Definition of dy and dx?

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Narius
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Definition of dy and dx?

Postby Narius » Fri Oct 16, 2009 7:14 pm UTC

So I'm assuming this is a fairly basic Calc concept, but I'm still not sure I entirely get it.
dy/dx=y', the first derivative of a function y.
But dy/dx is not just a set symbol, because you can split it up by multiplying by dx.

So what exactly do the values dy and dx by themselves represent mathematically?

And if this question has been answered somewhere in these forums, please give me a link.
I tried a search, but typing dy dx in the search bar naturally pulls up a whole lot of forum links.
Thanks.

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Re: Definition of dy and dx?

Postby BlackSails » Fri Oct 16, 2009 7:50 pm UTC

dx is a very small change in x. Very very small.

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Re: Definition of dy and dx?

Postby mike-l » Fri Oct 16, 2009 7:51 pm UTC

For the purposes of starting calculus, it IS just a symbol. It's convenient notation because a lot of rules work the way you might think:

eg the chain rule:
[math]\frac{dg}{dx} = \frac{dg}{df}\frac{df}{dx}[/math]
or the fundamental theorem:
[math]\int \frac{dy}{dx} dx = \int dy = y[/math]

Intuitively, they are the differences in the y and x coordinates between two very close points on the graph of the function.

After some more math, you can assign more precise meanings to these things, but that comes later.
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Re: Definition of dy and dx?

Postby BlackSails » Fri Oct 16, 2009 7:55 pm UTC

Woods defines differentials in the following way.

The differential of an independant variable is equal to the increment of the variable; that is, dx=x_0-x_1

The differential of a function y=f(x) is the principal part of the increment of y and is given by dy=f'(x)dx.

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Re: Definition of dy and dx?

Postby Narius » Fri Oct 16, 2009 8:04 pm UTC

Thanks for the responses!
So for all calc uses for the time being, dy/dx can just be considered as a symbol? I just want to make sure I'm understand the stuff completely.

and I've seen the definition of dy=f'(x)dx in a lesson on linear approximation and differentials; unfortunately, though that may be the ultimate explanation, it just doesn't make sense to me.

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Re: Definition of dy and dx?

Postby Macbi » Fri Oct 16, 2009 8:21 pm UTC

Be aware that it's somtimes written as [imath]\frac{d}{dx}y[/imath]. So that you might see something like [imath]\frac{d}{dx}(x^2-3x)=2x-3[/imath]
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Re: Definition of dy and dx?

Postby Ended » Fri Oct 16, 2009 8:37 pm UTC

Narius wrote:it just doesn't make sense to me.
Don't worry, it's not just you. The notation in differential calculus is generally pretty confusing*. For now, you can just treat dy/dx, dy, dx as symbols, with the understanding that:

  • dx is a "small change in x"
  • dy is the "corresponding small change in y"
  • dy/dx is a function of x
  • you can "multiply by dx or cancel the dx's" in certain cases, but this is a kind of notational shortcut (but does have a deeper meaning).
__________
*
Spoiler:
Case in point: the triple product rule
[math]f(x,y,z)=0 \rightarrow \left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y = -1[/math]
For me at least, the notation completely fails to convey where the minus sign came from.
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Re: Definition of dy and dx?

Postby BlackSails » Fri Oct 16, 2009 9:01 pm UTC

I like the notation. It implies the equality of mixed partials.

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Re: Definition of dy and dx?

Postby antonfire » Fri Oct 16, 2009 10:21 pm UTC

It also implies various incorrect versions of the chain rule in higher dimensions.
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Re: Definition of dy and dx?

Postby RogerMurdock » Fri Oct 16, 2009 10:40 pm UTC

Ended wrote:
Narius wrote:it just doesn't make sense to me.

  • you can "multiply by dx or cancel the dx's" in certain cases, but this is a kind of notational shortcut (but does have a deeper meaning).


