Chapter 14, Linear Regression Video Solutions, Applied Numerical Methods with MATLAB for Engineers and Scientists | Numerade (2024)

Using the same approach as was employed to derive Eqs. (14.15) and (14.16), derive the least-squares fit of the following model:

$$
y=a_1 x+e
$$

That is, determine the slope that results in the least-squares fit for a straight line with a zero intercept. Fit the following data with this model and display the result graphically.
$$
\begin{array}{cccccccccc}
\hline \boldsymbol{x} & 2 & 4 & 6 & 7 & 10 & 11 & 14 & 17 & 20 \\
\boldsymbol{y} & 4 & 5 & 6 & 5 & 8 & 8 & 6 & 9 & 12 \\
\hline
\end{array}
$$

Beyond the examples in Fig. 14.13, there are other models that can be page 395 linearized using transformations. For example,

$$
y=\alpha_4 e^{\beta_4 x}
$$

Linearize this model and use it to estimate $\alpha_4$ and $\beta_4$ based on the following data. Develop a plot of your fit along with the data.
$$
\begin{array}{llllllllll}
\hline \boldsymbol{x} & 0.1 & 0.2 & 0.4 & 0.6 & 0.9 & 1.3 & 1.5 & 1.7 & 1.8 \\
\boldsymbol{y} & 0.75 & 1.25 & 1.45 & 1.25 & 0.85 & 0.55 & 0.35 & 0.28 & 0.18
\end{array}
$$

Rather than using the base-e exponential model [Eq. (14.22)], a common alternative is to employ a base-10 model:

$$
y=\alpha_5 10^\beta g^x
$$

When used for curve fitting, this equation yields identical results to the base-e version, but the value of the exponent parameter $\left(\beta_5\right)$ will differ from that estimated with Eq. (14.22) $\left(\beta_1\right)$. Use the base-10 version to solve Prob. 14.9. In addition, develop a formulation to relate $\beta_1$ to $\beta_5$.

On average, the surface area $A$ of human beings is related to weight $W$ and height $H$. Measurements on a number of individuals of height 180 cm and different weights $(\mathrm{kg})$ give values of $A\left(\mathrm{~m}^2\right)$ in the following table:
$$
\begin{array}{ccccccccc}
\hline \boldsymbol{W}(\mathrm{kg}) & 70 & 75 & 77 & 80 & 82 & 84 & 87 & 90 \\
\boldsymbol{A}\left(\mathbf{m}^2\right) & 2.10 & 2.12 & 2.15 & 2.20 & 2.22 & 2.23 & 2.26 & 2.30 \\
\hline
\end{array}
$$

Show that a power law $A=a W^b$ fits these data reasonably well. Evaluate the constants $a$ and $b$, and predict what the surface area is for a $95-\mathrm{kg}$ person.

An investigator has reported the data tabulated below for an experiment to determine the growth rate of bacteria $k$ (per d) as a function of oxygen concentration $c(\mathrm{mg} / \mathrm{L})$. It is known that such data can be modeled by the following equation:

$$
k=\frac{k_{\max } c^2}{c_s+c^2}
$$

where $c_s$ and $k_{\text {max }}$ are parameters. Use a transformation to linearize this equation. Then use linear regression to estimate $c_s$ and $k_{\max }$ and predict the growth rate at $c=$ $2 \mathrm{mg} / \mathrm{L}$.
$$
\begin{array}{lllllc}
\hline \boldsymbol{c} & 0.5 & 0.8 & 1.5 & 2.5 & 4 \\
\boldsymbol{k} & 1.1 & 2.5 & 5.3 & 7.6 & 8.9 \\
\hline
\end{array}
$$

You perform experiments and determine the following values of heat capacity $c$ at various temperatures $T$ for a gas:
$$
\begin{array}{ccccccc}
\boldsymbol{T} & -50 & -30 & 0 & 60 & 90 & 110 \\
\boldsymbol{c} & 1250 & 1280 & 1350 & 1480 & 1580 & 1700
\end{array}
$$

Use regression to determine a model to predict $c$ as a function of $T$.

