Chapter 3, Problem 1SE

a.

To determine

Find the uncertainty in $U=XY+Z$.

Answer to Problem 1SE

The uncertainty in $U=XY+Z$ is $\underset{\_}{{\sigma }_U=9}$.

Explanation of Solution

Given info:

The form of the measurements of the variables X, Y and Z are as follows: $X=25\pm 1$, $Y=5.0\pm 0.3$ and $Z=3.5\pm 0.2$.

Calculation:

The form of the measurements of a process is,

$\text{Measured}\text{value}\left(\mu \right)\pm \text{Standard deviation}\left(\sigma \right)$.

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The form of the measurements of the variable X is $X=25\pm 1$.

Here, the measured value or mean of the variable X is $X=25$ and the uncertainty in the variable X is ${\sigma }_X=1$.

The form of the measurements of the variable Y is $Y=5.0\pm 0.3$.

Here, the measured value or mean of the variable Y is $Y=5.0$ and the uncertainty in the variable Y is ${\sigma }_Y=0.3$.

The form of the measurements of the variable Z is $Z=3.5\pm 0.2$.

Here, the measured value or mean of the variable Z is $Z=3.5$ and the uncertainty in the variable Z is ${\sigma }_Z=0.2$.

Uncertainty:

The uncertainty of a process is determined by the standard deviation of the measurements. In other words it can be said that, measure of variability of a process is known as uncertainty of the process.

Therefore, it can be said that uncertainty is simply $\left(\sigma \right)$.

Standard deviation:

The standard deviation is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula for standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum x_i{}^2-\frac{{\left({\displaystyle \sum x_i}\right)}^2}{n}}}{n-1}}$.

From the properties of uncertainties for functions of one measurement it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements with uncertainties ${\sigma }_{X_1},{\sigma }_{X_2},\mathrm{...},{\sigma }_{X_n}$ and if $U=U\left(X_1,X_2,\mathrm{..},X_n\right)$ is a function of $X_1,X_2,\mathrm{...},X_n$ then the uncertainty in the variable U is ${\sigma }_U=\sqrt{{\left(\frac{\partial U}{\partial X_1}\right)}^2{\sigma }_{X_1}^2+{\left(\frac{\partial U}{\partial X_2}\right)}^2{\sigma }_{X_2}^2+\mathrm{....}+{\left(\frac{\partial U}{\partial X_n}\right)}^2{\sigma }_{X_n}^2}$.

Uncertainty in U:

The uncertainty of the random variable $U=XY+Z$ is,

$\begin{array}{c}{\sigma }_{U=XY+Z}=\sqrt{{\left(\frac{\partial U}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial U}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial U}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{\partial \left(XY+Z\right)}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial \left(XY+Z\right)}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial \left(XY+Z\right)}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(Y\right)}^2{\sigma }_X^2+{\left(X\right)}^2{\sigma }_Y^2+{\sigma }_Z^2}\\ =\sqrt{{\left(5\right)}^2\times {\left(1\right)}^2+{\left(25\right)}^2\times {\left(0.3\right)}^2+{\left(1\right)}^2\times {\left(0.2\right)}^2}\end{array}$

            $=9$

Thus, the uncertainty of the random variable $U=XY+Z$ is $\underset{\_}{{\sigma }_U=9}$.

b.

To determine

Find the uncertainty in $U=\frac{Z}{X+Y}$.

Answer to Problem 1SE

The uncertainty in $U=\frac{Z}{X+Y}$ is $\underset{\_}{{\sigma }_U=0.0078}$.

Explanation of Solution

Calculation:

From part (a), $X=25$, ${\sigma }_X=1$, $Y=5.0$, ${\sigma }_Y=0.3$ and $Z=3.5$, ${\sigma }_Z=0.2$.

Uncertainty in U:

The uncertainty of the random variable $U=\frac{Z}{X+Y}$ is,

$\begin{array}{c}{\sigma }_{U=\frac{Z}{X+Y}}=\sqrt{{\left(\frac{\partial U}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial U}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial U}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{\partial \left(\frac{Z}{X+Y}\right)}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial \left(\frac{Z}{X+Y}\right)}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial \left(\frac{Z}{X+Y}\right)}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{-Z}{{\left(X+Y\right)}^2}\right)}^2{\sigma }_X^2+{\left(\frac{-Z}{{\left(X+Y\right)}^2}\right)}^2{\sigma }_Y^2+{\left(\frac{1}{X+Y}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{-3.5}{{\left(25+5\right)}^2}\right)}^2\times {\left(1\right)}^2+{\left(\frac{-3.5}{{\left(25+5\right)}^2}\right)}^2\times {\left(0.3\right)}^2+{\left(\frac{1}{25+5}\right)}^2\times {\left(0.2\right)}^2}\end{array}$

$=0.0078$

Thus, the uncertainty of the random variable $U=\frac{Z}{X+Y}$ is $\underset{\_}{{\sigma }_U=0.0078}$.

c.

