Answer: (a) \(s=\int_{0}^{\frac{3}{5}} \sqrt{\cosh 2t}\,dt\).
(b) \(\sqrt{\cosh 2t}=1+t^2+\cdots\), so
\(s\approx \int_0^{3/5}(1+t^2)\,dt=\left[t+\frac13 t^3\right]_0^{3/5}=\frac{84}{125}=0.672\).
(a) For a curve given parametrically by \(x=x(t)\), \(y=y(t)\), the arc length from \(t=a\) to \(t=b\) is
\(s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)
Here
\(x=\cosh t,\qquad y=\sinh t,\)
so
\(\frac{dx}{dt}=\sinh t,\qquad \frac{dy}{dt}=\cosh t.\)
Hence
\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=\sinh^2 t+\cosh^2 t.\)
Using \(\cosh 2t=\cosh^2 t+\sinh^2 t\), this becomes
\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=\cosh 2t.\)
Therefore
\(s=\int_0^{3/5}\sqrt{\cosh 2t}\,dt.\)
(b) Let \(f(t)=\sqrt{\cosh 2t}=(\cosh 2t)^{1/2}.\)
We need the Maclaurin series up to and including the term in \(t^2\).
First,
\(f(0)=\sqrt{\cosh 0}=1.\)
Differentiate:
\(f'(t)=\frac12(\cosh 2t)^{-1/2}(2\sinh 2t)=\sinh 2t\,(\cosh 2t)^{-1/2}.\)
So
\(f'(0)=0.\)
Differentiate again:
\(f''(t)=2\sqrt{\cosh 2t}-\sinh^2 2t\,(\cosh 2t)^{-3/2}.\)
Hence
\(f''(0)=2.\)
Therefore the Maclaurin series is
\(f(t)=f(0)+f'(0)t+\frac{f''(0)}{2}t^2+\cdots=1+t^2+\cdots\)
So
\(\sqrt{\cosh 2t}\approx 1+t^2.\)
Thus
\(s\approx \int_0^{3/5}(1+t^2)\,dt=\left[t+\frac13 t^3\right]_0^{3/5}.\)
This gives
\(s\approx \frac35+\frac13\left(\frac35\right)^3=\frac35+\frac{9}{125}=\frac{84}{125}=0.672.\)
