- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.13 Exercises
- (a) Set up the integral which gives the length of the curve $y = c\cosh(x/c)$ from $x = 0$ to $x = a$ $(a > 0, c > 0).$ (b) Show that $c$ times the length of this curve is equal to the area of the region bounded by $y = c\cosh(x/c),$ the $x$-axis, the $y$-axis, and the line $x = a.$ (c) Evaluate this integral and find the length of the curve when $a = 2.$
- For (a), use the identity $\sinh^2 x = \cosh^2 x - 1.$
- (a) If the curve is described by the equation $y = c\cosh(x/c)$ then we can define its position $\mathbf{r}(x)$ as: $$ \mathbf{r}(x) = x\,\mathbf{i} + c\cosh\frac{x}{c}\mathbf{j} $$ With velocity $\mathbf{v}(x)$ and speed $v(x):$ $$ \begin{align*} \mathbf{v}(x) &= \mathbf{i} + \sinh\frac{x}{c}\mathbf{j} \\ \\ v(x) &= \sqrt{1 + \sinh^2(x/c)} \\ &= \sqrt{1 + \cosh^2(x/c) - 1} \\ &= \sqrt{\cosh^2(x/c)} \end{align*} $$ And since the hyperbolic cosine is positive for all real $x,$ $$ v(x) = \cosh\frac{x}{c} $$ Then, the arc length $s$ from $x = 0$ to $x = a$ is: $$ \begin{align*} s &= \int_0^a \cosh\frac{x}{c}\,dx \quad \blacksquare \end{align*} $$ (b) The area bounded by the curve $y = c\cosh(x/c),$ the $x$-axis, the $y$-axis, and the vertical line $x=a$ is defined by the integral: $$ A = c\int_0^a \cosh\frac{x}{c}\,dx $$ which is $c$ times the length $s.\quad \blacksquare$ (c) Evaluating the integral from (a) where $a = 2,$ we get: $$ \begin{align*} s_2 &= \int_0^2 \cosh\frac{x}{c}\,dx \\ &= c\sinh\frac{x}{c}\,\Biggr|_0^2 \\ &= c\sinh\frac{2}{c} \quad \blacksquare \end{align*} $$