Chapter 20, Problem 60P

Temperatures are measured at various points on a heatedplate (Table P20.60). Estimate the temperature at (a) $x=4,y=3.2$, and (b) $x=4.3,y=2.7$.

TABLE P20.60 Temperatures $\left(°\text{C}\right)$ at various points on a square heated plate.

	$x=0$	$x=2$	$x=4$	$x=6$	$x=8$
$y=0$	100.00	90.00	80.00	70.00	60.00
$y=2$	85.00	64.49	53.50	48.15	50.00
$y=4$	70.00	48.90	38.43	35.03	40.00
$y=6$	55.00	38.78	30.39	27.07	30.00
$y=8$	40.00	35.00	30.00	25.00	20.00

To determine

To calculate: The value of temperature at $x=4,y=3.2$ from the given table if the temperatures in $\left(°C\right)$ are recorded at various points on a heated plate.

	$x=0$	$x=2$	$x=4$	$x=6$	$x=8$
$y=0$	100.00	90.00	80.00	70.00	60.00
$y=2$	85.00	64.49	53.50	48.15	50.00
$y=4$	70.00	48.90	38.43	35.03	40.00
$y=6$	55.00	38.78	30.39	27.07	30.00
$y=8$	40.00	35.00	30.00	25.00	20.00

Answer to Problem 60P

Solution:

The value of temperature at $x=4,y=3.2$ is $43.368°C$.

Explanation of Solution

Given Information:

The data is provided as,

	$x=0$	$x=2$	$x=4$	$x=6$	$x=8$
$y=0$	100.00	90.00	80.00	70.00	60.00
$y=2$	85.00	64.49	53.50	48.15	50.00
$y=4$	70.00	48.90	38.43	35.03	40.00
$y=6$	55.00	38.78	30.39	27.07	30.00
$y=8$	40.00	35.00	30.00	25.00	20.00

Formula used:

The zero-order Newton’s interpolation formula:

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

The first-order/linear Newton’s interpolation formula:

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

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

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

Where,

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

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 calculate the temperature $x=4,y=3.2$, Newton interpolation is the perfect choice.

First use the linear interpolation formula and arrange the points as close to about $x=4$,

$f_1\left(y\right)=b_0+b_1\left(y-y_0\right)$

The values are,

$\begin{array}{c}{y}_0=4,f\left(y_0\right)=38.43\\ y_1=2,f\left(y_1\right)=53.5\\ y_2=6,f\left(y_2\right)=30.39\\ y_3=0,f\left(y_3\right)=80\end{array}$

And,

$\begin{array}{c}{y}_4=8,f\left(y_4\right)=30\\ y=3.2\end{array}$

First calculate $b_1$,

$\begin{array}{c}{b}_1=\frac{53.5-38.43}{2-4}\\ =-7.535\end{array}$

Put in above equation,

$\begin{array}{c}{f}_1\left(y\right)=38.43+\left(-7.535\right)\left(3.2-4\right)\\ =44.458\end{array}$

Similarly for quadratic interpolation,

$f_2\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)$

Now calculate $b_2$,

$\begin{array}{c}{b}_2=\frac{\frac{30.39-53.5}{6-2}-\frac{53.5-38.43}{2-4}}{6-4}\\ =\frac{-5.7775+7.535}{2}\\ =0.87875\end{array}$

Put in quadratic interpolation equation,

$\begin{array}{c}{f}_2\left(y\right)=44.458+\left(0.87875\right)\left(3.2-4\right)\left(3.2-2\right)\\ =44.458-0.8436\\ =43.6144\end{array}$

Now, do it for cubic interpolation by the use of the formula,

$f_3\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)+b_3\left(y-y_0\right)\left(y-y_1\right)\left(y-y_2\right)$

