I'm struggling to complete an exercise that gives a reparametrisaton to use to convert one 1st Fundamental form into that of the half-plane model of the hyperbolic plane. I've written out the exercise below followed by my various attempts:
If we reparametrize the fundamental form $I=du^2+cosh^2(u)dv^2$ by $V=e^vtanh(u)$, $W=\frac{e^v}{cosh(u)}$, then we have $$ I=\frac{dV^2+dW^2}{W^2}$$
I've tried differentiating $V$ and $W$ with respect to $u$ and $v$ respectively, rearranging for expressions of $du$ and $dv$ to get $du=\frac{dVcosh^2(u)}{e^v}$ and $dv=\frac{dWcosh(u)}{e^v}$ and then substituted these into $I=du^2+cosh^2(u)dv^2$ however I'm left with an extra factor of $cosh^2(u)$.
I also tried finding $\frac{dV}{dv}$ and $\frac{dW}{du}$ and doing the same but was then left with an extra factor of $coth^2(u)$.
Finally I tried adding each of my two expressions for $du$ and $dv$ to get expressions for $du$ and $dv$ in terms of both $dV$ and $dW$ however this gave me an extra factor of $\frac{cosh^4(u)}{4sinh^2(u)}$.
I think there must be an error in my understanding as I've checked my working so many times and can't see an obvious error myself. Any help would be much appreciated.
You need to write out your work. It looks like you're not doing a correct multivariable computation. You should start by writing out $$dV = e^v(\text{sech}^2 u\,du+\tanh u\,dv)$$ and $$dW = e^v\text{sech }u(-\tanh u\,du + dv).$$ Then compute $(dV)^2 + (dW)^2$ as a quadratic differential.