Timeline for Revolution plot with 2 variable
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 7, 2013 at 4:38 | comment | added | J. M.'s missing motivation♦ | This article has a nice discussion on the numerical parametrization of a curve. | |
| Dec 14, 2012 at 20:30 | comment | added | whuber | Nothing to be sorry about, Renaud. An upvote had been reversed, which made me concerned that someone had tested out this solution, found it wanting, and then downvoted it out of frustration. It turns out to have a more mundane explanation: that user has now been removed from SE, suggesting that their actions overall might have been found to be inconsistent with norms of reasonable behavior. | |
| Dec 14, 2012 at 19:35 | comment | added | RNB | It gave me exactly what I wanted, so I don't see why it had been downvoted (I should have answered to the answer you gave, I'm sorry). As I am far from understanding Mathematica, I cannot see a better alternative. | |
| Dec 14, 2012 at 19:14 | comment | added | whuber | I know this is a shot in the dark, but given the recent downvote, I would be grateful to learn what could be done to improve this answer. Are there limitations worth pointing out? (Probably.) Is the exposition unclear at any point? Is there a superior alternative? | |
| Dec 14, 2012 at 17:23 | history | edited | whuber | CC BY-SA 3.0 |
edited body
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| Dec 14, 2012 at 0:30 | vote | accept | RNB | ||
| Dec 13, 2012 at 23:29 | comment | added | whuber |
It's incredibly simple: paste your expression (without the initial 0==) in place of that to the right of := in the definition of f and you're all set.
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| Dec 13, 2012 at 23:16 | comment | added | RNB |
I understand the approach and it does seem to be more efficient, but I'm not quite sure how derivate the function the way you did if it gets more complex. For instance, 0==1/2 + x^2/2 + y^2/2 - x^2 Cos[(13 \[Pi])/64]^2 - 2 x Cos[(13 \[Pi])/64] Sin[(13 \[Pi])/64] - Sin[(13 \[Pi])/64]^2 is a simple hyperbola equation (which is what I'm working with right now). I can deriviate the equation by myself, but I don't understand how to do it the way you did.
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| Dec 13, 2012 at 3:49 | history | answered | whuber | CC BY-SA 3.0 |