Derivative of directional vector

  • Thread starter yecko
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  • #1
yecko
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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.

螢幕快照 2017-10-21 下午8.28.59.png


Homework Equations


solution:
螢幕快照 2017-10-21 下午8.24.25.png


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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Answers and Replies

  • #2
Orodruin
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as (∇f)(P0)= <1,1,1>,
It isn’t. Double check your computations.
 
  • #3
Ray Vickson
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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}##.
 
  • #4
Orodruin
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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.
 
  • #5
yecko
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o yes... thanks
 

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