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.Dhamnekar Winod said: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?
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 +dtopsquark said: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.
Well, without the coefficient functions h and g we aren't going to get anything! :)Dhamnekar Winod said: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?
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.Klaas van Aarsen said: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 $F(x,y)=axy+bx+cy+d$.
I believe it is the same as a potential function in physics, although a minus sign is typically incorporated, such as in $\mathbf E=-\nabla \phi$.topsquark said: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
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.Klaas van Aarsen said: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?Klaas van Aarsen said: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.
$\mathbf E$ 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).Dhamnekar Winod said: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?
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).topsquark said: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
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.Klaas van Aarsen said: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.
F(x,y) is a function that describes the relationship between two variables, x and y. It represents the output of a mathematical operation involving these variables.
To find F(x,y) for the given function, you need to substitute the values of x and y into the function and perform the necessary mathematical operations. This will give you the output of the function, which is F(x,y).
h(y) and g(x) are known as the component functions of the given function. They represent the individual effects of y and x on the overall function. For example, h(y) may represent the vertical movement of the function while g(x) may represent the horizontal movement.
Yes, you can use any variables in place of x and y as long as they are consistent throughout the function. For example, you can use a and b instead of x and y, but make sure to use them in the same order and follow the same mathematical operations.
There may be restrictions on the values of x and y depending on the given function. For example, some functions may have a domain and range that restrict the values of x and y to certain intervals. It is important to check for any restrictions before finding F(x,y) for a given function.