Question

Using differentials, find the approximate value of each of the following up to 3 places of decimal.
$(0.009)^{\frac{1}{3} }$

Answer 1

Answer:

Let y=x13

Let x=0.008

dx=0.001

So that x+dx=0.009

Now (x+Δx)13−x13

⇒(0.009)13−(0.008)13

⇒(0.009)13−0.2

∴(0.009)13−0.2+Δy-----(1)

Step 2:

Also dydxΔx is approximately equal to dy

dy=(dydx)Δx

=13x23Δx

=13(0.008)23×0.001

=13(0.2)2×0.001

=0.0013×0.04

=0.0010.12

=1120

=0.008

Step 3:

Approximate value of Δy=dy=0.008

Hence from Equation(1) we have

(0.009)13=0.2+0.008

=0.208