20080924, 04:27  #1 
7·47 Posts 
Difficult Differential
Alright, I have a quick question regarding a partial derivative.
Given nRT = (P + (an^2)/V^2)*(Vnb), 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. 
20080924, 16:40  #2 
∂^{2}ω=0
Sep 2002
República de California
2·3^{2}·619 Posts 
Assuming I've had sufficient coffee this morning...
nRT = (P + (an^2)/V^2)*(Vnb), 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)*(Vc). 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)*(Vc) + (b + 1/V^2)*(dVc), now just rearrange to isolate dV. You can do it with all the original constants carried through. 
20080924, 19:35  #3 
19·337 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!

20080924, 21:06  #4 
Aug 2002
Ann Arbor, MI
433 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)*(Vc) + (b + 1/V^2)*(dV) 
20080924, 21:39  #5  
∂^{2}ω=0
Sep 2002
República de California
2×3^{2}×619 Posts 
Quote:


20080925, 03:33  #6 
Aug 2002
Ann Arbor, MI
661_{8} Posts 
Considering I spend 14 hours a week tutoring college students in math, I better be damn good at this stuff.

20080925, 05:34  #7 
8168_{10} 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.

20080925, 19:31  #8 
∂^{2}ω=0
Sep 2002
República de California
25606_{8} 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! 
20080926, 13:07  #9 
"Lucan"
Dec 2006
England
1100101001010_{2} 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 
20081001, 21:24  #10  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


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