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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:
$\log_\text{e}10.02$ it being given that $\log_\text{e}10=2.3026$
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Answer

Consider the function $\text{y}=\text{f} (\text{x})=\log_\text{e}\text{x}$

Let:

x = 10

$\text{x}+\triangle \text{x}=10.02$

Then,

$\triangle\text{x} =0.02$

For x

$\text{y}\log_\text{e} 10=2.3026$

Let:

$\text{dx}=\triangle \text{x}=0.02$

Now, $\text{y}=\log_\text {e}\text{x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =10}=\frac{1}{10}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {10}\times0.02=0.002$

$\Rightarrow\text{y} =0.002$

$\therefore\log_\text {e}10.02=\text{y}+\triangle\text{y} =2.3046$

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