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Calculus Help

Anonymous Tue 01 Sep 00:37:23 2020 No.12071867 View Reply Original Report

Quoted By: >>12071978 >>12072061 >>12073910

I am fundamentally misunderstanding something integral to calculus.
I have applied the definition of a differential to a function of 2 variables, and I do some manipulations that lead to nonsense.
Perhaps I am misunderstanding something about differentials and derivatives or partial differentials and differentials, IDK.
Could someone tell me what rules I broke or what I am misunderstanding in pic related?

Furthur reading suggestions or links would be appreciated.
Thanks!

Anonymous

Anonymous Tue 01 Sep 2020 00:51:07 No.12071895 Report

Quoted By:

u=y

Anonymous

Anonymous Tue 01 Sep 2020 00:53:11 No.12071899 Report

Quoted By: >>12071945

are y and z meant to be functions of some other parameter?

Anonymous

Anonymous Tue 01 Sep 2020 01:14:57 No.12071945 Report

Quoted By: >>12072044

>>12071899
Potentially, would that make a difference?
My notation x = x(y,z) is to say that x is a function of y and z.
e.g. x = 2y + 6z or something along those lines.

Anonymous

Anonymous Tue 01 Sep 2020 01:28:57 No.12071978 Report

Quoted By: >>12072024

>>12071867
Once you apply the definition you can't go further: that's the differential of your function and that's it. When you divide by dz you're stretching the notation, but even then, the result is that dy/dz=0 (as y isn't a function of z), and you're left with the identity between the total derivative and the partial derivative, which makes sense only if z isn't a function of something else.

Anonymous

Anonymous Tue 01 Sep 2020 01:50:03 No.12072024 Report

Quoted By: >>12072149 >>12073728

>>12071978
Im confused when you say 'stretching the notation'? When you solve a difeq by separation of variables, you are shifting around dz's by dividing and multiplying, so what is wrong with doing it here?

Anonymous

Anonymous Tue 01 Sep 2020 02:03:56 No.12072044 Report

Quoted By:

>>12071945
x is a function of y and z. so you can take partial derivatives of x with respect to y or z. but at that point you are done. there is nothing to "chain"

Anonymous

Anonymous Tue 01 Sep 2020 02:10:07 No.12072061 Report

Quoted By:

>>12071867
$\partial x\neq dx$
You can't use one to fraction cancel the other.

Anonymous

Anonymous Tue 01 Sep 2020 02:53:19 No.12072149 Report

Quoted By: >>12072159

>>12072024
There are some things that you can do, and some things it's better not to do. From a purely mathematical standpoint juggling around the dx is a bit an abuse of the notation if done too carelessly, but it isn't usually a problem. The problem here is that you're trying to squeeze more informations than the equation can give you. When you separate the variables in a differential equation what you're doing is just regrouping the derivatives with the functions they can derive, but here when you move the dz you're just rearranging an expression into something that doesn't make formal sense: dx/dz is an abuse of notation because x isn't function of just z, that's why partial derivatives exist and have their own symbol.

Anonymous

Anonymous Tue 01 Sep 2020 02:58:07 No.12072159 Report

Quoted By:

>>12072149
It just clicked with me that dx/dz is not valid. Thanks for pointing that out.

Anonymous

Anonymous Tue 01 Sep 2020 03:02:51 No.12072169 Report

Quoted By:

Since the post is already here, and I am curious. Could I use the definition of the differential (used in going from eq. 0 to eq. 1 ) to get the d^2x ?

Anonymous

Anonymous Tue 01 Sep 2020 15:37:21 No.12073728 Report

Quoted By:

>>12072024
also what you do when you "shift around" the dz can be always done in a rigorous way without doing it, it's only a mnemonic notation thing that facilitates the resolution

Anonymous

Anonymous Tue 01 Sep 2020 16:54:27 No.12073910 Report

Quoted By: >>12073970

>>12071867
This is just the chain rule. We assume y is also a function of z if we don't know it explicitly. If y is independent of z, dy/dz will equal 0 and that term will go away. Hence:

$\frac{dx(y(z), z)}{dz}=\frac{\partial x(y(z), z)}{\partial [y(z)]} \frac{d [y(z)]}{dz} + \frac{\partial x(y(z), z)}{\partial z} = \frac{\partial x}{\partial y} \frac{d y}{dz} + \frac{\partial x}{\partial z}$

The "devide both sides by dz" explanation, which is a "trick" that happens to work, to me, is more confusing honestly

Anonymous

Anonymous Tue 01 Sep 2020 17:21:00 No.12073970 Report

Quoted By: >>12073972

>>12073910
btw, just to be clear:
$\frac{\partial x}{\partial y} \frac{d y}{dz} = \left( \frac{\partial x}{\partial y} \right) \cdot \left( \frac{d y}{dz} \right)<script type="math/tex"><br /><br />You can't just cancel anything out here.$

Anonymous

Anonymous Tue 01 Sep 2020 17:22:07 No.12073972 Report

Quoted By:

>>12073970
$\frac{\partial x}{\partial y} \frac{d y}{dz} = \left( \frac{\partial x}{\partial y} \right) \cdot \left( \frac{d y}{dz} \right)$

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