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Need with 3d graph question

  1. Nov 28, 2003 #1
    z=f(x,y)= (cos y) - (cos x)

    u are at the origin and want to hike to point (4pie,0,0). U want to get to 4pie,0,0 by hiking a route that alway keeps you at the same elevation. what is the lenght of this route?

    I ploted the graph and I already know the route that will keep me at the same elevation from origin to 4pie,0,0 but how do I find the lenght of that route??

    As for the second part

    Your second route always moves along the gradient. Determine such a route of minimal lenght, assuming you start hiking in the positive x direction. What is the length? If you cannot find and exact answer, then determine an upper bound and a lower bound bbetween which the actual lenght must lie.

    I have no clue on doing the second part.
    I need some serious help because this will determine my final grade so be clear and help me as much as u can. Thanks
  2. jcsd
  3. Nov 28, 2003 #2
    I edited my response to be less helpful, since this appears to be not just ordinary homework, but a take-home final exam or something.

    You use the arc length formula. I'm sure you can find it in your notes or textbook.

    If you travel along some parametrized path (x(t),y(t)), then you'll have to solve the differential equations,

    [tex]\nabla z \propto \dot{x} \hat{x} + \dot{y} \hat{y}[/tex]

    subject to the initial condition [itex]\dot{y}_{|t=0} = 0[/itex].
    Last edited: Nov 28, 2003
  4. Nov 28, 2003 #3
    it is an ec assigment that worth quite a lot, not a take home exam.

    why do u have to use the arc formula? The path is not on the arc but on the path between the circle.

    My teacher told me to do integration to get the answer. I haven't done differential equation yet so...

    don't be shy with giving hints :wink:
    Last edited: Nov 28, 2003
  5. Nov 28, 2003 #4
    "Arc" means the same thing as "path", in this context: the arc-length formula gives the length of any path.

    You solve the differential equation by integrating. If you write down the relation that [itex]\dot{x}[/itex] is proportional to the x component of [itex]\nabla f[/itex] (by a possibly position-dependent proportionality factor), and similarly for [itex]\dot{y}[/itex] (by the same factor), then you'll get some equations involving derivatives of x and y which you will have to integrate to get x and y. (I'd recommend first trying t=x to simplify the parametrization.)
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