Directional derivatives for altitude

Click For Summary
SUMMARY

The discussion focuses on finding the direction to maintain a constant altitude while standing at the point (-100, -100, 360) on a hill described by the equation z = 500 - 0.006x² - 0.008y². The key concept is that the directional derivative must equal zero, which leads to the calculation of the gradient vector ∇f = < -0.012x, -0.016y >. At the specified point, this results in ∇f = < 1.2, 1.6 >. The solution involves finding a unit vector u, such as u = < 1, -0.75 >, that satisfies the condition of the directional derivative being zero.

PREREQUISITES
  • Understanding of directional derivatives and their significance in multivariable calculus.
  • Familiarity with gradient vectors and their computation.
  • Knowledge of vector operations, particularly dot products.
  • Basic skills in solving equations involving vectors and magnitudes.
NEXT STEPS
  • Study the properties of directional derivatives in multivariable functions.
  • Learn how to compute gradient vectors for different functions.
  • Explore unit vectors and their applications in directional movement.
  • Investigate angular bearings and their relevance in directional calculations.
USEFUL FOR

Students in calculus, particularly those studying multivariable calculus, as well as educators and anyone interested in the applications of gradient vectors and directional derivatives in real-world scenarios.

reddawg
Messages
46
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
22
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K