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

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.

[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!! 🎉

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-zzri]

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

[thomasolson]

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 <x/sqrt(x^2-(y^2-z^2)), y/sqrt(x^2-(y^2-z^2)), z/sqrt(x^2-(y^2-z^2))>. 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.