Question

Using differentials, find the approximate values of the following:
(vii) $\frac{1}{(2.002{)}^2}$

Answer 1

Let $y=\frac{1}{x^2}\Rightarrow \frac{dy}{dx}=\frac{-2}{x^3}$
And x = 2 and $\Delta x=0.002$
Then,
$\Delta y=\frac{1}{(x+\Delta x{)}^2}-\frac{1}{x^2}$
$\Rightarrow \Delta y=\frac{1}{(2+0.002{)}^2}-\frac{1}{2^2}$
$\Rightarrow \Delta y=\frac{1}{(2.002{)}^2}=0.25+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx \frac{-2}{x^3}\times \Delta x$
$\Rightarrow \Delta y\approx \frac{-2}{x^3}\times \left(0.002\right)\approx -0.0005$
From equation(1)
$\frac{1}{(2.002{)}^2}\approx 0.25-0.0005$
$\therefore \frac{1}{(2.002{)}^2}\approx 0.2495$