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f(x,y) = (X^2y)/x^2+y^2 for (x,y) not equal to (0,0)

and 0 for (x,y) equal to (0,0)

unit vector = <-sqrt(3)/2, 1/2>

When I find the partial derivatives of the function at (0,0), they both equal zero, so I conclude that the gradient vector at (0,0) = <0, 0> Obviously any dot product with this vector will be zero, and yet when we used the limit definition of a directional derivative in class, we got that the direction vector at f(0,0) = 3/8.

According to the other definition of a directional derivative, it should = the dot product of the gradient vector and unit vector.

I'm not entirely sure what it means for the gradient vector to = a zero vector, but I think it just means that the function increases or decreases at the same rate in whatever direction you move from that point? So it should be like a max or min?

So if a zero gradient vector says you move at the same rate in any direction from that point, does the definition of the direction derivative being equal to the dot product of gradient vector and unit vector become meaningless? And is that because a zero vector has no direction/has arbitrary direction, and so will be orthogonal to ANY unit vector in ANY direction at that point?

But given that, you still have a rate of change in the function from that point, so you have to conclude that the directional derivative still exists, and must be found using the definition involving limits?

Obviously I'm a little cloudy on the geometric meaning of this stuff too..