Consider the surface and point given below:-
Surface: f(x,y)= 4-x2-2y2
a) Find the gradient of f.
b) Let C' be the path of steepest descent on the surface beginning at P and let C be the projection of C' on the xy-plane. Find an equation of C in the xy-plane.
1) ∇f = <fx , fy>
The Attempt at a Solution
a) ∇f = <fx , fy> = <-2x, -4y>
b)Descent means -∇f = <2x,4y>
Subbing in P(1,1,1)
-∇f = <2,4> ⇔ <1,2>
Unit vector for descent, u = (1/√5) <1,2>
,where <1,2> is the direction vector of the descent gradient.
I am really stuck here. I am not sure what they mean by the projection on the xy-plane. So are we moving from xyz to xy dimensions? And can I am not sure if I should use the projection formulas in this case or use derivatives to get the projection. Please help!!