I've been told this by multiple teachers before, but I'm not yet advanced enough to know what that deeper meaning is. I'm assuming you're talking about when you find error, you "multiply" by dx to move it to the correct side of the equation, or at least that's what it appears you're doing. Do you have any quick explanation or wikipedia links? I don't even know what to search for really.

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Re: Definition of dy and dx?

Postby mmmcannibalism » Fri Oct 16, 2009 11:47 pm UTC

I look at dy and dx as delta y and delta x where the delta is very very tiny.
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Re: Definition of dy and dx?

Postby Ended » Sat Oct 17, 2009 12:45 am UTC

RogerMurdock wrote:
Ended wrote:
  • you can "multiply by dx or cancel the dx's" in certain cases, but this is a kind of notational shortcut (but does have a deeper meaning).
I've been told this by multiple teachers before, but I'm not yet advanced enough to know what that deeper meaning is. I'm assuming you're talking about when you find error, you "multiply" by dx to move it to the correct side of the equation, or at least that's what it appears you're doing. Do you have any quick explanation or wikipedia links? I don't even know what to search for really.

The dx things are called differentials. I've tried to give a sort-of-coherent explanation below (spoilered for wall of text), apologies if it's pitched too low or high.
Spoiler:
There are essentially two things going on. The first (1) is a notational convenience. The second (2) is to do with more advanced maths.

(1)

The derivative y' = dy/dx is usually defined as the limit of the fraction
[math]\frac{\delta y}{\delta x} = \frac{y(x+\delta x) -y(x)}{\delta x}[/math]
as [imath]\delta x \to 0[/imath] (basically, the value that this fraction approaches as you take the value [imath]\delta x[/imath] very small).

Now the fraction [imath]\frac{\delta y}{\delta x}[/imath] is just a ratio of two things and can be manipulated just like any fraction. In particular, let's suppose [imath]\delta x[/imath] is very small, so the fraction is a good approximation to the derivative y'. Then the following 'error estimate' holds true:
[math]\frac{\delta y}{\delta x} \approx y'[/math]
[math]\rightarrow \delta y \approx y' \delta x[/math]
This is fine. The trouble is that if we try to take the limit as [imath]\delta x \to 0[/imath], then [imath]\delta y \to 0[/imath] as well so we get 0 = 0 which is not very useful. Nevertheless we can take a kind of formal limit and still write
[math]dy = y' dx[/math]
even though we haven't really defined what dx and dy mean on their own.

The reason this is useful is because a lot of simple calculus identities are derived with limit arguments, and the behaviour of the real quantities [imath]\delta x, \delta y[/imath] is exactly mirrored by the behaviour of the formal quantities dx, dy. So we can skip the limiting arguments and just manipulate the formal quantities.


(2)

If you do more calculus you will see that the derivative can be generalized to functions in multi-dimensional spaces. In this context the differential of f is often thought of as function which gives you a linear function at each point of your domain, where the linear function is the closest approximation to f at that point.

Let's see how it works in one dimension. For a function f(x), we define the differential df as a function which takes a value p and returns f'(p)*x, which is a function (with argument x).

Now suppose we have f(x) = x. Then f'(x) = 1, so by our definition, df is always just the identity function 1x. And since f(x)=x, we have df = dx = 1x.

Therefore, going back to our definition, we can say that for a general function f, the differential df at any point p is the same as the function f'(p)*x which is the same as f'(p)dx. Therefore

[math]df(p) = f'(p) dx[/math]

at any point p, where the equality is an equality between functions of x. So we're back to dy = y' dx, but this time without any weird pseudo-limits. This approach generalises very powerfully. Someone who knows more differential geometry than me can comment further (or correct my mistakes :))
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Re: Definition of dy and dx?

Postby RogerMurdock » Sat Oct 17, 2009 3:31 am UTC

Ended wrote:
RogerMurdock wrote:
Ended wrote:
  • you can "multiply by dx or cancel the dx's" in certain cases, but this is a kind of notational shortcut (but does have a deeper meaning).
I've been told this by multiple teachers before, but I'm not yet advanced enough to know what that deeper meaning is. I'm assuming you're talking about when you find error, you "multiply" by dx to move it to the correct side of the equation, or at least that's what it appears you're doing. Do you have any quick explanation or wikipedia links? I don't even know what to search for really.

