Defintion of Total Differential

[Bashyboy]

Hello,

I provided a snap-shot of the definition of a Total Differential, that my textbook provides. I am having difficulty grasping what this new quantity represents. Could someone help me?

 

[lurflurf]

The (total) differential is a linearization of a function.
This means that given the (in general not linear) function Δz=f(x+dx,y+dy)-f(x,y)
dz is the nearest linear function having the form
dz(x,y;dx,dy)=u(x,y) dx + v(x,y) dy
where as it turns out u and v are the partial derivatives of f
u=f_x
v=f_y

Geometrically dy represents the plane tangent to f at the point (x,y)

The differential is the linear part of the relation between variables

One applications is that if dx and dy are small dz and Δz are approximately equal so one can be used to estimate the other. For example consider the volume of a torus

V(r,R)=(pi r^2)(2pi R)
dV=(2pi r)(2pi R)dr+(pi r^2)(2pi)dR
so we can estimate
V(1.01,10.1)-V(1,10)~(2pi*1)(2pi*10).01+(pi*1^2)(2pi).1~5.92176...
while
V(1.01,10.1)-V(1,10)=(pi*1.01^2)(2pi*10.1)-(pi*1^2)(2pi*10)~5.98118...

 

[verty]

I interpret these as the total rate of change given the rates of change of each variable.

Let f(x,y) = 2xy. If x changes at a rate of dx/dt and y changes at a rate of dy/dt, then f changes at a rate of

##df/dt = 2y \; dx/dt + 2x \; dy/dt##

If you leave off the dt's, you get

##df = 2y \; dx + 2x \; dy##

You can think of the df, dx, dy as having a hidden dt underneath. Dividing by dx, we get

##df/dx = 2y + 2x \; dy/dx##

This is consistent because, by the chain rule, df/dx = df/dt dt/dx = df/dt / (dx/dt). You can use these differentials like they are rates, with impunity really.

I defer to how individual courses or lecturers define these. As a default interpretation, this works.

 

[WannabeNewton]

This is to supplement what lurflurf said (because this is also how I think of the differential-if you ever end up going on to more advanced calculus then you will thank yourself for thinking of it in this geometric way i.e. via tangent planes because the abstraction in advanced calculus will be exactly based off of this geometric interpretation in terms of tangent planes): http://www.centerofmath.org/mc_pdf/sec2_3.pdf

 

[Stephen Tashi]

I provided a snap-shot of the definition of a Total Differential, that my textbook provides. I am having difficulty grasping what this new quantity represents.

Your difficulty is justified because the definition from the textbook merely states some terms that it does not define and specifies some notation for those undefined terms. When it comes to defining differentials in a calculus course, you are on your own to a great extent. Some good ideas have been suggested by other posters. (See also https://www.physicsforums.com/showthread.php?t=701423)

In contrast to differentials, definitions of concepts such as [itex] Lim_{x \rightarrow a } f(x) = L [/itex] are given precise definitions in most texts and students are expected to know them and use them rigorously. Various hypocrisies are traditional in the teaching of mathematics. You don't get to complain about the definition of differentials as long as the textbook doesn't give any problems where you have to do a detailed proof that involves them.

 

[lurflurf]

Again dz is the unique function such that

$$\lim_{(dx,dy) \rightarrow 0} \frac{|dz-\Delta z|}{|(dx,dy)|}=\lim_{(dx,dy) \rightarrow 0} \frac{|dz-\Delta z|}{\sqrt{dx^2+dy^2}}=0$$
or by the chain rule
$$dz=\left. \dfrac{d}{dt}f(x+t \, dx,y+t \, dy) \right|_{t=0} \\
=\left. \left( \dfrac{\partial}{\partial (x+t \, dx)} f(x+t \, dx,y+t \, dy) \right) \dfrac{d}{dt}t \, dx \right|_{t=0}+\left. \left( \dfrac{\partial}{\partial (y+t \, dy)} f(x+t \, dx,y+t \, dy) \right) \dfrac{d}{dt}t \, dy \right|_{t=0} \\ = \dfrac{\partial}{\partial x} f(x,y) \, dx + \dfrac{\partial}{\partial y} f(x,y) \, dy $$

All the epsilon-delta aficionados can just relax.

 

[CubicFlunky77]

All the epsilon-delta aficionados can just relax.

LOL! Your first explanation was terrific and your second post complemented the point quite well. Good job :).