mersenneforum.org Difficult Differential
 Register FAQ Search Today's Posts Mark Forums Read

 2008-09-24, 04:27 #1 Unregistered   24×3×89 Posts 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.
 2008-09-24, 16:40 #2 ewmayer ∂2ω=0     Sep 2002 República de California 9,791 Posts 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.
 2008-09-24, 19:35 #3 Unregistered   1F0416 Posts 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!
 2008-09-24, 21:06 #4 Kevin     Aug 2002 Ann Arbor, MI 43310 Posts 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)
2008-09-24, 21:39   #5
ewmayer
2ω=0

Sep 2002
República de California

9,791 Posts

Quote:
 Originally Posted by Kevin 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. :)

 2008-09-25, 03:33 #6 Kevin     Aug 2002 Ann Arbor, MI 433 Posts Considering I spend 14 hours a week tutoring college students in math, I better be damn good at this stuff.
 2008-09-25, 05:34 #7 Unregistered   6,397 Posts 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.
 2008-09-25, 19:31 #8 ewmayer ∂2ω=0     Sep 2002 República de California 9,791 Posts 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!
 2008-09-26, 13:07 #9 davieddy     "Lucan" Dec 2006 England 11001001100112 Posts 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
2008-10-01, 21:24   #10
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by ewmayer 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)

 Similar Threads Thread Thread Starter Forum Replies Last Post Random Poster Math 2 2010-07-18 22:31 Oddball Lounge 2 2010-05-06 02:18 CRGreathouse Math 3 2009-08-25 14:11 Damian Math 31 2008-10-03 02:11 hyh1048576 Puzzles 6 2003-07-27 06:46

All times are UTC. The time now is 08:20.

Wed Oct 21 08:20:17 UTC 2020 up 41 days, 5:31, 0 users, load averages: 1.00, 1.30, 1.36