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Differentials, Errors and Approximations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentials, Errors and Approximations5 Marks
Question
Using differentials, find the approximate values of the following:
$\frac{1}{(2.002)^2}$
✓

Answer

consider the function $\text{y}=\text{f} (\text{x})=\frac{1}{\text{x}^2}$ Let:
x = 2 $\text{x}+\triangle \text{x}=2.002$Then,
$\triangle\text{x}=- 0.002$ For x = 2 $\text{y}=\frac{1} {2^2}=\frac{1}{4}$ Let:
$\text{dx}=\triangle \text{x}=0.002$Now, $\text{y}=\frac{1} {\text{x}^2}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{2}{\text{x} ^3}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=2} =\frac{1}{4}$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1}{4}\times- 0.002=-0.005$ $\Rightarrow\triangle \text{y}=-0.0005$ $\therefore\frac{1} {(2.002)^2}=\text{y}+\triangle\text{y} =0.2495$

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