Question

Show that if $\mathit{w}\mathbf{=}\mathit{x}\mathit{y}$or,$\mathit{w}\mathbf{=}\mathit{x}\mathbf{/}\mathit{y}$then$\mathbf{\left(}\mathbf{10.14}\mathbf{\right)}$ givesthe convenient formula for relative error

$\frac{{\mathbf{r}}_{\mathbf{w}}}{\mathbf{w}}\mathbf{=}\sqrt{{\mathbf{\left(}\frac{{\mathbf{r}}_{\mathbf{x}}}{\mathbf{x}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\frac{{\mathbf{r}}_{\mathbf{y}}}{\mathbf{y}}\mathbf{\right)}}^{\mathbf{2}}}$

Short answer

Required expressions is,

$\frac{r_w}{w}=\sqrt{{\left(\frac{r_x}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}$

Step-by-step solution

Given Information 

Use values of w as mentioned below.

$\begin{array}{c}w=xy\\ w=x/y\end{array}$

Definition of Confidence interval. 

The probability that a parameter will fall between two values around the mean is represented by a confidence interval.

Calculate  rw.

Use relative error formula.

$\begin{array}{c}{r}_i=I{\sigma }_{mi}\mathrm{...}\left(1\right)\\ {\sigma }_{mi}=\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\sigma }_{me}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\sigma }_m^2}\\ r_w=\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\sigma }_{ms}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\sigma }_{mw}^2}\\ =\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\left(I{\sigma }_{me}\right)}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\left(I{\sigma }_{mw}\right)}^2}\mathrm{...}\left(2\right)\end{array}$

Write equationas,

$r_w=\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\left(r_z\right)}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\left(r_y\right)}^2}\mathrm{...}\left(3\right)$

Differentiate according to given function. 

Assume,$w=xy$

$\begin{array}{c}{r}_w=\sqrt{y^2{\left(r_x\right)}^2+x^2{\left(r_y\right)}^2}\\ =\sqrt{{\left(\frac{x^2}{x^2}\right)}^2{\left(r_x\right)}^2+\left(\frac{y^2}{y^2}\right)x^2{\left(r_y\right)}^2}\\ =\sqrt{{\left(\frac{r_x}{x}\right)}^2{y}^2{x}^2+{\left(\frac{r_y}{y}\right)}^2{y}^2{x}^2}\\ =xy\sqrt{{\left(\frac{r_x}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}\end{array}$

On further simplification,

$r_w=w\sqrt{{\left(\frac{r_x}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}$

Assume,$w=x/y$

$\begin{array}{c}{r}_w=\sqrt{{\left(\frac{1}{y}\right)}^2{\left(r_x\right)}^2+{\left(\frac{x}{y^2}\right)}^2{\left(r_y\right)}^2}\\ =\sqrt{\left(\frac{x^2}{y^2}\right)\cdot \frac{1}{x^2}{\left(r_x\right)}^2+\left(\frac{x^2}{y^2}\right)\frac{1}{y^2}{\left(r_y\right)}^2}\\ =\sqrt{{\left(\frac{r_x}{x}\right)}^2\frac{x^2}{y^2}+{\left(\frac{r_y}{y}\right)}^2\frac{x^2}{y^2}}\\ =\frac{x}{y}\sqrt{{\left(\frac{r_z}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}\end{array}$

On further simplification,

$r_w=w\sqrt{{\left(\frac{r_z}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}$

Hence, required expressions is,

$\frac{r_w}{w}=\sqrt{{\left(\frac{r_x}{x}\right)}^2+{\left(\frac{r_y}{y}\right)}^2}$