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#1 2006-10-04 14:41:37

mikau
Member
Registered: 2005-08-22
Posts: 1,504

the nasty old chain rule

since I have to take calc again in school, I'm trying to work on understanding more theroughly this time around.

a particles displacement is given by the function  p(t) its velocity is given by v(t)  and its acceleration a(t)

Now we know p'(t) = v(t) and v'(t) = a(t)

show that a(t) = v(t) dv/dp

now using leibnitz notation, its a walk in the park. a(t) = v'(t) = dv/dt   and v(t) = p'(t) = dp/dt

so dv/dt = dp/dt * dv/dp
dv/dt = dv/dt

No problemo. But I think the trouble with leibnitz notation is it allows you to fairly complex problems without really knowing whats going on. Course that could be its advantage as well, but what if we use newtons notation only..

dv/dp = ???   After some though it seems to me,  this is the derivative of
v(p(t)) but with respect to what variable? Using the chain rule, the derivative is v'(p(t))p'(t)  but I'm not sure if this is the derivative with respect to t or p(t). Rather confusing stuff....

Liebnitz notation does make sense to me but it bothers me I cant' seem to get it right the other way.

What bothers me is in dv/dp  is entirely dependant on dt. When we use newtons notation f'(t) the portion dt is sort of stuck there. We can't use an expression involving just dt.

I may be just confusing myself more.. :-(

Last edited by mikau (2006-10-04 14:54:35)


A logarithm is just a misspelled algorithm.

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#2 2006-10-04 18:35:14

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: the nasty old chain rule

Yeah, the chain rule is a bit confusing. I'll try to give an explanation, but I really should be in bed right now, so it'll have to be quick!

As you said, a(t)  = v'(p(t))p'(t), but you have to be careful what you mean by the primes - they aren't all the same. The original expression that you had was:

a = dv/dp * dp/dt

As you said, this is correct. I know this will make the math majors wince, but you really can think of the dp's as cancelling on the right hand side. So the prime on v is a derivative with respect to p and the prime on the p is a derivative with respect to t. If you're confused about how to use this formula, here's an example:

Say
v(p) = 4p+7
p(t) = 9t^2

Then a(t) = (4)*(18t) = 72t

To check this, you could solve the problem without the chain rule by writing v(t) = 4p + 7 = 4(9t^2)+7 = 36t^2 + 7
So a(t) = dv/dt = 72t. You get the same answer either way.

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#3 2006-10-05 01:44:03

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: the nasty old chain rule

Hi Mikau!  I don't understand any of this, but I wish I did.
My calculus understanding goes back to 1980's.
If you find any good reading on this subject, post the links please!


igloo myrtilles fourmis

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#4 2006-10-05 12:18:19

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: the nasty old chain rule

I don't entirely understand it myself, John E. I have no trouble differentiating, but I'm trying to theroughly understand precisly what is happening in each step. And thats the tricky thing. I don't know of any particularly good articles on this subject. But google "caclulus "the chain rule" " and you should find at least a simple explanation somewhere's.

Anyways, thanks fgarb, yeah canceling the differentials makes me wince as well which is why I posted this. The problem with newtons notation is it uses f'(x) to define dy/dx, there is no way to express just dy or dx.


A logarithm is just a misspelled algorithm.

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#5 2006-10-05 12:49:24

polylog
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Registered: 2006-09-28
Posts: 162

Re: the nasty old chain rule

I think a truly fundamental understanding of the chain rule would require writing it out using the limit definition, and then writing the definition of that limit using the delta-epsilon definition of limit.

Which should be equal to:


Now, how to prove that with delta-epsilon is beyond me, I always hated those proofs, but I think that would give a sufficiently deep understanding.

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#6 2006-10-05 13:03:47

polylog
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Registered: 2006-09-28
Posts: 162

Re: the nasty old chain rule

So the statement to prove would be:

If:

Then:

Which would be the fundamental 'from first principles' proof of the chain rule. smile

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#7 2006-10-05 13:04:56

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: the nasty old chain rule

good stuff, polylog. I have seen the delta-epsilon proof of the chain rule and it does make sense. Its not as hard as some D-E proofs I've seen. My confusion arises when I have two functions like v(t) and p(t) and we have to find dv/dp. How does this look in newtons notation and what does it represent? My guess is it means the derivative of v(p(t)) which would be a(p(t)) v(t)  but what the heck is a(p(t))? Hey! I think I just finally figured out what my question was!

Last edited by mikau (2006-10-05 13:15:39)


A logarithm is just a misspelled algorithm.

