Thread: finding the length of the curve

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

Use the arc length formula:

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

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

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

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

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

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

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

Hello, Aleksandra12!

Another approach . . .

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

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

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

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

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

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

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

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

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