Directional Derivatives and Gradient question

[srkambbs]

Homework Statement

Consider the surface and point given below:-
Surface: f(x,y)= 4-x²-2y²
Point: P(1,1,1)

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.

Homework Equations

1) ∇f = <f_x , f_y>
2)

3)

The Attempt at a Solution

a) ∇f = <f_x , f_y> = <-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!

 

[LCKurtz]

I think that sometimes in this type of problem it helps to visualize a physical model. Suppose you have a thin metal plate and your $f(x,y) =4 - x^2-2y^2$ represents the temperature at the point $(x,y)$, so the temperature at $(1,1)$ is $1$. You are looking for the path to follow to cool off quickest. You have calculated that the direction to go at each point is $-\nabla f = \langle 2x,4y \rangle$. From this you can conclude that the slope of the desired curve at $(x,y)$ is $\frac{dy}{dx} =\frac {4y}{2x}=\frac {2y}{x}$. This is a simple first order differential equation. Do you know how to solve it? If so, the solution through $(1,1)$ is what you are looking for.