# Thread: Find length of arc

1. ## Find length of arc

Find the length of the curve for the specified portion:
$\displaystyle x=\tanh t$, $\displaystyle y=sech t$; between $\displaystyle A(t=0)$ and $\displaystyle B(t=1)$. What is the curve called?

I know the formula:
$\displaystyle \displaystyle s=\int \left(\left[\frac{dx}{dt}\right]^2+\left[\frac{dy}{dt}\right]^2\right)^{\frac{1}{2}} dt$

$\displaystyle \frac{dx}{dt}=sech^2 t$. $\displaystyle \frac{dy}{dt}=-sech t\tanh t$
$\displaystyle \displaystyle s=\int^1_0 \left(\left[sech^2 t\right]^2+\left[\tanh tsech t\right]^2\right)^{\frac{1}{2}} dt$
$\displaystyle \displaystyle s=\int^1_0 sech t\left(sech^2 t+\tanh^2 t)^{\frac{1}{2}}$
$\displaystyle \displaystyle s=\int^1_0 sech t (1) dt$
$\displaystyle \displaystyle s=\int^1_0 \frac{2e^t}{e^{2t}+1} dt$
Substituting $\displaystyle u= e^t$
$\displaystyle du= e^t dt$
$\displaystyle \displaystyle s=2\int^e_1\frac{1}{u^2+1}du$
$\displaystyle \displaystyle s=2[\arctan u]^e_1$

Answer gives $\displaystyle \arcsin (\tanh 1)$ which i calculated to be the same, but i don't know how they got it in that form.
Thanks

2. Hello, arze!

$\displaystyle \text{Find the length of the curve: }\begin{Bmatrix}x &=& \tanh t \\ y &=& \text{sech }t \end{Bmatrix}\:\text{ from }t = 0\text{ to }t = 1.$;

$\displaystyle \displaystyle s=\int^1_0 \text{sech }\!t\,dt \;=\;2\arctan e^t$

$\displaystyle \text{Answer gives }\arcsin (\tanh 1),\text{ which i calculated to be the same,}$
$\displaystyle \text{ but i don't know how they got it in that form.}$

I'll take a guess . . . but it's pretty far-fetched.

Identity: .$\displaystyle \text{sech}^2t \:=\:1 - \tanh^2\!t$

. . . . . . . $\displaystyle \text{sech }\!t \:=\:\sqrt{1 - \tanh^2\!t} \quad\Rightarrow\quad \dfrac{\text{sech }\!t}{\sqrt{1-\tanh^2\!t}} \:=\:1$

We have: .$\displaystyle \displaystyle \int \text{sech }\!t\,dt$

Multiply by $\displaystyle \frac{\text{sech }\!t}{\sqrt{1-\tanh^2\!t}}: \;\;\displaystyle \int \frac{\text{sech }\!t}{1}\cdot\frac{\text{sech }\!t}{\sqrt{1-\tanh^2\!t}}\,dt \;=\;\int \frac{\text{sech}^2t\,dt}{\sqrt{1-\tanh^2\!t}}$

Now let $\displaystyle u \,=\,\tanh t$