Results 1 to 8 of 8

Math Help - Another Maximum Error Problem

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Another Maximum Error Problem

    Hi
    Can someone tell me why my answer is incorrect.
    The area of the circle segment is given by
    A=\frac{1}{2}r^2(\theta - sin(\theta)

    The radius r is measured with a maximum percentage error of 0.2% and \theta (the angle subtended at the circle's center) is measured as 45 degrees with a maximum error of 0.1 degrees. Find the approximately the maximum percentage error in the calculated area.

    This is my attempt on the question:
    \partial r = \frac{0.2r}{100}

    \partial \theta = \frac{0.1\pi}{180}

    \frac{\partial A}{\partial r} = r\theta-rsin(\theta}

    \frac{\partial A}{\partial r} = \frac{1}{2}r^2-\frac{1}{2}r^2cos(\theta)

    \partial A = [r(\theta-sin(\theta)](\frac{0.2r}{180})+[\frac{1}{2}r^2(1-cos(\theta)](\frac{0.1\pi}{180})

    \partial A =  [r(\frac{45\pi}{180}-sin(\frac{45\pi}{180})](\frac{0.2r}{180})+[\frac{1}{2}r^2(1-cos(\frac{45\pi}{180})](\frac{0.1\pi}{180})
    <br />
\partial A =[\frac{\pir^2}{3600}-\frac{sin(\frac{\pi}{4}}{900}]+[\frac{\pir^2}{3600}-\frac{\pir^2}{3600}cos(\frac{\pi}{4}]
    <br />
\partial A = 3.425879759*10^-4

    A = \frac{1}{2}r^2(\frac{45\pi}{180}-sin(\frac{45\pi}{180}))
    A = 0.0391456911

    Maximum error is \frac{3.425879759*10^-4}{0.0391456911} * 100 = 0.88%

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone tell me why my answer is incorrect.
    The area of the circle segment is given by
    A=\frac{1}{2}r^2(\theta - sin(\theta)

    The radius r is measured with a maximum percentage error of 0.2% and \theta (the angle subtended at the circle's center) is measured as 45 degrees with a maximum error of 0.1 degrees. Find the approximately the maximum percentage error in the calculated area.

    This is my attempt on the question:
    \partial r = \frac{0.2r}{100}

    \partial \theta = \frac{0.1\pi}{180}

    \frac{\partial A}{\partial r} = r\theta-rsin(\theta}

    \frac{\partial A}{\partial r} = \frac{1}{2}r^2-\frac{1}{2}r^2cos(\theta)

    \partial A = [r(\theta-sin(\theta)](\frac{0.2r}{180})+[\frac{1}{2}r^2(1-cos(\theta)](\frac{0.1\pi}{180})
    \delta A \approx  \dfrac{\partial}{\partial r} A(r,\theta)\  \delta r+\dfrac{\partial}{\partial \theta} A(r,\theta)\  \delta \theta

    \delta A \approx [r(\theta-\sin(\theta)](0.002r)+[\frac{1}{2}r^2(1-\cos(\theta)](\frac{0.1\pi}{180})

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    so now i sub in \theta = 45 to get my result for \partial A
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Paymemoney View Post
    so now i sub in \theta = 45 to get my result for \partial A
    You want the maximum percentage error, so you want:

    \dfrac{\delta A}{A} \times 100 \%

    which conveniently cancels out the r^2 terms

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2008
    Posts
    509
    This is what i got, however its incorrect.

    = \frac{\partial A}{A} * 100 \%

    = \frac{0.0887124277r^2}{22.07454824r^2} * 100 \%

    = 0.0041
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Paymemoney View Post
    This is what i got, however its incorrect.

    = \frac{\partial A}{A} * 100 \%

    = \frac{0.0887124277r^2}{22.07454824r^2} * 100 \%

    = 0.0041
    Well I get 1.05%.

    Do you have all the angles in radians?

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Yes i made it in radians,
    After doing the calculatiobn again i still get the same answer as before.
    This is how i worked it out:
    \parital A = (r(45-sin45))(0.002r)+(0.5r^2(1-cos45))(\frac{0.1\pi}{180})
    \partial A =0.0887124277
    A = 0.5r^2(45-sin45)
    A =
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Here are my calculations (the r's can be dropped as they cancel):

    Code:
    >
    >
    >theta=pi/4
         0.785398 
    >
    >deltaA=(theta-sin(theta))*0.002+(1/2)*(1-cos(theta))*(0.1*pi/180)
       0.00041218 
    >
    >
    >A=(1/2)*(theta-sin(theta))
        0.0391457 
    >
    >deltaA/A*100
          1.05294 
    >
    and given the prescission of the errors we should round this to 1 digit, that is 1%

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding Maximum Error Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 20th 2010, 10:28 PM
  2. Maximum Error Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 6th 2010, 05:05 PM
  3. Maximum Possible Error Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 2nd 2010, 11:41 PM
  4. Relative Error & Maximum Error
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 10th 2010, 02:40 AM
  5. [SOLVED] Maximum error
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 1st 2010, 03:26 PM

Search Tags


/mathhelpforum @mathhelpforum