Problem with a directional derivative calculation

In summary: It seems to me that you are saying that the function is not differentiable at (0,0) - is that what you are saying?I'm not sure what you are asking, but it seems that you are asking about whether the function is differentiable at the point (0,0). The answer is that the function is not differentiable at (0,0).
  • #1
Amaelle
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Homework Statement
the directional derivative of the function f(x,y) in respect to the point(0,0) according to the vector v(1,2) is equal to
Relevant Equations
directional derivative
this is the function

1604746632373.png

and this is the solution in which the definition has been used
1604746740222.png


my question is
Why we can not use the traditional approach? I mean calculation the partial derivative which equals 0 in our case? And doing the dot product with the vector v (after normalizing it)

many thanks in advance
 
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  • #2
The traditional approach ( i suppose you mean to calculate ##\nabla f (0,0)\cdot \vec{v}## )seems more complex to me.

But i am afraid there is another problem here too: The traditional approach can be used only when the function is differentiable at the point of interest, but i am afraid this function isn't differentiable at (0,0) (from a quick look I see that even the partial derivatives at (0,0) don't exist).

So if indeed the partial derivatives don't exist at (0,0) then we just can't use the traditional approach anyway. I hope I didn't blunder on this, can you show me how you calculate the partial derivatives at (0,0) cause I get infinities...
 
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  • #3
The function is not even continuous at ##(0,0)##.
 
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  • #4
Thanks a million for your answer, it made thing clearer now :)
 
  • #5
The answer given still makes no sense. It ignores the 2/t term in taking the limit.
 
  • #6
haruspex said:
It ignores the 2/t term in taking the limit
I see two 2/t term, one positive and one negative and they cancel each other.
 
  • #7
Delta2 said:
I see two 2/t term, one positive and one negative and they cancel each other.
Ah, yes, ok.
But the answer is still wrong. t is an arbitrary parameter. It makes no sense to divide by t; need to consider the length in the given direction.
Given that we know the limit of the function is zero, making it continuous, in the given direction, it is simpler to switch to polar:
##f=5\sin(\theta)\cos(\theta)+3r\cos(\theta)##
Diff wrt r and plug in the value of theta.
 
  • #8
haruspex said:
Ah, yes, ok.
But the answer is still wrong. t is an arbitrary parameter. It makes no sense to divide by t; need to consider the length in the given direction.
That's the definition of the directional derivative, one divides by ##t## there. By the way I see it ,##t## is the length in the direction of ##\vec{v}##.

In the normalized version you divide also by ##|\vec{v}|##.
 
  • #9
Delta2 said:
That's the definition of the directional derivative
So the answer depends on the magnitude of ##\vec v##? Strange, but I stand corrected.
 
  • #10
haruspex said:
So the answer depends on the magnitude of ##\vec v##? Strange, but I stand corrected.
Well yes i guess the non normalized version depends on the magnitude of the vector, though i can't imagine how we can prove that the dependency is always linear.
 
  • #11
There are two different conventions, one that ##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v}##, and the other that ##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \left(\vec{v}/|\vec{v}| \right)##. I was taught that the first convention is more common (##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v}##), but that you should be aware of the other convention.
 
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  • #12
etotheipi said:
There are two different conventions, one that ##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v}##, and the other that ##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \left(\vec{v}/|\vec{v}| \right)##. I was taught that the first convention is more common (##\nabla_{\vec{v}} f(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v}##), but that you should be aware of the other convention.
According to wikipedia (and i think also according to my vector calculus notes of 199x) neither of the above is the definition of the directional derivative. The above we can prove from the definition, if the function f is differentiable.
 
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  • #13
Delta2 said:
According to wikipedia (and i think also according to my vector calculus notes of 199x) neither of the above is the definition of the directional derivative. The above we can prove from the definition, if the function f is differentiable.

In Physics, everything is differentiable :wink:
 
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  • #14
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1. What is a directional derivative?

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

2. What is the formula for calculating a directional derivative?

The formula for calculating a directional derivative of a function f(x,y) at a point (x0,y0) in the direction of vector v = ai + bj is Dvf(x0,y0) = afx(x0,y0) + bfy(x0,y0), where fx and fy are the partial derivatives of f with respect to x and y, respectively.

3. What is the significance of a directional derivative?

A directional derivative is important in many fields of science and engineering, as it allows us to understand how a function changes in a specific direction. It is particularly useful in optimization problems, where we want to find the direction of steepest ascent or descent of a function.

4. What are the common sources of error in a directional derivative calculation?

The most common sources of error in a directional derivative calculation are incorrect input values, mistakes in the formula or calculation, and rounding errors. It is important to double-check all input values and calculations to ensure accuracy.

5. How can I check my directional derivative calculation for accuracy?

To check the accuracy of a directional derivative calculation, you can use the definition of a directional derivative and compare it to your calculated value. Additionally, you can use software or online tools that can calculate directional derivatives to verify your result.

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