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

[@pioneeringproofs]

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!! 🎉 IMPORTANT: in this video, I only proved this thing for a 2D slice of a 3D object! But this extends to a 3D slice of a 4D object and beyond as well! P.S. Sorry about the whispering at the end of the video; I had to be quiet since I was recording late at night.

[@Kiwi-zz8ri]

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]

Happy to help!

[@thomasolson7447]

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]

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]

@@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]

​@@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!

[@newmantwine1224]

Bro, how is it that you can't even construct a basic proof that if the dot product of two vectors is 0 then the 2 vectors are perpendicular to one another. All you proved is that if 2 vectors are perpendicular to one another their dot product is zero.

[@pioneeringproofs]

Because I wasn’t a math major - Engineer/scientist here

[@pioneeringproofs]

Also, I only proved this for a 3D level curve, and I’m pretty sure this extends to 4D, 5D, etc. level curves, right? The point is, I’m no expert, but I try, and it’s up to people to decide whether my attempt was good or bad. But good point!

[@ilikegeorgiabutiveonlybeen6705]

instant sub

[@Khashayarissi-ob4yj]

With luck and With regards

[@Cdictator]

Math genius??

[@Cdictator]

By the way, you helped me understand this question which has bothered me for a long time.

[@noodle7788]

@@Cdictator dude has a massive brain

[@pioneeringproofs]

Thanks, I’m happy to help!!