finding the length of the curve

Aleksandra12 · Aug 8th 2007, 05:03 PM

Find the length of the curve. Give annswer to 2 decimal places

X=e^t+e^(-t), Y=5-2t, 0 < t < 5

Please help!

Thank you

**galactus** · Aug 8th 2007, 05:14 PM

Use the arc length formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy} {dt})^{2}}dt$

$x(t)=e^{t}+e^{-t}=2cosh(t)$

$\frac{dx}{dt}=2sinh(t)$

$\frac{dy}{dt}=-2$

$\int_{0}^{5}\sqrt{4(sinh^{2}(t)+1)}$

But $sinh^{2}(t)+1=cosh^{2}(t)$

$2\int_{0}^{5}cosh(t)dt$

**Soroban** · Aug 8th 2007, 07:02 PM

Hello, Aleksandra12!

Another approach . . .

Find the length of the curve. Give annswer to 2 decimal places

. . $\begin{array}{ccc}x & =& e^t+e^{\text{-}t} \\<br /> y & = & 5-2t\end{array}\quad0 \leq t \leq 5$

Formula: . $L \:=\:\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

We have: . $\begin{array}{ccc}\frac{dx}{dt} & = & e^t - e^{\text{-}t} \\ \frac{dy}{dt} & = & \text{-}2\end{array}$

Then: . $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \:=\:\left(e^t - e^{\text{-}t}\right)^2 + (\text{-}2)^2 \:=\:e^{2t} -2 + e^{\text{-}2t} + 4$

. . $= \:e^{2t} + 2 + e^{\text{-}2t} \:=\:\left(e^t + e^{\text{-}t}\right)^2$

. . Hence: . $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \:=\:\sqrt{\left(e^t + e^{\text{-}t}\right)^2} \:=\:e^t + e^{\text{-}t}$

And we have: . $L \;=\;\int^5_0\left(e^t + e^{\text{-}t}\right)\,dt \;=\;\left(e^t - e^{\text{-}t}\right)\bigg]^5_0 \;=\;\left(e^5 - e^{\text{-}5}\right) - \left(e^0 - e^0\right)$

. . $= \;e^5 - e^{\text{-}5} \;=\;\frac{e^{10} - 1}{e^5} \;\approx\;148.41$

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