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

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentials, Errors and Approximations5 Marks
Question
Using differentials, find the approximate values of the following:
$\sqrt{0.082}$
✓

Answer

Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$Let:
$\text{x}=0.0841$
$\text{x}+\triangle \text{x}=0.082$Then,
$\triangle\text{x}=0.0021$For $\text{x}=0.0841$
$\text{y}=\sqrt{0.0841}=0.29$Let:
$\text{dx}=\triangle \text{x}=-0.0021$Now, $\text{y}=(\text{x})^{\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=0.0841}=\frac{1}{0.58}=\frac{50}{29}$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{50}{29}\times(-0.0021)=-0.0036$ $\Rightarrow\triangle\text{y} =-0.0036$ $\therefore\sqrt{0.082}=\text{y}+\triangle\text{y}=0.2864$

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