00:01
In this exercise, we have a rectangle that has a length of 5 .10 plus or minus 0 .01 centimeters and a width of 1 .90 plus or minus 0 .01 centimeters.
00:14
Our going question a is to find both the area and the uncertainty that is related to this area.
00:21
So the area is just the length times the width.
00:26
So this is 5 .10 centimeters times 1 .90 centimeters.
00:37
And this is equal to 9 .69 centimeters.
00:45
This is the area.
00:47
In order to find the uncertainty of the area, what i'm going to do is to find the maximum value of the area, the maximum possible value.
00:57
I'm also going to find the minimum value, a min, and i'm going to divide the result by 2, and this will be the uncertainty of the area.
01:12
So let's start with the maximum value of the area.
01:16
The maximum value is reached when the length is maximum, that is when the length is l plus delta l, the maximum width is w plus delta w.
01:33
This is 5 .10 plus 0 .01, so that's 5 .11 centimeters times 1 .90 plus 0 .01, which is 1 .91 centimeters.
01:48
Delta l and delta w are just the uncertainties in the length and the width, as i'm writing up here.
02:04
So, obtain that the maximum area is 9 .76 centimeters.
02:11
The minimum area happens when both the length and width are minimum...