Refer to the above figure. Assuming the length of the 3 edges of triangle are $r_0,z_0,\xi_0$. And we have $\xi=\sqrt{r^2+z^2}$ (Eqn.1)and $\xi_0=\sqrt{r_0^2+z_0^2}$. The normal vector on the hypotenuse is $\hat{n}$, and the components are $n_r=\frac{z_0}{\xi_0}$ and $n_z=-\frac{r_0}{\xi_0}$
Meanwile$\frac{\xi}{r}=\frac{\xi_0}{r_0}=\frac{1}{\cos{\theta}}$(Eqn.2) and $\frac{\xi}{z}=\frac{\xi_0}{z_0}=\frac{1}{\sin{\theta}}$(Eqn.3).
Now I have a funtion $T=T(r,z)$ on the hypotenuse, and I need to calculate $f=\frac{\partial{T}}{\partial{z}}n_r-\frac{\partial{T}}{\partial{r}}n_z=\frac{\partial{T}}{\partial{\xi}}\frac{\partial{\xi}}{\partial{z}}n_r-\frac{\partial{T}}{\partial{\xi}}\frac{\partial{\xi}}{\partial{r}}n_z$ (Eqn.4)
If using Eqn.1, then $\frac{\partial{\xi}}{\partial{z}}=\frac{z}{\sqrt{r^2+z^2}}$ (Eqn.1*a)and $\frac{\partial{\xi}}{\partial{r}}=\frac{r}{\sqrt{r^2+z^2}}$(Eqn.1*b) and
$f=\frac{\partial{T}}{\partial{\xi}}(\frac{z_0z}{\xi_0\sqrt{r^2+z^2}}+\frac{r_0r}{\xi_0\sqrt{r^2+z^2}})$. (Eqn.5)
If using Eqn.2 and Eqn.3, $\frac{\partial{\xi}}{\partial{z}}=\frac{\xi_0}{z_0}$ (Eqn.6)and$\frac{\partial{\xi}}{\partial{r}}=\frac{\xi_0}{r_0}$(Eqn.7) and
$f=\frac{\partial{T}}{\partial{\xi}}(\frac{\xi_0z_0}{z_0\xi_0}+\frac{\xi_0r_0}{r_0\xi_0})=2\frac{\partial{T}}{\partial{\xi}}$(Eqn.8)
It looks to me that Eqn.6 and Eqn.7 are right. but Eqn.8 is wrong. because when $\theta$ goes to 90degree,it should be $f=\frac{\partial{T}}{\partial{z}}=\frac{\partial{T}}{\partial{\xi}}$
So what is the right expression for $f$, and why?
Thanks in advance!
ok, Eqn.6/7 are wrong. they are not the partial derivative.