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Sketch the gradient vector for the function

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Untitled.png


    2. Relevant equations



    3. The attempt at a solution
    Ok so I think I know how to get the direction. It's going to be perpendicular to the tangent of the level curve and pointing in the direction where f(x,y) is increasing. So on the graph that was provided it will point in negative y and positive x.

    I am completely lost on finding the length though.
     
  2. jcsd
  3. Mar 1, 2014 #2

    Dick

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    The length of the gradient is going to proportional to the rate at which f(x,y) is increasing as you move along the direction of the gradient. Can't you approximate that by the distance between the level curves and the amount f(x,y) increases between them?
     
  4. Mar 1, 2014 #3
    The distance between level curves -2 and -1 = 1

    Isn't that how much f(x,y) increases?
     
  5. Mar 1, 2014 #4

    Dick

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    Right. So the magnitude of the gradient is going be approximately 1/(distance between the level curves in the direction of the gradient). The denominator will depend on where you are on the level curves.
     
  6. Mar 1, 2014 #5
    Ok so in a simple rise/run, that means 1 is my rise, what is my run?

    My professor has the right answer as "approx 2".
     
  7. Mar 1, 2014 #6

    Dick

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    At (4,6)? The run is the perpendicular distance between the level curves at that point. What's that "approx"?
     
  8. Mar 1, 2014 #7
    I guess it stands for "approximately".

    This just isn't working out at all.. I'm so confused :/
     
  9. Mar 1, 2014 #8

    Dick

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    I guessed what "approx" meant, I was just asking you what your guess for the run was around the point (4,6). What's your guess for the distance between the level curves -1 and -2 near that point by reading off the graph? It is just a rise/run guesstimate.
     
  10. Mar 1, 2014 #9
    Well my guess is that the rise is 1.

    As for the run that's where I get lost. Do i estimate in the same direction as the gradient, or in the X or Y direction? What's the deal here?
     
  11. Mar 1, 2014 #10

    Dick

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    In the same direction as the gradient, i.e. perpedicular to the level curves. Looks to me like about 1/2 is a good guess around (4,6). Don't you agree?
     
  12. Mar 1, 2014 #11
    Ah it makes sense in that case. I would have guessed that it would be in that direction too but I wasn't sure. I guess I'm just not really grasping the concept. I realize that a gradient is just essentially a derivative in any direction but for some reason I'm having a hard time wrapping my head around this.
    Anyway, thanks a bunch for your patience!
     
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