Question

31-36Use a linear approximation (or differentials) to estimate the given number.

34. \(\sqrt {100.5} \)

Short answer

The required value is: \(\sqrt {100.5} = 10.025\)

Step-by-step solution

Differentials

The differentials can be defined as the Del operators in calculus that can be helpful to determine the approximated value of any function as:

\(dy = \left\{ {f'\left( x \right)} \right\}dx\)

Calculating values using differential operators:

The given number is:

\(f\left( x \right) = \sqrt {100.5} = \sqrt x = {x^{\frac{1}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{say}}} \right)\)

Now, solving the linear approximation at \(a = 100\):

\(\begin{aligned}dy & = \left\{ {f'\left( a \right)} \right\}dx\\dy & = \left\{ {\left( {\frac{1}{2}{a^{ - \frac{1}{2}}}} \right)\left| {_{a = 100}} \right.} \right\}\left( {0.5} \right)\\ & = \frac{1}{2}\left( {\frac{1}{{10}}} \right)\left( {0.5} \right)\\ & = \frac{1}{{40}}\end{aligned}\)

So, for \(x = 100.5\):

\(\begin{aligned}\sqrt {100.5} & = f\left( {100} \right) + \frac{1}{{40}}\\ & = \sqrt {100} + \frac{1}{{40}}\\ & = 10 + \frac{1}{{40}}\\ & = 10.025\end{aligned}\)

Hence, the required value is:

\(\sqrt {100.5} = 10.025\)