Question

From (10.1) find $\frac{\mathbf{\partial }\mathbf{\theta }}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\left(\frac{1}{r}\right)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }$and show that$\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{\ne }\frac{\mathbf{\partial }\mathbf{\theta }}{\mathbf{\partial }\mathbf{z}}$. Note carefully that $\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{\theta }}$means that$\mathit{r}$ and $\mathit{\varphi }$are constant, but $\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{\theta }}$means that and are constant. (See Chapter 4, Example 7.6 for further discussion.)

Short answer

Thus,the required value is$\frac{\partial x}{\partial \theta }\ne \frac{\partial \theta }{\partial x}$

Step-by-step solution

Step 1:Determine the function.

$\begin{array}{l}x=r\sin \theta \cos \varphi \\ y=r\sin \theta \sin \varphi \\ z=r\cos \theta \end{array}$

Also,

$r^2=x^2+y^2+z^2$and$\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$

The objective is to determine $\frac{\partial \theta }{\partial x}$and to show that$\frac{\partial x}{\partial \theta }\ne \frac{\partial \theta }{\partial x}$ differentiate$x=r\sin \theta \cos \varphi$with respect to $\theta$.

$\frac{\partial x}{\partial \theta }=r\cos \theta \cos \varphi$take r and $\varphi$as constants.

Thus, the value of $\frac{\partial x}{\partial \theta }$is$\frac{\partial x}{\partial \theta }=r\cos \theta \cos \varphi \mathrm{...}\left(1\right)$

Determine the differentiation.

Differentiate$\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$with respect to.

$\begin{array}{c}\frac{\partial \theta }{\partial x}=\frac{\partial \theta }{\partial r}\frac{\partial r}{\partial x}\\ =\frac{z}{r^2\sqrt{1-\frac{z^2}{r^2}}}\frac{x}{\sqrt{x^2+y^2+z^2}}\\ =\frac{z}{r\sqrt{r^2-z^2}}\frac{x}{r}\Rightarrow (r=\sqrt{x^2+y^2+z^2})\\ =\frac{zx}{r^2\sqrt{r^2-z^2}}\end{array}$

Substitute x and z in the equation,

$\begin{array}{c}\frac{\partial \theta }{\partial x}=\frac{zx}{r^2\sqrt{r^2-z^2}}\\ \frac{\partial \theta }{\partial x}=\frac{\left(r\cos \theta \right)\left(r\sin \theta \cos \varphi \right)}{r^2\sqrt{r^2-{\left(r\cos \theta \right)}^2}}\\ \frac{\partial \theta }{\partial x}=\frac{r^2\cos \theta \sin \theta \cos \varphi }{r^2\sqrt{r^2-r^2{\cos }^2\theta }}\end{array}$

So, we get,

$\begin{array}{c}\frac{\partial \theta }{\partial x}=\frac{\cos \theta \sin \theta \cos \varphi }{\sqrt{r^2(1-{\cos }^2\theta )}}\\ \frac{\partial \theta }{\partial x}=\frac{\cos \theta \sin \theta \cos \varphi }{\sqrt{r^2{\sin }^2\theta }}\Rightarrow ({\cos }^2\theta +{\sin }^2\theta =1)\\ =\frac{\cos \theta \sin \theta \cos \varphi }{r\sin \theta }\\ =\frac{\cos \theta \cos \varphi }{r}\end{array}$

Thus, the value of $\frac{\partial \theta }{\partial x}$is$\frac{\partial \theta }{\partial x}=\frac{1}{r}\cos \theta \cos \varphi \mathrm{...}\left(2\right)$

From equation (1) and (2) observe that,

$\frac{\partial x}{\partial \theta }\ne \frac{\partial \theta }{\partial x}$