Question

Using differentials, find the approximate values of the following:
(iv) $\sqrt{401}$

Answer 1

Let $y=\sqrt{x}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}$
And x =400 and $\Delta x=1$
Then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}$
$\Rightarrow \Delta y=\sqrt{400+1}-\sqrt{400}$
$\Rightarrow \Delta y=\sqrt{401}-20$
$\Rightarrow \sqrt{401}=20+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{2\sqrt{x}}\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{2\sqrt{400}}\times 1\approx 0.025$
From equation(1)
$\sqrt{401}\approx 20+0.025$
$\therefore \sqrt{401}\approx 20.025$