Question

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

$\sqrt{49.5}$

Answer 1

Consider $y=\sqrt{x}$.

Let $x=49$ and $\Delta x=0.5$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{49.5}-\sqrt{49}=\sqrt{49.5}-7$

$\Rightarrow \sqrt{49.5}=7+\Delta y$

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

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(0.5\right)$ [as $y=\sqrt{x}$]

$=\frac{1}{2\sqrt{49}}\left(0.5\right)=\left(0.5\right)=\frac{1}{14}\left(0.5\right)=0.035=\Delta y$

Hence, the approximate value of $\sqrt{49.5}$ is $7+.035=7.035$