Proof of Why Gradient of a Function is Perpendicular to its Level Curves

Рет қаралды 7,010

2 жыл бұрын

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.

Пікірлер: 13

@pioneeringproofs 11 ай бұрын

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

@Kiwi-zz8ri Жыл бұрын

Thank you! I’ve was wondering why the gradient vector was orthogonal when I was finding the equation of tangent planes and this helped

@pioneeringproofs Жыл бұрын

Happy to help!

@thomasolson7447 Жыл бұрын

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.

@pioneeringproofs Жыл бұрын

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?

@thomasolson7447 Жыл бұрын

@@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).

@pioneeringproofs Жыл бұрын

@@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!