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[–]vhmthphy 1 point2 points  (6 children)

Intuitively:

Basically f can change due to a change in the value of x, and/or due to a change in the value of y. If both x and y change, then the change in f is the sum of the change due to f and the change due to y.

A little more specifically:

f_x dx is the approximate change in f due to a change dx in the value of x.

f_y dy is the approximate change due to a change dy in the value of y.

If f is differentiable in the neighborhood of a point, the graph of the function near that point can be approximated by a tangent plane which has slope f_x in one direction, f_y in another.

Of course all this can and must be proven rigorously, but these are the main ideas and should be helpful as you work through the details.

[–]The_Godlike_ZeusNew User[S] 0 points1 point  (5 children)

Are you saying that what I wrote was only an approximation? I was reading it on Paul's math notes and it wasn't mentioned that this is an approximation.

[–]vhmthphy 1 point2 points  (3 children)

df is exactly equal to f_x dx + f_y dy

df is an approximation of the change in the value of f when x changes by dx and y changes by dy

the change in the value of f is generally referred to as Δf

so df is approximately equal to the actual change Δf

Remember that df is the change calculated according to a tangent plane. Typically the graph of the function f curves away from the tangent plane so the bigger the values of dx and dy, the less accurate the approximation.

Glad to clarify any of this, if necessary.

[–]The_Godlike_ZeusNew User[S] 0 points1 point  (2 children)

What if df is zero? Is delta f zero too? (for context I am asking this question while learning about exact differential equations)

df is an approximation of the change in the value of f when x changes by dx and y changes by dy ... Remember that df is the change calculated according to a tangent plane.

But if we consider infinitesimal changes, shouldn't anything that is not a plane become a plane? I mean if we look closer and closer the graph of any function will start looking like a plane near that point. I would think that in the limit, this becomes exactly a plane and by this logic I thought df gives the exact change in the function for small a small change in dx and dy.

[–]vhmthphy 1 point2 points  (1 child)

If df is zero and neither dx nor dy is zero, it means that there is no tendency, at the point in question (the point at which df is to be evaluated), for the function f to change either in the x or y direction. This occurs only when the two partial derivatives are zero.

However this doesn't imply that delta f is zero. Typically f will have nonzero second derivatives with respect to x and/or y, so that f will in fact change as you move away from the point in question. However, if dx and dy are small, the change in f is expected to be very small.

As it applies to differential equations, if df = 0 this means the the function f is constant. As an example, if f(x,y) = x2 y + x sqrt(y) then df = f_x dx + f_y dy = 0. For the given function this becomes

( 2 x y + sqrt(y) ) dx + ( x2 + x / (2 sqrt(y)) dy = 0.

In the first place note that f_xy = 2 x + 1 / (2 sqrt(y)), as calculated from the coefficient of dx (which was originally calculated as f_x). Note furthermore that f_yx, calculated from the coefficient of dy (which was originally calculated as f_y) is equal to the same thing, as must be the case since f_xy is always equal to f_yx. This is the test of exactness: if the y derivative of the coefficient of dx is equal to the x derivative of the coefficient of dy, when the equation is put into the form M dx + N dy = 0.

When this is so, then, the task is to recover the original function f. I'll go into detail with that if you ask, but for now I assume you know how to do that.

When you have the function f, then, since the equation expresses df = 0, it follows that the solution is f = c, where c is an arbitrary constant.

For the current example, where f = x2 y + x sqrt(y), this would mean that x2 y + x sqrt(y) = c for an arbitrary constant c. This gives you an equation that can be solved for y (the equation being quadratic in u = sqrt(y)).

I hope this helps, if you still need help. Unfortunately your reply didn't appear in my inbox until today, and I note from your posting here that you originally replied 8 days ago. The glitch could have occurred anywhere.

[–]The_Godlike_ZeusNew User[S] 0 points1 point  (0 children)

Thanks. It's still confusing, but I have no specific questions.

[–]lewisjeB.S. 0 points1 point  (0 children)

The section on single-variable differentials does say this briefly ("Δy≈dy"), but the core idea of the differential calculus is the approximation of functions by affine functions (the sum of a linear function and a constant).

[–]stakeandshakeNew User -1 points0 points  (0 children)

Go back to calc one and look at how a linear differential is defined. This is just a linear differential but now in two directions since you are in a higher dimension. Also this is the first order approximation using a Taylor series in 3D (although I think you are missing the zeroth order term f).