Question

Q. Calculate the arc–length of catenary γ(t) = (t, cosh t) starting at the point (0, 1).

Answer 1

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Answer on Question #70831 – Math – Geometry

Question

Calculate the arc–length of catenary $\gamma(t) = (t, \cosh t)$ starting at the point $(0, 1)$.

Solution

A parametrized curve in the plane is a differentiable function

$$\gamma(t) = (t, \cosh t),$$

where $t \geq 0$.

Then for any $t > 0$ we define its arclength from 0 to $t$ to be

$$s(t) = \int_0^t \|\gamma'(u)\| \, du$$

We have $\gamma'(u) = (1, \sinh t)$ and

$$\|\gamma'(u)\| = \sqrt{(1)^2 + (\sinh u)^2} = \sqrt{(\cosh u)^2} = \cosh u$$ $$\begin{array}{l} s(t) = \int_0^t \|\gamma'(u)\| \, du = \int_0^t \cosh u \, du = \sinh u \Big|_0^t = \sinh(t) - \sinh(0) = \\ = \sinh(t), t > 0 \end{array}$$

**Answer**: the arc – length is $s(t) = \sinh(t), t > 0$.