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#8 2006-10-05 13:11:57

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: the nasty old chain rule

Perhaps the chain rule will make more sense if you think about it this way. Imagine you put together a computer program to calculate derivatives. One way to do this would be to find the slope of your function at a given point. That is, for sufficiently small deltas:

[align=center]

[/align]

This can be made arbitrarily accurate at your point by shrinking delta so long as y(x) is well behaved near x. If it has infinities or other troublesome discontinuities within delta of x then this equality breaks down. When you're dealing with division instead of true derivatives cancelation makes more sense. Now say y and x both depend on t. Then,

[align=center]

[/align]

And you can make your substitutions:

[align=center]

[/align]

Does this seem reasonable? Makes sense to me anyway. This argument should work for the vast majority of functions you're likely to encounter in real world applications.

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#9 2006-10-05 13:20:11

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: the nasty old chain rule

Oh, I see.. so it's the derivative of v(t) with respect to the 'variable' p(t), and p(t) itself depends on t.

I guess in the prime notation this would be "v'(t) " where the prime is understood to mean "differentiate with respect to whatever v depends on", in this case p(t) ... I think the prime notation is limited in such cases, since for example say i want to write   d/ds f(t) = 0. The prime notation doesn't have a way to indicate partial derivatives.

As for the physical significance... I guess a(p(t)) means acceleration is dependent on a position function which is dependent on time?

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#10 2006-10-05 13:20:38

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: the nasty old chain rule

A similar argument can be made for the multivariable generalization of course. Say we can express a function as f(t1, ..., tn) or as f(x). Logically, changing x a small amount should change each t a corresponding small amount, which is related to the change in f. So, assuming that f is well behaved in each dimension we could make a computer program that uses this:

[align=center]

[/align]

And make the derivative substitutions:

[align=center]

[/align]

And this is a more general version of the chain rule that is very useful in certain situations.

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#11 2006-10-05 13:20:51

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: the nasty old chain rule

It does seem reasonable, and it makes sense. But there must be a way of doing it without liebnitz notation. So the first step would be to convert expressions like dv/dp into an exression inolving only newtons notation. But thats not as easy as I thought.

Last edited by mikau (2006-10-05 13:21:06)


A logarithm is just a misspelled algorithm.

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#12 2006-10-05 13:26:21

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: the nasty old chain rule

Polylog - that's right, the chain rule is used when you can either write a function as depending on one set of variables or as another set. You want to work with the function expressed using the first set of variables, but you want to take a derivative with respect to a variable in the second set - say if you had some function that was natural to look at in polar coordinates and you wanted to find d/dx of it.

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#13 2006-10-05 13:33:15

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: the nasty old chain rule

Personally I think using the prime notation (it that what liebnitz notation is?) when there's more than one variable involved is just asking for trouble. It's confusing enough without ambiguous notation to get in the way.

Btw, anyone know how to write partial derivatives in latex? Lower case deltas don't quite cut it. smile

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#14 2006-10-05 13:41:09

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: the nasty old chain rule

It's "\partial" I think .
Here is a test :

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#15 2006-10-05 13:41:10

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: the nasty old chain rule

polylog wrote:

Oh, I see.. so it's the derivative of v(t) with respect to the 'variable' p(t), and p(t) itself depends on t.

I guess in the prime notation this would be "v'(t) " where the prime is understood to mean "differentiate with respect to whatever v depends on", in this case p(t) ... I think the prime notation is limited in such cases, since for example say i want to write   d/ds f(t) = 0. The prime notation doesn't have a way to indicate partial derivatives.

As for the physical significance... I guess a(p(t)) means acceleration is dependent on a position function which is dependent on time?

Generally yeah, they're both ultimately dependant on t, but p(t) is a position function which takes a given time and output something different, a(t) is an acceleration function of time, both take time and output distance. Using v(p(t)) means we're interpeting a position or displacement as time, and thats definitly wrong. No, I think dV/dP is not the derivative of V(p(t)) using liebnitz notation you can multiply above and below by 1/dt and get dV/dt over dP/dt, this can be replaced with their equivelents in prime notation and get v'(t) / P'(t) which is equal to a(t) / v(t)  but again, we had to first resort to good ol' leibnitz by multiplying above and below by 1/dt. There must be a way to do it independant of liebnitz. In otherwords, what would newton do? lol...

Last edited by mikau (2006-10-05 13:41:59)


A logarithm is just a misspelled algorithm.

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#16 2006-10-05 13:43:37

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: the nasty old chain rule

Indeed! Although, wasn't the prime notation due to Lagrange?

And I see your point about the correct interpretation.

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