The discussion revolves around finding the unit vector that is perpendicular to the level curve of the function f(x,y) = x²y - 10xy - 9y² at the point (2,-1). The subject area includes concepts from multivariable calculus, particularly related to gradients and level curves.
The discussion is active, with participants confirming that the gradient evaluated at the specified point is relevant to finding the perpendicular vector. There is a clarification of the concept of level curves, which seems to enhance understanding among participants.
Some participants express uncertainty about the definition of a level curve and its application in this context, indicating a need for further exploration of foundational concepts.
Yes. Is that a problem?Bestphysics112 said:Wouldn't this just be the gradient of f(x,y) evaluated at (2,-1) then normalized?
No problem at all. I just wasn't sure what a level curve was.Orodruin said:Yes. Is that a problem?
Oh that makes much more sense. Thank you.Orodruin said:A level curve of a function is the curve such that the function takes a fixed value. For example, the level curves of f=x^2+y^2 are circles.