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

1. Nov 28, 2011

MeMoses

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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.

2. Nov 28, 2011

Dick

Yes, the gradient is the tangent to the curve. And yes, finding the curve involves solving differential equations. Can you start it out?

3. Nov 28, 2011

MeMoses

does it start out as dy/dx=-2x-3y?

4. Nov 28, 2011

Dick

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. Nov 28, 2011

MeMoses

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. Nov 28, 2011

Dick

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. Nov 28, 2011

MeMoses

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. Nov 28, 2011

Dick

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?