Question

Verify (7.16) in three ways:

(a) Differentiate equations (7.6). (b)

(b) Take differentials of (7.5) and solve for$\mathit{d}\mathit{r}\mathbf{\text{and}}\mathit{d}\mathit{\theta }$.

(c) Find ${\mathit{A}}^{\mathbf{-}\mathbf{1}}$in (7.15) from A in (7.13); note that this is (b) in matrix notation.

Short answer

The required differentials are:

$$\begin{gathered} dr=\frac{xdx+ydy}{r} \\ d\theta =\frac{xdy-ydx}{r^2} \end{gathered}$$

Step-by-step solution

Given data: 

The given equations are: $r=\sqrt{x^2+y^2}and\theta ={\tan }^{-1}\left(\frac{y}{X}\right)$.

Define the chain Rule 

Consider the function:

$v=v\left(x,y\right)$

Consider the independent variables for the abovefunction as

$x=x\left(s,t\right)andy=y\left(s,t\right)$

Then, the chain rule for the function is:

$\begin{array}{l}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{s}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{x}}\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{s}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{y}}\frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{s}}\\ \frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{x}}\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{t}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{y}}\frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{t}}\end{array}$

Differentiate the given equation

Differentiate $r=\sqrt{x^2+y^2}$as follows:

role="math" localid="1664193126679" $$\begin{gathered} dr=\frac{1}{2\sqrt{x^2+y^2}}\left(2xdx+2ydy\right) \\ =\frac{1}{\sqrt{x^2+y^2}}\left(xdx+ydy\right) \\ =\frac{\left(xdx+ydy\right)}{r} \end{gathered}$$

Similarly, differentiate role="math" localid="1664190579105" $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$as follows:

role="math" localid="1664193907448" $\begin{array}{l}d\theta =\frac{1}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}d\left(\frac{y}{x}\right)\\ =\frac{x^2}{x^2+y^2}·\frac{xdy-ydx}{x^2}\\ =\frac{xdy-ydy}{r^2}\end{array}$

Hence, the required differentials are:

$$\begin{gathered} dr=\frac{xdx+ydy}{r} \\ d\theta =\frac{xdy-ydx}{r^2} \end{gathered}$$