Now calculate $b_3$,

$\begin{array}{c}{b}_3=\frac{\frac{\left(\frac{80-30.39}{0-6}-\frac{30.39-53.5}{6-2}\right)}{0-2}-\frac{\left(\frac{30.39-53.5}{6-2}-\frac{53.5-38.43}{2-4}\right)}{6-4}}{0-4}\\ =\frac{\frac{\left(-8.26833+5.7775\right)}{-2}-\frac{\left(-5.7775+7.535\right)}{2}}{-4}\\ =\frac{\left(1.2454\right)-\left(0.87875\right)}{-4}\\ =-0.0916625\end{array}$

Put in cubic interpolation equation,

$\begin{array}{c}{f}_3\left(y\right)=43.6144-0.0916625\left(3.2-4\right)\left(3.2-2\right)\left(3.2-6\right)\\ =43.6144-0.2463888\\ =43.3680112\end{array}$

And, the error is calculated as,

$\begin{array}{c}\text{Error}=f_2\left(y\right)-f_1\left(y\right)\\ =43.6144-44.458\\ =-0.8436\end{array}$

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for $y=3.2$ as,

Order	$f\left(3.2\right)$	Error
0	38.43	6.028
1	44.458	$-0.8436$
2	43.6144	$-0.2464$
3	43.368	$0.112442$
4	43.48045

Since the minimum error for order third, therefore, it can be concluded that the value of temperature at $x=4,y=3.2$ is $43.368$.

To determine

To calculate: The value of temperature at $x=4.3,y=2.7$ from the given table if the temperatures in $\left(°C\right)$ are recorded at various points on a heated plate.

	$x=0$	$x=2$	$x=4$	$x=6$	$x=8$
$y=0$	100.00	90.00	80.00	70.00	60.00
$y=2$	85.00	64.49	53.50	48.15	50.00
$y=4$	70.00	48.90	38.43	35.03	40.00
$y=6$	55.00	38.78	30.39	27.07	30.00
$y=8$	40.00	35.00	30.00	25.00	20.00

Answer to Problem 60P

Solution:

The value of temperature at $x=4.3,y=2.7$ is $46.14°C$.

Explanation of Solution

Given Information:

The data is provided as,

	$x=0$	$x=2$	$x=4$	$x=6$	$x=8$
$y=0$	100.00	90.00	80.00	70.00	60.00
$y=2$	85.00	64.49	53.50	48.15	50.00
$y=4$	70.00	48.90	38.43	35.03	40.00
$y=6$	55.00	38.78	30.39	27.07	30.00
$y=8$	40.00	35.00	30.00	25.00	20.00

Formula used:

The zero-order Newton’s interpolation formula:

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

The first-order/linear Newton’s interpolation formula:

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

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

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

Where,

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

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 calculate the temperature $x=4.3,y=2.7$, Newton interpolation is the perfect choice.

Since, this is a two-dimensional interpolation, therefore one way is to use cubic interpolation along the y direction for specific values of x and then go along the x direction for values of y obtained from the previous analysis.

First use the linear interpolation formula and arrange the points as close to about $x=2$,

$f_1\left(y\right)=b_0+b_1\left(y-y_0\right)$

The values are,

$\begin{array}{c}{y}_0=2,f\left(y_0\right)=64.49\\ y_1=4,f\left(y_1\right)=48.90\\ y_2=6,f\left(y_2\right)=38.78\\ y_3=0,f\left(y_3\right)=90.00\end{array}$

And,

$\begin{array}{c}{y}_4=8,f\left(y_4\right)=35.00\\ y=2.7\end{array}$

First calculate $b_1$,

$\begin{array}{c}{b}_1=\frac{48.9-64.49}{4-2}\\ =-7.795\end{array}$

Put in above equation,

$\begin{array}{c}{f}_1\left(y\right)=64.49+\left(-7.795\right)\left(2.7-2\right)\\ =59.0335\end{array}$

Similarly for quadratic interpolation,

$f_2\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)$

Now calculate $b_2$,

$\begin{array}{c}{b}_2=\frac{\frac{38.78-48.90}{6-4}-\frac{48.90-64.49}{4-2}}{6-2}\\ =\frac{-5.06+7.795}{4}\\ =0.68375\end{array}$

