Problem with a directional derivative calculation

In summary, the gradient approach does not work because the partial derivatives are not continuous at (0,0).
  • #1
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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  • #2
The partial deivatives are not continuous: for [itex]x= 0[/itex] and [itex]y \neq 0[/itex] you have [itex]\frac{\partial f}{\partial x} = \frac{y^4}{y^4} = 1[/itex] which does not tend to 0 as [itex]y \to 0[/itex].
 
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  • #3
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?
 
  • #4
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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  • #5
yes thank you, understanding things takes times :) but thank you for easing the pain
 
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  • #6
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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  • #7
thanks a million!
 
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1. What is a directional derivative?

A directional derivative is a measure of the rate of change of a function in a specific direction. It is used to calculate the slope of a function along a given direction, similar to how a regular derivative calculates the slope at a single point.

2. How is a directional derivative calculated?

To calculate a directional derivative, you first need to find the gradient of the function at the point of interest. Then, you need to take the dot product of the gradient vector and the unit vector in the direction you want to find the derivative. The result is the directional derivative.

3. What is the difference between a directional derivative and a regular derivative?

A directional derivative calculates the slope of a function in a specific direction, while a regular derivative calculates the slope at a single point. In other words, a directional derivative takes into account the direction of the slope, while a regular derivative only considers the slope at a specific point.

4. When is a directional derivative useful?

A directional derivative is useful when studying the behavior of a function in a specific direction. It can be used to determine the maximum rate of change of a function in a particular direction, or to find the direction in which the function has the steepest slope.

5. What can cause a problem with a directional derivative calculation?

There are several potential causes for a problem with a directional derivative calculation. These can include errors in the calculation process, undefined points or discontinuities in the function, or using an incorrect or non-unit vector for the direction of interest.

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