# Derivative of directional vector

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## 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.

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

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Staff Emeritus
Homework Helper
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as (∇f)(P0)= <1,1,1>,
It isn’t. Double check your computations.

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## 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}##.

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