Consider x=(e^t)+(e^-t), y=5-2t, 0<t<3. Find the length of the curve.
Recall the arc length integral:
$\displaystyle = \int_{ {t }_{i } }^{ {t }_{f } } \left | r'(t) \right | dt$
Where,
$\displaystyle r'(t) = \sqrt[ ]{x'(t)^{2}+y'(t)^{2} } $
So we have:
$\displaystyle x(t) = e^{t}+e^{-t} = 2cosh(t)$
$\displaystyle y(t) = 5 - 2t$.
Then,
$\displaystyle x'(t) = 2sinh(t)$
$\displaystyle y'(t) = -2$.
So,
$\displaystyle r'(t) = \sqrt[ ]{4 \sinh ^2(t) + 4} = 2 \cosh (t)$.
Then,
Arc Length = $\displaystyle \int_{0}^{3} 2 \cosh (t) = 2 \sinh (3) - 2 \sinh (0) = 2 \sinh (3) \approx 20.04 $
Check out hyperbolic cosine from wikipedia: Hyperbolic function - Wikipedia, the free encyclopedia