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

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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.

Last updated date: 21st Sep 2023

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