Timeline for Why is the gradient normal?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2021 at 20:19 | comment | added | Steven Gubkin | Oh, this is me. I wonder why I had this alt account. Can it be merged? | |
| S Sep 7, 2020 at 22:46 | history | suggested | jsr24 | CC BY-SA 4.0 |
grammar and a couple clarifications for h and p
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| Sep 7, 2020 at 18:42 | review | Suggested edits | |||
| S Sep 7, 2020 at 22:46 | |||||
| Jan 20, 2014 at 7:48 | comment | added | Dirk | on a fine scale you do not see that "zero increase" is normal to "steepest increase". Compare this with a smoother setting like skiing (where the piste is usually somehow smooth and the scale is coarser due to the length of the ski): If you want to stand still on a slope you have to adjust you ski normal to the steepest descent direction. | |
| Jan 20, 2014 at 7:45 | comment | added | Dirk | I think this is somehow the best answer, as it stresses the fact that the derivative is defined as a linear map, i.e. the function is approximated by a hyperplane and this uniformly in the direction. Then the conclusion is clear: The directions of "zero increase" on a hyperplane are normal to the directions of steepest increase. If you relax the definition of derivative and only speak about directional derivatives then nothing of this sort is true anymore. This reflects the experience you make while hiking in the mountains: Since the underlying area is usually not differentiable (cont) | |
| Mar 20, 2013 at 16:51 | history | answered | Steve | CC BY-SA 3.0 |