Problem with a directional derivative calculation

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Homework Help Overview

The discussion revolves around the calculation of directional derivatives, specifically comparing two approaches: the limit definition and the dot product of the gradient vector with the direction vector. Participants are exploring why these methods yield different results in certain cases.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of using the dot product method versus the limit method, particularly in relation to the continuity of partial derivatives. Questions are raised about the conditions under which each method is valid.

Discussion Status

There is an ongoing exploration of the relationship between differentiability and the validity of the directional derivative methods. Some participants have provided insights regarding the continuity of partial derivatives and their impact on the use of the gradient approach.

Contextual Notes

It is noted that the partial derivatives are not continuous at the point of interest, which raises questions about the differentiability of the function at that point. This context is crucial for understanding the limitations of the methods being discussed.

Amaelle
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Homework Statement
calculate the directional derivative
Relevant Equations
directional derivative
Good day

I have a problem regarding the directional derivative (look at the example below)

1605103234654.png


in this example, we try to find the directional derivatives according to the two approaches ( the definition with the limit and the dot product of the vector gradient and the vector direction)
in this example, we got totally different results?

why?

many thanks in advance!
 
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The partial deivatives are not continuous: for x= 0 and y \neq 0 you have \frac{\partial f}{\partial x} = \frac{y^4}{y^4} = 1 which does not tend to 0 as y \to 0.
 
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Thanks a lot for your prompt answer
so If I understand well, the dot product method works only if the function is differentiable? otherwise, we should only use the limit method?
 
Amaelle said:
in this example, we try to find the directional derivatives according to the two approaches ( the definition with the limit and the dot product of the vector gradient and the vector direction)
in this example, we got totally different results?

why?

many thanks in advance!

What your book is doing is calculating the derivative along a parameterised curve through ##(0,0)##:$$g(t) = f(t\cos \theta, t\sin \theta) = t\cos \theta \ \sin^2 \theta$$ Which is well-defined. And, trivially: $$g'(t) = \cos \theta \ \sin^2 \theta$$
 
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yes thank you, understanding things takes times :) but thank you for easing the pain
 
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Amaelle said:
Thanks a lot for your prompt answer
so If I understand well, the dot product method works only if the function is differentiable? otherwise, we should only use the limit method?
Yes. A function of many variables is differentiable if and only if all the partial derivatives exist and are continuous . But in this case as post #2 shows the partial derivatives are not continuous at (0,0) hence the function is not differentiable at (0,0) hence you can't apply the gradient approach for calculating the directional derivative.
 
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thanks a million!
 
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