- #1
Superdemongob
- 8
- 0
1. Question:
Find the directional derivatives of f(x, y, z) = x2+2xyz−yz2 at (1, 1, 2) in the directions parallel to the line (x−1)/2 = y − 1 = (z−2)/-3.
2. Solution:
We have ∇f = (2x + 2yz)i + (2xz - z2)j + (2xy - 2yz)k.
Therefore, ∇f(1, 1, 2) = 6i - 2k.
The given line is parallel to the vector v = (2, 1, -3).
The corresponding unit vectors are u = ± 1/||v|| and v = (±1/√14)(2, 1, -3).
For the directional derivatives we find f'u±(1, 1, 2) = ∇f(1, 1, 2)dot(u) = ±18/√14
3. My Questions:
The only part of this that I have no clue about is how do they get the vector v from the information given.
Could someone please explain how they find that direction vector?
I get that since the lines are parallel the direction vector is the same but how does one find the directional vector?
Any help is greatly appreciated.
Thanks.
Find the directional derivatives of f(x, y, z) = x2+2xyz−yz2 at (1, 1, 2) in the directions parallel to the line (x−1)/2 = y − 1 = (z−2)/-3.
2. Solution:
We have ∇f = (2x + 2yz)i + (2xz - z2)j + (2xy - 2yz)k.
Therefore, ∇f(1, 1, 2) = 6i - 2k.
The given line is parallel to the vector v = (2, 1, -3).
The corresponding unit vectors are u = ± 1/||v|| and v = (±1/√14)(2, 1, -3).
For the directional derivatives we find f'u±(1, 1, 2) = ∇f(1, 1, 2)dot(u) = ±18/√14
3. My Questions:
The only part of this that I have no clue about is how do they get the vector v from the information given.
Could someone please explain how they find that direction vector?
I get that since the lines are parallel the direction vector is the same but how does one find the directional vector?
Any help is greatly appreciated.
Thanks.