The dx things are called differentials. I've tried to give a sort-of-coherent explanation below (spoilered for wall of text), apologies if it's pitched too low or high.
Spoiler:
There are essentially two things going on. The first (1) is a notational convenience. The second (2) is to do with more advanced maths.

(1)

The derivative y' = dy/dx is usually defined as the limit of the fraction
[math]\frac{\delta y}{\delta x} = \frac{y(x+\delta x) -y(x)}{\delta x}[/math]
as [imath]\delta x \to 0[/imath] (basically, the value that this fraction approaches as you take the value [imath]\delta x[/imath] very small).

Now the fraction [imath]\frac{\delta y}{\delta x}[/imath] is just a ratio of two things and can be manipulated just like any fraction. In particular, let's suppose [imath]\delta x[/imath] is very small, so the fraction is a good approximation to the derivative y'. Then the following 'error estimate' holds true:
[math]\frac{\delta y}{\delta x} \approx y'[/math]
[math]\rightarrow \delta y \approx y' \delta x[/math]
This is fine. The trouble is that if we try to take the limit as [imath]\delta x \to 0[/imath], then [imath]\delta y \to 0[/imath] as well so we get 0 = 0 which is not very useful. Nevertheless we can take a kind of formal limit and still write
[math]dy = y' dx[/math]
even though we haven't really defined what dx and dy mean on their own.

The reason this is useful is because a lot of simple calculus identities are derived with limit arguments, and the behaviour of the real quantities [imath]\delta x, \delta y[/imath] is exactly mirrored by the behaviour of the formal quantities dx, dy. So we can skip the limiting arguments and just manipulate the formal quantities.


(2)

If you do more calculus you will see that the derivative can be generalized to functions in multi-dimensional spaces. In this context the differential of f is often thought of as function which gives you a linear function at each point of your domain, where the linear function is the closest approximation to f at that point.

Let's see how it works in one dimension. For a function f(x), we define the differential df as a function which takes a value p and returns f'(p)*x, which is a function (with argument x).

Now suppose we have f(x) = x. Then f'(x) = 1, so by our definition, df is always just the identity function 1x. And since f(x)=x, we have df = dx = 1x.

Therefore, going back to our definition, we can say that for a general function f, the differential df at any point p is the same as the function f'(p)*x which is the same as f'(p)dx. Therefore

[math]df(p) = f'(p) dx[/math]

at any point p, where the equality is an equality between functions of x. So we're back to dy = y' dx, but this time without any weird pseudo-limits. This approach generalises very powerfully. Someone who knows more differential geometry than me can comment further (or correct my mistakes :))


Thanks for the information, I understand the first approach but I struggled with the df and whatnot (I failed to see exactly what the second part had proven), but I'm going read through the wikipedia article and do a bit more research now that I know what they are called alone.

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Re: Definition of dy and dx?

Postby HenryS » Sat Oct 17, 2009 5:04 am UTC

In the context of a calculus class I think differentials are a terrible idea. Everything you can do with them you can do with linear approximation, and much less confusingly. In the class I'm teaching right now, we did exactly that.

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Re: Definition of dy and dx?

Postby MSTK » Sat Oct 17, 2009 9:01 am UTC

Like many things in mathematics, I did not understand the concepts of dx and dy until they were applied to Physics.

It all made sense, then -- dy and dx were infinitesimals. In this context, that infinitesimals have this definition: the smallest non-zero positive number. Or, the closest you can get to zero while still having a non-zero difference.
You sort have to let this sink in for a while to get this. Think about exactly how big this has to be. And then, in that context, think of how you would compare and do math with infinitesimals.

