Directional derivative question

Click For Summary
SUMMARY

The discussion centers on calculating the directional derivative of the function f(x,y) = x/(1+y) at the point (0,0) in the direction of the vector (i-j). The gradient ∇f(x,y) is correctly derived as (1/(y+1), -x/(y+1)^2). The directional derivative D_u is computed using the dot product of the gradient and the direction vector, yielding D_u = (y+x+1)/(y+1)^2. The confusion arises regarding the presence of √2 in the denominator, which is clarified by noting that the direction vector must be normalized to a unit vector.

PREREQUISITES
  • Understanding of directional derivatives in multivariable calculus
  • Familiarity with gradient vectors and their properties
  • Knowledge of unit vectors and vector normalization
  • Proficiency in using mathematical software like Wolfram Alpha for verification
NEXT STEPS
  • Study the concept of unit vectors and how to normalize direction vectors
  • Learn about the properties of gradients in multivariable functions
  • Explore directional derivatives in various contexts, including optimization
  • Practice using Wolfram Alpha for solving multivariable calculus problems
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable functions and directional derivatives, as well as anyone seeking to clarify concepts related to gradients and vector analysis.

julz127
Messages
13
Reaction score
0

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.
 
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   Reactions: 1 person

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K