]> Exercise 2

# Two Dimensional Transformation, Exercise 2

## Question

Answer the following questions by using the transformation of coordinates  $\left(x,y\right)⇔\left(\beta ,\gamma \right)$  :

$\begin{array}{l}x=\gamma a\text{ }\mathrm{cosh}\beta \\ y=\gamma b\text{ }\mathrm{sinh}\beta \end{array}$

where  a and b  are positive constants, and  x, y, a, b  have physical dimensions of length.

1. What is the domain of (x,y) subject to this transformation? (hint: $\text{tanh}\beta$)

2. Obtain the absolute value of the Jacobian $|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}|$ in terms of β and γ !
3. Consider the region enclosed between the positive x branch of the hyperbola

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ and the straight line $x={x}_{0}>a$
Obtain the corresponding curves after transforming to $\left(\beta ,\gamma \right)$ coordinates.
4. Make a sketch of the regions in question 3 !
5. Consider the region (of question 3) as a planar body with density (mass per area)

$\sigma ={\sigma }_{0}\frac{\frac{x}{{x}_{0}}}{\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}$ where ${\sigma }_{0}$ is a positive constant.
Use the $\left(\beta ,\gamma \right)$ variables, in order to calculate the mass M of the body!
Check the physical dimensions!

## Reminder

... The change of two variables ... is expressed by:

 $\begin{array}{l}x=x\left(u,v\right)\\ y=y\left(u,v\right)\end{array}\right\}$ (3.2.3.3)

We'll require also one to one correspondance, between the two sets of variable pairs: (u,v) and (x,y), which is essential for changing variables of integration. This requirement confines the change of variables to a transformation of coordinates ....

If we express the (u,v) variables as a set of new Cartesian coordinates, in order to integrate in this plane, we have to know the magnification factor μ of a rectangular element of "area" ΔuΔv, mapped to the corresponding element ΔxΔy from the original plane:

 $\begin{array}{l}F\left(x,y\right)\Delta x\Delta y=F\left[x\left(u,v\right),y\left(u,v\right)\right]\mu \left(u,v\right)\Delta u\Delta v\\ \text{or}\text{ }\Delta x\Delta y=\mu \left(u,v\right)\Delta u\Delta v\end{array}\right\}$ (3.2.3.5)

... the magnification factor μ ..... is the absolute value of the Jacobian :

 $\Delta x\Delta y=|\frac{D\left(x,y\right)}{D\left(u,v\right)}|\Delta u\Delta v=|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}|\Delta u\Delta v$ (3.2.3.6)

and therefore the balance of the physical dimensions is kept.

## Part 1

Solution of question 1.

1. The transformation yields
$\frac{y}{x}=\frac{b}{a}\mathrm{tanh}\beta$
but since $-1<\mathrm{tanh}\beta <1$ we obtain $-\frac{b}{a}<\frac{y}{x}<\frac{b}{a}$

## Part 2

Solution of question 2.

1.   $\begin{array}{l}|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}|=|\frac{\partial x}{\partial \beta }\frac{\partial y}{\partial \gamma }-\frac{\partial x}{\partial \gamma }\frac{\partial y}{\partial \beta }|=\\ =|ab\gamma \left({\mathrm{sinh}}^{2}\beta -{\mathrm{cosh}}^{2}\beta \right)|=ab|\gamma |\end{array}$

## Parts 3-4

Solution of question 3.

1. The substitution of $\begin{array}{l}x=\gamma a\text{ }\mathrm{cosh}\beta \\ y=\gamma b\text{ }\mathrm{sinh}\beta \end{array}$ into $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ yields ${\gamma }^{2}\left({\mathrm{cosh}}^{2}\beta -{\mathrm{sinh}}^{2}\beta \right)={\gamma }^{2}=1$
For positive x, one have to use the positive value of γ , yielding finally

$\gamma =1$

2. the substitution of $x={x}_{0}$ into $x=\gamma a\text{ }\mathrm{cosh}\beta$ yields $\gamma =\frac{{x}_{0}}{a}\frac{1}{\text{ }\mathrm{cosh}\beta }$ with $\frac{{x}_{0}}{a}>1$ as required.

## Part 5

Solution of question 4.

1. See the Fig. Sketch for exercise 2.

## Parts 6-9

Solution of question 5.

