Proof of Why Gradient of a Function is Perpendicular to its Level Curves
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- Опубліковано 6 лип 2022
- In this video, I take the familiar multivariable/vector calculus idea that the gradient of a 3D function is orthogonal to its level curves, prove this idea, and then try to extend the proof to higher dimensions of space.
It looks like this is my most popular video so far! I wasn’t really expecting this, but I’m glad it seemed to help so many people!! 🎉
P.S. Sorry about the whispering at the end of the video; I had to be quiet since I was recording late at night. And I made a small mistake in displaying the brackets in the equation in that one slide toward the beginning, but the overall idea that I conveyed remains accurate despite this minor error
Thank you! I’ve was wondering why the gradient vector was orthogonal when I was finding the equation of tangent planes and this helped
Happy to help!
instant sub
With luck and With regards
Shouldn't those tangents be longer? The head part is extended to where the origin vector would conect if it had continued beyond the radius. Can I take a gradient from . I haven't made up my mind if the sqrt should be x^2-(y^2-z^2) or x^2-y^2-z^2. But I guess it doesn't really matter at this point. It's Pi time btw, 3:14 am.
Ah yes, Pi time, the optimal time to communicate about such matters. I’m also writing this at Pi time. Anyway, I think you’re right but the overall concept should still be correct (I think). And I’m curious about this vector field you’re talking about, I’ll have to graph it and see. I’m not too familiar with tensors (at least not yet), but I think you’ll get a 2nd order tensor of some kind when you take the gradient of it?
@@pioneeringproofs I don't think anything can graph that. When there is a negative square root, there is a change between x and y-z axis. Half the values would be complex and the computer wouldn't know what to do with them. You would have to program it. This is cosh and sinh on the complex 2d plane nonsese converted to 3d. . This is x^2-y^2-z^2=1 graph.If the square root is negative, you pretend you're on the complex plane x/sqrt(x^2-y^2-z^2)+sqrt(y^2+z^2)/sqrt(x^2-y^2-z^2).
@@thomasolson7447 There always seems to be these subtle details that end up making things more complicated than originally expected, especially in special cases like these. But it seems like an interesting problem to solve - Good luck!
Math genius??
By the way, you helped me understand this question which has bothered me for a long time.
@@Cdictator dude has a massive brain
Thanks, I’m happy to help!!