Finding direction where the rate of change is fastest

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Homework Help Overview

The problem involves a climber navigating the surface of a mountain, modeled by a specific function, to reach a dropped bottle at a lower elevation. The climber's current position and the bottle's position are provided, along with the need to determine the fastest direction to descend to a specific elevation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the concept of the gradient and its implications for direction and rate of change. There is confusion regarding the calculation of the gradient and how to interpret its direction in relation to the problem context.

Discussion Status

Some participants have provided insights into the nature of the gradient as a vector and its significance in determining direction. There is ongoing exploration of how to express direction, with multiple interpretations being considered regarding the gradient's components and their relation to compass directions.

Contextual Notes

Participants are navigating the complexities of three-dimensional gradients and their application to a two-dimensional plane, with some ambiguity regarding the desired form of direction (e.g., unit vector vs. compass direction).

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Homework Statement


surface of mountain is modeled by h(x,y)=5000-0.001x^2-.004y^2. Climber is at (500,300,4390) and he drops bottle which is at (300,450,4100)
What is his elevation rate of change if he heads for the bottle?
In what direction should he proceed to reach z=4100 feet the fastest, so that he can be on a constant elevation to reach his bottle?

Homework Equations





The Attempt at a Solution


i solved for first question and got -.334 where i just found gradient and multiplied by unit vector
and for the 2nd question I am little confused.
 
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The gradient is a vector and contains two pieces of information- its "length" and its direction. In which direction does the gradient here point?
 
is the direction just gradient of (500,300) which is <-1,-2.4>
or do i find the difference of the 2 points and find gradient of (200,-150) = <-.6,-1.2>
 
I'm sorry, I assumed from your saying that you had found the gradient that you knew what "gradient" meant!

The gradient of the function f(x,y,z) is the vector
[tex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/tex]
Isn't that what you did? The "direction" you want to go is the direction of that vector. It may be, what you wrote was ambiguous, that you want a compass direction, not including the "downward" part. If that is the case, find the angle that
[tex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}[/tex] makes with "north", the positive y-axis.
 
HallsofIvy said:
I'm sorry, I assumed from your saying that you had found the gradient that you knew what "gradient" meant!

The gradient of the function f(x,y,z) is the vector
[tex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/tex]
Isn't that what you did? The "direction" you want to go is the direction of that vector. It may be, what you wrote was ambiguous, that you want a compass direction, not including the "downward" part. If that is the case, find the angle that
[tex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}[/tex] makes with "north", the positive y-axis.

yea i did found the gradient which was <-1,-2.4> i just wasnt sure how to get direction from this.
 
What do you mean by "direction"? A direction in three dimensions is given by either a unit vector or by the "direction cosines" (the cosines of the angles a line in that direction makes with each coordinate axis which is the same as the components of the unit vector).

But, as I said, you may want the two dimensional compass direction the person should take.
 

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