Why Does dV Equal ∂V/∂x(dx) + ∂V/∂y(dy)?

[Tony Hau]

So in my lecture notes on Differential Equations, it states that a first order ODE is exact if A(x,y)dx + B(x,y)dy = 0 and ∂A/∂y = ∂B/∂x. Okay I accept this definition.

Then, there is a sentence like this:
Our goal is to find the function V(x,y) satisfying
Adx + Bdy = dV = ∂V/∂x(dx) + ∂V/∂y(dy), where A(x,y) =∂V/∂x and B(x,y) =∂V/∂y.
I am confused here. Why dV = ∂V/∂x(dx) + ∂V/∂y(dy)? I think you can say that dx cancels with ∂x and dy cancels with ∂y and so it is the total change in infinitesimal V in both x and y direction. Is my concept correct? I am okay with the idea that dx cancels with dx but still, I am not really convinced that dx cancels with ∂x.

 

[Infrared]

The intuition is basically that if you change $x$ by a small amount $dx$, then $V$ changes by the amount $\frac{\partial V}{\partial x}dx$ to first order (just think about the definition of partial derivative), and similarly for $y$. So if you change both $x$ and $y$ simultaneously, then you get your formula for $dV$.

A little more rigorously, the multivariate chain rule tells you that if $x$ and $y$ are functions of $t$, then $\frac{dV}{dt}=\frac{\partial V}{\partial x}\frac{dx}{dt}+\frac{\partial V}{\partial y}\frac{dy}{dt}.$ This is essentially the same formula you have, written with derivatives instead of differentials.

 

[PeroK]

So in my lecture notes on Differential Equations, it states that a first order ODE is exact if A(x,y)dx + B(x,y)dy = 0 and ∂A/∂y = ∂B/∂x. Okay I accept this definition.

Then, there is a sentence like this:
Our goal is to find the function V(x,y) satisfying
Adx + Bdy = dV = ∂V/∂x(dx) + ∂V/∂y(dy), where A(x,y) =∂V/∂x and B(x,y) =∂V/∂y.
I am confused here. Why dV = ∂V/∂x(dx) + ∂V/∂y(dy)? I think you can say that dx cancels with ∂x and dy cancels with ∂y and so it is the total change in infinitesimal V in both x and y direction. Is my concept correct? I am okay with the idea that dx cancels with dx but still, I am not really convinced that dx cancels with ∂x.

You can check out the idea with an example function. E.g. let
$$V(x, y) = xy^2 + x^3y$$
Then we have:
$$V(x + dx, y + dy) = (x + dx)(y + dy)^2 + (x + dx)^3(y + dy)$$
$$ = (x + dx)(y^2 + 2ydy) + (x^3 + 3x^2dx)(y + dy)$$
$$ = xy^2 + y^2dx + 2xydy + x^3y + x^3dy + 3x^2ydx$$
$$ = V(x, y) + (y^2 + 3x^2y)dx + (2xy + x^3)dy$$
$$ = V(x, y) + \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial y}dy$$
Where we dropped higher order differential terms.

I would get away from the idea that $dx$ and $\partial x$ "cancel" and think in terms of the multivariable chain rule.

 

[PeroK]

PS the idea of the exact differential is to work this in reverse. Suppose we start with:
$$dV = (y^2 + 3x^2y)dx + (2xy + x^3)dy$$
Then, knowing that in general we have:
$$dV = \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial y}dy$$
We can equate these two and look for $V(x, y)$, such that:
$$\frac{\partial V}{\partial x} = y^2 + 3x^2y, \ \text{and} \ \frac{\partial V}{\partial y} = 2xy + x^3$$

 

[Gaussian97]

Well, I think you should look at your notes in multivariable calculus, the equation you have is basically the definition of $df$, which can be defined as
$$df = \vec{\nabla} f \cdot d\vec{x}$$
or at least, any other definition should imply this relation very easily.

 

[Tony Hau]

Well, I think you should look at your notes in multivariable calculus, the equation you have is basically the definition of $df$, which can be defined as
$$df = \vec{\nabla} f \cdot d\vec{x}$$
or at least, any other definition should imply this relation very easily.

We didn't have such relationship. My professor in last semester just said that if we held x constant in f(x,y) and differentiate w.r.t y, we have ∂f(x,y)/∂y. I think the concept of ∂f(x,y) alone is rather loosely defined in mathematics and is more 'physics'. It is like the concept of 'cancelling dx with dx', which we do frequently in physics but not in maths.

 

[Tony Hau]

Thanks for all your great answers! Especially the first few answers! I get the clear concept now.

 

[PeroK]

Thanks for your reply, but why is there a V(x,y) in the dV term? What does that mean?

By definition:
$$V(x +dx, y+dy) = V(x, y) + dV$$
That's what $dV$ means. It's the incremental change in $V(x, y)$ from incremental changes in $x$ and $y$.

 

[Tony Hau]

By definition:
$$V(x +dx, y+dy) = V(x, y) + dV$$
That's what $dV$ means. It's the incremental change in $V(x, y)$ from incremental changes in $x$ and $y$.

Got it! Thanks for your help!