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- anonymous

If r=6cm and h=2cm, give or take 0.1cm, which is the expected maximum percentage error in calculating V= pie*(r^2)*h

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- anonymous

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- anonymous

Apply the propagation of error to the equation of volume. Are you familiar with it?

- anonymous

no

- anonymous

could you show me?

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- anonymous

Sure. Actually, for worst case, its a complete differential. My mistake. But here we go.
Percent errors come in the form of dr/r, which in our case is .1/6, and dh/h is .1/2. We are looking for dV/V.
Do you know any easy way to get to this result?

- anonymous

no, could you go through it quickly all the way to the end please?

- anonymous

Sure. Take the natural log of both sides, giving you:
\[lnV = \ln(\pi r ^{2} h) = \ln \pi + 2 \ln r + \ln h\]
Can you take the derivative of this?

- anonymous

(1/pie)+(2/r)+(1/h)?

- anonymous

Close. Each variable is subject to change, however.

- anonymous

oh wait is it 2/r?

- anonymous

\[dV/V = 2dr/r + dh/h\]
ln pi is a number, so it falls away when a derivative is taken.

- anonymous

Since we are looking for the worst case, you take the absolute value of each term. This doesn't effect this set up at all, however.
So we know dr/r and dh/h. So now we plug in and get our answer for dV/V.

- anonymous

so answer is 6%?

- anonymous

2*(.1/6) + (.1/2) = .083333 = 8.33%

- anonymous

ok thank you very much

- anonymous

Anytime. Take care!

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