Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentials, Errors and Approximations5 Marks
Question
Using differentials, find the approximate values of the following:
$\sqrt{401}$
✓
Answer
Consider the function $\text{y}=\text{f}(\text{x})=\sqrt{\text{x}}$
Let:
x = 400
$\text{x}+\triangle\text{x}=401$
Then,
$\triangle\text{x}=1$
For x = 400
$\text{y}=\sqrt{400}=20$
Let:
$\text{dx}=\triangle\text{x}=1$
Now, $\text{y}=\sqrt{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=400}=\frac{1}{40}$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{40}\times1=\frac{1}{40}$
$\Rightarrow\triangle\text{y}=\frac{1}{40}=0.025$
$\therefore\sqrt{401}=\text{y}=\triangle\text{y}=20.025$
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