Question

Using differential find the approximate value of $\sqrt{0.037}$

Answer 1

Take $y=\sqrt{x}$

Let $x=0.04$ and $\Delta x=-0.003$,

Then, $\Delta y=\sqrt{x+\Delta x}-\sqrt{x}$

$\Delta y=\sqrt{0.037}-\sqrt{0.04}$

$\sqrt{0.037}=\Delta y+0.2$

Now, $dy$ is approximately equal to $Δy$ and is given by

$dy=\left(\frac{dy}{dx}\right)\Delta x$

$=\frac{1}{2\sqrt{x}}(-0.003)$

$=\frac{1}{2\sqrt{0.04}}(-0.003)$

$=-0.0075$

Thus, the approximate value of $\sqrt{0.037}$ is $0.2-0.0075=0.1925$

Hence, this is the answer.