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Determine the area, calculate the basis vectors and determine the inner product

Karl Karlsson

New member
Sep 13, 2020
2
A coordinate system with the coordinates s and t in \(\displaystyle R^2\) is defined by the coordinate transformations: \(\displaystyle s = y/y_0\) and \(\displaystyle t=y/y_0 - tan(x/x_0)\) , where \(\displaystyle x_0\) and \(\displaystyle y_0\) are constants.

a) Determine the area that includes the point (x, y) = (0, 0) where the coordinate system
is well defined. Express the area both in the Cartesian coordinates (x, y) and in
the new coordinates (s, t).

b) Calculate the tangent basis vectors \(\displaystyle \vec E_s\) and \(\displaystyle \vec E_t\) and the dual basis vectors \(\displaystyle \vec E^s\) and \(\displaystyle \vec E^t\)

c)Determine the inner products \(\displaystyle \vec E_s\cdot\vec E^s\), \(\displaystyle \vec E_s\cdot\vec E^t\), \(\displaystyle \vec E_t\cdot\vec E^s\) and \(\displaystyle \vec E_t\cdot\vec E^t\)

My attempt:
a) Since \(\displaystyle tan(x/x_0)\) is not defined for \(\displaystyle x=\pm\pi/2\cdot x_0\) I assume x must be in between those values therefore \(\displaystyle -\pi/2\cdot x_0 < x < \pi/2\cdot x_0\) and y can be any real number. Is this the correct answer on a)?

b) I can solve x and y for s and t which gives me \(\displaystyle y=y_0\cdot s\) and \(\displaystyle x=x_0\cdot arctan(s-t)\). \(\displaystyle \vec E_s = \frac {x_0} {1 + (s-t)^2}\cdot\vec e-x + y_0\cdot\vec e_y\) and \(\displaystyle \vec E_t = - \frac { x_0} { 1 + (s-t)^2}\cdot\vec e_x\). I get the dual basis vectors from \(\displaystyle \vec E^s = \frac {1} {y_0}\cdot\vec e_y\) and \(\displaystyle \vec E^t = \frac {1} {y_0}\cdot\vec e_y - \frac {1} {x_0(1+(x/x_0)^2)}\cdot\vec e_x\) , is this the correct approach?

c) It was here that I really started to question if i had done correct on a and b since I get \(\displaystyle \vec E_s\cdot \vec E^s = 1 \)and\(\displaystyle \vec E_t\cdot \vec E^s = 0\), this feels correct but then i get by just plugging in \(\displaystyle \vec E_t\cdot \vec E^t = \frac {x_0} {(1+(s-t)^2)(1+arctan(s-t)^2)} \)and \(\displaystyle \vec E_s\cdot \vec E^t = 1-\frac {1} {(1+(s-t)^2)(1+arctan(s-t)^2)}\). Is this really correct? Because it feels like it is not correct.

Thanks in advance!
 

Alan

Member
Jul 21, 2012
59