# What is the arc length of the curve given by r(t)= (9sqrt(2),e^(9t),e^(-9t)) on  t in [3,4]?

Mar 12, 2018

A zeroth-order approximation gives $2 \sinh \left(36\right) - 2 \sinh \left(27\right)$ units.

#### Explanation:

r(t)=(9sqrt2, e^(9t), e^(−9t))

r'(t)=(0, 9e^(9t), -9e^(−9t))

Arc length is given by:

$L = {\int}_{3}^{4} \sqrt{0 + 81 {e}^{18 t} + 81 {e}^{- 18 t}} \mathrm{dt}$

Simplify:

$L = 9 {\int}_{3}^{4} \sqrt{{e}^{18 t} + {e}^{- 18 t}} \mathrm{dt}$

Apply the identity $\cosh x = \frac{1}{2} \left({e}^{x} + {e}^{-} x\right)$:

$L = 9 {\int}_{3}^{4} \sqrt{2 \cosh \left(18 t\right)} \mathrm{dt}$

Apply the identity $\cosh 2 x = 2 {\cosh}^{2} x - 1$:

$L = 9 \sqrt{2} {\int}_{3}^{4} \sqrt{2 {\cosh}^{2} \left(9 t\right) - 1} \mathrm{dt}$

Factor out the larger piece:

$L = 18 {\int}_{3}^{4} \cosh \left(9 t\right) \sqrt{1 - \frac{1}{2} {\sech}^{2} \left(9 t\right)} \mathrm{dt}$

For $t \in \left[3 , 4\right]$, $\frac{1}{2} {\sech}^{2} \left(9 t\right) < 1$. Take the series expansion of the square root:

$L = 18 {\int}_{3}^{4} \cosh \left(9 t\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{1}{2} {\sech}^{2} \left(9 t\right)\right)}^{n}\right\} \mathrm{dt}$

Isolate the $n = 0$ term:

$L = 18 {\int}_{3}^{4} \cosh \left(9 t\right) \mathrm{dt} + 18 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{1}{2}\right)}^{n} {\int}_{3}^{4} {\sech}^{2 n - 1} \left(9 t\right) \mathrm{dt}$

A zeroth-order approximation gives:

$L \approx 2 {\left[\sinh \left(9 t\right)\right]}_{3}^{4}$

Hence:

$L \approx 2 \sinh \left(36\right) - 2 \sinh \left(27\right)$