5
$\begingroup$

I'm trying to derive the definitions of hyperbolic functions with this image in mind, where $a := \cosh u$, $b := \sinh u$, and $u = 2A$.

I have $$2A = 2\int_{0}^b \sqrt{1+y^2} \ \mathrm{d}y = b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)$$ and therefore $$\begin{align} \cosh u &= \cosh\left(\;b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)\;\right) = \sqrt{b^2 + 1} \\ \sinh u &= \sinh\left(\;b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)\;\right) = b \end{align}$$

The goal here is to solve for $u$ in terms of $b$ and then derive the usual formulas, but that's proven to be impossible (unless I made a mistake, which I think might be the case here). Where can I go from here?

Also, I'm using the derivation shown in this guide.


EDIT: The correct integral is $$2A = 2\int_{0}^b \left( \sqrt{1+y^2} - \frac{a}{b}y \ \mathrm{d}y \right) = \log\left(b + \sqrt{b^2 + 1}\right)$$ so $$b = \cfrac{e^{2u} - 1}{2e^u}$$ and from here the rest follows trivially.

$\endgroup$
5
$\begingroup$

Don't forget that the region $A$ is sliced off by the line $x=\frac ab y$, so you should amend your integral to be $$ 2A = 2\int_{0}^b \left(\sqrt{1+y^2}-\frac ab y\right) dy.\tag1$$ But since $(a,b)$ lies on the hyperbola, we have $a=\sqrt{b^2+1}$. Plugging this in to (1) and evaluating the integral, you'll end up with $$u=2A=\log(b+\sqrt{b^2+1}).\tag2$$ (Note that this amendment amounts to subtracting off double the area of a right triangle with legs $a$ and $b$.)

Your next step is to solve (2) for $b$ in terms of $u$.

$\endgroup$
  • $\begingroup$ Knew I had made a mistake somewhere. Damn it. Thanks! $\endgroup$ – Matheus Andrade Feb 2 '18 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.