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Homework Help: Differential calculus problem

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data

    the height of a certain hill is given by


    wher y is the distance north and x is the distance east of south hadley

    a)Where is the top of the hill

    b) how high is the hill

    c) how steep is the hill at a point 1 mile north

    2. Relevant equations

    3. The attempt at a solution



    using linear system of algebra I find that my coordinates on top of the hill are


    b) just plug in the coordinates you got in part a, into the h function; answer: h(x=-26,y=-3)

    c) the slope would just be the gradient of h , which is: [tex]\nabla[/tex]h=(dh/dx)x-hat+(dh/dy)y-hat=-7x-hat+11y-hat


    direction is just: cos(theta)=[tex]\nabla[/tex]h dot dl/(
    |[tex]\nabla[/tex]h||dl|), you take the inverse of cos(theta)
  2. jcsd
  3. Aug 28, 2009 #2


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    In the problem it is given that the x is the distance east of south.
    Will it make any difference on the co-ordinates of the tip of the hill?
  4. Aug 28, 2009 #3


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    Your equations look correct, but your result (-26,-3) is not....check your algebra.

    Right idea

    Check your math on that again....you are missing a factor of 20(!) and your x-component is incorrect.
  5. Aug 28, 2009 #4


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    No, "x is the distance East of South Hadley"....South Hadley being the name of a town :wink:
  6. Aug 28, 2009 #5


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    To be fair, this isn't a very good characterization of the topography of the region. =( On what interval is this function valid? I get the feeling they're aiming for Mount Holyoke, simply because it's close to where the peak of the function is, though.
  7. Aug 29, 2009 #6

    I did check my math and I continue to come up with: [tex]\nabla[/tex]h=(y-3x-9)x-hat+(x-4*y+14)y-hat, with x=1,y=1
  8. Aug 29, 2009 #7


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    Look at your original expression for 'h'...you are missing the factor of 10 in front and also another factor of 2 that you seem to have divided by....in part (a) it didn't matter, because you were setting it equal to zero, but here it does matter.

    Also, when you plug in (1,1) to your above expression you do not get (-7,11).
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