Chapter 20, Problem 48P

Dynamic viscosity of water $\mu \left({10}^{-3}\text{N}\cdot {\text{s/m}}^2\right)$ is related to temperature $\left(°c\right)$ in the following manner:

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

(a) Plot these data.

(b) Use interpolation to predict $\mu$ at $T=7.5°C$.

(c) Use polynomial regression to fit a parabola to these data in order to make the same prediction.

To determine

To graph: The given data if $\mu$ is dynamic viscosity of water $\left({10}^{-3}\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)$ and $T$ is the temperature in Celsius.

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Explanation of Solution

Given Information:

The data is,

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Graph:

The plot can be easily made with the help of excel as shown below,

Step 1. First put the data in the excel as shown below,

Step 2. Now click on insert and then scatter plot.

Thus, the scatter plot is,

Interpretation:

From the plot, this can be interpreted that the dynamic viscosity of water decreases as the temperature increases.

To determine

To calculate: The value of $\mu$ at $T=7.5°C$ by interpolation if $\mu$ is dynamic viscosity of water $\left({10}^{-3}\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)$ and $T$ is the temperature in Celsius.in Celsius.

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Answer to Problem 48P

Solution:

The value of $\mu$ at $T=7.5°C$ can be predicted as $1.406\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$.

Explanation of Solution

Given Information:

The data is,

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Formula used:

The zero-order Newton’s interpolation formula:

$f_0\left(x\right)=b_0$

The first-order Newton’s interpolation formula:

$f_1\left(x\right)=b_0+b_1\left(x-x_0\right)$

The second- order Newton’s interpolating polynomial is given by,

$f_2\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)$

The n th-order Newton’s interpolating polynomial is given by,

$f_n\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)+\cdots +b_n\left(x-x_0\right)\left(x-x_1\right)\cdots \left(x-x_{n-1}\right)$

Where,

$\begin{array}{c}{b}_0=f\left(x_0\right)\\ b_1=f\left[x_1,x_0\right]\\ b_2=f\left[x_2,x_1,x_0\right]\\ ⋮\\ b_2=f\left[x_n,\cdots ,x_2,x_1,x_0\right]\end{array}$

The first finite divided difference is,

$f\left[x_i,x_j\right]=\frac{f\left(x_i\right)-f\left(x_j\right)}{x_i-x_j}$

And, the n th finite divided difference is,

$f\left[x_n,x_{n-1},\mathrm{...},x_1,x_0\right]=\frac{f\left[x_n,x_{n-1},\mathrm{...},x_1\right]-f\left[x_{n-1},\mathrm{...},x_1,x_0\right]}{x_n-x_0}$

Calculation:

To solve for $T=7.5°C$, Newton interpolation will be best for this data.

Assume $x=T$,

First, order the provided value as close to 7.5 as below,

$x_0=5,x_1=10,x_2=0,x_3=20,x_4=30$ and $x_5=40$.

Therefore,

$\begin{array}{c}f\left(x_0\right)=1.519\times {10}^{-3}\\ f\left(x_1\right)=1.307\times {10}^{-3}\\ f\left(x_2\right)=1.787\times {10}^{-3}\\ f\left(x_3\right)=1.002\times {10}^{-3}\end{array}$

And,

$\begin{array}{l}f\left(x_4\right)=0.7975\times {10}^{-3}\\ f\left(x_5\right)=0.6529\times {10}^{-3}\end{array}$

The first divided difference is,

$\begin{array}{c}f\left[x_1,x_0\right]=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ =\frac{1.307\times {10}^{-3}-1.519\times {10}^{-3}}{10-5}\\ =-0.040\times {10}^{-3}\end{array}$

Thus, the first degree polynomial value can be calculated as,

$f_1\left(x\right)=b_0+b_1\left(x-x_0\right)$

Put $b_0=1.519\times {10}^{-3},b_1=-0.040\times {10}^{-3},x=7.5\text{and}{x}_0=5$ in above equation,

$\begin{array}{c}{f}_1\left(x\right)=1.519\times {10}^{-3}+\left(-0.040\times {10}^{-3}\right)\left(7.5-5\right)\\ =1.419\times {10}^{-3}\end{array}$

Solve for other values as,

$\begin{array}{c}f\left[x_2,x_1\right]=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\\ =\frac{1.787\times {10}^{-3}-1.307\times {10}^{-3}}{0-10}\\ =\frac{0.480\times {10}^{-3}}{-10}\\ =-0.0480\times {10}^{-3}\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2\right]=\frac{f\left(x_3\right)-f\left(x_2\right)}{x_3-x_2}\\ =\frac{1.002\times {10}^{-3}-1.787\times {10}^{-3}}{20-0}\\ =-0.03925\times {10}^{-3}\end{array}$

Similarly, $f\left[x_4,x_3\right],f\left[x_5,x_4\right]$ can be calculated.

