Use Differentials to Approximate Cube Root of 28

In this tutorial we shall be concerned with the differentials of independent and dependent variables. Some applications of differential will be discussed.
Use differentials to approximate the value of \sqrt[3]{{28}}
The nearest number to 28 whose perfect cube root can be taken is 27, so let us consider that x = 27 and \delta x = dx = 1.
Now consider

y =  \sqrt[3]{x} = {x^{\frac{1}{3}}}\,\,\,\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right)


\begin{gathered} y + \delta y = {\left( {x + \delta x}  \right)^{\frac{1}{3}}} \\ \Rightarrow {\left( {x + \delta x}  \right)^{\frac{1}{3}}} = y + \delta y \approx {\left( x \right)^{\frac{1}{3}}}  + dx\,\,\,\,\,\,\,\therefore y = {\left( x \right)^{\frac{1}{3}}},\delta y  \approx dy\,\,\,\,\,{\text{ -  -  - }}\left( {{\text{ii}}} \right) \\ \end{gathered}

Taking differential of equation (i), we have

dy  = d{\left( x \right)^{\frac{1}{3}}} = \frac{1}{3}{x^{\frac{1}{3} - 1}}dx = \frac{1}{3}{x^{  - \frac{2}{3}}}dx


Putting this value in equation (ii), we have

\begin{gathered} {\left( {x + \delta x} \right)^{\frac{1}{3}}}  \approx {x^{\frac{1}{3}}} + \frac{1}{3}{x^{ - \frac{2}{3}}}dx \\ \Rightarrow {\left( {27 + 1}  \right)^{\frac{1}{3}}} \approx {\left( {27} \right)^{\frac{1}{3}}} +  \frac{1}{3}{\left( {27} \right)^{ - \frac{2}{3}}}\left( 1  \right)\,\,\,\,\,\,\,\because x = 27,\,dx = 1 = \delta x \\ \Rightarrow {\left( {28}  \right)^{\frac{1}{3}}} \approx {\left( {{3^3}} \right)^{\frac{1}{3}}} +  \frac{1}{3}{\left( {{3^3}} \right)^{ - \frac{2}{3}}} = 3 + \frac{1}{3}\left(  {{3^{ - 2}}} \right) \\ \Rightarrow \sqrt[3]{{28}} \approx 3 +  \frac{1}{3} \times \frac{1}{9} = 3 + \frac{1}{{27}} = \frac{{81 + 1}}{{27}} =  \frac{{82}}{{27}} = 3.037 \\ \end{gathered}

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