Timeline for Orthogonal Trajectories Using Polar Coordinates. Correct Calculations, Two Different Answers?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 17:48 | answer | added | Narasimham | timeline score: 0 | |
| S Jun 9, 2020 at 13:15 | history | suggested | Math2718 | CC BY-SA 4.0 |
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| Jun 9, 2020 at 13:13 | review | Suggested edits | |||
| S Jun 9, 2020 at 13:15 | |||||
| Mar 12, 2017 at 5:36 | vote | accept | The Pointer | ||
| Mar 10, 2017 at 20:01 | history | edited | The Pointer | CC BY-SA 3.0 |
edited tags
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| Mar 10, 2017 at 19:38 | history | edited | The Pointer | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:26 | history | edited | The Pointer | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:17 | answer | added | Ted Shifrin | timeline score: 5 | |
| Mar 10, 2017 at 19:06 | comment | added | The Pointer | @TedShifrin I've updated the OP for clarity. | |
| Mar 10, 2017 at 19:04 | history | edited | The Pointer | CC BY-SA 3.0 |
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| Mar 10, 2017 at 19:00 | comment | added | The Pointer | @TedShifrin Yes, they've done this by converting the original equation of the family of curves, which was $x^2 + y^2 = 2Cx$, into polar coordinates. | |
| Mar 10, 2017 at 18:59 | answer | added | Rafa Budría | timeline score: 1 | |
| Mar 10, 2017 at 18:58 | comment | added | Ted Shifrin | I don't follow either of these. $dr/d\theta$ is not the slope of the curve in the $xy$-plane. Do they mean orthogonal trajectory in the $r\theta$-plane? | |
| Mar 10, 2017 at 18:51 | history | asked | The Pointer | CC BY-SA 3.0 |