1. The density expressed by the $\left(\beta ,\gamma \right)$ variables is

$\sigma \text{=}{\sigma }_{\text{0}}\frac{\frac{x}{{x}_{0}}}{\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}=\frac{{\sigma }_{\text{0}}\gamma a\text{ }\mathrm{cosh}\beta }{{x}_{0}{\gamma }^{2}}=\frac{{\sigma }_{\text{0}}a}{{x}_{0}}\frac{\mathrm{cosh}\beta }{\gamma }$
and the integral of the mass is

$\begin{array}{l}M=\underset{{\beta }_{\mathrm{min}}}{\overset{{\beta }_{\mathrm{max}}}{\int }}\text{d}\beta \underset{{\gamma }_{\mathrm{min}}\left(\beta \right)}{\overset{{\gamma }_{\mathrm{max}}\left(\beta \right)}{\int }}|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}|\sigma \left(\beta ,\gamma \right)\text{d}\gamma \text{=}\\ \text{=}ab\frac{{\sigma }_{\text{0}}a}{{x}_{0}}\underset{{\beta }_{\mathrm{min}}}{\overset{{\beta }_{\mathrm{max}}}{\int }}\text{d}\beta \underset{{\gamma }_{\mathrm{min}}\left(\beta \right)}{\overset{{\gamma }_{\mathrm{max}}\left(\beta \right)}{\int }}\gamma \frac{\mathrm{cosh}\beta }{\gamma }\text{d}\gamma \text{=}\\ \text{=}\frac{{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\underset{{\beta }_{\mathrm{min}}}{\overset{{\beta }_{\mathrm{max}}}{\int }}\mathrm{cosh}\beta \text{ }\text{d}\beta \underset{{\gamma }_{\mathrm{min}}\left(\beta \right)}{\overset{{\gamma }_{\mathrm{max}}\left(\beta \right)}{\int }}\text{d}\gamma \end{array}$

2. The limits of the integral over γ are according to question 3

$\left\{\begin{array}{l}{\gamma }_{\mathrm{min}}=1\\ {\gamma }_{\mathrm{max}}=\frac{{x}_{0}}{a}\frac{1}{\mathrm{cosh}\beta }\text{ }\text{with}\text{ }\frac{{x}_{0}}{a}>1\end{array}$
The limits for the integration over β are obtained from the intersection of  ${\gamma }_{\mathrm{max}}$  with  ${\gamma }_{\mathrm{min}}$

$\begin{array}{l}\mathrm{cosh}{\beta }_{0}=\frac{{x}_{0}}{a}\\ \left\{\begin{array}{l}{\beta }_{\mathrm{min}}=-{\beta }_{0}=-\text{acosh}\left(\frac{{x}_{0}}{a}\right)\\ {\beta }_{\mathrm{max}}={\beta }_{0}=\text{acosh}\left(\frac{{x}_{0}}{a}\right)\end{array}\end{array}$

3. The solution of the integral of part 6 is therefore

$\begin{array}{l}M=\frac{{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\underset{-{\beta }_{0}}{\overset{{\beta }_{0}}{\int }}\mathrm{cosh}\beta \left(\frac{{x}_{0}}{a}\frac{1}{\mathrm{cosh}\beta }-1\right)\text{d}\beta =\\ =\frac{{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\underset{-{\beta }_{0}}{\overset{{\beta }_{0}}{\int }}\text{d}\beta \left(\frac{{x}_{0}}{a}-\mathrm{cosh}\beta \right)=\\ =\frac{{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\left(\frac{2{\beta }_{0}{x}_{0}}{a}-2\mathrm{sinh}{\beta }_{0}\right)=\\ =\frac{2{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\left(\frac{{\beta }_{0}{x}_{0}}{a}-\sqrt{{\mathrm{cosh}}^{2}{\beta }_{0}-1}\right)=\\ =\frac{2{\sigma }_{\text{0}}{a}^{2}b}{{x}_{0}}\left(\frac{{\beta }_{0}{x}_{0}}{a}-\sqrt{\frac{{x}_{0}^{2}}{{a}^{2}}-1}\right)=\\ =2{\sigma }_{\text{0}}ab\left(\text{acosh}\left(\frac{{x}_{0}}{a}\right)-\sqrt{1-\frac{{a}^{2}}{{x}_{0}^{2}}}\right)\end{array}$

4. The expression inside the brackets is dimensionless. The dimensions of σ0 are mass/area and of the product ab - area. Therefore the physical dimensions are balanced.

## Score

By parts.
Parts 1,2,3,4,6,7,8,9 (all except 5) are worth 1 point each.
Part 5 is worth 2 points.

By questions.
Questions 1 and 2 are worth 1 point each.
Questions 3 and 4 are worth 2 points each.
Question 5 is worth 4 points.