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Hiker climbing, find the curve of projection of path on xy plane

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    a.) A hiker is climbing a mountain whose height is z = 1000 - 2x**2 - 3y**2. When he is at the point (1,1,995) in what direction should he move in order to ascent as rapidly as possible?
    b.) If he continues along a path of steepest ascent, obtain the equation of the curve which gives the projection of his path in the xy plane


    2. Relevant equations



    3. The attempt at a solution
    Part a is simply the gradient of the height equation which is (-4,-6) which simplifies to (-2,-3). What exactly do I need to find the equation of the projection of the path though? As I'm typing this I think it might involve a differential equation since I think the gradient I found might be the tangent of the projected curve, but I'm not sure. Any help is appreciated, Thanks.
     
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  3. Nov 28, 2011 #2

    Dick

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    Yes, the gradient is the tangent to the curve. And yes, finding the curve involves solving differential equations. Can you start it out?
     
  4. Nov 28, 2011 #3
    does it start out as dy/dx=-2x-3y?
     
  5. Nov 28, 2011 #4

    Dick

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    No. The gradient is a vector, isn't it? What vector is it? Think of the tangent vector as (dx/dt,dy/dt) where t is a time parameter.
     
  6. Nov 28, 2011 #5
    Ok one last question. So I set it up as dx/dt=-2x and dy/dt=-3y and solved to x=e**(-2t)*c and y=e**(-3t)*c respectfully. However do I need to solve for c in anyway and is it relevant since its in both equations? Could I just set c as 1 and be done with? The only reason I see that as a solution because then when t=0 I get (1,1) which was the only point I was given initially. Is this correct?
     
  7. Nov 28, 2011 #6

    Dick

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    Yes, you should reach the conclusion that both c's are 1. Now can you figure out a way to eliminate the t and write the curve only in terms of x and y?
     
  8. Nov 28, 2011 #7
    Well the current parametric equation should do, but I could solve for t in one equation and just substitute into the other. Thanks for all the help too
     
  9. Nov 28, 2011 #8

    Dick

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    You're welcome. But it's so easy to get rid of the parameter. You can tell x^3=y^2 just by looking at it, can't you?
     
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