Assignment: Pick any exercise in Chapter 12 of the calculus textbook. Post the solution or your attempt. Discuss any issues you might have had.
Exercises 12.4 page 720
It is ‘common sense’ that it is far better to measure a long distance with a long measuring tape rather than a short one. A measured distance D can be viewed as a product of the length l of a measuring tape times the number of n of time it was used. For instance, using a 3’ tape 10 times gives a length of 30’. To measure the same distance with a 12’ tape, we would use the tape 2.5 times (i.e., 30 = 12 * 2.5.) Thus D = nl.
Suppose each time a measurment is taken with the tape, the recorded distance is within 1/16" of the acutal distance. (i.e.,, dl = 1/16" = 0.005ft). Using differntials, show why common sense proves correct that it is better to use a long tape to measure long distances.
For this question, I’m seeking to minimize the total amount of error over a process of n measurements. The example in the book endorses that I should treat ‘n’ continuously, the example they give is that measuring 30 feet with a 12-foot tape should be measuring 2.5 times and not 3. The error per measurment is defined as dl = 0.005.
The general formula for the total differential is
\[dz = f_x(x,y)dx + f_y(x,y)dy\]
In this case, x and y are n and l, with z = D and dz is the total error. Substiting in for D = nl:
\[dD = l \; dn + n \; dl = l \; dn + 0.005n\]
The total differential dD is equal to l dn + 0.005n. This means that the error will increase as n approaches infinity. In the case of measuring with a 12ft tape, relative total error is dl/l = .042%. The relative error in D corresponds to the formula
\[\frac {dD}{D} = \frac{l \; dn + 0.005n}{ln} = \frac {1}{n} dn + \frac{dl}{l} = \frac {1}{n} dn + .00042 \] At a minimum n = 1 measurments the relative total error is 0.042 and dn/n approaches infinity as n increases.