Directional derivatives for altitude

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To maintain a constant altitude at the point (-100, -100, 360) on the hill defined by the equation z=500-0.006x²-0.008y², the directional derivative must be zero. This requires finding a vector u such that the dot product of the gradient vector ∇f and u equals zero. The gradient vector is calculated as ∇f = <1.2, 1.6> at the given coordinates. A valid direction for u can be <1, -0.75>, which satisfies the condition for a zero directional derivative. The discussion emphasizes that multiple solutions exist, all pointing in similar directions, and u should be a unit vector for clarity.
reddawg
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Homework Statement


Suppose you are standing at the point (-100,-100,360) on a hill that has the shape of the graph of z=500-0.006x2-0.008y2.

In what direction should you head to maintain a constant altitude?

Homework Equations


Duf = ∇f\bulletu
formula for directional derivative


The Attempt at a Solution


Here's my idea. If the directional derivative at the point specified is zero, then that's equivalent to maintaining constant altitude. Therefore, I must find the corresponding gradient vector ∇f when the directional derivative is zero right? I have no idea how to carry this out however.
 
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reddawg said:

Homework Statement


Suppose you are standing at the point (-100,-100,360) on a hill that has the shape of the graph of z=500-0.006x2-0.008y2.

In what direction should you head to maintain a constant altitude?

Homework Equations


Duf = ∇f\bulletu
formula for directional derivative


The Attempt at a Solution


Here's my idea. If the directional derivative at the point specified is zero, then that's equivalent to maintaining constant altitude. Therefore, I must find the corresponding gradient vector ∇f when the directional derivative is zero right? I have no idea how to carry this out however.

Yes, find the gradient vector. But you can't choose what it is, it's fixed by the problem statement. What you need to find is a vector u such that the directional derivative is zero.
 
Ok, I'm confused on how to go about this.

0 = ∇f\bulletu

You say ∇f is fixed:

∇f = <zx,zy>

= < -0.012x, -0.016y >

= < 1.2, 1.6 > Subs. coordinates given

Then solve for u?
 
reddawg said:
Ok, I'm confused on how to go about this.

0 = ∇f\bulletu

You say ∇f is fixed:

∇f = <zx,zy>

= < -0.012x, -0.016y >

= < 1.2, 1.6 > Subs. coordinates given

Then solve for u?

Yes, write u=<ux, uy>. Since the dot product of u with the gradient is 0 you should be able to find a relation between ux and uy. You won't be able to find a unique u. But the question is only asking for a direction.
 
So u could just equal < 0 , 0 > ?
 
reddawg said:
So u could just equal < 0 , 0 > ?

It could. That means you just stay in the same place and don't go anywhere. But the question is asking for a direction in which you could move and not change altitude. Find another solution.
 
Oh I get it.

I'll use u = < 1 , -0.75 > , that comes out to zero.

Thanks Dick.
 
reddawg said:
Oh I get it.

I'll use u = < 1 , -0.75 > , that comes out to zero.

Thanks Dick.

Right and you're welcome. Actually there are lots of solutions but they all point in either that direction or the opposite direction.
 
u is supposed to be a unit vector. So take your u and divide by its magnitude.
 
  • #10
Chestermiller said:
u is supposed to be a unit vector. So take your u and divide by its magnitude.

Or they might want an angular bearing. Hard to say. reddawg might know based on previous questions.
 

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