January 5th, 2017, 07:59 AM  #1 
Newbie Joined: Jan 2017 From: Canada Posts: 6 Thanks: 1  Cancelling differentials
I ran into the following formula while reviewing the covariant/contravariant base vectors: (âˆ‚x/âˆ‚u)(âˆ‚u/âˆ‚x) + (âˆ‚y/âˆ‚u)(âˆ‚u/âˆ‚y) + (âˆ‚z/âˆ‚u)(âˆ‚u/âˆ‚z) = âˆ‚u/âˆ‚u = 1 Can someone walk me through the first step here that gives "âˆ‚u/âˆ‚u". Why isn't it that you cancel the differentials âˆ‚x, âˆ‚y, and âˆ‚z which leaves you with: âˆ‚u/âˆ‚u+âˆ‚u/âˆ‚u+âˆ‚u/âˆ‚u = 3âˆ‚u/âˆ‚u = 3? 
January 5th, 2017, 11:25 AM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011  Quote:
$\dfrac{du}{dt} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial t} + \dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial t} + \dfrac{\partial u}{\partial z}\dfrac{\partial z}{\partial t}$ so letting $t=u$ $\dfrac{du}{du} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial u} + \dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial u} + \dfrac{\partial u}{\partial z}\dfrac{\partial z}{\partial u}$ the expression on the right is what you are given. differentials don't cancel like algebraic terms. You have to "cancel" them in terms of the chain rule.  
January 5th, 2017, 12:52 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,170 Thanks: 869 
It is not "cancelling" strictly speaking, it is the "chain rule".

January 5th, 2017, 02:07 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
$\displaystyle du=\frac{\partial u}{\partial x}dx +\frac{\partial u}{\partial y}dy +\frac{\partial u}{\partial z}dz$ and divide by du. Interpretation: u=u(x,y,z) > x=x(u,y,z), y=y(x,u,z), z=z(x,y,u) 
January 5th, 2017, 05:45 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,309 Thanks: 2443 Math Focus: Mainly analysis and algebra 
As has been said above, you can't just cancel differentials in the expression you gave. But nobody has given the correct reason. $\frac{\partial y}{\partial x}$ is not a ratio of differentials. It is the partial differential operator with respect to $x$ (that is $\frac{\partial}{\partial x}$) acting on $y$. So the real reason that you can't cancel differentials in your expression is that there aren't any! Of course, we use the notation that we do because in many instances there are theoretical reasons why "cancelling differentials" produces a valid result. In this case it is just the chain rule for multivariable calculus as stated above. Last edited by v8archie; January 5th, 2017 at 05:51 PM. 
January 5th, 2017, 08:13 PM  #6  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
 
January 6th, 2017, 05:08 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  differentials
Definition: $\displaystyle \frac{\partial u}{\partial x}=Lim\frac{u(x+\Delta x,y,z)}{\Delta x} \equiv \frac{\mathrm{d} u_{y,z}}{\mathrm{d} x}$ $\displaystyle \frac{\partial u}{\partial x}dx=\frac{\mathrm{d} u_{y,z}}{\mathrm{d} x}dx =du_{y,z}$, cancelling dx, where $\displaystyle du_{y,z}$ is a differential (partial) by definition, which is exactly analogous to dy in dy=f'(x)dx. $\displaystyle du = du_{y,z} + du_{x,z} + du_{x,y}$, or $\displaystyle du(dx,dy,dz) = du(dx) + du(dy) + du(dz)$ which says the total differential of u is the differential of u holding y,z fixed plus the differential of u holding x,z fixed, plus the differential of u holding x,y fixed. This is not the same as du = du + du + du =3du. The differentials on the right are not the same as the differentials on the left. Another way of interpreting it is that the function du of the variables dx.dy,dz is equal to the (different) function du of the variable dx, plus the (different) function du of the variable dy, plus the (different) function du of the variable dz. 
January 6th, 2017, 07:02 PM  #8 
Newbie Joined: Jan 2017 From: Canada Posts: 6 Thanks: 1 
Thanks everyone for the input on this. It appears that the solution here is based on the idea of already knowing the answer. I understand the chain rule and how to start with du/dt ... however is there a process that starts with (âˆ‚x/âˆ‚u)(âˆ‚u/âˆ‚x) + (âˆ‚y/âˆ‚u)(âˆ‚u/âˆ‚y) + (âˆ‚z/âˆ‚u)(âˆ‚u/âˆ‚z) and gets âˆ‚u/âˆ‚u? Or is this the common solution to calculus problems of "guessing" what the answer might be and seeing if the reverse process works. 
January 6th, 2017, 07:09 PM  #9 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out.  
January 7th, 2017, 08:48 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
Guess you missed this: Quote:
 

Tags 
cancelling, differentials 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Justification for cancelling terms in limits?  Mr Davis 97  Calculus  7  August 14th, 2015 12:58 PM 
How to justify the cancelling of variables in a rational expression?  Mr Davis 97  Calculus  3  February 19th, 2015 05:16 AM 
Differentials  delphine cormier  Calculus  1  August 17th, 2014 08:04 AM 
Cancelling factors.  zengjinlian  Probability and Statistics  2  April 4th, 2014 02:35 PM 
Differentials cos(57°)  math999  Calculus  1  May 5th, 2013 09:04 PM 