Using Differentials to Approximate the Square Root of 49.5

In this tutorial we shall develop the differentials to approximate the value of \sqrt {49.5} .

The nearest number to 49.5 whose square root can be taken is 49, so let us consider that x = 49 and \delta x = dx = 0.5.

Now consider

y = \sqrt x \,\,\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Differentiating equation (i) with respect to x, we have

\begin{gathered} \frac{{dy}}{{dx}} = \frac{d}{{dx}}\sqrt x \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\,\,\,\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right) \\ \end{gathered}

Taking the differential of equation (ii), we get

\Rightarrow dy = \frac{1}{{2\sqrt x }}dx

Using the values x = 49 and dx = 0.5, we have

\begin{gathered} dy = \frac{1}{{2\sqrt {49} }}\left( {0.5} \right) = \frac{{0.5}}{{2\left( 7 \right)}} \\ \Rightarrow dy = 0.0357 \\ \end{gathered}

Now

\begin{gathered} \sqrt {49.5} = y + dy \\ \Rightarrow \sqrt {49.5} = \sqrt x + dy \\ \Rightarrow \sqrt {49.5} = \sqrt {49} + 0.0357 = 7 + 0.0357 \\ \Rightarrow \sqrt {49.5} = 7.0357 \\ \end{gathered}