Using differentials, find the approximate value of $$\sqrt {49.5} $$.

Question

Using differentials, find the approximate value of  $$\sqrt {49.5} $$.

Vedantu Content Team · Accepted Answer

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 7.036Note: 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.

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

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

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