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

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?

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!

• Thank you very much. Surely you and other users you have helped me. Last help into my question. Please why $(1)\implies (2), (3)$ and $(4)$? – Sebastiano Dec 14 '19 at 9:58
• @Sebastiano Shouldn't it be the other way around? Note that (2),(3), and (4) are consequences from the definition of $x,y,z$ by taking partial derivatives. You then use the relations (2),(3) and (4) for computing the determinant given in (1). The result is $r^3\sin\theta$ as I sketched in my answer. – Pythagoras Dec 14 '19 at 20:41
• @Pythagoras I have taken the partial derivatives. Infact being $x=f\cdot g \cdot h$ using the partial derivates (supposing only for $\theta$) I should have 3 terms and not 2 terms. – Sebastiano Dec 14 '19 at 20:47
• Using your example, the last factor $h$ does not depend on $\theta$, so the derivative is zero. They are treating $\theta,\phi$ as independent variables. So a function of $\phi$ is like a constant with respect to the variable $\theta$. – Pythagoras Dec 14 '19 at 20:54
• @Pythagoras I was and actually in tilt for a lot of work. I have not considered that: $x_\theta=f_\theta \cdot g \cdot h+f \cdot g_\theta \cdot h + f \cdot g \cdot h_\theta$ where $\phi=h_\theta=0$. You're right and thank you very much again. – Sebastiano Dec 15 '19 at 21:50

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

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