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anonymous
 5 years ago
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
anonymous
 5 years ago
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
 5 years ago
Best ResponseYou've already chosen the best response.0Apply the propagation of error to the equation of volume. Are you familiar with it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. 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
 5 years ago
Best ResponseYou've already chosen the best response.0no, could you go through it quickly all the way to the end please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Close. Each variable is subject to change, however.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[dV/V = 2dr/r + dh/h\] ln pi is a number, so it falls away when a derivative is taken.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since 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
 5 years ago
Best ResponseYou've already chosen the best response.02*(.1/6) + (.1/2) = .083333 = 8.33%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thank you very much
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