- #1
Dell
- 590
- 0
when calculating the mean square error i have been using the differential,
if a length measured is x=2 and the error mx=+- 0.005
then x=2+-0.005
if i have x+y where y=3, my=+-0.02
mx+y=[tex]\sqrt{my2+mx2}[/tex]
mx*y=[tex]\sqrt{(y*mx)2+(x*my)2}[/tex]
but if i have x^2 does this work the same
for example if the area of a rectangle is x*2x can i say 2x2
m2x2=4x*mx
if a length measured is x=2 and the error mx=+- 0.005
then x=2+-0.005
if i have x+y where y=3, my=+-0.02
mx+y=[tex]\sqrt{my2+mx2}[/tex]
mx*y=[tex]\sqrt{(y*mx)2+(x*my)2}[/tex]
but if i have x^2 does this work the same
for example if the area of a rectangle is x*2x can i say 2x2
m2x2=4x*mx