Question

Using differentials, find the approximate values of the following:
(xii) $\frac{1}{\sqrt{25.1}}$

Answer 1

Let $y=\frac{1}{\sqrt{x}}\Rightarrow \frac{dy}{dx}=-\frac{1}{2x^{\frac{3}{2}}}$
And x =25 and $\Delta x=0.1$
Then,
$\Delta y=\frac{1}{\sqrt{x+\Delta x}}-\frac{1}{\sqrt{1}}$
$\Rightarrow \Delta y=\frac{1}{\sqrt{25+0.1}}-\frac{1}{\sqrt{25}}$
$\Rightarrow \Delta y=\frac{1}{\sqrt{25.1}}-0.2$
$\Rightarrow \frac{1}{\sqrt{25.1}}=0.2+\Delta y$ ..(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx -\frac{1}{2x^{\frac{3}{2}}}\times \Delta x$
$\Rightarrow \Delta y\approx -\frac{1}{2(25{)}^{\frac{3}{2}}}\times 0.1$
$\Rightarrow \Delta y\approx -\frac{0.1}{2\times 125}\approx -0.0004$
From equation (1)
$\frac{1}{\sqrt{25.1}}\approx 0.2-0.0004$
$\therefore \frac{1}{\sqrt{25.1}}\approx 0.1996$