How to Calculate the Partial Derivative of a Vector in Spherical Coordinates?

[ytht100]

I have the following equations:
$$\left\{ \begin{array}{l}<br /> x = \sin \theta \cos \varphi \\<br /> y = \sin \theta \cos \varphi \\<br /> z = \cos \theta<br /> \end{array} \right.$$

Assume $$\vec r = (x,y,z)$$, which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate $$\frac{{\partial \vec r}}{{\partial z}}$$, could you please tell me if you have any hint?

I have Googled the questions a lot with different terms but can't find an answer that I am sure of. Many thanks for your attention!

Attempt 1: The problem seems related to coordinate transformation between spherical and cartesian coordinates.

Attempt 2: The problem seems related to "The Cartesian partial derivatives in spherical coordinates" shown here: http://mathworld.wolfram.com/SphericalCoordinates.html.

 

[PeroK]

$$r = (x,y,z)$$, which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate $$\frac{{\partial r}}{{\partial z}}$$

First, you have:

$\vec{r} = (x, y, z)$

And

$r = \sqrt{x^2 + y^2 + z^2}$

$x, y, z$ are independent variables, so are unrelated to each other. But, both $\vec{r}$ and $r$ can be treated as functions of $x, y, z$ and differentiated.

 

[ytht100]

First, you have:

$\vec{r} = (x, y, z)$

And

$r = \sqrt{x^2 + y^2 + z^2}$

$x, y, z$ are independent variables, so are unrelated to each other. But, both $\vec{r}$ and $r$ can be treated as functions of $x, y, z$ and differentiated.

1, Thank you for reminding me to write $${\vec r}$$ as vector, I have made revisions above.
However, x, y, and z are NOT independent variables, for example $${x^2} + {y^2} + {z^2} = 1$$. If they are independent, the problem is very easy because $$\frac{{d{\vec r}}}{{dz}} = \frac{{d(x,y,z)}}{{dz}} = (0,0,1)$$
2, the physical problem at my hand tells me they are NOT independent.

 

[zwierz]

which problem do you really solve? I am afraid the current statement merely does not make sense

 

[ytht100]

which problem do you really solve? I am afraid the current statement merely does not make sense

I am trying to solve a ray tracing problem. That is all I can say. Other information may not appropriate on this forum. Could you please tell me which part doesn't make sense? Thank you very much!

 

[zwierz]

the partial derivative by its definition uses some coordinate of a coordinate system. Coordinate system consists of independent coordinates

 

[PeroK]

I am trying to solve a ray tracing problem. That is all I can say. Other information may not appropriate on this forum. Could you please tell me which part doesn't make sense? Thank you very much!

You effectively have a function of two independent variables, with $r=1$. But, to get a partial derivative wrt $z$ you must be explicit about what the other variable is. For example, you could have:

$\vec{r} = (x, \sqrt{1-x^2-z^2}, z)$ or $\vec{r} = (\sqrt{1-y^2-z^2}, y, z)$ or $\vec{r} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ where $z = \cos{\theta}$

 

[ytht100]

You effectively have a function of two independent variables, with $r=1$. But, to get a partial derivative wrt $z$ you must be explicit about what the other variable is. For example, you could have:

$\vec{r} = (x, \sqrt{1-x^2-z^2}, z)$ or $\vec{r} = (\sqrt{1-y^2-z^2}, y, z)$ or $\vec{r} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ where $z = \cos{\theta}$

Thanks indeed! You clear up my mind!