# Concept of differential

## Main Question or Discussion Point

Hey Mathwonk, just a quickie. When I think of a differential, it is written as y=y(t), and dy= dy/dt * dt. When I see this, I think of it as we want to deal with only the variable y. But y(t) is a function of t. So we set y=y(t), so that y is a variable by itself. As a consequence, we usually want to integrate with resepct to y. But if we take the derivative of y(t), we get dy/dt. But we want the change in y, not y(t) with respect to t, which is why we multiply by an increment of dt. This leaves a change in dy only. I know dy/dt is not a fraction, but In a sense, this is what is happening. Is this interpretation wrong? I also know that dy/dt is the slope, and multiplying it by dt, gives you the value dt away, but it seems like they are two ways of looking at the same thing.

Thanks,

Cyrus

mathwonk
Homework Helper
a differential is something is independent of the coordinates. i.e. it is rigged up to give the same answer for a path integral no matter how you parametrize the path integral.

i.e. given a path, how would you integrate a function over it? you would subdivide the path into smaller pieces, then multiply the value of the function at an endpoint of one piece by the "length" of that piece, add up and take limits,..

but what if there is no measure of "length" for paths (as on a manifold)? then you would not know what to multiply the value of your function by.

so instead we integrate "differentials". these assign a number not to a point on the path, but to a point and a tangebntnvector at that point.

now if we parametrize the path by running along it at a certain speed, we can say that the length of the path between two points is the time it took to traverse that piece.

but that changes when we go faster, i.e. going faster makes the path look shorter. but since we are integrating a differential, the differential sees the velocity vector of the parametrization, and gives a bigger number on a longer velocity vector, so it comes out even.

now if we already have a coordinate system in our space then we could use that to measure lengths of tangenht svectors. i.e. we call dx the differential that projects a vector on the x axis and takes that length.

but some other differential might act diferently, so we multiply dx by some coefficicne function like f(x)dx to get our mroe general differential.

but then suppose someone changes the coordiantes from x to t. then dt is a different measure for vectors. if dt sees a vector ir does not give the same lnegth as dx does.

rather the product (dx/dt) dt does give the same length as dx does on eacxh vector.;

so injtegratinhg (dx/dt) dt over any path will give the same result as integarting dx will.

sorr this isn't so great, but these little devils are a little confusing.

e.g. consider the path to be the unit circle in the x, y plane.

If I map the unit t interval onto the circle by sending t to (cos(2pi t), sin(2pi t)), then at the point t = 0, the velocity vector is the vertical vector (0, 2pi) in the x,y plane.

this means the parametrization by polar coordinates maps the unit vector from the t interval onto this vector of length 2pi in the x, y plane.

on the other hand, if we paramnewtrize the circle by the map sending u to (cos(u),sin(u)) we get velocity vectors of length 1, but the time interval requires u=0 to u=2pi to get all the way around.

if we integrate the differential (-ydx + xdy)/(x^2 + y^2), in these two cases, it transforms into either 2pi dt, integrated from t=0 to t=1, or into du, integrated from u=0 to u = 2pi.

we get the same thing both ways, namely 2pi.

anyway, the reason dx = (dx/dt) dt, is that these two give the same answer on all velocity vectors.

but what if there is no measure of "length" for paths (as on a manifold)? then you would not know what to multiply the value of your function by.
Sorry Im not smart enough to know what a manifold is. Your way above my mathematical abilities, or lackthere of.

mathwonk
Homework Helper
i'm just saying you cannot integrate a function over a circle without knowing how to measure length on the circle, but you can integrate a differential.

a circle is an example of a manifold. a manifold is a curved space that is smooth everywhere.

mathwonk
Homework Helper
the reason dx = dx/dt dt, is the same reason that the number of feet in a certain length, equals the number of inches in that same lnegth times the number of feet in an inch, i.e. times 1/12.

i.e. if t is inches and x is feet then dx/dt = 1/12 = no. feet per inch, and

dx = no. feet, and dt = no. inches, and so dx = (dx/dt) dt.

mathwonk said:
the reason dx = dx/dt dt, is the same reason that the number of feet in a certain length, equals the number of inches in that same lnegth times the number of feet in an inch, i.e. times 1/12.

i.e. if t is inches and x is feet then dx/dt = 1/12 = no. feet per inch, and

dx = no. feet, and dt = no. inches, and so dx = (dx/dt) dt.
Ok good, so I did have the concept right then. I thought of it as the rate of change of x W.R.T t, but we want only the change in x, so we multiply by change in t, to get just the change in x.

Your post makes sense, thanks MathWonk, you never cease to amaze me. Last edited:
mathwonk