The following data were taken from a stirred tank reactor for the reaction $A \rightarrow$ $B$. Use the data to determine the best possible estimates for $k_{01}$ and $E_1$ for the following kinetic model:

$$
-\frac{d A}{d t}=k_{01} e^{-E_1 / R T} A
$$

where $R$ is the gas constant and equals $0.00198 \mathrm{kcal} / \mathrm{mol} / \mathrm{K}$.
$$
\begin{array}{llcccc}
-\boldsymbol{d} \boldsymbol{A} / \boldsymbol{d} \boldsymbol{t} \text { (moles} / \mathbf{L} / \mathrm{s}) & 460 & 960 & 2485 & 1600 & 1245 \\
\boldsymbol{A} \text { (moles/L) } & 200 & 150 & 50 & 20 & 10 \\
\boldsymbol{T}(\mathbf{K}) & 280 & 320 & 450 & 500 & 550
\end{array}
$$

Concentration data were collected at 15 time points for the polymerization reaction:

$$
x A+y B \rightarrow A_x B_y
$$

We assume the reaction occurs via a complex mechanism consisting of many steps. Several models have been hypothesized, and the sum of the squares of the residuals had been calculated for the fits of the models of the data. The results are shown below. Which model best describes the data (statistically)? Explain your choice.
$$
\begin{array}{lccc}
& \text { Model A } & \text { Model B } & \text { Model C } \\
\hline \boldsymbol{S}_r & 135 & 105 & 100 \\
\begin{array}{c}
\text { Number of Model } \\
\text { Parameters Fit }
\end{array} & 2 & 3 & 5
\end{array}
$$

Below are data taken from a batch reactor of bacterial growth (after lag phase was over). The bacteria are allowed to grow as fast as possible for the first 2.5 hours, and then they are induced to produce a recombinant protein, the production of which slows the bacterial growth significantly. The theoretical growth of bacteria can be described by

$$
\frac{d X}{d t}=\mu X
$$

where $X$ is the number of bacteria, and $\mu$ is the specific growth rate of the bacteria during exponential growth. Based on the data, estimate the specific growth rate of the bacteria during the first 2 hours of growth and during the next 4 hours of growth.
$$
\begin{array}{cccccccc}
\begin{array}{c}
\text { Time, } \\
\mathbf{h}
\end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\begin{array}{c}
\text { [Cells] } \\
\mathbf{g} / \mathbf{L}
\end{array} & 0.100 & 0.335 & 1.102 & 1.655 & 2.453 & 3.7025 .460
\end{array}
$$

The population $p$ of a small community on the outskirts of a city grows rapidly over a 20 -year period:
$$
\begin{array}{cccccc}
\boldsymbol{t} & 0 & 5 & 10 & 15 & 20 \\
\boldsymbol{p} & 100 & 200 & 450 & 950 & 2000
\end{array}
$$

As an engineer working for a utility company, you must forecast the population 5 years into the future in order to anticipate the demand for power. Employ an exponential model and linear regression to make this prediction.

Andrade's equation has been proposed as a model of the effect of temperature on viscosity:

$$
\mu=D e^{B / T_a}
$$

where $\mu=$ dynamic viscosity of water $\left(10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2\right), T_a=$ absolute temperature (K), and $D$ and $B$ are parameters. Fit this model to the following data for water $T$ is in ${ }^{\circ} \mathrm{C}$ and $\mu$ is in $10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2$ :
$$
\begin{array}{lcccccc}
\boldsymbol{T} & 0 & 5 & 10 & 20 & 30 & 40 \\
\boldsymbol{\mu} & 1.787 & 1.519 & 1.307 & 1.002 & 0.7975 & 0.6529
\end{array}
$$

