Derivative of directional vector

[yecko]

Homework Statement

Find the unit vectors along which the given functions below increase and decrease most rapidly at P₀ . Then find the derivatives of the functions in these directions.

Homework Equations

solution:

The Attempt at a Solution

why are the derivatives’ values along these two directions are 2*sqrt(3)?
as (∇f)(P₀)= <1,1,1>, absolute value of it = sqrt(3)
why is the solution multiplied 2 for it?
thanks

 

[Orodruin]

as (∇f)(P0)= <1,1,1>,

It isn’t. Double check your computations.

 

[Ray Vickson]

Homework Statement

Find the unit vectors along which the given functions below increase and decrease most rapidly at P₀ . Then find the derivatives of the functions in these directions.

View attachment 213475

Homework Equations

solution:
View attachment 213474

The Attempt at a Solution

why are the derivatives’ values along these two directions are 2*sqrt(3)?
as (∇f)(P₀)= <1,1,1>, absolute value of it = sqrt(3)
why is the solution multiplied 2 for it?
thanks

You are correct: the directional derivative is, indeed, $2 \sqrt{3}$. The reason is that $\nabla f(1,1,1) = \langle 2,2,2 \rangle$ is a vector of length $\sqrt{3 \times 4} = 2 \sqrt{3}$.

 

[Orodruin]

You are correct: the directional derivative is, indeed, $2 \sqrt{3}$.

The OP thinks the directional derivative is $\sqrt{3}$ and $\nabla f = \langle 1,1,1\rangle$, so he is not correct. This is why I suggested him to check his computation of the gradient in #2.

 

[yecko]

o yes... thanks