# Thread: Find length of arc

1. ## Find length of arc

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

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

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

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

2. Hello, arze!

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

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

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

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

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

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

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

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

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