The basic formula is ¦¤y ¡Ö y' ¦¤x
y ' = 1 / ( 2 ¡Ì x )
122 = 121 + 1
¡Ì 121 = 11
So let x = 121 , y = 11 and ¦¤x = 1
y ' = 1 / ( 2 ¡Ì x ) = 1 / ( 2 * 11 ) = 1 / 22 = 0.0455
¦¤y ¡Ö y' ¦¤x = ( 1 / 22 ) * 1 = 1 / 22 = 0.0455
¡Ì122 ¡Ö y + ¦¤y
¡Ì122 ¡Ö 11 + 0.0455
¡Ì122 ¡Ö 11.0455
Use the differential to approximate the quantity to four decimal places.
√(122)
2 answers
y = x^.5
which means
x = y^2
dy/dx = .5 x^-.5 which is .5/y guessed
y2 = y1 + (.5/y1)(error in x)
y2 = y1 + (.5/y1)(122-y1^2)
start with y1 = one tenth of 122, or 12 for example then error x =122 - 144 approx = -22
y2 = 12 + (.5/12)(-22) = 12 -.917 = 11.08
so our next y1 is 11.08
and again
y2 = y1 + (.5/y1)(122-y1^2)
y2 = 11.08 + (.5/11.08)(122-122.84)
y2 = 11.08 - .0379
y2 or new y1 = 11.042
well keep going using that new y1
the real answer is about 11.04536102
which means
x = y^2
dy/dx = .5 x^-.5 which is .5/y guessed
y2 = y1 + (.5/y1)(error in x)
y2 = y1 + (.5/y1)(122-y1^2)
start with y1 = one tenth of 122, or 12 for example then error x =122 - 144 approx = -22
y2 = 12 + (.5/12)(-22) = 12 -.917 = 11.08
so our next y1 is 11.08
and again
y2 = y1 + (.5/y1)(122-y1^2)
y2 = 11.08 + (.5/11.08)(122-122.84)
y2 = 11.08 - .0379
y2 or new y1 = 11.042
well keep going using that new y1
the real answer is about 11.04536102