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

In summary: Yes, that's true. So the curve in the xy plane is given by the equation x^3=y^2. In summary, the conversation discusses a hiker climbing a mountain and determining the direction to ascent as rapidly as possible. It also involves finding the equation of the projected path in the xy plane using differential equations. The solution involves solving for the constants and eliminating the parameter to obtain the equation x^3=y^2.
  • #1
MeMoses
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.
 
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  • #2
Yes, the gradient is the tangent to the curve. And yes, finding the curve involves solving differential equations. Can you start it out?
 
  • #3
does it start out as dy/dx=-2x-3y?
 
  • #4
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.
 
  • #5
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
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?
 
  • #7
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
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?
 

1. What is the curve of projection of the hiker's path on the xy plane?

The curve of projection of the hiker's path on the xy plane is the two-dimensional representation of the hiker's actual path in three-dimensional space. It is the shadow of the hiker's path when projected onto the xy plane.

2. How is the curve of projection calculated?

The curve of projection is calculated by taking the x and y coordinates of the hiker's position at different time intervals and plotting them on the xy plane. The points are then connected to create a continuous curve.

3. What factors can affect the shape of the curve of projection?

The shape of the curve of projection can be affected by factors such as the terrain and elevation of the hiker's path, the hiker's speed and direction of movement, and any obstacles or changes in direction along the path.

4. Can the curve of projection be used to determine the hiker's speed or distance traveled?

No, the curve of projection only shows the hiker's path in two dimensions and does not provide information about the hiker's speed or distance traveled. These would need to be calculated using other methods, such as measuring the time it takes the hiker to reach specific points along the path.

5. How can the curve of projection be used in hiking or outdoor activities?

The curve of projection can be used to visualize the hiker's path and identify any potential obstacles or challenges along the way. It can also be used to plan and track the hiker's progress and compare different routes or hiking strategies.

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