What is the Correct Directional Derivative for Vector w in the Given Scenario?

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Homework Help Overview

The discussion revolves around the concept of directional derivatives in the context of vector calculus, specifically focusing on the correct interpretation and calculation of the directional derivative for a given vector w.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of the directional derivative, questioning whether to use a unit vector or not. There is discussion about the implications of different definitions of directional derivatives and the significance of the notation used in the problem.

Discussion Status

The conversation is ongoing, with participants providing insights into the definitions and interpretations of directional derivatives. Some guidance has been offered regarding the use of unit vectors and the potential variations in definitions across different texts.

Contextual Notes

There is mention of differing definitions of directional derivatives in various texts, which may affect the interpretation of the problem. Participants are encouraged to refer to their specific textbook for clarity on the definitions being used.

DottZakapa
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i compute the partial derivative, the vector that i have to use the one in the text or
w=(2/(5^(1/2)), 1/(5^(1/2)))

using the last one i get minus square root of five , if i don't divide by the norm the answer should be B.
i don't understand what D means
 
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DottZakapa said:
View attachment 255446
i compute the partial derivative, the vector that i have to use the one in the text or
w=(2/(5^(1/2)), 1/(5^(1/2)))

using the last one i get minus square root of five , if i don't divide by the norm the answer should be B.
i don't understand what D means

A direction derivative should be taken wrt a unit vector, so ##-\sqrt{5}## looks right. Unless ##\frac{\partial f}{\partial \vec v}## means something else?

The expression in ##D## is the magnitude of the gradient of ##f## at that point. Sometimes ##||u||## is used instead of ##|u|## for the "norm" or magnitiude of vector.
 
PeroK said:
A direction derivative should be taken wrt a unit vector, so ##-\sqrt{5}## looks right. Unless ##\frac{\partial f}{\partial \vec v}## means something else?

The expression in ##D## is the magnitude of the gradient of ##f## at that point. Sometimes ##||u||## is used instead of ##|u|## for the "norm" or magnitiude of vector.

There are authors who do not ask for this. For example Munkres in "Analysis on manifolds" who defines the directional derivative for any non-zero vector.

Also, I believe it is better to use ##\Vert v \Vert## (Use \Vert v \Vert as tex command) instead of ##|| v ||## (formats a bit more nicely imo).
 
so according to the text what shall i consider ?
 
DottZakapa said:
so according to the text what shall i consider ?

Depends on your definition of directional derivative. So please give us the definition your book provides.

If it is w.r.t. a unit vector it is ##-\sqrt{5}##, otherwise ##-5## is correct.
 
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