1
$\begingroup$

I am doing an exercise in which I am given a function $z(x,y)$ and I am introduced the "change of variables" $x=u\cos v$ and $y=u\sin v$. I am asked to calculate $\frac{\partial z}{\partial u}$ and applying chain rule I have done $\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v$ but with respect to second order partial derivative, it could be obtained like this?:

$\frac{\partial^2 z}{\partial u^2}=\frac{\partial }{\partial u}(\frac{\partial z}{\partial u})=\frac{\partial }{\partial x}(\frac{\partial z}{\partial u})\frac{\partial x}{\partial u}+\frac{\partial }{\partial y}(\frac{\partial z}{\partial u})\frac{\partial y}{\partial u}=\frac{\partial }{\partial x}(\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v)\frac{\partial x}{\partial u}+\frac{\partial }{\partial y}(\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v)\frac{\partial y}{\partial u}$

And if that is correct, when doing the derivatives, if for example $\frac{\partial x}{\partial v}=-u\sin v$, then we have that $\frac{\partial v}{\partial x}=-\frac{1}{u\sin v}$? Thanks for your help.

$\endgroup$
1
  • $\begingroup$ The formula for $\frac{\partial^\color{red}{2}z}{\partial u^2}$ is correct. What has this to do with $\frac{\partial x}{\partial v}$ ? $\endgroup$
    – Kurt G.
    Nov 2, 2022 at 16:02

2 Answers 2

1
$\begingroup$

We can use the notation $z_{x}=\frac{\partial z}{\partial x}$.

We have, \begin{align*} z_{uu}&=(z_{x}x_{u}+z_{y}y_{u})_{u}\\ &=(z_{x})_{u}x_{u}+z_{x}x_{uu}+(z_{y})_{u}y_{u}+z_{y}y_{uu}\\ &=\underbrace{(z_{xx}x_{u}+z_{xy}y_{u})}_{(z_{x})_{u}}x_{u}+z_{x}x_{uu}+\underbrace{(z_{yx}x_{u}+z_{yy}y_{u})}_{(z_{y})_{u}}y_{u}+z_{y}y_{uu}. \end{align*}Since $z$ is suppose to be $ C^{2}$, then $z_{xy}=z_{yx}$, so $$z_{uu}=z_{xx}x^{2}_{u}+2z_{xy}x_{u}y_{u}+z_{yy}y^{2}_{u}+z_{x}x_{uu}+z_{y}y_{uu}$$ Now, notice that $$x_{u}=(u+\cos v)_{u}=1,\quad y_{u}=(u+\sin v)_{u}=1, x_{uu}=0, \quad y_{uu}=0$$

$\endgroup$
0
$\begingroup$

There are a number of issues here. Differentials can be used algebraically, but only with improved notation. With the standard notation, you must apply the chain rule. This is doubly true of partial differentials.

Partial differentials cannot be flipped the same way that total differentials can, because the numerator and the denominator have different meanings. The $\partial v$ in $\frac{\partial v}{\partial x}$ is a DIFFERENT $\partial v$ than the one in $\frac{\partial x}{\partial v}$.

If you wanted a notation that was more amenable to algebraic manipulations such as these, see the paper by Fite and myself, "Total and Partial Differentials as Algebraically Manipulable Entities."

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .