How to find F(x,y) for $f(x,y)= h(y)\hat{i} + g(x)\hat{j}?$

[WMDhamnekar]

Let g(x) and h(y) be differentiable functions, and let f(x, y) = h(y)i+ g(x)j. Can f have a potential F(x, y)? If so, find it. You may assume that F would be smooth. (Hint: Consider the mixed partial derivatives of F.)

How to answer this question? What is the answer to this question?

 

[topsquark]

Let g(x) and h(y) be differentiable functions, and let f(x, y) = h(y)i+ g(x)j. Can f have a potential F(x, y)? If so, find it. You may assume that F would be smooth. (Hint: Consider the mixed partial derivatives of F.)

How to answer this question? What is the answer to this question?

It's not going to happen. The curl of f(x, y) isn't going to be 0. As I just mentioned on MHF the curl has to be zero or a potential function won't exist.

-Dan

 

[WMDhamnekar]

It's not going to happen. The curl of f(x, y) isn't going to be 0. As I just mentioned on MHF the curl has to be zero or a potential function won't exist.

-Dan.

But the author has given the hint to consider the mixed partial derivative of F. Author has given the answer as follows: F(x,y)= axy + bx + cy +d

I don't understand how did the author compute that answer?

 

[topsquark]

But the author has given the hint to consider the mixed partial derivative of F. Author has given the answer as follows: F(x,y)= axy + bx + cy +d

I don't understand how did the author compute that answer?

Well, without the coefficient functions h and g we aren't going to get anything! :)

The definition of a potential function is what I have given you above. At least for Physics. Perhaps Calculus has a slightly different interpretation. Is it possible that you are trying to show that you are trying to construct a function F(x, y) such that \(\displaystyle \dfrac{ \partial F }{ \partial x} = \dfrac{ \partial F}{ \partial y}\) ? That would be to find a function F(x, y) that has an exact derivative. That's the only thing I can think of.

-Dan

 

[I like Serena]

As a point of notation, $F_x$ is the partial derivative of $F$ with respect to $x$.

Suppose the potential function $F(x,y)$ exists.
Then $F_x=h(y)$ and $F_y=g(x)$.
If we take the mixed partial derivative, we get $F_{xy}=h'(y)=g'(x)$.

So we can check if $h'(y)=g'(x)$. If they are different, then there is no smooth potential function.
Alternatively, we have $\operatorname{curl} f=\nabla\times f=g'(x)-h'(y)$ and we know that there is a potential function $F$ iff the curl is zero, which boils down to the same thing.

To find $F$, we can calculate $F(x,y)=\int h(y)\,dx=h(y)x + C_1(y)$ and $F(x,y)=\int g(x)\,dy=g(x)y + C_2(x)$.
When we compare terms, we can deduce that $F$ must be of the form $F(x,y)=axy+bx+cy+d$.

 

[topsquark]

As a point of notation, $F_x$ is the partial derivative of $F$ with respect to $x$.

Suppose the potential function $F(x,y)$ exists.
Then $F_x=h(y)$ and $F_y=g(x)$.
If we take the mixed partial derivative, we get $F_{xy}=h'(y)=g'(x)$.

So we can check if $h'(y)=g'(x)$. If they are different, then there is no smooth potential function.
Alternatively, we have $\operatorname{curl} f=\nabla\times f=h'(y)-g'(x)$ and we know that there is a potential function $F$ iff the curl is zero, which boils down to the same thing.

To find $F$, we can calculate $F(x,y)=\int h(y)\,dx=h(y)x + C_1(y)$ and $F(x,y)=\int g(x)\,dy=g(x)y + C_2(x)$.
When we compare terms, we can deduce that $F$ must be of the form $F(x,y)=axy+bx+cy+d$.

Okay, I see what's up now. I've done this before but it wasn't called a "potential." (I forget what it was called.) This is not quite the same thing as a potential function in Physics, though there are similarities.

How are you defining \(\displaystyle \nabla \times f\) in general here? I've never seen a 2D version of curl.

-Dan

 

[I like Serena]

Okay, I see what's up now. I've done this before but it wasn't called a "potential." (I forget what it was called.) This is not quite the same thing as a potential function in Physics, though there are similarities.

How are you defining \(\displaystyle \nabla \times f\) in general here? I've never seen a 2D version of curl.

-Dan

I believe it is the same as a potential function in physics, although a minus sign is typically incorporated, such as in $\mathbf E=-\nabla \phi$.

It is not given that f is a 2D function. It could be 3D with a k component that is 0. I've interpreted it like that for convenience. Curl is indeed only defined in 3D.

 

[topsquark]

I believe it is the same as a potential function in physics, although a minus sign is typically incorporated, such as in $\mathbf E=-\nabla \phi$.

It is not given that f is a 2D function. It could be 3D with a k component that is 0. I've interpreted it like that for convenience. Curl is indeed only defined in 3D.

Oh! I see. What you meant to say was that \(\displaystyle \nabla \times (h(y) \hat{i} + g(x) \hat{j} ) = (g'(x) - h'(y)) \hat{k} = - (h'(y) - g'(x)) \hat{k}\). You left the vector off the end and I thought you were defining something in 2D.

-Dan

 

[WMDhamnekar]

I believe it is the same as a potential function in physics, although a minus sign is typically incorporated, such as in $\mathbf E=-\nabla \phi$.

It is not given that f is a 2D function. It could be 3D with a k component that is 0. I've interpreted it like that for convenience. Curl is indeed only defined in 3D.

I am not Physics expert. That's why I ask you what is E ? I think E is a conservative vector field. Isn't it?

 

[I like Serena]

I am not Physics expert. That's why I ask you what is E ? I think E is a conservative vector field. Isn't it?

$\mathbf E$ is the electric field, which is indeed a conservative vector field in electro statics (not generally). $\phi$ is the electric potentential measured in volts (V).
The formula is called Coulomb's Law.

 

[I like Serena]

Oh! I see. What you meant to say was that \(\displaystyle \nabla \times (h(y) \hat{i} + g(x) \hat{j} ) = (g'(x) - h'(y)) \hat{k} = - (h'(y) - g'(x)) \hat{k}\). You left the vector off the end and I thought you were defining something in 2D.

-Dan

Yeah, I've left out the k-vector in the result, similar to how we write $g=-10 m/s^2$ for the gravitational vector field. We write a minus sign with the understanding that it is with respect to a vector that points upward (the k-vector).
Hmm... I see I accidentally used the wrong sign. As you are pointing out, it should be $g'(x) - h'(y)$. Fixed now in that post.

 

[topsquark]

Yeah, I've left out the k-vector in the result, similar to how we write $g=-10 m/s^2$ for the gravitational vector field. We write a minus sign with the understanding that it is with respect to a vector that points upward (the k-vector).
Hmm... I see I accidentally used the wrong sign. As you are pointing out, it should be $g'(x) - h'(y)$. Fixed now in that post.

No, that's fine. You were pointing out why the condition on the functions is what it is. I just didn't recognize it, that's all.

-Dan