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Vector Calc - Directional Derivative Question

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Igor, the inchworm, is crawling along graph paper in a magnetic field. The intensity of the field at the point (x,y) is given by M(x,y) = 4x^2 + y^2 + 5000. If Igor is at the point (8,6), describe the curve along which he should travel if he wishes to reduce the field intensity as rapidly as possible.


    2. Relevant equations
    Gradient M = <8x, 2y>
    To minimize, one must travel opposite the gradient

    3. The attempt at a solution
    I got the gradient as <8x, 2y> (just the partial derivatives as a vector). Also, at (8,6), the gradient is <64, 12>, which is in the direction of the unit vector 16i + 3j/sqrt(16^2 +3^2). So, the direction he wants to travel is the negative version of that vector. However, the problem wants me to arrive at the formula (y^3 = 27x towards the origin). How do I go from the vector to the curve?

    This TEX is driving me crazy. If anyone can help me by editing my post, that would be great.

    Thanks in advance!
     
  2. jcsd
  3. Oct 4, 2007 #2

    HallsofIvy

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    Science Advisor

    You are correct that <64, 12> gives the direction he wants to go at that time. But as soon as he has moved slightly it will change. What you are saying is that he wants to move so that dx/dt= -8x and dy/dt= -2y. It's not difficult to solve both of those for x and y as functions of t but you don't need the time dependence- you only need the curve itself:
    [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{1}{4}\frac{y}{x}[/tex].
    That's a simple separable equation. Separate the variables and integrate.
     
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