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

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentials, Errors and Approximations4 Marks
Question
Using differentials, find the approximate values of the following:
$\sqrt{26}$
✓

Answer

Let $\text{x}=25,\\\text{x}+ \triangle\text{x}=26$

$\triangle\text{x}=26- 25$

$=-1$

Let $\text{y}=\sqrt{\text {x}}$

$\frac{\text{dy}} {\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=\frac{1} {2\sqrt{25}}$

$=\frac{1}{10}$

$=0.1$

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$

$=(0.01)(1)$

$=0.01$

$\sqrt{26}=\text{y}+ \triangle\text{y}$

$=\sqrt{\text {x}}+0.01$

$\sqrt{25}+0.1$

$\sqrt{26}=5.1$

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