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Simple Gradient Question (funtion of two variables)

  • Thread starter xWaffle
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  • #1
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Homework Statement


Find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point.

g(x,y) = x2/2 - y2/2; (√2, 1)


Homework Equations


∇f = (∂f/∂x)i + (∂f/∂y)j


The Attempt at a Solution


∇g = <x, -y>
∇g(√2, 1) = <√2, -1>
_______________________

Am I done with this solution? Or is there more I need to put for the gradient at the point?

I'm not really sure if this needs to be reduced to a single number or not, I'm guessing that has to do with my lack of understand of the gradient itself. I thought I was supposed to have a direction vector, is it implied the direction vector is just u = i + j if it is not specified?

I also have no idea how to draw what it's asking. I can't visualise any of this.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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No, that is the gradient.

You are confusing this with the derivative "in the direction of unit vector v". That is equal to [itex]v\cdot \nabla f[/itex].
 
  • #3
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Yes, the gradient you calculated is correct.

In order to visualize the gradient in your problem,
depict a two dimensional graph and pick out arbitrary (x,y) points i.e. (2, 3) (0,1) (-1,0) (-1,-1) … and plug those into ∇(g). Use these corresponding vectors to depict the vector field. For example, at (1,0) the resulting vector would be <1,0>. Likewise…

(2,3) <2, -3>
(0,1) <0,-1>
(-1,0) <-1,0>
(-1,-1) ….
…. ….


To draw each vector, begin with the tail at the chosen (x,y) coordinates and move accordingly. For the first vector, begin at (2,3) then move in the positive x direction 2 units, then move down the negative y direction 3 units – there’s your endpoint. As you draw more vectors out, you should start to get an idea of the field’s behavior.
 

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