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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:
$\sqrt{49.5}$
✓

Answer

Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =49$
$\text{x}+\triangle \text{x}=49.5$
Then,
$\triangle\text{x}=0.5$
For $\text{x}=49$
$\text{y}=\sqrt{49}=7$
Let:
$\text{dx}=\triangle \text{x}=0.5$
Now, $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =49}=\frac{1}{14}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {14}\times0.5=0.0357$
$\Rightarrow\triangle \text{y} =0.0357$
$\therefore\sqrt{49.5} =\text{y}+\triangle\text{y} =7.0357$

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