Timeline for A tricky tractrix question about vertical tangents
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2015 at 20:37 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Corrected some careless errors in the integration of X
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| Nov 13, 2015 at 12:41 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added a remark about the algebraic nature of the tractrix curve.
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| Nov 9, 2015 at 14:12 | comment | added | Robert Bryant | @Wolfgang: As to your first question, I don't know anything about that because I haven't looked at that earlier question. I'll have a look when I have a moment. As for the second, I was just checking that you had noticed the shift in coordinates. No, it doesn't change the problem, I was just thrown off for a bit by seeing you write the equation for the line as $x+y = r$, when that didn't agree with your description later. I realized that the difference was probably that we were using different coordinate systems, and I wanted to make sure that this was the cause. | |
| Nov 9, 2015 at 14:06 | comment | added | Wolfgang | Yes I know about the cusp happening later. Do you think, like somebody had suggested in between in a comment now deleted, that the optimal solution of the original problem MUST use an 'entire' tractrix (i.e. ending at a cusp)? Re your 2nd remark: Yes I have seen that your $X(0)$ is shifted by $-h$ compared with my coordinates. But does that mean your results refer to a different construction? I hope not! | |
| Nov 9, 2015 at 13:41 | comment | added | Robert Bryant | @Wolfgang: The value of $r$ for which the curve first turns vertical exactly when it reaches the line $x+y = 0$ is $r\approx 0.8250033$. However, that curve does not have a cusp at that point; the cusp happens further along. By the way, I hope that you have noticed that I have normalized so that the 'handle' drawing the point along is moving along the curve $x^2+y^2 = r^2$, not $x^2 + (y-h)^2 = r^2$, as you have drawn it in your diagram. | |
| Nov 9, 2015 at 13:41 | comment | added | Wolfgang | Thank you again! I have incorporated that in the other thread. | |
| Nov 9, 2015 at 13:40 | vote | accept | Wolfgang | ||
| Nov 9, 2015 at 13:03 | comment | added | Robert Bryant | @Wolfgang: Yes, that value turns out to be about $r=0.8250033$. | |
| Nov 9, 2015 at 12:42 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added some information about the periods and the structure of the curve.
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| Nov 9, 2015 at 10:00 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added information about how to continue the curve past the cusps
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| Nov 9, 2015 at 9:48 | comment | added | Wolfgang | Thank you very much! I guess that is about the best one can say. Now as you have been able to find numerically the radius which makes the tractrix stop at $\frac\pi2$, can you also find numerically for which $r$ we have $x+y=r$ for the cusp $(x,y)$? This would be the less wasteful way to keep space for inserting a whole quandrant in my original construction: the bottom end of the vertical tangent lying on the diagonal line of slope $-1$ through the center of the circle (as BTW is roughly the case in my drawing above). | |
| Nov 9, 2015 at 1:57 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 27 characters in body
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| Nov 8, 2015 at 22:19 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 160 characters in body
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| Nov 8, 2015 at 22:06 | history | answered | Robert Bryant | CC BY-SA 3.0 |