Timeline for Finding normals to a curve through a given point
Current License: CC BY-SA 3.0
28 events
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Jun 11, 2013 at 17:32 | vote | accept | Raymond Ghaffarian Shirazi | ||
Jun 11, 2013 at 15:37 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 3.0 |
edited title
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Jun 11, 2013 at 15:35 | answer | added | Sarah Aria | timeline score: 1 | |
Jun 11, 2013 at 12:16 | history | reopened |
m_goldberg Oleksandr R. Silvia rm -rf♦ |
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Jun 11, 2013 at 12:12 | comment | added | Silvia | @RaymondGhaffarianShirazi I've edited the question according to your comment and voted for reopen. Please don't be hesitate to correct any possible mistake in case I misunderstood you. | |
Jun 11, 2013 at 12:09 | history | edited | Silvia | CC BY-SA 3.0 |
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Jun 11, 2013 at 11:04 | comment | added | Raymond Ghaffarian Shirazi | sol2 are the coefficients fired to the equation U[r,o], Ei is dataset fited to equation U[r,o] before, so the dots outside of line are Ei. | |
Jun 11, 2013 at 0:39 | history | edited | m_goldberg | CC BY-SA 3.0 |
General cleanup
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Jun 10, 2013 at 21:16 | comment | added | Silvia |
What is Ei and sol2 ?
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Jun 10, 2013 at 19:49 | review | Reopen votes | |||
Jun 11, 2013 at 12:16 | |||||
Jun 10, 2013 at 19:34 | history | edited | Raymond Ghaffarian Shirazi | CC BY-SA 3.0 |
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Jun 10, 2013 at 18:54 | vote | accept | Raymond Ghaffarian Shirazi | ||
Jun 11, 2013 at 17:32 | |||||
S Jun 10, 2013 at 18:28 | history | closed |
Silvia Jens Ajasja Yves Klett Sjoerd C. de Vries |
too localized | |
S Jun 10, 2013 at 18:28 | comment | added | Sjoerd C. de Vries | possible duplicate of How can I get even steps on x and y axes in a plot? | |
Jun 10, 2013 at 17:43 | comment | added | Silvia |
I think you already found the actual perpendicular line. You just thought you haven't because a plot with aspect ratio $\neq 1$ misled you.. If you try Plot[y[x],{x,0,5},Prolog->{PointSize[.02],Through[{Point,Line}[{{4,3},{x,y[x]}/.x->3.85977}]]},AspectRatio->Automatic] , you'll see what I mean.
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Jun 10, 2013 at 17:33 | answer | added | george2079 | timeline score: 4 | |
Jun 10, 2013 at 17:31 | comment | added | Daniel Lichtblau | What @Silvia is saying is that the line through that point (probably) is perpendicular, and simply does not look so. This happens because aspect ratio !=1 will distort scale. | |
Jun 10, 2013 at 17:16 | comment | added | Raymond Ghaffarian Shirazi | actually no it's not silvia. it's about finding actual perpendicular line I don't really care about aspect ratio. as I mentioned in the title. | |
Jun 10, 2013 at 16:53 | review | Close votes | |||
Jun 10, 2013 at 18:28 | |||||
Jun 10, 2013 at 16:48 | comment | added | Silvia | I voted to close as I think your real question is merely about plot with specific aspect ratio, which in my opinion is a please-read-the-documentation question and too localized (also might be duplicate). | |
Jun 10, 2013 at 16:36 | comment | added | Silvia | Have you tried the option I mentioned above? Does it look right with that option? | |
Jun 10, 2013 at 16:31 | comment | added | Raymond Ghaffarian Shirazi | yes the plot doesn't look perpendicular to me if you know how you can plot it better please let me know too | |
Jun 10, 2013 at 16:11 | comment | added | Silvia |
I'm not following you. It looks to me the right answer indeed. Or do you mean the plot looks not perpendicular? Have you tried Plot with option AspectRatio -> Automatic ?
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Jun 10, 2013 at 16:08 | history | edited | Silvia | CC BY-SA 3.0 |
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Jun 10, 2013 at 15:53 | comment | added | Raymond Ghaffarian Shirazi | no it's not a line I just made this simple code for testing. | |
Jun 10, 2013 at 15:32 | comment | added | J. M.'s missing motivation♦ | If your "curve" is a line, then you don't need optimization machinery; there's a nice geometric method for that purpose... | |
Jun 10, 2013 at 15:31 | history | edited | J. M.'s missing motivation♦ |
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Jun 10, 2013 at 15:28 | history | asked | Raymond Ghaffarian Shirazi | CC BY-SA 3.0 |