Directional Derivative Solved Question: Explanation Needed Please

[Superdemongob]

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.

 

[tiny-tim]

Hi Superdemongob!

… the line (x−1)/2 = y − 1 = (z−2)/-3.
…
The given line is parallel to the vector v = (2, 1, -3).
…
The only part of this that I have no clue about is how do they get the vector v from the information given.

the (x-1) (y-1) and (z-2) are only there to make sure it goes through (1,1,2)

if it went through (0,0,0) instead, the line would be x/2 = y/1 = z/-3,

which is the vector v = (2, 1, -3) ! ​

 

[Superdemongob]

I get what you mean but how do you know that that is the directional vector?

Like what is the method for finding out?

If the question had the line 2(x−1) = y − 1 = -3(z−2) then how would the directional vector be different?



Sorry but I'm really trying to understand this.

 

[tiny-tim]

Hi Superdemongob!

If the question had the line 2(x−1) = y − 1 = -3(z−2) then how would the directional vector be different?

In that case, the line would still go through (1,1,-2),

but the direction would be parallel to the direction 2x = y = -3z,

to which the solution is (x,y,z) = (k/2,k,-k/3) for any value of k,

ie the line containing the vector (1/2,1,-1/3) ​

 

[Superdemongob]

I think I finally get it.

You take the inverse of the coefficients and that is your directional vector?

 

[tiny-tim]

yes,

and you can check it works because …

it's the only answer that gives you 1/1 = 1/1 = 1/1 ! ​

 

[Superdemongob]

thank you so so much.

i'm studying for a test and this helps a LOT.

thanks again.