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# dX notation...

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Can someone please help with understanding dX, whatever notation. I know that if you have a expression 3x^2, the derivative is 6x, and is called d/dX. Why d/dX? Also if you have y = (same expression) the derivative becomes dY/dX. What does this mean? Even further, when you integrate, specifically for my physics class. When you have something like instantaneous velocity(V) it''s equation is dx/dt (derivative of position function). Then when we are given that formula, my physics teacher uses the dx/dt like a variable and says velocity times dt is dx ( V * dT = dX). From V = dX/dT to V * dT = dX Then he integrates both sides: §(V * dT) = §(dX) The integral of dX(or any other variable is just the variable). Just before the final step(or answer) he gets: T * §(v) = X Basically, all we end up doing is integrating the velocity function. Can anyone shed some light? And if anyone has been so generous as to answering that, when discuss work and engery, we have force(F) is the derivative of work(W) over the displacement(X). F = dW/dX; Then to find work we get: dW = F * dX; then §(dW) = §(F * dx) then W = §(F) * X Then comes finding the integral of force...we do that, but then the X part just disapppears. That doesn''t make sense to me. Answers? Thanks a great great deal to anyone who answers By the way, this is a Physics AP course at my high school, equivalent to a college Physics BC course, focused on mechanics. "Ogun''s Laughter Is No Joke!!!" - Ogun Kills On The Right, A Nigerian Poem.

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quote:
Original post by EbonySeraph
Can someone please help with understanding dX, whatever notation. I know that if you have a expression 3x^2, the derivative is 6x, and is called d/dX. Why d/dX? Also if you have y = (same expression) the derivative becomes dY/dX. What does this mean?

Wow, someone did a poor job teaching you derivatives. It's not your fault, and it's good that you're seeking answers. I'll just tackle one aspect as it's late and I have homework of my own.

When doing power derivatives, the derivative of a given polynomial equation is the formula for the slope of the graph of the equation at a given point. For example, as you said, the derivative of f(x)=3x^2 is 6x. This means that the slope of a point (yeah, just a sec...) on the graph of f(x) is 6x. Now, as you're asking, "How can a point have a slope?" This really means that the slope from this point to the "next" point is 6x, but here's something interesting: dX is really tiny, as it represents the smallest amount you can move along the x-axis (or the smallest amount you want to consider); dY is also really tiny, since it represents the amount you move along the y-axis per step along the x-axis; BUT when you say dY/dX you get a perfectly reasonable number (in this case 6x), because you're in reality saying "the smallest difference between two solutions of this equation" i.e. the slope .

Hmm. that's a terrible and rushed explanation...hope someone else has the time to answer better.

Peace,
ZE.

//email me.//zealouselixir software.//msdn.//n00biez.//

[edited by - zealouselixir on October 20, 2002 11:45:46 PM]

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Although this is a homework question, it is clear that EbonySeraph has been working hard to figure this out on his own, and isn''t simply asking for an answer. So I''ll allow the thread to continue. I don''t mind talking about the origin of the d?/dx notation, fundamental theorem of calculus, definition of derivative. But I really don''t want to get into specifics of the equations used in the AP physics course. EbonySeraph, please use what you learn here to help you communicate better with your teacher and fellow students.

That said, I wonder if that AP Physics course was designed to be taken in parallel with an AP Calculus course. It does seem that the teacher neglected to teach the basic calculus that you''re seeming to need in the physics course. You seem to have not picked up on the fundamental theorem of calculus (which would help explain the integration of V = dx/dt to get a formula for x, for example), and the limit definition of a derivative (which would explain the d?/dx notation).

You can think of the d?/dx notation as being kind of short-hand for delta_whatever/delta_x, where the delta''s are relative changes in the numerator an denominator. The greek symbol capital delta (not sure what the HTML code is for that but its a triangle with one edge along the bottom and a vertex at the top) is often used to represent a finite (not infinitesimal) change in a variable. Since the English language word for that greek symbol starts with a ''d'' the calculus notation uses ''d'' (at least that is one logical reason for the use of ''d'' in derivative notation).

