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  1. #1
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    gradient

    The height of a hill (in feet) is given by  h(x,y) = 10(2xy - 3x^2-4y^2-18x+28y+12) , where  y is the distance north,  x is the distance east of South Hadley.

    (a) Where is the top of the hill located.

    So  \nabla h = (20y - 60x - 180)\bold{\hat{x}} + (20x-80y + 280) \bold{\hat{j}} . Then what? Set it equal to  0 ?



    (b) How high is this hill?
    This is the magnitude of  \nabla h ?



    (c) How steep is the slope (in feet per mile) at a point  1 mile north and one mile east of South Hadley? In what direction is the slope steepest at that point?

    Plug in value in gradient and find magnitude?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    The height of a hill (in feet) is given by  h(x,y) = 10(2xy - 3x^2-4y^2-18x+28y+12) , where  y is the distance north,  x is the distance east of South Hadley.

    (a) Where is the top of the hill located.

    So  \nabla h = (20y - 60x - 180)\bold{\hat{x}} + (20x-80y + 280) \bold{\hat{j}} . Then what? Set it equal to  0 ?

    Mr F says: Yes. I get x = 2, y = 15.


    (b) How high is this hill?
    This is the magnitude of  \nabla h ?

    Mr F says: Calculate h(2, 15).


    (c) How steep is the slope (in feet per mile) at a point  1 mile north and one mile east of South Hadley? In what direction is the slope steepest at that point?

    Mr F says: The first part doesn't make sense without specifying the direction you want to calculate the steepness. In which case you'd calculate the directional derivative at (1, 1).

    For the second part, note that \nabla h is the direction of maximum steepness.


    Plug in value in gradient and find magnitude?
    ..
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