T
How can a velocity be both constant and periodic unless it is merely constant?
 
  • #6
jbriggs444 said:
How can a velocity be both constant and periodic unless it is merely constant?
I'm confused by that too
 
  • #7
Sorry, I mean speed. The magnitude doesn't change but the direction does. It's basically a one dimensional oscillation that creates a two dimensional oscillation.
 
  • #8
Samky said:
Sorry, I mean speed. The magnitude doesn't change but the direction does. It's basically a one dimensional oscillation that creates a two dimensional oscillation.
I still can't picture the motion. An object can travel in a circle with constant speed, and if you take one component of its motion then you will have a one-dimensional oscillator. However, if you have a one-dimensional oscillator it must turn around 180 degrees at its maximum displacement. To do that, it must get to zero speed before turning around.

Do you mean it has a constant angular frequency?
 
  • #9
It's an up/down motion with constant speed and alternating sign, like a frictionless ball bouncing between two walls.
 
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  • #10
mfb said:
It's an up/down motion with constant speed and alternating sign, like a frictionless ball bouncing between two walls.
Thank you!
 
  • #11
I've got an outline of a solution. Are you competent at vectors and derivatives?
 
  • #12
I'm student-level at vectors and derivatives. I've done basic vector calculus and all that. I'm confident I can solve simple problems involving these things.
 
  • #13
Here's the first two hints:
  • If a particle moves directly at another particle, mathematically how is this represented?
  • You'll also need to write out an equation for the motion of particle 1, which will be used later.
 

1. What is the general form of the equation of a line?

The general form of the equation of a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

2. How do I find the slope of a line?

The slope of a line can be found by using the formula (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are any two points on the line.

3. What is the point-slope form of the equation of a line?

The point-slope form of the equation of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.

4. How do I find the equation of a line given two points?

To find the equation of a line given two points, first find the slope using the formula (y2 - y1) / (x2 - x1). Then, plug the slope and one of the points into the point-slope form of the equation: y - y1 = m(x - x1).

5. Can the equation of a line be written in different forms?

Yes, the equation of a line can be written in different forms such as slope-intercept form (y = mx + b), standard form (Ax + By = C), and point-slope form (y - y1 = m(x - x1)). These forms are all equivalent and can be converted to one another.

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