Question

Using Differentials, Find The Approximate Value Of The Following Up To $3$ Places Of Decimal.

${(0.009)}^{\frac{1}{3}}$

Answer 1

Given: ${(0.009)}^{\frac{1}{3}}$

Consider $y=x^{\frac{1}{3}}$

Let $x=0.008$

$△x=0.001$

Then

$\begin{array}{rcl}∆y& =& {(x+∆x)}^{\frac{1}{3}}-x^{\frac{1}{3}}\\ ∆y& =& 0.{009}^{\frac{1}{3}}-0.2\\ 0.{009}^{\frac{1}{3}}& =& 0.2+∆y\\ & & \\ & & \end{array}$

Now, $\mathrm{dy}$ is approximately equal to $\mathrm{\Delta y}$ and is given by,

$$\begin{gathered} \mathrm{dy}=\left(\frac{dy}{dx}\right)∆x \\ =\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}}(∆x) \end{gathered}$$

As $y=x^{\frac{1}{3}}$

So,

$$\begin{gathered} \mathrm{dy}=\frac{1}{3\times 0.04}(0.001) \\ =\frac{0.001}{0.12} \\ =0.008 \end{gathered}$$

Hence, the approximate value of $$\begin{gathered} {(0.009)}^{\frac{1}{3}}=0.2+0.008 \\ =0.208 \end{gathered}$$