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Math Help - finding the length of the curve

  1. #1
    Aleksandra12
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    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
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    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
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  3. #3
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    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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