Solve Line Integral Question | Get Math Help from Physics Forums

[WMDhamnekar]

TL;DR Summary

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.)

I don't have any idea about how to use the hint given by the author.
Author has given the answer to this question i-e F(x,y) = axy + bx + cy +d.
I don't understand how did the author compute this answer.

Would any member of Physics Forums enlighten me in this regard?

Any math help will be accepted.

 

[malawi_glenn]

Because
$- \dfrac{\partial F}{\partial x} = h(y)$
then
$\dfrac{\partial^2 F}{\partial x^2} = 0$

And we also have
$\dfrac{\partial^2 F}{\partial x \partial y } = \dfrac{\partial^2 F}{\partial y \partial x }$

If you just want to show that the potential exists, then show that $\vec \nabla \times \vec f = 0$

 

[WMDhamnekar]

Because
$- \dfrac{\partial F}{\partial x} = h(y)$
then
$\dfrac{\partial^2 F}{\partial x^2} = 0$

And we also have
$\dfrac{\partial^2 F}{\partial x \partial y } = \dfrac{\partial^2 F}{\partial y \partial x }$

Why is the sign negative for $\frac{\partial{F}}{\partial{x}}= h(y)?$ Would you explain that ?

 

[malawi_glenn]

Why is the sign negative for $\frac{\partial{F}}{\partial{x}}= h(y)?$ Would you explain that ?

Isn't the potential F(x,y) defined as $\vec f = - \vec \nabla F$ ?

 

[WMDhamnekar]

Isn't the potential F(x,y) defined as $\vec f = - \vec \nabla F$ ?

In the following corollary 4.6 $\vec f \not= -\vec \nabla F$ Why?

 

[malawi_glenn]

In the following corollary 4.6 $\vec f \not= -\vec \nabla F$ Why?
View attachment 303550

Matter of convetion I guess, I have always used $\vec f = - \vec \nabla F$ as the definition as potential. But it does not matter, the math is the same, just follow your books conventions. And also -0 = 0 so...