Doubt about the use of partial derivatives: what's the solution?

Question

I should calculate this mixed product, where $\overline r=\overrightarrow{OP}$,

$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi}),$$ to get with the determinant,

$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi})=\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}=\color{red}{r^3\sin \theta}.\tag 1$$

If

$$\begin{cases} x=r(\theta,\varphi)\sin \theta\cos \varphi\\ y=r(\theta,\varphi)\sin \theta\sin \varphi\\ z=r(\theta,\varphi)\cos \theta \end{cases}$$

why are they true these systems?

$$\begin{cases} x_\theta(\theta,\varphi)=r_\theta\sin \theta\cos\varphi+r\cos \theta\cos \varphi\\ x_\varphi(\theta,\varphi)=r_\varphi\sin \theta\cos\varphi-r\sin\theta\sin \varphi \end{cases} \tag 2$$

$$\begin{cases} y_\theta(\theta,\varphi)=r_\theta\sin \theta\sin\varphi+r\cos \theta\sin \varphi\\ y_\varphi(\theta,\varphi)=r_\varphi\sin \theta\sin\varphi+r\sin\theta\cos \varphi \end{cases} \tag 3$$

$$\begin{cases} z_\theta(\theta,\varphi)=r_\theta\cos\theta-r\sin \theta\\ z_\varphi(\theta,\varphi)=r_\varphi\cos\theta \end{cases} \tag 4$$

With $(2), (3), (4)$ I have not get $r^3\sin \theta$ with the determinant $(1)$. Please, can I have your attention and help?

Answer 1

A strategy suggestion to calculate by hand this determinant:

First, I'll use condensed notations: $r$ instead of $r(\theta,\varphi)$. Next, observe each column is the sum of two column vectors $$\begin{bmatrix}x\\x_\theta\\x_\varphi\end{bmatrix}=\underbrace{\begin{bmatrix}r\sin\theta\cos\varphi\\r_\theta\sin\theta\cos\varphi\\r_\varphi\sin\theta\cos\varphi\end{bmatrix}}_{\textstyle C_x}+\underbrace{\begin{bmatrix}0\\r\cos\theta\cos\varphi\\-r\sin\theta\sin\varphi\end{bmatrix}}_{\textstyle D_x}=\sin\theta\cos\varphi\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}+r\begin{bmatrix}0\\\cos\theta\cos\varphi\\-\sin\theta\sin\varphi\end{bmatrix}$$ and similarly for the other columns.

Now the determinant, with these notations becomes, by the multilinearity properties of determinants \begin{align}\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}&=\begin{vmatrix}C_x{+}D_x & C_y {+}D_y& C_z{+}D_z\\ \end{vmatrix}\\&=\bigl|C_x\enspace C_y\enspace C_z \bigr|+\bigl|D_x\enspace C_y\enspace C_z \bigr|+\bigl|C_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace C_y\enspace D_z \bigr|\\[1ex] &\quad+\bigl| C_x\enspace D_y\enspace C_z \bigr|+\bigl|D_x\enspace D_y\enspace C_z \bigr|+\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace D_z \bigr| \end{align} Each of $C_x, C_y, C_z$ is collinear to the column vector $\;\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}$, so any of the determinants in the sum which comprises two of these column vectors is $0$, further $\bigl| D_x\enspace D_y\enspace D_z \bigr|$ has a row of $0$s. Therefore there remains $$\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix} =\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl|D_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace C_z \bigr|, $$ which can easily be computed by hand.

Hope this will help!

Answer 2

Equations (2), (3) and (4) are true because of the product rule for derivatives.

Answer 3

Hint: write out the determinant in the matrix form.

Step 1: Multiply the first row by $-r_{\theta}/r$ and add it to the second row. Do similar thing with respect to the third row. You will see that all the terms involving $r_{\theta}$ and $r_{\phi}$ will be canceled.

Step 2: Extract the factors $r,r,r\sin\theta$ outside from the first, second and third row, respectively. This explains the factor $r^3\sin\theta$.

Step 3: Work on the remaining determinant by expanding with respect to the third row. It can be easily checked that the result is $\sin^2\phi+\cos^2\phi=1$, hence the result