Question

About how much (in percent) does an error of 1% in a and b affect${\mathit{a}}^{\mathbf{2}}{\mathit{b}}^{\mathbf{3}}$ ?

Short answer

The answer is $c_{re}=5\%.$

Step-by-step solution

Explanation of Solution

Provided:

The given equation is $a^2{b}^3$ an error in a and b are,$a=1\%,b=1\%$

Approximation by differentials

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$, find the value of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when.

As an example, the derivative of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$with respect toxwill be employed.

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{=}\left(f\left(x\right)\right)$ is Change in y with respect to change in x as $\mathit{d}\mathit{x}\mathbf{\Rightarrow }\mathbf{0}$.

If the value of $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx , is vanishingly small, one can derive the change in the value of the function $y=f\left(\mathrm{x}\right)$corresponding to the change dx in x . In practice, however, the concept of vanishingly small is not possible.

Calculation

Consider the relative errors for two of the three variables, differentials may be used to calculate the relative error in$c=a^2{b}^3\left(c_{re}\right)$ .

It will, however, need to first extract the relationship's differentials:

Since,

$c=a^2{b}^3$

Taking log on both sides,

$\mathrm{Inc}=2\mathrm{In}\mathrm{a}+3\mathrm{Inb}$

Here,

$\frac{dc}{c}=2\frac{da}{a}+3\frac{db}{b}$

Substitute the values,

$$\begin{gathered} C_{re}=2\times 0.01+3\times 0.01 \\ =0.05 \end{gathered}$$

Hence,$c_{re}=5\%$