Use differentials to approximate `sqrt(25. 2)`

Options

A. 5.01
B. 5.02
C. 5.03
D. none of these

Answer

B

Explanation

Consider the function `y=f(x)=sqrt(x)`.
Let x=25 and `x +Deltax=25.2. Then, Deltax = 25.2-25 =0.2` For x = 25, we have `y=sqrt(25)=5" "["Putting" " "x=25 " in "y=sqrt(x)]`
Let `dx= Delta x = 0.2` Now,
`y=sqrt(x) rArr dy/dx=-(1)/(2sqrt(x)) rArr(dy/dx)_(x=25) =1/(2(5))=1/10`
`:." " dy=dy/dxdx`
`rArr" "dy=1/10(0.2)=0.02 rArr Delta" " y= 0.02" "[:' Delta y cong dy]`
Hence, `sqrt(25.2)= y + Deltay=5 +0.02=5.02`