Lets look at the definition of a derivative, which says that given a function f(x), the derivative df/dx is defined to be:

df/dx = limit(as h->0) of [f(x+h) - f(x)]/[(x+h) - x]

Here, h is some finite number, positive or negative and typically fairly small in value.

If you look at the right-hand side, you have in the numerator
f(x+h) - f(x), which is a change in f, a "delta"_f, that is measured relative to a change in x, a "delta"_x, given in the denominator as (x+h) - x. So really, its like:

df/dx = limit(as h->0) of delta_f/delta_x = change in f/change in x. Remind you of the old slope of a 2D line equation? Same thing, but in general derivatives can deal with curves.

Now, what''s that limit about? Well, those d''s in the derivative are NOT finite delta''s actually. They are infinitesimal delta''s. This is a very hard thing to imagine. Both the numerator and denominator become infinitely small as h->0 (i.e., as h approaches the value of 0), but the ratio df/dx does NOT become 0/0. It approaches something that may be a finite, well-defined number, based on the rate at which df and dx approach zero. If they do not approach zero at the same rate, then df/dx will not be the undefined 0/0 when h->0. The derivative formula for polynomials such as x2 can be derived quite easily to prove this, and there was another recent thread (yours?) where this was proven.

You also asked about the separation of the numerator and denominator to do integration, e.g., if dy/dx = f, then you can separate to get dy = f*dx and then integrate each side to find a formula for y. Well if you think about the df, dy, d?, dx as being the delta''s I mentioned, ignoring the fact that they are infinitesimally small, then it kind of makes sense. Why can''t you just multiply both sides by the denominator of dy/dx? Well, in many cases you can and that''s what''s happening. dy, dx are just treated like algebraic variables.

Integration refers to the calculation of anti-derivatives, which are basically the functions that, once you apply the limit definition of derivative, lead to the derivative you started with. For polynomial functions, which is the type of thing you''re finding in your AP physics course (and a first-level college physics course, as you mentioned), integrating to find the anti-derivatives is quite easy and you can do it by observation or simple formula.

The individual numerator and denominator df, dy, dx, those are called "differentials." You may have heard of something called differential equations and that''s sort of where that name comes from.

As to the specifics of doing the integration of V = dx/dt to get an equation for x, and the integration of F = dw/dx to get work, I really ask that you speak to your teacher or classmates. As I mentioned before, I don''t mind clarifying some theory, given that you''ve been struggling yourself to understand. But I don''t really want to get into the specific details of your class.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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As I wake up in the morning seeing these two replies, I graciously thank both of you a lot for your help. I can''t ready all of it now(I''ll finish in school) but I should have mentioned, I am taking a AP Calculus BC course too. In my PreCalc class last year, and so far this year, when we take derivatives, we use f(x) = some function, and it''s derivative is noted as f''(x) - f prime of X. The only time we used d?/d? notation was for related rates - which I utterly failed(not literally) a test on, and implicit derivations. Taking a derivate is easy - it''s the notation thats killing me. I''ll read the rest of grhodes_at_work post at school as I don''t have time now.

Thanks again, both of you.

"Ogun''s Laughter Is No Joke!!!" - Ogun Kills On The Right, A Nigerian Poem.

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Just one final comment. The f', f' notation is shorthand. Really, when f = f(x), f' by definition is df/dx and f' by definition is d2f/dx2. The superscripts indicate, hopefully obviously, first, second, n-th derivatives. They confuse the notation further since the superscript in the numerator comes after the d and before the variable, but in the denominator the superscript comes after the variable.

There is another notation, common in the area of differential equations, which I prefer. This notation uses subscripts to indicate differentiation. The notation looks like this:

fx = df/dx

fxx = d2f/dx2

fxy = d2f/dxdy (2nd partial derivative)

The subscripts just leave no doubt as to what the derivative is, but that notation is much more compact than d/dx.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

[edited by - grhodes_at_work on October 21, 2002 2:02:58 PM]

[edited by - grhodes_at_work on October 21, 2002 2:04:04 PM]

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Graham has provided most of the answers you require on the understanding of derivative notation and origins... I just want to point something out that may help you understand what your teacher was doing with the physics problems.

