Question

Explain, in terms of linear approximations or differentials, why the approximation is reasonable.38. \(\sqrt {4.02} \approx 2.005\)

Short answer

The approximation \(\sqrt {4.02} \approx 2.005\) is reasonable.

Step-by-step solution

Linear Approximation

TheLinear approximation is basically an expanded representation of differentials in calculus that can be expressed as:

\(L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\)

Examining the function using approximation method:

Let we have:

\(y = f\left( x \right) = \sqrt x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{say}}} \right)\)

On differentiating, we have:

\(f'\left( x \right)= \frac{1}{{2\sqrt x }}\)

Now, solving at\(x = 4\):

\(\begin{aligned}{l}f\left( 4 \right)&= 2\\f'\left( 4 \right)&= \frac{1}{4}\end{aligned}\)

So, the linear approximation at\(x = 4.02\)will be:

\(\begin{aligned}{c}f\left( {4.02} \right)&= f\left( 4 \right) + f'\left( 4 \right)\left( {4.02 - 4} \right)\\&= 2 + \frac{1}{4}\left( {0.02} \right)\\&= 2.005\end{aligned}\)

Hence, the approximation \(\sqrt {4.02} \approx 2.005\) is reasonable.