Results 1 to 6 of 6

Thread: Find circumference of cardioid

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Find circumference of cardioid

    The cardioid $\displaystyle r=a(1-\sin\theta)$; find the length of its circumference.

    The formula for polar equations:
    $\displaystyle \displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta$

    $\displaystyle \frac{dr}{d\theta}=-a\cos\theta$
    $\displaystyle \displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}} d\theta$
    $\displaystyle \displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra c{1}{2}}$
    $\displaystyle \displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta$

    $\displaystyle (1-\sin\theta)^{1/2}$ is what stumps me, how do i convert that to something that can be integrated?
    Thanks
    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 arze View Post
    The cardioid $\displaystyle r=a(1-\sin\theta)$; find the length of its circumference.

    The formula for polar equations:
    $\displaystyle \displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta$

    $\displaystyle \frac{dr}{d\theta}=-a\cos\theta$
    $\displaystyle \displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}} d\theta$
    $\displaystyle \displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra c{1}{2}}$
    $\displaystyle \displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta$

    $\displaystyle (1-\sin\theta)^{1/2}$ is what stumps me, how do i convert that to something that can be integrated?
    Thanks
    integrate Sqrt[1 - Sin[x]] - Wolfram|Alpha

    Click on Show steps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Quote Originally Posted by mr fantastic View Post
    Ok, so i did that, and got:
    $\displaystyle 2a\sqrt{2} [2\sqrt{\sin\theta+1}]^{\pi}_0$
    which gives me 0.
    I tried adding the integrals between $\displaystyle \frac{\pi}{2}$ and 0, and $\displaystyle \pi$ and $\displaystyle \frac{\pi}{2}$ they still add to 0.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, arze!

    The cardioid $\displaystyle r=a(1-\sin\theta)$; find the length of its circumference.

    $\displaystyle \displaystyle s \;=\;2\sqrt{2}\,a\int^{\pi}_0\left(1-\sin\theta\right)^{\frac{1}{2}}d\theta$

    We have: .$\displaystyle \sqrt{1-\sin\theta}$

    Multiply by $\displaystyle \frac{\sqrt{1+\sin\theta}}{\sqrt{1+\sin\theta}}:\; \;\dfrac{\sqrt{1-\sin\theta}}{1}\cdot\dfrac{\sqrt{1+\sin\theta}}{\s qrt{1+\sin\theta}} \;=\;\dfrac{\sqrt{1-\sin^2\theta}}{\sqrt{1+\sin\theta}} $

    . . . . . . . . . . . . . $\displaystyle =\;\dfrac{\sqrt{\cos^2\theta}}{\sqrt{1+\sin\theta} } \;=\;\dfrac{\cos\theta}{\sqrt{1+\sin\theta}} $


    $\displaystyle \text{The integral becomes: }\;\displaystyle 2\sqrt{2}\,a\int (1 + \sin\theta)^{-\frac{1}{2}}\,\cos\theta\,d\theta $
    $\displaystyle \text}Let }\,u \:=\:1 + \sin\theta $

    Got it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    yes, got it!
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    ok, now i have another problem. between pi and 0, u has the same values, what values should i use? for the angles?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of a cardioid
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Nov 15th 2010, 06:56 PM
  2. Consider the the cardioid
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jul 27th 2010, 07:11 AM
  3. Replies: 2
    Last Post: May 5th 2010, 11:55 PM
  4. cardioid
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 2nd 2008, 05:36 AM
  5. Replies: 2
    Last Post: Apr 11th 2008, 02:03 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum