Question

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

33. \(\sqrt(3){{1001}}\)

Short answer

The required value is: \(\sqrt(3){{1001}} = 10.003\)

Step-by-step solution

Differentials

The calculus uses basic mathematical operators termed as differentials to calculated approximate value of any function expressed as:

\(dy = \left\{ {f'\left( a \right)\left| {_{a = x}} \right.} \right\}dx\)

Calculating values using differential operators:

The givennumber is:

\(f\left( x \right) = \sqrt(3){{1001}} = \sqrt(3){x} = {x^{\frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{say}}} \right)\)

Now, solving the linear approximation at \(a = 1000\):

\(\begin{aligned}dy & = \left\{ {f'\left( a \right)} \right\}dx\\dy & = \left\{ {\left( {\frac{1}{3}{a^{ - \frac{2}{3}}}} \right)\left| {_{a = 1000}} \right.} \right\}\left( 1 \right)\\ & = \frac{1}{3}\left( {\frac{1}{{100}}} \right)\\ & = \frac{1}{{300}}\end{aligned}\)

So, for \(x = 1001\):

\(\begin{aligned}\sqrt(3){{1001}} & = f\left( {1000} \right) + \frac{1}{{300}}\\ & = \sqrt(3){{1000}} + \frac{1}{{300}}\\ & = 10 + \frac{1}{{300}}\\ & = 10.003\end{aligned}\)

Hence, the required value is:

\(\sqrt(3){{1001}} = 10.003\)