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January 5th, 2017, 08:59 AM   #1
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Cancelling differentials

I ran into the following formula while reviewing the covariant/contravariant base vectors:

(∂x/∂u)(∂u/∂x) + (∂y/∂u)(∂u/∂y) + (∂z/∂u)(∂u/∂z) = ∂u/∂u = 1

Can someone walk me through the first step here that gives "∂u/∂u".

Why isn't it that you cancel the differentials ∂x, ∂y, and ∂z which leaves you with:

∂u/∂u+∂u/∂u+∂u/∂u = 3∂u/∂u = 3?
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January 5th, 2017, 12:25 PM   #2
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Quote:
Originally Posted by photogo01 View Post
I ran into the following formula while reviewing the covariant/contravariant base vectors:

(∂x/∂u)(∂u/∂x) + (∂y/∂u)(∂u/∂y) + (∂z/∂u)(∂u/∂z) = ∂u/∂u = 1

Can someone walk me through the first step here that gives "∂u/∂u".

Why isn't it that you cancel the differentials ∂x, ∂y, and ∂z which leaves you with:

∂u/∂u+∂u/∂u+∂u/∂u = 3∂u/∂u = 3?
it's a bit clearer if you find $\dfrac {du}{dt}$ first and then let $t = u$

$\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.
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January 5th, 2017, 01:52 PM   #3
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It is not "cancelling" strictly speaking, it is the "chain rule".
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January 5th, 2017, 03:07 PM   #4
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$\displaystyle du=\frac{\partial u}{\partial x}dx +\frac{\partial u}{\partial y}dy
+\frac{\partial u}{\partial z}dz$

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)
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January 5th, 2017, 06:45 PM   #5
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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.
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Last edited by v8archie; January 5th, 2017 at 06:51 PM.
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January 5th, 2017, 09:13 PM   #6
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Quote:
Originally Posted by romsek View Post
it's a bit clearer if you find $\dfrac {du}{dt}$ first and then let $t = u$

$\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}$
What are x,y,and z functions of? t? Then it's dx/dt, dy/dt, dz/dt
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January 6th, 2017, 06:08 AM   #7
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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.
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January 6th, 2017, 08:02 PM   #8
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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 (∂x/∂u)(∂u/∂x) + (∂y/∂u)(∂u/∂y) + (∂z/∂u)(∂u/∂z) and gets ∂u/∂u?

Or is this the common solution to calculus problems of "guessing" what the answer might be and seeing if the reverse process works.
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January 6th, 2017, 08:09 PM   #9
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Quote:
Originally Posted by photogo01 View Post
Thanks everyone for the input on this.
It appears that the solution here is based on the idea of already knowing the answer.
All math is easy once you know how
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January 7th, 2017, 09:48 AM   #10
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Quote:
Originally Posted by photogo01 View Post
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 (∂x/∂u)(∂u/∂x) + (∂y/∂u)(∂u/∂y) + (∂z/∂u)(∂u/∂z) and gets ∂u/∂u?

Or is this the common solution to calculus problems of "guessing" what the answer might be and seeing if the reverse process works.
The solution is based on learning basic principles (not jargon), and, ok, in the process coming across this type of equation. There are times when you guess a solution.

Guess you missed this:

Quote:
Originally Posted by zylo View Post
$\displaystyle du=\frac{\partial u}{\partial x}dx +\frac{\partial u}{\partial y}dy
+\frac{\partial u}{\partial z}dz$

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)
Missed something in previous post so I'll clean it up here.

Quote:
Originally Posted by zylo View Post
Definition:
$\displaystyle \frac{\partial u}{\partial x}=Lim\frac{u(x+\Delta x,y,z)-u(x,y,z)}{\Delta x} $
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