Timeline for Finding normals to a curve through a given point
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2013 at 14:25 | comment | added | Corey Kelly | Why not check mathematically if they are perpendicular, rather than by eye? Evaluate the slopes m1 and m2 at the intersection point and check that m1 == -1/m2 | |
| Jun 10, 2013 at 19:27 | comment | added | Raymond Ghaffarian Shirazi | I used you command AspectRatio -> Automatic but still in the curve i was working on actually it seems not perpendicular | |
| Jun 10, 2013 at 18:54 | vote | accept | Raymond Ghaffarian Shirazi | ||
| Jun 11, 2013 at 17:32 | |||||
| Jun 10, 2013 at 18:07 | comment | added | J. M.'s missing motivation♦ |
Of course, since the base curve whose normal you're seeking is a line, you can do this: -(x - 4)/0.36 + 3.
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| Jun 10, 2013 at 18:02 | history | edited | george2079 | CC BY-SA 3.0 |
added 1 characters in body
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| Jun 10, 2013 at 17:33 | history | answered | george2079 | CC BY-SA 3.0 |