Put in quadratic interpolation equation,

$\begin{array}{c}{f}_2\left(y\right)=59.0335+\left(0.68375\right)\left(2.7-2\right)\left(2.7-4\right)\\ =59.0335-0.6222125\\ =58.4112875\end{array}$

Now, do it for cubic interpolation by the use of the formula,

$f_3\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)+b_3\left(y-y_0\right)\left(y-y_1\right)\left(y-y_2\right)$

Now calculate $b_3$,

$\begin{array}{c}{b}_3=\frac{\frac{\left(\frac{90.00-38.78}{0-6}-\frac{38.78-48.90}{6-4}\right)}{0-4}-\frac{\left(\frac{38.78-48.90}{6-4}-\frac{48.90-64.49}{4-2}\right)}{6-2}}{0-2}\\ =\frac{\frac{\left(-8.5366+5.06\right)}{-6}-\frac{\left(-5.06+7.535\right)}{4}}{-2}\\ =\frac{\left(0.57943\right)-\left(0.61875\right)}{-2}\\ =0.01965833\end{array}$

Put in cubic interpolation equation,

$\begin{array}{c}{f}_3\left(y\right)=58.4112875+0.01965833\left(2.7-2\right)\left(2.7-4\right)\left(2.7-6\right)\\ =58.4112875+0.059033\\ =58.47032\end{array}$

And, the error is calculated as,

$\begin{array}{c}\text{Error}=f_2\left(y\right)-f_1\left(y\right)\\ =58.4112875-59.0335\\ =-0.6222125\end{array}$

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for $y=2.7$ as,

Order	$f\left(2.7\right)$	Error
0	64.49	$-5.4565$
1	59.0335	$-0.6225$
2	58.411	$0.05932$
3	58.47032

Now, do this for $x=4$,

$f_1\left(y\right)=b_0+b_1\left(y-y_0\right)$

The values are,

$\begin{array}{c}{y}_0=4,f\left(y_0\right)=38.43\\ y_1=2,f\left(y_1\right)=53.5\\ y_2=6,f\left(y_2\right)=30.39\\ y_3=0,f\left(y_3\right)=80\end{array}$

And,

$\begin{array}{c}{y}_4=8,f\left(y_4\right)=30\\ y=2.7\end{array}$

First calculate $b_1$,

$\begin{array}{c}{b}_1=\frac{53.5-38.43}{2-4}\\ =-7.535\end{array}$

Put in above equation,

$\begin{array}{c}{f}_1\left(y\right)=38.43+\left(-7.535\right)\left(2.7-4\right)\\ =48.2255\end{array}$

Similarly for quadratic interpolation,

$f_2\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)$

Now calculate $b_2$,

$\begin{array}{c}{b}_2=\frac{\frac{30.39-53.5}{6-2}-\frac{53.5-38.43}{2-4}}{6-4}\\ =\frac{-5.7775+7.535}{2}\\ =0.87875\end{array}$

Put in quadratic interpolation equation,

$\begin{array}{c}{f}_2\left(y\right)=48.2255+\left(0.87875\right)\left(2.7-4\right)\left(2.7-2\right)\\ =48.2255-0.7996625\\ =47.4253375\end{array}$

Now, do it for cubic interpolation by the use of the formula,

$f_3\left(y\right)=b_0+b_1\left(y-y_0\right)+b_2\left(y-y_0\right)\left(y-y_1\right)+b_3\left(y-y_0\right)\left(y-y_1\right)\left(y-y_2\right)$

Now calculate $b_3$,

$\begin{array}{c}{b}_3=\frac{\frac{\left(\frac{80-30.39}{0-6}-\frac{30.39-53.5}{6-2}\right)}{0-2}-\frac{\left(\frac{30.39-53.5}{6-2}-\frac{53.5-38.43}{2-4}\right)}{6-4}}{0-4}\\ =\frac{\frac{\left(-8.26833+5.7775\right)}{-2}-\frac{\left(-5.7775+7.535\right)}{2}}{-4}\\ =\frac{\left(1.2454\right)-\left(0.87875\right)}{-4}\\ =-0.0916625\end{array}$

Put in cubic interpolation equation,

$\begin{array}{c}{f}_3\left(y\right)=47.4253375-0.0916625\left(2.7-4\right)\left(2.7-2\right)\left(2.7-6\right)\\ =47.4253375-0.2752624875\\ =47.1505625\end{array}$

Similarly, for $x=6\text{and}x=8$, the above process can be repeated and the following values is obtained.

