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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:
$\sqrt{0.48}$
✓

Answer

Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =0.49$
$\text{x}+\triangle \text{x}=0.48$
Then,
$\triangle\text{x}=- 0.01$
For $\text{x}=0.49$
$\text{y}=\sqrt{0.49} =0.7$
Let:
$\text{dx}=\triangle \text{x}=0.01$
Now  $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt {\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =0.49}=\frac{1}{1.4}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {1.4}\times(-0.01)=0.007143$
$\Rightarrow\triangle \text{y} =0.007143$
$\therefore\sqrt{0.48} =\text{y}+\triangle\text{y} =0.693$

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