Propagation of Error

This worksheet involves propagation of error. Recall that propagation of error is just the idea that a value that is calculated from other values that have error will also have error, right? So this is the method we use in this class. So the problem is: The sides of a small rectangular box are measured to be 1.80 +/- 0.01 cm, 2.05 +/- 0.02 cm, and 3.1 +/- 0.1 cm. Calculate it's volume and uncertainty in cm^3 and report the result in proper form. So I'm going to draw a picture, just so that it's more clear in my head. So here's a small box, not necessarily to scale. Let's label some of these sides, we've got 3.1, maybe 2.05, and 1.8. So we could call those a length, a width, and a height. So maybe the length is (3.1 +/- 0.1) cm, maybe the width is (2.05 +/- 0.02) cm, and maybe the height is (1.80 +/- 0.01) cm. Ok, so we're asked to find the volume and the error in that volume, so this little delta here is the error in. So let's start by finding the volume. You may recall that the volume of a rectangular box is simply length times width times height. So in this step we can ignore these errors; the 0.1, the 0.02, the 0.01; we're just dealing with the 3.1, the 2.05, and the 1.8. If you do that math you end up with 11.439 cm^3. So that's the volume, and we just wrote down all of the digits that the calculator gave us, so we should ask ourselves: how many of these digits can we trust, right? Because I know how much I can trust the length, the width, and the height, so now I need to know how many of these digits are meaningful and how many of them are just buried in the error. So we'll find the error. So the error in the volume - the procedure we've used is to assume that the errors in the length and the width and the height conspired to give us the largest wrong answer. So we called that V_largest. Once we have V_largest, if we subtract the actual V, that difference is a reasonable estimate of this error. So let's find V_largest. (Maybe I'll slide this up a little bit). Let's find V_largest. The procedure here is simply to again determine from the length, the width, and the height, the errors in these, how they could conspire to give us the largest wrong answer. In this case you're just going to add the error to each of these values. Remember, that's not always the case, but it happens to be the case this time. So that's 3.1 + 0.1, 2.05 + 0.02, and 1.80 + 0.01, and if you add those up and then multiply, you end up with 11.9894 cm^3. So now we can use that largest value and the actual value, find a difference and come up with the error. So we'll plug those in here: 11.9894 - 11.439, and you end up with 0.55044 cm^3. That's the error. Ok, so we could write our result as V = (11.439 +/- 0.55044) cm^3. But again, there are lots of digits here and not all of them are meaningful, so we're going to write this in what we call Proper Form, and the way we do this is the following. We can round the error to 1 significant figure, because error is inherently uncertain, so we can only trust one digit in the error. So we're going to round this 5 to a 6 so it's 0.6 cm^3, and now we use the error to guide our rounding of the answer. The error is in the tenths place, so we're going to round our answer to the tenths place. Ok, so it looks like we're going to round this 0.439 down to just a 0.4, so in proper form our volume is (11.4 +/- 0.6) cm^3