# Directional derivatives

1. Feb 24, 2014

### reddawg

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

2. Relevant equations
Duf = ∇f$\bullet$u
formula for directional derivative

3. 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.

2. Feb 24, 2014

### Dick

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.

3. Feb 24, 2014

### reddawg

0 = ∇f$\bullet$u

You say ∇f is fixed:

∇f = <zx,zy>

= < -0.012x, -0.016y >

= < 1.2, 1.6 > Subs. coordinates given

Then solve for u?

4. Feb 24, 2014

### Dick

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.

5. Feb 24, 2014

### reddawg

So u could just equal < 0 , 0 > ?

6. Feb 24, 2014

### Dick

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.

7. Feb 24, 2014

### reddawg

Oh I get it.

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

Thanks Dick.

8. Feb 24, 2014

### Dick

Right and you're welcome. Actually there are lots of solutions but they all point in either that direction or the opposite direction.

9. Feb 24, 2014

### Staff: Mentor

u is supposed to be a unit vector. So take your u and divide by its magnitude.

10. Feb 24, 2014

### Dick

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