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

  • Thread starter MeMoses
  • Start date
  • #1
129
0

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
619
Yes, the gradient is the tangent to the curve. And yes, finding the curve involves solving differential equations. Can you start it out?
 
  • #3
129
0
does it start out as dy/dx=-2x-3y?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
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.
 
  • #5
129
0
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?
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
619
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?
 
  • #7
129
0
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
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
619
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?
 

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

  • Last Post
Replies
2
Views
5K
Replies
5
Views
600
  • Last Post
Replies
5
Views
1K
Replies
3
Views
926
Replies
7
Views
1K
  • Last Post
Replies
1
Views
30K
Replies
1
Views
4K
Replies
3
Views
2K
Replies
8
Views
8K
Top