11. If the variation in 2 is k corresponding to variation h in x near x=3 then
√(1+x±h)=2±k.
Squaring: 1+x±h=4±4k+k². When x=3, h=±4k+k².
When k=1, h=5 and -3, and x=(3-3,3+5)=(0,8) correspond to (1,3) which are (2-1,2+1).
When k=0.0002, h=±0.0008+0.00000004, that is to 4 decimal places h=±0.0008.
So x=(3-0.0008,3+0.0008)=(2.9992,3.0008).
12. Let x be the length of each wire then x² is the area 100±0.06 sq in. If the variation for x is h, then:
(x+h)²=100±0.06=100(1±0.06/100)=100(1±0.0006)
x+h=10√(1±0.0006)=10(1±0.0006)^½=10(1±0.0003) approx=10±0.003.
Whenever x=10, h=±0.003. So the variation in the length of each piece of wire is less than 0.003 in.
Another way to solve this: x²+2hx+h²=100±0.06, so, putting x=10, 20h=0.06 if h is small. So h=0.003 in. approx.