Partial Derivative Signs Through Level Curves

[1question]

Homework Statement

Question 2 from http://math.berkeley.edu/~mcivor/math53su11/solutions/hw6solution.pdf here.

I do not understand b) and e). How do I think of the slope with respect to y?

Homework Equations

The Attempt at a Solution

I do know that the partial derivatives are the slopes of the tanget lines. I do know that f_x < 0 as the slope is negative there (at point P), but I do not understand how the slop in terms of y looks.

 

[Office_Shredder]

When you say "the slope", it's not clear what you mean because there is a slope in every direction from P for this function. If you did the correct procedure to figure out the sign of f_x you should be able to do the exact same thing to figure out f_y - why don't you elaborate on how you did f_x?

 

[1question]

When you say "the slope", it's not clear what you mean because there is a slope in every direction from P for this function. If you did the correct procedure to figure out the sign of f_x you should be able to do the exact same thing to figure out f_y - why don't you elaborate on how you did f_x?

Well, the way I solved for f_x was as follows: draw a cross section of the function's 2D hill image (similar to y=1/x). Then place my point p. As the line tangent through P has a negative slope, f_x is negative. Actually, I would appreciate help with getting f_xx as well - I don't understand how "b/c the x-slope becomes less negative as we move right" correlates with f_xx. If f_x is the tangent to the curve, then what is f_xx geometrically? Answering that would help me a lot.

EDIT: Just googled the second derivative. Now I remember that it tells us the concavity of the quadratic. In this case, is that why f_xx is positive? Is that the entire solution to part c) ?

Thanks.