# Max error and relative error

• Oct 10th 2009, 02:31 PM
mybrohshi5
Max error and relative error
The circumference of a sphere was measured to be http://webwork.asu.edu/webwork2_file...63e9778881.png cm with a possible error of http://webwork.asu.edu/webwork2_file...5b5ed35541.png cm. Use differentials to estimate the maximum error in the calculated surface area.

Estimate the relative error in the calculated surface area.

This is what i tried but i got it wrong?

Circumference of a sphere = 2πr
so r = 11.61831

SA=4πr^2

SA=1696.2731
SAmax=1845.4147

Maximum error = 149.1416 ? this is wrong

Relative error = change in SA / SA
f'sa = 8πr so change in SA = 8π(11.61831)(.5) = 145.9999

relative error = 145.9999/1696.2731 = .08607

can i get some help with what to do because i guess i am doing it wrong (Crying)

thank you
• Oct 11th 2009, 11:21 AM
Media_Man
Use differentials
What you are doing is correct, but you seem to have made some arithmetic errors. Consider:

$\displaystyle C=2\pi r=73\pm .5=C\pm\Delta C$

$\displaystyle S=4\pi r^2=\frac{C^2}{\pi}$

So $\displaystyle S=\frac{73^2}{\pi}=1696.27$ but can be in the range $\displaystyle \frac12\left(\frac{73.5^2}\pi-\frac{72.5^2}\pi\right)=\pm23.24$

Here is how to reach the same answer using differentials:

$\displaystyle S=\frac{C^2}{\pi}$

$\displaystyle dS=\frac{2C}{\pi} dC=\frac{2*73}{\pi}(.5)=\pm23.24$

So the surface area is $\displaystyle S=1696.27\pm23.24$, with a relative error of 1.37%
• Oct 11th 2009, 12:23 PM
mybrohshi5
That was very helpful and clearly showed what i did wrong.

i tried to get help from my professor but he just confused me more and your work was much more clear than his.

thanks again (Happy)