Manning's formula for a rectangular channel can be written as

$$
Q=\frac{1}{n_m} \frac{(B H)^{5 / 3}}{(B+2 H)^{2 / 3}} \sqrt{S}
$$

where $Q=$ flow $\left(\mathrm{m}^3 / \mathrm{s}\right), n_m=$ a roughness coefficient, $B=$ width (m), $H=\operatorname{depth}(\mathrm{m})$, and $S=$ slope. You are applying this formula to a stream where you know that the width $=20 \mathrm{~m}$ and the depth $=0.3 \mathrm{~m}$. Unfortunately, you know the roughness and the slope to only a $\pm 10 \%$ precision. That is, you know that the roughness is about 0.03 with a range from 0.027 to 0.033 and the slope is 0.0003 with a range from 0.00027 to 0.00033 . Assuming uniform distributions, use a Monte Carlo analysis with $n=$ 10,000 to estimate the distribution of flow.

A Monte Carlo analysis can be used for optimization. For example, the trajectory of a ball can be computed with

$$
y=\left(\tan \theta_0\right) x-\frac{g}{2 v_0^2 \cos ^2 \theta_0} x^2+y_0
$$

where $y=$ the height (m), $\theta_0=$ the initial angle (radians), $v_0=$ the initial velocity $(\mathrm{m} / \mathrm{s}), g=$ the gravitational constant $=9.81 \mathrm{~m} / \mathrm{s}^2$, and $y_0=$ the initial height $(\mathrm{m})$. Given $y_0=1 \mathrm{~m}, v_0=25 \mathrm{~m} / \mathrm{s}$, and $\theta_0=50^{\circ}$, determine the maximum height and the corresponding $x$ distance (a) analytically with calculus and (b) numerically with Monte Carlo simulation. For the latter, develop a script that generates a vector of 10,000 uniformly distributed values of $x$ between 0 and 60 m . Use this vector and Eq. (P14.35) to generate a vector of heights. Then, employ the max function to determine the maximum height and the associated $x$ distance.

Stokes Settling Law provides a means to compute the settling velocity of spherical particles under laminar conditions

$$
v_s=\frac{g}{18} \frac{\rho_s-\rho}{\mu} d^2
$$

where $v_s=$ the terminal settling velocity $(\mathrm{m} / \mathrm{s}), g=$ gravitational acceleration $(=9.81$ $\left.\mathrm{m} / \mathrm{s}^2\right), \rho=$ the fluid density $\left(\mathrm{kg} / \mathrm{m}^3\right), \rho_s=$ the particle density $\left(\mathrm{kg} / \mathrm{m}^3\right), \mu=$ the dynamic viscosity of the fluid $\left(\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$, and $d=$ the particle diameter (m). Suppose that you conduct an experiment in which you measure the terminal settling velocities of a number of $10-\mu \mathrm{m}$ spheres having different densities,
$$
\begin{array}{lllllllllll}
\rho_s, \mathrm{~kg} / \mathrm{m}^2 & 1500 & 1600 & 1700 & 1800 & 1900 & 2000 & 2100 & 2200 & 2300 \\
v_s, 10^{-3} \mathrm{~m} / \mathrm{s} & 1.03 & 1.12 & 1.59 & 1.76 & 2.42 & 2.51 & 3.06 & 3 & 3.5
\end{array}
$$
(a) Generate a labeled plot of the data. (b) Fit a straight line to the data with linear regression (polyfit) and superimpose this line on your plot. (c) Use the model to predict the settling velocity of a $2500 \mathrm{~kg} / \mathrm{m}^3$ density sphere. (d) Use the slope and the intercept to estimate the fluid's viscosity and density.

