Gradient Vector Problem: Steepness and Slope Direction?

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The discussion revolves around solving a gradient vector problem related to the steepness and direction of a hill defined by a specific elevation equation. For part (a), the gradient was calculated as ∇f(x,y,z) = 0.5i - 0.25j, and there was confusion about finding the angle between this gradient and the z-axis. It was clarified that the angle of steepness can be determined using the relationship between the gradient and the change in elevation. For part (b), the gradient indicates the direction of the steepest ascent, and it was suggested that understanding the gradient's components can help determine the slope's direction. The conversation emphasizes the importance of correctly interpreting the gradient in relation to the problem's requirements.
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Homework Statement


For a hill the elevation in meters is given by z=10 + .5x +.25y + .5xy - .25x^2 -.5y^2, where x is the distance east and y is the distance north of the origin.

a.) How steep is the hill at x=y=1 i.e. what is the angle between a vector perpendicular to the hill and the z axis?

b.)In which compass direction is the slope at x=y=1 steepest? Indicate whether the angle you provide is the angle measured in the standard way in the counter-clockwise direction from the x-axis (east) or whether it is the compass azimuth.



Homework Equations





The Attempt at a Solution


a.) So I got the gradient to be ∇f(x,y,z)= .5i - .25j...simply by the definition of the gradient. So i figured to figure the angle between the gradient and the z axis i could use the formula cosθ= (A dotted into B)/(lAl*lBl) but wouldn't be the z axis be the unit vector in the k hat direction making the cosθ=0...which can't be right. Any help on what to do? Also, I was given an equation that says ∇f(x,y,z) perpendicular to a 3 dimensional surface f(x,y,z) = constant. But I don't know how to use that equation to generate an answer.

b.)The gradient gives you the direction of the quickest altitude ascension (at least for this problem) so how do I use what I calculated for the gradient to answer this part?

Thanks for any help!
 
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bmb2009 said:
a.) So I got the gradient to be ∇f(x,y,z)
Think of the altitude as a potential in 2 dimensions, just x and y. So ∇ is ?
 
grad z gives you the x-y direction of steepest ascent, but now you have to ask yourself: when going in that direction, how much does dz change when you go a distance ds in the aforederived direction.

Think about how the x and y components of the gradient at your point (1,1) tell you what dz/ds would be.

P.S. the angle the problem asks for happens to be (90 deg - the steepness angle). Also, once you did (a) you already have (b). In fact, (b) should come first ... at least the way I look at it. Probably you were not supposed to solve for grad z for part (a) ... I don't know.
 
Dear Rude Man,

Very nice job of doping it out. That's exactly the way I would have done it, including doing part (b) first.

Chet
 
Chestermiller said:
Dear Rude Man,

Very nice job of doping it out. That's exactly the way I would have done it, including doing part (b) first.

Chet

Thanks Chet.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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