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Directional Derivatives and Gradient question

  • Thread starter srkambbs
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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 = <fx , fy>
2)
ad92a17b245e925a2bff2e4444a520af.png

3)
864157ee7a7f4f55c1a7ce58dfb5cbb1.png


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!!
 

Answers and Replies

  • #2
LCKurtz
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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.
 

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