Calculations of Significant Figures

[playgames]

A rectangular plate has a length of (21.3\pm0.2) cm and a
width of (9.80\pm0.1) cm. Find the area of the plate and the
uncertainty in the calculated area.

Solution
Area = lw = (21.3\pm0.2 cm) X (9.80\pm0.1 cm)
\cong(21.3 X 9.80 \pm 21.3 X 0.1 \pm 0.2 X 9.80) cm^2
\cong(209 \pm4) cm^2

Because the input data were given to only three significant
figures, we cannot claim any more in our result. Do you see
why we did not need to multiply the uncertainties 0.2 cm and
0.1 cm?

 

[tiny-tim]

welcome to pf!

hi playgames! welcome to pf!

(try using the X² icon just above the Reply box )

Solution
Area = lw = (21.3\pm0.2 cm) X (9.80\pm0.1 cm)
\cong(21.3 X 9.80 \pm 21.3 X 0.1 \pm 0.2 X 9.80) cm^2
\cong(209 \pm4) cm^2

Do you see why we did not need to multiply the uncertainties 0.2 cm and
0.1 cm?

because it's too small ever to be of significance …

0.2 * 0.1 = 0.02, which is two orders of magnitude smaller than anything else

(if you've done calculus, this is similar to writing (x + dx)(y + dy) = xy + xdy + ydx, and ignoring the dxdy as being "second-order")

the 21.3 X 9.80 is "ordinary", the 21.3 X 0.1 and 0.2 X 9.80 are "first-order" of smallness, and the missing 0.2 X 0.1 is "second-order" of smallness