$\begin{array}{l}T\left(x=2,y=2.7\right)=58.47032\\ T\left(x=4,y=2.7\right)=47.1505625\\ T\left(x=6,y=2.7\right)=42.74770188\\ T\left(x=8,y=2.7\right)=46.5\end{array}$

Now for the calculation for $x=4.3$,

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

The values are,

$\begin{array}{c}{x}_0=4,f\left(y_0\right)=47.15\\ x_1=6,f\left(y_1\right)=42.74\\ x_2=2,f\left(y_2\right)=58.47\\ x_3=8,f\left(y_3\right)=46.5\end{array}$

And,

$x=4.3$

First calculate $b_1$,

$\begin{array}{c}{b}_1=\frac{42.74-47.15}{6-4}\\ =-2.205\end{array}$

Put in above equation,

$\begin{array}{c}{f}_1\left(x\right)=47.15+\left(-2.205\right)\left(4.3-4\right)\\ =46.4885\end{array}$

Similarly for quadratic interpolation,

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

Now calculate $b_2$,

$\begin{array}{c}{b}_2=\frac{\frac{58.47-42.74}{2-6}-\frac{42.74-47.15}{6-4}}{2-4}\\ =\frac{-3.9325+2.205}{-2}\\ =0.86375\end{array}$

Put in quadratic interpolation equation,

$\begin{array}{c}{f}_2\left(x\right)=46.4885+\left(0.86375\right)\left(4.3-4\right)\left(4.3-6\right)\\ =46.4885-0.4405125\\ =46.0479875\end{array}$

Now, do it for cubic interpolation by the use of the formula,

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

Now calculate $b_3$,

$\begin{array}{c}{b}_3=\frac{\frac{\left(\frac{46.5-58.47}{8-2}-\frac{58.47-42.74}{2-6}\right)}{8-6}-\frac{\left(\frac{58.47-42.74}{2-6}-\frac{42.74-47.15}{6-4}\right)}{2-4}}{8-4}\\ =\frac{\frac{\left(-1.995+3.9325\right)}{2}-\frac{\left(-3.9325-2.205\right)}{-2}}{4}\\ =\frac{\left(0.96875\right)-\left(3.06875\right)}{4}\\ =-0.525\end{array}$

Put in cubic interpolation equation,

$\begin{array}{c}{f}_3\left(x\right)=46.4885-0.525\left(4.3-4\right)\left(4.3-6\right)\left(4.3-2\right)\\ =46.4885-0.348075\\ =46.140425\end{array}$

And, the error is calculated as,

$\begin{array}{c}\text{Error}=f_2\left(y\right)-f_1\left(y\right)\\ =46.0479875-46.4885\\ =-0.4405125\end{array}$

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized for $x=4.3$ as,

Order	$f\left(4.3\right)$	Error
0	47.15	$-0.6615$
1	46.4885	$-0.4405125$
2	46.0479875	$0.0924375$
3	46.140425

Hence, the value of temperature at $x=4.3,y=2.7$ is $46.14°C$.

This problem can also be solved with MATLAB as it contains the predefined function interp2.

The MATLAB code is as shown below,

% the given table values are stored in variable Z.

Z=[10090807060;

8564.4953.548.1550;

7048.938.4335.0340;

5538.7830.3927.0730;

4035302520];

% The X values are given.

X=[02468];

% The Y values are given.

Y=[02468];

The output in the command window is,

For more accuracy, the result can also be obtained from the bicubic interpolation as shown below,

Finally, the interpolation can also be implemented with the use of splines as shown below,

Hence, it can be concluded that the result is similar to that obtained from the calculation.