Question

Using differentials, find the approximate value of each of the following up to 3 places of decimal (i) 25.3 (ii)49.5 (iii)0.6(iv)(0.009)1/3(v)(0.999)1/10(vi)(15)1/4(vii)(26)1/3(viii)(225)1/4(ix)(82)1/4(x)(401)1/2(xi)(0.0037)1/2(xii)(26.57)1/3(xiii)(81.5)1/4(xiv)(3.968)1/2(xv)(32.15)1/5

Answer 1

(i)√25.3

Consider y=√x.Let x=25 and Δx=0.3

Then,

Δy,√x+Δx-√x=√25.3-√25=√25.3-5

=√25.3=Δy+5

Now, dy is approximately equal to ∆y and is given by,

dy=(dy.dx)Δx=1/2√25(0.3)=0.03

Hence, the approximate value of √25.3 is 0.03 + 5 = 5.03.

(ii) √49.5

Consider y=√x. Let x = 49 and ∆x = 0.5. Then,

Δy=√x+Δx-√x=√0.6-1

√0.6=1+Δy

Now, dy is approximately equal to ∆y and is given by

dy=(dy/dx)Δx=1/√x(Δx)

Hence, the approximate value of √0.6 is 1 + (−0.2) = 1 − 0.2 = 0.8

Now, dy is approximately equal to ∆y and is given by,

dy=(dy/dx)∆x

=1/5x(2)4(0.15)

=0.15/80=0.00187

Hence, the approximate value of (32.15)^1/5 is 2 + 0.00187 = 2.00187.