Mathematical Immaturity

14.7 Exercises

• The identity $\cosh^2 \theta - \sinh^2 \theta = 1$ for hyperbolic functions suggests that the hyperbola $x^2/a^2 - y^2/b^2 = 1$ may be represented by the parametric equations $x = a\cosh \theta,$ $y = b\sinh \theta,$ or what amounts to the same thing, by the vector equation $r = a\cosh \theta\mathbf{i} + b\sinh \theta\mathbf{j}.$ When $a = b = 1,$ the parameter $\theta$ may be given a geometric interpretation analogous to that which holds between $\theta,$ $\sin \theta,$ and $\cos \theta$ in the unit circle shown in Figure 14.7(a). Figure 14.7(b) shows one branch of the hyperbola $x^2 - y^2 = 1.$ If the point $P$ has coordinates $x = \cosh \theta$ and $y = \sinh \theta,$ prove that $\theta$ equals twice the area of the sector $OAP$ shaded in the figure. [Hint: Let $A(\theta)$ denote the area of sector $OAP.$ Show that $$ \begin{align*} \\ A(\theta) = \frac{1}{2}\cosh \theta \sinh \theta - \int_1^{\cosh \theta} \sqrt{x^2 - 1}\,dx \end{align*} $$ Differentiate to get $A'(\theta) = \frac{1}{2}.$]
• Since the curve in question is a hyperbola, we know that each $P$ on the curve can be written as $(x, y) = \left(\cosh \theta, \sinh \theta \right).$ As such, the area of the right triangle with vertices $O, P, (x, 0)$ is $\frac{1}{2}bh = \frac{1}{2} \cosh \theta \sinh \theta.$ Now, to find the area of the sector $OAP,$ we must subtract from the right triangle $O, P, (x, 0)$ the area under the curve $y = \sqrt{x^2 - 1}$ from $x = A$ to $x = \cosh \theta.$ But $A$ is a point on the hyperbola desctibed by $x^2 - y^2 = 1,$ since $A$ touches the $x$-axis, thus $A$ must be $(1, 0).$ This difference, written as a function of $\theta$ is $$A(\theta) = \frac{1}{2} \cosh \theta \sinh \theta - \int_1^{\cosh \theta} \sqrt{x^2 - 1}\, dx$$ Taking the derivative with respect to $\theta$ gives us: $$\displaylines{A'(\theta) = \frac{1}{2}\left(\sinh^2 \theta + \cosh^2 \theta\right) - \left(\sinh \theta \sqrt{\cosh^2 \theta - 1}\right)\\ = \frac{1}{2}\left(\sinh^2 \theta + \cosh^2 \theta\right) - \left(\sinh \theta \sqrt{\sinh^2 \theta}\right)\\ = \frac{1}{2}\left(\cosh^2 \theta- \sinh^2 \theta\right)\\ = \frac{1}{2}}$$ This implies, however, that the area $A(\theta)$ is $$A(\theta) = \int_0^{\theta}\frac{1}{2}\,dt = \frac{\theta}{2}$$ Or in other words, that the angle $\theta$ is twice the area $A(\theta)$ of the sector $OAP.$ This completes the proof.