Chapter 15: Q5P (page 775)

Show that if $w = x y$ or, $w = x / y$ then $(10.14)$ givesthe convenient formula for relative error
$\frac{r_{w}}{w} = \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

Short Answer

Expert verified

Required expressions is,

$\frac{r_{w}}{w} = \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

Step by step solution

Given Information

Use values of w as mentioned below.

$\begin{matrix} w = x y \\ w = x / y \end{matrix}$

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.

$\begin{matrix} r_{i} = I σ_{m i} ... (1) \\ σ_{m i} = \sqrt{{(\frac{\partial w}{\partial x})}^{2} σ_{m e}^{2} + {(\frac{\partial w}{\partial y})}^{2} σ_{m}^{2}} \\ r_{w} = \sqrt{{(\frac{\partial w}{\partial x})}^{2} σ_{m s}^{2} + {(\frac{\partial w}{\partial y})}^{2} σ_{m w}^{2}} \\ = \sqrt{{(\frac{\partial w}{\partial x})}^{2} {(I σ_{m e})}^{2} + {(\frac{\partial w}{\partial y})}^{2} {(I σ_{m w})}^{2}} ... (2) \end{matrix}$

Write equationas,

$r_{w} = \sqrt{{(\frac{\partial w}{\partial x})}^{2} {(r_{z})}^{2} + {(\frac{\partial w}{\partial y})}^{2} {(r_{y})}^{2}} ... (3)$

Differentiate according to given function.

Assume, $w = x y$

$\begin{matrix} r_{w} = \sqrt{y^{2} {(r_{x})}^{2} + x^{2} {(r_{y})}^{2}} \\ = \sqrt{{(\frac{x^{2}}{x^{2}})}^{2} {(r_{x})}^{2} + (\frac{y^{2}}{y^{2}}) x^{2} {(r_{y})}^{2}} \\ = \sqrt{{(\frac{r_{x}}{x})}^{2} y^{2} x^{2} + {(\frac{r_{y}}{y})}^{2} y^{2} x^{2}} \\ = x y \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}} \end{matrix}$

On further simplification,

$r_{w} = w \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

Assume, $w = x / y$

$\begin{matrix} r_{w} = \sqrt{{(\frac{1}{y})}^{2} {(r_{x})}^{2} + {(\frac{x}{y^{2}})}^{2} {(r_{y})}^{2}} \\ = \sqrt{(\frac{x^{2}}{y^{2}}) \cdot \frac{1}{x^{2}} {(r_{x})}^{2} + (\frac{x^{2}}{y^{2}}) \frac{1}{y^{2}} {(r_{y})}^{2}} \\ = \sqrt{{(\frac{r_{x}}{x})}^{2} \frac{x^{2}}{y^{2}} + {(\frac{r_{y}}{y})}^{2} \frac{x^{2}}{y^{2}}} \\ = \frac{x}{y} \sqrt{{(\frac{r_{z}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}} \end{matrix}$

On further simplification,

$r_{w} = w \sqrt{{(\frac{r_{z}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

Hence, required expressions is,

$\frac{r_{w}}{w} = \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

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Show that if $w = x y$ or, $w = x / y$ then $(10.14)$ givesthe convenient formula for relative error
$\frac{r_{w}}{w} = \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$

Short Answer

Step by step solution

Given Information

Definition of Confidence interval.

Calculate rw.

Differentiate according to given function.

One App. One Place for Learning.

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Show that if w=xyor,w=x/ythen(10.14) givesthe convenient formula for relative errorrww=(rxx)2+(ryy)2

Short Answer

Step by step solution

Given Information

Definition of Confidence interval.

Calculate rw.

Differentiate according to given function.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Atoms and Radioactivity

Work Energy and Power

Dynamics

Energy Physics

Electricity and Magnetism

Study anywhere. Anytime. Across all devices.

Show that if $w = x y$ or, $w = x / y$ then $(10.14)$ givesthe convenient formula for relative error
$\frac{r_{w}}{w} = \sqrt{{(\frac{r_{x}}{x})}^{2} + {(\frac{r_{y}}{y})}^{2}}$