Results 1 to 6 of 6

Thread: Length of Cycloid

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    128

    Length of Cycloid

    Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

    We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

    Assume a is a constant:

    $\displaystyle f(t)=x=at-a\sin{t}$

    $\displaystyle g(t)=y=a-a\cos{t}$

    $\displaystyle 0\leq{t}\leq{2\pi}$

    $\displaystyle f'(t)=a-a\cos{t}$

    $\displaystyle g'(t)=a\sin{t}$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}|$ $\displaystyle dt$

    $\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle \sin{t}$ for $\displaystyle t $ $\displaystyle \epsilon$ $\displaystyle [0,\pi]$

    $\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle -\sin{t}$ for $\displaystyle t $ $\displaystyle \epsilon$ $\displaystyle [\pi,2\pi]$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t}$ $\displaystyle dt$ + $\displaystyle \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|(-\cos{t})]_0^\pi$ + $\displaystyle \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}$

    $\displaystyle L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))$

    $\displaystyle L=4\sqrt{2}|a|$

    It's evident that I should not have a $\displaystyle \sqrt{2}$, instead I should have just 2 outside the integral. Right?
    Last edited by elizsimca; Apr 18th 2008 at 09:26 PM. Reason: Interval Error
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by elizsimca View Post
    Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

    We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

    Assume a is a constant:

    $\displaystyle f(t)=x=at-a\sin{t}$

    $\displaystyle g(t)=y=a-a\cos{t}$

    $\displaystyle 0\leq{t}\leq{2\pi}$

    $\displaystyle f'(t)=a-a\cos{t}$

    $\displaystyle g'(t)=a\sin{t}$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}}$ $\displaystyle dt$

    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}|$ $\displaystyle dt$

    $\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle \sin{t}$ for $\displaystyle t $ $\displaystyle \epsilon$ $\displaystyle [0,\pi]$

    $\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle -\sin{t}$ for $\displaystyle t $ $\displaystyle \epsilon$ $\displaystyle [\pi,2\pi]$

    $\displaystyle L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t}$ $\displaystyle dt$ + $\displaystyle \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t}$ $\displaystyle dt$

    $\displaystyle L=\sqrt{2}|a|(-\cos{t})]_0^\pi$ + $\displaystyle \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}$

    $\displaystyle L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))$

    $\displaystyle L=4\sqrt{2}|a|$

    It's evident that I should not have a $\displaystyle \sqrt{2}$, instead I should have just 2 outside the integral. Right?
    No time now. If no-one else replies I'll take a look later.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    128
    It's okay Mr. F.
    It would have been up 45 mins ago, but I decided to try my hand at LaTeX so everyone could see the work I had already done.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Errors

    Elizsimca
    You were correct up to here:
    $\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$
    but you made an error in the next line:
    $\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-{\color{red}2a^2\cos{t}}+a^2\cos^2{t}+a^2\sin^2{t} }$
    So
    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos{t}}$
    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}$
    $\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}$
    $\displaystyle L=|a|\int_{0}^{2\pi}\sqrt{2(1-\cos{t})}$
    $\displaystyle L=|a|\int_{0}^{2\pi}\sqrt{2(2\sin^2(\frac{t}{2}))}$
    $\displaystyle L=2|a|\int_{0}^{2\pi}\sqrt{\sin^2(\frac{t}{2})}$
    And from there, you should be able to get the correct answer.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    Posts
    128
    Oh what a dumb mistake!! Thank you I super appreciate it!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by elizsimca View Post
    Oh what a dumb mistake!! Thank you I super appreciate it!
    And we appreciate you being so dilligent in showing all your working. Latex can be tricky. Well done - as you can see, it made it very easy to get a good answer.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Surface area of a cycloid
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 26th 2011, 02:39 AM
  2. Finding a parametric Equation for a Cycloid
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 8th 2010, 05:31 PM
  3. Area of cycloid using integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jul 7th 2009, 06:26 AM
  4. Surface Area of a Cycloid
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 4th 2008, 02:38 AM
  5. cycloid generated by a circle
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 23rd 2008, 12:25 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum