1. 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:
$\partial r = \frac{0.2r}{100}$

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

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

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

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

$\partial A = [r(\frac{45\pi}{180}-sin(\frac{45\pi}{180})](\frac{0.2r}{180})+[\frac{1}{2}r^2(1-cos(\frac{45\pi}{180})](\frac{0.1\pi}{180})$
$
\partial A =[\frac{\pir^2}{3600}-\frac{sin(\frac{\pi}{4}}{900}]+[\frac{\pir^2}{3600}-\frac{\pir^2}{3600}cos(\frac{\pi}{4}]$

$
\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

2. Originally Posted by Paymemoney
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

3. so now i sub in $\theta = 45$ to get my result for $\partial A$

4. Originally Posted by Paymemoney
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

5. This is what i got, however its incorrect.

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

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

$= 0.0041$

6. Originally Posted by Paymemoney
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

After doing the calculatiobn again i still get the same answer as before.
This is how i worked it out:
$\parital A = (r(45-sin45))(0.002r)+(0.5r^2(1-cos45))(\frac{0.1\pi}{180})$
$\partial A =0.0887124277$
$A = 0.5r^2(45-sin45)$
$A =$

8. 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