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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:
$25^{\frac{1}{3}}$
✓

Answer

Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})^{\frac{1}{3}}$

Let:

$\text{x}=27$

$\text{x}+\triangle \text{x}=25$

Then,

$\triangle\text{x}=-2$

For $\text{x}=27$

$\text{y}=(27)^{\frac{1}{3}}=3$

Let:

$\text{dx}=\triangle \text{x}=-2$

Now, $\text{y}=(\text{x})^{\frac{1}{3}}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{3(\text{x})^{\frac{2}{3}}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=27}=\frac{1}{27}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1}{27}\times(-2)=0.07407$

$\Rightarrow\triangle\text{y} =0.07407$

$\therefore(25)^{\frac{1}{3}}=\text{y}+\triangle\text{y}=2.9259$

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