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Directional derivative question

  • Thread starter julz127
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Homework Statement


rate of change of [itex] f(x,y) = \frac{x}{(1+y)} [/itex] in the direction (i-j) at the point (0,0)


Homework Equations





The Attempt at a Solution


[itex] ∇f(x,y) = \frac{1}{(y+1)}\hat{i} - \frac{x}{(y+1)^2}\hat{j} [/itex]

[itex]D_u = ( f_x, f_y) \bullet ( 1, -1 )[/itex]

[itex]D_u = \frac{(y+x+1)}{(y+1)^2} [/itex]

Wolfram and the answer sheet is telling me that there should be a [itex]\sqrt{2}[/itex] in the denominator, but I can't figure out where it comes from, thanks.
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


rate of change of f(x,y) = x/(1+y) in the direction (i-j) at the point (0,0)


Homework Equations





The Attempt at a Solution


grad(f(x,y)) = 1/(y+1)i - x/(y+1)^2j

Du = ( f_x, f_y ) dot ( 1, -1 )

Du = (y+x+1)/(y+1)^2

Wolfram and the answer sheet is telling me that there should be a sqrt(2) in the denominator, but I can't figure out where it comes from, thanks.
The vector you want to dot the grad with should be a unit vector pointed in the direction i-j. That's what 'in the direction' means.
 

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