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Rate of change on a hill

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose you are climbing a hill whose shape is given by the equation:
    [itex] z = 1400 − 0.005x^2 − 0.01y^2 [/itex]
    where x, y, and z are measured in meters, and you are standing at a point with coordinates (120, 80, 1264). The positive x-axis points east and the positive y-axis points north.

    In which direction is the slope largest?
    What is the rate of ascent in that direction?
    At what angle above the horizontal does the path in that direction begin? (Round your answer to two decimal places.)

    2. The attempt at a solution

    The direction of the gradient is the direction in which the slope is largest
    We know that
    [tex] \nabla z = (-0.01,-0.02) [/tex]
    [tex] \nabla z(120,80) = (-1.2,-1.6) [/tex]

    The rate of ascent at this direction would be given as:
    [tex] \sqrt{1.2^2+1.6^2} = 2 [/tex]

    and this given a corresponding angle of [itex] \tan^{-1} 2 = 63.43 [/itex] degrees. Is this correct?

    The system into which I have put in this answer says im wrong :(
     
  2. jcsd
  3. Apr 13, 2013 #2

    haruspex

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    Looks right to me. Does it reject all three answers or only certain ones?
     
  4. Apr 13, 2013 #3


    It worked out in the end. i was calculating it with 1.8 instesd of 1.6 silly me!

    But thanks for your confirmation!
     
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