Directional derivative question

julz127
Messages
13
Reaction score
0

Homework Statement


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


Homework Equations





The Attempt at a Solution


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

D_u = ( f_x, f_y) \bullet ( 1, -1 )

D_u = \frac{(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.
 
Last edited:
Physics news on Phys.org
julz127 said:

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.
 
  • Like
Likes 1 person
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top