First:
quote:
Original post by EbonySeraph
When you have something like instantaneous velocity(V) it''s equation is dx/dt (derivative of position function).

This is a key point. The velocity v is an instanateous velocity (rate of change of displacement with respect to time) and is really just a shorthand way of writing dx/dt. Think of dx as representing the instantaneous change in displacement (but not with respect to a change in any other variable).

quote:
Original post by EbonySeraph
Then when we are given that formula, my physics teacher uses the dx/dt like a variable and says velocity times dt is dx ( V * dT = dX).

Each of the infinitesimals dx and dt can be replaced by the limit of the finite differences delta_x and delta_t and can thus be treated as variables. This is a very ''loose'' explanation and the one taught to students up to about 3rd year university level. If you take a subject on advanced geometrical methods the truth comes out... but isn''t necessary unless you want to go on to study manifolds.

quote:
Original post by EbonySeraph
From
V = dX/dT to V * dT = dX

Then he integrates both sides:

§(V * dT) = §(dX)

The integral of dX(or any other variable is just the variable). Just before the final step(or answer) he gets:

T * §(v) = X

The best way to think of this problem is that by definition, the displacement of the object is the integral (over t) of the instantaneuous velocity function V(t). So that §V(t)dt = X. Everything else he did was just to get to this definition from the definition of instantaneous velocity... in other words, yes, it''s a circular argument he has applied... but probably because he wanted to show the equivalence of the two definitions.

Cheers,

Timkin

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Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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If anyone would care to refresh my memory...

is dy/dx "change in y over change in x", OR "the derivative of y with respect to x", or are they equivalent?

(I can''t wait until I take multivariable calculus.. linear algebra is nice, but calculus is more interesting IMHO)

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quote:
Original post by DyDx
is dy/dx "change in y over change in x", OR "the derivative of y with respect to x", or are they equivalent?

The latter...

They''re only equivalent if you take the limit (as change in x goes to zero) of change in y over change in x.

Timkin

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quote:
Original post by EbonySeraph, original response by Timkin
Then when we are given that formula, my physics teacher uses the dx/dt like a variable and says velocity times dt is dx ( V * dT = dX).

Each of the infinitesimals dx and dt can be replaced by the limit of the finite differences delta_x and delta_t and can thus be treated as variables. This is a very ''loose'' explanation and the one taught to students up to about 3rd year university level. If you take a subject on advanced geometrical methods the truth comes out... but isn''t necessary unless you want to go on to study manifolds.

Something else I''d like to say on the matter. When using substitution methods (such as the famous u-substitution method) in Calculus to find, say, the antiderivative of a function, it really helps if you think of dx and dy as variables. As Timkin said, "the truth comes out...", implying that they aren''t really variables per say, but if you think of them as such, you can then go on and perform simple algebraic manipulations to help you picture what you should be substituting where. In your case, these manipulations help you relate the calculus aspect to the physics aspect.

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Your teacher seems to be using the same techniques to solve those simple integration problems as the ones that we used in Differential Equations class to solve the intro functions.

Physics equations *usually* use the DiffEq approach, like the "spring" equation involves a hefty equation, and orbital path is easy to write in DiffEq form, but it's its own nightmare to solve.

He's probably just used to using that method for both DiffEq and typical Integrals.

I tried to work out why the X disappeared, and... well... I have no idea anymore. I purged tricky details about DiffEq and Calculus after graduating College.

I thought that Work = Force * Distance... so the X should still be there...?

8(

I usually think of d/dX as "you derived an unnamed function with respect to X" and dY/dX as "you derived a function named Y with respect to X"...

d2Y/dX2 "you took the second derivative of Y, with respect to X twice"

makes more sense when you look at:

d2Y/dXdY "you took two derivatives of Y, one with X, one with Y"
(I forgot which one was derived first, but it's not the point.)

Does that help any??

Oh yeah, and don't worry much about when you can call it "the change in" or "with respect to"... all bets are off when you have to integrate them.

[edited by - Nypyren on October 22, 2002 5:03:55 AM]

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