Timeline for Orthogonal Trajectories Using Polar Coordinates. Correct Calculations, Two Different Answers?
Current License: CC BY-SA 3.0
27 events
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| Mar 12, 2017 at 15:15 | comment | added | The Pointer | @RafaBudría What do you mean? I'm just asking you to explain why you put the $r$ in the term $\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}$. As I said, more explanation is needed to understand your answer. | |
| Mar 12, 2017 at 9:20 | comment | added | Rafa Budría | Don't flame, please | |
| Mar 12, 2017 at 0:07 | comment | added | The Pointer | @RafaBudría How did you know where to put the $r$ in $\frac{\mathrm d\mathbf v}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat r}+\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}$? You put it in the $\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}$ term but why? You don't say how you got the original vector $v$... As you can see you did not explain all steps... | |
| Mar 11, 2017 at 14:02 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 11, 2017 at 13:53 | comment | added | Rafa Budría | Ah, ok. It's the expression of the tangent vector and then using the fact that a vector is orthogonal to another if one has the components of the other but interchanged and one of the components with sign changed too $(a,b)·(b,-a)=0\implies$ orthogonal. Check again my post. | |
| Mar 11, 2017 at 7:30 | comment | added | Rafa Budría | It's the same as when we write $\dfrac{\mathrm dy/\mathrm dt}{\mathrm d x/\mathrm dt}=\dfrac{\mathrm dy}{\mathrm dx}=g(y,x)$ in cartesian, it's the way to get the derivative from the parametric equation to the explicit one. | |
| Mar 10, 2017 at 22:55 | comment | added | Rafa Budría | You can follow the idea applied to your problem at the end of the answer. There can be seen we have the derivative of $r$ wrt $\theta$ as funtion of $t$.Then, we can eliminate $t$ and obtain an expression with $r$ and $\theta$ (only $\theta$ in your problem) and solve the ODE. | |
| Mar 10, 2017 at 21:21 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 21:13 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 21:03 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 21:01 | comment | added | Rafa Budría | I've just added at the end the manipulations with your equation. | |
| Mar 10, 2017 at 20:57 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 20:57 | comment | added | The Pointer | @RafaBudría I think you are correct: To get the orthogonal trajectories in polar form, we must take the negative reciprocal of $\dfrac{dr}{d\theta} * \dfrac{1}{r}$, which gives us $\dfrac{-d\theta}{dr} * r$. However, I am struggling to understand your calculations. Can you please elaborate further by showing your intermediate calculations? | |
| Mar 10, 2017 at 20:29 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 20:23 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:30 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:27 | comment | added | Rafa Budría | The circle case has a ugly zero and is not directy tractable by $dr/d\theta$, but it's the same as a vertical line in cartesian where you cannot get the quotient. | |
| Mar 10, 2017 at 19:22 | comment | added | Rafa Budría | I am possibily mistaken, but my point is that is not the slope in fact, but you can anyway procceed. If you consider the curve parametrized I get a vector, the perpendicular of wich interchanges components and the sign of one of them. | |
| Mar 10, 2017 at 19:19 | comment | added | Ted Shifrin | But it's not orthogonal in the $xy$-plane!! Is it? Why? My counterexample doesn't fit your formula, does it? | |
| Mar 10, 2017 at 19:19 | comment | added | Rafa Budría | Not the slope. I am possibly mistaken, but I think you need only the orthogonal to the tangent and you get so. | |
| Mar 10, 2017 at 19:19 | history | edited | Rafa Budría | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:19 | comment | added | Ted Shifrin | Can you please explain why the slope is $\dfrac1r\dfrac{dr}{d\theta}$? I gave a counterexample in my answer. Am I being stooopid? | |
| Mar 10, 2017 at 19:16 | comment | added | Rafa Budría | You are right. But, anyway, the book get the right equation. | |
| Mar 10, 2017 at 19:16 | history | undeleted | Rafa Budría | ||
| Mar 10, 2017 at 19:05 | history | deleted | Rafa Budría | via Vote | |
| Mar 10, 2017 at 19:02 | comment | added | Ted Shifrin | Again I complain that this is not the slope of the curve in the $xy$-plane. | |
| Mar 10, 2017 at 18:59 | history | answered | Rafa Budría | CC BY-SA 3.0 |