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Math Help - slope of hill

  1. #1
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    slope of hill

    The height of a certain hill (in meters) is given by
    h(x,y)
    = 10 ( 2x y - 3 x^2- 4 y^2- 18 x + 28 y + 12),
    where
    y is the distance (in kilometers) North, x the distance East

    (a) How steep is the slope (in meters per kilometer) at a point 1 km north and 1 km east and in what direction is the slope the steepest at that point?

    I know this has something to do with partial derivatives, but I am not 100% sure of the step proces..

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  2. #2
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    Quote Originally Posted by sk1001 View Post
    The height of a certain hill (in meters) is given by


    h(x,y)
    = 10 ( 2x y - 3 x^2- 4 y^2- 18 x + 28 y + 12),
    where y is the distance (in kilometers) North, x the distance East

    (a) How steep is the slope (in meters per kilometer) at a point 1 km north and 1 km east and in what direction is the slope the steepest at that point?


    I know this has something to do with partial derivatives, but I am not 100% sure of the step proces..

    (a) \nabla f \cdot \hat{l} evaluated at (1, 1) where \hat{l} is a unit vector in the given direction.

    (b) \nabla f evaluated at (1, 1).

    I suggest you revise your notes on the directional derivative.
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  3. #3
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    I've calculated grad (http://www.mathhelpforum.com/math-he...fb9e51b8-1.gif), should I dot with the unit vector before evaluating, or should I evaluate grad at 1,1 then dot with unit vector?
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  4. #4
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    Quote Originally Posted by sk1001 View Post
    I've calculated grad (http://www.mathhelpforum.com/math-he...fb9e51b8-1.gif), should I dot with the unit vector before evaluating, or should I evaluate grad at 1,1 then dot with unit vector?
    I'd first evaluate \nabla f at (1, 1) ....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    I'd first evaluate \nabla f at (1, 1) ....
    I tried and returned an answer of 0, somehow I'm not convinced this is correct!

    I found:
    partial derivative of h with respect to x to be 20y - 60x - 180
    and
    partial derivative of h with respect to y to be 20x - 80y + 280

    Evaluated at (1,1) these are -220 and 220 respectively.

    Then the at (1,1) is (1/sqrt(2), 1/sqrt(2)).

    Dotting gives (-220)(1/sqrt(2)) + (220)(1/sqrt(2)) yielding 0!

    Where have I gone wrong?
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