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Classical Dynamics prob, need help please.

  1. Oct 2, 2005 #1
    The question is as follows: The height of a hill (meters) is given by [z=(2xy)-(3x^2)-(4y^2)-(18x)+(28y)+12], where x is the distance east, y is the distance north of the origin. a). where is the top of the hil (x,y,z) and how high is it (z=?)? b). How steep is the hill at x=y=1, that is, what is the angle between a vector perpendicular to the hill and the z axis? c). In which compass direction is the slope at x=y=1 the steepest?

    okay, i know part a)., which is to get the derivative of dz/dx and dz/dy to get your x and y maximum values, then plug them back into the equation for z, giving you the location and the height of the hill at the steepest point.

    The part i am having trouble with is b). how am i supposed to draw a vector that is perpendicular to the hill and calculate hte angle it makes with the z-axis? i thought of plugging in the x and y values of x=y=1 into the z equation, and this would give you z at that point, then create a vector normal to that point, but how do u find the angle (theta) made with the z-axis.

    I also figured out part c). which is in a south east direction.

    Please help, homework is due soon, thank you. :surprised
  2. jcsd
  3. Oct 3, 2005 #2


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    Think of f(x,y,z)= constant (here that would be (2xy)-(3x^2)-(4y^2)-(18x)+(28y)+12- z= 0) as "level surface" for f. Then the gradient of f (the vector whose components are fx, fy, fz) is perpendicular to the surface (evaluated at (1,1,37)). You can find the angle it makes with the z-axis by taking its dot product with (0, 0, 1).
  4. Oct 5, 2005 #3
    thank u :smile:
  5. Aug 30, 2007 #4
    this is a major bump, but it is similar to a problem i have. would someone else explain this?

    where did (1,1,37) come from?

    i think it is easy, but i am just drawing a huge blank...

    and then the gradient of f is (fx, fy, fz) where fx is your dz/dx you needed to find to solve part a? and fy is your dz/dy, then fz would be 1? or am i totally wrong.

    thanks for any help! I am new to this site, it is a great help
  6. Sep 12, 2007 #5
    In this case, you don't have to find fz. Z itself is f(x,y). The value for z comes from plugging in the values for x and y into the equation which gives you z.
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