Question

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

Answer 1

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.