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If I drew a level curve of a given function, how would one "loosely" determine at what point/area on that level curve the gradient vector is $(0,-1)$? What's the general idea here?

The level curve I am currently looking at is shaped like the number 8, but tilted a bit to the right, lying in the 1st and 4th quadrant of the xy-plane: $\textit{8}$. It tells me that the gradient vector is $(0,-1)$ at the top of the 8.... why?

The answer is D in the following::

enter image description here

Joaneael
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  • Do you have a picture or an expression you can share with us? – Arthur Nov 06 '14 at 12:04
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    Pic added. The answer is supposedly D. – Joaneael Nov 06 '14 at 12:07
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    Does it have something to do with the fact that the gradient is ortogonal on the level curve? – Joaneael Nov 06 '14 at 12:12
  • Yes, the gradient is orthogonal to the level curve, and the only two marked places where the level cure moves in the right direction for the gradient to be $(0, y)$ for any $y$ are A and D. Why isn't A a correct answer? – Arthur Nov 06 '14 at 12:26
  • @Arthur: Perhaps A is excluded because the curve self-intersects there, so no simple statement can be made about "the gradient". – hardmath Nov 06 '14 at 12:58
  • @hardmath There are plenty of surfaces which have a well-defined gradient everywhere and still a self-intersecting level curve like this at some point. Just note that the gradient has to be orthogonal to both of the incoming curves. – Arthur Nov 06 '14 at 13:01
  • @Arthur: Can we exclude A based on what you just wrote, given that one incoming curve is vertical and one is horizontal? – hardmath Nov 06 '14 at 13:10
  • @hardmath Yes. It means that if the gradient is well defined at A, then it is zero. You often see level curves behave like that at saddle points. – Arthur Nov 06 '14 at 13:41

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