Classical Dynamics prob, please.

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The discussion centers on solving a problem involving the height of a hill defined by a mathematical equation. For part a), the approach involves calculating the derivatives to find the maximum height and coordinates. Part b) requires determining the angle between the normal vector to the hill and the z-axis, which can be found using the gradient of the function. The steepest slope direction at the point (1,1) is identified as southeast. Clarifications are sought regarding the gradient components and the origin of the coordinates (1,1,37).
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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.
 
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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).
 
thank u :smile:
 
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
 
SEMonChron said:
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

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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