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Math Help - Having Trouble Finding The Length Of A Parametric Curve

  1. #1
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    Having Trouble Finding The Length Of A Parametric Curve

    Hey everyone. I've been trying to find the length of this parametric curve, but haven't had much luck.

    x=cos{t}
    y=t+sin{t}
    0\leq t \leq \pi

    After finding \frac{dy}{dt} and \frac{dx}{dt} and substituting them into the formula for curve length, I got:

    \int^\pi_0 \sqrt{2+2cost}\ dt

    After this, I tried u-substitution, but couldn't figure out how to eliminate the resultant t:

    u = 2+2cost

    du = -2sint\ dt

    dt = \frac{du}{-2sint}

    I consulted Wolfram Alpha as a last result, but couldn't figure out how they got the \int \frac{1}{\sqrt{4-u}}\ du

    Help would be greatly appreciated!
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  2. #2
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    Re: Having Trouble Finding The Length Of A Parametric Curve

    Multiply top and bottom by the conjugate.
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  3. #3
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    Re: Having Trouble Finding The Length Of A Parametric Curve

    Hello, Algebrah!

    I've been trying to find the length of this parametric curve.

    . . \begin{Bmatrix}x \:=\:\cos t \\ y\:=\:t+\sin t  \end{Bmatrix} \quad  0\:\leq\:t\:\leq\:\pi


    After finding \tfrac{dy}{dt} and \tfrac{dx}{dt} and substituting, I got:

    . . L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt \;=\; \int^\pi_0 \sqrt{2+2\cos t}\ dt . . Good!

    We have: . \sqrt{2(1+\cos t)} \;=\; \sqrt{4\cdot \tfrac{1\:+\:\cos t}{2}} \;=\;\sqrt{4\cos^2\tfrac{t}{2}} \;=\;2\cos\tfrac{t}{2}

    Thanks from Algebrah
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  4. #4
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    Re: Having Trouble Finding The Length Of A Parametric Curve

    Quote Originally Posted by Soroban View Post
    Hello, Algebrah!


    We have: . \sqrt{2(1+\cos t)} \;=\; \sqrt{4\cdot \tfrac{1\:+\:\cos t}{2}} \;=\;\sqrt{4\cos^2\tfrac{t}{2}} \;=\;2\cos\tfrac{t}{2}

    This is beautiful, but the OP wanted to know how Wolfram got the particular substitution they did. I expect it was by multiplying top and bottom by a particular conjugate, but when I do I get this...

    \displaystyle \begin{align*} \int_0^{\pi}{\sqrt{2 + 2\cos{t}}\,dt} &= \int_0^{\pi}{\frac{\sqrt{2 + 2\cos{t}}\,\sqrt{2 - 2\cos{t}}}{\sqrt{2 - 2\cos{t}}}\,dt} \\ &= \int_0^{\pi}{\frac{\sqrt{4 - 4\cos^2{t}}}{\sqrt{2 - 2\cos{t}}}\,dt} \\ &= \int_0^{\pi}{\frac{2\sin{t}}{\sqrt{2 - 2\cos{t}}}\,dt} \end{align*}

    And now the substitution \displaystyle \begin{align*} u = 2 - 2\cos{t} \implies du = 2\sin{t}\,dt \end{align*} is appropriate.
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  5. #5
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    Re: Having Trouble Finding The Length Of A Parametric Curve

    Sorry, I turned off my computer last night and then proceeded to solve the problem using the conjugate method.

    I also really appreciate Soroban's method, which is a nice use of the power-reducing formula.

    Thanks a lot for the help!
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