Question

Using differentials, find the approximate values of the following:
(ii) $(0.009{)}^{\frac{1}{3}}$

Answer 1

Let $y=x^{\frac{1}{3}}\Rightarrow \frac{dy}{dx}=\frac{1}{3x^{\frac{2}{3}}}$
where, x = 0.008 & $\Delta x=0.001$
Now, we know
$\Delta y=(x+\Delta x{)}^{\frac{1}{3}}-x^{\frac{1}{3}}$
$\Rightarrow \Delta y=(0.008+0.001{)}^{\frac{1}{3}}-(0.008{)}^{\frac{1}{3}}$
$\Rightarrow \Delta y=(0.009{)}^{\frac{1}{3}}-0.2$
$\Rightarrow (0.009{)}^{\frac{1}{3}}=0.2+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{3x^{\frac{2}{3}}}\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{3(0.008{)}^{\frac{2}{3}}}\times 0.001$
$\Rightarrow \Delta y\approx \frac{0.001}{0.12}$
$\Rightarrow \Delta y\approx 0.00833$
From equation (1)
$(0.009{)}^{\frac{1}{3}}\approx 0.2+0.00833$
$\therefore (0.009{)}^{\frac{1}{3}}\approx 0.20833$