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Directionnal Derivative

  1. Oct 11, 2006 #1
    This is a problem thats been bugging me. Suppose you are on the surface M(x,y)=3*x^2+y^2+5000 that describes a magnetic field, you are at the point (8,6), what is the curve along which you should travel so as to minimize field intensity as rapidly as possible? I am told I need to solve a differential equation, I am thinking I need to find the minimal directional derivative and somehow from that find the equation of the line that represents, and integrate it to get a curve. But I am at a loss as to how to start. Someone help?
     
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  3. Oct 11, 2006 #2

    quasar987

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    You know that [itex]-\nabla M(x,y)[/itex] is the direction you need to go into in order for your field to decrease the most rapidly. And given a parametrized curve [itex]\gamma (t) = (x(t),y(t))[/itex], the speed [itex]\dot{\gamma}[/itex], is the direction in which the curve is "going". So if you'd have a curve whose speed is [itex]-\nabla M(x,y)[/itex], you'd have a curve that follows the path of fastest decrease, would you not?
     
  4. Oct 11, 2006 #3
    Sorry I don't quite follow, the negative of the directional derivative is the directional I need to go, but what is the parametrized curve? Is the equation of setting up -M(x,y)=(x(t),y(t))?
     
  5. Oct 11, 2006 #4

    quasar987

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    A curve and a "parametrized curve" are practically the same thing. A curve is a function (typically smooth) [itex]\gamma : (t_1, t_2)\rightarrow \mathbb{R}^2[/itex], [itex]\gamma(t)=(x(t),y(t))[/itex].

    This is a what a curve is. The other way to define a curve is as a "level curve", i.e. by finding a function f(x,y) such that f(x,y)=constant gives a certain relation btw x and y.
     
  6. Oct 11, 2006 #5

    quasar987

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    And just as the derivative of a single valued function f(x) gives the tangant to the curve, the derivative of the curve, typically noted [itex]\dot{\gamma}[/itex]*, gives the tangeant to the curve, i.e. the direction where the curve is "heading".


    *[tex]\dot{\gamma} =\frac{d}{dt}\gamma(t) = \left(\frac{dx(t)}{dt},\frac{dy(t)}{dt} \right)[/tex]
     
    Last edited: Oct 11, 2006
  7. Oct 11, 2006 #6
    I am still a bit confused but I have something like this - (xn,yn) = (8,6)+[(-3x,2y)/sqrt(36x^2+4y^2)]*[v,w] where (xn,yn) is the new point after moving some distance, [v,w] the unit vector and [(-3x,2y)/sqrt(36x^2+4y^2)] the gradient divided by its length to give you the unit gradient, but how do I solve this?
     
  8. Oct 11, 2006 #7

    quasar987

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    I don't understand what you've done here. What do you not understand about the method I suggested?
     
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