# Using differentials, find the approximate value of \[\sqrt {49.5} \].

Last updated date: 16th Mar 2023

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Answer

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Hint: The differentiation of \[x\] is represented by \[dx\] is defined by \[dx = x\] where \[x\] is the minor change in \[x\]. The differential of \[y\] is represented by \[dy\] is defined by \[dy = \dfrac{{dy}}{{dx}}x\]. As \[x\] is very small compared to \[x\], so \[dy\] is the approximation of \[y\]. Hence the increment in \[y\] corresponding to the increment in \[x\], denoted by \[\Delta y\], is given by \[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\].

Complete step-by-step answer:

Let \[y = \sqrt x \] where \[x = 49\& \Delta x = 0.5\]

Since \[y = \sqrt x \]

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

Now,

\[

\Delta y = \dfrac{{dy}}{{dx}}\Delta x \\

\Rightarrow \Delta y = \dfrac{1}{{2\sqrt x }}\left( {0.5} \right) \\

\Rightarrow \Delta y = \dfrac{1}{{2\sqrt {49} }} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{1}{{2 \times 7}} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{1}{{14}} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{{0.5}}{{14}} \\

\therefore \Delta y = 0.036 \\

\]

Also,

\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\]

By substituting the above values, we have

\[

\Delta y = \sqrt {x + \Delta x} - \sqrt x {\text{ }}\left[ {\because f\left( x \right) = y = \sqrt x } \right]{\text{ }} \\

0.036 = \sqrt {49 + 0.5} - \sqrt {49} \\

{\text{0}}{\text{.036}} = \sqrt {49.5} - 7 \\

\sqrt {49.5} = 0.036 + 7 \\

\therefore \sqrt {49.5} = 7.036{\text{ }} \\

\]

Thus, the approximate value of \[\sqrt {49.5} \] is 7.036

Note: In this problem we have solved the approximation up to 3 places of decimals. And also, we have rounded off the decimal numbers up to three places. We use differentiation to find the approximate values of the certain quantities.

Complete step-by-step answer:

Let \[y = \sqrt x \] where \[x = 49\& \Delta x = 0.5\]

Since \[y = \sqrt x \]

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

Now,

\[

\Delta y = \dfrac{{dy}}{{dx}}\Delta x \\

\Rightarrow \Delta y = \dfrac{1}{{2\sqrt x }}\left( {0.5} \right) \\

\Rightarrow \Delta y = \dfrac{1}{{2\sqrt {49} }} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{1}{{2 \times 7}} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{1}{{14}} \times 0.5 \\

\Rightarrow \Delta y = \dfrac{{0.5}}{{14}} \\

\therefore \Delta y = 0.036 \\

\]

Also,

\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\]

By substituting the above values, we have

\[

\Delta y = \sqrt {x + \Delta x} - \sqrt x {\text{ }}\left[ {\because f\left( x \right) = y = \sqrt x } \right]{\text{ }} \\

0.036 = \sqrt {49 + 0.5} - \sqrt {49} \\

{\text{0}}{\text{.036}} = \sqrt {49.5} - 7 \\

\sqrt {49.5} = 0.036 + 7 \\

\therefore \sqrt {49.5} = 7.036{\text{ }} \\

\]

Thus, the approximate value of \[\sqrt {49.5} \] is 7.036

Note: In this problem we have solved the approximation up to 3 places of decimals. And also, we have rounded off the decimal numbers up to three places. We use differentiation to find the approximate values of the certain quantities.

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