Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentials, Errors and Approximations5 Marks
Question
Using differentials, find the approximate values of the following:
$\sqrt{25.02}$
✓
Answer
consider the function $\text{y}=\text{f}(\text{x})=\sqrt{\text{x}}$
Let:
x = 25
$\text{x}+\triangle\text{x}=25.02$
Then
$\triangle\text{x}=0.02$
For x = 25
$\text{y}=\sqrt{25}=5$
Let:
$\text{dx}+\triangle\text{x}=0.02$
Now, $\text{y}=\sqrt{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=25}=\frac{1}{10}$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{10}\times0.02=0.002$
$\Rightarrow\triangle\text{y}=0.002$
$\therefore\text{y}+\triangle\text{y}=5.002$
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