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

#### Karl Karlsson

##### New member
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.