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Math Help - cycloid parametrically and surface area

  1. #1
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    cycloid parametrically and surface area

    Hi I'm new here, this is my first post I hope you don't mind if I ask 2 questions I'm stuck at.

    1. a cycloid is given parametrically by x = t-sint, y= 1 - cost
    find the arc length between t = 0 and t = 2 pie

    2. find the surface area when the curve y = (square root of)2x + 1 x E [1, 7] is rotated by 360 degrees about the x axis
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  2. #2
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    Hello, graeme87!

    Welcome aboard!
    Here's the first one . . .


    1. A cycloid is given parametrically by: . \begin{array}{ccc} x &=& t-\sin t \\ y&=& 1 - \cos t\end{array}

    Find the arc length between t = 0 and t = 2\pi
    Formula: . L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \,dt


    We have: . \begin{array}{ccc}\dfrac{dx}{dt} &=& 1 -\cos t \\ \\[-3mm] \dfrac{dy}{dt} &=& \sin t \end{array}


    \text{Then: }\;\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \;=\;(1\;-\;\cos t)^2 + \;\sin^2\!t \quad\Rightarrow\quad 1 \;-\; 2\cos t \;+\; \underbrace{\cos^2\!t + \sin^2\!t }_{\text{This is 1}}

    . . . . . . = \;2 - 2\cos t \;=\;2(1 - \cos t) \;=\;4\left(\frac{1-\cos t}{2}\right) \;=\;4\sin^2\tfrac{t}{2}

    Hence: . \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \;=\;2\sin\tfrac{t}{2}


    We have: . L \;=\;\int^{2\pi}_0 \!\!2\sin\tfrac{t}{2}\,dt \;=\;-4\cos\tfrac{t}{2}\,\bigg]^{2\pi}_0 \;=\;\bigg(-4\cos\pi\bigg) - \bigg(-4\cos 0\bigg)

    . . . . . . = \;(\text{-}4)(\text{-}1) - (\text{-}4) \;=\; 4 + 4 \;=\;\boxed{8}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    That is a surprising answer . . .

    We have a circle . . . and we are rolling it.

    We would expect to see \pi in the arc length.

    Instead, the answer is an integer.


    Fact: The arc length of one arch of a cycloid is equal to the perimeter
    . . . . of the square circumscribed about the generating circle.


    Fact: The area under one arch of a cycloid
    . . . . is 3 times the area of the generating circle.
    Code:
                            * * *...
                     *  *           *::*..
                *     *               *:::::* 
            *        *                 *::::::::*
         *                              :::::::::::*
       *            *        πrē        *::::::::::::*
            πrē                         ::::: πrē::::::
     *               *                 *:::::::::::::::*
                      *               *::::::::::::::::::
                        *           *::::::::::::::::::::
    * - - - - - - - - - - - * * * - - - - - - - - - - - *

    That shark-fin-shaped region is equal to the generating circle.

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  3. #3
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    Thank you, you're help is very much appreciated!
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