How do I find the directional derivative?

In summary, the conversation discusses finding the directional derivative of the function f(x,y) = ln sqrt(x^2+y^2) at a point (x,y) that is not equal to (0,0) and towards the origin. The gradient of the function is given as grad f(x,y) = 1/(x^2+y^2)*(xi+yj), and the formula for finding the directional derivative is also mentioned. The solution to the problem is (x^2+y^2)^(-1/2).
  • #1
mnky9800n
4
0
Find the directional derivative of f(x,y)= ln sqrt(x^2+y^2) at (x,y) does not = (0,0) toward the origin


I found the gradient: grad f(x,y)=1/(x^2+y^2)*(xi+yj) but I have no idea what to do after that.
 
Physics news on Phys.org
  • #2
mnky9800n said:
Find the directional derivative of f(x,y)= ln sqrt(x^2+y^2) at (x,y) does not = (0,0) toward the origin


I found the gradient: grad f(x,y)=1/(x^2+y^2)*(xi+yj) but I have no idea what to do after that.

How are you given the direction? The directional derivative of f(x,y) in the direction of the unit vector [itex]\vec{u}[/itex] is [itex]\nabla f\cdot \vec{u}[/itex].

The directional derivative of of f(x,y) in the direction of any vector [itex]\vec{v}[/itex] can be done by finding a unit vector in that direction first: [itex]\vec{v}/|\vec{v}|[/itex] and is [itex]\nabla f\cdot \vec{v}/|\vec{v}|[/itex].

If the direction is at angle [itex]\theta[/itex] with the x-axis (whether the point at which you are finding the derivative is (0,0) or not), then a unit vector in that direction is [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex] and so the directional derivative is
[tex]\nabla f(x,y)\cdot (cos(\theta)\vec{i}+ sin(\theta)\vec{j})= \frac{\partial f}{\partial x}cos(\theta)+ \frac{\partial f}{\partial y}sin(\theta)[/tex]

Of course, if the angle [itex]\theta[/itex] is with the y-axis, just swap "sine" and "cosine".
 
  • #3
When re-examining the problem I solved for grad f*u/|u| which gives (xi+yj)/[sqrt(x^2+y^2)]*[1/(x^2+y^2)*(xi+yj)] which when solved gives: (x^2+y^2)^(-1/2) which I believe is the correct answer.
 

1. What is the formula for calculating the directional derivative?

The formula for calculating the directional derivative is given by Dvf(x,y) = ∇f(x,y) ⋅ v, where Dvf(x,y) represents the directional derivative in the direction of the unit vector v, ∇f(x,y) represents the gradient of the function f(x,y), and ⋅ represents the dot product.

2. How do I find the unit vector in the desired direction?

The unit vector in the desired direction can be found by dividing the vector representing the direction by its magnitude. This will give a vector with the same direction but with a magnitude of 1, making it a unit vector.

3. Can the directional derivative be negative?

Yes, the directional derivative can be negative. It represents the rate of change of a function in a specific direction, so it can be positive, negative, or zero depending on the direction and the function itself.

4. What does the directional derivative tell us?

The directional derivative tells us the rate of change of a function in a specific direction. It can be interpreted as the slope of a tangent line to the function's graph in the direction of the unit vector v.

5. Are there any practical applications of directional derivatives?

Yes, directional derivatives have practical applications in fields such as physics, engineering, and economics. They can be used to optimize functions, analyze motion and forces, and predict changes in a system based on a specific direction.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
518
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
6
Views
505
Replies
9
Views
661
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
636
  • Calculus and Beyond Homework Help
Replies
8
Views
829
Replies
7
Views
466
  • Calculus and Beyond Homework Help
Replies
2
Views
459
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top