finding the length of the curve

Aleksandra12 · August 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** · August 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** · August 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: . $L \;=\;\int^5_0\left(e^t + e^{\text{-}t}\right)\,dt \;=\;\left(e^t - e^{\text{-}t}\right)\bigg]^5_0 \;=\;\left(e^5 - e^{\text{-}5}\right) - \left(e^0 - e^0\right)$

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

Math Help - finding the length of the curve

LinkBack

Thread Tools

Display

finding the length of the curve

Similar Math Help Forum Discussions

Finding arc length along the curve

Finding the length of the curve (where it gets undefined at t=0)

Finding the length of a curve ??

Finding the length of a curve parametrized by...

two Q's: finding center of mass of a bounded region, and finding length of curve