Question

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

Answer 1

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

Answer 2

no

Answer 3

could you show me?

Answer 4

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?

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

Close. Each variable is subject to change, however.

Answer 9

oh wait is it 2/r?

Answer 10

\[dV/V = 2dr/r + dh/h\]

ln pi is a number, so it falls away when a derivative is taken.

Answer 11

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.

Answer 12

so answer is 6%?

Answer 13

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

Answer 14

ok thank you very much

Answer 15

Anytime. Take care!