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For whatever reason I am incapable of doing this problem.

The circumference of a sphere was measured to be 84 cm with a possible error of .5 cm a) use differentials to estimate the max error in calculated surface area. What is the relative error?

I know that the formula will be $A=4 \pi r^2$ and that the radius is $r= \frac{84}{2\pi}$

know that I will find the error by using differentials where I get

$\frac {dx}{dy} = f \prime (x)$

$ dy = dx f \prime (x)$

$dy=.05 (8\pi r)$ is the same as $dy = .5 (8\pi\frac{84}{2\pi})$

which is the wrong answer. I have no idea what I am doing wrong besides that I suck at math.

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  • $\begingroup$ You want $\Delta A = \frac {dA}{dc}\Delta c$ where $c$ is the circumference. I don't know what $f$, $x$ and $y$ are $\endgroup$ – Ross Millikan Mar 20 '12 at 22:50
  • $\begingroup$ f is a function. $\endgroup$ – toby yeats Mar 20 '12 at 22:58
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The .05 is wrong, but the final answer you get looks OK to me (for the maximum error; for the relative error, it seems to me we did a problem of that sort very recently, you could go look for it). Why do you say it's wrong?

EDIT: I think I made the same mistake you made. $.5$ is the possible error in the circumference but you've put it into your answer where you need the possible error in the radius. Since $C=2\pi r$, $dC=2\pi dr$, so $dr=dC/(2\pi)=(.5)/(2\pi)$. Put that number for what you are calling $dx$, and you should be OK.

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  • $\begingroup$ The answer is $\frac{84}{\pi}$ and I can't get that. $\endgroup$ – toby yeats Mar 21 '12 at 1:01
  • $\begingroup$ Ahhh I get it now, I can't believe I spent so long on this problem and I couldn't get it. I don't know why I have such a problem with math like this. $\endgroup$ – toby yeats Mar 21 '12 at 2:45

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