Gradient Vector Problem: Steepness and Slope Direction?

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Homework Help Overview

The problem involves determining the steepness and slope direction of a hill described by a given elevation function in terms of two variables, x and y, representing distances in a coordinate system. The specific questions focus on calculating the angle between a vector perpendicular to the hill and the z-axis, as well as identifying the compass direction of the steepest slope at a particular point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the gradient and its implications for understanding the steepness of the hill. There are questions about how to relate the gradient to the angle with the z-axis and how to interpret the results for the steepest slope direction.

Discussion Status

Some participants have provided insights into the relationship between the gradient and the steepness of the hill, suggesting that the angle of steepness can be derived from the gradient. There is an ongoing exploration of how to approach the problem, with some participants indicating that the order of solving parts (a) and (b) could be reconsidered.

Contextual Notes

Participants are navigating the complexities of applying the gradient in a three-dimensional context and are questioning the assumptions related to the definitions of steepness and direction in relation to the coordinate axes.

bmb2009
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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.
 

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