Question

How do you estimate the quantity using the Linear Approximation and find the error using a calculator ${\left( {15.8} \right)^{\dfrac{1}{4}}}$?

Answer 1

In order to find the linear approximation of the given quantity, we should first know what Linear approximation is. Linear approximation is a method that is used to calculate the approx value of a quantity, as sometimes it becomes difficult to find the actual value.

Formula used:
$\% {\text{ error formula = }}\left| {\dfrac{{{\text{estimated value - actual value}}}}{{{\text{actual value}}}} \times {\text{100}}} \right|$
$If{\text{ f}}\left( x \right) = {x^n}{\text{ then f'}}\left( x \right) = n{x^{n - 1}}$

Complete answer: We are given a quantity ${\left( {15.8} \right)^{\dfrac{1}{4}}}$,we need to find the approximation quantity.
From the linear approximation formula, we know that:
The linear approximation formula for $f\left( x \right) = a$ about $x = a$ , that is numerically represented as:
$f\left( x \right) \approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)$ ……(1)
Considering $f\left( x \right) = {x^{\dfrac{1}{4}}}$ …..(2)
Differentiating both the sides with respect to $x$:
Since, from the differential rules, we know that:
$If{\text{ f}}\left( x \right) = {x^n}{\text{ then f'}}\left( x \right) = n{x^{n - 1}}$
Using this in equation 2, we get:
$f'\left( x \right) = \dfrac{1}{4}{x^{ - \dfrac{3}{4}}}$
Similarly, if we substitute $x = a$ in the above equation, we get:
$f'\left( a \right) = \dfrac{1}{4}{a^{ - \dfrac{3}{4}}} = \dfrac{1}{{4{a^{\dfrac{3}{4}}}}}$
Substituting this value in the equation (1):
And, so the linear approximation for $f\left( x \right)$ is:
$f\left( x \right) \approx f\left( a \right) + \dfrac{{\left( {x - a} \right)}}{{4{a^{\dfrac{3}{4}}}}}$
Since, we know that $16$ is the most close term to $15.8$, so we are taking $a = 16$ and substituting it in the above equation, and we get:
$f\left( x \right) \approx {16^{\dfrac{1}{4}}} + \dfrac{{\left( {x - 16} \right)}}{{{{4.16}^{\dfrac{3}{4}}}}}$
We know that ${2^4} = 16 \Rightarrow {16^{\dfrac{3}{4}}} = {\left( 2 \right)^3} = 8$, so substituting this in the above equation, we get:
$f\left( x \right) \approx 2 + \dfrac{{\left( {x - 16} \right)}}{{4 \times 8}}$
$\Rightarrow f\left( x \right) \approx 2 + \dfrac{{\left( {x - 16} \right)}}{{32}}$
Substituting the value of $x = 15.8$:
$f\left( {15.8} \right) \approx 2 + \dfrac{{\left( {15.8 - 16} \right)}}{{32}}$
Solving the brackets:
$f\left( {15.8} \right) \approx 2 - \dfrac{{0.2}}{{32}}$

$$f\left( {15.8} \right) \approx 2 - 0.00625 \\\ f\left( {15.8} \right) \approx 1.99375 \\\$$

Which gives us the estimated approximate values as $f\left( {15.8} \right) = {\left( {15.8} \right)^{\dfrac{1}{4}}} \approx 1.98375$.
Now, for the comparison between the approximate value and the calculated value, we need to know the calculator value.
Calculator Result:
Using a calculator, we find:
$f\left( {15.8} \right) = 1.99372048$
Now, we need to calculate the % error, for that we use the % error formula, which is:
$\% {\text{ error formula = }}\left| {\dfrac{{{\text{estimated value - actual value}}}}{{{\text{actual value}}}} \times {\text{100}}} \right|$
Substituting the value of estimated and calculated value, and we get:
$\% {\text{ error formula = }}\left| {\dfrac{{1.975{\text{ - 1}}{\text{.99372048}}}}{{1.99372048}} \times {\text{100}}} \right|$
Solving the values, we get:
$\% {\text{ error formula = }}\left| {\dfrac{{1.975{\text{ - 1}}{\text{.99372048}}}}{{1.99372048}} \times {\text{100}}} \right|$
$\Rightarrow \% {\text{ error formula = }}\left| { - \dfrac{{0.01872048}}{{1.99372048}} \times {\text{100}}} \right|$
$\Rightarrow \% {\text{ error formula = }}\left| { - \dfrac{{0.01872048}}{{1.99372048}} \times {\text{100}}} \right|$
$\Rightarrow \% {\text{ error formula = }}\left| { - 0.0093897215 \times {\text{100}}} \right|$
$\Rightarrow \% {\text{ error formula = }}\left| { - 0.93897215} \right|$
$\Rightarrow \% {\text{ error formula = }}0.93897215 \\\ \Rightarrow \% {\text{ error formula }} \approx {\text{ 1}} \\\$
Therefore, the estimated value is approx $1\%$ of the estimated value.

Note:
Remember, to use the $\approx$ instead of ‘=’ in the ways of finding the estimated value, as it is the approximate value, that means it can be equal to the value or not.
Always preferred to solve step by step for ease, rather than solving at once, because in error approximation value there is a huge chance of error.