Calculate the dual basis and tangent basis vectors

Karl Karlsson
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Homework Statement
A coordinate system with the coordinates s and t in ##R^2## is defined by the coordinate transformations: ## s = y/y_0## and ##t=y/y_0 - tan(x/x_0)## , where ##x_0## and ##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 ##\vec E_s## and ##\vec E_t## and the dual basis vectors##\vec E^s## and ##\vec E^t##

c)Determine the inner products ##\vec E_s\cdot\vec E^s##, ##\vec E_s\cdot\vec E^t##, ##\vec E_t\cdot\vec E^s## and ##\vec E_t\cdot\vec E^t##
Relevant Equations
## s = y/y_0## and ##t=y/y_0 - tan(x/x_0)## , where ##x_0## and ##y_0## are constants.
a) Since ##tan(x/x_0)## is not defined for ##x=\pm\pi/2\cdot x_0## I assume x must be in between those values therefore ##-\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 ##y=y_0\cdot s## and ##x=x_0\cdot arctan(s-t)##. ##\vec E_s = \frac {x_0} {1 + (s-t)^2}\cdot\vec e-x + y_0\cdot\vec e_y## and ##\vec E_t = - \frac { x_0} { 1 + (s-t)^2}\cdot\vec e_x##. I get the dual basis vectors from ##\vec E^s = \frac {1} {y_0}\cdot\vec e_y## and ##\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 ##\vec E_s\cdot \vec E^s = 1## and##\vec E_t\cdot \vec E^s = 0##, this feels correct but then i get by just plugging in ##\vec E_t\cdot \vec E^t = \frac {x_0} {(1+(s-t)^2)(1+arctan(s-t)^2)}## and ##\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!
 
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Does anyone know if above is correct or not?
 
Does anybody know?
 
I am still confused about this
 
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