To determine

Find the uncertainty in $U=\sqrt{X\left(\ln Y+Z\right)}$.

Answer to Problem 1SE

The uncertainty in $U=\sqrt{X\left(\ln Y+Z\right)}$ is $\underset{\_}{{\sigma }_U=0.32}$.

Explanation of Solution

Calculation:

From part (a), $X=25$, ${\sigma }_X=1$, $Y=5.0$, ${\sigma }_Y=0.3$ and $Z=3.5$, ${\sigma }_Z=0.2$.

Uncertainty in U:

The uncertainty of the random variable $U=\sqrt{X\left(\ln Y+Z\right)}$ is,

$\begin{array}{c}{\sigma }_{U=\sqrt{X\left(\ln Y+Z\right)}}=\sqrt{{\left(\frac{\partial U}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial U}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial U}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{\partial \left(\sqrt{X\ln Y+XZ}\right)}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial \left(\sqrt{X\ln Y+XZ}\right)}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial \left(\sqrt{X\ln Y+XZ}\right)}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{\ln Y+Z}{2\sqrt{X\left(\ln Y+Z\right)}}\right)}^2{\sigma }_X^2+{\left(\frac{X}{2Y\sqrt{X\left(\ln Y+Z\right)}}\right)}^2{\sigma }_Y^2+{\left(\frac{X}{2\sqrt{X\left(\ln Y+Z\right)}}\right)}^2{\sigma }_Z^2}\\ =\left(\sqrt{\begin{array}{l}{\left(\frac{\ln \left(5\right)+3.5}{2\sqrt{25\left(\ln \left(5\right)+3.5\right)}}\right)}^2\times {\left(1\right)}^2+{\left(\frac{25}{2\times 5\sqrt{25\left(\ln \left(5\right)+3.5\right)}}\right)}^2\times {\left(0.3\right)}^2\\ +{\left(\frac{25}{2\sqrt{25\left(\ln \left(5\right)+3.5\right)}}\right)}^2\times {\left(0.2\right)}^2\end{array}}\right)\end{array}$

$=0.32$

Thus, the uncertainty of the random variable $U=\sqrt{X\left(\ln Y+Z\right)}$ is $\underset{\_}{{\sigma }_U=0.32}$.

d.

To determine

Find the uncertainty in $U=X{e}^{Z^2-2Y}$.

Answer to Problem 1SE

The uncertainty in $U=X{e}^{Z^2-2Y}$ is $\underset{\_}{{\sigma }_U=361.41}$.

Explanation of Solution

Calculation:

From part (a), $X=25$, ${\sigma }_X=1$, $Y=5.0$, ${\sigma }_Y=0.3$ and $Z=3.5$, ${\sigma }_Z=0.2$.

Uncertainty in U:

The uncertainty of the random variable $U=X{e}^{Z^2-2Y}$ is,

$\begin{array}{c}{\sigma }_{U=X{e}^{Z^2-2Y}}=\sqrt{{\left(\frac{\partial U}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial U}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial U}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(\frac{\partial \left(X{e}^{Z^2-2Y}\right)}{\partial X}\right)}^2{\sigma }_X^2+{\left(\frac{\partial \left(X{e}^{Z^2-2Y}\right)}{\partial Y}\right)}^2{\sigma }_Y^2+{\left(\frac{\partial \left(X{e}^{Z^2-2Y}\right)}{\partial Z}\right)}^2{\sigma }_Z^2}\\ =\sqrt{{\left(e^{Z^2-2Y}\right)}^2{\sigma }_X^2+{\left(-2X\times e^{Z^2-2Y}\right)}^2{\sigma }_Y^2+{\left(2XZ\times e^{Z^2-2Y}\right)}^2{\sigma }_Z^2}\\ =\left(\sqrt{\begin{array}{l}{\left(e^{{\left(3.5\right)}^2-2\times 5}\right)}^2\times {\left(1\right)}^2+{\left(-2\times 25\times e^{{\left(3.5\right)}^2-2\times 5}\right)}^2\times {\left(0.3\right)}^2\\ +{\left(2\times 25\times 3.5\times e^{{\left(3.5\right)}^2-2\times 5}\right)}^2\times {\left(0.2\right)}^2\end{array}}\right)\end{array}$

                   $=361.41$

Thus, the uncertainty of the random variable $U=X{e}^{Z^2-2Y}$ is $\underset{\_}{{\sigma }_U=361.41}$.