To me, they were just a cute idea in calculus. Neat, but honestly, you never really need to understand them. So here, I shall introduce the concept of the continuous distribution, from physics, where differentials and infinitesimals dance around and play with each other and with other numbers:

(spoilered for wall of text)

Spoiler:
The equation of center of mass for two points on a straight line is pretty easily derived. Let's say you have a point at x=1 and x=4, both of mass 1g. The center of mass would then be the average, (1+4)/2, at x=3, right?
However, let's say that the two points were of different masses. Let's say one is 2g and the other is 1g. That is, one is twice as "significant" as the other. How should we calculate this?
If the answer isn't immediately obvious, we can say "Hey, that's just like putting two 1g points at x=1, and having a three-way average!" [imath]\mathbf{R} = \frac{1+1+3}{3} ; \mathbf{R} = \frac{(2)*1 + (1)*3}{3} \approx 1.7[/imath]
From this, we seem to be able to get a general formula for the center of mass on two points A and B on a straight line: [imath]\mathbf{R} = \frac{m_a(x_a) + m_b(x_b)}{m_a+m_b}[/imath] It doesn't take that much to extend this to a group of, well...as many points as you need.
[math]\mathbf{R} = \, \frac{m_a(x_a) + m_b(x_b) + m_c(x_c) \ldots}{m_a+m_b +m_c \ldots}[/math]
[math]\mathbf{R} = \, { \sum m_i \mathbf{x}_i \over \sum m_i }[/math]
-- that is, the sum of all masses*positions over the sum of all masses. Pretty neat, eh?

That's cool and all, but we're here to do calculus. Discrete masses and x's aren't important to us. We're here to do what calculus was invented for: work with infinitesimals! So here is a typical application in the form of continuous distribution:

Imagine a bar 5 cm long, but it's not exactly an even bar. As you go from x=0 to x=5, it gets heavier. That is, it gets denser. What you have is a bar that's light on one side and heavier on the other. The equation for the density of this bar is [imath]\lambda(x) = 2x[/imath]. So, for example, at x=2, the density is 4. At x=0, the density is 0. At x =5, the density is 10.
Find the center of mass


Huh. That's interesting. If you were a silly person unenlightened with the ways of calculus, you could say "Hm. Well why don't I just split that bar into 6 equidistant points at x=0, x=1, etc. and then find the center of mass from those 5 points?"
It doesn't exactly work that way. For one, how would you find the total mass? And, even still, how could you find the masses of the individual points?

Here's the beauty of the differential approach: consider it as the center of mass of infinitely many points, each with a length of dx.

Image


So now we need to find a function for the mass of dx. Obviously, this mass would be getting higher the more right you go and...
Oh wait, what? How can an infinitely tiny length have mass? Let's see how it all works out.

-- We're trying to find length as a function of mass. What do we know that links length and mass? A little thing called density.
now, we know density = mass/length. Let's call our density [imath]\lambda[/imath]. Let's see what we can do to solve for mass in terms of length:
[imath]\lambda = m/l[/imath]
[imath]m = \lambda \, l[/imath]
There! We got one. Now, one final step to make this relevant: we need the mass when l = dx. How can we turn l into dx? Differentiation, my friend. Let's differentiate both sides over dx...
[imath]\frac{dm}{dx} = \lambda \, \frac{dl}{dx} = \lambda \, (1)[/imath]
[imath]dm = \lambda \, dx[/imath] -- and, using [imath]\lambda = 2x[/imath] (given)
[imath]dm = 2x \, dx[/imath]

Looks like we now have the mass of dx. True to what we thought, the higher x is, the bigger the mass is. Now there's something else of note here -- something that will probably lend light on what differentials actually are:
the differential length is defined in terms of differential mass
Did you see that? Differentials and "normal" numbers are on completely different scales, on (literally) infinite orders of magnitude. It makes no sense to give a "mass" of a practically-zero length. However, you here have a practically-zero mass for a practically-zero length. Also note that there is a relation between the two! That some differentials are bigger than others. For example, when x > 1, the value of dm is actually bigger than the value of dx. That is, one practically-zero number is bigger than another practically-zero number.

