anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Apply the propagation of error to the equation of volume. Are you familiar with it?
anonymous
  • anonymous
no
anonymous
  • anonymous
could you show me?

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anonymous
  • 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
  • anonymous
no, could you go through it quickly all the way to the end please?
anonymous
  • 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
  • anonymous
(1/pie)+(2/r)+(1/h)?
anonymous
  • anonymous
Close. Each variable is subject to change, however.
anonymous
  • anonymous
oh wait is it 2/r?
anonymous
  • anonymous
\[dV/V = 2dr/r + dh/h\] ln pi is a number, so it falls away when a derivative is taken.
anonymous
  • 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
  • anonymous
so answer is 6%?
anonymous
  • anonymous
2*(.1/6) + (.1/2) = .083333 = 8.33%
anonymous
  • anonymous
ok thank you very much
anonymous
  • anonymous
Anytime. Take care!

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