If $x=r\cosh (t)$ and $y=r\sinh (t)$, $y=y(r,t)$ and $x=x(r,t)$ but are $r$ and $t$ functions of $x$ and $y$

Question

If it is acceptable to write $r$ and $t$ and functions of $x$ and $y$, why is this the case and also, how do I find $(\frac{\partial r}{\partial x})$, $(\frac{\partial r}{\partial y})$, $(\frac{\partial t}{\partial x})$ and $(\frac{\partial t}{\partial y})$?

Answer 1

Hint:

$x^2-y^2=r^2$ and $\dfrac{y}{x}=\tanh t$

Answer 2

It depends on $x,y$.

Here is the locus of $t \mapsto (\cosh t, \sinh t)$ (note that the map is injective):

Let $L = \{ (\cosh t, \sinh t) \}_t$, and let $C = \cup_r rL$ be the non convex cone generated by taking all lines that pass through the origin and $L$. It is easy to see that $C= \{(x,y) | |y| < |x| \} \cup \{0\}$.

If $(x,y) \in C \setminus \{0\}$, then it is easy to see that there is a unique ${1 \over r}$ such that ${1 \over r}(x,y) \in L$ and a unique $t$ such that ${1 \over r}(x,y) = (\cosh t, \sinh t)$. That is, $(x,y) = r (\cosh t, \sinh t)$. Since $(r,t) \mapsto r (\cosh t, \sinh t)$ has an invertible Jacobian, the inverse function theorem shows that we can write $r,t$ as differentiable functions of $x,y$.

Or, directly: Using the usual $\sinh,\cosh$ identities, we have $x^2-y^2=r^2$, $t= { 1\over 2} \ln {x+y \over x-y}$, and since $\operatorname{sgn} x = \operatorname{sgn}r$, we have $r = \operatorname{sgn} x \sqrt{x^2-y^2}$.

It follows that the functions $(x,y) \mapsto t(x,y), r(x,y)$ are differentiable.