Directional Derivative at an Angle with a 3d Gradient

In summary, the directional derivative using ##f\left(x,y,z\right)=xy+z^2## at the point (4, 2, 1) in the direction of a vector making an angle of ##\frac{3π}{4}## with ##\nabla f(4, 2, 1)## is equal to the dot product of the gradient of f and the unit vector in the given direction, which is equal to the magnitude of the gradient times the magnitude of the unit vector times the cosine of the angle between them. This value is independent of the specific orientation of the vector in that direction.
  • #1
Drakkith
Mentor
23,082
7,479

Homework Statement


Find the directional derivative using ##f\left(x,y,z\right)=xy+z^2## at the point (4, 2, 1) in the direction of a vector making an angle of ##\frac{3π}{4}## with ##\nabla f(4, 2, 1)##.

Homework Equations


##f\left(x,y,z\right)=xy+z^2##

The Attempt at a Solution


I found the gradient of ##f(4, 2, 1)## as ##\langle 2,4,2 \rangle##.
Now I'm not sure what to do next. I have a 3d vector, but they only give me one angle, so I don't know which orientation the new unit vector is at.

It's taken me about 30 min to type all this up (I keep going back and adding and deleting stuff), and I've got to go to an exam for another class, but I just now thought of this: I think that if I knew which direction I had to rotate I could solve this by finding a vector perpendicular to the gradient vector and a vector opposite of the gradient vector and then adding them together to get a vector at 3π/4 from the gradient vector. Knowing that would allow me to get the unit vector I need to solve this.
 
Physics news on Phys.org
  • #2
Drakkith said:

Homework Statement


Find the directional derivative using ##f\left(x,y,z\right)=xy+z^2## at the point (4, 2, 1) in the direction of a vector making an angle of ##\frac{3π}{4}## with ##\nabla f(4, 2, 1)##.

Homework Equations


##f\left(x,y,z\right)=xy+z^2##

The Attempt at a Solution


I found the gradient of ##f(4, 2, 1)## as ##\langle 2,4,2 \rangle##.
Now I'm not sure what to do next. I have a 3d vector, but they only give me one angle, so I don't know which orientation the new unit vector is at.

It's taken me about 30 min to type all this up (I keep going back and adding and deleting stuff), and I've got to go to an exam for another class, but I just now thought of this: I think that if I knew which direction I had to rotate I could solve this by finding a vector perpendicular to the gradient vector and a vector opposite of the gradient vector and then adding them together to get a vector at 3π/4 from the gradient vector. Knowing that would allow me to get the unit vector I need to solve this.
There is a shorter way than we discussed in that recent similar 2D problem. Use$$
D_{\hat u}f(P) = \nabla f(P)\cdot \hat u = |\nabla f(P)||\hat u|\cos\theta$$
 
  • #3
LCKurtz said:
There is a shorter way than we discussed in that recent similar 2D problem. Use$$
D_{\hat u}f(P) = \nabla f(P)\cdot \hat u = |\nabla f(P)||\hat u|\cos\theta$$

Wait... does that mean the directional derivative is the same at all orientations of 3π/4?

Also, I got an answer of ##-2\sqrt{3}##, which is correct. Thanks Kurtz.
 
  • #4
Drakkith said:
Wait... does that mean the directional derivative is the same at all orientations of 3π/4?
Assuming you mean all orientations from the gradient direction, then yes it does. And you could have done the earlier problem this way too. For some reason in that other problem I got the impression you were to find a vector in the new direction first.
 
  • #5
LCKurtz said:
Assuming you mean all orientations from the gradient direction, then yes it does. And you could have done the earlier problem this way too. For some reason in that other problem I got the impression you were to find a vector in the new direction first.

That was my impression too. I thought I was supposed to find a vector first. :frown:
 

FAQ: Directional Derivative at an Angle with a 3d Gradient

1. What is a directional derivative at an angle with a 3D gradient?

A directional derivative at an angle with a 3D gradient is a measure of how a function changes in a specific direction in a three-dimensional space. It takes into account both the slope of the function in that direction (represented by the gradient) and the angle at which the direction is measured.

2. How is a directional derivative at an angle with a 3D gradient calculated?

The directional derivative at an angle with a 3D gradient is calculated using the dot product of the gradient and a unit vector in the direction of interest. This dot product represents the rate of change of the function in that direction.

3. What is the difference between a directional derivative and a partial derivative?

A directional derivative measures the rate of change of a function in a specific direction, while a partial derivative measures the rate of change of a function with respect to only one variable while keeping all other variables constant.

4. Why is the directional derivative at an angle with a 3D gradient important?

The directional derivative at an angle with a 3D gradient is important in many applications, particularly in fields such as physics, engineering, and economics. It allows us to understand how a function changes in a particular direction, which can be useful in optimizing processes or predicting future outcomes.

5. Can the directional derivative at an angle with a 3D gradient be negative?

Yes, the directional derivative at an angle with a 3D gradient can be negative. This indicates that the function is decreasing in the direction of interest. A positive directional derivative indicates that the function is increasing in that direction, and a value of zero indicates that the function is not changing in that direction.

Back
Top