Metric for the construction of Mercator map

[Whitehole]

Homework Statement

The familiar Mercator map of the world is obtained by transforming spherical coordinates θ , ϕ to coordinates x , y given by
$x = \frac{W}{2π} φ, y = -\frac{W}{2π} log (tan (\frac{Θ}{2}))$

Show that $ds^2 = Ω^2(x,y) (dx^2 + dy^2)$ and find $Ω$

Homework Equations

$g'_{ρσ} = g_{μν} (\frac{∂x^μ}{∂x'^ρ}) (\frac{∂x^ν}{∂x'^σ})$

In spherical coordinates $ds^2 = dΘ^2 + \sin^2(Θ)dφ^2$ with $r=1$

The Attempt at a Solution

So in this case, I have to find $g_{xx}$ and $g_{yy}$, they should be the same since $Ω^2(x,y)$ is factored out.

By isolating $Θ$ and $φ$, I got

$Θ=2\arctan(e^\frac{-2πy}{W})$ and $φ = \frac{2πx}{W}$

$g_{xx} = g_{ΘΘ} \frac{∂Θ}{∂x} + g_{φφ} \frac{∂φ}{∂x} = \frac{2π}{W}\sin^2(2\arctan(e^\frac{-2πy}{W}))$

$g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y}$

For this part, $y = -\frac{W}{2π} log (tan (\frac{Θ}{2}))$, manipulate so that $tan (\frac{Θ}{2}) = e^\frac{-2πy}{W}$, then differentiate with respect to

y so that $\frac{∂Θ}{∂y} = \frac{-4πe^\frac{-2πy}{W}}{Wsec^2(\frac{Θ}{2})}$

After some manipulation I ended up

$g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y} = \frac{-2π}{Wcosh(\frac{2πy}{W})}$

According to the book I'm reading the answer should be

$Ω = \frac{2π}{Wcosh(\frac{2πy}{W})}$

Can anyone help me find out what is wrong?

 

[George Jones]

$g_{xx} = g_{ΘΘ} \frac{∂Θ}{∂x} + g_{φφ} \frac{∂φ}{∂x}$

$g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y}$

Are you sure that this is correct?

 

[Whitehole]

Are you sure that this is correct?

$g_{xx} = g_{ΘΘ} (\frac{∂Θ}{∂x})^2 + g_{φφ} (\frac{∂φ}{∂x})^2$
$g_{yy} = g_{ΘΘ} (\frac{∂Θ}{∂y})^2 + g_{φφ} (\frac{∂φ}{∂y})^2$

Forgot the squared! How dumb. Thanks!