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Math Help - Find length of arc

  1. #1
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    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
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  2. #2
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    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
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