Exponential Functions and Models

Classification of Mathematical Functions through Tabular Analysis

In mathematical modeling, identifying the type of function that represents a set of data is a fundamental skill. This involves examining the relationship between the independent variable $x$ and the dependent variable $y$. In Problem 24, four distinct tables are provided to determine if they represent exponential, linear, or quadratic functions, or if they do not fit these standard categories.

Table (a) displays the coordinates $(0, 2)$, $(1, 5)$, $(2, 8)$, and $(3, 11)$. By analyzing the first differences of the $y$ values, we see that $5 - 2 = 3$, $8 - 5 = 3$, and $11 - 8 = 3$. Since the rate of change is constant, the data is modeled by a linear function, which can be represented by the slope-intercept form $y = mx + b$, where $m = 3$ and $b = 2$.

Table (b) presents the pairs $(-1, 44)$, $(0, 11)$, $(1, 2.75)$, and $(2, 0.6875)$. To determine the nature of this relationship, we check the ratio between successive $y$ values. We find that $\frac{11}{44} = 0.25$, $\frac{2.75}{11} = 0.25$, and $\frac{0.6875}{2.75} = 0.25$. Because there is a constant ratio of $0.25$, this set of data is modeled by an exponential function. Specifically, it represents exponential decay since the ratio is less than $1$.

Table (c) includes the values $(-2, 0)$, $(-1, -1)$, $(0, 0)$, and $(1, 3)$. Calculating the first differences yields $-1 - 0 = -1$, $0 - (-1) = 1$, and $3 - 0 = 3$. Since these are not constant, we check the second differences: $1 - (-1) = 2$ and $3 - 1 = 2$. Because the second differences are constant, this table is modeled by a quadratic function, characterized by the form $f(x) = ax^2 + bx + c$.

Table (d) provides the coordinates $(-2, -2)$, $(-1, -1)$, $(0, 2)$, and $(1, 4)$. The first differences are $1$, $3$, and $2$. The second differences are $2$ and $-1$. There is neither a constant difference nor a constant ratio (as $\frac{-1}{-2} = 0.5$ while $\frac{2}{-1} = -2$). Therefore, this table is categorized as none of the standard functions mentioned.

Fundamental Formulas for Exponential Growth and Decay

Exponential functions are used to model quantities that change at a rate proportional to their current value. The formulas provided for these calculations are essential for solving real-world problems involving population changes, financial interest, and physics.

The formula for exponential growth is expressed as $f(t) = A(1+r)^t$. In this equation, $f(t)$ represents the final amount after a specific time, $A$ is the initial amount present at time $t = 0$, $r$ is the growth rate expressed as a decimal, and $t$ is the time elapsed. The term $(1+r)$ is known as the growth factor.

The formula for exponential decay is expressed as $f(t) = A(1-r)^t$. Similar to the growth formula, $A$ is the initial amount and $t$ is the time. However, $r$ represents the decay rate as a decimal. The term $(1-r)$ is the decay factor, which must be between $0$ and $1$ for the quantity to decrease over time.

Applications of Exponential Models: Population and Finance

In Problem 25, we apply the exponential decay formula to a biological population. In the year $2000$, the deer population in a local forest was approximately $1100$. The population is decreasing at a rate of $4\%$. To write the expression for the population five years later, we identify $A = 1100$, $r = 0.04$, and $t = 5$. The resulting expression is $1100(1 - 0.04)^5$, which simplifies to $1100(0.96)^5$. Calculating this value: $0.96^5 \approx 0.81537$. Multiplying by $1100$ gives a deer population of approximately $896.9$, or roughly $897$ deer after five years.

In Problem 26, the context shifts to finance and debt. Joe borrows $\$500$ at an interest rate of $8\%$. To represent the amount of money $f(t)$ that Joe will owe after $t$ years, we use the growth formula because interest increases the debt over time. Here, $A = 500$ and $r = 0.08$. The resulting equation is $f(t) = 500(1 + 0.08)^t$, which simplifies to $f(t) = 500(1.08)^t$.

Problem 27 considers an investment scenario. Mary invests $\$2000$ at an interest rate of $0.6\%$ compounded annually. One must be careful to convert the percentage to a decimal: $0.6\% = 0.006$. Using the growth formula where $A = 2000$ and $r = 0.006$, the equation representing the amount of money in the account after $t$ years is $f(t) = 2000(1 + 0.006)^t$, or $f(t) = 2000(1.006)^t$.

Radioactive Decay and Half-Life Calculations

Problem 28 introduces a specific radioactive decay model: $y = A \times 2^{-\frac{t}{200}}$. In this equation, $y$ is the final amount, $A$ is the initial mass in grams, and $t$ is the time in years. The denominator in the exponent, $200$, indicates that the substance has a half-life of $200$ years (since the amount is halved when $t = 200$ and the exponent becomes $-1$).

If the initial amount $A$ is $9000$ grams, and we need to find the remaining amount after $400$ years, we substitute these values into the equation: $y = 9000 \times 2^{-\frac{400}{200}}$. Simplifying the exponent gives $y = 9000 \times 2^{-2}$. Mathematically, $2^{-2}$ is equivalent to $\frac{1}{2^2}$ or $\frac{1}{4}$. Therefore, the calculation is $9000 \times 0.25$, which results in $2250$ grams. Thus, after $400$ years, $2250$ grams of the radioactive element will remain.

Identifying Mathematical Models from Sample Data Sets

Problem 29 requires identifying the correct equation for a provided table of values. The table lists time in hours as $x$ and population as $y$. The data points are $(0, 5)$, $(1, 10)$, $(2, 20)$, $(3, 40)$, $(4, 80)$, $(5, 160)$, and $(6, 320)$.

Observing the relationship, at $t=0$, the population is $5$, meaning the initial value or y-intercept is $5$. As the time increases by $1$ hour, the population doubles: $5 \times 2 = 10$, $10 \times 2 = 20$, $20 \times 2 = 40$, and so on. This indicates an exponential growth model with a base of $2$. The general form is $y = a(b)^x$. Substituting our values, $a = 5$ and $b = 2$, we get the equation $y = 5(2)^x$. This matches option (d) in the provided multiple-choice list. Option (a) $y = 2x + 5$ is incorrect because the growth is not linear. Option (b) $y = 2^x$ is incorrect because it has an initial value of $1$ (where $2^0=1$). Option (c) $y = 2x$ is a linear model and does not fit the starting population or the growth rate.