Vectors and deciding which direction to move

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If you're given a point, with coordinates (0.8,0.7,0.129), and its elevation is given by f(x,y)=(e^(-0.5(x^2+y^2))*sin(pi*x+2pi*cos(5y)). How would you determine which direction to move in, where the elevation would not change?
 
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129), and its elevation is given by f(x,y)=(e^(-0.5(x^2+y^2))*sin(pi*x+2pi*cos(5y)). How would you determine which direction to move in, where the elevation would not change?
By solving the equation:

{change of elevation} = 0
 
So set f(x,y)=0, and solve for x and y?
Or set the gradient equal to 0, and solve for x and y?
What would I do after that point?
 
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No, not f(x,y)= 0, that's not the "change in elevation". Remember that the gradient always points in the direction of greatest increase and is perpendicular to the direction in which the derivative is 0.
 
That is what I thought. But I'm still lost as far as what to do here...
The gradient is just the partial derivatives with respect to x and y. If those were equal to zero then there would be no change in elevation, or am I understanding this completely wrong? :cry:
 
Am I onto something with my previous post, or am I getting no where? :cry:
 
ScienceMonkey said:
That is what I thought. But I'm still lost as far as what to do here...
The gradient is just the partial derivatives with respect to x and y. If those were equal to zero then there would be no change in elevation, or am I understanding this completely wrong? :cry:
Yes, you are completely wrong! If you take the gradient and set it equal to 0, then you would be finding the point at which the surface is "level"- going in any direction makes no change in elevation.

But here you are given the point (0.8,0.7,0.129) and are asked to find a direction in which there is no elevation change. As I said before, that is the direction that is perpendicular to the gradient. Find the gradient of the given function, at (0.8, 0.7) (z is the value of the function. The gradient is a 2 dimensional vector) and then determine a vector that is perpendicular to that. (Hint: a vector perpendicular to
ai+ bj is bi- aj since their dot product is 0.)
 
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