Beyond the examples in Fig. 14.13, there are other models that can be linearized using transformations. For example, the following model applies to thirdorder chemical reactions in batch reactors

$$
c=c_0 \frac{1}{\sqrt{1+2 k c_0^2 t}}
$$

where $c=$ concentration $(\mathrm{mg} / \mathrm{L}), c_0=$ initial concentration $(\mathrm{mg} / \mathrm{L}), k=$ reaction rate $\left(\mathrm{L}^2 /\left(\mathrm{mg}^2 \mathrm{~d}\right)\right)$, and $t=$ time (d). Linearize this model and use it to estimate $k$ and $c_0$ based on the following data. Develop plots of your fit along with the data for both the transformed, linearized form and the untransformed form.
$$
\begin{array}{ccccccccc}
\boldsymbol{t} & 0 & 0.5 & 1 & 1.5 & 2 & 3 & 4 & 5 \\
\boldsymbol{c} & 3.26 & 2.09 & 1.62 & 1.48 & 1.17 & 1.06 & 0.9 & 0.85
\end{array}
$$

In Chap. 7 we presented optimization techniques to find the optimal values of one- and multi-dimensional functions. Random numbers provide an alternative means to solve the same sorts of problems (recall Prob. 14.35). This is done by repeatedly evaluating the function at randomly selected values of the independent variable and keeping track of the one that yields the best value of the function being optimized. If a sufficient number of samples are conducted, the optimum will eventually be located. In their simplest manifestations, such approaches are not very efficient. However, they do have the advantage that they can detect global optimums for functions with lots of local optima. Develop a function that uses random numbers to locate the maximum of the humps function

$$
f(x)=\frac{1}{(x-0.3)^2+0.01}+\frac{1}{(x-0.9)^2+0.04}-6
$$

in the domain bounded by $x=0$ to 2 . Here is a script that you can use to test your function
```
clear,clc,clf, format compact
xmin=0;xmax=2;n=1000
xp=linspace(xmin,xmax,200); yp=f(xp);
plot(xp,yp)
[xopt, fopt]=RandOpt(@f,n,xmin,xmax)
```

Using the same approach as described in Prob. 14.38, develop a page 399 function that uses random numbers to determine the maximum and corresponding $x$ and $y$ values of the following two-dimensional function

$$
f(x, y)=y-x-2 x^2-2 x y-y^2
$$

in the domain bounded by $x=-2$ to 2 and $y=1$ to 3. The domain is depicted in Fig. P14.39. Notice that a single maximum of 1.25 occurs at $x=-1$ and $y=1.5$. Here is a script that you can use to test your function

```
clear,clc,format compact
xint=[-2;2];y int=[1;3];n=10000;
[xopt,yopt, fopt]=RandOpt2D(@fxy,n,xint,yint)
```

The table below shows the 2015 world record times and holders for outdoor running. Note that all but the 100 m and the marathon $(42,195 \mathrm{~m})$ are run on oval tracks.

Fit a power model for each gender and use it to predict the record time for a half marathon $(21,097.5 \mathrm{~m})$. Note that the actual records for the half marathon are 3503 s (Tadese) and 3909 s (Kiplagat) for men and women, respectively.
$$
\begin{array}{rrlrl}
\text { Event (m) } & \text { Time } \mathbf{( s )} & \text { Men Holder } & \text { Time } \mathbf{( s )} & \text { Women Holder } \\
\hline 100 & 9.58 & \text { Bolt } & 10.49 & \text { Griffith-Joyner } \\
200 & 19.19 & \text { Bolt } & 21.34 & \text { Griffith-Joyner } \\
400 & 43.18 & \text { Johnson } & 47.60 & \text { Koch } \\
800 & 100.90 & \text { Rudisha } & 113.28 & \text { Kratochvilova } \\
1000 & 131.96 & \text { Ngeny } & 148.98 & \text { Masterkova } \\
1500 & 206.00 & \text { El Guerrouj } & 230.07 & \text { Dibaba } \\
2000 & 284.79 & \text { El Guerrouj } & 325.35 & \text { O'Sullivan } \\
5000 & 757.40 & \text { Bekele } & 851.15 & \text { Dibaba } \\
10.000 & 1577.53 & \text { Bekele } & 1771.78 & \text { Wang } \\
20,000 & 3386.00 & \text { Gebrselassie } & 3926.60 & \text { Loroupe } \\
42.195 & 7377.00 & \text { Kimetto } & 8125.00 & \text { Radcliffe }
\end{array}
$$