Question

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

32. \(\frac{1}{{4.002}}\)

Short answer

The required value is: \(\frac{1}{{4.002}} = 0.249875\)

Step-by-step solution

Differentials

The differentials are basically mathematical operators that can be explained mathematically as:

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

Calculating values using differential operators:

The givennumber is:

\(f\left( x \right) = \frac{1}{{4.002}} = \frac{1}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{say}}} \right)\)

Now, solving the linear approximation at \(a = 4\):

\(\begin{aligned}dy & = \left\{ {f'\left( a \right)} \right\}dx\\dy & = \left\{ {\left( { - \frac{1}{{{a^2}}}} \right)\left| {_{a = 4}} \right.} \right\}\left( {0.002} \right)\\ & = - \frac{1}{{16}}\left( {0.002} \right)\\ & = - \frac{1}{{8000}}\end{aligned}\)

So, for \(x = 4.002\):

\(\begin{aligned}\frac{1}{{4.002}} & = f\left( 4 \right) - \frac{1}{{8000}}\\ & = \frac{1}{4} - \frac{1}{{8000}}\\ & = \frac{{1999}}{{8000}}\\ & = 0.249875\end{aligned}\)

Hence, the required value is:

\(\frac{1}{{4.002}} = 0.249875\)