January 5th, 2017, 08:59 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, 12:25 PM  #2  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 817  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, 01:52 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,824 Thanks: 752 
It is not "cancelling" strictly speaking, it is the "chain rule".

January 5th, 2017, 03:07 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
$\displaystyle du=\frac{\partial u}{\partial x}dx +\frac{\partial u}{\partial y}dy +\frac{\partial u}{\partial z}dz$ 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, 06:45 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,037 Thanks: 2343 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 06:51 PM. 
January 5th, 2017, 09:13 PM  #6  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  Quote:
 
January 6th, 2017, 06:08 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  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, 08:02 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, 08:09 PM  #9 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,413 Thanks: 482 Math Focus: Yet to find out.  
January 7th, 2017, 09:48 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  Quote:
Guess you missed this: Quote:
 

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