Anyways, moving on, from this we can find the total mass of the bar. How? We add up infinitely tiny amounts of masses, from x=0 to x=5. If you're unfamiliar with this concept, it's called Integration.
[math]M = \int_{x=0}^{x=5} dm[/math]
(or -- the total mass is equal to the sum of the infinitely many dm's from x=0 to x=1)
We can substitute in what we found dm to equal -- 2x dx.
[math]M = \int_0^5 2x \, dx[/math]
(or, the total mass is equal to the sum of the infinitely many 2x dx's from 0 to 5)

Following very, very basic integration rules, like the equivalent of the power rule, we get M = 25. So we now know our total mass is 25.

That's the bottom of our Center-of-Mass equation. Now what's the numerator?

Recall that it is the sum of m*x for every particle. Well, in our case, we have an infinite amount of particles. We can see this as easily being dm * x, or x dm. And we know dm = 2x dx. So we can substitute everything in...
[math]\int_0^5 (x) (2x \, dx)[/math]
(the sum of infinitely many [x*(2x dx)]'s -- which are [x*dm]'s -- from 0 to 5)

Applying simple integration rules, we get 250/3, or 83.3~.

Now, final step! Our center of mass is the quotient of these two numbers. 83.3~ / 25 = 3.3~

So, our center of mass is at x = 3.333...

That makes sense, right? If our x goes from 0 to 5 and is heavier on the 5 side, you'd expect it to find the center of mass on the left side of the bar. And now, we know exactly where it is.


Anyways, hopefully from seeing dx's and other differentials and infinitessimals being actually used and interacting with each other instead of just other normal numbers, you can start to see what they exactly are, really.

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Re: Definition of dy and dx?

Postby MostlyHarmless » Sat Oct 17, 2009 9:33 am UTC

If you're really curious, dy and dx are differential forms. (Specifically, one forms). Don't worry if this goes way over your head. It's way over mine too, and I've been studying it.

The basic idea is that calculus is a way to talk about infinitely small changes. One of the best ways to describe an infinitely small change is to say where it starts and what direction it goes. We call this pair a tangent vector. Derivatives depend on these two pieces of information (the point you start at and the direction in which you move). In particular, derivatives should be linear functions, so we restrict ourselves to linear functions of tangent vectors. These linear functions are called cotangent vectors. Differential forms like dx, dy and df are special collections of these cotangent vectors.

Spoiler:
For a little less hand waving (very little less): Sufficiently nice spaces (like the real numbers, but let's call an arbitrary one M) have several associated vector spaces. One of these is called the tangent space, or TM. Every point in the tangent space is a vector starting from a point in M. These vectors are just like tangent lines that you learned about in calculus.

Another one of these spaces is the cotangent space, or T*M. Every point in the cotangent space is a linear function from TM to the real numbers. That means that a point in T*M takes tangent vectors as input and gives real numbers as output. A differential form (like dx or dy or df) is a section (think of it as a slice) of the space T*M.


Edit: To sum up, because I realized that that may not have had a very clear ending, differentials are functions that take two pieces of information as input: where you start (a point in M) and what direction you go in (a vector in M) and give you a number. This number is basically how much you change when you move a tiny (infinitely tiny, in fact) distance in that direction. The notions of tangent spaces and cotangent spaces are there so that we can talk precisely about "moving an infinitely tiny distance".
Last edited by MostlyHarmless on Sat Oct 17, 2009 12:12 pm UTC, edited 1 time in total.

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Re: Definition of dy and dx?

Postby Ended » Sat Oct 17, 2009 11:06 am UTC

RogerMurdock wrote:Thanks for the information, I understand the first approach but I struggled with the df and whatnot (I failed to see exactly what the second part had proven)
It doesn't really prove anything, it's just a way of defining things like dx so that they behave like we expect them to. But it's unnecessary for a lot of purposes.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola


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