Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a. Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

The equation of the general linear model is defined as:

\({\bf{y}} = X\beta + \in \)

Here, \({\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is the parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is the residual vector.

(a)

The given equation is \(y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}\) , and the given data sets are \(\left( {10,21.34} \right)\), \(\left( {11,20.68} \right)\), \(\left( {12,20.05} \right)\), \(\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

Write the Design matrix, observational vector, and the parameter vector for the given equation and data set by using the information given in step 1.

Design matrix:

\(X = \left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)\)

Observational vector:

\({\bf{y}} = \left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)\)

And the parameter vector for the given equation is,

\(\beta = \left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right)\)

Thus, the above values are the best fit for the given data set and equation.

The normal equation is given by,

\({X^T}X\beta = {X^T}{\bf{y}}\)

The general least-squares equation is given by \(y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}\), and to find the associated least-squares curve, the values of \({M_A},{M_B}\) are required, so find the values of \({M_A},{M_B}\) by using normal equation.

By using the obtained information from step 2, the normal equation will be,

\(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}{\bf{y}}\)

That implies;

\(\left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)\)

Use the following steps to find the associated values for the obtained data in MATLAB.

1. Enter the data \(\left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)\).
2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right)\) is \(\left( {\begin{aligned}{19.94}\\{10.10}\end{aligned}} \right)\).

Now, substitute the obtained values into \(y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}\).

\(y = 19.94{e^{ - 0.02t}} + 10.10{e^{ - 0.07t}}\)

So, the required equation is \(y = 19.94{e^{ - 0.02t}} + 10.10{e^{ - 0.07t}}\).