Hiker climbing, find the curve of projection of path on xy plane

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Homework Help Overview

The problem involves a hiker climbing a mountain described by the equation z = 1000 - 2x² - 3y². The discussion focuses on determining the direction of steepest ascent and finding the projection of the hiker's path onto the xy plane.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the gradient of the height equation and its relation to the tangent vector of the projected curve. There are attempts to set up differential equations based on the gradient, with questions about the relevance of constants in the solutions.

Discussion Status

The discussion is ongoing, with participants providing guidance on setting up equations and exploring the relationship between the parametric equations. There is an acknowledgment of the need to eliminate the parameter to express the curve in terms of x and y.

Contextual Notes

Participants are navigating the implications of the gradient and its representation as a vector, as well as the initial conditions provided by the problem statement.

MeMoses
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Homework Statement


a.) A hiker is climbing a mountain whose height is z = 1000 - 2x**2 - 3y**2. When he is at the point (1,1,995) in what direction should he move in order to ascent as rapidly as possible?
b.) If he continues along a path of steepest ascent, obtain the equation of the curve which gives the projection of his path in the xy plane


Homework Equations





The Attempt at a Solution


Part a is simply the gradient of the height equation which is (-4,-6) which simplifies to (-2,-3). What exactly do I need to find the equation of the projection of the path though? As I'm typing this I think it might involve a differential equation since I think the gradient I found might be the tangent of the projected curve, but I'm not sure. Any help is appreciated, Thanks.
 
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Yes, the gradient is the tangent to the curve. And yes, finding the curve involves solving differential equations. Can you start it out?
 
does it start out as dy/dx=-2x-3y?
 
MeMoses said:
does it start out as dy/dx=-2x-3y?

No. The gradient is a vector, isn't it? What vector is it? Think of the tangent vector as (dx/dt,dy/dt) where t is a time parameter.
 
Ok one last question. So I set it up as dx/dt=-2x and dy/dt=-3y and solved to x=e**(-2t)*c and y=e**(-3t)*c respectfully. However do I need to solve for c in anyway and is it relevant since its in both equations? Could I just set c as 1 and be done with? The only reason I see that as a solution because then when t=0 I get (1,1) which was the only point I was given initially. Is this correct?
 
MeMoses said:
Ok one last question. So I set it up as dx/dt=-2x and dy/dt=-3y and solved to x=e**(-2t)*c and y=e**(-3t)*c respectfully. However do I need to solve for c in anyway and is it relevant since its in both equations? Could I just set c as 1 and be done with? The only reason I see that as a solution because then when t=0 I get (1,1) which was the only point I was given initially. Is this correct?

Yes, you should reach the conclusion that both c's are 1. Now can you figure out a way to eliminate the t and write the curve only in terms of x and y?
 
Well the current parametric equation should do, but I could solve for t in one equation and just substitute into the other. Thanks for all the help too
 
MeMoses said:
Well the current parametric equation should do, but I could solve for t in one equation and just substitute into the other. Thanks for all the help too

You're welcome. But it's so easy to get rid of the parameter. You can tell x^3=y^2 just by looking at it, can't you?
 

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