mersenneforum.org  

Go Back   mersenneforum.org > New To GIMPS? Start Here! > Homework Help

Reply
 
Thread Tools
Old 2008-09-24, 04:27   #1
Unregistered
 

11011101102 Posts
Default Difficult Differential

Alright, I have a quick question regarding a partial derivative.

Given nRT = (P + (an^2)/V^2)*(V-nb), find the partial V respect to T. Isolating V prior to taking the partial derivative seems to be notoriously difficult unless I am missing something obvious. My original plan was to expand, set to 0, factor and then try and use the quadratic equation, eliminate the extraneous root. However, when trying to do this, it is impossible to factor so that your ony terms are V^2, V and a constant- you cannot completely factor out a cubic V. Perhaps I am missing something.

Can anyone provide some advice or hints to lead me on the right path? Thanks.
  Reply With Quote
Old 2008-09-24, 16:40   #2
ewmayer
2ω=0
 
ewmayer's Avatar
 
Sep 2002
República de California

22×3×72×19 Posts
Default

Assuming I've had sufficient coffee this morning...

nRT = (P + (an^2)/V^2)*(V-nb), find the partial V respect to T

Rescaling to lump all the constants into a minimal set [for the purpose of the partial w.r.to T, I presume that everything but V and T are constants]:

T = (b + 1/V^2)*(V-c). Differentiate both side w.r.to T to get, using dV as shorthand for "partial of V w.r.to T":

1 = (b - 2*dV/V^3)*(V-c) + (b + 1/V^2)*(dV-c), now just rearrange to isolate dV.

You can do it with all the original constants carried through.
ewmayer is offline   Reply With Quote
Old 2008-09-24, 19:35   #3
Unregistered
 

258310 Posts
Default

Thanks, this appears to be similar to just using implicit differentiation and then isolating partial V respect to T. An explicit solution is possible, but only through use of the Cubic Formula (much worse than the quadratic equation). This is because each appearance of V in the actual function is actually v(t), or a function of T, therefore it must be treated as such. Thank you for the help!
  Reply With Quote
Old 2008-09-24, 21:06   #4
Kevin
 
Kevin's Avatar
 
Aug 2002
Ann Arbor, MI

6618 Posts
Default

The method ewmayer used is called implicit differentiation, if you haven't seen it before.

http://en.wikipedia.org/wiki/Implici...ifferentiation

Also, ewmayer screwed up a few derivatives in the final expression =P. It should read

1 = ( -2*dV/V^3)*(V-c) + (b + 1/V^2)*(dV)
Kevin is offline   Reply With Quote
Old 2008-09-24, 21:39   #5
ewmayer
2ω=0
 
ewmayer's Avatar
 
Sep 2002
República de California

1117210 Posts
Default

Quote:
Originally Posted by Kevin View Post
The method ewmayer used is called implicit differentiation, if you haven't seen it before.

http://en.wikipedia.org/wiki/Implici...ifferentiation

Also, ewmayer screwed up a few derivatives in the final expression =P. It should read

1 = ( -2*dV/V^3)*(V-c) + (b + 1/V^2)*(dV)
Thanks for being my "caffeine conscience, Kevin - apparently I'd forgotten how to differentiate a constant - tricky stuff, that. :)
ewmayer is offline   Reply With Quote
Old 2008-09-25, 03:33   #6
Kevin
 
Kevin's Avatar
 
Aug 2002
Ann Arbor, MI

43310 Posts
Default

Considering I spend 14 hours a week tutoring college students in math, I better be damn good at this stuff.
Kevin is offline   Reply With Quote
Old 2008-09-25, 05:34   #7
Unregistered
 

24×3×53 Posts
Default

Yes, at first I was trying for an explicit solution, which is impossible unless you use the very unwieldy and ugly Cubic Formula. However, I did not realize that you could differentiate implicitly, which can be done in only a few easy steps, going back to the good old Calc 1 days. The differential is rather easy once you know the correct method to use. Thank you for pointing me on the right path.
  Reply With Quote
Old 2008-09-25, 19:31   #8
ewmayer
2ω=0
 
ewmayer's Avatar
 
Sep 2002
República de California

22×3×72×19 Posts
Default

A related useful differentiation technique is Logarithmic Differentiation.

The above site has a whole slew of "Calc I tricky differentiation techniques" - just go up 2 directories to see the whole list.

Glad we were able to help!
ewmayer is offline   Reply With Quote
Old 2008-09-26, 13:07   #9
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

I know some members of this forum are allergic to physics,
but Van der Waals equation (which this is) is quite instructive
to derive, as an example of the principle that
rate of change of momentum of a system = total external force.
"a" accounts for molecular attraction, and "b" for their finite size.

David
davieddy is offline   Reply With Quote
Old 2008-10-01, 21:24   #10
Orgasmic Troll
Cranksta Rap Ayatollah
 
Orgasmic Troll's Avatar
 
Jul 2003

641 Posts
Default

Quote:
Originally Posted by ewmayer View Post
A related useful differentiation technique is Logarithmic Differentiation.

The above site has a whole slew of "Calc I tricky differentiation techniques" - just go up 2 directories to see the whole list.

Glad we were able to help!
For easier navigation: www.calculus.org (inconsequential fun fact: the other managing editor wrote one of my recommendation letters for grad school)
Orgasmic Troll is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
A non-linear differential equation Random Poster Math 2 2010-07-18 22:31
A rather difficult lottery Oddball Lounge 2 2010-05-06 02:18
Calculating a difficult sum CRGreathouse Math 3 2009-08-25 14:11
Why is RH so difficult to prove? Damian Math 31 2008-10-03 02:11
Geometry puzzle(difficult) hyh1048576 Puzzles 6 2003-07-27 06:46

All times are UTC. The time now is 08:34.

Thu Jun 4 08:34:51 UTC 2020 up 71 days, 6:07, 0 users, load averages: 1.15, 1.22, 1.27

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.