Timeline for The class of uniformly accelerated curves and surfaces
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2014 at 19:44 | vote | accept | Mikhail Gaichenkov | ||
| Aug 3, 2014 at 15:31 | comment | added | Mikhail Gaichenkov | Let us continue this discussion in chat. | |
| Aug 3, 2014 at 14:52 | comment | added | Robert Bryant | @MikhailGaichenkov: ...and yet, there must be a mistake because your curve cannot satisfy $x'(t)^2 + y'(t)^2 = t^2$, as you claim. | |
| Aug 3, 2014 at 14:16 | comment | added | Mikhail Gaichenkov | I added apicture to show initial 'physical' effect - the picture explains why x=t. I double checked my calculations, but still cannot find an error. | |
| Aug 3, 2014 at 12:42 | comment | added | Robert Bryant | @MikhailGaichenkov: It seems that, in your Example 2, you are claiming that the curve $$\bigl(x(t),y(t)\bigr) = (\cosh(t){-}1, \ \tfrac14\sinh(2t){-}\tfrac12t)$$ satisfies $x'(t)^2+y'(t)^2 = t^2$ (since you were solving with $A=1$ and $B=0$), but this is obviously not true since $x'(t) = \sinh(t) > t$ when $t>0$, so we must have $x'(t)^2+y'(t)^2 > t^2$. Can you check your calculations? I'm afraid that you are using the variable $t$ in two different senses, and this might have confused you. | |
| Aug 3, 2014 at 10:24 | comment | added | Mikhail Gaichenkov | I added more comments in eg 2 to see if I understand you correctly. Could you look at notes in eg. 2 please. If I confirm, can you show how we get the curve of eg. 2 from your answer? ( I would postpone "surface" question untill confirmation with curves). | |
| Aug 3, 2014 at 8:46 | comment | added | Robert Bryant | @MikhailGaichenkov: I don't understand what you mean by 'crossing line'. Do you mean the curve of intersection of the two surfaces? If so, I don't see how you are getting a parametrization of the intersection curve in order to speak of it being 'uniformly accelerated'. Finally, do you agree, in the case of curves, that you don't need to speak of moving lines or rays, only the velocity of a parametrized curve? I assumed that this was so in my answer, but you didn't confirm this. If the parametrization of the curve is all that matters to you, could you confirm this? | |
| Aug 2, 2014 at 17:13 | comment | added | Mikhail Gaichenkov | Dear Robert, Thank you, let me explain how I see the 'uniformly accelerated surface": the crossing line of the 'uniformly accelerated surface" and a predefined surface moves uniformly accelerated. For eg. let's extend eg 2 of my questions: a plane moves along abscissa axis and crosses the uniformly accelerated surface according to the lines of the example above. I hope my explanation is clear ( I could draw a picture). | |
| Aug 2, 2014 at 10:13 | history | answered | Robert Bryant | CC BY-SA 3.0 |