The second divided difference is,

$\begin{array}{c}f\left[x_2,x_1,x_0\right]=\frac{f\left[x_2,x_1\right]-f\left[x_1,x_0\right]}{x_2-x_0}\\ =\frac{-0.0480\times {10}^{-3}-1.419\times {10}^{-3}}{0-5}\\ =0.2934\times {10}^{-3}\end{array}$

Thus, the second degree polynomial value can be calculated as,

$f_2\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)$

Put $b_0=1.519\times {10}^{-3},b_1=-0.040\times {10}^{-3},b_2=0.2934\times {10}^{-3},x_1=10,x=7.5\text{and}{x}_0=5$ in above equation,

$\begin{array}{c}{f}_2\left(x\right)=1.519\times {10}^{-3}+-0.040\times {10}^{-3}\left(7.5-5\right)+0.2934\times {10}^{-3}\left(7.5-5\right)\left(7.5-10\right)\\ =1.406\times {10}^{-3}\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2,x_1\right]=\frac{f\left[x_3,x_2\right]-f\left[x_2,x_1\right]}{x_3-x_1}\\ =\frac{-0.03925\times {10}^{-3}+0.0480\times {10}^{-3}}{20-10}\\ =0.000875\times {10}^{-3}\end{array}$

The third divided difference is,

$\begin{array}{c}f\left[x_3,x_2,x_1,x_0\right]=\frac{f\left[x_3,x_2,x_1\right]-f\left[x_2,x_1,x_0\right]}{x_3-x_0}\\ =\frac{0.000875\times {10}^{-3}-0.2934\times {10}^{-3}}{20-5}\\ =-0.0195\times {10}^{-3}\end{array}$

Thus, the third degree polynomial value can be calculated as,

$f_3\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)+b_3\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)$

Put this in above equation,

$b_0=1.519\times {10}^{-3},b_1=-0.040\times {10}^{-3},b_2=0.2934\times {10}^{-3},b_3=-0.0195\times {10}^{-3},x_2=0,x_1=10,x=7.5\text{and}{x}_0=5$

Therefore,

$\begin{array}{c}{f}_3\left(x\right)=1.519\times {10}^{-3}+-0.040\times {10}^{-3}\left(7.5-5\right)+0.2934\times {10}^{-3}\left(7.5-5\right)\left(7.5-10\right)\\ -0.0195\times {10}^{-3}\left(7.5-5\right)\left(7.5-10\right)\left(7.5-0\right)\\ =1.4067\times {10}^{-3}\end{array}$

And, the error is calculated as,

$\begin{array}{c}\text{Error}=f_2\left(x\right)-f_1\left(x\right)\\ =1.406\times {10}^{-3}-1.419\times {10}^{-3}\\ =-7\times {10}^{-6}\end{array}$

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for $T=7.5$ as,

Order	$f\left(7.5\right)$	Error
0	$1.519\times {10}^{-3}$	$-0.00011$
1	$1.419\times {10}^{-3}$	$-7\times {10}^{-6}$
2	$1.406\times {10}^{-3}$	$7.66\times {10}^{-7}$
3	$1.406\times {10}^{-3}$	$9.18\times {10}^{-8}$
4	$1.406\times {10}^{-3}$	$5.81\times {10}^{-9}$
5	$1.406\times {10}^{-3}$

Since the minimum error for order four, therefore, it can be concluded that the value of $\mu$ at $T=7.5°C$ is $1.406\times {10}^{-3}$.

To determine

To calculate: The value of $\mu$ at $T=7.5°C$ by fitting the data into polynomial regression if $\mu$ is dynamic viscosity of water $\left({10}^{-3}\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)$ and $T$ is the temperature in Celsius.

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Answer to Problem 48P

Solution:

The value of $\mu$ at $T=7.5°C$ can be predicted as $1.4269\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$.

Explanation of Solution

Given Information:

The data is,

T	0	5	10	20	30	40
$\mu$	1.787	1.519	1.307	1.002	0.7975	0.6529

Calculation:

This problem can be easily solved with the help of excel as shown below,

Step 1. First put the data in the excel as shown below,

Step 2. Now click on insert and then scatter plot.

Step 3. Click on layout, Trendline, more trendline options, polynomial and then display equation on chart.

Step 4. Click on display R-squared value on chart as shown below,

Thus, the final plot is shown as below,

Therefore, the regression plot equation of the data will be,

$\mu =0.0005T^2-0.0495T+1.7672$

Put $T=7.5$ in above equation,

$\begin{array}{c}\mu =0.0005{\left(7.5\right)}^2-0.0495\left(7.5\right)+1.7672\\ =1.4269\times {10}^{-3}\end{array}$

Hence, the value of $\mu$ at $T=7.5°C$ can be predicted as $1.4269\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$.