Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.

$(0.0037{)}^{\frac{1}{2}}$

Answer 1

Consider $y=x^{\frac{1}{2}}$. Let $x=0.0036$ and $\Delta x=\mathrm{0.0001.}$

Then,

$\Delta y=(x+\Delta x{)}^{\frac{1}{2}}-(x{)}^{\frac{1}{2}}=(0.0037{)}^{\frac{1}{2}}-(.0036{)}^{\frac{1}{2}}=(0.0037{)}^{\frac{1}{2}}-0.06$

$\Rightarrow (0.0037{)}^{\frac{1}{2}}=0.06+\Delta y$

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

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{2}}$]

$=\frac{1}{2\sqrt{.0036}}\left(.0001\right)=\frac{1}{2\times .06}\left(.0001\right)=.00083$

Hence, the approximate value of $(0.0037{)}^{\frac{1}{2}}$ is $.06+.00083=0.06083\approx 0.061$